Duffie_Beckman Solutions to Problems
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problem solutions for duffie and backman...
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Solutions Manual for Problems in Appendix A of
SOLAR ENGINEERING OF THERMAL PROCESSES Third Edition
J. A. Duffie (Deceased) and W. A. Beckman
* John Wiley & Sons, Inc., New York (2006)
SOLUTIONS MANUAL for
SOLAR ENGINEERING OF THERMAL PROCESSES
Third edition
CONTENTS Introduction
iii
Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 14 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24
1 24 74 96 113 139 161 176 185 191 213 240 242 249 251 265 286 297 306 320
Course Material
326
ii
SOLUTIONS MANUAL for
SOLAR ENGINEERING OF THERMAL PROCESSES, 3rd Ed. John A. Duffie (deceased) and William A. Beckman
This manual includes solutions to the problems in Appendix A of the third edition of Solar Engineering of Thermal Processes, published by John Wiley & Sons, New York (2006). These solutions have been done with the program EES, Engineering Equation Solver, (available from F-Chart Software, on the web at fchart.com). The equations, tables, solutions, and comments have been translated into a word processor and edited for inclusion in the manual. The equation statements and functions are to a degree the result of the programming style used in EES, but they should be no mystery to those who have done any programming. The EES functions (called the SETP library) useful for solving solar energy problems are freely available at fchart.com under EES "add-ons" The nomenclature in this manual is close to that in the book. Where differences have been required or are convenient, obvious variable names have been used, such as Dec for declination but sometimes δ. There are two exceptions in the matter of style. First, EES allows subscripts in variable names but they show as subscripts only in the Formatted Equations Window and are indicated as in the following examples: I_T is substituted for IT, R_b for Rb, I_o for Io, and A_c for Ac, etc. Second, variables that are ratios can not be shown with a slash "/" as it denotes a division operation; the reverse slash is used in variable names that are ratios, such as: FR'\FR for FR'/FR, and Id\I for the diffuse fraction Id/I. The equation, table, and figure numbers are those in the book. Thermophysical property data are usually obtained from the EES built-in functions in but sometimes are from Appendix E. Monthly average meteorological data are from Appendix G or from various web sites. The EES program allows preprogrammed routines that begin with the word FUNCTION, PROCEDURE, SUBPROGRAM or MODULE and are terminated with the word END. Routines that are not part of the SETP library must always appear at the beginning of the iii
program. The use of preprogrammed routines is a personal choice; the equations can be typed in the solutions or the routines can be used. In the early solutions the equations are typically typed in but in later problem the preprogrammed routines are used extensively. The following EES code is the SETP routine for the extraterrestrial radiation on a horizontal surface. The names of all SETP routines end with an underline (e.g., H_zero_). Text between squiggly brackets or between double quotes is for information only. However, a special case exists when the left squiggly bracket on the next line following the function name is followed by a dollar sign. This text is used as help. Text between squiggly brackets does not appear in the Formatted Equation Window. Routines such as this often call other routines (e.g., Dec_). Units are particularly important in all of engineering and are included in the SETP functions. Here units are assigned to the two constants f [deg] and g [MJ/m^2]. The units conversion function in EES Convert(UnitsFrom, UnitsTo) is one of EES’s most useful tools and is illustrated here where the units degrees are converted to radians with Convert(deg, rad). Units cannot be assigned to variables in the Equations Window. The units of SunSetHrAng is set to degrees in the EES program but “[deg]” is only a reminder that the units are degrees. Finally, the value returned by the function must be set using the function name. Function H_Zero_(n,Lat) {Equation 1.10.3} {$H_Zero_ Equation 1.10.3: Calculates the day's extraterrestrial radiation H_Zero. Uses Fctn. Dec_ for Declination. All angles are in degrees and n is the day number. H_Zero is in MJ/m2.} Dec = Dec_(n) "[deg]" SunSetHrAng = arccos(-tan(Lat)*tan(Dec)) "[deg]" f=0.9863[deg] g=37.6[MJ/m^2] H_Zero_ = g*(1+.033*cos(f*n))*(cos(Lat)*cos(Dec)*sin(SunSetHrAng)+Convert(deg, rad)*SunSetHrAng*sin(Lat)*sin(Dec)) "[MJ/m^2]" END
The solution to Problem 1.10 uses this function and is as follows: "!1.10 Calculation of extraterrestrial radiation on horizontal surface" “Knowns:” month=2; day=8; lat=45[deg]; G_sc=1367[W/m^2] "Table 1.6.1" Dec = 23.45[deg]*sin(.9863*(284+n)) "Eqn 1.6.1" SunSetHrAng = arccos(-tan(Lat)*tan(Dec)) "Eqn 1.6.10" n=nDay_(month, day) "We could use Equation 1.10.3 the day's extraterrestrial radiations: H_o=(24*3600[s]*G_sc)/pi*(1+0.033*cos(360[deg]*n/365)) *(cos(lat)*cos(dec)& *sin(SunSetHrAng)+(pi*SunSetHrAng/180[deg]*Sin(lat)*sin(dec)))*Convert(J, MJ) "
iv
"But we will use the SETP function:” H_o = H_Zero_(n, Lat)" "To find the monthly average daily radiation we will use the SETP function with the average day in February." n_ave=AveDay_(month) "Table 1.6.1" H_bar_o=H_Zero_(n_ave, Lat) "To find the energy in the hour 10 to 11 use Equation 1.10.4" Time=10:30 "note that this is the midpoint of the hour" HrAng=(Time-12[hr])*15[deg/hr]; HrAng2=HrAng+7.5[deg]; HrAng1=HrAng-7.5[deg] I_o = (12*3600[s]*G_sc)/pi*(1+.033*cos(360[deg]*n/365))*(cos(Lat)*cos(Dec)*(sin(HrAng2)& -sin(HrAng1))+Convert(deg, rad)*(HrAng2-HrAng1)*sin(Lat)*sin(Dec))*Convert(J, MJ) "We could have used the SETP function I_o=I_Zero_(n, Lat, HrAng1, HrAng2) SOLUTION day=8
Dec=-15.52 [deg]
G_sc=1,367 [W/m^2]
HrAng=-22.5 [deg]
HrAng1=-30 [deg]
HrAng2=-15 [deg]
H_bar_o=17.63 [MJ/m^2]
H_o=15.84 [MJ/m^2]
I_o=2.214 [MJ/m^2]
lat=45 [deg]
month=2
n=39 [-]
n_ave=47 [-]
SunSetHrAng=73.88 [deg]
Time=10:30:00 [hr]
There are alternative methods of solving some problems. For example, calculation of an hour’s radiation on sloped surfaces can be done with any of the four methods shown in Section 2.15 and 2.16. In some instances the problem statements call for use of specific methods. In others the choice is left open. In a few cases we have included solutions by more than one method. Students may complain that some data is not included in the problem, such as ground reflectance. We leave it up the student to choose reasonable values. Typical values of 0.2 are used unless there is snow cover then 0.7 seems reasonable. Most of these problems have been used as homework or examination problems in courses at the University of Wisconsin. Many of them have arisen from our research at the Solar Energy Laboratory, from practical experience, or from suggestions of others. We appreciate the assistance of the many people who have contributed ideas and suggestions on which these problems and their solutions are based. For those who teach courses in solar energy to engineering students, we have included on the final pages of this manual a typical outline of a one-semester course we teach to our fourth-year undergraduate and first year graduate student classes. Madison, Wisconsin, April1, 2011
William A. Beckman
v
P01 01
Equations Knowns: σ = sigma#
(1)
Dsun = 1.39 × 109 [m]
(2)
Dearth = 1.27 × 107 [m]
(3)
RSunEarth = 1.495 × 1011 [m]
(4)
TSun = 5777 [K]
(5)
Calculate the emitted solar energy 2 Areasun = π · Dsun ;
(6)
4 EmitSol = σ · Areasun · TSun
(7)
Calculate the fraction intercepted by the earth F raction = Areaearth,proj /Area1au Areaearth,proj = π ·
(8)
2 Dearth 4
(9)
2 Area1au = 4 · π · RSunEarth
(10)
Estimate solar constant SolarConstant = F raction · EmitSol/Areaearth,proj
(11)
Solution Area1au = 2.809 × 1023 m2 Dearth = 1.270 × 107 [m] F raction = 4.510 × 10−10 [-] SolarConstant = 1365 W/m2
Areaearth,proj = 1.266 × 1014 m2 Dsun = 1.390 × 109 [m] RSunEarth = 1.495 × 1011 [m] TSun = 5777 [K]
1
Areasun = 6.070 × 1018 m2 EmitSol = 3.833 × 1026 [W] σ = 5.670 × 10−8 W/m2 -K4
P01 02
Equations Start with Solar Constant = 1367 W/m2 from Problem 1.1. SolarConstV enus will be inversely proportional to the square of the distance from sun: SolarConst .722 SolarConst = 1367 W/m2 SolarConstV enus =
(1) (2)
Solution SolarConst = 1367 W/m2
SolarConstV enus = 2637 W/m2
2
P01 03
Equations From Table 1.3.1. f0to0.5 = 0.216
(1)
f0to2.0 = 0.941
(2)
Therefore the fraction between 0.5 and 2.0 is: f0.5to2.0 = f0to2.0 − F0to0.5
(3)
Solution f0.5to2.0 = 0.725 [-]
f0to0.5 = 0.216 [-]
f0to2.0 = 0.941 [-]
3
P01 04
Equations This problem is essentially solved in Table 1.3.1b. The wavelength divisions for the increments are the wavelengths given in the table for F0 -λ of 0.1, 0.2, 0.3, etc. The energy midpoints are the wavelengths in the table for 0.05, 0.15, 0.25, etc. The first increment of a tenth is in the wavelength range of 0 to 0.416, and the wavelength at the energy midpoint of the increment is 0.364. Use the ‘Insert/Modify an Array’ in the Edit menu to simplify data input. λ0 = 0.300;
λ1 = .416;
λ2 = .489;
λ6 = .849;
λ7 = 1.008;
λ8 = 1.244;
λ3 = .561;
λ4 = .638;
λ9 = 1.654;
λmid,1 = .364;
λmid,2 = .455;
λmid,3 = .525;
λmid,6 = .787;
λmid,7 = .923;
λmid,8 = 1.113;
λ5 = .731
λ10 = 8.000;
λmid,4 = .599; λmid,9 = 1.412;
(1) (2)
λmid,5 = .682 λmid,10 = 2.117
(3) (4)
So, the range of wavelengths for each of the 10 bands is: duplicate i = 1, 10
(5)
λrange,i = λi − λi−1
(6)
end
(7)
Solution
Arrays Row
λi
λmid,i
λrange,i
0 1 2 3 4 5 6 7 8 9 10
0.300 0.416 0.489 0.561 0.638 0.731 0.849 1.008 1.244 1.654 8.000
0.364 0.455 0.525 0.599 0.682 0.787 0.923 1.113 1.412 2.117
0.116 0.073 0.072 0.077 0.093 0.118 0.159 0.236 0.410 6.346
4
P01 05
Equations Dec = 23.45 [deg] · sin (.9863 · (284 + n))
Eqn. 1.6.1
(1)
CosIncAng = sin (Dec) · sin (Lat) · Cos(Slope) − sin (Dec) · Cos(Lat) · sin (Slope) · Cos(Surf AzAng)+ Cos(Dec) · Cos(Lat) · Cos(Slope) · Cos(HourAngle)+ Cos(Dec) · sin (Lat) · sin (Slope) · Cos(Surf AzAng) · Cos(HourAngle) + Cos(Dec) · sin (Slope) · sin (Surf AzAng) · sin (HourAngle)
Eqn 1.6.2
θ = arccos (CosIncAng)
(3)
All data and results are in the Parametric Table We could have used the SETP function to find θ: θ = arcCos(CosIncAng(Lat, n, HourAngle, Slope, SurfAzAng))
Table 1 Run
n
1 2 3 4 5
41 41 41 41 41
Lat [deg] 43.3 43.3 43.3 43.3 43.3
HourAngle [deg] 30 30 30 30 30
(2)
Surf AzAng [deg] 0 0 40 0 90
Slope [deg] 0 60 60 90 90
5
Dec [deg] -14.9 -14.9 -14.9 -14.9 -14.9
θ [deg] 64.4 28.9 8.0 40.4 61.1
P01 06
Equations Use of Equation of Time (Eqn 1.5.3) in Equation 1.5.2 EqT ime = 229.2 [min]·(.000075 + .001868 · Cos(B) − .032077 · sin (B) − .014615 · Cos(2 · B) − .04089 · sin (2 · B))(1)
B = (n − 1) · 360
[deg] 365
Eqn 1.5.3
n = nDay (month, day)
(2)
Use either the SETP function or Table 1.6.1 to find the day number)
(3)
The difference between standard and solar time is: hr Solar − Standard = (4 [min/deg] · (Long Std − Long Loc ) + EqT ime) · 0.016666667 min
Eqn 1.5.2
(4)
The difference between daylight time and standard time is one hour Daylight − Standard = 1 [hr]
(5)
Note: In part d the time of 10:00 AM is solar time as nothing else is specified.
Table 1 Run 1 2 3 4
month [-] 2 2 7 7
day [-] 3 3 31 31
n [-] 34 34 212 212
B [deg] 32.55 32.55 208.1 208.1
Long Loc [deg] 101 116 70.5 90
Long Std [deg] 105 105 75 90
6
EqT ime [min] -13.49 -13.49 -6.549 -6.549
Solar [hr] 14:02:31 13:02:31 12:00:00 10:00:00
Standard [hr] 14:00:00 14:00:00 11:48:33 10:06:33
Daylight [hr] 15:00:00 15:00:00 12:48:33 11:06:33
P01 07
Equations Determine the sunset hour angle andday length for various conditions Dec = 23.45 [deg] · sin (.9863 · (284 + n))
Eqn 1.6.1a
SunSetHrAngle = arccos (− tan (Lat) · tan (Dec)) DayLength = 2 [hr] ·
Eqn 1.6.10
arccos (− tan (Lat) · tan (Dec)) 15 [deg]
n = nDay (month, day)
(1)
Eqn 1.6.11
Table 1.6.1 or SETP function
The more exact declination equation, 1.6.1b can be used dec=(0.006918-0.399912*cos(B)+0.070257*sin(B)-0.000758*Cos(2*B) +0.000907*sin(2*B) Eqn 1.6.1b B=(n-1)*360[deg]/365 The day lengths using the two equations differ by 2 minutes. For most solar calculations this difference can be ignored.
Dec with Eqn 1.6.1a month
day
n
1 2 3 4 5 6 7 8
1 3 7 3 1 3 7 3
1 22 1 16 1 22 1 16
1 81 182 75 1 81 182 75
Lat [deg] 22.5 22.5 22.5 22.5 43.3 43.3 43.3 43.3
Dec [deg] -23.01 -0.00 23.12 -2.42 -23.01 -0.00 23.12 -2.42
DayLength [hr] 10:38 12:00 13:21 11:51 8:51 12:00 15:09 11:41
SunSetHrAngle [deg] 79.9 90.0 100.2 89.0 66.4 90.0 113.7 87.7
Lat [deg] 22.5 22.5 22.5 22.5 43.3 43.3 43.3 43.3
Dec [deg] -22.71 0.01 23.52 -2.33 -22.71 0.01 23.52 -2.33
DayLength [hr] 10:40 12:00 13:23 11:52 8:54 12:00 15:13 11:42
SunSetHrAngle [deg] 80.0 90.0 100.4 89.0 66.8 90.0 114.2 87.8
Dec with Eqn 1.6.1b Run
month
day
n
1 2 3 4 5 6 7 8
1 3 7 3 1 3 7 3
1 22 1 16 1 22 1 16
1 81 182 75 1 81 182 75
7
(3) (4)
The average day for February is the 22. The average day could be found from Table 1.6.1 or SETP function, AveDay=AveDay(month )
Run
(2)
P01 08
Equations 1.8 Calculation of θbeam for tracking surfaces. Use parametric table for the two times. Lat = 27;
month = 3;
Day = 5;
n = nDay (month, day)
could use Table 1.6.1 to find n
HourAngle = (T ime − 12hr ) · 15 [deg/hr] Dec = 23.45 [deg] · sin (.9863 · (284 + n))
(1) (2)
1.6.1
(3)
CosZenithAngle = Cos(Lat) · Cos(Dec) · Cos(HourAngle) + sin (Lat) · sin (Dec)
1.6.5
(4)
ZenithAngle = arccos (CosZenithAngle)
(5)
For horizontal E-W axis, use Equation 1.7.2a: 2 2 θEW = arccos 1 − (cos (Dec)) · (sin (HourAngle))
(6)
For horizontal N-S axis, use Equation 1.7.3a: .5 2 2 2 (cos (ZenithAngle)) + (cos (Dec)) · (sin (HourAngle)) θN S = arccos
(7)
For polar axis, use Equation 1.7.5a: θP olar = arccos (cos (Dec))
(8)
Table 1 Run 1 2
time [hr] 9 12
Dec [deg] -6.765 -6.765
HourAngle [deg] -45 0
ZenithAngle [deg] 55.1 33.77
θEW [deg] 59.54 0
8
θN S [deg] 25.07 33.77
θP olar [deg] 6.765 6.765
P01 09
Equations Calculate Rb from Equation 1.8.2. Use parametric table for the two conditions. Dec = 23.45 [deg] · sin (.9863 · (284 + n)) n = nDay (month, day)
1.6.1
(1)
Table 1.6.1
(2)
HrAng = (T ime − 12hr ) · 15 [deg/hr]
(3)
CosZen = Cos(Lat) · Cos(Dec) · Cos(HrAng) + sin (Lat) · sin (Dec)
1.6.5
(4)
CosT heta = sin (Dec) · sin (Lat) · Cos(Slope) − sin (Dec) · Cos(Lat) · sin (Slope) · Cos(Surf AzAng) + Cos(Dec)· Cos(Lat) · Cos(Slope) · Cos(HrAng) + Cos(Dec) · sin (Lat) · sin (Slope) · Cos(Surf AzAng) · Cos(HrAng) + Cos(Dec) · sin (Slope) · sin (Surf AzAng) · sin (HrAng) Rb = CosT heta/CosZen
(5)
1.8.2
(6)
We could use the SETP function to find CosTheta CosTheta = CosTheta(Lat, Dec, HrAng, Slope, SurfAzAngle)
Table 1 Run 1 2
Run 1 2
Lat [deg] 43 43
month
day
n
3 3
5 5
64 64
Slope [deg] 60 45
Surf AzAng [deg] 0 15
Rb [-] 1.589 1.413
9
T ime [hr] 14:30 10:30
HrAng [deg] 37.5 -22.5
Dec [deg] -6.765 -6.765
CosT heta [-] 0.7878 0.8343
CosZen [-] 0.4958 0.5906
P01 10
Equations 1.10 Calculation of extraterrestrial radiation on horizontal surface n = nDay (month, day) month = 2; day = 8; lat = 45; Gsc = 1367 W/m2 ; Dec = 23.45 [deg] · sin (.9863 · (284 + n))
Table 1.6.1
1.6.1
SunSetHrAng = arccos (− tan (Lat) · tan (Dec))
(1) (2)
1.6.10
(3)
From Equation 1.10.3 the day’s extraterrestrial radiations is: Ho =
24 · 3600 [s] · Gsc · (1 + 0.033 · Cos(360 [deg] · n/365)) π
MJ SunSetHrAng · Cos(lat) · Cos(dec) · sin (SunSetHrAng) + π · · sin (lat) · sin (dec) · 1 × 10·6 (4) 180 [deg] J We could have used the SETP function HZero = HZero, (n, Lat) To find the monthly average daily radiation we will use the SETP function with the average day in February. nave = AveDay (month)
Table 1.6.1
(5)
¯ o = HZero, (nave , Lat) H
(6)
To find the energy in the hour 10 to 11 use Equation 1.10.4 T ime = 10 : 30
note that this is the midpoint of the hour
HrAng = (T ime − 12hr ) · 15 [deg/hr] ;
Io =
(7)
HrAng2 = HrAng + 7.5 [deg] ;
HrAng1 = HrAng − 7.5deg
(8)
12 · 3600 [s] · Gsc · (1 + .033 · Cos(360 [deg] · n/365)) · (Cos(Lat) · Cos(Dec) π
rad ·6 MJ ·(sin (HrAng2) − sin (HrAng1))+ 0.017453293 ·(HrAng2 − HrAng1)·sin (Lat)·sin (Dec)· 1 × 10 (9) deg J We could have used the SETP function Io =IZero, (n, Lat, HrAng1, HrAng2)
Solution
HrAng2 = −15 [deg]
Gsc = 1367 W/m2 ¯ o = 17.63 MJ/m2 H
HrAng = −22.5 [deg] Ho = 15.84 MJ/m2
lat = 45 [deg]
month = 2
n = 39 [-]
day = 8
Dec = −15.52 [deg]
HrAng1 = −30 [deg] Io = 2.214 MJ/m2 nave = 47 [-]
SunSetHrAng = 73.88 [deg] 10 T ime = 10 : 30 : 00 [hr]
P01 11
Equations 1.11 Calculation of shading by a building ELL As the ell is high compared to the window, the Object Altitude Angle varies from 0 to 90. An equation for the Object Azimuth Angle(which has a single value) is: ObjAzAng = − (90 [deg] − (arctan (20.2/10) + 15 [deg]))
(1)
When the Solar Azimuth Angle (γ s ) is less than ObjAzAng the corner of the window is shaded. So, for the mean day of each month we need to find the time (i.e., the hour angle, ω) at which the ObjAzAng is equal to γ s. We will use the EES Parametric Table to work through the 12 months. lat = 40
(2)
nave = AveDay (month) 284 + nave dec = 23.45 [deg] · sin 360 · 365
(3) (4)
θz = arccos (cos (lat) · cos (dec) · Cos (ω) + sin (lat) · sin (dec)) cos (θz ) · sin (lat) − sin (Dec) γs = − arccos sin (θz ) · Cos (lat) T ime = −
ω · 60 [min/hr] 15 [deg/hr]
(5) (6) (7)
Here Time is the number of minutes before solar noon; this makes an easier to understand plot.
11
These equations are diffucult to solve for the hour angle ω unless very good guesses are provided. An alternative method of solution is to define an error as the absolute difference between γ s and the ObjAzAng and minimize this error. error = abs (γs − ObjAzAng)
(8)
The Object Azimuth Angle could have been plotted on a solar position plot for Latitude 40. The times when shading ceases is determined for the time of year from this plot. . The plot shows the number of minutes before solar noon at which the sun is incident on the east edge of the window as a function of month.
Results Run
month
1 2 3 4 5 6 7 8 9 10 11 12
1 2 3 4 5 6 7 8 9 10 11 12
dec [deg] -20.92 -12.95 -2.418 9.415 18.79 23.09 21.18 13.45 2.217 -9.599 -18.91 -23.05
nave 17 47 75 105 135 162 198 228 258 288 318 344
γs [deg] -11.34 -11.34 -11.34 -11.34 -11.34 -11.34 -11.34 -11.34 -11.34 -11.34 -11.34 -11.34
θz [deg] 61.7 53.7 43.0 31.0 21.5 17.2 19.1 26.9 38.3 50.3 59.7 63.9
ω [deg] -10.68 -9.352 -7.711 -5.896 -4.373 -3.621 -3.959 -5.255 -7.005 -8.825 -10.34 -11.06
minutes before solar noon
12
error [deg] 9.08E-4 1.48E-4 2.02E-3 6.01E-5 1.87E-3 6.34E-5 4.74E-5 3.26E-4 8.44E-4 6.14E-3 2.63E-4 2.53E-4
T ime [min] 43 37 31 24 17 14 16 21 28 35 41 44
P01 12
Equations 1.12
Calculation of shading by an overhang
Since the system is symmetrical about the N-S line, it is necessary to calculate only points on one side (e.g., west). Use a coordinate system with the origin at the center of the window. The Z axis is vertical, the X axis is towards the west (i.e., left in Fig 14.4.1) and the Y axis is towards the south. A general point on the overhang is x, y, z. For example, the corner of the overhang is at x=(W+E)/2, y=P, and z=H/2+G. The azimuth angle is arcTan(x/y) and the altitude angle is arcTan(z/sqrt(x2 +y2 )) Expressions for the relationships of distances and angles for any point on the overhang are as follows: W = 3.25;
E = 0.5;
γ = arctan (x/y); α = arctan
z p
x2
P = 0.75;
H = 1.75;
G = 0.25
γeast = −γ !
(1) (2) (3)
+ y2
Set up a parametric table for γ and α with various values of x, y, and z around the boarder of the overhang. Both positive and negative values of γ were placed in the parametric table for plotting purposes. The parametric table is saved as an EES LKT
13
table. The program, SolarPositionDiagram.EES reads this lookup table and overlays the values on a solar postion diagram for a latitude of 34 degrees. The figure shows the shading boundary plotted on a solar position diagram. Area above the red line are shaded. The shading diagram ends at plus or minus 90 degrees since the sun is then behind the window. $SaveTable ’Parametric’
’shading.lkt’
/N
Part b: From the diagram we see that all of February (the 4th month from the bottom) is unshaded and all of July (the second month from the top) is shaded. In August (the 4th month from the top) the center is shaded after 8am and before 4pm.
Plot 1
14
P01 13
Equations 1.13 Calculation of solar information We will use the EES SETP functions for the calculations. See Key variables in Solutions window. Knowns month = 2
(1)
day = 16
(2)
γ = 0 [deg]
(3)
Lat = 38.3 [deg]
(4)
Long = 104.6 [deg]
(5)
n = AveDay (month)
Or from Table 1.6.1
(6)
a) Day Length ωs = SunsetHrAng (Lat, n) Hrday = ωs ·
2 15 [deg/hr]
Equation 1.6.10
(7)
Equation 1.6.11
(8)
b) Declination - assumed to be at noon. δ = dec (n)
Equation 1.6.1a
δha = dec (−n)
(9)
Equation 1.6.1b - high accuracy declination
15
(10)
At sunrise and sunset the equivalent n is ωs nsunrise = n − 15 [deg/hr] · 24 [hr] nsunset = n +
(11)
ωs 15 [deg/hr] · 24 [hr]
(12)
δchange = dec (nsunrise ) − dec (nsunset )
(13)
c) change in θ throughout the day for various slopes. Use a parametric table with ω ranging from -75 to+75. Slope ranging between 0 and 90. θ = arccos (cosIncAng (Lat, n, ω, β, γ))
Equation 1.6.2
(14)
d) Rb vs ω for β=60 - use a second parametric table ω -75 to +75. Rb = RBeam, (Lat, n, ω, β, γ)
Equation 1.8.1
(15)
β = 60
(16)
e) What is Ho? Ho = HZero, (n, Lat)
Equation 1.10.3
(17)
f) Io throughout the day - use arrays with ω ranging from -75 to +75. Use center of the hour. We could have used a third parametric table. duplicate i = 7, 18
(18)
ωi = 15 · (i − 12) − 7.5
(19)
Io,i = IZero, (n, Lat, ωi − 7.5 [deg] , ωi + 7.5 [deg])
(20)
end
(21)
g) Find solar radiation in band 0.64 to 1.1 micro meters. From Table 1.3.1a f0,0.64 = 0.402
(22)
f0,1.1 = 0.744
(23)
∆f = f0,1.1 − f0,0.64
(24)
Solar1.1,0.64 = ∆f · Gsc#
(25)
Note that the solar constant, Gsc is a built-in constant. h) 10AM solar time is what local clock time. Long std = 105
(26)
SolarT Ime = 10
(27)
hr SolarT Ime − StdT ime = (4 [min/deg] · (Long std − Long) + EqnT ime (n)) · 0.016666667 min
(28)
Solution Variables in Main program
16
β = 60 [deg] ∆f = 0.342 [-] γ = 0 [deg] Long = 104.6 [deg] nsunrise = 46.78 [-] Rb = 5.377 θ = 71.5 [deg]
day = 16 δha = −12.61 [deg] Hrday = 10.6 [hr] Long std = 105 [deg] nsunset = 47.22 [-] SolarT Ime = 10 [hr]
δ = −12.95 [deg] f0,0.64 = 0.402 Ho = 21.51 MJ/m2 month = 2 ω = 75 [deg] Solar1.1,0.64 = 467.5 W/m2
Arrays Row 7 8 9 10 11 12 13 14 15 16 17 18
Io,i 2 MJ/m 0.04501 0.7697 1.638 2.346 2.847 3.107 3.107 2.847 2.346 1.638 0.7697 0.04501
ωi [deg] -82.5 -67.5 -52.5 -37.5 -22.5 -7.5 7.5 22.5 37.5 52.5 67.5 82.5
theta vx omega Run 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
θ [deg]
β [deg] 0 0 0 0 0 0 0 0 0 0 0 15 15 15 15 15 15 15 15 15 15 15
ω [deg] -75 -60 -45 -30 -15 0 15 30 45 60 75 -75 -60 -45 -30 -15 0 15 30 45 60 75
17
δchange = −0.1487 [deg] f0,1.1 = 0.744 Lat = 38.3 [deg] n = 47 [-] ωs = 79.53 [deg] StdT ime = 10.21 [hr]
23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66
30 30 30 30 30 30 30 30 30 30 30 45 45 45 45 45 45 45 45 45 45 45 60 60 60 60 60 60 60 60 60 60 60 90 90 90 90 90 90 90 90 90 90 90
-75 -60 -45 -30 -15 0 15 30 45 60 75 -75 -60 -45 -30 -15 0 15 30 45 60 75 -75 -60 -45 -30 -15 0 15 30 45 60 75 -75 -60 -45 -30 -15 0 15 30 45 60 75
Rb vs omega Run 1 2 3 4 5
θ [deg] -75 -74 -73 -72 -71
Rb 5.377 4.626 4.106 3.724 3.432
18
6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59
-70 -69 -68 -67 -66 -65 -64 -63 -62 -61 -60 -59 -58 -57 -56 -55 -54 -53 -52 -51 -50 -49 -48 -47 -46 -45 -44 -43 -42 -41 -40 -39 -38 -37 -36 -35 -34 -33 -32 -31 -30 -29 -28 -27 -26 -25 -24 -23 -22 -21 -20 -19 -18 -17
3.201 3.015 2.861 2.731 2.621 2.526 2.444 2.372 2.308 2.251 2.2 2.154 2.113 2.075 2.041 2.009 1.98 1.954 1.929 1.907 1.885 1.866 1.848 1.83 1.815 1.8 1.786 1.772 1.76 1.748 1.737 1.727 1.717 1.708 1.7 1.691 1.684 1.676 1.669 1.663 1.657 1.651 1.645 1.64 1.635 1.63 1.626 1.622 1.618 1.614 1.611 1.607 1.604 1.602
19
60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113
-16 -15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37
1.599 1.596 1.594 1.592 1.59 1.588 1.587 1.585 1.584 1.583 1.582 1.581 1.58 1.58 1.58 1.579 1.579 1.579 1.58 1.58 1.58 1.581 1.582 1.583 1.584 1.585 1.587 1.588 1.59 1.592 1.594 1.596 1.599 1.602 1.604 1.607 1.611 1.614 1.618 1.622 1.626 1.63 1.635 1.64 1.645 1.651 1.657 1.663 1.669 1.676 1.684 1.691 1.7 1.708
20
114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151
38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75
1.717 1.727 1.737 1.748 1.76 1.772 1.786 1.8 1.815 1.83 1.848 1.866 1.885 1.907 1.929 1.954 1.98 2.009 2.041 2.075 2.113 2.154 2.2 2.251 2.308 2.372 2.444 2.526 2.621 2.731 2.861 3.015 3.201 3.432 3.724 4.106 4.626 5.377
21
theta vs omega
Rb vs omega
22
Io vs omega
23
P02 01
Equations 2.01
Determination of irradiation from strip-chart records.
The areas under the curves of Figure 2.5.1 are estimated by graphical integration. Using hour time increments, i.e., the time spans between the vertical lines on the plots, estimates for the hour increments are as shown in the EES LookUp table. The sum of the column can be found by first converting the LookUp table values in to arrays: duplicate i = 1, 12
(1)
Cleari = Lookup(i, ‘Clear day’ )
(2)
Cloudy i = Lookup(i, ‘Cloudy day’ )
(3)
end
(4)
The sum of each row is found by: ClearDay = Sum(Cleari , i = 1, 12)
(5)
CloudyDay = Sum(Cloudy i , i = 1, 12)
(6)
In the units usually used, these are 18.0 and 5.7 MJ/m2, respectively. MJ/m2 ClearDayM J = ClearDay · 0.0036 W · hr/m2 MJ/m2 CloudyDayM J = CloudyDay · 0.0036 W · hr/m2
(7)
(8)
For such a simple set of calculations this problem is easier done by hand. Howver, if this processes were to be repeated for a year, some sort of program would be needed.
Solution ClearDay = 5010
ClearDayM J = 18.04
CloudyDay = 1640
Arrays Row 1 2
Clear2i W/m 40 190
Cloudy2 i W/m 10 30
24
CloudyDayM J = 5.904
3 4 5 6 7 8 9 10 11 12
360 530 640 720 720 650 530 360 200 70
60 120 200 210 210 300 270 120 90 20
25
P02 02
Equations 2.02 Division of a terrestrial solar spectrum into 10 equal increments using Table 2.6.1. This problem is essentially solved in Table 2.6.1. The wavelength divisions for the increments are the wavelengths given in the table for energy bands of 0.1, 0.2, 0.3, etc. The energy midpoints are the wavelengths in the table for 0.05, 0.15, 0.25, etc. For the first one-tenth:increment, the wavelength range is 0.300 to 0.479 and the energy midpoint wavelength is 0.434.
26
P02 03
Equations ¯ from average sunshine hour data, n 2.03 Estimation of H ¯ , from Table 2.7.1 Use the Parametric table for the two months. a = 0.30;
b = 0.34;
n = AveDay (month)
Lat = 43 [deg]
(1)
EES function for Table 1.6.1
(2)
Dec = 23.45 · sin (.9863 · (284 + n))
Eqn 1.6.1a
(3)
SunSetHrAng = arccos (− tan (Lat) · tan (Dec))
Eqn 1.6.10
(4)
Gsc = Gsc# Built-in EES constant for the solar constant (5) 2 ¯ Zero = 1 × 10·6 MJ/m · Gsc · 3600 · 24 · (1 + .033 · Cos(.9863 · n)) · (Cos(Lat) · Cos(Dec) · sin (SunSetHrAng) H J/m2 π +0.01745 · SunSetHrAng · sin (Lat) · sin (Dec))
Eqn 1.10.3
AvgDayLngth = (2/15) · arccos (− tan (Lat) · tan (Dec)) ¯ H ¯ Zero = a + b · n H/ ¯ /AvgDayLngth
(6)
Eqn 1.6.11
(7)
Eqn 2.7.2
(8)
¯ of 6.4 and 23.0 for January and June, compared to estimates of 6.2 and 22.0 shown in the table. Note: Appendix G gives Hs
Solution a = 0.3 Gsc = 1367 W/m2 month = 6
AvgDayLngth = 15.12 ¯ = 22.0 MJ/m2 H n = 162
b = 0.34 ¯ Zero = 41.77 MJ/m2 H n ¯ = 10.1
Dec = 23.09 [deg] Lat = 43 [deg] SunSetHrAng = 113.4 [deg]
Table 1 Run
month
n ¯
n
1 2
1 6
4.5 10.1
17 162
Dec [deg] -20.92 23.09
SunSetHrAng [deg] 69.12 113.4
27
¯ HZero2 MJ/m 13.37 41.77
AvgDayLngth 9.22 15.12
¯ H 2 MJ/m 6.2 22.0
P02 04
Equations 2.4a
Calculation of hourly clearness index with Equation 2.9.3 and diffuse fraction with Equation 2.10.1 n = 9; HourAngle1 = −15 [deg] ; HourAngle2 = 0 [deg] ; Lat = 40 [deg] ; I = 0.402 MJ/m2
(1)
Io = IZero, (n, Lat, HourAngle1, HourAngle2)
(2)
kT = I/Io
EES function or Eqn 1.10.4
Eqn 2.9.3
fd = Idif f /I (kT )
(3)
EES function or Eqn 2.10.1 or Fig 2.10.2
(4)
Id /I = fd
(5)
b Calculation of a day’s clearness index with Equation 2.9.2 and daily diffuse fraction from Eqn 2.11. H = 4.48 MJ/m2
(6)
Ho = HZero, (n, Lat)
(7)
KT,day = H/Ho
EES function or Eqn 2.11.1 or Fig 2.11.2.
Eqn 2.9.2
(8)
Fd,day = Hdif f /H (Lat, n, KT,day )
(9)
Solution fd = 0.984 HourAngle2 = 0 [deg] Io = 2.331 MJ/m2 n = 9 [-]
Fd,day = 0.880 Ho = 14.37 MJ/m2 kT = 0.1724
H = 4.48 MJ/m2 I = 0.402 MJ/m2 KT,day = 0.3117
28
HourAngle1 = −15 [deg] Id = 0.3958 Lat = 40 [deg]
P02 05
Equations 2.5
Calculation of monthly beam and diffuse using Equation 2.12.1
¯ = 18.9; H
Lat = 42.8;
n = AveDay (month) ¯ o = HZero, (n, lat) H ¯ T = H/ ¯ H ¯o K
month = 3
(1)
EES Function or Table 1.6.1
(2)
EES Function or Eqn 1.10.3
(3)
Eqn 2.9.1
(4)
¯T Fdif f = Hdif f bar/Hbar Lat, n, K
EES Function or Eqn 2.12.1 or Fig 2.12.2
(5)
¯ d /H ¯ Fdif f = H
(6)
¯ =H ¯b + H ¯d H
(7)
Solution Fdif f = 0.2289 ¯ o = 26.13 MJ/m2 H n = 75
¯ = 18.9 MJ/m2 H ¯ T = 0.7234 [-] K
¯ b = 14.57 MJ/m2 H Lat = 42.8 [deg]
29
¯ d = 4.327 MJ/m2 H month = 3
P02 06
Equations 2.6 Estimate radiation on tilted surface using various methods starting with daily total radiation. Lat = 43 [deg] ; n = 20; H = 8.0 MJ/m2
(1)
hrAng = −22.5 [deg] ;
(2)
slope = 60 [deg] ;
Surf AzAng = 0
a) total radiation in hour 10 to 11. rT = rt, (hrAng, lat, n)
EES function or Eqn 12.13.2 or Figure 2.13.1
I = H · rT
(3) (4)
b) beam and diffuse radiation in hour 10 to 11. This problem can be done two ways. Method 1 - find beam and diffuse for the day and then for the hour. Ho = HZero, (n, lat) KT,day = H/Ho
EES function or Eqn 1.10.3 or Fig 1.10.1 or Table 1.10.1
Eqn 2.9.2
Fdif f,day = Hdif f /H (Lat, n, KT,day )
(5) (6)
EES function or Eqn 2.11.2 or Fig 2.11.2
30
(7)
Fdif f,day = Hd /H
(8)
H = H d + Hb
(9)
rd = rd, (hrAng, Lat, n) rd = Id,1 /Hd
EES function or Eqn 2.13.4 or Fig2.13.2
(10)
Eqn2.13.3
I = Ib,1 + Id,1
(11)
I is from part a
(12)
Method 2 - find beam and diffuse for the hour Io = IZero, (n, Lat, HrAng − 7.5deg , HrAng + 7.5 [deg])
(13)
kT,hour = I/Io
(14)
Eqn 2.9.3
Fdif f,hour = Idif f /I (kT,hour )
EES function or Eqn 2.10.2 or Fig 2.10.3
(15)
Fdif f,hour = Id,2 /I
(16)
I = Id,2 + Ib,2
(17)
I is from part a
We will use method 2 results since it relies on one less relationship. Ib = Ib,2 ;
Id = Id,2
(18)
c) find Rb Rb = CosT heta/CosZen
Eqn 1.8.1
(19)
CosT heta = cosIncAng (Lat, n, HrAng, Slope, Surf AzAng) CosZen = CosZenAng (Lat, n, HrAng)
EES function or Eqn 1.6.1
EES function or Eqn 1.6.5
(20) (21)
d) Find radiation on tilted surface if all treated as beam. IT,allBeam = Rb · I
(22)
e) Estimation of IT assuming diffuse plus ground-reflected radiation are together isotropic IT,isotropic = Rb · Ib + Id
(23)
f) Estimation of IT assuming isotropic sky (Liu & Jordan model), Equation 2.15.1 Fcs =
1 + Cos(slope) 2
Fcs + Fcg = 1;
ρg,nosnow = 0.2;
(24) ρg,snow = 0.7
(25)
IT,LJ,nosnow = Ib · Rb + Id · Fcs + ρg,nosnow · I · Fcg
(26)
IT,LJ,snow = Ib · Rb + Id · Fcs + ρg,snow · I · Fcg
(27)
g) Estimation of IT with Hay & Davies model, Equation 2.16.4 AA = Ib /Io
(28)
IT,HD,nosnow = (Ib + AA · Id ) · Rb + Id · (1 − AA) · Fcs + ρg,nosnow · I · Fcg
(29)
31
IT,HD,snow = (Ib + AA · Id ) · Rb + Id · (1 − AA) · Fcs + ρg,snow · I · Fcg
(30)
h) Estimation of IT with the HDKR model, Equation 2.16.7 p f = Ib /I
(31)
3 IT,HDKR,nosnow = (Ib + AA · Id ) · Rb + Id · (1 − AA) · Fcs · 1 + f · sin (slope/2) + ρg,nosnow · I · Fcg
(32)
3 IT,HDKR,snow = (Ib + AA · Id ) · Rb + Id · (1 − AA) · Fcs · 1 + f · sin (slope/2) + ρg,snow · I · Fcg
(33)
i) Estimation of IT by the Perez model, Equation 2.16.14 IT,P erez,nosnow = Ib ·Rb +Id ·(1 − F 1)· IT,P erez,snow = Ib ·Rb +Id ·(1 − F 1)· a = max (0, cosT heta)
(1 + Cos(Slope)) 1 − Cos(Slope) +Id ·F 1·a/b+Id ·F 2·sin (Slope)+I·ρg,snow · (35) 2 2
Eqn 2.16.9
b = max (Cos(85), cosZen) =
(1 + Cos(Slope)) 1 − Cos(Slope) +Id ·F 1·a/b+Id ·F 2·sin (Slope)+I·ρg,nosnow · (34) 2 2
Eqn 2.16.9
Id +Ib /cosZen Id
(36) (37) 3
+ 5.535 · 10−6 · (arccos (cosZen)) 3 1 + 5.535 · 10−6 · (arccos (cosZen))
Eqn 2.16.10
(38)
=2.24. The coefficients from Table 2.16.1 are then: f 11 = 0.873;
f 12 = −0.392;
f 13 = −0.362;
f 22 = −0.462;
f 21 = 0.226;
f 23 = 0.001
(39)
Note that m=1/cosZen from Eqn 1.5.1 and Ion =Io /CosZen so that ∆ = Id /Io
Eqn 2.16.11
(40)
F 1 = max (0, (f 11 + f 12 · ∆ + π · arccos (cosZen) · f 13/180)) F 2 = f 21 + f 22 · ∆ + π · arccos (cosZen) · f 23/180
Eqn 2.16.12
Eqn 2.16.13
(42)
Solution Variables in Main program a = 0.93 CosT heta = 0.93 = 2.24 f 11 = 0.873 F 2 = 0.1061 f 23 = 0.001 Fdif f,day = 0.3927 hrAng = −22.5 [deg] 2 Ho = 13.75 MJ/m Ib,1 = 0.7459 MJ/m2 2 Id,1 = 0.4596 MJ/m 2 IT,allBeam = 2.83 MJ/m IT,HD,nosnow = 2.34 MJ/m2 IT,LJ,nosnow = 2.05 MJ/m2 IT,P erez,snow = 2.54 MJ/m2 Lat = 43 [deg] ρg,snow = 0.7 rT = 0.1507
AA = 0.3399 CosZen = 0.3965 f = 0.7514 f 12 = −0.392 f 21 = 0.226 Fcg = 0.25 Fdif f,hour = 0.4353 [-] Hb = 4.858 MJ/m2 2 I = 1.205 MJ/m Ib,2 = 0.6807 MJ/m2 Id,2 = 0.5248 MJ/m2 2 IT,HDKR,nosnow = 2.36 MJ/m IT,HD,snow = 2.49 MJ/m2 IT,LJ,snow = 2.20 MJ/m2 KT,day = 0.5817 n = 20 Rb = 2.346 32 slope = 60 [deg]
(41)
b = 0.3965 ∆ = 0.262 F 1 = 0.3492 f 13 = −0.362 f 22 = −0.462 Fcs = 0.75 2 H = 8 MJ/m Hd = 3.142 MJ/m2 Ib = 0.6807 MJ/m2 Id = 0.5248 MJ/m2 Io = 2.003 MJ/m2 2 IT,HDKR,snow = 2.51 MJ/m 2 IT,isotropic = 2.12 MJ/m IT,P erez,nosnow = 2.39 MJ/m2 kT,hour = 0.6019 [-] ρg,nosnow = 0.2 rd = 0.1463 [-] Surf AzAng = 0 [deg]
P02 07
Equations 2.7
Calculation of hour’s radiation on a south-facing vertical surface, starting with daily total.
We have 4 model choices: Liu and Jordan, Hay and Davies, HDKR and Perez. They all follow the same basic procedure. Here we will solve in detail using the Liu and Jordan model. HourAngle = 7.5; day = 22; month = 12; H = 8.80 MJ/m2 ; Lat = 43; GrRef = 0.7 (1) HrAng1 = −15;
Slope = 90;
n = nDay (month, day)
HrAng2 = 0;
HrAng = −7.5;
Surf AzAng = 0;
ρg = 0.7
EES function or Table 1.6.1
(2) (3)
Method A: Find the hour‘s total radiation with Equation 2.13.2, calculate I o for that hour, and then estimate the beam and diffuse compone s horizontal radiation. This provides the basis for estimating radiation on the vertical surface with Equation 2.15.1. rT = rt, (hrAng, lat, n)
EES function or Eqn 2.10.1
(4)
I = H · rT
(5)
Io = IZero, (n, Lat, HrAng1, HrAng2) kt = I/Io
EES function or Eqn 1.10.3 or Table 1.10.1 or Fig 1.10.1
Eqn 2.9.3
fd = Idif f /I (kt )
(6) (7)
EES function or Eqn 2.10.1
(8)
Id = I · fd
(9)
I = Ib + Id
(10)
CosT heta = cosIncAng (Lat, n, HrAng, Slope, Surf AzAng) CosZenith = CosZenAng (Lat, n, HrAng) Rb = CosT heta/CosZenith
EES function or Eqn 1.6.2
EES function or Eqn 1.6.5
Eqn 1.8.2
(11) (12) (13)
On the tilted surface using Liu and Jordan method: Ib,T = Rb · Ib Id,sky = Id ·
(14)
1 + Cos(slope) 2
Id,ground = I · ρg ·
(15)
1 − Cos(slope) 2
(16)
33
The total radiation on the vertical surface is then: IT = Ib,T + Id,sky + Id,ground
(17)
We could use the EES functions for L&J, H&D and HDKR models: IT,LJ = IT,LJ, (I, Lat, n, slope, Surf AzAng, ρg , HrAng)
(18)
IT,HD = IT,HD, (I, Lat, n, slope, Surf AzAng, ρg , HrAng)
(19)
IT,HDKR = IT,HDKR, (I, Lat, n, slope, Surf AzAng, ρg , HrAng)
(20)
Note that the L&J model is very conservative while the H&D and the HDKR models give similar results. Method B: Divide the day‘s radiation into beam and diffuse, using Equation 2.11.1, and then get the hour’ s diffuse from the day‘s diffuse using Equation 2.13.4. Equation 2.13.2 gives the hour’ s total from the day‘s total, and the hour’ s beam is then the difference between the hour‘s total and the hour’ s diffuse.
Solution CosT heta = 0.9113 GrRef = 0.7 HrAng1 = −15 [deg] Ib,T = 3.114 2 Io = 1.992 MJ/M IT,LJ = 3.812 MJ/m2 n = 356 Slope = 90 [deg]
CosZenith = 0.3939 H = 8.8 MJ/m2 HrAng2 =0 [deg] Id = 0.266 MJ/m2 IT = 3.812 MJ/m2 kt = 0.8092 ρg = 0.7 Surf AzAng = 0 [deg]
day = 22 HourAngle = 7.5 [deg] I = 1.612 MJ/m2 2 Id,ground = 0.5642 MJ/m IT,HD = 4.138 MJ/m2 Lat = 43 [deg] Rb = 2.314
34
fd = 0.165 HrAng = −7.5 [deg] 2 Ib = 1.346 MJ/m 2 Id,sky = 0.133 MJ/m IT,HDKR = 4.152 MJ/m2 month = 12 rT = 0.1832
P02 08
Equations 2.8
Calculation of daily average radiation on sloped surfaces
a) Using the Liu & Jordan model, Equation 2.19.1 month = 3;
Lat = 35.1;
n = AveDay (month)
¯ T = H/ ¯ H ¯o K
ρg = .2;
KT bar = .68;
¯ = 20.06 MJ/m2 H
EES function or Table 1.6.1
¯b = R ¯ BEAM, (n, Lat, slope) R ¯ o = HZero, (n, lat) H
Slope = 35;
(1) (2)
Eqn 2.19.3
(3)
EES function or Eqn 10.1.3
(4)
Eqn 2.9.1
(5)
¯T fdif f use = Hdif f bar/Hbar Lat, n, K
EES function or Eqn 2.12.1 or Fig 2.12.2
(6)
¯ d /H ¯ fdif f use = H
(7)
¯ =H ¯b + H ¯d H
(8)
¯ T,LJ = H ¯b · R ¯b + H ¯ d · 1 + Cos(slope) + H ¯ · ρg · 1 − Cos(slope) H 2 2
Eqn 2.19.1
(9)
This result could have been found from the single EES function. ¯ T,LJ =H ¯ T,LJ ( H, ¯ Lat, n, Slope, ρg ) H b) Using the Klein-Theilacker model, Equation 2.20.4a, et seq: ¯ T,KT = H ¯ T,KT, H, ¯ Lat, n, Slope, ρg H EES function or Eqn 2.20.4
(10)
Solution fdif f use = 0.2698 ¯ o = 29.58 MJ/m2 H KT bar = 0.68 ρg = 0.2
¯ = 20.06 MJ/m2 H ¯ T,KT = 23.78 MJ/m2 H Lat = 35.1 [deg] ¯ b = 1.281 R
¯ b = 14.65 MJ/m2 H ¯ T,LJ = 24.05 MJ/m2 H month = 3 Slope = 35 [deg]
35
¯ d = 5.412 MJ/m2 H ¯ T = 0.6781 K n = 75
P02 09
Equations 2.9 Monthly variation of radiation on surfaces of various orientations. Lat = 35.1;
ρg = 0.2;
Surf AzAng = 0
(1)
Monthly average solar radiation is found in Appendix G. The following Arrays were entered using the Edit/Insert or Modify Array menu ¯ 1 = 11.54; H
¯ 2 = 15.23; H
¯ 7 = 28.25; H
¯ 8 = 26; H
slope1 = 0;
slope2 = 20;
¯ 3 = 20.06; H ¯ 9 = 22.39; H slope3 = 35;
¯ 4 = 25.30; H ¯ 10 = 17.56; H
¯ 5 = 28.82; H ¯ 11 = 12.87; H
slope4 = 50;
slope5 = 90
¯ 6 = 30.41 H
(2)
¯ 12 = 10.53 H
(3) (4)
The EES duplicate statement will be used with arrays to simplify the calculations. : duplicate i = 1, 12
(5)
monthi = i
(6)
duplicate j = 1, 5
(7)
¯ ¯ ¯ H T,i, j = HT,KT, Hi , Lat, AveDay (i) , Slopej , ρg end j
(8) (9)
end i
(10)
The results are shown in the plot. The variable month was used for the abcissa. This solution is based on the Klein-Theilacker model; a solution based on the Liu and Jordan model would be similar to that of Problem 2.8a.
Arrays
36
Row 1 2 3 4 5 6 7 8 9 10 11 12
¯ Hi 2 MJ/m 11.54 15.23 20.06 25.30 28.82 30.41 28.25 26.00 22.39 17.56 12.87 10.53
slopei [deg] 0 20 35 50 90
¯ HT,i,12 MJ/m 11.5 15.1 19.9 25.2 28.7 30.4 28.2 25.9 22.2 17.4 12.8 10.5
¯ HT,i,22 MJ/m 16.12 19.36 22.94 26.48 28.27 28.97 27.29 26.43 24.75 21.58 17.67 15.16
¯ HT,i,32 MJ/m 18.57 21.3 23.78 25.84 26.29 26.32 25.09 25.27 25.08 23.31 20.18 17.68
Plot 1
37
¯ HT,i,42 MJ/m 19.93 22.03 23.32 23.86 23.04 22.48 21.75 22.84 24.04 23.73 21.51 19.17
¯ HT,i,52 MJ/m 17.84 17.93 16.25 13.28 10.46 9.218 9.581 11.76 15.41 18.46 18.88 17.57
monthi 1 2 3 4 5 6 7 8 9 10 11 12
P02 10
Equations 2.10
Estimation of monthly radiation on a vertical surface.
This solution is based on the Liu & Jordan method, Equation 2.19.1; the Klein-Theilacker method, Equation 2.20.4 could also be used. See Problem 2.8. M onth = 1;
Lat = 36;
n = AveDay (month) ¯ o = HZero, (n, Lat) H ¯ T = H/ ¯ H ¯o K
Slope = 90;
ρg = 0.2;
¯ T = 0.47 K
(1)
EES function or Table 1.6.1
(2)
EES function or Eqn 1.10.3 or Fig 1.10.1 or Table 1.10.1
(3)
Eqn 2.9.1
(4)
¯T Fd = Hdif f Bar/Hbar Lat, n, K
EES function or Eqn 2.12.2 or Fig 2.12.2
(5)
¯ d /H ¯ Fd = H
(6)
¯ =H ¯b + H ¯d H
(7)
dec = dec (n)
EES function or Eqn 1.6.1
ωs = arccos (− tan (Lat) · tan (dec))
(8)
Eqn 1.6.10 or EES function
(9)
ωs0 = min (arccos (− tan (Lat) · tan (Dec)), arccos (− tan (Lat − Slope) · tan (Dec))) ¯ b = Cos(lat − slope) · Cos(dec) · sin (ωs0 ) + (π/180) · ωs0 · sin (lat − slope) · sin (dec) R Cos(lat) · Cos(dec) · sin (ωs ) + (π/180) · ωs · sin (lat) · sin (dec) ¯ T,LJ = H ¯b · R ¯b + H ¯ d · 1 + Cos(slope) + H ¯ · ρg · 1 − Cos(slope) H 2 2
Eqn 2.19.3b Eqn 2.19.3a
Eqn 2.19.1
The Klein-Theilacker method is simple when the EES functions are used. ¯ T,KT = H ¯ T,KT, H, ¯ Lat, n, Slope, ρg H
Fd = 0.4213 ¯ o = 17.66 MJ/m2 H Lat = 36 [deg] ωs0 = 73.88 [deg]
¯ = 8.302 MJ/m2 H ¯ T,KT = 11.4 MJ/m2 H M onth = 1 ρg = 0.2
38
(11)
(12)
(13)
Solution dec = −20.92 [deg] ¯ d = 3.498 MJ/m2 H ¯ T = 0.47 K ωs = 73.88 [deg] Slope = 90 [deg]
(10)
¯ b = 4.805 MJ/m2 H ¯ T,LJ = 12.1 MJ/m2 H n = 17 ¯ b = 1.976 R
P02 11
Equations 2.11
Estimation of October radiation on a sloped surface.
This solution is based on the Liu & Jordan method, Equation 2.19.1 and an assumption of GrRef = 0.2. We also had to assume the monthly diffuse fraction (Eqn 1.9.1) is valid. Lat = 43;
Slope = 58;
n = AveDay (month) ¯ o = HZero, (n, Lat) H ¯ T = H/ ¯ H ¯o K
ρg = 0.2;
¯ T = 0.50; K
month = 10
(1)
EES function or Table 1.6.1
(2)
EES function or Eqn 1.10.3
(3)
Eqn 1.9.1
(4)
¯T Fd = HDIF F BAR/HBAR Lat, n, K
EES function or Eqn 2.12.1
(5)
¯ d /H ¯ Fd = H
(6)
¯ =H ¯b + H ¯d H
(7)
¯b = R ¯ BEAM, (n, Lat, slope) R
EES function or Eqn 2.19.3
¯ · ρg · 1 − Cos(slope) ¯T = H ¯b · R ¯b + H ¯ d · 1 + Cos(slope) + H H 2 2 ¯=H ¯ T /H ¯ R
(8) Eqn 2.19.1
Eqn 2.19.2
(10)
Solution Fd = 0.3911 ¯ o = 20.9 MJ/m2 H month = 10 ¯ b = 1.816 R
¯ = 10.45 MJ/m2 H ¯ T = 15.17 MJ/m2 H n = 288 Slope = 58 [deg]
(9)
¯ b = 6.363 MJ/m2 H ¯ T = 0.5 K ρg = 0.2
39
¯ d = 4.087 MJ/m2 H Lat = 43 [deg] ¯ = 1.452 R
P02 12
Equations 2.12
Calculation of average radiation on sloped surfaces using Liu & Jordan, and Klein & Theilacker
Lat = 43; a)
Slope = 90;
ρg = 0.2;
¯ T = 0.54; K
¯ = 21.96; H
month = 7
(1)
Using the Liu & Jordan model, Equation 2.19.1:
n = AveDay (month)
EES function or Table 1.6.1
(2)
¯ o = HZero, (n, Lat) H
EES function or Equation 1.10.3 ¯T Eqn 2.12.1 Fd = HDIF F BAR/HBAR Lat, n, K
(3)
¯ d /H ¯ Fd = H
(5)
¯ =H ¯b + H ¯d H
(6)
¯b = R ¯ BEAM, (n, Lat, slope) R
(4)
EES function or Eqn 2.19.3
¯ T,LJ = H ¯b · R ¯b + H ¯ d · 1 + Cos(slope) + H ¯ · ρg · 1 − Cos(slope) H 2 2
(7) Eqn 2.19.2
(8)
b)
Using the Klein-Theilacker model, Equation 2.20.4a, et seq. ¯ T,KT = H ¯ T,KT, H, ¯ Lat, n, Slope, ρg H EES function or Eqns 2.20.4 et seq.
(9)
Solution Fd = 0.3917 ¯ o = 40.56 MJ/m2 H Lat = 43 ¯ b = 0.2482 R
¯ = 21.96 MJ/m2 H ¯ T,KT = 10.1 MJ/m2 H month = 7 Slope = 90
¯ b = 13.36 MJ/m2 H ¯ T,LJ = 9.813 MJ/m2 H n = 198
40
¯ d = 8.602 MJ/m2 H ¯ T = 0.54 K ρg = 0.2
P02 13
Equations 2.13
Hourly radiation on a horizontal surface for a standard clear day using Hottel’s method.
The solution will use a parametric table. Arrays could have been used. hrAng=-52.5 Use for checking - put hrand in Parametric Table for final results. Lat = 46.5;
A = 0.432;
n = nDay (month, day) dec = dec (n)
month = 12;
Day = 23
(1)
EES function or Table 1.6.1
(2)
EES function or Eqn 1.6.1
(3)
CosZenith = Cos(Lat) · Cos(Dec) · Cos(HrAng) + sin (Lat) · sin (Dec)
Eqn 1.6.5
(4)
τb = a0 + a1 · exp (−k/cosZenith) Eqn 2.8.1a 2 a∗0 = 0.4237 − 0.008216 1/km2 · (6 [km] − A)
Eqn 2.8.1b
(6)
2 a∗1 = 0.5055 + 0.00595 1/km2 · (6.5 [km] − A)
Eqn 2.8.1c
(7)
2 k ∗ = 0.2711 + 0.01858 1/km2 · (2.5 [km] − A)
Eqn 2.8.1d
(8)
r0 = a0 /a∗0 ; r0 = 1.03;
r1 = a1 /a∗1 ; r1 = 1.01;
τd = 0.271 − 0.294 · τb τd = Id /I0
(5)
rk = k/k ∗ rk = 1.00
(9)
Table 2.8.1
(10)
Eqn 2.8.5
(11)
Just below Eqn 2.8.5
(12)
I0 = IZero, (n, Lat, HrAng − 7.5deg , HrAng + 7.5 [deg])
EES function or Eqn 1.10.4
(13)
Ib = I0 · τb
(14)
I = Ib + Id
(15)
Solution A = 0.432 [km] a∗1 = 0.7246 hrAng = −7.5 [deg] Id = 0.245 MJ/m2 month = 12 rk = 1
a0 = 0.1741 CosZenith = 0.3377 I = 0.985 MJ/m2 k = 0.3506 n = 357 τb = 0.4332
a1 = 0.7318 Day = 23 I0 = 1.707 MJ/m2 k ∗ = 0.3506 r0 = 1.03 τd = 0.1436
Table 1
41
a∗0 = 0.169 dec = −23.43 [deg] Ib = 0.740 MJ/m2 Lat = 46.5 [deg] r1 = 1.01
Run 1 2 3 4
hrAng [deg] -52.5 -37.5 -22.5 -7.5
τb
τd
0.1931 0.3148 0.3971 0.4332
0.2142 0.1785 0.1543 0.1436
I0 2 MJ/m 0.4825 1.073 1.491 1.707
Ib 2 MJ/m 0.093 0.338 0.592 0.740
42
Id 2 MJ/m 0.103 0.192 0.230 0.245
I 2 MJ/m 0.197 0.529 0.822 0.985
P02 14
Equations 2.14
Calculation of hourly beam radiation on a horizontal, north-south axis tracking surface.
Lat = 38;
day = 7;
month = 1
(1)
The following arrays were added using the Edit/Insert Modify Array menu item. Hour1 = 8.5;
Hour2 = 9.5;
Hour7 = 14.5;
Hour3 = 10.5;
Hour4 = 11.5;
Hour5 = 12.5;
Hour6 = 13.5
Hour8 = 15.5
Ibn,1 = 0.35;
Ibn,2 = 0.70;
Ibn,7 = 1.80;
Ibn,8 = 1.42
HrAng 1 = −52.5; HrAng 6 = 22.5;
(2) (3)
Ibn,3 = 2.66;
Ibn,4 = 3.05;
Ibn,5 = 3.30;
Ibn,6 = 3.19
(4) (5)
HrAng 2 = −37.5; HrAng 7 = 37.5;
HrAng 3 = −22.5;
HrAng 4 = −7.5;
HrAng 5 = 7.5
HrAng 8 = 52.5
n = nDay (month, day) 284 + n Dec = 23.45 [deg] · sin 360 · 365
(6) (7) (8)
Eqn 1.6.1
(9)
duplicate i = 1, 8
(10)
CosZenithAnglei = Cos(Lat) · Cos(Dec) · Cos(HrAng i ) + sin (Lat) · sin (Dec) Eqn 1.6.5 q 2 2 2 CosT hetaN S,i = (cosZenithAnglei ) + (Cos(Dec)) · (sin (HrAng i )) Eqn 1.7.3a
(11)
Ib,i = Ibn,i · CosT hetaN S,i
(13)
end
(14)
Solution day = 7 month = 1
(12)
Dec = −22.42 [deg] n=7
Lat = 38 [deg]
43
Arrays Row
Houri
1 2 3 4 5 6 7 8
8.5 9.5 10.5 11.5 12.5 13.5 14.5 15.5
HrAng i [deg] -52.5 -37.5 -22.5 -7.5 7.5 22.5 37.5 52.5
Ibn,i 2 MJ/m 0.35 0.70 2.66 3.05 3.30 3.19 1.80 1.42
CosT hetaN S,i
CosZenithAnglei
0.762 0.659 0.563 0.502 0.502 0.563 0.659 0.762
0.209 0.343 0.438 0.487 0.487 0.438 0.343 0.209
44
Ib,i 2 MJ/m 0.27 0.46 1.50 1.53 1.66 1.80 1.19 1.08
P02 15
Equations 2.15
Estimation of diffuse and beam components using clearness indexes
Lat = 45;
HrAng1 = −30 [deg] ;
month = 2; a)
HrAng2 = −15 [deg] ;
H = 10.80 MJ/m2 ;
day = 8;
I = 1.57 MJ/m2
¯ = 8.67 MJ/m2 H
(1) (2)
Use Equation 2.10.1 for hourly calculation:
n = nDay (month, day)
(3)
Io = IZero, (n, Lat, HrAng1, HrAng2) kT,hr = I/Io
EES function or Eqn 1.10.4
Eqn 2.9.3
fd = Idif f /I (kT,hr )
(4) (5)
EES function or Eqn 2.10.1 or Figure 2.10.2
(6)
Id /I = fd
(7)
I = Id + Ib
(8)
b Estimation of beam and diffuse components for a day. Ho = HZero, (n, Lat) KT,day = H/Ho
EES function or Eqn 1.10.3 or Figure 1.10.1 or Table 1.10.1
Eqn 2.9.2
fd,day = Hdif f /H (Lat, n, KT,day )
(9) (10)
EES function or Eqn 2.11.1 or Figure 2.11.2
(11)
fd,day = Hd /H
(12)
H = H d + Hb
(13)
c) Estimation of average day’s beam and diffuse. nave = AveDay (month) ¯ o = HZero, (nave , Lat) H ¯ T = H/ ¯ H ¯o K
(14) EES function or Eqn 1.10.3 or Figure 1.10.1 or Table 1.10.1
Eqn 2.9.1
¯T F¯d = HDIF F BAR/HBAR Lat, nave , K
(15) (16)
EES function or Eqn 2.12.1 or Figure 2.12.2
(17)
¯ d /H ¯ F¯d = H
(18)
¯ =H ¯b + H ¯d H
(19)
45
Solution day = 8 H = 10.8 MJ/m2 ¯ = 8.67 MJ/m2 H Hd = 2.089 MJ/m2 Id = 0.3617 MJ/m2 kT,hr = 0.7091 nave = 47
F¯d = 0.3992 HrAng1 = −30 [deg] ¯ b = 5.209 MJ/m2 H Ho = 15.84 MJ/m2 Io = 2.214 MJ/m2 Lat = 45 [deg]
fd = 0.2304 HrAng2 = −15 [deg] ¯ d = 3.461 MJ/m2 H I = 1.57 MJ/m2 ¯ T = 0.4918 K month = 2
46
fd,day = 0.1934 Hb = 8.711 MJ/m2 ¯ o = 17.63 MJ/m2 H Ib = 1.208 MJ/m2 KT,day = 0.6818 n = 39
P02 16
Equations 2.16
Estimation of an hour’s radiation on south facing surfaces with three sky models
HrAng 1 = −30; slope = 60;
HrAng 2 = −15;
HrAng = −22.5;
Lat = 45;
ρg = 0.7;
I = 1.57;
month = 2;
day = 8
(1)
Surf AzAng = 0
(2)
The first part, the estimation of the beam-diffuse distribution, and some angles are common to all of the models. n = nDay (month, day)
(3)
Io = IZero, (n, Lat, HrAng 1 , HrAng 2 ) kT = I/Io
EES function or Eqn 1.10.4
(4)
Eqn 2.9.3
fd = Idif f /I (kT )
(5)
EES function or Eqn 2.10.1 or Fig 2.10.2
(6)
fd = Id /I
(7)
I = Ib + Id
(8)
Dec = 23.45 [deg] · sin 360 ·
284 + n 365
Eqn 1.6.1 or EES function
(9)
CosT heta = Cos(lat − slope) · Cos(dec) · Cos(hrAng) + sin (lat − slope) · sin (dec) CosZenith = Cos(lat) · Cos(dec) · Cos(hrAng) + sin (lat) · sin (dec)
Eqn 1.6.7a or EES function(10)
Eqn 1.6.5 or EES function
(11)
θ = arccos (CosT heta)
(12)
θz = arccos (CosZenith)
(13)
Rb = CosT heta/CosZenith
Eqn 1.8.1 or EES function
(14)
a) Use of Liu & Jordan model, Equation 2.15.1 Ib,T,LJ = Ib · Rb Id,T,LJ = Id ·
beam by Liu and Jordan
1 + Cos(slope) 2
Ig,T,LJ = I · ρg ·
1 − Cos(slope) 2
(15)
diffuse by Liu and Jordan ground reflected by Liu and Jordan
47
(16) (17)
IT,LJ = Ib,T,LJ + Id,T,LJ + Ig,T,LJ
(18)
The total radiation of the sloped surface and could have been easily found from the EES function: IT,LJx = IT,LJ, (I, Lat, n, Slope, Surf AzAng, ρg , HrAng)
(19)
b) Estimation of hourly radiation on sloped surface using the HDKR model, Equation 2.16.7 Ai = Ib /Io p f = Ib /I
Eqn 1.16.3
(20)
Eqn 2.16.7
(21)
Ib,T,HDKR = (Ib + Id · Ai ) · Rb
(22)
1 + Cos(slope) 2 Id,T,HDKR = Id · (1 − Ai ) · · 1 + f · sin (slope/2) 2 1 − Cos(slope) Ig,T,HDKR = I · ρg · 2 IT,HDKR = Ib,T,HDKR + Id,T,HDKR + Ig,T,HDKR
(23) (24) (25)
The total radiation of the sloped surface and could have been easily found from the EES function: IT,HDKRx = IT,HDKR, (I, Lat, n, Slope, Surf AzAng, ρg , HrAng)
(26)
(c Estimation of radiation on sloped surface using the Perez model, Eqn 2.16.14 a = max (0, CosT heta)
Eqn 2.16.9
b = max (Cos(85), CosZenith)
(27)
Eqn 2.16.9
(28)
Calculate to use to get f11, f12, etc from Table 2.16.1: Id +Ib /CosZenith + 5.535 · 10−6 1/deg3 · θz3 Id = (1 + (5.535 · 10−6 ) [1/deg3 ] · θz3 )
Eqn 2.16.10
(29)
is 4.11. The coefficients from Table 2.16.1 are: f 11 = 1.132;
f 12 = −1.237;
f 13 = −0.412;
f 21 = 0.288;
f 22 = −0.823;
f 23 = .056
(30)
Equation 2.16.11 reduces to: ∆ = Id /Io
(31)
F1 = max 0,
f 11 + f 12 · ∆ + π · θz ·
F2 = f 21 + f 22 · ∆ + π · θz ·
f 23 180 [deg]
f 13 180 [deg]
Eqn 2.16.12
Eqn 2.16.13
Ib,T,P erez = Ib · Rb (1 + Cos(Slope)) Id,T,P erez = Id · (1 − F1 ) · + F1 · a/b + F2 · sin (Slope) 2 1 − Cos(Slope) 2 = Ib,T,P erez + Id,T,P erez + Ig,T,P erez
(32) (33) (34) (35)
Ig,T,P erez = I · ρg ·
(36)
IT,P erez
(37)
Solution Variables in Main program
48
a = 0.9291 CosZenith = 0.4403 = 4.106 f 13 = −0.412 F1 = 0.4706 HrAng 1 = −30 [deg] Ib,T,HDKR = 2.97 MJ/m2 2 Id,T,HDKR = 0.15 MJ/m 2 Ig,T,LJ = 0.27 MJ/m IT,HDKRx = 3.38 MJ/m2 kT = 0.7091 [-] ρg = 0.7 θ = 21.7 [deg]
Ai = 0.5457 day = 8 f = 0.8773 f 21 = 0.288 F2 = 0.216 HrAng 2 = −15 [deg] Ib,T,LJ = 2.55 MJ/m2 2 Id,T,LJ = 0.27 MJ/m 2 Ig,T,P erez = 0.27 MJ/m IT,LJ = 3.10 MJ/m2 Lat = 45 [deg] Rb = 2.11 θz = 63.88 [deg]
Key Variables Ig,T,LJ = 0.27 MJ/m2 Id,T,LJ = 0.27 MJ/m2 Ib,T,LJ = 2.55 MJ/m2 2 IT,LJ = 3.10 MJ/m Ig,T,HDKR = 0.27 MJ/m2 Id,T,HDKR = 0.15 MJ/m2 Ib,T,HDKR = 2.97 MJ/m2 IT,HDKR = 3.39 MJ/m2 Ig,T,P erez = 0.27 MJ/m2 Ib,T,P erez = 2.55 MJ/m2 Id,T,P erez = 0.57 MJ/m2 IT,P erez = 3.39 MJ/m2
a a a a b b b b c c c c
49
b = 0.4403 Dec = −15.52 [deg] f 11 = 1.132 f 22 = −0.823 fd = 0.2304 [-] I = 1.57 MJ/m2 Ib,T,P erez = 2.55 MJ/m2 Id,T,P erez = 0.57 MJ/m2 2 Io = 2.21 MJ/m IT,LJx = 3.10 MJ/m2 month = 2 slope = 60 [deg]
CosT heta = 0.9291 ∆ = 0.1634 f 12 = −1.237 f 23 = 0.056 HrAng = −22.5 [deg] 2 Ib = 1.208 MJ/m Id = 0.3617 MJ/m 2 Ig,T,HDKR = 0.27 MJ/m2 IT,HDKR = 3.39 MJ/m2 IT,P erez = 3.39 MJ/m2 n = 39 Surf AzAng = 0 [deg]
P02 17
Equations 2.17
From an hour’s radiation on horizontal surface, get beam radiation on tracking surface
HrAng 1 = −30; slope = 60;
HrAng 2 = −15;
HrAng = −22.5;
Lat = 45;
ρg = 0.7;
I = 1.57;
month = 2;
day = 8
(1)
Surf AzAng = 0
(2)
The first part, the estimation of the beam-diffuse distribution, and some angles are common to all of the models. n = nDay (month, day)
(3)
Io = IZero, (n, Lat, HrAng 1 , HrAng 2 ) kT = I/Io
EES function or Eqn 1.10.4
(4)
Eqn 2.9.3
fd = Idif f /I (kT )
(5)
EES function or Eqn 2.10.1 or Fig 2.10.2
(6)
fd = Id /I
(7)
I = Ib + Id
(8)
Dec = 23.45 [deg] · sin 360 ·
284 + n 365
Eqn 1.6.1 or EES function
CosZenith = Cos(lat) · Cos(dec) · Cos(hrAng) + sin (lat) · sin (dec)
(9) Eqn 1.6.5 or EES function
θz = arccos (CosZenith)
(10) (11)
.5 2 2 CosT hetaN S = cosZenith2 + (Cos(Dec)) · (sin (HrAng))
Eqn 1.7.3a or EES function
(12)
θN S = arccos (CosT hetaN S )
(13)
Ib,N S = Ib · CosT hetaN S
(14)
Solution CosT hetaN S = 0.5743 fd = 0.2304 2 I = 1.57 MJ/m Io = 2.214 MJ/m2 n = 39 θN S = 54.95 [deg]
cosZenith = 0.4403 HrAng = −22.5 [deg] Ib = 1.208 MJ/m2 kT = 0.7091 ρg = 0.7 θz = 63.88 [deg]
day = 8 HrAng 1 = −30 [deg] Ib,N S = 0.694 MJ/m2 Lat = 45 [deg] slope = 60 [deg]
50
Dec = −15.52 [deg] HrAng 2 = −15 [deg] Id = 0.3617 MJ/m2 month = 2 Surf AzAng = 0 [deg]
P02 18
Equations 2.18
Calculation of monthly average beam, diffuse, ground-reflected, and total radiation on sloped surface
Lat = 43;
Slope = 60
(1)
Monthly average radiation data are from Appendix G. The arrays were entered using ‘Insert/modify array’ in the Edit menu. ¯ 1 = 6.44; ¯ 2 = 9.89; ¯ 3 = 12.86; ¯ 4 = 16.05; ¯ 5 = 21.36; H H H H H ¯ 7 = 22.58; H
¯ 8 = 20.33; H
¯ 9 = 14.59; H
¯ 10 = 10.48; H
ρg,1 = 0.7;
ρg,2 = 0.7;
ρg,3 = 0.4;
ρg,4 = 0.2;
ρg,7 = 0.2;
ρg,8 = 0.2;
ρg,9 = 0.2;
ρg,10 = 0.2;
¯ 6 = 23.04 H
¯ 11 = 6.37; H
ρg,5 = 0.2; ρg,11 = 0.4;
¯ 12 = 5.74 H
ρg,6 = 0.2 ρg,12 = 0.7
(2) (3) (4) (5)
This solution uses the Liu and Jordan model duplicate i = 1, 12
(6)
ni = AveDay (i)
(7)
Ho,i = HZero, (ni , Lat) ¯ ¯ T,i = Hi K Ho,i
EES function or Eqn 1.10.3
Eqn 2.9.1
(9)
¯ T,i Fd,i = HDIF F BAR/HBAR Lat, ni , K
EES function or Eqn 2.12.1 or Figure 2.12.1
¯ d,i = H ¯ i · Fd,i H ¯ b,i = R ¯ BEAM, (ni , Lat, Slope) R
(8)
(10) (11)
EES function or Eqn 2.19.3
(12)
¯i = H ¯ b,i + H ¯ d,i H
(13)
¯ T,b,i = H ¯ b,i · R ¯ b,i H
(14)
¯ T,d,i = H ¯ d,i · 1 + Cos(Slope) H 2 1 − Cos(Slope) ¯ T,g,i = H ¯ i · ρg,i · H 2 ¯ ¯ ¯ ¯ HT,i = HT,b,i + HT,d,i + HT,g,i
(15) (16) (17)
end
(18)
This problem is easily done by the following call to an EES procecdure. Note that the EES procedure returns the three components. Duplicate i=1,12 n[i]=AveDay(i ) ¯ T,LJ,p, (H[i], ¯ ¯ T,b,i , H ¯ T,d,i , H ¯ T,g,i , H ¯ T,i ) CALL H Lat, n[i], Slope, ρg,i : H End
51
Arrays Row
ni
1 2 3 4 5 6 7 8 9 10 11 12
17 47 75 105 135 162 198 228 258 288 318 344
Row 1 2 3 4 5 6 7 8 9 10 11 12
¯ Hi 2 MJ/m 6.44 9.89 12.86 16.05 21.36 23.04 22.58 20.33 14.59 10.48 6.37 5.74
¯ Hb,i 2 MJ/m 3.81 6.26 7.27 8.77 13.03 14.25 14.08 12.84 8.42 6.40 3.45 3.39
¯ b,i R [] 2.788 2.043 1.419 0.960 0.711 0.617 0.657 0.844 1.214 1.813 2.565 3.062
ρg,i 0.7 0.7 0.4 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.4 0.7
Ho,i 2 MJ/m 13.36 18.81 26.03 33.78 39.42 41.78 40.56 35.92 28.80 20.90 14.62 11.91
¯ HT,b,i2 MJ/m 10.61 12.80 10.31 8.42 9.26 8.79 9.25 10.84 10.22 11.60 8.85 10.39
¯ T,i K [] 0.482 0.526 0.494 0.475 0.542 0.551 0.557 0.566 0.507 0.501 0.436 0.482
¯ HT,d,i2 MJ/m 1.98 2.72 4.19 5.46 6.25 6.59 6.38 5.62 4.63 3.06 2.19 1.76
Fd,i [] 0.409 0.367 0.435 0.453 0.390 0.381 0.377 0.368 0.423 0.390 0.459 0.409
¯ Hd,i 2 MJ/m 2.63 3.63 5.59 7.28 8.33 8.79 8.50 7.49 6.17 4.08 2.92 2.35
¯ HT,g,i2 MJ/m 1.13 1.73 1.29 0.80 1.07 1.15 1.13 1.02 0.73 0.52 0.64 1.00
52
¯ HT,i 2 MJ/m 13.71 17.25 15.79 14.69 16.58 16.53 16.76 17.47 15.58 15.18 11.67 13.15
P02 19
Equations 2.19
Estimation of monthly average radiation on a vertical, south-facing surface
Lat = 39.8;
Slope = 90
¯ 1 = 6.63; H
¯ 2 = 9.77; H
¯ 7 = 23.36; H
(1) ¯ 3 = 12.97; H
¯ 8 = 20.5; H
¯ 4 = 17.2; H
¯ 9 = 16.5; H
¯ 5 = 21.17; H
¯ 10 = 12.13; H
ρg,1 = 0.7;
ρg,2 = 0.7;
ρg,3 = 0.4;
ρg,4 = 0.2;
ρg,7 = 0.2;
ρg,8 = 0.2;
ρg,9 = 0.2;
ρg,10 = 0.2;
¯ 6 = 23.8 H
¯ 11 = 7.68; H
ρg,5 = 0.2; ρg,11 = 0.4;
(2)
¯ 12 = 5.57 H
(3)
ρg,6 = 0.2
(4)
ρg,12 = 0.7
(5)
a Use EES Procedures for both the Liu & Jordan (Section 2.19) and Klein & Theilacker (Section 2.20) methods, . (The detailed L & J solution is essentially the same as that of Problem 2.18.) duplicate i = 1, 12
(6)
ni = AveDay (i)
(7)
¯ T,LJ,p, H ¯ i , Lat, ni , Slope, ρg,i : H ¯ T b,LJ,i , H ¯ T d,LJ,i , H ¯ T,g,LJ,i , H ¯ T,LJ,i call H
¯ T,KT,p, H ¯ i , Lat, ni , Slope, ρg,i : H ¯ T b,KT,i , H ¯ T d,KT,i , H ¯ T,g,KT,i , H ¯ T,KT,i call H end
Lat = 39.8 [deg]
Slope = 90 [deg]
Arrays
1 2 3 4 5
(9) (10)
Solution
Row
(8)
¯ Hi 2 MJ/m 6.6 9.8 13.0 17.2 21.2
ρg,i
ni
0.7 0.7 0.4 0.2 0.2
17 47 75 105 135
¯ HT b,LJ,i MJ/m2 8.20 8.99 6.68 4.68 3.17
¯ HT d,LJ,i MJ/m2 1.53 2.04 2.97 3.72 4.21
53
¯ HT,g,LJ,i MJ/m2 2.32 3.42 2.59 1.72 2.12
¯ HT,LJ,i2 MJ/m 12.05 14.45 12.24 10.13 9.50
6 7 8 9 10 11 12
23.8 23.4 20.5 16.5 12.1 7.7 5.6
Row
¯ HT b,KT,i MJ/m2 7.60 8.45 6.49 4.88 3.42 2.80 3.28 5.01 7.48 9.41 8.55 6.57
1 2 3 4 5 6 7 8 9 10 11 12
0.2 0.2 0.2 0.2 0.2 0.4 0.7
162 198 228 258 288 318 344
¯ HT d,KT,i MJ/m2 1.53 2.04 2.97 3.72 4.21 4.34 4.21 3.82 3.17 2.40 1.64 1.39
2.57 3.03 4.74 7.48 9.92 9.17 7.12
4.34 4.21 3.82 3.17 2.40 1.64 1.39
2.38 2.34 2.05 1.65 1.21 1.54 1.95
¯ H T,g,KT,i MJ/m2 2.32 3.42 2.59 1.72 2.12 2.38 2.34 2.05 1.65 1.21 1.54 1.95
¯ HT,KT,i MJ/m2 11.45 13.91 12.05 10.32 9.75 9.52 9.83 10.88 12.30 13.02 11.73 9.91
54
9.29 9.58 10.62 12.30 13.53 12.35 10.46
P02 20
Equations 2.20
Estimation of effect of finite horizontal diffuse reflector on IT
Ib = 1.28;
Id = 0.33;
p = 2.1 [m] ;
I = 1.61;
m = 4.2 [m] ;
Rb = 2.31;
n = 2.1 [m]
ρg = 0.2;
Slope = 90;
dimensions of window and reflector - see Figure 2.17.1
(1) (2)
ρr,a = 0.85
diffuse reflectance for case a
(3)
ρr,b = 0.70
diffuse reflectance for case b
(4)
Ac = m · n
(5)
Ar = p · m
(6)
Equation 2.17.1 is for the case of a horizontal reflector, and is applicable here. In this solution, the diffuse radiation on the reflector is assumed to be reduced to half of that on an unobscured horizontal plane, on the assumption that the building of which the window is part is large in extent. It is also assumed that the diffuse radiation on the ground beyond the reflector is not significantly affected by the building. (These two assumptions tend to offset.) View factor algebra Fc,s + Fc,r + Fc,g = 1 Ac · Fc,r = Ar · Fr,c Fc,s =
Sum of view factors from collector to surroundings must be 1.0 Reciprosity relationship
(7) (8)
1 + Cos(slope) 2
(9)
For ψ=90, n/m=2.1/4.2=0.5, p/m=2.1/4.2=0.5 then from Figure 2.17.2a: Fr,c = 0.25
(10)
The radiation on the reflector is Ar *Ir =Ar * Ib +As *Id *Fs -r. But with reciprosity As *Fs,r =Ar *Fr,s so the radiation on the receiver is: Ir = Ib + Id · Fr,s Fr,s + Fr,c = 1
(11) Sum of view factors from reflector to surroundings must be 1.0
55
(12)
Finally, the radiation on the window for case a with reflector reflectance ρr,a is: IT,a = Ib · Rb + Id · Fc,s + Ir · ρr,a · Fc,r + I · ρg · Fc,g
Eqn 2.17.1
(13)
And for case b with reflector reflectance ρr,b is: IT,b = Ib · Rb + Id · Fc,s + Ir · ρr,b · Fc,r + I · ρg · Fc,g
(14)
Note that the difference is not very large.
Solution Ac = 8.82 m2 Fr,c = 0.25 Ir = 1.528 p = 2.1 [m] Slope = 90 [deg]
Ar = 8.82 m2 Fr,s = 0.75 IT,a = 3.53 MJ/m2 ρg = 0.2
Fc,g = 0.25 2 I = 1.61 MJ/m IT,b = 3.47 MJ/m2 ρr,a = 0.85
56
Fc,r = 0.25 Ib = 1.28 MJ/m2 m = 4.2 [m] ρr,b = 0.7
Fc,s = 0.5 Id = 0.33 MJ/m2 n = 2.1 [m] Rb = 2.31
P02 21
Equations 2.21
Calculation of beam radiation on the aperture of tracking collectors Lat = 40; month = 2; day = 20; Ibn,AM = 2.31 MJ/m2 ; Ibn,P M = 1.67 MJ/m2
(1)
hrAng AM = −7.5 [deg] ;
(2)
hrAng P M = 52.5 [deg]
n = nDay (month, day)
(3)
δ = dec (n)
(4)
a)
Tracking on horizontal EW axis:
For the AM .5 2 2 CosT hetaEW AM = 1 − (Cos(δ)) · (sin (HrAng AM ))
Eqn 1.7.2a
IT,b,EW,AM = CosT hetaEW AM · Ibn,AM
(5) (6)
For the PM .5 2 2 CosT hetaEW P M = 1 − (Cos(δ)) · (sin (HrAng P M ))
Eqn 1.7.2a
IT,b,EW,P M = CosT hetaEW P M · Ibn,P M b)
(7) (8)
Tracking on horizontal NS axis:
For the AM CosZenithAngleAM = Cos(Lat) · Cos(δ) · Cos(HrAng AM ) + sin (Lat) · sin (δ) Eqn 1.6.5 .5 2 2 2 CosT hetaN S AM = (cosZenithAngleAM ) + (Cos(δ)) · (sin (HrAng AM )) Eqn 1.7.3a
(10)
IT beamN S,AM = CosT hetaN S AM · Ibn,AM
(11)
(9)
For the PM CosZenithAngleP M = Cos(Lat) · Cos(δ) · Cos(HrAng P M ) + sin (Lat) · sin (δ) Eqn 1.6.5 .5 2 2 2 CosT hetaN S P M = (cosZenithAngleP M ) + (Cos(δ)) · (sin (HrAng P M )) Eqn 1.7.3a
(12)
IT beamN S,P M = CosT hetaN S P M · Ibn,P M
(14)
57
(13)
Solution CosT hetaEW AM = 0.9918 CosT hetaN S P M = 0.8435 day = 20 hrAng P M = 52.5 [deg] 2 IT beamN S,AM = 1.45 MJ/m IT,b,EW,P M = 1.05 MJ/m2 n = 51
CosT hetaEW P M = 0.6292 CosZenithAngleAM = 0.615 δ = −11.58 [deg] Ibn,AM = 2.31 MJ/m2 IT beamN S,P M = 1.41 MJ/m2 Lat = 40 [deg]
58
CosT hetaN S AM = 0.6282 CosZenithAngleP M = 0.3278 hrAng AM = −7.5 [deg] 2 Ibn,P M = 1.67 MJ/m IT,b,EW,AM = 2.29 MJ/m2 month = 2
P02 22
Equations 2.22
Calculation of utilizable energy
$Arrays off I¯1 = 1.52;
I¯2 = 1.15;
I¯3 = 0.68;
(1)
Use Figure 2.22.3, which is specific to Blue Hill For the hour pair 11-12 and 12-1: Xc,1 = 0.72/1.52
(2)
φ1 = 0.65
(3)
From Fig 2.18.3 at Xc =0.47
For hour pair 10-11 and 1-2 at Xc =0.63: Xc,2 = 0.72/1.15;
φ2 = 0.56
(4)
For hour pair 9-10 and 2-3 at Xc =1.06: Xc,3 = 0.72/0.68;
φ3 = 0.35
(5)
N = N umDay (1)
(6)
The monthly average utilizable energy for these 3 hour pairs is: U E = 2 · N · Sum I¯i · φi , i = 1, 3
(7)
Solution ¯ = 1.52 MJ/m2 I[1] φ[1] = 0.65 Xc,1 = 0.47
¯ = 1.15 MJ/m2 I[2] φ[2] = 0.56 Xc,2 = 0.63
¯ = 0.68 MJ/m2 I[3] φ[3] = 0.35 Xc,3 = 1.06
59
N = 31 U E = 115.9 MJ/m2
P02 23
Equations 2.23
Calculation of utilizability and utilizable energy using generalized utilizability charts, Figure 2.23.1.
The major steps are calculation of month’s diffuse fraction, I¯T , Xc , Rb , and phi. ¯ T = 0.49; K ρg = 0.6;
¯ = 8.67; H
Ic,1 = 536;
Surf AzAng = 0;
n = AveDay (month)
Ic,2 = 322;
HrAng = 22.5;
Lat = 45
month = 2;
(1) Slope = 60
EES function or Table 1.6.1
(2) (3)
Find the monthly average daily diffuse fraction ¯T fd = HDIF F BAR/HBAR Lat, n, K
(4)
¯ d /H ¯ = fd H
(5)
Find radiation in the given hour. rT = rt, (HrAng, Lat, n) rd = rd, (hrAng, Lat, n)
EES function or Eqn 2.13.2
(6)
EES function or Eqn 2.13.4
(7)
¯d Id = rd · H
(8)
¯ I = rT · H
(9)
I = Ib + Id
(10)
Find radiation on the tilted surface Rb = RBeam, (Lat, n, HrAng, Slope, Surf AzAng)
EES function or Eqn 1.8.2
1 + Cos(slope) 1 − Cos(slope) I¯T = Ib · Rb + Id · + I · ρg · 2 2
Liu and Jordan model, Eqn 2.15.1
Find the two dimensionless critical radiation levels MJ/m2 0.0036 W·hr/m2 Xc1 = Ic,1 · 1 [hr] · Eqn2.22.3 - note needed conversion factor ¯IT
60
(11) (12)
(13)
Xc2
MJ/m2 0.0036 W·hr/m2 = Ic,2 · 1 [hr] · ¯IT
Eqn2.22.3 - note needed conversion factor
(14)
¯b To use Figures 2.23.1 we need to find R ¯b = R ¯ BEAM, (n, Lat, Slope) R
EES function or Eqn 2.19.3
(15)
¯ b = 2.17 and Xc =0.96 and Xc =0.58. Reading Figures 2.23.1 b and c and interpolating for R Reading and interpolating the graphs is difficult. φ1 = 0.31;
φ2 = 0.54
(16)
Finally, the utilizable energies are: Nday = N umDay (month)
(17)
U E 1 = 2 · Nday · φ1 · I¯T
from statement just under Eqn 2.22.2
(18)
U E 2 = 2 · Nday · φ2 · I¯T
from statement just under Eqn 2.22.2
(19)
We could use Eqns 2.23.5 - 2.23.8 to find utilizability δ = dec (n) ¯h = R
I¯T ¯ rT · H
¯ T · rT /rd κ ¯T = K
(20) Eqn 2.23.8 Eqn 2.23.7 - note this is kappabar,T not a lower case k since Kbar,T was already used.
Xm = 1.85 + 0.169 ·
g=
Xm − 1 (2 − Xm )
(21)
¯h Cos(slope) κ ¯T R + 0.0696 · − 0.981 · 2 κ ¯ 2T κ ¯ 2T Cos(δ)
Eqn 2.25.5c
Eqn 2.23.5b
(22) (23)
(24)
Since Xm > 2 we use the following for φ q 2 φ1×10 qn = abs (g) − g 2 + (1 + 2 · g) · (1 − Xc1 /Xm )
Eqn 2.23.5a
(25)
q 2 g 2 + (1 + 2 · g) · (1 − Xc2 /Xm )
Eqn 2.23.5a
(26)
φ2×10 qn = abs (g) −
From the above equation (with Xc1 and Xc2 ) are φ1 = 0.34 and φ2 = 0.54.
Solution δ = −12.95 ¯ = 8.67 MJ/m2 H I¯T = 2.00 MJ/m2 κ ¯ T = 0.5082 n = 47 φ2 = 0.54 ¯ b = 2.17 R Slope = 60 [deg] Xc1 = 0.964 [-]
fd = 0.401 [-] ¯ d = 3.48 MJ/m2 H Ic,1 = 536 MJ/m2 ¯ T = 0.49 K Nday = 28 φ2eqn = 0.54 ¯ h = 1.63 R Surf AzAng = 0 [deg] Xc2 = 0.579 [-]
g = −2.9 2 I = 1.23 MJ/m Ic,2 = 322 MJ/m2 Lat = 45 [deg] φ1 = 0.31 ρg = 0.6 rd = 0.1365 U E 1 = 34.7 MJ/m2 Xm = 2.526
61
HrAng =22.5 [deg] Ib = 0.75 MJ/m2 Id = 0.47 MJ/m2 month = 2 φ1eqn = 0.34 Rb = 1.94 rT = 0.1416 U E 2 = 60.5 MJ/m2
P02 24
Equations 2.24
Calculation of daily utilizability and month’s utilizable energy ¯ = 28.7; month = 12; Lat = −34; Itc = 124 W/m2 ; H ρg = .2 HourAngleN oon = 0; Slope = 20; MJ/m2 IT,c = IT c · 1 [hr] · 0.0036 W · hr/m2
Surf AzAng = 180
Southern hemi facing N
(1) (2) (3)
¯T Find K n = AveDay (month) ¯ o = HZero, (n, Lat) H ¯ T = H/ ¯ H ¯o K
EES function or Table 1.6.1
(4)
EES function or Eqn 1.10.3
(5)
Eqn 2.9.1
(6)
¯ Find radiation on collector and R ¯T f¯d = HDIF F BAR/HBAR Lat, n, K
EES function or Eqn 2.12.1 or Fig 2.12.2
(7)
¯ d /H ¯ = f¯d H
(8)
¯ =H ¯b + H ¯d H
(9)
¯b = R ¯ BEAM, (n, Lat, Slope) R
(10)
¯T = H ¯b · R ¯b + H ¯ d · 1 + Cos(slope) + ρg · H ¯ · 1 − Cos(slope) H 2 2
(11)
The following could have been used instead of the previous five equations ¯ T =H ¯ T,LJ, (H, ¯ Lat, n, Slope, ρg ) H ¯=H ¯ T /H ¯ R
(12)
Find Rn HrAng noon = 0 [deg]
(13)
rt,n = rt, (HrAng noon , Lat, n)
EES function or Eqn 2.13.2 or Fig 2.13.1
(14)
rd,n = rd, (HrAng noon , Lat, n)
EES function or Eqn 2.13.4 or Fig 2.13.2
(15)
62
Rb,n = RBeam, (Lat, n, HrAng noon , Slope, Surf AzAng) ¯ H = H;
¯T KT = K
EES function or Eqn 1.6.7
A day that has the same total radiation as the monthly average.
fd = Hdif f /H (Lat, n, KT )
EES function or Eqn 2.11.1
(16) (17) (18)
fd = Hd /H
(19)
H = Hb + Hd rd,n · Hd 1 + Cos(slope) 1 − Cos(slope) rd,n · Hd · Rb,n + · + ρg · Rn = 1 − rt,n · H rt,n · H 2 2
(20) Eqn 2.24.2
(21)
Find dimensionless critical level ¯c = X
IT,C ¯ rT,n · Rn · H
(22)
FInd φ¯ ¯ T + 4.031 · K ¯ T2 a = 2.943 − 9.271 · K
(23)
¯ T − 3.602 · K ¯ T2 b = −4.345 + 8.853 · K
(24)
¯ T + 2.936 · K ¯ T2 c = −0.170 − 0.306 · K ¯ · X ¯c + C · X ¯2 φ¯ = exp a + b · Rn /R c
(25) (26)
φ¯could have been found from the known data using the EES function. ¯ H, ¯ n, Lat, Slope, ρg , IT c φ¯f unction = Φ Note that ITc is in W/m2
(27)
The utilizable energy is then: ¯ T · φ¯ U E = N umDay (month) · H
(28)
Solution a = −1.382 fd = 0.3355 Hb = 19.1 MJ/m2 ¯ o = 44.1 MJ/m2 H IT,c = 0.446 MJ/m2 month = 12 ρg = 0.2 rd,n = 0.113 Surf AzAng = 180 [deg]
b = −0.1106 H = 28.7 MJ/m2 ¯ = 28.7 MJ/m2 H ¯ HT = 26.99 ¯ T = 0.650 K n = 344 [-] ¯ = 0.9403 R Rn = 1.001 U E = 678 MJ/m2
c = 0.873 HourAngleN oon = 0 ¯ b = 20.26 H Hd = 9.6 MJ/m2 KT = 0.650 φ¯ = 0.810 ¯ b = 0.9194 R rt,n = 0.123 ¯ c = 0.1265 X
63
f¯d = 0.2941 HrAng noon = 0 [deg] ¯ d = 8.442 H Itc = 124 W/m2 Lat = −34 [deg] φ¯f unction = 0.810 Rb,n = 1.006 Slope = 20 [deg]
P02 25
Equations 2.25
Calcuation of beam radiation on a window under a long overhang using the shading plane concept Ibn = 800 · 3600 · 10−6 ;
Lat = 38;
Slope2 = 144.5; H = 2 [m] ;
Slope1 = 105.6;
W = 3 [m] ;
HrAng = 22.5
(1)
Surf AzAng = 0
(2)
G = 0.5 [m] ;
P = 0.7 [m]
(3)
Use to set up problem. Month and day will be set in the parametric table. The angle ψ 1 and ψ 2 for shading planes 1 and 2 are: P ψ1 = arctan G+H
(4)
ψ2 = arctan (P/G)
(5)
The slopes for these two planes are then: slope1 = ψ1 + 90 [deg]
(6)
slope2 = ψ2 + 90 [deg]
(7)
The areas of the two planes are: A1 = W ·
G+H Cos(ψ1 )
(8)
A2 = W ·
G Cos(ψ2 )
(9)
Use Equation 1.6.7a to calculate CosTheta to get the angles of incidence of beam radiation on these planes, and also to constrain CosTheta to positive values. n = nDay (month, day) dec = dec (n)
EES function or Table 1.6.1
EES function or Eqn 1.6.1
(10) (11)
cosT heta1 = max (0, Cos(Lat − Slope1 ) · Cos(Dec) · Cos(HrAng) + sin (Lat − Slope1 ) · sin (Dec))
Eqn 1.6.7(12)
cosT heta2 = max (0, Cos(Lat − Slope2 ) · Cos(Dec) · Cos(HrAng) + sin (Lat − Slope2 ) · sin (Dec))
Eqn 1.6.7(13)
64
Beam radiation on window: Eb,window = Ibn · (A1 · CosT heta1 − A2 · CosT heta2 )
(14)
We need to estimate the total radiation knowing only the bean normal radiation. CosZen = CosZenAng (Lat, n, HrAng)
(15)
Ib = Ibn · CosZen
(16)
Io = IZero, (n, Lat, HrAng − 7.5deg , HrAng + 7.5 [deg])
(17)
The following four eqns have to be solved simultaneously - this is time consuming by hand. I = Ib + Id
(18)
fd = Idif f /I (kT )
(19)
fd = Id /I
(20)
KT = I/Io
(21)
Use the Liu and Jordan model for finding the radiation on the two planes. ρg = 0.2
assumed for both Jan and April since we do not know the location
IT,1 = IT,LJ, (I, Lat, n, Slope1 , Surf AzAng, ρg , HrAng)
(22)
EES function or Eqn 2.15.1
(23)
IT,2 = IT,LJ, (I, Lat, n, Slope2 , Surf AzAng, ρg , HrAng)
(24)
Ewindow = A1 · IT,1 − A2 · IT,2
(25)
Note that the total radiation can be large compared to the beam radiation.
Table 1 Run
month
day
1 2
1 4
17 15
Run
kT
fd
1 2
0.7228 0.7359
0.2119 0.1966
Ib 2 MJ/m 1.326 2.359
Id 2 MJ/m 0.3563 0.5771
cosT heta2
cosT heta1
0.09782 0
I 2 MJ/m 1.682 2.936
0.6585 0.1954
Eb,window [MJ] 14.04 4.38
Io 2 MJ/m 2.327 3.989
IT,1 2 MJ/m 2.24 1.146
IT,2 2 MJ/m 0.62 0.5862
Ewindow
65
15.85 7.415
P02 26a
Equations 2.26a Solar radiation calculations Due to the nature of this problem it will be solved in 2 EES programs. First put hourly data into a lookup table Lonstd = 105 [deg] ; Lat = 38.3 [deg] ;
Lon = 104.8 [deg] ; Rotation = 15 [deg/hr] ;
month = 2;
day = 16;
Surf AzAng = 0 [deg]
66
ρg = 0.4
(1) (2)
hours = 24 [hr] ;
hour = 1 [hr] ;
zero = 0 [deg]
(3)
n = nDay (month, day)
(4)
a) For the hour 10-11 AM what is kT ? slope = 60 [deg]
(5)
I = Lookup(‘Data’ , 5, ‘I’ ) · hour · 0.0036
MJ/m2 W · hr/m2
(6)
Io = IZero, (n, Lat, HrAng1, HrAng2)
(7)
HrAng1 = −30 [deg]
(8)
Hour angle at 10AM
HrAN g2 = −15 [deg]
Hour angle at 11AM
(9)
kT = I/Io
(10)
b) What is IT ? What are the beam, diffuse and ground-reflected components of IT ? MJ/m2 Ib,n = Lookup(‘Data’ , 5, ‘I bn’ ) · hour · 0.0036 MJ/m2 W · hr/m2 HrAng3 = −22.5 [deg]
at the middle of the hour
(12)
Cos(θ) = CosIncAng (Lat, n, HrAng3, zero, Surf AzAng) Ib = Ib,n · Cos(θ) I = Ib + Id
(11)
convert Ib,n to Ib
incidence angle for horizontal radiation
(13) (14)
find Id
(15)
IT = Ib,T + Id,T + Ig,T
(16)
Rb = RBeam, (Lat, n, HrAng3, Slope, Surf AzAng)
(17)
Ib,T = Ib · Rb
(18)
Id,T = Id ·
1 + Cos(slope) 2
Ig,T = ρg · I ·
(19)
1 − Cos(slope) 2
(20)
c) Estimate beam normal radiation from I and compare the value with that given in the table. Use your estimated value of Ib,n and estimate IT. Compare to the value obtained in question 2. Id,estimated = I · Idif f /I (kT )
EES function or Eqn 2.10.1
(21)
I = Ib,estimated + Id,estimated
(22)
Ib,n,estimated · Cos(θ) = Ib,estimated
(23)
errorbn =
Ib,n − Ib,n,estimated · |100 %| Ib,n
(24)
67
Using the Liu and Jordan model IT,estimated = Ib,estimated · Rb + Id,estimated · errortotal =
1 − Cos(slope) 1 + Cos(slope) + ρg · I · 2 2
IT − IT,estimated · |100 %| IT
(25) (26)
IT from part b is 2.334. From part c with an estimated diffuse fraction IT is 2.316 for an ‘error’ of less than 1%. Do not expect such good agreement in general. d) Plot Io and kT versus time. Can do with arrays or Parametric table - use arrays. duplicate j = 1, 10
(27)
Ij = Lookup(‘Data’ , j, ‘I’ ) · hour · 0.0036
MJ/m2 W · hr/m2
(28)
hourj = Lookup(‘Data’ , j, ‘hour’ )
(29)
HrAng j = (hourj − 12) · Rotation
(30)
Io,j = IZero, kT,j =
n, Lat, HrAng j − 7.5, HrAng j + 7.5
(31)
Ij Io,j
(32)
end
(33)
$TabStops 3 cm
Solution Variables in Main program day = 16 hours = 24 [hr] I = 1.721 MJ/m2 2 Ib,n,estimated = 1.728 MJ/m 2 Id,T = 0.5396 MJ/m IT,estimated = 2.316 MJ/m2 Lonstd = 105 [deg] Rotation = 15 [deg/hr] θ = 55.41 [deg]
errorbn = 2.024 [%] HrAng1 =−30 [deg] 2 Ib = 1.001 MJ/m 2 Ib,T = 1.622 MJ/m Ig,T = 0.1721 MJ/m2 kT = 0.604 month = 2 Rb = 1.62 zero = 0 [deg]
Key Variables kT = 0.604 2 IT = 2.334 MJ/m 2 Ib,T = 1.622 MJ/m Id,T = 0.5396 MJ/m2 Ig,T = 0.1721 MJ/m2 Ib,n,estimated = 1.728 MJ/m2 IT,estimated = 2.316 MJ/m2 errorbn = 2.024 [%] errortotal = 0.7555 [%]
a b b b b c c c c
68
errortotal = 0.7555 [%] HrAng2 = −15 [deg] Ib,estimated = 0.9811 MJ/m2 Id = 0.7195 MJ/m2 Io = 2.847 MJ/m2 Lat = 38.3 [deg] n = 47 slope = 60 [deg]
hour = 1 [hr] HrAng3 = −22.5 [deg] Ib,n = 1.764 MJ/m2 Id,estimated = 0.7397 MJ/m2 IT = 2.334 MJ/m2 Lon = 104.8 [deg] ρg = 0.4 Surf AzAng = 0 [deg]
Arrays Row 1 2 3 4 5 6 7 8 9 10
Ii 2 MJ/m 0.2592 0.7848 1.462 1.811 1.721 2.293 1.584 1.667 1.188 0.5976
houri
HrAng i
7.5 8.5 9.5 10.5 11.5 12.5 13.5 14.5 15.5 16.5
-67.5 -52.5 -37.5 -22.5 -7.5 7.5 22.5 37.5 52.5 67.5
Io,i 2 MJ/m 0.770 1.638 2.346 2.847 3.107 3.107 2.847 2.346 1.638 0.770
kT,i 0.337 0.479 0.623 0.636 0.554 0.738 0.556 0.710 0.725 0.776
Data Row
hour
1 2 3 4 5 6 7 8 9 10
7.5 8.5 9.5 10.5 11.5 12.5 13.5 14.5 15.5 16.5
I 2 W/m 72 218 406 503 478 637 440 463 330 166
Ibn 2 W/m 254 377 741 705 490 853 485 739 756 838
Tamb [C] 2.8 5.4 8.0 10.6 12.1 13.5 15.0 14.8 14.6 14.4
Io and kT vs time
69
P02 26b
Equations 2.26 (parts e - i) Solar radiation calculations Due to the nature of this problem it will be solved in 2 EES programs. This program is solved pressing F3 or from ‘solve table’ in the calculate menu. The lookup table from the first part is also used here. Lonstd = 105 [deg] ; Lat = 38.3 [deg] ; hours = 24 [hr] ;
Lon = 104.8 [deg] ; Rotation = 15 [deg/hr] ; hour = 1 [hr] ;
month = 2;
day = 16;
ρg = 0.4
Surf AzAng = 0 [deg]
zero = 0 [deg]
(1) (2) (3)
n = nDay (month, day)
(4)
e) For slopes of 30, 45, 60 and 75 degrees, plot IT vs. time. Use the L&J method with Id estimated as in part c. Set up four parametric tables (one for each slope) with hour angle from -67.5 to +67.5. The hourly values of radiation are assumed to be numerically equal to the average for the hour MJ/m2 Ib,n = Lookup(‘Data’ , T ableRun#, ‘I bn’ ) · hour · 0.0036 W · hr/m2
(5)
CosZen = CosZenAng (Lat, n, HrAng)
(6)
Ib = Ib,n · CosZen
(7)
MJ/m2 I = Lookup(‘Data’ , T ableRun#, ‘I’ ) · hour · 0.0036 W · hr/m2
(8)
I = Ib + Id
(9)
Finds Id
Rb = RBeam, (Lat, n, HrAng, Slope, Surf AzAng) IT = Ib · Rb + Id ·
Eqn 1.8.1
1 + Cos(slope) 1 − Cos(slope) + I · ρg · 2 2
Eqn 2.15.1
(10) (11)
f) If the selection of slope is based on maximizing incident radiation, what slope would you select for this day? What would you think might be the answer to this question if a summer’s day had been used in the calculations? It does not matter very much but 60 degrees is the highest. g) What are H, KT and Ho? H is the sum of I over all hours. MJ/m2 H = Sumlookup(‘Data’ , ‘I’ ) · 0.0036 W · hr/m2
(12)
70
Ho = HZero, (n, Lat)
(13)
KT = H/Ho
(14)
h) Calculate HT and its components for a slope of 60 degrees. HT = Sumparametric(‘Slope=60’ , ‘I T’ )
(15)
¯ T and what are its components for the month of February a slope of 60 degrees? The monthly average i) What is H radiation for February is 13.30 MJ/m2. Use the Liu and Jordan method. ¯ = 13.3 MJ/m2 H From Appendix G (16) ¯ T,LJ,p, H, ¯ Lat, n, Slope, ρg : H ¯ T beam,LJ , H ¯ T dif f,LJ , H ¯ T GrRef,LJ , H ¯ T,LJ call H (17) $TabStops 1 cm
Solution Variables in Main program CosZen = 0.1537 hours = 24 [hr] ¯ T dif f,LJ = 2.853 MJ/m2 H HT = 21.19 MJ/m2 Id = 0.1338 MJ/m2 Lon = 104.8 [deg] ρg = 0.4 Surf AzAng = 0 [deg] Key Variables H = 13.37 MJ/m2 Ho = 21.51 MJ/m2 KT = 0.622 HT = 21.19 MJ/m2 ¯ T GrRef,LJ = 1.33 MJ/m2 H ¯ T beam,LJ = 17.07 MJ/m2 H ¯ T dif f,LJ = 2.853 MJ/m2 H ¯ T,LJ = 21.26 MJ/m2 H
day = 16 HrAng = 67.5 [deg] ¯ T GrRef,LJ = 1.33 MJ/m2 H 2 I = 0.60 MJ/m IT = 1.46 MJ/m2 Lonstd = 105 [deg] Rotation = 15 [deg/hr] zero = 0 [deg]
g g g h i i i i
Slope=30 Run 1 2 3 4 5 6 7 8 9 10
Slope [deg] 30 30 30 30 30 30 30 30 30 30
HrAng [deg] -67.5 -52.5 -37.5 -22.5 -7.5 7.5 22.5 37.5 52.5 67.5
I 2 MJ/m 0.26 0.78 1.46 1.81 1.72 2.29 1.58 1.67 1.19 0.60
IT 2 MJ/m 0.43 1.09 2.19 2.57 2.26 3.26 2.09 2.39 1.82 1.16
71
H = 13.37 MJ/m2 ¯ = 13.3 MJ/m2 H 2 ¯ HT,LJ = 21.26 MJ/m Ib = 0.4638 MJ/m2 KT = 0.622 month = 2 Rb = 2.793
hour = 1 [hr] ¯ T beam,LJ = 17.07 MJ/m2 H Ho = 21.51 MJ/m2 Ib,n = 3.017 MJ/m2 Lat = 38.3 [deg] n = 47 Slope = 60 [deg]
Slope=45 HrAng [deg] -67.5 -52.5 -37.5 -22.5 -7.5 7.5 22.5 37.5 52.5 67.5
I 2 MJ/m 0.26 0.78 1.46 1.81 1.72 2.29 1.58 1.67 1.19 0.60
IT 2 MJ/m 0.48 1.17 2.39 2.76 2.38 3.50 2.21 2.57 2.00 1.35
HrAng [deg] -67.5 -52.5 -37.5 -22.5 -7.5 7.5 22.5 37.5 52.5 67.5
I 2 MJ/m 0.26 0.78 1.46 1.81 1.72 2.29 1.58 1.67 1.19 0.60
IT 2 MJ/m 0.51 1.20 2.44 2.79 2.37 3.53 2.21 2.62 2.07 1.46
Slope [deg] 75 75 75 75 75 75 75 75 75 75
HrAng [deg] -67.5 -52.5 -37.5 -22.5 -7.5 7.5 22.5 37.5 52.5 67.5
I 2 MJ/m 0.26 0.78 1.46 1.81 1.72 2.29 1.58 1.67 1.19 0.60
IT 2 MJ/m 0.51 1.16 2.36 2.67 2.25 3.38 2.10 2.52 2.02 1.48
Row
hour
1 2 3
7.5 8.5 9.5
I 2 W/m 72 218 406
Ibn 2 W/m 254 377 741
Tamb [C] 2.8 5.4 8.0
Run 1 2 3 4 5 6 7 8 9 10
Slope [deg] 45 45 45 45 45 45 45 45 45 45
Slope=60 Run 1 2 3 4 5 6 7 8 9 10
Slope [deg] 60 60 60 60 60 60 60 60 60 60
Slope=75 Run 1 2 3 4 5 6 7 8 9 10
Data
72
4 5 6 7 8 9 10
10.5 11.5 12.5 13.5 14.5 15.5 16.5
503 478 637 440 463 330 166
705 490 853 485 739 756 838
10.6 12.1 13.5 15.0 14.8 14.6 14.4
IT vs time for various slopes
73
P03 01
Equations 3.1
Calculation of blackbody emissive power
Arrays are used but a Parametric Table could have been used. We can use the EES built-in constants, C1# and C2# λ1 = 10;
T1 = 1000
(1)
λ2 = 5;
T2 = 400
(2)
λ3 = 1;
T3 = 6000
(3)
duplicate i = 1, 3 Eλ,b,i =
λ5i · exp
(4) C1# C2# λi ·Ti
−1
(5)
end
(6)
To show all variables in the Solutions Window use: $Arrays off The results are consistent with Figure 3.4.1.
Solution Key Variables T [1] = 1000 [K] Eλ,b,1 = 1164 W/m2 -micrometer λ[1] = 10 [micrometer] T [2] = 400 [K] λ[2] = 5 [micrometer] Eλ,b,2 = 90 W/m2 -micrometer T [3] = 6000 [K] λ[3] = 1 [micrometer] Eλ,b,3 = 3.742 × 107 W/m2 -micrometer
a a a b b b c c c
74
P03 02
Equations 3.2
Determination of the fraction of blackbody radiation in a wavelength range
This can be done by computing the fraction of the energy in the spectrum from zero to the two wavelengths with Equation 3.6.4 and taking the difference. Table 3.6.1a could have been used to find the two fractions. T = 300 [K]
(1)
λ1 = 8 [micrometer]
(2)
λ2 = 14 [micrometer]
(3)
f2 = BlBodyF ract (λ2 · T )
EES function or Eqn 3.6.4
(4)
f1 = BlBodyF ract (λ1 · T )
EES function or Eqn 3.6.4
(5)
f = f2 − f1
(6)
Solution f = 0.3757 λ2 = 14 [micrometer]
f1 = 0.1402 T = 300 [K]
f2 = 0.516
λ1 = 8 [micrometer]
75
P03 03
Equations 3.3 We could use EES to integrate Eqn 3.6.3 or we could use Eqn 3.6.4 and sum for 10 terms. The second method is used for the SETP EES functions. Here we will do the integration using the EES integral function. SubprogramBBf (LT 1, LT 2 : f ) Z
LT 2
f = (c1#/sigma#) · LT 1
(1)
1 5 LT · (exp (C2#/LT ) − 1)
dLT
(2)
end
(3)
LT 1 = 1000
(4)
LT 2 = 5000
(5)
call BBf (LT 1, LT 2 : f )
(6)
For the given conditions the table gives f=0.6334 and the integral gives 0,6335. Note that a subprogram was not necessary but was requested in the problem statement. The solution could have been: LT1=1000 LT2=5000 f=(c1#/sigma#)*Integral(1/(LT5 *(exp(C2#/LT)-1)), LT, LT1, LT2)
Solution Variables in Subprogram BBf f = 0.6335
LT = 5000 [micrometer]
LT 1 = 1000 [micrometer]
76
LT 2 = 5000 [micrometer]
P03 04
Equations 3.4
Calculation of radiant energy transport between large parallel plates:
T1 = 500; qrad = σ ·
1 = 0.45;
T2 = 300;
T14 − T24 (1/1 + 1/2 − 1)
qrad = hrad · (T1 − T2 )
2 = 0.20;
σ = sigma#
Eqn 3.8.4
Eqn 3.10.1
Solution 1 = 0.45 σ = 5.670 × 10−8 W/m2 -K4
2 = 0.2 T1 = 500 [K]
hrad = 2.478 W/m2 -K T2 = 300 [K]
77
qrad = 495.7 W/m2
P03 05
Equations 3.5
Calculation of a heat loss coefficient for combined wind and radiation loss for a plate facing up.
Tamb = ConvertTemp(C, K, 10) TP late = ConvertTemp(C, K, 50) TDewP oint = 3 [C] T ime = 0 = 0.88 hwind = 25 W/m2 ·K Estimate an equivalent sky temperature using the Berdahl & Martin method, Equation 3.9.2. Note the unusual units in this Equation. .25 2 Tsky = Tamb · 0.711 + 0.0056 [1/C] · TDewP oint + 0.000073 1/C2 · (TDP ) + 0.013 · Cos(15 [deg/hr] · T ime) Radiation heat transfer is given by Equation 3.9.1. Divide this heat transfer by (TP late - Tamb ) to get the radiation coefficient based on the ambient temperature. hRad = sigma# · ·
4 TP4 late − Tsky TP late − Tamb
Add the convection coefficient to get the total loss coefficient. Uoverall = hRad + hwind
Solution = 0.88 Tamb = 283.1 [K] Uoverall = 32.7 W/m2 -K
hRad = 7.657 W/m2 -K TDewP oint = 3 [C]
78
hwind = 25 W/m2 -K TP late = 323.1 [K]
T ime = 0 [hr] Tsky = 262.7 [K]
P03 06
Equations 3.6
Calculation of convection heat transfer in a collector.
The Rayleigh number is found from Eqn 3.11.2 Ra = g# · beta0 · (T1 − T2 ) ·
l3 µ·α
The Nusselt Number is found from Eqn 3.11.4 1.6
N us = 1 + 1.44 ·
(sin (1.8 · min (Slope, 75))) 1 − 1708 · Ra · Cos(min (Slope, 75))
! ·A+B
where A and B must be constrained to be positive or zero and for slopes greater than 75, use slope = 75: 1708 A = max 0, 1 − Ra · Cos(min (Slope, 75)) !! .333 Cos(min (Slope, 75)) B = max 0, Ra · −1 5830 N us = hconv · l/k Qconv = hconv · (T1 − T2 ) T 1 + T2 T¯ = 2 beta0 = β Air, T = T¯ ;
µ = µ air, T = T¯ ;
ρ = ρ air, T = T¯, p = Patm ; T1 = 100;
T2 = 50;
k = λ air, T = T¯
Cp = cp air, T = T¯ ;
α=
k ρ · Cp
Patm = 101.3 [kPa]
The three conditions are done in a Parametric Table.
Table 1 Run 1 2 3
l [m] 0.020 0.050 0.020
Slope [deg] 0 0 45
N us [-] 2.794 5.135 2.422
Ra [-] 19030 297343 19030
hconv Qconv2 W/m2 -K W/m 4.07 203.7 3.00 149.8 3.53 79 176.6
P03 07
Equations 3.7
Calculation of equilibrium temperatures of plates exposed to solar radiation
a For a plate in space with surface normal to the solar radiation an energy balance on the plate (which absorbes solar radiation on one side and emits radiation from both) is: 4 Gsc# · αsolar = 2 · Sigma# · · Tplate,a
From Table 4.7.1 the solar absorptance is: αsolar = 0.35 Assume the emittance of copper, from Table 4.7.1,can be approximated by a parabola 2 = a + b · Tplate,a + c · Tplate,a 2
0.041 = a + b · 338 [K] + c · 338 [K] ;
2
0.036 = a + b · 463 [K] + c · 463 [K] ;
2
0.039 = a + b · 803 [K] + c · 803 [K]
b For a plate in the atmosphere with surface normal to the solar radiation and a zenith angle of 90 deg an energy balance on the plate (which absorbes solar radiation on one side, exchanges radiation from the top side with the sky and from the bottom side with the ground, and loses energy by convection from both sides to the atmosphere) is: 2 b = a + b · Tplate,b + c · Tplate,b
Gsc# · τatm · αsolar = Sigma# · b · τatm = 0.8;
Tamb = 298 [K] ;
4 4 4 4 Tplate,b − Tsky + Tplate,b − Tground + 2 · hwind · (Tplate,b − Tamb )
Tsky = 288 [K] ;
hwind = 23 W/m2 ·K ;
Tground = 288 [K]
The convection term in the energy balance dominates so it is clear that the exact value of the emittance is not critical.
Solution a = 0.07095 [-] = 0.03428 Tamb = 298 [K] Tsky = 288 [K]
αsolar = 0.35 b = 0.0428 Tground = 288 [K]
b = −0.0001241 [1/K] hwind = 23 W/m2 -K Tplate,a = 592 [K]
80
c = 1.050 × 10−7 1/K2 τatm = 0.8 Tplate,b = 306 [K]
P03 08
Equations 3.8
Energy balance calculations on a hypothetical instrument for measuring solar radiation.
Knowns hconv = 16 W/m2 ·K ; disk = 0.95;
Tdisk,b = ConvertTemp(C, K, 15);
τcover = 0.90;
αb = 0.95;
Tdisk,w = ConvertTemp(C, K, 5);
cover = 0.88
αw = 0.35
a
An energy balance on each disk, assuming that the disks and cover act like parallel plates for radiation exchange: ! 4 4 Tdisk,b − Tcover + hconv · (Tdisk,b − Tcover ) G · τcover · αb = 2 · Sigma# · (1/disk + 1/cover − 1) G · τcover · αw = 2 ·
! 4 4 Tdisk,w − Tcover Sigma# · + hconv · (Tdisk,w − Tcover ) (1/disk + 1/cover − 1)
81
b
Heat transfer coefficient from disk to cover: 2 2 Udisk,cover,b = 2 · Sigma# · Tdisk,b + Tcover · Udisk,cover,w = 2 · Sigma# · c
2 Tdisk,w
+
2 Tcover
Tdisk,b + Tcover + hconv (1/disk + 1/cover − 1)
Tdisk,w + Tcover + hconv · (1/disk + 1/cover − 1)
An expression for incident radiation
Incident solar energy, G, is expressed in terms of the temperature difference between black and white disks. Note that the first two equations are sufficient to solve for G; so the new variable G1 is introduced.
For the black disk: G1*τ cover *αb = Udisk,cover,b *(Tdisk,b -Tcover ) For the white disk: G1*τ cover *αw = Udisk,cover,w *(Tdisk,w -Tcover ) Solving for G1 G1 =
Tdisk,b − Tdisk,w (αb /Udisk,cover,b − αw /Udisk,cover,w ) · τcover
d
Calculation of incident radiation
In the days before EES, the nonlinear radiation term would have been linearized using the second two equations. The solution would then involve guessing the cover temperature, finding the U’s, substituting into the equation of part c and then checking the guess of the cover temperature using one of the commented equations of pary c.. With EES we do not need to worry about the solution technique - let EES do it. Also note that the ambient temperature was not used except maybe to get a reasonable guess of Tcover.
Solution αb = 0.95 G = 753.0 W/m2 Tcover = 272.2 [K] Udisk,cover,w = 39.95 W/m2 -K
αw = 0.35
cover = 0.88 hconv = 16 W/m2 -K
disk = 0.95
G1 = 753.0 W/m2 Tdisk,b = 288.2 [K]
Tdisk,w = 278.2 [K]
Udisk,cover,b = 40.4 W/m2 -K
82
τcover = 0.9 [-]
P03 09
Equations 3.9
Convection loss coefficient and convection heat transfer in a collector
$UnitSystem SI
Mass
Deg
Tu = ConvertTemp(C, K, 110 [C]);
L = 0.020 [m] ;
∆T = Tu − Tl ;
k = λ Air, T = T¯ ;
kPa
K kJ
Tl = ConvertTemp(C, K, 60 [C]);
α=
β = 1/T¯
Slope = 60 [deg]
µ = µ Air, T = T¯ ;
Cp = cp Air, T = T¯ ;
Tl + Tu T¯ = ; 2
k ρ · Cp · 1000
ρ = ρ Air, T = T¯, P = 101.3 [kPa]
;
ν = µ/ρ
J kJ
We can use the SETP functions to fing the Nusselt Number N u# = N uF latP late (Slope, Tu , Tl , L) or we can evaluate the Rayleigh number and then find the Nusselt number. Ra# =
g# · β · ∆T · L3 ν·α 1.6
0
N u# = 1 + 1.44 ·
(sin (1.8 · SlopeL)) 1 − 1708 · Ra# · Cos(SlopeL)
! ·A+B
SlopeL = min (Slope, 75 [deg]) Slope is constrained to be between 0 and 75 1708 A = max 0, 1 − Ra# · Cos(SlopeL) !! .333 Cos(SlopeL) B = max 0, Ra# · −1 5830 As expected, both Nusselt numbers are identical. N u# = h · L/k;
q = h · ∆T
Solution A = 0.7985 ∆T = 50 [K] ν = 0.00002149 m2 /s ρ = 0.9854 kg/m3 Tu = 383.2 [K]
α = 0.00003006 m2 /s h = 3.092 W/m2 -K
B = 0.1328
β = 0.002792 [1/K]
Cp = 1.009 [kJ/kg-K]
k = 0.02988 [W/m-K]
L = 0.02 [m] q = 154.6 W/m2 T¯ = 358.2 [K]
µ = 0.00002117 [Pa-s]
0
N u# = 2.069 [-]
N u# = 2.069 [-]
Slope = 60 [deg]
SlopeL = 60 [deg] 83
Ra# = 16, 957 [-] Tl = 333.2 [K]
P03 10
Equations 3.10
Effect of a slat-type honeycomb on the convection heat transfer of Problem 3.9
$UnitSystem SI
Mass
Deg
Tu = ConvertTemp(C, K, 110 [C]);
∆T = Tu − Tl ;
L = 0.020 [m] ;
k = k Air, T = T¯ ;
kPa
K kJ
Tl = ConvertTemp(C, K, 60 [C]);
Tl + Tu T¯ = ; 2
β = 1/T¯
Slope = 60 [deg]
µ = µ Air, T = T¯ ; k ρ · Cp · 1000
Cp = cp Air, T = T¯ ;
α=
SlatSpace = 10 [mm] ;
P lSpace = 20 [mm]
ρ = ρ Air, T = T¯, P = 101.3 [kPa]
;
J kJ
ν = µ/ρ
AspectRatio = SlatSpace/P lSpace From Figure 3.12.2 at an AspectRatio of 0.5: C1 = 0.113;
C2 = 1.00
We need to evaluate the Rayleigh number and then find the Nusselt number. Ra# =
g# · β · ∆T · L3 ν·α
From Eqn 3.12.1 N u# = max 1.1 · C1 · C2 · Ra#0.28 , 1 N u# = h · L/k q = h · ∆T
Solution α = 0.00003006 m2 /s
AspectRatio = 0.5
β = 0.002792 [1/K]
C1 = 0.113
C2 = 1
Cp = 1.009 [kJ/kg-K]
∆T = 50 [K]
k = 0.02988 [W/m-K]
L = 0.02 [m]
N u# = 1.90 [-] ρ = 0.9854 kg/m3 Tl = 333.2 [K]
P lSpace = 20 [mm]
µ = 0.00002117 [Pa-s] q = 141.9 W/m2
h = 2.839 W/m2 -K ν = 0.00002149 m2 /s
SlatSpace = 10 [mm] Tu = 383.2 [K]
Slope = 60 [deg]
84
Ra# = 16, 957 [-] T¯ = 358.2 [K]
P03 11
Equations Heat transfer coefficients for air flowing in a rectangular duct . $UnitSystem SI
Mass
Deg
kPa
C kJ
Knowns, properties and definitions Length = 2.0 [m] ;
width = 1 [m] ;
k = k (air, T = Tair ) ;
Tair = 25 [C] ;
Af low = width · depth;
µ = µ (air, T = Tair ) ;
P = 2 · (width + depth) ;
P r# = Pr (air, T = Tair )
Dh =
4 · Af low P
m=ρ*A ˙ f low *V and Re#=ρ*V*Dh /µ so that when ρ is eliminated we have Re# =
m ˙ · Dh µ · Af low
Use a parametric table for the three situations. From Eqn 3.14.7 1.2
N u# = 4.9 + .0606 ·
(Re# · P r# · Dh /Length)
.7
1 + .0909 · (Re# · P r# · Dh /Length) · P r#.17
N u# = h · Dh /k For situation 3 the Reynolds number is in the transition region. For conservative design use the laminar flow results. The turblent heat transfer coefficient from Eqn 3.14.6 is 7.26 W/m2 -K
Solution Af low = 0.015 m2 Length = 2 [m] P r# = 0.7281
depth = 0.015 [m] µ = 0.00001849 [kg/m-s] Re# = 2, 558
Dh = 0.02956 [m] m ˙ = 0.024 [kg/s] Tair = 25 [C]
Table 1 Run 1 2 3
depth [m] 0.015 0.0075 0.015
m ˙ [kg/s] 0.012 0.012 0.024
Re# 1,279 1,289 2,558
ν# [-] 5.815 5.367 6.625
h W/m2 -K 5.019 9.197 5.718
85
h = 5.718 W/m2 -K N u# = 6.625 [-] width = 1 [m]
k = 0.02551 [W/m-K] P = 2.03 [m]
P03 12
Equations 3.12
Calculation of pressure drop in a packed bed using the Dunkle & Ellul correlation
$UnitSystem SI Length = 2.0; T = 40 [C] ;
Mass
Deg
kPa
K kJ C
M assF lowRate = 1.1 [kg/s] ;
Area = 12 m2 ;
DenAir = ρ (Air, T = T, p = 101.3) ;
P artDiam = 0.02 [m]
V isc = µ (air, T = T )
M assV el = M assF lowRate/Area Equation 3.16.5 or use the SETP function for pressure drop. ∆p = P resDrop (Length, M assV el, P artDiam, T )
Solution Area = 12 m2 M assF lowRate = 1.1 [kg/s] V isc = 1.918 × 10−5 [kg/m-s]
∆p = 29.31 [Pa] M assV el = 0.09167 kg/s-m2
86
DenAir = 1.127 kg/m3 P artDiam = 0.02 [m]
Length = 2 [m] T = 40 [C]
P03 13
Equations 3.13
Estimation of particle size to get a pressure drop of 55 Pa for the circumstances of Problem 3.12
$UnitSystem SI
Mass
Deg
kPa
K kJ C
Note that with EES it is only necessary to delete the particle diameter in problem 3.12 and add the pressure drop. Length = 2.0; M assF lowRate = 1.1 [kg/s] ; Area = 12 m2 ; ∆p = 55 [Pa] T = 40 [C] ;
DenAir = ρ (Air, T = T, p = 101.3) ;
V isc = µ (air, T = T )
M assV el = M assF lowRate/Area Equation 3.16.5 or use the SETP function for pressure drop. ∆p = P resDrop (Length, M assV el, P artDiam, T ) An alternative solution is to use Equation 3.16.4 by McCorquodale et al. V F = 0.45 ∆p = Length ·
M assV el2 DenAir · P artDiamM cC
· (1 − V F ) ·
1.5 1.5 V isc · 4.74 + 166 · (1 − V F ) · · V F 3/2 V F 3/2 M assV el · P artDiamM cC
Solution Area = 12 m2 Length = 2 [m] P artDiam = 0.01321 [m] V F = 0.45 [-]
∆p = 55 [Pa] M assF lowRate = 1.1 [kg/s] P artDiamM cC = 0.01032 [m] V isc = 1.918 × 10−5 [kg/m-s]
87
3 DenAir = 1.127 kg/m M assV el = 0.09167 kg/s-m2 T = 40 [C]
P03 14
Equations 3.14
Calculation of upward heat loss terms for a collector.
$UnitSystem SI
Mass
Deg
kPa
K kJ
Data: Tplate = ConvertTemp(C, K, 100); plate = 0.10; a
Slope = 50;
σ = sigma#;
g = g#;
Tsurr = ConvertTemp(C, K, 10) L = .025 [m]
Radiation Heat Transfer
Qrad = σ · b
4 4 − Tcover Tplate 1/plate + 1/cover − 1
Radiation coefficient
hrad = c
cover = 0.88;
Tcover = ConvertTemp(C, K, 31.5);
Qrad Tplate − Tcover
Convection Heat Transfer
N u# = N uF latP late (Slope, Tcover , Tplate , L) Tmean =
Equation 3.11.14
Tplate + Tcover 2
k = k (Air, T = Tmean ) N u# = hconv · L/k Qconv = hconv · (Tplate − Tcover ) Instead of using the NuFlatPlate EES function we could have used either Figure 3.11.1 or Eqn 3.11.4 along with the following: β = 1/Tmean Cp = cp (Air, T = Tmean )
88
ρ = ρ (Air, T = Tmean , P = 101.3) ν=
µ (Air, T = Tmean ) ρ
α=
k ρ · Cp · 1000
Ra = d
J kJ
g · β · (Tplate − Tcover ) · L3 ν·α Radiation from cover to surroundings
4 4 QradCover,Surr = cover · σ · Tcover − Tsurr
We can determine QconvCover,Surr since loss is either convection or radiation. QconvCover,Surr = Qconv + Qrad − QradCover,Surr
Solution α = 0.00002717 m2 /s cover = 0.88 [-] hconv = 3.616 W/m2 -K L = 0.025 [m] Qconv = 247.7 W/m2 QradCover,Surr = 109.1 W/m2 σ = 5.670 × 10−8 W/m2 -K4 Tmean = 338.9 [K]
β = 0.002951 [1/K] plate = 0.1 [-] 2 hrad = 0.8797 W/m 2 -K ν = 0.00001953 m /s QconvCover,Surr = 198.9 W/m2 Ra = 58, 368 [-] Slope = 50 [deg] Tplate = 373.1 [K]
89
Cp = 1.007 [kJ/kg-K] g = 9.807 m/s2 k = 0.0285 [W/m-K] N u# = 3.172 2 Qrad = 60.26 W/m ρ = 1.041 kg/m3 Tcover = 304.7 [K] Tsurr = 283.1 [K]
p03 15
Equations 3.15
Estimation of wind convection coefficients
$UnitSystem SI
Mass
Deg
kPa
K kJ
Data: Length = 10 [M] ;
W idth = 2.5 [M] ;
TSurr = ConvertTemp(C, K, 10);
V el = 5.0 [m/s]
Tcover = ConvertTemp(C, K, 31.5)
Evaluate properties at the mean air tempeature. Tprop =
Tsurr + Tcover ; 2
µ = µ (‘air’ , T = Tprop ) ; a
k = k (‘air’ , T = Tprop ) ;
ρ = ρ (‘air’ , T = Tprop , P = 101.3)
P r = Pr (air, T = Tprop )
For a free standing colector:
LChar = 4 · Length ·
W idth ; 2 · (Length + W idth)
N u#wind = 0.86 · Re1/2 · P r1/3
Re = LChar · V el · ρ/µ
Eqn 3.15.1
N u#wind = hwind,a · Lchar /k Note: The Re is 1.31*106 , which is slightly beyond the range of the experiments on which Equation 3.15.1 is based. With no satisfactory alternative, use it. b
For a flush-mounted collector on a building
V olume = 564;
hwind,b
= max 5 W/m2 ·K , 8.6 ·
!
V el.6 V olume1/3
.4
Eqn 3.15.10
Note: Equations 3.15.9 and 3.15.10 are not dimensionally correct unless the 8.6 has units of (W/m2 -K)/(m0.2 /s0.6 ). For empirical equations like this it is best to just turn off unit checking for that equation.
Solution hwind,a = 5.587 W/m2 -K LChar = 4 [m] Re = 1.313 × 106 [-] TSurr = 283.1 [K]
hwind,b = 9.706 W/m2 -K µ = 0.00001829 [kg/m-s] ρ = 1.201 kg/m3 V el = 5 [m/s]
90
k = 0.02519 [W/m-K]
Length = 10 [M]
N u#wind = 887 [-] Tcover = 304.7 [K] V olume = 564 m3
P r = 0.7291 [-] Tprop = 293.9 [K] W idth = 2.5 [M]
P03 16
Equations 3.16
Effectiveness-NTU calculations for a counterflow heat exchanger
$UnitSystem SI
Mass
Deg
kPa
K kJ
Data: m ˙ H2O = 3.75 [kg/s] ;
U A = 2.1 × 105 [W/K] ;
Tin,H2O = 49 [C] ;
Tin,Glycol = 65 [C] ;
Cp,Glycol = 3780 [J/kg ·K] ;
Ccold = m ˙ H2O · Cp,H2O Chot = m ˙ Glycol · Cp,Glycol Cmin = min (Ccold , Chot ) Cmax = max (Ccold , Chot ) N T U = U A/Cmin C ∗ = Cmin /Cmax Ef f =
1 − exp (−N T U · (1 − C ∗ )) 1 − C ∗ · exp (−N T U · (1 − C ∗ ))
Q = Ef f · Cmin · (Tin,Glycol − Tin,H2O ) Q = Ccold · (Tout,H2O − Tin,H2O ) Q = Chot · (Tin,Glycol − Tout,Glycol ) Run a parametric table with m ˙ glycol reanging from 0 to 20. The solution is shown for the specific case of : m ˙ glycol = 5
Solution Ccold = 15, 713 [W/K] Cp,Glycol = 3, 780 [J/kg-K] m ˙ Glycol = 5.0 [kg/s] Tin,Glycol = 65 [C] U A = 210, 000 [W/K]
Chot = 18, 900 [W/K] Cp,H2O = 4, 190 [J/kg-K] m ˙ H2O = 3.75 [kg/s] Tin,H2O = 49 [C]
91
Cmax = 18, 900 [W/K] C ∗ = 0.8313 [-] N T U = 13.37 [-] Tout,Glycol = 51.956419181 [C]
Cmin = 15, 713 [W/K] Ef f = 0.981 [-] Q = 246, 524 [W] Tout,H2O = 64.69 [C]
Tout vs mdotg lycol
92
P03 17
Equations 3.17
Heat transfer from the hot absorber plate to the ambient.
$UnitSystem SI
Mass
Deg
kPa
K kJ
Data: p = 0.2;
g = 0.88;
P lateSp = 0.030 [m] ; a
slope = 60 [deg] ; σ = sigma#;
hwind = 7 W/m2 ·K ;
Tsky = Tamb ;
Tamb = ConvertTemp(C, K, 15)
Tplate = ConvertTemp(C, K, 85)
(1) (2)
Heat loss from collector
We can use the SETP function for Eqn 3.11.4 to find hplate,cover N u# = N uF latP late (Slope, Tcover , Tplate , P lateSp)
(3)
N u# = hplate,cover · P lateSp/k Tcover + Tplate k = k air, T = 2
(4) (5)
The various energy flows through the cover system are: qconv,p,c = hplate,cover · (Tplate − Tcover ) qrad,p,c = σ ·
(6)
4 4 Tplate − Tcover 1/p + 1/g − 1
(7)
qwind = hwind · (Tcover − Tamb ) 4 4 qrad,sky = σ · g · Tcover − Tsky
(8)
(9)
The cover temperature is found from an energy balance on the cover. qconv,p,c + qrad,p,c = qrad,sky + qwind
(10)
93
The heat loss through the cover is then: qloss = qconv,p,c + qrad,p,c
(11)
b Heat loss coefficient as a function of plate temperature is found by commenting out the equation above that sets the plate temperature to 85 C and setting the plate temperature (in C) in a parametric table. Tplate = ConvertTemp(C, K, TplateC )
(12)
qloss = U · (Tplate − Tamb )
(13)
c What energy transfers are we neglecting in this analysis of a collector? back losses through the insulation. edge losses two-dimensional losses in the corners.
Solution g = 0.88 [-] k = 0.02806 [W/m-K] qloss = 236.5 W/m2 σ = 5.670 × 10−8 W/m2 -K4 Tplate = 358.2 [K]
p = 0.2 [-] N u# = 3.235 [-] qrad,p,c = 83.03 W/m2
hplate,cover = 3.026 W/m2 -K P lateSp = 0.03 [m] qrad,sky = 101.6 W/m2
hwind = 7 W/m2 -K qconv,p,c = 153.5 W/m2 qwind = 134.9 W/m2
slope = 60 [deg] TplateC = 85 [C]
Tamb = 288.2 [K] Tsky = 288.2 [K]
Tcover = 307.4 [K] U = 3.379 W/m2 -K
Table 1 Run 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
TplateC [C] 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105
Tcover [K] 289.0 290.2 291.4 292.7 294.0 295.4 296.8 298.3 299.7 301.2 302.7 304.3 305.8 307.4 309.0 310.7 312.3 314.0
U W/m2 -K 2.112 2.425 2.595 2.716 2.813 2.897 2.971 3.039 3.103 3.163 3.220 3.274 3.327 3.379 3.429 3.479 3.528 3.576
94
U vs Tplate
95
P04 01
Equations 4.1
Calculation of equilibrium temprerature of a surface in space.
Assume that the 6000 K radiation is normal to the surface and that there is no loss other than radiation from the surface. Tsun = 6000 [K] ; S = 1367 W/m2 Use a Parametric table to specify the critical wavelength. ρ0, λ = 0.1;
ρλ,∞ = 0.9
Fraction of the sun’s energy between zero and λ fo, λ = BlBodyF ract (λ · Tsun )
From the EES SETP library.
ρsun = fo, λ · ρ0, λ + (1 − fo, λ) · ρλ,∞ αsun = 1 − ρsun Fraction of the IR radiation from surface at Tsurf ace fo, λsurf = BlBodyF ract (λ · Tsurf ) ρsurf = fo, λsurf · ρ0, λ + (1 − fo, λsurf ) · ρλ,∞ αsurf = 1 − ρsurf αsurf = surf Energy balance on surface assuming back is insulated. 4 αsun · S = surf · sigma# · Tsurf
If we did this by hand we would need to guess a surface temperature, find the fraction of the IR radiation in each band, find a temperature from the energy balance and repeat the process until the guess and calculated temperatures are equal.
Table 1 Run 1 2 3
λ [micrometer] 1 2 3
fo,lambda
fo,lambda,surf
αsun
surf
0.7378 0.9451 0.9808
3.322E-7 0.005337 0.049
0.690 0.856 0.885 96
0.100 0.104 0.139
Tsurf [K] 638.7 667.0 625.7
P04 02
Equations 4.2
Calculation of absorptance and emittance for a surface with a step change in spectral reflectance
$UnitSystem SI
Mass
Deg
kPa
K kJ
Data: Tsun = 5777 [K] ; αλ,1 = 0.95;
Tsurf ace = ConvertTemp(C, K, 150)
αλ,2 = 0.05;
(1)
λ = 1.8 [micrometer]
(2)
fraction of solar radiation below λ from Eqn 3.6.3. fsol = BlBodyF ract (λ · Tsun )
(3)
From Eqn 4.6.4 for two bands α = αλ,1 · fsol + αλ,2 · (1 − fsol )
(4)
fraction of IR radiation below λ from Eqn 3.6.3. fIR = BlBodyF ract (λ · Tsurf ace )
(5)
From Eqn 4.6.1 for two bands and using αλ = λ from Eqn 4.4.5 = αλ,1 · fIR + αλ,2 · (1 − fIR )
(6)
Solution α = 0.880 fIR = 0.000007637 Tsurf ace = 423.1 [K]
αλ,1 = 0.95 fsol = 0.9219
αλ,2 = 0.05 λ = 1.8 [micrometer]
97
= 0.050 Tsun = 5, 777 [K]
p04 03
Equations 4.3
Calculation of absorptance and emittance of a surface from spectral reflectance data.
$UnitSystem SI
Mass
Deg
kPa
K kJ
a For the extraterrestrial solar spectrum. Using 10 equally spaced intervals from Table 1.3.1b, the wavelength at center points are: 0.364, 0.455, 0.525, 0.599, 0.682, 0.787, 0.923, 1.113, 1.412, 2.117. Reading corresponding reflectance values from figure 4.8.3 ρ1 = 0.04;
ρ2 = 0.04;
ρ3 = 0.04;
ρ4 = 0.04;
ρ5 = 0.04
ρ6 = 0.04;
ρ7 = 0.05;
ρ8 = 0.06;
ρ9 = 0.10;
ρ10 = 0.32
αextraterrestrial = 1 −
Sum(ρi , i = 1, 10) 10
b For the standard terrestrial solar spectrum. Using 10 equally spaced intervals from Table 2.6.1, the appropriate wavelenghts are: 0.426, 0.508, 0.581, 0.653, 0.732, 0.822, 0.929, 1.080, 1.300, 1.974. Note that the numbers differ little from part a. Reading reflectance values from figure 4.8.3: r11 = 0.04;
r12 = 0.04;
r13 = 0.04;
r14 = 0.04;
r15 = 0.04
r16 = 0.04;
r17 = 0.05;
r18 = 0.06;
r19 = 0.08;
r110 = 0.31
αterrestrial = 1 −
Sum(r1i , i = 1, 10) 10
c Emittance of the surface at 350 C. Need to use radiation tables to generate the intervals. Using 10 points from Table 3.6.1b. The appropriate wavelenghts are: 3.02, 3.93, 4.65, 5.38, 6.15, 7.08, 8.23, 9.87, 12.6, 20.1. Reading reflectance values from Figure 4.8.3 r21 = 0.59;
r22 = 0.72;
r23 = 0.77;
r24 = 0.84;
r25 = 0.87
r26 = 0.90;
r27 = 0.91;
r28 = 0.93;
r29 = 0.94;
r210 = 0.94
=1−
Sum(r2i , i = 1, 10) 10
Solution αextraterrestrial = 0.923
αterrestrial = 0.926
= 0.159
98
Estimated r2[9] and r2[10]
p04 04
Equations 4.4
Calculation of emittance for a real surface at two different temperatures.
$UnitSystem SI
Mass
Deg
kPa
K kJ
Data Ta = ConvertTemp(C, K, 175) Tb = ConvertTemp(C, K, 30) a
At 175 C
Use the 10 equal increments( from either Table 3.6.1b or as calculated here), and from the reflectance curve read the reflectances at the characteristic (midpoint) wavelengths. Note that it was necessary to limit λ between 0 and 100. duplicate i = 1, 10 i − 0.5 10
fa,i =
fa,i = BlBodyF ract (λa,i · Ta ) end Reading reflectances from curve C of Figure 4.8.2 are: r11 = .32;
r12 = .63;
r13 = .66;
r18 = .90;
r19 = .92;
r110 = .88
175 = 1 − b
r14 = .81;
r15 = .84;
Sum(r1i , i = 1, 10) 10
At 30 C
Repeat with new temperature. duplicate i = 1, 10 fb,i =
i − 0.5 10
fb,i = BlBodyF ract (λb,i · Tb ) end
99
r16 = .87;
r17 = .88
The corresponding reflectances from curve C of Figure 4.8.2 are: r21 = .69;
r22 = .82;
r23 = .87;
r24 = .88;
r28 = .91;
r29 = .88;
r210 = .87
estimated
30 = 1 −
r25 = .90;
r26 = .91;
r27 = .92
Sum(r2i , i = 1, 10) 10
Note that it would have been useful to read the reflectances from the curve in small increments and place the values in a lookup table. Software exists to automatically digitize curves.
Solution 175 = 0.229
30 = 0.135
Ta = 448.1 [K]
Tb = 303.2 [K]
Arrays Row 1 2 3 4 5 6 7 8 9 10
fa,i [-] 0.05 0.15 0.25 0.35 0.45 0.55 0.65 0.75 0.85 0.95
λa,i [micrometer] 4.21 5.46 6.47 7.47 8.56 9.84 11.45 13.72 17.51 27.80
r1i [-] 0.32 0.63 0.66 0.81 0.84 0.87 0.88 0.9 0.92 0.88
fb,i [-] 0.05 0.15 0.25 0.35 0.45 0.55 0.65 0.75 0.85 0.95
λb,i [micrometer] 6.22 8.07 9.56 11.04 12.65 14.54 16.93 20.28 25.89 41.10
100
r2i [-] 0.69 0.82 0.87 0.88 0.9 0.91 0.92 0.91 0.88 0.87
P04 05
Equations 4.5 Calculation of the α/ ratio for a surface in space. $UnitSystem SI
Mass
Deg
kPa
K kJ
Data λ,1 = 0.20; λ1 = 0.6;
λ,2 = 0.75; λ2 = 2.6;
λ,3 = 0.90
(1)
λ3 = 100
(2)
αλ,1 = 1 − λ,1
(3)
αλ,2 = 1 − λ,2
(4)
αλ,3 = 1 − λ,3
(5)
T = 350 [K]
(6)
From Table 1.3.1 for solar radiation the fraction in each band is: fsolar,1 = 0.351
(7)
fsolar,2 = 0.968 + 0.2 · (0.981 − 0.968)
(8)
fsolar,3 = 1
(9)
∆f solar,1 = 0.351
(10)
∆f solar,2 = fsolar,2 − fsolar,1
(11)
∆f solar,3 = fsolar,3 − fsolar,2
(12)
From Eqn 4.6.4 α = αλ,1 · ∆f solar,1 + αλ,2 · ∆f solar,2 + αλ,3 · ∆f solar,3
(13)
The fraction of the infrared radiation from 0 to λ is found from Table 3.6.1a or from the EES SETP function. fIR,1 = BlBodyF ract (λ1 · T )
(14)
101 1
fIR,2 = BlBodyF ract (λ2 · T )
(15)
fIR,3 = BlBodyF ract (1000 [micrometer] · T )
Note that we extended the data beyond the 100 micrometer limit.(16)
∆f IR,1 = fIR,1
(17)
∆f IR,2 = fIR,2 − fIR,1
(18)
∆f IR,3 = fIR,3 − fIR,2
Note that almost all of the energy is in band 3.
(19)
From Eqn 4.6.1 = λ,1 · ∆f IR,1 + λ,2 · ∆f IR,2 + λ,3 · ∆f IR,3
(20)
Ratio Ratio = α/
(21)
Solution α = 0.4386 ∆fIR,1 = 9.091 × 10−26 [-] ∆fsolar,2 = 0.6196 [-] λ,2 = 0.75 fIR,3 = 0.9997 λ1 = 0.6 [micrometer] T = 350 [K]
αλ,1 = 0.8 ∆fIR,2 = 0.0001006 [-] ∆fsolar,3 = 0.0294 [-] λ,3 = 0.9 fsolar,1 = 0.351 [-] λ2 = 2.6 [micrometer]
αλ,2 = 0.25 ∆fIR,3 = 0.9996 [-] = 0.8997 fIR,1 = 9.091 × 10−26 fsolar,2 = 0.9706 [-] λ3 = 100 [micrometer]
102 2
αλ,3 = 0.1 ∆fsolar,1 = 0.351 [-] λ,1 = 0.2 fIR,2 = 0.0001006 fsolar,3 = 1 [-] Ratio = 0.4875
P04 06
Equations 4.6 Calculation of solar absorptance as a function of angle of incidence. function alpha/alphan (θ)
(1)
Equation 4.11.1 alpha/alphan = 1 − 1.5879 × 10−3 · θ + 2.731 × 10−4 · θ2 − 2.3026 × 10−5 · θ3 + 9.0244 × 10−7 · θ4 − 1.8000 × 10−8 · θ5 + 1.7734 × 10−10 · θ6 − 6.9937 × 10−13 · θ7
(2)
end
(3)
Use 10 equally spaced intervals from Table 2.6.1. The appropriate wavelenghts (i.e., midpoints) are: 0.426, 0.508, 0.581, 0.653, 0.732, 0.822, 0.929, 1.080, 1.300, 1.9745. Reading reflectance values at these wavelengths for curve C of Figure 4.8.2: ρ1 = 0.04;
ρ2 = 0.08;
ρ3 = 0.10;
ρ4 = 0.07;
ρ5 = 0.04
(4)
ρ6 = 0.10;
ρ7 = 0.14;
ρ8 = 0.22;
ρ9 = 0.28;
ρ10 = 0.19
(5)
From Eqn 4.6.4 αn = 1 −
Sum(ρi , i = 1, 10) 10
(6)
Use Equation 4.11.1 to find α at angles of 30, 45 and 60 degrees. α30 = αn · alpha/alphan (30)
(7)
α45 = αn · alpha/alphan (45)
(8)
α60 = αn · alpha/alphan (60)
(9)
Solution Variables in Main program α30 = 0.86 [-] ρ[4] = 0.07
α45 = 0.85 [-] ρ[5] = 0.04
α60 = 0.81 [-] ρ[6] = 0.1
αn = 0.874 ρ[7] = 0.14
103
ρ[1] = 0.04 ρ[8] = 0.22
ρ[2] = 0.08 ρ[9] = 0.28
ρ[3] = 0.1 ρ[10] = 0.19
P04 07a
Equations 4.7
Determine collector solar and IR radiation properties. This problem is best done in 4 parts.
function P lanck (λ, T ) C1 = C1#; P lanck =
λ5
(1)
C2 := C2#
(2)
C1 C2 −1 · exp λ·T
(3)
end P lanck
(4)
Subprogramemittance (T : )
(5)
T otal = sigma# · T 4 R 100 [micrometer] P lanck (λ, T ) · Interpolate(‘Curve C Fig 4 8 2’ , ‘lambda’ , ‘rho’ , ‘lambda’ = λ) dλ 0.01 [micrometer] = T otal
(6)
104
(7)
end emittance a
(8)
What is the absorptance for solar radiation assuming the sun is a blackbody at 5777?
Tsun = 5777 [K]
(9)
4 T otal = sigma# · Tsun
(10)
wavelength range 0.01 to 100 Z 100 E0.01to100 = P lanck (λ, Tsun ) · Interpolate(‘Curve C Fig 4 8 2’ , ‘lambda’ , ‘rho’ , ‘lambda’ = λ) dλ
(11)
0.01
αsolar,0.01to100 = 1 − E0.01to100 /total
(12)
wavelength range 0.041 to 25 Z 25 P lanck (lambda1, Tsun )·Interpolate(‘Curve C Fig 4 8 2’ , ‘lambda’ , ‘rho’ , ‘lambda’ = lambda1) dLambda1(13 E0.041to25 = 0.041
αsolar,0.041to25 = 1 − E0.041to25 /total
(14)
The energy beyond 25 micrometers is very small. The energy below 0.41 micrometers is about 14% of the total. It isfortuitous that the two integrals give the same value to 3 significant figures.
Solution Variables in Main program αsolar,0.01to100 = 0.887 λ = 100 [micrometer]
αsolar,0.041to25 = 0.887 Lambda1 = 25 [micrometer]
Curve CF ig482 Row 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
λ [micrometer] 0.01 0.41 0.4844 0.5195 0.5313 0.5547 0.6016 0.6367 0.6953 0.7773 0.8359 0.8711 0.9297 0.9883 1.012 1.21 1.309
ρ 0.04327 0.04327 0.05288 0.07692 0.09135 0.09615 0.09135 0.04808 0.03846 0.03846 0.0625 0.09615 0.1202 0.1346 0.1531 0.2343 0.2677
105
6 2 E0.01to100 = 7.139 × 10 W/m T otal = 6.315 × 107 W/m2
E0.041to25 = 7.141 × 106 W/m2 Tsun = 5, 777 [K]
18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58
1.454 1.692 1.927 2.209 2.399 2.591 2.783 3.122 3.515 3.911 4.207 4.596 4.786 5.079 5.421 5.857 6.629 7 7.866 8.925 9.984 11.14 12.68 13.55 14.7 16.91 18.45 19.42 19.9 20.09 20.86 21.25 21.63 22.21 22.5 23.07 23.75 23.94 24.52 25 100
0.2819 0.2483 0.1859 0.1187 0.0898 0.09443 0.09428 0.1323 0.242 0.3901 0.4903 0.5522 0.5281 0.5901 0.6664 0.7187 0.7659 0.7933 0.8317 0.8606 0.8702 0.8846 0.8894 0.9086 0.9183 0.9135 0.9135 0.9086 0.899 0.9183 0.9135 0.9039 0.899 0.899 0.9086 0.9039 0.8846 0.8846 0.8846 0.8798 0.8798
106
Plot 1
107
P04 07b
Equations 4.7
Determine collector solar and IR radiation properties. This problem is best done in 4 parts.
function P lanck (λ, T ) C1 = C1#; P lanck =
(1)
C2 := C2#
(2)
C1 C2 λ5 · exp λ·T −1
(3)
end P lanck
(4)
Subprogramemittance (T : )
(5)
T otal = sigma# · T 4
(6)
R 100 [micrometer] =
0.01 [micrometer]
P lanck (λ, T ) · Interpolate(‘Curve C Fig 4 8 2’ , ‘lambda’ , ‘rho’ , ‘lambda’ = λ) dλ (7)
T otal
end emittance b.
(8)
Repeat problem a but assume the solar radiation is distributed as in Table 2.6.1.
We will use 10 increments. The variable limit are the bounds of the bands from Table 2.6.1. limit1 = .28;
limit2 = .47;
limit7 = .871;
limit8 = 1.012;
limit3 = .545;
limit4 = .617;
limit9 = 1.201;
limit5 = .691;
limit10 = 1.588;
limit11 = 5
duplicate i = 1, 10 R limiti+1 ρi =
limiti
(9) (10) (11)
Interpolate(‘Curve C Fig 4 8 2’ , ‘lambda’ , ‘rho’ , ‘lambda’ = λi ) dλi limiti+1 − limiti
end αsolar,b = 1 −
limit6 = .777
(12) (13)
Sum(ρj , j = 1, 10) 10
(14)
Part a used a black body sun to estimate the solar absorptance as 0.887 whereas using the solar distribution of Table 2.6.1 the solar absorptance is estimated to be 0.876. Part of this difference can be attributed to the different methods applied for integration.
Solution Variables in Main program αsolar,b = 0.876
108
P04 07c
Equations 4.7
Determine collector solar and IR radiation properties. This problem is best done in 4 parts.
function P lanck (λ, T ) C1 = C1#; P lanck =
λ5
(1)
C2 := C2#
(2)
C1 C2 · exp λ·T −1
(3)
end P lanck
(4)
Subprogramemittance (T : )
(5)
T otal = sigma# · T 4 R 100 [micrometer] P lanck (λ, T ) · Interpolate(‘Curve C Fig 4 8 2’ , ‘lambda’ , ‘rho’ , ‘lambda’ = λ) dλ 0.01 [micrometer] = T otal
(6)
end emittance
(8)
c
(7)
What is the emittance for long-wave radiation at 325 K?
TIR = 325 [K]
(9)
4 T otalIR = sigma# · TIR
(10)
T $ = ‘Curve C Fig 4 8 2’
(11)
wavelength range 0.01 to 100 Z 100 IR0.01to100 = P lanck (lambda2, TIR ) · Interpolate(T $, ‘lambda’ , ‘rho’ , ‘lambda’ = lambda2) dLambda2(12) 0.01
IR,0.01to100 = 1 − IR0.01to100 /totalIR
(13)
wavelength range 0.041 to 25 Z 25 IR0.041to25 = P lanck (lambda3, TIR )·Interpolate(T $, ‘lambda’ , ‘rho’ , ‘lambda’ = lambda3) dLambda3(14) 0.041
IR,0.041to25 = 1 − IR0.041to25 /totalIR
(15)
There is a very large difference between the two IR emittance values (0.376 vs 0.262) since about 14% of the energy is beyond 25 micrometers.
Solution Variables in Main program
109
IR,0.01to100 = 0.3755 Lambda2 = 100 [micrometer] TIR = 325 [K]
IR,0.041to25 = 0.2615 Lambda3 = 25 [micrometer]
110
IR0.01to100 = 395 W/m2 T $ = ‘Curve C Fig 4 8 2’
2 IR0.041to25 = 467.1 W/m T otalIR = 632.5 W/m2
P04 07d
Equations 4.7
Determine collector solar and IR radiation properties. This problem is best done in 4 parts.
function P lanck (λ, T ) C1 = C1#; P lanck =
(1)
C2 := C2#
(2)
C1 C2 λ5 · exp λ·T −1
(3)
end P lanck
(4)
Subprogramemittance (T : )
(5)
T otal = sigma# · T 4
(6)
R 100 [micrometer] =
0.01 [micrometer]
P lanck (λ, T ) · Interpolate(‘Curve C Fig 4 8 2’ , ‘lambda’ , ‘rho’ , ‘lambda’ = λ) dλ T otal
end emittance
(7)
(8)
d Plot the emittance versus surface temperature for a range of temperatures from 300 K to 800 K. Since EES has a limit on the number of integration variables that can be in the main program, an integral function is used. duplicate k = 1, 11
(9)
TK = 250 + k · 50
(10)
call Emittance (TK : k )
(11)
end
(12)
A second order curve fit does not fo a very good job. A third order does much better epsilon[i]=-0.51218632 + 0.00637126087*T[i] - 0.0000106300682*T[i]2 + 5.39253381E-09*T[i]3
Solution Variables in Main program
Arrays Row
i
Ti
1 2 3 4 5
0.5821 0.654 0.6856 0.6931 0.6857
300 350 400 450 500
111
6 7 8 9 10 11
0.6689 0.6463 0.6205 0.5932 0.5655 0.5383
550 600 650 700 750 800
epsilon vs T
112
P05 01
Equations Calculate the reflectance of one glass surface. function r (n, θ1 )
(1)
If(θ1180 ) then
(2)
r=0
(3)
else
(4)
If(θ1 = 0) then r=
n−1 n+1
(5)
2 (6)
else
(7)
θ2 = arcsin
rperp =
sin (θ1 ) n
sin (θ2 − θ1 )
2
sin (θ2 + θ1 )
2
(8)
(9)
2
rpar = r=
tan (θ2 − θ1 )
(10)
2
tan (θ2 + θ1 )
rperp + rpar 2
(11)
endif
(12)
endif
(13)
end
(14)
knowns n = 1.526;
θa = 10;
ra = r (n, θa ) ;
θb = 30;
rb = r (n, θb ) ;
θc = 50;
θd = 70
rc = r (n, θc ) ;
(15)
rd = r (n, θd )
Solution Variables in Main program n = 1.526 θa = 10 [deg]
ra = 0.04338 θb = 30 [deg]
rb = 0.04494 θc = 50 [deg]
rc = 0.06147 θd = 70 [deg]
113
rd = 0.1758
(16)
P05 02
Equations Transmittance of three covers. function τ (index, Nc , θ1 )
(1)
If(θ1 < 0)or (θ1 > 180) then
(2)
τ =0
(3)
else
(4)
If(θ1 = 0) then rperp =
(5)
index − 1 index + 1
2 (6)
rpar = rperp
(7)
else
(8)
θ2 = arcsin
rperp =
sin (θ1 ) index
sin (θ2 − θ1 )
2
sin (θ2 + θ1 )
2
(9)
(10)
2
rpar =
tan (θ2 − θ1 )
(11)
2
tan (θ2 + θ1 )
endif
(12)
τ = 1/2 ·
1 − rpar 1 − rperp + 1 + (2 · Nc − 1) · rpar 1 + (2 · Nc − 1) · rperp
endif
(13) (14)
end
(15)
knowns n = 1.526;
Nc = 3;
τa = τ (n, Nc , θa ) ;
θa = 10;
θb = 70
τb = τ (n, Nc , θb ) ;
(16) The results agree with Fig 5.1.3
Solution Variables in Main program n = 1.526 θa = 10 [deg]
Nc = 3 θb = 70 [deg]
τa = 0.7862
τb = 0.5326
114
(17)
P05 03
Equations Transmittance of two covers at two angles. function T ransIncAng (N cov, IncAng, KL, Ref rInd) sin (IncAng) Ref rAng = arcsin Ref rInd If(incAng ≤ 0) then 2 Ref rind − 1 r= ; Ref rind + 1
(1) (2) (3)
τref =
1−r 1 + (2 · N cov − 1) · r
(4)
else
(5)
rperp = rpara =
sin (Ref rAng − IncAng) sin (Ref rAng + IncAng)
2
tan (Ref rAng − IncAng) tan (Ref rAng + IncAng)
(6) 2 (7)
τperp =
1 − rperp 1 + (2 · N cov − 1) · rperp
(8)
τpara =
1 − rpara 1 + (2 · N cov − 1) · rpara
(9)
τref =
τperp + τpara 2
(10)
endif
(11)
τabs = exp −N cov ·
KL Cos(Ref rAng)
(12)
T ransIncAng = τabs · τref
(13)
end
(14)
Knowns K = 20 [1/m] ;
L = 0.002 [mm] ;
Rindex = 1.526;
τa = T ransIncAng (N, θa , K · L, Rindex) ;
N = 2;
θa = 0 [deg] ;
θb = 50 [deg]
τb = T ransIncAng (N, θb , K · L, Rindex)
Solution Variables in Main program
115
(15) (16)
K = 20 [1/m] τa = 0.7814
L = 0.002 [mm] τb = 0.7446
N =2 θa = 0 [deg]
Rindex = 1.526 θb = 50 [deg]
116
P05 04
Equations τ -α product for two covers at two angles. We will use SETP functions TransIncAng and Absn. Knowns N = 2;
θa = 25;
θb = 60;
KL = 0.037;
αn = 0.96;
p = 0.96;
index = 1.526
(1)
From footnote 2 of Section 5.5 ρd = 0.22
(2)
Part a τa = T ransIncAng (N, θa , KL, index)
(3)
αa = αn · Abs/Absn (θa )
(4)
T auAlphaP roducta =
τa · αa (1 − (1 − αa ) · ρd )
Eqn 5.5.1
(5)
Part b τb = T ransIncAng (N, θb , KL, index)
(6)
αb = αn · Abs/Absn (θb )
(7)
T auAlphaP roductb =
τb · α b (1 − (1 − αb ) · ρd )
Eqn 5.5.1
(8)
Solution αa = 0.9475 index = 1.526 T auAlphaP roducta = 0.750 θa = 25 [deg]
αb = 0.8922 KL = 0.037 T auAlphaP roductb = 0.634 θb = 60 [deg]
117
αn = 0.96 N =2 τa = 0.7828
p = 0.96 ρd = 0.22 τb = 0.6936
P05 05
Equations Estimate transmittance for diffuse sky and ground reflected radiation We will use the SETP functions IncAngEffGrRef, IncAngEffDiff and TransIncAng Knowns N = 1;
KL = 0.037;
index = 1.526;
slopea = 45;
slopeb = 90
(1)
Part a angleg,a = IncAngEf f GrRef (Slopea )
(2)
angled,a = IncAngEf f Dif f (Slopea )
(3)
τg,a = T ransIncAng N, angleg,a , KL, index
(4)
τd,a = T ransIncAng N, angled,a , KL, index
(5)
Part a angleg,b = IncAngEf f GrRef (Slopeb )
(6)
angled,b = IncAngEf f Dif f (Slopeb )
(7)
= T ransIncAng N, angleg,b , KL, index
(8)
τd,b = T ransIncAng N, angled,b , KL, index
(9)
τg,b
Solution angled,a = 56.49 [deg] KL = 0.037 τd,b = 0.81
angled,b = 59.33 [deg] N =1 τg,a = 0.70
angleg,a = 69.45 [deg] slopea = 45 τg,b = 0.81
118
angleg,b = 59.81 [deg] slopeb = 90
index = 1.526 τd,a = 0.83
P05 06
Equations function T ransIncAng (N cov, IncAng, KL, Ref rInd) sin (IncAng) Ref rAng = arcsin Ref rInd
(1) (2)
If(incAng ≤ 0) then r=
(3)
Ref rind − 1 Ref rind + 1
2 ;
τref =
1−r 1 + (2 · N cov − 1) · r
(4)
else
(5)
rperp = rpara =
sin (Ref rAng − IncAng) sin (Ref rAng + IncAng)
2
tan (Ref rAng − IncAng) tan (Ref rAng + IncAng)
(6) 2 (7)
τperp =
1 − rperp 1 + (2 · N cov − 1) · rperp
(8)
τpara =
1 − rpara 1 + (2 · N cov − 1) · rpara
(9)
τref =
τperp + τpara 2
(10)
endif
(11)
T ransIncAng = exp −N cov ·
KL Cos(Ref rAng)
· τref
(12)
end
N = 1;
(13)
L = 0.002;
K = 10;
Angle = 58;
τa = T ransIncAng (N, angle, K · L, indexa ) ;
indexa = 1.6;
indexb = 1.40
Knowns
τb = T ransIncAng (N, angle, K · L, indexb )
Solution Variables in Main program Angle = 58
indexa = 1.6
indexb = 1.4
K = 10
119
L = 0.002
N =1
τa = 0.819
τb = 0.865
(14) (15)
P05 07
Equations Calculate transmittance and compare exact and approximate methods. function T ransIncAng (N cov, IncAng, KL, Ref rInd) sin (IncAng) Ref rAng = arcsin Ref rInd
(1) (2)
If(incAng ≤ 0) then r=
(3)
Ref rind − 1 Ref rind + 1
2 ;
τref =
1−r 1 + (2 · N cov − 1) · r
(4)
else
(5)
rperp = rpara =
sin (Ref rAng − IncAng) sin (Ref rAng + IncAng)
2
tan (Ref rAng − IncAng) tan (Ref rAng + IncAng)
(6) 2 (7)
τperp =
1 − rperp 1 + (2 · N cov − 1) · rperp
(8)
τpara =
1 − rpara 1 + (2 · N cov − 1) · rpara
(9)
τref =
τperp + τpara 2
(10)
endif
(11)
τabs = exp −N cov ·
KL Cos(Ref rAng)
(12)
T ransIncAng = τabs · τref
(13)
end
(14)
Knowns L = 0.0025;
K = 25;
θ = 55;
index = 1.526;
N1 = 1;
120
N2 = 2
(15)
Part a - single surface sin (θ) θ2 = arcsin Index rP erp = rP ar = r=
sin (θ2 − θ) sin (θ2 + θ)
tan (θ2 − θ) tan (θ2 + θ)
(16)
2 (17)
2 (18)
rP erp + rP ar 2
(19)
Part b using the approximate method τ1 = T ransIncAng (N1 , θ, K · L, index)
(20)
Part c using the approximate method τ2 = T ransIncAng (N2 , θ, K · L, index)
(21)
Part d - calculate single cover transmittance by the exact method L τAbs = exp −K · Eqn 5.2.2 Cos(θ2 ) ! 1 − rP2 erp 1 − rP erp τP erp = τAbs · Eqn 5.3.1 · 2 1 + rP erp 1 − (rP erp · τAbs )
τP ar = τAbs ·
1 − rP ar 1 + rP ar
τ1×10 xact =
τP ar + τP erp 2
·
(23)
!
1 − rP2 ar 1 − (rP ar · τAbs )
(22)
2
Eqn 5.3.1
(24)
(25)
The two methods of calculation differ by 0.001 out of 0.808 or a little more than 0.1%
Solution Variables in Main program index = 1.526 rP ar = 0.0003372 τP ar = 0.928
K = 25 rP erp = 0.1472 τP erp = 0.6883
L = 0.0025 τ1 = 0.809 θ = 55 [deg]
N1 = 1 τ1Exact = 0.808 θ2 = 32.47 [deg]
121
N2 = 2 τ2 = 0.686
r = 0.074 τAbs = 0.9286
P05 08
Equations Estimate absorbed radiation for an inclined collector. function T ransIncAng (N cov, IncAng, KL, Ref rInd) sin (IncAng) Ref rAng = arcsin Ref rInd
(1) (2)
If(incAng ≤ 0) then r=
(3)
Ref rind − 1 Ref rind + 1
2 ;
τref =
1−r 1 + (2 · N cov − 1) · r
(4)
else
(5)
rperp = rpara =
sin (Ref rAng − IncAng) sin (Ref rAng + IncAng)
2
tan (Ref rAng − IncAng) tan (Ref rAng + IncAng)
(6) 2 (7)
τperp =
1 − rperp 1 + (2 · N cov − 1) · rperp
(8)
τpara =
1 − rpara 1 + (2 · N cov − 1) · rpara
(9)
τref =
τperp + τpara 2
(10)
endif
(11)
τabs = exp −N cov ·
KL Cos(Ref rAng)
(12)
T ransIncAng = τabs · τref
(13)
end
(14)
Knowns N = 1;
KL = 0.037;
IbT = 4.1 MJ/m2 ;
index = 1.526; Id = 0.4 MJ/m2 ;
θb = 25 [deg] ; β = 25 [deg] ;
122
α = 0.95; ρg = 0.2
I = 3 MJ/m2
(15) (16)
For the beam radiataion τb = T ransIncAng (N, θb , KL, index)
(17)
From Figure 5.4.1 at 45 degrees or Eqn 5.4.2 θd = 57
(18)
τd = T ransIncAng (N, θd , KL, index)
(19)
From Figure 5.4.1 at 45 degrees or Eqn 5.4.1 θg = 69
(20)
τg = T ransIncAng (N, θg , KL, index)
(21)
From Equation 5.2.2 T auAlphab = 1.01 · α · τb
(22)
T auAlphad = 1.01 · α · τd
(23)
T auAlphag = 1.01 · α · τg
(24)
From Equation 5.9.1 Sb = IbT · T auAlphab Sd = Id · T auAlphad ·
(25) 1 + Cos(β) 2
Sg = ρg · I · T auAlphag ·
(26)
1 − Cos(β) 2
(27)
S = Sb + Sd + Sg
(28)
Solution Variables in Main program
ρg = 0.2
β = 25 [deg] Id = 0.4 MJ/m2 S = 3.787 MJ/m2
I = 3 MJ/m2 KL = 0.037 Sb = 3.467 MJ/m2
index = 1.526 N =1 Sd = 0.3014 MJ/m2
Sg = 0.01906 MJ/m2 τb = 0.8812 θd = 57 [deg]
T auAlphab = 0.8455 τd = 0.824 θg = 69
T auAlphad = 0.7906 τg = 0.7068
T auAlphag = 0.6781 θb = 25 [deg]
α = 0.95 IbT = 4.1 MJ/m2
123
P05 09
Equations Estimate absorbed radiation for a collector. We will use EES SETP functions TransIncAng, Absn , and IDIF F
.
Knowns Lat = 48;
β = 63;
Io = 1.58 MJ/m2 ; a
θb = 36.2;
n = 1;
Zenith = 71.5;
Index = 1.526;
ρg = 0.4;
αn = 0.955
KL = 0.013;
I = 1.03 MJ/m2
(1) (2)
For beam radiation
τb = T ransIncAng (N, θb , KL, index)
(3)
αb = αn · Abs/Absn (θb )
(4)
Need diffuse reflectance of the cover for use in Equation 5.5.1. The equivalent diffuse angle from Figure 5.4.1 is 59 degrees. Note that this is NOT the same angle as required for the diffuse radiation in part b. τd59 = T ransIncAng (N, 59, KL, index) T auAlphad59 =
(5)
τd59 T ransIncAng (n, 59, KL, 1)
(6)
5.3.4 ρd59 = T auAlphad59 − τd59
(7)
5.3.6 T auAlphab = τb ·
αb 1 − (1 − αb ) · ρd59
(8)
Using Equation 5.5.2, TauAlphab =1.01*τ b *αb =0.853 rather than 0.845
124
b
For the diffuse radiation
The equivalent angle for diffuse sky radiation on a collector sloped at an angle of 63 degrees is 57 from Fig 5.4.1. θd = 57
(9)
τd = T ransIncAng (n, θd , KL, index)
(10)
αd = αn · Abs/Absn (θd )
(11)
αd T auAlphad = τd · 1 − (1 − αd ) · ρd59
(12)
Using Equation 5.5.2, TauAlphad =1.01*τ d *αd =0.772 rather than 0.766 c
For the ground reflected radiation
The equivalent angle for ground reflected radiation on a collector sloped at an angle of 63 degrees is 64 from Fig 5.4.1. θg = 64
(13)
τg = T ransIncAng (n, θg , KL, index)
(14)
αg = αn · Abs/Absn (θg )
(15)
αg T auAlphag = τg · 1 − (1 − αd ) · ρd59
(16)
d
The absorbed solar radiation
kT = I/Io Id = I · IDIF F/I (kT ) ; Rb =
(17) Ib = I − Id
(18)
Cos(θb ) Cos(Zenith)
(19)
Using Equation 5.9.1, the diffuse sky model: Sb = T auAlphab · Ib · Rb
(20)
1 + Cos(β) 2 1 − Cos(β) Sg = T auAlphag · I · ρg · 2 S = Sb + Sd + Sg Sd = T auAlphad · Id ·
(21) (22) (23)
Using Equation 5.9.2, the HDKR sky model, the results are S=1.99 rather than 1.75. This difference results from the choice of sky models and more than offsets ifferences resulting from choosing Equation 5.5.2 to evaluate TauAlpha. Ai = Ib /Io p f = Ib /I
(24)
Sbh = T auAlphab · Rb · (Ib + Id · Ai ) 1 + Cos(β) 3 Sdh = T auAlphad · Id · (1 − Ai ) · 1 + f · (sin (β/2)) · 2 1 − Cos(β) Sgh = T auAlphag · I · ρg · 2 Sh = Sbh + Sdh + Sgh
(26)
(25)
Solution
125
(27) (28) (29)
αb = 0.937 Ai = 0.4369 Index = 1.526 KL = 0.013
αd = 0.9009 β = 63 [deg] Ib = 0.6903 MJ/m2 kT = 0.652
αg = 0.8642 f = 0.8186 Id = 0.3397 MJ/m2 Lat = 48 [deg]
ρd59 = 0.021 Sh = 1.99 MJ/m2 Sdh = 0.12 MJ/m2
ρg = 0.4
Rb = 2.543
Sb = 1.48 MJ/m2 Sg = 0.07734 MJ/m2
Sbh = 1.80 MJ/m2 Sgh = 0.08 MJ/m2
T auAlphad = 0.766 τd = 0.848 θd = 57 [deg]
T auAlphad59 = 0.857 τd59 = 0.836 θg = 64 [deg]
T auAlphag = 0.6876 τg = 0.7939 Zenith = 71.5 [deg]
126
αn = 0.955 I = 1.03 MJ/m2 Io = 1.58 MJ/m2 n=1 S = 1.75 MJ/m2 Sd = 0.1891 MJ/m2 T auAlphab = 0.843 τb = 0.899 θb = 36.2 [deg]
P05 10
Equations Determine a months absorbed radiation. We will use EES SETP functions TransIncAng, Abs/Absn and HDIF F BAR /HBAR. Knowns Lat = 43;
slope = 90;
n = 1;
, Index = 1.526;
KL = 0.013;
ρg = 0.6;
αn = 0.93
(1)
(From Appendix G) ¯ = 6.44; H
¯ t = 0.47 K
(2)
First need to find average for the three components. For beam radiation: θb = 40
(3)
τb = T ransIncAng (n, θb , KL, Index)
(4)
αb = αn · Abs/Absn (θb )
(5)
Need ρd of the cover for use in Equation 5.5.1. The equivalent diffuse angle from Figure 5.4.1 is 59 degrees -& this is generally NOT the same angle as required for the diffuse sky radiation. τd59 = T ransIncAng (n, 59, KL, Index) T auAlphad59 =
τd59 T ransIncAng (n, 59, KL, 1)
ρd59 = T auAlphad59 − τd59 T auAlphab = τb ·
(6) Eqn 5.3.4
Eqn 5.3.6
(7) (8)
αb 1 − (1 − αb ) · ρd59
(9)
For the diffuse radiation: θd = 59
(10)
The equivalent angle for diffuse sky radiation on a vertical collector is 59 from Fig 5.4.1 the same angle as used in some calculations in part a. τd = τd59
(11)
127
αd = αn · Abs/Absn (θd ) T auAlphad = τd ·
(12)
αd 1 − (1 − αd ) · ρd59
(13)
For the ground reflected radiation: From Fig 5.4.1 the equivalent angle for ground reflected radiation on a vertical collector is exactly the same as for diffuse sky radiation. T auAlphag = T auAlphad
(14)
Need to break total into beam and diffuse - use Erbs’ correlation for winter - i.e. ω s is less than 81.4 - use 50 ¯d = H ¯ · Hdif f Bar/HBar Lat, n, K ¯T H
(15)
¯b = H ¯ −H ¯d H
(16)
¯ b = 2.62 R
(17)
a)
From Figure 2.19.1d
The absorbed solar radiation components using Equation 5.10.2, the diffuse sky model.
¯b · R ¯b S¯b = T auAlphab · H
(18)
¯ d · 1 + Cos(slope) S¯d = T auAlphad · H 2
(19)
¯ · ρg · 1 − Cos(slope) S¯g = T auAlphag · H 2
(20)
b)
The total radiation
S¯ = S¯b + S¯d + S¯g c)
(21)
¯T The average transmittance-absorptance product: Need to find H
¯ bT = H ¯b · R ¯b H
(22)
¯ dT = H ¯ d · 1 + Cos(slope) H 2
(23)
¯ gT = H ¯ · ρg · 1 − Cos(slope) H 2
(24)
¯T = H ¯ bT + H ¯ dT + H ¯ gT H
(25)
¯ ¯ H ¯T T auAlpha = S/
(26)
Solution αb = 0.9102 ¯ b = 3.727 MJ/m2 H ¯ gT = 1.932 MJ/m2 H ¯ t = 0.47 K ρg = 0.6 S¯b = 7.97 MJ/m2 ¯ T auAlpha = 0.7941 τb = 0.8951 θd = 59 [deg]
¯ = 6.44 MJ/m2 H ¯ dT = 1.357 MJ/m2 H KL = 0.013 ρd59 = 0.02137 S¯ = 10.37 MJ/m2
αd = 0.869 ¯ bT = 9.765 MJ/m2 H ¯ T = 13.05 MJ/m2 H Lat = 43 [deg] ¯ b = 2.62 R S¯d = 0.9881 MJ/m2
αn = 0.93 ¯ d = 2.713 MJ/m2 H Index = 1.526 n=1 slope = 90 [deg] S¯g = 1.407 MJ/m2
T auAlphab = 0.8163
T auAlphad = 0.7284 τd = 0.8358
T auAlphad59 = 0.8572 τd59 = 0.8358
T auAlphag = 0.7284 θb = 40 [deg]
128
P05 11
Equations Transmittance of cover with wavelength-dependent properties function P lanck(λ, T ) P lanck =
λ5 · exp
(1)
C1# C2# λ·T
−1
(2)
end
(3)
Although only two temperatures were requested, 12 temperatures are evaluated, including 45 C and 145 C. A plot shows the source temperature dependence of the IR transmittance. At 318 K, 95% of the blackbody radiation is between 5.22and 33.3 micrometers. At 418 K, 95% of the blackbody radiation is between 4.0 and 25.4 micrometers. Therefore, at a minimum, the tedlar transmittances needs to cover the range from 4 to 33 micrometers. However, Figure 5.7.2 only covers the range out to 15 mircometers. The transmittance beyond 15 will be assumed to be 0.8. The free program, Plot Digitizer, was used to generate the Lookup table. duplicate i = 1, 12
(4)
T C i = −5 + (i − 1) · 50
(5)
Ti = ConvertTemp(C, K, T C i ) R 100 P lanck(λ, Ti ) · Interpolate1(‘tau vs lambda’ , ‘lambda’ , ‘tau’ , ‘lambda’ = λ) dλ τi = 2.5 sigma# · Ti4
(6)
end
(7) (8)
Note that at 45 C and 145 C the transmittances are nearly the same
Arrays
129
Row 1 2 3 4 5 6 7 8 9 10 11 12
Ti [K] 268.2 318.2 368.2 418.2 468.2 518.2 568.2 618.2 668.2 718.2 768.2 818.2
τi [-] 0.670 0.649 0.645 0.651 0.663 0.676 0.689 0.700 0.707 0.709 0.708 0.705
T Ci [C] -5 45 95 145 195 245 295 345 395 445 495 545
tau vs lambda Row 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35
λ [micrometer] 2.5 2.823 2.857 3.064 3.373 3.399 3.509 3.856 4.062 4.235 4.475 4.751 5.06 5.371 5.921 6.301 6.919 7.018 7.185 7.293 7.533 7.605 7.71 7.917 8.018 8.12 8.294 8.675 8.804 9.011 9.181 9.492 9.767 10.08 10.36
τ 0.9726 0.9725 0.8903 0.9175 0.8489 0.2529 0.7598 0.9103 0.8554 0.9169 0.8483 0.9235 0.8479 0.9231 0.8475 0.9226 0.799 0.4839 0.1207 0.4769 0.3877 0.6205 0.7643 0.7368 0.5997 0.4901 0.6612 0.7638 0.1746 0.2019 0.05114 0.1332 0.09191 0.3109 0.4272
130
36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52
10.84 11.18 11.63 11.84 11.94 12.42 12.56 12.67 12.88 13.12 13.36 13.84 14.12 14.67 14.98 15 100
0.4681 0.4062 0.543 0.3648 0.1455 0.3165 0.5425 0.789 0.8231 0.8024 0.8708 0.7883 0.8224 0.781 0.774 0.8 0.8
tau vs lambda
tau vs T
131
P05 12
Equations Energy absorbed in a room for a month ¯ We will use EES SETP functions TransIncAng, abs/absn and HdiffBar/H. This is essentially the same as Problem 5.10. Knowns Lat = 43;
slope = 90;
¯ = 6.44; H
k¯t = 0.47
n = 2;
Index = 1.526;
KL = 0.0125;
ρg = 0.7;
αn = 0.92
From Appendix G for January
(1) (2)
First need to find average for the three components. For beam radiation: θb = 40
(3)
τb = T ransIncAng (n, θb , KL, Index)
(4)
αb = αn · Abs/Absn (θb )
(5)
Need ρd of the cover for use in Equation 5.5.1. The equivalent diffuse angle from Figure 5.4.1 is 59 degrees this is generally NOT the same angle as required for the diffuse sky radiation. τd59 = T ransIncAng (n, 59, KL, Index) T auAlphad59 =
τd59 T ransIncAng (n, 59, KL, 1)
ρd59 = T auAlphad59 − τd59 T auAlphab = τb ·
(6) Eqn 5.3.4
Eqn 5.3.6
(7) (8)
αb 1 − (1 − αb ) · ρd59
(9)
For the diffuse radiation: θd = 59
(10)
The equivalent angle for diffuse sky radiation on a vertical collector is 59 from Fig 5.4.1 the same angle as used in some calculations in part a. τd = τd59
(11)
132
αd = αn · Abs/Absn (θd ) T auAlphad = τd ·
(12)
αd 1 − (1 − αd ) · ρd59
(13)
For the ground reflected radiation: From Fig 5.4.1 the equivalent angle for ground reflected radiation on a vertical collector is exactly the same as for diffuse sky radiation. T auAlphag = T auAlphad
(14)
Need to break total into beam and diffuse - use Erbs’ correlation for winter - i.e. ω s is less than 81.4 - use 50 ¯d = H ¯ · Hdif f Bar/HBar Lat, n, K ¯T H
(15)
¯b = H ¯ −H ¯d H
(16)
¯ b = 2.62 R
(17)
a)
From Figure 2.19.1d
The absorbed solar radiation components using Equation 5.10.2, the diffuse sky model.
¯b · R ¯b S¯b = T auAlphab · H
(18)
¯ d · 1 + Cos(slope) S¯d = T auAlphad · H 2
(19)
¯ · ρg · 1 − Cos(slope) S¯g = T auAlphag · H 2
(20)
b)
The total radiation
S¯ = S¯b + S¯d + S¯g c)
(21)
¯T The average transmittance-absorptance product: Need to find H
¯ bT = H ¯b · R ¯b H
(22)
¯ dT = H ¯ d · 1 + Cos(slope) H 2
(23)
¯ gT = H ¯ · ρg · 1 − Cos(slope) H 2
(24)
¯T = H ¯ bT + H ¯ dT + H ¯ gT H
(25)
¯ ¯ H ¯T T auAlpha = S/
(26)
Solution αb = 0.9004 ¯ b = 3.727 MJ/m2 H ¯ gT = 2.254 MJ/m2 H k¯t = 0.47 ρg = 0.7 S¯b = 7.19 MJ/m2 ¯ T auAlpha = 0.7111 τb = 0.8144 θd = 59 [deg]
αd = 0.8597 ¯ bT = 9.765 MJ/m2 H ¯ T = 13.38 MJ/m2 H Lat = 43 [deg] ¯ b = 2.62 R S¯d = 0.87 MJ/m2
αn = 0.92 ¯ d = 2.713 MJ/m2 H Index = 1.526 n=2
T auAlphad = 0.6436 τd = 0.7448
T auAlphad59 = 0.7818 τd59 = 0.7448
slope = 90 [deg] S¯g = 1.45 MJ/m2
133
¯ = 6.44 MJ/m2 H ¯ dT = 1.357 MJ/m2 H KL = 0.0125 ρd59 = 0.03704 S¯ = 9.51 MJ/m2 T auAlphab = 0.736 T auAlphag = 0.6436 θb = 40 [deg]
P05 13
Equations 5.13 Daily solar absorbed by a collector. We will use the SETP EES functions as much as possible. The Boulder-Week data does not indicate whether it is based on clock time or solar time. If we assume time is clock time we find that kT exceeds 1.0 at 4:30 pm. If we plot Ih versus clock time we see that the data is not symmetrical around solar noon on the 10th and 14th, the two clear days (See Plot 1). However, for Boulder (meridian = standard meridian) solar time minus clock time is -6.7 minutes (i.e., -1.68 degrees of hour angle) ,which is both small and in the wrong direction. Looking at Plot 1, we need to shift the hour angle by about +5 degrees (about +20 minutes) . Finally, the peak radiation between 11 and 12 on the 13th has a kT of 0.97 which is very high. The conclusion is that this data is not very good. We will add a time shift of +20 minutes in order to solve this problem. . knowns αn = 0.85;
KL = 0.031;
Ref rInd = 1.526;
lat = 40 [deg] ;
M onth = 1;
Day = 10;
Long = 105 [deg] ;
Surf AzAng = 0;
ρg = 0.6;
shif t = 5 [deg]
nCov = 1(1) (2)
a) we will use 4 parametric tables, one for each slope. nday = nDay (M onth, Day) slope = Stringval(T ableN ame$)
(3) Note the use of the TableName# function to determine the slope
(4)
The Boulder data starts at 1am on Jan 8 LookU pRow = (Day − 8) · 24 + T ableRun# + 7
Row 1 in LookTable starts at 8am of Day.
hour = Lookup(‘Boulder Week’ , LookU pRow, ‘hour’ ) − 0.5 [hr]
134
(5)
Start at midpoint of hour between 7 and 8am .(6)
HrAng = (hour − 12hr ) · 15 [deg/hr] − Shif t
here we used the empirically found Shift
(7)
CosT heta = cosIncAng (Lat, nDay, HrAng, Slope, Surf AzAng)
(8)
θ = arccos (CosT heta)
(9)
α = αn · abs/absn (θ)
(10)
MJ Ih = Lookup(‘Boulder Week’ , LookU pRow, ‘I’ ) · 0.001 kJ
(11)
I0 = IZero, (nDay, lat, HrAng − 7.5deg , HrAng + 7.5 [deg])
(12)
kT = Ih /I0
(13)
Id = Ih · Idif f /I (kT )
(14)
IT = IT,HDKR, (Ih , Lat, nDay, slope, Surf AzAng, ρg , HrAng)
This function uses Eqns 2.10.1 and 2.16.7 (15)
ST = ST,HDKR, (Ih , lat, nDay, slope, Surf AzAng, ρg , HrAng, Ncov , KL, Ref rInd, αn )
(16)
The general shape of ST and IT versus hour angle are similar. For this day the values at slopes 60 and 75 degrees are nearly equal. b) the average transmittance-absorptance. taave = ST /IT
(17)
In the range of greatest importance (e.g.,3 hours before and after noon) the values differ by about 10% . Early in the morning and late in the afternoon the values drops considerably. c) addition of a second cover. We added a column in Parametric table 45 for the absorbed solar radiation with two covers. We have added a sum column option under the main menu Options/Preferences. slope2 = 45
(18)
ncov,2 = 2
(19)
ST,2 = ST,HDKR, (Ih , lat, nDay, slope2 , Surf AzAng, ρg , HrAng, Ncov,2 , KL, Ref rInd, αn )
(20)
The absorbed radiation for one cover was 17.6 MJ/m2 and for two covers was 15.6 MJ/m2. Although the absorbed radiation was reduced the thermal losses will also be reduced. It is not possible to tell if the second cover is appropriate d) Use a higher absorptance black paint absorber surface. αblack = 0.96
(21)
ST,black = ST,HDKR, (Ih , lat, nDay, slope, Surf AzAng, ρg , HrAng, Ncov , KL, Ref rInd, αblack )
(22)
The black absorber had an absorbed solar of 19.8 MJ/m2 whereas the selective surface had an absorber radiation value of 17.6 MJ/m2. Again it is not possible to to tell if this increased absorbed solar radiation more than offsets the increase in IR radiation losses that will occur with the flat black absorber. We need information from Chapter 6 to fully answer parts c and d.
30
135
Run 1 2 3 4 5 6 7 8 9 10 11
hour [hr] 7.5 8.5 9.5 10.5 11.5 12.5 13.5 14.5 15.5 16.5 17.5
HrAng [deg] -72.5 -57.5 -42.5 -27.5 -12.5 2.5 17.5 32.5 47.5 62.5 77.5
θ [deg] 77.9 64.8 52.6 41.9 34.3 32.1 36.3 45.2 56.5 69.1 82.4
kT [-] 0.640 0.592 0.734 0.799 0.789 0.790 0.786 0.747 0.643 0.471 0.400
hour [hr] 7.5 8.5 9.5 10.5 11.5 12.5 13.5 14.5 15.5 16.5 17.5
HrAng [deg] -72.5 -57.5 -42.5 -27.5 -12.5 2.5 17.5 32.5 47.5 62.5 77.5
θ [deg] 71.9 58.1 44.5 31.6 20.9 17.2 24.0 35.8 49.0 62.7 76.6
kT [-] 0.640 0.592 0.734 0.799 0.789 0.790 0.786 0.747 0.643 0.471 0.400
hour [hr] 7.5 8.5 9.5 10.5 11.5 12.5 13.5 14.5 15.5 16.5 17.5
HrAng [deg] -72.5 -57.5 -42.5 -27.5 -12.5 2.5 17.5 32.5 47.5 62.5 77.5
θ [deg] 67.0 53.4 39.6 25.7 11.8 3.1 16.5 30.3 44.2 58.0 71.5
kT [-] 0.640 0.592 0.734 0.799 0.789 0.790 0.786 0.747 0.643 0.471 0.400
hour [hr] 7.5 8.5
HrAng [deg] -72.5 -57.5
θ [deg] 63.7 51.4
kT [-] 0.640 0.592
Id MJ/m2 0.012 0.191 0.208 0.259 0.297 0.308 0.286 0.251 0.269 0.147 0.003
Ih 2 MJ/m 0.033 0.419 1.047 1.570 1.805 1.872 1.733 1.352 0.775 0.205 0.004
Id MJ/m2 0.012 0.191 0.208 0.259 0.297 0.308 0.286 0.251 0.269 0.147 0.003
Ih 2 MJ/m 0.033 0.419 1.047 1.570 1.805 1.872 1.733 1.352 0.775 0.205 0.004
Id MJ/m2 0.012 0.191 0.208 0.259 0.297 0.308 0.286 0.251 0.269 0.147 0.003
Ih 2 MJ/m 0.033 0.419 1.047 1.570 1.805 1.872 1.733 1.352 0.775 0.205 0.004
Id MJ/m2 0.012 0.191
Ih 2 MJ/m 0.033 0.419
α 0.60 0.76 0.81 0.83 0.83 0.84 0.83 0.83 0.80 0.73 0.46
IT 2 MJ/m 0.008 1.016 2.193 2.987 3.282 3.371 3.183 2.618 1.610 0.446 0.003
ST 2 MJ/m 0.005 0.608 1.519 2.166 2.409 2.480 2.330 1.880 1.078 0.250 0.002
taave [-] 0.625 0.599 0.692 0.725 0.734 0.736 0.732 0.718 0.670 0.561 0.658
IT 2 MJ/m 0.009 1.242 2.598 3.467 3.771 3.864 3.665 3.052 1.911 0.539 0.003
ST 2 MJ/m 0.006 0.821 1.870 2.551 2.799 2.877 2.714 2.234 1.347 0.339 0.002
taave [-] 0.625 0.661 0.720 0.736 0.742 0.745 0.740 0.732 0.705 0.628 0.658
IT 2 MJ/m 0.011 1.398 2.852 3.747 4.044 4.138 3.938 3.310 2.105 0.606 0.003
ST 2 MJ/m 0.007 0.963 2.076 2.770 3.017 3.092 2.931 2.436 1.512 0.401 0.002
taave [-] 0.635 0.689 0.728 0.739 0.746 0.747 0.744 0.736 0.718 0.662 0.657
IT 2 MJ/m 0.012 1.473
ST 2 MJ/m 0.008 1.028
taave [-] 0.644 0.698
45 Run 1 2 3 4 5 6 7 8 9 10 11
α 0.70 0.80 0.83 0.84 0.84 0.84 0.84 0.83 0.82 0.78 0.62
60 Run 1 2 3 4 5 6 7 8 9 10 11
α 0.75 0.81 0.83 0.84 0.85 0.85 0.84 0.84 0.83 0.80 0.70
75 Run 1 2
136
α 0.77 0.82
3 4 5 6 7 8 9 10 11
9.5 10.5 11.5 12.5 13.5 14.5 15.5 16.5 17.5
-42.5 -27.5 -12.5 2.5 17.5 32.5 47.5 62.5 77.5
39.2 27.3 17.0 13.1 20.1 31.2 43.3 55.5 67.7
0.734 0.799 0.789 0.790 0.786 0.747 0.643 0.471 0.400
0.208 0.259 0.297 0.308 0.286 0.251 0.269 0.147 0.003
1.047 1.570 1.805 1.872 1.733 1.352 0.775 0.205 0.004
Plot 1
IT vs Omega
137
0.83 0.84 0.84 0.85 0.84 0.84 0.83 0.81 0.74
2.936 3.808 4.083 4.172 3.982 3.375 2.178 0.640 0.003
2.138 2.807 3.032 3.104 2.951 2.479 1.567 0.433 0.002
0.728 0.737 0.743 0.744 0.741 0.734 0.720 0.676 0.656
ST vs omega
taave vs omega
138
P06 01
Equations function UT op,Klein, (Tpl , Tamb , Slope, W indCoef f , Emitt, N cov) If Unitsystem(‘SI’ ) then
(1) (2)
If Unitsystem(‘C’ ) then
(3)
T $ = ‘C’
(4)
Tpl0 = ConvertTemp(C, K, Tpl )
(5)
Tamb0 = ConvertTemp(C, K, Tamb )
(6)
else
(7)
T $ = ‘K’
(8)
Tpl0 = Tpl
(9)
Tamb0 = Tamb
(10)
endif
(11)
else
(12)
call Error(‘Units must be set to SI to use the function U TOP klein. Your plate temperature is xxxF1 R or F.’ , Tpl )(13) endif
(14) 2 f = 1 + 0.089 m ·K/W · W indCoef f − 0.1166m2 −K/W · W indCoef f · Emitt · (1 + 0.07866 · N cov) (15)
[K] e = 0.430 · 1 − 100 Tpl0
(16)
If Slope < 70 then
(17)
slope0 = Slope
(18)
else
(19)
slope0 = 70
(20)
endif
(21)
c = 520 · 1 − 0.000051 1/deg2 · slope02
(22)
139
w=
Tpl0 − Tamb0 N cov + f
(23)
If w ≤ 0 then w0 = 0.1 else w0 = w e x = (w0 ) · c/Tpl0
(24) (25)
2 2 y = sigma# · (Tpl0 + Tamb0 ) · Tpl 0 + Tamb0
(26)
−1 2 · N cov + f − 1 + 0.133 · Emitt − N cov + z = Emitt + 0.00591 m2 ·K/W · N cov · W indCoef f 0.88 −1 UT op,Klein,= 1 m2 ·K/W · N cov/x + 1/W indCoef f +
y z [W/m2 ·K]
(27)
(28)
end
(29)
Calculation of top loss coefficient for flat-plate collector Knowns. Note that the slope used in Figure 6.4.4 is 45. Slope = 45 [deg] ; b = 0.1; a)
a = 0.95;
Tamb,b = 40 [C] ;
Tamb,a = 10 [C] ; Tpl,b = 100 [C] ;
Tpl,a = 60 [C] ;
hw,a = 5 W/m2 ·K ;
hw,b = 20 W/m2 ·K ;
Ncov,b = 1
Ncov,a = 2(30) (31)
Non-selective black absorber
Ut,a = UT op,Klein, (Tpl,a , Tamb,a , Slope, hw,a , a , Ncov,a )
(32)
From Fig 6.4.4a Ut,a =3.0 b)
Selective absorber
Ut,b = UT op,Klein, (Tpl,b , Tamb,b , Slope, hw,b , b , Ncov,b )
(33)
FromFig 6.4.4f Ut,b =4.1
Solution Variables in Main program a = 0.95 Ncov,b = 1 Tpl,b = 100 [C]
b = 0.1 Slope = 45 [deg] Ut,a = 3.03 W/m2 -K
hw,a = 5 W/m2 -K Tamb,a = 10 [C] Ut,b = 3.78 W/m2 -K
140
hw,b = 20 W/m2 -K Tamb,b = 40 [C]
Ncov,a = 2 Tpl,a = 60 [C]
P06 02
Equations Calculation of overall loss coefficient for flat-plate collector. We could use the SETP EES functions but we will do this in detail. Knowns. The temperature units are set to C so we need to be careful where K is needed. Tplate = 100 [C] ; W indSp = 6.5;
Tamb = 25 [C] ; plate = 0.11;
kIns = 0.034 [W/m ·K] ;
P lateSpacing = 0.025 [m] ;
HouseV ol = 300 m3 ;
InsulT hick = 0.070 [m] ;
glass = 0.88
Slope = 43 [deg] ;
σ = sigma#;
(1) Patm = 101.3 [kPa]
Tdp = 20 [C] ;
Wind heat transfer coefficient from Equation 3.15.9 W indSp.6 hwind = max 5, 8.6 · HouseV ol0.4/3
T ime = 12 [hr]
(2) (3)
(4)
Sky temp from Equation 3.9.2 .25 2 Tsky = Tamb · 0.711 + 0.0056 · Tdp + 0.000073 · Tdp + 0.013 · Cos(15 · time)
(5)
Tsky,K = ConvertTemp(C, K, Tsky )
(6)
Back loss from Equation 6.4.10 Ub = kIns/InsulT hick
(7)
Qback = Ub · (TP late − Tamb )
(8)
Use Equation 3.11.4 to determine the heat transfer coefficient between absorber and cover. Tplate,K = ConvertTemp(C, K, Tplate )
(9)
Tcover,K = ConvertTemp(C, K, Tcover )
(10)
Tplate + Tcover T¯ = 2
(11)
141
k = k Air, T = T¯
(12)
1 v Air, T = T¯, P = Patm µ Air, T = T¯ ν= ρ J ¯ Cp = cp Air, T = T · 1000 kJ ρ=
α=
(13)
(14)
(15)
k ρ · Cp
(16)
∆T = Tplate − Tcover
(17)
P lateSpacing 3 ConvertTemp C, K, T¯ · ν · α ! 1.6 sin (1.8 · Slope) 1708 1 − 1708 · · max 0, 1 − Ra · Cos(Slope) Ra · Cos(Slope)
Ra = g# · ∆T ·
(18)
a = 1.44 ·
(19)
b = max 0,
Cos(Slope) Ra · 5830
!
1/3
−1
(20)
N usselt = 1 + a + b
(21)
N usselt = hcp,c · P lateSpacing/k
(22)
Use Equation 6.4.1 for losses from absorber to cover glass. Qtop = hcp,c · (Tplate − Tcover ) + σ ·
4 4 Tplate,K − Tcover,K 1/plate + 1/glass − 1
(23)
Equation 6.4.2 for losses from glass to surroundings 4 4 Qtop = hwind · (Tcover − Tamb ) + σ · glass · Tcover,K − Tsky,K
(24)
The definition of an overall loss coefficient is: Qtop + Qback = UL · (Tplate − Tcover )
(25)
Solution a = 1.307 ∆T = 60.01 [C] HouseV ol = 300 m3 kIns = 0.034 [W/m-K] Patm = 101.3 [kPa] ρ = 1.028 kg/m3 Tamb = 25 [C] Tdp = 20 [C] Tsky,K = 297.1 [K]
α = 0.0000278 m2 /s glass = 0.88 2 hwind = 12.36 W/m 2 -K ν = 0.00001995 m /s 2 Qback = 36.43 W/m −8 σ = 5.670 × 10 W/m2 -K4 T¯ = 70 [C] Tplate = 100 [C] Ub = 0.4857 W/m2 -K
142
b = 0.8231 plate = 0.11 InsulT hick = 0.07 [m] N usselt = 3.13 [-] Qtop = 276.4 W/m2 Slope = 43 [deg] Tcover = 39.99 [C] Tplate,K = 373.2 [K] UL = 5.214 W/m2 -C
Cp = 1, 007 [J/kg-K] hcp,c = 3.606 W/m2 -K k = 0.02881 [W/m-K] P lateSpacing = 0.025 [m] Ra = 48, 304 T ime = 12 [hr] Tcover,K = 313.1 [K] Tsky = 23.93 [C] W indSp = 6.5 [m/s]
P06 03
Equations We will use the $If directive to solve for parts a and b. Comment out one of the following two equations: P art$ = ‘a’
(1)
Part a, Verfication of convective heat transport shown in Figure 6.4.3(a). $UnitSystem SI kPa K kJ Knowns Tplate = 100;
Ta = 10;
P lateSpacing = 0.025;
T K plate = ConvertTemp(C, K, Tplate ); plate = 0.95;
Slope = 45;
glass = 0.88;
Tsky = Ta
T K a = ConvertTemp(C, K, Ta );
σ = sigma#;
hwind = 10;
g = g#
(2)
T K sky = ConvertTemp(C, K, Tsky )(3) (4)
Heat transfer from absorber to cover glass: $IF Part$=’a’ then N u# = N uF latP late (Slope, T K cover , T K plate , P lateSpacing)
(5)
$Else ρ = ρ (AIR, T = T K ave , P = 101.3) ν=
(6)
µ (AIR, T = T K ave ) ρ
(7)
J Cp = cp (AIR, T = T K ave ) · 1000 kJ α=
(8)
k/Cp ρ
(9)
Ra = g · (1/T K ave ) · (T K plate − T K cover ) · C1 = 0.088;
P lateSpacing 3 ν·α
C2 = 1.0
(From Figure 3.12.2) N u# = max 1.1 · C1 · C2 · Ra0.28 , 1
(10) (11) (12)
$EndIf
143
T K ave =
T K plate + T K cover 2
(13)
k = k (air, T = T K ave )
(14)
N u# = hp,c · P lateSpacing/k
(15)
Qconv = hp,c · (T K plate − T K cover )
(16)
Qrad,P late,Cover = σ ·
T K 4plate − T K 4cover 1/plate + 1/glass − 1
(17)
Heat transfer from cover to surroundings: Qwind = hwind · (T K cover − T k a ) Qrad,Sky = σ · glass · T K 4cover − T K 4sky
(18)
(19)
Energy Balance on cover: Qrad,P late,Cover + Qconv − Qwind − Qrad,Sky = 0
(20)
Definition of top loss coeffficient: Ut =
Qwind + Qrad,Sky T Kplate − T Ka
(21)
Tcover = ConvertTemp(K, C, T K cover )
(22)
In part a here are small differences between these calculated values and those in Figure 6.4.3 (which was done using a simple calculator). Part b, Effects of adding convection suppression slats to Problem 6.3a. The only thing that changes between parts a and b is the calculation of the Nusselt Number. The top loss coefficient, Ut , in part a is 6.60 W/m2 and in part b is 6.13. The convection suppression device provided a 7% reduction.
Solution glass = 0.88 hwind = 10 W/m2 -K P lateSpacing = 0.025 [m] Qwind = 382.3 W/m2 T K ave = 347.3 [K] Ta = 10 [C] Ut = 6.599 W/m2 -K
plate = 0.95 k = 0.0291 [W/m-K] Qconv = 178.1 W/m2 σ = 5.670 × 10−8 W/m2 -K4
g = 9.807 m/s2 N u# = 2.956
Slope = 45
T K a = 283.1 [K]
T K cover = 321.4 [K] Tcover = 48.23 [C]
T K plate = 373.1 [K] Tplate = 100 [C]
T K sky = 283.1 [K] Tsky = 10 [C]
144
Qrad,P late,Cover = 415.8 W/m2
hp,c = 3.441 W/m2 -K P art$ = ‘a’ Qrad,Sky = 211.6 W/m2
P06 03 b
Solution α = 0.00002823 m2 /s glass = 0.88 hwind = 10 W/m2 -K
C1 = 0.088 plate = 0.95 k = 0.02901 [W/m-K]
P art$ = ‘b’
P lateSpacing = 0.025 [m] Qwind = 357 W/m2
Qrad,Sky = 195 W/m2 σ = 5.670 × 10−8 W/m2 -K4 T K cover = 318.8 [K] Tcover = 45.7 [C]
Slope = 45 T K plate = 373.1 [K] Tplate = 100 [C]
145
C2 = 1 g = 9.807 m/s2 ν = 0.00002024 m2 /s Qconv = 120.3 W/m2
Cp = 1, 008 [J/kg-K] hp,c = 2.215 W/m2 -K N u# = 1.908
Ra = 42, 094 [-]
Qrad,P late,Cover = 431.7 W/m2 ρ = 1.02 kg/m3
T K a = 283.1 [K] T K sky = 283.1 [K] Tsky = 10 [C]
T K ave = 346 [K] Ta = 10 [C] Ut = 6.133 W/m2 -K
P06 04
Equations Calculation of collector top and overall loss coefficients There are three methods to solve this problem: Equation 6.4.9; Figure 6.4.4; or the detailed method of Example 6.4.3. We will use all three. $UnitSystem SI, kJ, K Knowns Slope = 45 [deg] ; Tamb = 10 [C] ;
pl = 0.95; Tpl =, 110 [C] ;
hwind = 10 W/m2 ·K ; Tsky = Tamb ;
Ncov = 2;
Space = 0.025 [m] ;
P lSpc1,c2 = 0.025 [m] ;
kinsulation = 0.035;
P lSpp,c1 = 0.025 [m] ;
P = P o#(2)
P lateLength = 2 [m]
(3)
P lateW idth = 1 [m]
(4)
Convert temperatures to K. T 0 amb = ConvertTemp(C, K, Tamb )
(5)
T 0 pl = ConvertTemp(C, K, Tpl )
(6)
T 0 c1 = ConvertTemp(C, K, Tc1 )
(7)
T 0 c2 = ConvertTemp(C, K, Tc2 )
(8)
T 0 sky = ConvertTemp(C, K, Tsky )
(9)
HT from cover 2 to sky c1 = 0.88;
c2 = 0.88
(10)
qc2,s = qw + qr0 c2,s
(11)
qw = hwind · (T 0 c2 − T 0 amb ) 4 4 qr0 c2,s = sigma# · T 0 c2 − T 0 sky · c2
(12) (13)
HT from cover 1 to cover 2 qc1,c2 = qc0 c1,c2 + qr0 c1,c2
(14)
146
l = 0.0
qc0 c1,c2 = hc0 c1,c2 · (T 0 c1 − T 0 c2 )
(15)
hc0 c1,c2 = N uF latP late Slope, T 0 c2 , T 0 c1 , P lSpc1,c2 · Condc1,c2 /P lSpc1,c2 condc1,c2 = k Air, T =
T 0 c1 + T 0 c2 2 4
qr0 c1,c2 = sigma# ·
(16)
(17)
4
T 0 c1 − T 0 c2 1/c1 + 1/c2 − 1
(18)
HT from plate to cover 1 qp,c1 = qc0 p,c1 + qr0 p,c1
(19)
qc0 p,c1 = hc0 p,c1 · (T 0 pl − T 0 c1 )
(20)
hc0 p,c1 = N uF latP late Slope, T 0 c1 , T 0 pl , P lSpp,c1 · condp,c1 /P lSpp,c1
(21)
condp,c1
T 0 pl + T 0 c1 = k Air, T = 2 4
qr0 p,c1 = sigma# ·
(22)
4
T 0 pl − T 0 c1 1/pl + 1/c1 − 1
(23)
qtop = qp,c1
(24)
qp,c1 = qc1,c2
(25)
qc1,c2 = qc2,s
(26)
Utop =
qtop 0 − T0 Tpl amb
(27)
Using the SETP function, we get the same value for the top loss coefficient. Reading Utop From Figure 6.4.4b, Ut =4.0 W/m2K Ub = kinsulation /l
(28)
UL = Utop + Ub
(29)
Solution Condc1,c2 = 0.02787 [W/m-K] pl = 0.95 kinsulation = 0.035 [W/m-K] P lateLength = 2 [m] qc1,c2 = 401.5 W/m2 qp,c1 = 401.5 W/m2 qtop = 401.5 W/m2 Tamb = 10 [C] Tsky = 10 [C]
condp,c1 = 0.03051 [W/m-K] hc0 c1,c2 = 3.298 W/m2 -K l = 0.05 [m] P lateW idth = 1 [m]2 qc2,s = 401.5 W/m 2 qr0 c1,c2 = 265.5 W/m qw = 264.1 W/m2 Tc1 = 77.64 [C] T 0 amb = 283.1 [K]
T 0 pl = 383.2 [K] Utop = 4.015 W/m2 -K
T 0 sky = 283.1 [K]
147
c1 = 0.88 hc0 p,c1 = 2.964 W/m2 -K Ncov = 2 P lSpc1,c2 = 0.025 [m] 2 qc0 c1,c2 = 136 W/m qr0 c2,s = 137.4 W/m2 Slope = 45 [deg] Tc2 = 36.41 [C] T 0 c1 = 350.8 [K] Ub = 0.7 W/m2 -K
c2 = 0.88 hwind = 10 W/m2 -K P = 101.3 [kPa] P lSpp,c1 = 0.025 [m] qc0 p,c1 = 95.91 W/m2 qr0 p,c1 = 305.6 W/m2 Space = 0.025 [m] Tpl = 110 [C] T 0 c2 = 309.6 [K] UL = 4.715 W/m2 -K
P06 05
Equations 6.5
Calculation of collector efficiency factor F’ and flow factor FR
Knowns Cb = 1 × 109 ;
UL = 8; hf i = 300;
δ = .0005;
Di = 0.020;
D = 0.020;
mdot/Ac = 0.012;
W = 0.100
Cp = 4180
(1) (2)
The temperature is not given - assume: T = 75 [C]
(3)
k = k (M aterial$, T = T ) r UL m= 6.5.4a k·δ
(4)
mL = m · F =
(5)
W −D 2
tanh (mL) mL
(6) 6.5.12
(7)
1/UL
F0 = W·
1 UL ·(D+(W −D)·F )
F lowP ar = mdot/Ac ·
+ 1/Cb +
1 π·Di·hf i
6.5.18
(8)
Cp UL · F 0
(9)
F 00 = F lowP ar · (1 − exp (−1/F lowP ar))
(10)
FR = F 0 · F 00
(11)
Table 1 Run
M aterial$
1 2 3
Copper Aluminum Carbon Steel
k [W/m-K] 397.1 237.9 58.67
mL [-] 0.2539 0.328 0.6606
F [-] 0.9791 0.9656 0.8761
F0 0.9439 0.9339 0.8677
148
FR [-] 0.8762 0.8677 0.8104
P06 06
Equations 6.6
Calculation of F’ from UL and plate dimensions
Knows Di = 0.010;
D = 0.020;
kSolder = 20;
W = 0.16;
W idthSolder = 0.005;
hf i = 300;
k = 385;
UL = 3;
del = .001
T hickSolder = 0.002
(1) (2)
Use Equation 6.5.18: Cb = kSolder · W idthSolder/T hickSolder r UL m= 6.5.4a k · del mL = m · F =
6.5.16
(4)
W −D 2
tanh (mL) mL
(5) 6.5.12
(6)
1/UL
F0 = W·
(3)
1 UL ·(D+(W −D)·F )
+ 1/Cb +
1 π·Di·hf i
6.5.18
(7)
Solution Cb = 50 [W/m-C] F = 0.9875 [-] kSolder= 20 [W/m-C] UL = 3 W/m2 -C
D = 0.02 [m] F 0 = 0.9332 [-] m = 2.791 [1/m] W = 0.16 [m]
del = 0.001 [m] hf i = 300 W/m2 -C mL = 0.1954 [-] W idthSolder = 0.005 [m]
149
Di = 0.01 [m] k = 385 [W/m-C] T hickSolder = 0.002 [m]
P06 07
Equations 6.7
Derivation of F’ for type (b) air heater
The energy balance equation is: Qu = Ac · (S − UL · (Ti − Ta ))
(1)
Useful energy transfer to air by convection is: Qu /Ac = h1 · (T1 − Tf ) + h2 · (T2 − Tf )
(2)
With no back losses: h2 · (T2 − Tf ) = hr · (T1 − T2 )
(3)
Solve for T2 T2 =
hr · T1 + h2 · Tf hr + h2
(4)
Substitute (4) in (2) h2 · hr · (T1 − Tf ) Qu /Ac = h1 + h2 + hr
(5)
Solve for T1 T1 = Tf + Qu /Ac ·
1 h1 +
! (6)
h2 ·hr (h2 +hr )
Substitute (6) in (1) Qu /Ac = S − UL ·
Tf − Ta − (Qu /Ac ) ·
!!
1 h1 +
h2 ·hr (h2 +hr )
150
(7)
Solve for Qu /Ac Qu /Ac =
1
1+
UL h ·h h1 + h 2+hrr
· (S − UL · (Ti − Ta ))
(8)
2
Then Qu /Ac = F 0 · (S − UL · (Ti − Ta )) where F0 =
(9)
1 1+
UL h ·h h1 + h 2+hrr
(10)
2
151
P06 08
Equations 6.8
Calculation of collector output from two parameter model
Use Equation 6.7.6 - and watch units. The radiation (in MJ/m2 ) is not a rate but is total energy for the hour. Knowns IT = 3.4 MJ/m2 ;
Qu/A,c
FRU,L = 6.3 W/m2 ·C ;
FRta = 0.83;
Ti = 56 [C] ;
Ta = 14 [C] ;
·6 MJ = FRta · IT − FRU,L · (Ti − Ta ) · hr · 1 × 10 J
Solution FRta = 0.83 Qu /Ac = 1.869 MJ/m2
FRU,L = 6.3 W/m2 -C Ta = 14 [C]
hr = 3, 600 [s] Ti = 56 [C]
152
IT = 3.4 MJ/m2
hr = 3600 [s]
P06 10
Equations 6.10
Calculation of a day’s collector output
Knowns: UL = 5.2 W/m2 ·C ;
FR = 0.92;
Ti = 35 [C] ;
hr = 1
Others are set in the table.
Equation 6.7.6 - units are MJ. Allow only positive values of Qu \Ac. MJ/hr 2 , 0 MJ/m Qu/A,c = max FR · S − UL · (Ti − Ta ) · hr · 0.0036 W We can find the daily output two ways. Under the Menu Options/Preferences we have turned on the sum column. An alternative is to use the SumParametric command and sum from row 1 to the current row solving the table. We cannot sum the whole column at one time since many cells have not yet been filled in. SumQ = Sumparametric(‘Table 1’ , ‘Q uA c’ , 1, T ableRun#)
Solution FR = 0.92 SumQ = 12.15 MJ/m2
hr = 1 [hr] Ta = −4 [C]
Qu /Ac = 0.00 MJ/m2 Ti = 35 [C]
Table 1 Run
Hour
1 2 3 4 5 6 7 8 9 10 Sum
7 8 9 10 11 12 13 14 15 16 115
S 2 MJ/m 0.01 0.40 1.90 2.85 3.02 3.42 3.21 1.54 1.07 0.52 17.94
Ta [C] -3 0 4 5 7 9 11 5 1 -4 35
Qu \Ac2 MJ/m 0.00 0.00 1.21 2.11 2.30 2.70 2.54 0.90 0.40 0.00 12.15
153
S = 0.52 MJ/m2 UL = 5.2 W/m2 -C
P06 11
Equations 6.11
Calculation of FR for an air heater.
$UnitSystem SI, C, J, kPa Calculate the dimensionless flow parameter φ and F”. Knowns: F low/Ac = 10 l/s ·m2 ;
F 0 = 0.75;
UL = 6.0 W/m2 ·K
1 v (AIR, T = 20, P = 101.3) m3 mdot/Ac = F lowAc · ρ · 0.001 l ρ=
Cp = cp (AIR, T = 20) φ=
mdot/Ac · Cp UL · F 0
F 00 = φ · (1 − exp (− (1/φ))) FR = F 00 · F 0
Solution Cp = 1, 004 [J/kg-K] F 00 = 0.835 UL = 6 W/m2 -K
2 F low/Ac = 10 l/s-m mdot/Ac = 0.01204 kg/s-m2
154
FR = 0.6263 φ = 2.687
F 0 = 0.75 ρ = 1.204 kg/m3
P06 12
Equations 6.12
Calculation of useful gain from no-flow temperature
Knowns F RU L = 5.5 W/m2 ·C ;
Ti = 30 [C] ;
Tp = 118 [C]
Equation 6.7.6 - when the pump is off the absorbed energy equals thermal losses: 0=FR *S-FRU,L *(Tp -Ta ). Therefore, when the pump is on Qu \Ac =Q”u =FRU,L *(Tp -Ta )-FRU,L *(Ti -Ta ) or: Q00 u = F RU L · (Tp − Ti )
Solution F RU L = 5.5 W/m2 -C
Q00 u = 484 W/m2
Ti = 30 [C]
155
Tp = 118 [C]
P06 13
Equations 6.13
Calculation of mean plate and fluid temperatures.
Knowns: m ˙ = 0.015 kg/s ·m2 ; Cp = 4190 [J/kg ·C] ;
UL = 4 W/m2 ·C ; Ti = 55 [C] ;
F 0 = 0.9;
T auAlphae = 0.85
IT = 1000 W/m2 ;
Ta = 15 [C] ;
Dust = 1 − 0.015
Calculate the dimensionless flow parameter, φ, and FR with Equation 6.7.5: φ=m ˙ ·
Cp F 0 · UL
F 00 = φ · (1 − exp (−1/φ))
Eqn 6.7.5
FR = F 00 · F 0 Qu = FR · (IT · T auAlphae · Dust − UL · (Ti − Ta )) 1 − FR T¯plate = Ti + Qu · UL · FR
Eqn 6.9.4
1 − F 00 T¯f luid = Ti + Qu · UL · FR
Eqn 6.9.2
Eqn 6.7.6
Solution Cp = 4, 190 [J/kg-C] F 00 = 0.9719 Qu = 592.4 W/m2 T¯plate = 76.21 [C]
Dust = 0.985 IT = 1, 000 W/m2 T auAlphae = 0.85 Ti = 55 [C]
FR = 0.8747 m ˙ = 0.015 kg/s-m2 Ta = 15[C] UL = 4 W/m2 -C
156
F 0 = 0.9 φ = 17.46 T¯f luid = 59.76 [C]
P06 14
Equations 6.14
Calculation of critical radiation level UL = 4 W/m2 ·C ; F 0 = 0.9; T auAlphae = 0.85;
Ti = 55 [C] ;
See Section 6.8. The critical radiation level is that at which the Qu is zero is: 0 = ItCrit · T auAlphae · Dust − UL · (Ti − Ta )
Solution Dust = 0.985 Ta = 15 [C]
F 0 = 0.9 Ti = 55 [C]
2 ItCrit =191.1 W/m UL = 4 W/m2 -C
157
T auAlphae = 0.85
Ta = 15 [C] ;
Dust = 1 − 0.015
P06 15
Equations 6.15
Estimation of no-flow plate temperature
Knowns: UL = 4 W/m2 ·C ;
T auAlphae = 0.85;
Ti = 55 [C] ;
Ta = 15 [C] ;
It = 1000 W/m2 ;
Dust = 1 − 0.015
Use Equation 6.7.6, with Q = 0. The value of FR does not matter. Qu =FR *(IT *TauAlphae*Dust-UL *(Tplate -Ta )) Qu =0 0 = (IT · T auAlphae · Dust − UL · (Tplate − Ta )) Figure 6.4.4 indicates that UL rises as plate temperature increases. The calculation above is based on the assumption that UL stays at 4 W/m2C. Thus the UL used is low and the estimate of Tplate is high.
Solution Dust = 0.985 Ti = 55 [C]
It = 1, 000 W/m2 Tplate = 224.3 [C]
T auAlpha e = 0.85 UL = 4 W/m2 -C
158
Ta = 15 [C]
P06 16
Equations 6.16
Calculation of effective τ *α product for diffuse radiation on a collector.
We will solve part a using the basic equations and part b using the SETP functions. Knowns: N = 1;
KLa = 0.0524;
KLb = 0.0125;
Ref Index = 1.526;
αn = 0.9
Part a Considering all of the radiation to be diffuse then the equivalent surface slope is zero. For diffuse radiation from Figure 5.4.1the angle is 59. slope = 0 θd = 59.7 [deg] − 0.1388 · Slope + 0.001497 [1/deg] · Slope2 Abs/Absn = 1.0000 − 1.5879 × 10−3 [1/deg] · θd + 2.7314 × 10−4 1/deg2 · θd2 − 2.3026 × 10−5 1/deg3 · θd3 + 9.0244 × 10−7 1/deg4 · θd4 − 1.8000 × 10−8 1/deg5 · θd5 + 1.7734 × 10−10 1/deg6 · θd6 − 6.9937 × 10−13 1/deg7 · θd7 α = αn · Abs/Absn Find the transmittance N cov = 1 Ref rAng = arcsin REF perp = REF para =
sin (θd ) Ref Index
sin (Ref rAng − θd ) sin (Ref rAng + θd )
Eqn 5.1.4 2
tan (Ref rAng − θd ) tan (Ref rAng + θd )
Eqn 5.1.1 2 Eqn 5.1.2
T RAN Sperp =
1 − REF perp 1 + (2 · N cov − 1) · REF perp
T RAN Spara =
1 − REF para 1 + (2 · N cov − 1) · REF para
159
T RAN Sperp + T RAN Spara Eqn 5.1.9 2 KLa Eqn 5.2.2 T RAN Sabs = exp −N cov · Cos(Ref rAng) T RAN Sref =
τ = T RAN Sabs · T RAN Sref ρd = 0.15
Eqn 5.3.4
Using the approximation in footnote 2 of Section 5.5.
T auAlphaP rod,a = τ ·
α 1 − (1 − α) · ρd
Eqn 5.5.1
If we use the EES SETP function T auAlphaP rod,a2 = T auAlphaP rod (N cov, θd , KLa , Ref Index, αn ) According to comments near Eqn 6.10.10a, T auAlphae,a = 1.02 · T auAlphaP rod,a Part b Here we will use the SETP EES function. KL is small so absorption is small and we can use Eqn 6.10.10b T auAlphaP rod,b = T auAlphaP rod (N cov, θd , KLb , Ref Index, αn ) T auAlphae,b = 1.01 · T auAlphaP rod,b
Solution Abs/Absn = 0.931 KLb = 0.0125 REF para = 0.001174 slope = 0 [deg] T auAlphaP rod,a = 0.680 T RAN Sabs = 0.9384
α = 0.8379 N =1 REF perp = 0.1828 τ = 0.7923 T auAlphaP rod,a2 = 0.680 T RAN Spara = 0.9977
160
αn = 0.9 N cov = 1 Ref rAng = 34.46 [deg] T auAlphae,a = 0.694 T auAlphaP rod,b = 0.714 T RAN Sperp = 0.6909
KLa = 0.0524 Ref Index = 1.526 ρd = 0.15 T auAlphae,b = 0.7212 θd = 59.7 [deg] T RAN Sref = 0.8443
bill2
Equations 7.1
Calculation of collector efficiency factor and collector flow factor for a concentrator with cylindrical receive
$UnitSystem SI C J, kPa Knowns Dout = 0.054 [m]
(1)
Din = 0.045 [m]
(2)
m ˙ = 0.0168 [kg/s]
(3)
Length = 3.1 [m]
(4)
Apertrue = 0.3 [m]
(5)
Tave = 100 [C]
(6)
P = 300 [kPa]
(7)
UL = 7.5 W/m2 ·K
(8)
Properties: The pipe material in not known but the results are not a strong functions of the pipe thermal conductivity. kwall = k (‘Carbon steel AISI1010’ , Tave )
(9)
µ = µ (SteamIAP W S , T = Tave , P = P )
(10)
P r# = Pr (SteamIAP W S , T = Tave , P = P )
(11)
k = k (SteamIAP W S , T = Tave , P = P )
(12)
Cp = cp (SteamIAP W S , T = Tave , P = P )
(13)
Dhyd = Dout − Din
(14)
From Equation 3.14.3
161
1
Re# = Dhyd *Vel*Density/µ = m*D ˙ ˙ hyd /(µ*AXsect ) = 4*m/(µ*π*(D out -Din )) Re# = 4 ·
m ˙ µ · π · (Dout + Din )
(15)
LoverD = Length/Dhyd
(16)
RelRough = 0
(17)
smooth tubes
call P ipeF lowN (Re#, P r#, LoverD, RelRough : N usseltT , N usseltH , f )
(18)
The flow in the annulus is laminar. We might have used Figure 3.14.1 to estimate the convection heat transfer coefficient in the annulus but this figure is for constant wall temperature conditions. The abscissa is: X = Re# · P r# · Dhyd /Length
(19)
At X = 3.89, from Figure 3.14.1 ν is about 3.9 - this is close to NusseltT found above, providing confidence in the function. hf i · Dhyd /k = N usseltH F0 = (1/UL ) +
Dout hf i ·Din
The system is subject to a constant heat flux.
1/U L + Dout ·
ln (Dout /Din ) 2·kwall
Eqn 7.3.10
ARec = π · Dout · Length Y =m ˙ ·
(21)
(22)
Cp ARec · UL · F 0
(23)
FR = F 0 · Y · (1 − exp (−1/Y ))
Eqn 7.3.13
(24)
Solution Apertrue = 0.3 [m] Dhyd = 0.009 [m] f = 0.08704 hf i = 371.9 W/m2 -K Length = 3.1 [m] m ˙ = 0.0168 [kg/s] P = 300 [kPa] RelRough = 0 X = 3.893
(20)
ARec = 0.5259 m2 Din = 0.045 [m] FR = 0.9497 k = 0.6792 [W/m-K] LoverD = 344.4 N usseltH = 4.928 [-] P r# = 1.749 Tave = 100 [C] Y = 18.4
Cp = 4215 [J/kg-K] Dout = 0.054 [m] F 0 = 0.9758 kwall = 60.1 [W/m-K] µ = 0.0002818 [kg/m-s] N usseltT = 3.897 [-] Re# = 766.7 [-] UL = 7.5 W/m2 -K
162
2
P07 02
Equations 7.2.
Redo Examples 7.3.1 and 7.3.2 with emittance of the tube at 0.18.
$UnitSystem SI K J kPa Knowns from Example 7.3.1 V = 5 [m/s] ;
Dcover = 0.090 [m] ;
cover = 0.88;
tube = 0.18;
Ta = 283 [K] ;
Dcover,i = Dcover − 0.008m ;
Dtube = 0.060 [m] ;
Ttube = 473 [K] ;
σ = sigma#
kglass = 1.4 [W/m ·k]
Tsky = 275 [K]
(1) (2) (3)
Knowns from Example 7.3.2 L = 10 [m] ;
W = 2.5 [m] ;
Cp = 3260 [J/kg ·K] ;
m ˙ = 0.0537 [kg/s] ;
ktube = 16 [W/m ·k] ;
Dtube,i = Dtube − 0.010m ;
Tf luid,in = ConvertTemp(C, K, 156.9 [C])
hf i = 300 W/m2 ·K
S = 430 W/m2
(4) (5) (6)
Guess the cover temperatue to develop the governing equations. Tcover =290 [K] Properties of air - assume atmospheric pressue is 101.3 [kPa] Tcover + Ta T¯ = 2
(7)
patm = 101.3 [kPa]
(8)
µ = µ Air, T = T¯
(9)
ρ = ρ Air, T = T¯, P = Patm k = k Air, T = T¯
(10) (11)
Re# = ρ · V · Dcover /µ N u# = 0.30 · Re#.6
(12) Eqn 3.15.12
(13)
N u# = hw · Dcover /k
(14)
163
Convection and radiation heat loss from cover. Acover = π · Dcover · L
(15)
4 4 Qloss = Acover · hw · (Tcover − Ta ) + σ · cover · Tcover − Tsky
(16)
Find temperature inside tube. Qloss =
((2 · π · kglass · L) · (Tcover,i − Tcover )) ln (Dcover /Dcover,i )
(17)
Radiation from tube to inside cover must equal heat loss from cover. At this point it is necessary to update guesses and then comment out the initial guess for cover temperatue. Atube = π · Dtube · L Qloss = Atube · σ ·
(18)
T4 − T4 tube cover,i cover 1/tube + 1− · Dtube /Dcover,i cover
(19)
The collector loss coefficient (based upon the tube temperatue and absorber area) is then found from Eqn 7.3.1 Qloss = UL · Atube · (Ttube − Ta )
(20)
Calculate F’: F0 = 1/UL +
Dtube (Dtube,i ·hf i )
+
1/U L Dtube 2·ktube
Eqn 7.3.10
(21)
· ln (Dtube /Dtube,i )
Calculate area of receiver, ARec : ARec = π · Dtube · L
(22)
Calculate unshaded area of aperture, AAper : AAper = (W − Dcover ) · L
(23)
Calculate the dimensionless flow rate DimFlowRate to use in calculation of F: DimF lowRate = m ˙ ·
Cp ARec · UL · F 0
F 00 = DimF lowRate · (1 − exp (−1/DimF lowRate))
(24) Eqn 7.13.13
(25)
FR = F 00 · F 0
(26)
Find the useful gain from Eqn 7.3.12. Qu = AAper · FR · (S − UL · (ARec /AAper ) · (Tf luid,in − Ta ))
(27)
Solution AAper = 24.1 m2 Cp = 3, 260 [J/kg-K] Dtube = 0.06 [m] FR = 0.9783 hw = 40.62 W/m2 -K L = 10 [m] patm = 101.3 [kPa] ρ = 1.235 kg/m3 T¯ = 285.8 [K] Tsky = 275 [K]
Acover = 2.827 m2 ARec = 1.885 m2 DimF lowRate = 41.3 [-] Dcover = 0.09 [m] Dtube,i = 0.05 [m] cover = 0.88 F 0 = 0.9902 F 00 = 0.988 [-] k = 0.02459 [W/m-K] kglass = 1.4 [W/m-k] µ = 0.0000179 [kg/m-s] m ˙ = 0.0537 [kg/s] Qloss = 813.3 [W] Q 164u = 9, 523 [W] S = 430 W/m2 σ = 5.670 × 10−8 W/m2 -K4 Tcover = 288.6 [K] Tcover,i = 289.4 [K] Ttube = 473 [K] UL = 2.271 W/m2 -K
Atube = 1.885 m2 Dcover,i = 0.082 [m] tube = 0.18 hf i = 300 W/m2 -K ktube = 16 [W/m-k] N u# = 148.7 [-] Re# = 31, 039 Ta = 283 [K] Tf luid,in = 430.1 [K] V = 5 [m/s]
P07 03
Equations 7.3
Estimation of the output of a full CPC
Knowns Surf Ref l = 0.85; Lat = 43;
β = 60;
KL = 0.0370; a
nr = 0.89;
Ibeam = 1.20 MJ/m2 ;
M onth = 1;
Ref rInd = 1.526;
day = 26;
Idif f use = 0.36 MJ/m2 ;
θc = 15
(1)
time = 10 : 30;
Surf AzAng = 0; Ncov = 1 (2) Ta = −5 [C] ; UL = 9.0 W/m2 ·C ; FR = 0.92 (3)
Tin = 55 [C] ;
Calculation of absorbed radiation
αb = 0.95
From Figure 4.11.1 at an angle of 21deg.
(4)
αd = 0.95
Assumed to be the same as for beam.
(5)
τcover,b = T ransIncAng (Ncov , θb , KL, Ref rInd)
(6)
τcover,d = T ransIncAng (Ncov , 60, KL, Ref rInd)
(7)
Angles ω = (T ime − 12 [hr]) · 15 [deg/hr]
(8)
n = nDay (month, Day)
(9)
Rbeam = RBeam, (Lat, n, ω, β, Surf AzAng)
(10)
δ = Dec (n)
(11)
θz = arccos (CosZenAng (Lat, n, ω))
(12)
θb = arccos (cosIncAng (Lat, n, ω, β, Surf AzAng))
(13)
γs = SolAzAng (Lat, n, ω)
(14)
165
Equation 7.7.2 is used to determine the ContFunct, F: ContF unct1 = If((β − θc ) , (arctan (tan (θz ) · Cos(γs ))) , 1, 1, 0)
(15)
ContF unct2 = If((arctan (tan (θz ) · Cos(γs ))) , (β + θc ) , 1, 1, 0)
(16)
ContF unct = ContF unct1 · ContF unct2
(17)
Use Equations 7.7.1 to calculate S: C=
1 sin (θc )
(18)
T ransCP C = Surf Ref lnr
(19)
AbsBeam = ContF unct · Ibeam · Rbeam · τcover,b · αb · T ransCP C
(20)
AbsDif f use = (Idif f use /C) · τcover,d · T ransCP C · T ransCP C · αd
(21)
AbsGrRef lected = 0
(Since (beta + thetac ) < 90)
(22)
S = AbsBeam + AbsDif f use + AbsGrRef lected
(23)
b Q00 u
Calculation of collector gain per unit area MJ/hr · m2 = FR · S − (UL /C) · 1 [hr] · (Tin − Ta ) · 0.0036 W/m2
(24)
Solution AbsBeam = 1.945 MJ/m2 αd = 0.95 ContF unct1 = 1 FR = 0.92 KL = 0.037 Ncov = 1 Ref rInd = 1.526 Surf Ref l = 0.85 θc = 15 [deg] Ta = −5 [C]
AbsDif f use = 0.05336 MJ/m2 β = 60 [deg] ContF unct2 = 1 γs = −23.45 [deg] Lat = 43 [deg] nr = 0.89 Rbeam = 2.235 τcover,b = 0.8821 θz = 65.39 [deg] Tin = 55 [C]
166
AbsGrRef lected = 0 MJ/m2 C = 3.864 [-] day = 26 Ibeam = 1.2 MJ/m2 M onth = 1 ω = −22.5[deg] S = 1.998 MJ/m2 τcover,d = 0.8051 time = 10.5 [hr] UL = 9 W/m2 -C
αb = 0.95 ContF unct = 1 δ = −19.03 [deg] Idif f use = 0.36 MJ/m2 n = 26 Q00 u = 1.375 MJ/m2 Surf AzAng = 0 θb = 21.48 [deg] T ransCP C = 0.8653
P07 04
Equations 7.4
Absorbed radiation for a truncated CPC array
$UnitSystem SI C J kPa Knowns ρsurf = 0.85;
nr = 0.90;
θc = 9 [deg] ;
C = 6.0;
τcover,b = 0.90;
Ib,n = 1.13 MJ/m2 ; Lat = 45;
Id = 0.37 MJ/m2 ;
β = 65 [deg]
τcover,d = 0.85
HrAng = −22.5 [deg]
(1) (2) (3)
Use the parametric table for the two dates. n = nday (month, day)
(4)
γs = SolAzAng (Lat, n, HrAng)
(5)
θz = arccos (CosZenAng (Lat, n, HrAng))
(6)
Dec = Dec (n)
(7)
θb = arccos (cos (Lat − β) · cos (Dec) · cos (HrAng) + sin (Lat − β) · sin (Dec))
(8)
Equation 7.7.2 is used to determine the ContFunct F ContF unct1 = If((β − θc ) , (arctan (tan (θz ) · Cos(γs ))) , 1, 1, 0)
(9)
ContF unct2 = If((arctan (tan (θz ) · Cos(γs ))) , (β + θc ) , 1, 1, 0)
(10)
ContF unct = ContF unct1 · ContF unct2
(11)
Equation 7.7.1 is used for the absorbed solar. r τCP C = ρnsurf
Eqn 7.7.4
(12)
AbsBeam = ContF unct · Ib,n · Cos(θb ) · τcover,b · τCP C · 0.88
(13)
AbsDif f use = (Id /C) · τcover,d · τCP C · 0.88
(14)
AbsGrRef lected = 0
(15)
Since (beta - thetac < 90)
S = AbsBeam + AbsDif f use + AbsGrRef lected
(16)
Solution
167
AbsBeam = 0 MJ/m2 C=6 day = 15 Ib,n = 1.13 MJ/m2 n = 74 τcover,b = 0.9 θc = 9 [deg]
AbsDif f use = 0.03985 MJ/m2 ContF unct = 0 Dec = −2.819 [deg] Id = 0.37 MJ/m2 nr = 0.9 τcover,d = 0.85 θz = 51.85 [deg]
AbsGrRef lected = 0 MJ/m2 ContF unct1 = 0 γs = −29.08 [deg] Lat = 45 [deg] ρsurf = 0.85 τCP C = 0.8639
β = 65 [deg] ContF unct2 = 1 HrAng = −22.5 [deg] month = 3 S = 0.04 MJ/m2 θb = 27.88 [deg]
Table 1 Run
month
day
1 2
1 3
2 15
Dec [deg] -22.93 -2.819
γs [deg] -21.89 -29.08
AbsBeam MJ/m2 0.7212 0
AbsDif f use MJ/m2 0.03985 0.03985
168
AbsGrRef lected MJ/m2 0 0
ContF unct
nr
1 0
0.9 0.9
S 2 MJ/m 0.76 0.04
P07 05
Equations 7.5
Performance of a linear concentrator with given focal zone flux distribution
Knowns M onth = 3; W = 1.8 [m] ;
day = 16;
Lat = 32 [deg] ;
L = 10.5 [m] ;
T ransAbsorptP rod = 0.78;
Wreceiver = 0.022 [m] ;
Tin = 120 [C] ;
Ref lconc = 0.87
(1)
hf i = 3000 W/m2 ·C
(2)
For linear concentrator on N-S horizontal axis, use Equation 1.7.3a Use the parametric table for time. n = nDay (month, Day)
(3)
Dec = Dec (n)
(4)
θz = arccos (CosZenAng (Lat, n, HrAng))
(5)
169
θ = arccos
2
2
2
(cos (θz )) + (cos (Dec)) · (sin (HrAng))
.5
Eqn 1.7.3a
(6)
The intercept factors IntFact must be determined. The image width with radiation normal to the aperture is 0.020m, from the figure. At times 9:30 and 11:30, when HrAngs are -37.5 and -7.5, the image widths are 0.020/cos(θ), and are 0.22 m and 0.024 m. These broadened images will be centered on the normal incidence images. From the plot, the intercept factors are 17/22 = 0.77 and 18/24 = 0.75, respectively. Areaaper = (W − Wreceiver ) · L
(7)
AbsRad = Areaaper · Ib,n · Cos(θ) · Ref lconc · IntF ract · T ransAbsorptP rod UL = 9.6 W/m2 ·C From Figure 7.13.2
(8) (9)
Use Equation 7.3.12 to estimate Qu. First estimate FR. Since liquid is being boiled, there is no temperature gradient in the flow direction, and FR = F’. Assume that all of the absorber is wetted by the boiling liquid. By analogy with Figure 6.14.1: F0 =
1 ; 1 + UL /hf i
FR = F 0
(10)
Areceiver = Wreceiver · L 3600 Qu = FR · AbsRad − UL · Areceiver · · (Tin − Tamb ) 106
(11) (12)
Solution AbsRad = 25.36 [MJ] Dec = −2.418 [deg] hf i = 3, 000 W/m2 -C Lat = 32 [deg] Ref lconc = 0.87 Tamb = 9 [C] Wreceiver = 0.022 [m]
Areaaper = 18.67 m2 F 0 = 0.9968 IntF ract = 0.77 M onth = 3 θ = 34.1 [deg] Tin = 120 [C]
Areceiver = 0.231 FR = 0.9968 Ib,n = 3.14 MJ/m2 n = 75 θz = 35.1 [deg] UL = 9.6 [W/m2-C]
day = 16 HrAng = −7.5 [deg] L = 10.5 [m] Qu = 24.4 [MJ] T ransAbsorptP rod = 0.78 W = 1.8 [m]
IntF ract
Qu [MJ] 16.43 24.4
Table 1 Run 1 2
HrAng [deg] -37.5 -7.5
Ib,n 2 MJ/m 2.06 3.14
Tamb [C] 2 9
θ [deg] 27.1 34.1
θz [deg] 49.5 35.1
0.75 0.77
170
AbsRad [MJ] 17.42 25.36
P07 06
Equations 7.6
Calculation of maximum concentration ratios for collectors on Venus.
Use Equations 7.2.5 and 7.2.6. Calculate θs for Venus at 67*106 miles from the sun. 4.32 · 105 θs = arcsin 67 · 106 Cmax,lin =
1 sin (θs )
(1) (2)
For linear concentrator use Equation 7.2.6: Cmax,circ =
1
(3)
2
(sin (θs ))
For circular concentrator use Equation 7.2.5:
Solution Cmax,circ = 24, 054
Cmax,lin = 155.1
θs = 0.3694
171
P07 07
Equations 7.7
Use of Incidence Angle Modifier for a CPC.
Knowns FRT A,n = .66;
FRU,L = 1.2 W/m2 ·C ;
β = 40 [deg] ;
lat = 38 [deg] ;
month = 3;
day = 21
γ=0
(1) (2)
At the equinox, with Slope nearly equal to the Latitude, the sun is in the plane perpendicular to the tubes. Since the sun travels 15 degrees/hr in this plane, the angles of incidence in the perpendicular plane will be approximately 0, 30, and 60 at noon, 2PM and 4PM. (The exact values are calculated using Equation 1.6.2.) Values of IncAngMod are read from the plot and entered in the table with the data. Incident angle n = nDay (month, day)
(3)
Cos(θ) = cosIncAng (Lat, n, HrAng, β, γ)
(4)
Since collector performance data are on the basis of the array area, use Equation 6.17.9. Qu/Areac = FRT A,n · IncAngM od · GT − FRU,L · (Ti − Ta )
(5)
Table 1 Run 1 2 3
Ti [C] 75.0 75.0 75.0
Ta [C] 20.0 22.0 22.0
GT 2 W/m 925 760 385
HrAng [deg] 0 30 60
θ [deg] 1.6 30.0 60.0
IncAngM od 1.00 1.13 0.91
172
Qu\Area c W/m2 545 503 168
P07 08
Equations 7.8
Output of linear parabolic collector for perfect and imperfect concentrators.
Knowns: Aperture = 1.36 [m] ;
F ocLength = 1.26 [m] ;
UL = 14 W/m2 ·C ;
Tin = 230 [C] ;
FR = 0.9;
T ransAbsP rod = .78;
Lat = 35 [deg] ;
Tamb = 7 [C] ;
Dec = 0 [deg] ;
InterceptF act = .95;
Ib,n = 2.14 MJ/m2 ;
L = 10 [m]
(1)
Ref lconc = .87;
(2)
HourAngle = 0 [deg]
(3)
Calculate RimAngle with Equation 7.9.2: RimAngle = arctan 8 ·
!
F ocLength/Aperture 2
16 · (F ocLength/Aperture) − 1
(4)
For the perfect reflector, use Figure 7.10.1b, at a rim angle of 30. For the intercept factor to be 0.95, Y/f*103 = 4.2. Wperf ect = 2 · 4.2 ·
F ocLength 103
(5)
For the imperfect reflector, use Figure 7.11.1 at σ = 0.15. It is necessary to graphically integrate to get intercept factor vs Y/f*103. This results in a value of Y/f*103 = 13 at an intercept factor of 0.95 Wimperf ect = 2 · 13 ·
F ocLength 103
(6)
Neglecting shading by receiver: AAperture = L · Aperture
(7)
ARec = L · Wimperf ect
(8)
hr = 1 [hr]
(9)
173
Calculate Qu : Qu = FR · AAperture · (Ib,n · InterceptF act · T ransAbsP rod · Ref lconc − MJ · (A /A ) · (T − T ) UL · hr · 0.0036 Rec Aperture in amb W · hr
(10)
Solution Aperture = 1.36 [m] F ocLength = 1.26 [m] InterceptF act = 0.95 [-] Qu = 13.57 [MJ] Tamb = 7 [C] Wperf ect = 0.01058 [m]
AAperture = 13.6 m2 FR = 0.9 [-] Ib,n = 2.14 MJ/m2 Ref lconc = 0.87 [-] Tin = 230 [C]
ARec = 0.3276 m2 HourAngle = 0 [deg] L = 10 [m] RimAngle = 30.2 [deg] UL = 14 W/m2 -C
174
Dec = 0 [deg] hr = 1 [hr] Lat = 35 [deg] T ransAbsP rod = 0.78 Wimperf ect = 0.03276 [m]
P07 09
Equations 7.9
Geometry of heliostat - central receiver collectors.
If the incident beam radiation is approximately notmal to the heliostat, and if there is no pointing error, the receiver dimension is the width of a beam from a point on the heliostat to to the receiver L meters away., plus the plus the largest linear dimension of the heliostat, W m. For L - 1000 m and W = 10 m, and with the half-angle subtended by the sun of 0.27 degrees: knowns: L = 1000 [m] ;
W = 10 [m]
(1)
RecDimension = 2 · L · sin (0.27) + W
(2)
If there is a pointing error of P degrees, it will add to the 0.27 half-angle in the same way as a dispersion angle. The relationship between the receiver dimension and the pointing error for L = 100 and W = 10 is: RecDimension0 = W + 2 · L · sin (0.27 + P )
(3)
so P = arcsin
RecDimension0 − W 2·L
− 0.27
(4)
Solution L = 1, 000 [m] W = 10 [m]
P = 0.6205
RecDimension = 19.42 [m]
175
RecDimension0 = 41.08 [m]
P08 01
Equations 8.1.
Track temperature changes on a fully mixed storage tank.
Loads and inputs are given as independent of tank temperature. This is not realistic; a real system will have inputs and outputs that are tank-temperature dependent, as shown, for example, in Chapter 10. Knowna: m = 500 [kg] ;
Tamb = 20 [C] ;
U As = 5.56 [W/C] ;
Cp = 4190 [J/kg ·C] ;
Ta = 20 [C] ;
hr = 1 [hr](1)
We will use two parametric tables. Note that in Row 1 we have entered the tank initial condition so, solutions start at row 2. Assume there are no constraints on temperature: $IF ParametricTable = ’Solution A’ J/hr J + Ts − Ts · m · Cp = (Qu − Load) · 1000000 − U As · (Ts − Ta ) · hr · 3600 MJ W Ts = Tablevalue(‘Solution A’ , tablerun# − 1, ‘Tplus s’ )
(2)
Used to extract the tank temperature for the next hour. (3)
$EndIf If these calculations were carried through more hours on a clear day, the tank temperature would go well over 100 C. The temperature can not go below 0 C unless the store goes to ice, which requires a different formulation. The temperature cannot exceed 100C unless the tank is pressurized (not common in solar systems). In practice the temperature of a store must be constrained by the characteristics of the storage medium or the mechanisms for adding or removing energy, or both. Solution B: Assume that temperature is constrained between 20 and 100, representing a minimum temperature for energy delivery and a maximum temperature for energy storage. Other constraints could be assumed. The constraints can be programmed thus: $If ParametricTable=’Solution B’ J/hr J + 3600 TsCalc − Ts · m · Cp = (Qu − Load) · 1000000 − U A · (T − T ) · hr · s s a MJ W + + Ts+0 = If TsCalc , 100, TsCalc , 100, 100 Ts+ = If Ts+0 , 20, 20, 20, Ts+0 Ts = Tablevalue(‘Solution A’ , tablerun# − 1, ‘Tplus s’ )
(4) (5) (6)
Used to extract the tank temperature for the next hour. (7)
$EndIf Here the temperature did not reach 100. Assuming reasonable values for Qul and Load for the following hours results in temperatures that are constrained to 100. Note also that during hours when the tank temperature is constrained the load must be met by other energy supplies. The ‘Load deficiencies’ for the four hours when Tplus s is at 20 are 8, 14, 18, and 21 MJ. Thus 61 MJ would have to be supplied from another source.
Solution
176
Cp = 4, 190 [J/kg-C] Qu = 75 [MJ/hr] Ts+ = 50.61
hr = 1 [hr] Ta = 20 [C] + TsCalc = 50.61
Load = 16 [MJ/hr] Tamb = 20 [C] Ts+0 = 50.61
Solution A Run
T ime
1 2 3 4 5 6 7 8 9 10 11 12 13
0 1 2 3 4 5 6 7 8 9 10 11 12
Load [MJ/hr]
Qu [MJ/hr]
Ts [C]
12 12 11 11 13 14 18 21 20 20 18 16
0 0 0 0 0 0 0 0 21 41 60 75
45.0 39.0 33.1 27.7 22.4 16.2 9.5 1.1 -8.8 -8.0 2.3 22.5
Load [MJ/hr]
Qu [MJ/hr]
Ts [C]
12 12 11 11 13 14 18 21 20 20 18 16
0 0 0 0 0 0 0 0 21 41 60 75
45.0 39.0 33.1 27.7 22.4 16.2 9.5 1.1 -8.8 -8.0 2.3 22.5
Tsplus 45 39.03 33.12 27.75 22.42 16.19 9.548 1.056 -8.787 -8.034 2.257 22.47 50.61
Solution B Run
T ime
1 2 3 4 5 6 7 8 9 10 11 12 13
0 1 2 3 4 5 6 7 8 9 10 11 12
Tsplus 45 39.03 33.12 27.75 22.42 20 20 20 20 20 20 22.47 50.61
177
m = 500 [kg] Ts = 22.5 [C] U As = 5.56 [W/C]
P08 02
Equations 8.2
Develop equations for a partially stratified, two node tank.
It can be assumed that water supplied from the mains at Tm will always enter the bottom node. The two node tank is shown in the diagram Window. Equation 8.4.1 becomes: F1c = If(To , T1 , 0, 0, 1)
(1)
Is the flow between nodes up or down? Fdown = If(F1c · m ˙ c, m ˙ L , 0, 0, 1);
Fup = 1 − Fdown
(2)
For node 1 the energy balance is: m1 ·
T1+ − T1 = (U A/Cp ) · (T 0 a − T1 ) + F1c · m ˙ c · To − m ˙ L · T1 − Fdown · (F1c · m ˙ c−m ˙ L ) · T1 ∆T
+ Fup · (F1c · m ˙ c−m ˙ L ) · T2
(3)
For node 2 the energy balance is: m2 ·
T2+ − T2 = (U A/Cp) · (T 0 a − T 2) + (1 − F1c ) · m ˙ c · To + Fdown · (F1c · m ˙ c−m ˙ L ) · T1 ∆T
− Fup · (F1c · m ˙ c−m ˙ L ) · T2
(4)
178
P08 03
Equations 8.3
Calculation of pressure drop across a packed bed store.
$UnitSystem SI C, kPa Knowns Length = 1.80 [m] ;
P artDiam = .0125 [m] ;
Patm = 101.3 [kPa] ;
Ta = 35 [C]
Af rontal = 14.8 m2 ;
V = 0.053 [m/s]
estimated
(1) (2)
ρ = ρ (air, p = Patm , T = Ta )
(3)
m ˙ = ρ · Af rontal · V
(4)
M assV el = m/A ˙ f rontal
(5)
∆P = P resDrop (Length, M assV el, P artDiam, Ta )
(6)
This design does not satisfy the minimum pressure drop of 55 Pa.
Solution Af rontal = 14.8 m2 m ˙ = 0.8984 [kg/s] Ta = 35 [C]
∆P = 29.98 [Pa] P artDiam = 0.0125 [m] V = 0.053 [m/s]
Length = 1.8 [m] Patm = 101.3 [kPa]
179
M assV el = 0.0607 kg/s-m2 ρ = 1.145 kg/m3
P08 04
Equations 8.4
Estimation of equivalent heat capacity of a phase change storage medium.
From Eqn 8.8.9: EquivHeatCap = mass · 1 × 10·6 MJ J
LatentHeat + 18.3 · 2 TP haseChange
1−
18.3 TP2 haseChange
! · Cp,Solid + 18.3 ·
Cp,Liquid TP2 haseChange
! (1)
Data on the paraffin wax are from Table 13.7.1: Cp,Solid = 2890;
Cp,Liquid = 2890
TP haseChange = 46.7 [C] ;
Assumed same as solid
LatentHeat = 209000 [J/kg] ;
(2) mass = 1300 [kg]
(3)
Solution Cp,Liquid = 2, 890 [J/kg-C] mass = 1, 300 [kg]
Cp,Solid = 2, 890 [J/kg-C] TP haseChange = 46.7 [C]
180
EquivHeatCap = 6.037 [MJ/C]
LatentHeat = 209, 000 [J/kg]
P08 05
Equations function Vcell, (I, F, Qm )
(1)
Battery Parameters from Eckstein thesis (1990) KDI = 29.3 [1/V]
(2)
IDI = 2.5 [amp]
(3)
mc = 0.864 [·]
(4)
md = 1.0 [·]
(5)
Rsc = 3
[ohm ·amp ·hr] Qm
Rsd = 0.5
(6)
[ohm ·amp ·hr] Qm
(7)
Gc = 0.08 [V]
(8)
Gd = 0.08 [V]
(9)
Esc = 2.25
(10)
Esd = 2.10
(11)
Qc = −0.035 · Qm
(12)
Qd = Qm /0.85
(13)
VDI = 1/KDI · ln Voc =
abs (I) +1 IDI
(14)
Esc + Esd 2
(15)
H =1−F If I < 0 then
(16) Discharge
(17)
Vcell,=V,oc − VDI − Gd · H + I · Rsd · 1 + md ·
H Qd /QM − H
181
(18)
else
Charge
(19)
Vcell,=V,oc + VDI − Gc · H + I · Rsc · 1 + mc ·
H Qc /QM − H
endif
(20) (21)
end Vcell,
(22)
The problem must be run in two parts, one charging and one discharging. The charging part is trial and error. Select a voltage until the charge time equals 11 hours. Must uncomment one of these two statements Charging$=‘True’ Charging$ = ‘False’
(23)
ChargeEf f iciency = 0.95
(24)
Imax = 11 [amp]
(25)
Ns = 6
(26)
Np = 1
(27)
Qm = 50 [amp ·hr]
(28)
Qm,cell = Qm /Ns
(29)
Vcell = Vcell, (Icell , F, Qm,cell )
Simple function to evaluate equations 8.10.1 and 8.10.2
(30)
Vbatt = Ns · Vcell
(31)
Ibatt = min (Icell · Np , Imax )
(32)
η = If(Ibatt , 0, 1, 1, ChargeEf f iciency)
eta = 1 when discharging and eta=0.95 when charging
F = Q/Qm
(33) (34)
Z
T imef inal
Ibatt · η dtime
Q = Qo +
(35)
T imeinitial
Z
T imef inal
Ibatt · Vbatt dtime
P =
(36)
T imeinitial
$IF Charging$=’True’ Qo = 10 [Amp ·hr]
(37)
T imeinitial = 0 [hr]
(38)
T imef inal = 11
(39)
Vbatt = 15.2
(40)
By trial and error a voltage of 15.2V will charge the battery in 11 hours.
182
$Else Rload = 2 [ohm]
(41)
Qo = 50 [Amp ·hr]
(42)
T imeinitial = 11 [hr]
(43)
T imef inal = 19.25 [hr] Vbatt = −Ibatt · Rload
By trial and error the battery lasts 19.25 hr.
(44)
Resistance load
(45)
$EndIF $IntegralTable time: 0.5,Q,F,I_batt, V_batt,P Find the cycle efficiency. The power to charge was 640 Wh and the power delivered discharge was 391 Wh. The cycle efficiency is: ηcycle = 391/640
(46)
Solution Variables in Main program ChargeEf f iciency = 0.95 F = 0.200 [-] Np = 1 Qm = 50 [amp-hr] time = 19.25 [hr] Vcell = 1.33 [V]
Charging$ = ‘False’ Ibatt = −3.989 [amp] Ns = 6 Qm,cell = 8.333 [Amp-hr] T imef inal = 19.25 [hr]
η = 1 [-] Icell = −3.989 [amp] P = −391 [W-hr] Qo = 50 [Amp-hr] T imeinitial = 11 [hr]
Integral Table Row 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
time [hr] 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 16.5 17 17.5 18 18.5 19
Q [amp-hr] 50.0 47.3 44.6 42.0 39.4 36.8 34.2 31.7 29.2 26.7 24.3 22.0 19.7 17.4 15.2 13.1 11.0
F [-] 1.000 0.946 0.892 0.839 0.787 0.735 0.684 0.634 0.584 0.535 0.487 0.439 0.393 0.348 0.304 0.261 0.220
Ibatt [amp] -5.430 -5.380 -5.327 -5.271 -5.213 -5.150 -5.084 -5.014 -4.938 -4.857 -4.769 -4.675 -4.573 -4.462 -4.341 -4.210 -4.066
Vbatt [V] 10.859 10.759 10.654 10.543 10.425 10.301 10.168 10.027 9.876 9.713 9.539 9.350 9.146 8.924 8.682 8.419 8.132
P [W-hr] 0 -29 -58 -86 -113 -140 -166 -192 -217 -241 -264 -286 -308 -328 -347 -366 -383
183
ηcycle = 0.6109 Imax = 11 [amp] Q = 10.0 [amp-hr] Rload = 2 [Ω] Vbatt = 7.979 [V]
18
19.25
10.0
0.200
-3.989
7.979
-391
F vs time (Problem 8.5)
184
P09 01
Equations 9.1 Calculation of water heating loads on a commercial building. $UnitSystem SI C J kPa Knowns U Aloop = 38.1 [W/C] ;
U Atank = 11 [W/C] ;
U seM,F = 610 [L/day] ;
Tave =
U seS,S = 210 [L/day] ;
Tcold = 12 [C] ;
Thot = 38 [C]
Troom = 20 [C] ;
L = 122 [m] ;
(1) D = 0.019 [m]
Thot + Tcold 2
(2)
(3)
Cp = cp (W ater, T = Tave , P = 200) 1/Liter ρ = ρ (W ater, T = Tave , P = 200) · 0.001 1/m3
(4) (5)
a Energy to heat water only: Loadhw = (U seM,F · 5 [day/week] + U seS,S
·6 MJ · 2 [day/week]) · ρ · Cp · (Thot − Tcold ) · 1 × 10 J
(6)
b Add piping and tank losses: Load = Loadhw + Losstank + Losspiping
(7)
MJ Losstank = U Atank · (Thot − Troom ) · 3600 [s/hr] · 24 [hr/day] · 7 [day/week] · 1 × 10·6 j
(8)
Losspiping
·6 MJ = U Aloop · (Thot − Troom ) · 3600 [s/hr] · 24 [hr/day] · 7 [day/week] · 1 × 10 j
185
(9)
c With night shut-down: If flow loop is shut down for 14 hours per day, assume that piping and water therein cools to 20 C. The energy loss from piping will then be losses for 10 hours per day, plus the heat to bring pipe and water therein back up to 38 C. MJ (10) LosspipingRed = U Aloop · (Thot − Troom ) · 3600 [s/hr] · 10 [hr/day] · 7 [day/week] · 1 × 10·6 J ·6 MJ ReheatEnergy = W aterM ass · Cp · 7 [1/week] · (Thot − Troom ) · 1 × 10 J
(11)
liter W aterM ass = π · D2 · L/4 · 1000 3 · ρ m
(12)
LoadRed = Loadhw + Losstank + LosspipingRed + ReheatEnergy
(13)
Solution Cp = 4183 [J/kg-K] Load = 910.8 [MJ/week] Losspiping = 414.8 [MJ/week] ReheatEnergy = 18.18 [MJ/week] Tcold = 12 [C] U Aloop = 38.1 [W/C] U seS,S = 210 [L/day]
D = 0.019 [m] Loadhw = 376.3 [MJ/week] LosspipingRed = 172.8 [MJ/week] ρ = 0.9971 [kg/liter] Thot = 38 [C] U Atank = 11 [W/C] W aterM ass = 34.49 [kg]
186
L = 122 [m] LoadRed = 687 [MJ/week] Losstank = 119.8 [MJ/week] Tave = 25 [C] Troom = 20 [C] U seM,F = 610 [L/day]
P09 02
Equations 9.2 Degree-day calculation of house heating load. Knowns Thouse = 19;
IntGen = 1.5 · 103 ;
U Ahouse = 335;
Tamb = −3
(1)
Use Equation 9.3.3 for balance temperature: Tbase = Thouse − IntGen/U Ahouse
(2)
Use Equation 9.3.4 for load rate in terms of Tbase : Loadrate = U Ahouse · (Tbase − Tamb )
(3)
Solution IntGen = 1500 [W] Tbase = 14.52 [C]
Loadrate = 5870 [W] Thouse = 19 [C]
Tamb = −3 [C] U Ahouse = 335 [W/C]
187
P09 03
Equations 9.3 Calculation of annual degree-days and loads for building of Problem 9.2. Use the degree-day method of Section 9.3. Note that from Example 9.3.1 σ yr = 10.6. We will use two dimensional arrays for this solution. We could have used three parametric tables. Checking Units shows a large number of problems. This is due to the empirical nature of some of the Equations. Knowns U Ah,1 = 112 [W/C] ; Tset = 19 [C] ;
U Ah,2 = 224 [W/C] ;
σyr = 10.6 [C] ;
U Ah,3 = 335 [W/C]
(1)
q˙ = 1500 [W]
(2)
T¯a,1..12 = [-8, -5, 1, 9, 14, 19, 22, 20, 15, 11, 2, -5]
(3)
duplicate i = 1, 12
(4)
Ni = N umDay (i)
(5)
σm,i = 1.45 [C] − 0.0290 · T¯a,i + 0.0664 · σyr
Eqn 9.3.10
(6)
end
(7)
duplicate k = 1, 3
(8)
Tbase,k = Tset −
q˙ U Ah,k
Eqn 9.3.3
(9)
duplicate i = 1, 12 hi, k =
(10)
Tbase,k − T¯a,i p σm,i · Ni · 1 [1/day]
DegDayAi, k = σm,i · Ni1.5 · ∆T i, k = Tbase,k − T¯a,i
Eqn 9.3.9
(11)
hi, k ln (cosh (1.698 · hi, k )) + + 0.2041 2 3.396
(12)
(13)
DegDayB i, k = Ni · ∆T i, k
(14)
DegDay 0 i, k = If ∆T i, k , 6 [C] , DegDayAi, k , DegDayAi, k , DegDayB i, k DegDay i, k = max DegDay 0 i, k , 0 [C ·day]
(15) (16)
188
MJ Li, k = U Ah,k · DegDay i, k · 24 [hr/day] · 3600 [s/hr] · 1 × 10·6 J
Eqn 9.3.6
(17)
end
(18)
GJ Lk = Sum(Lii, k , ii = 1, 12) · 0.001 MJ
(19)
end
(20)
For the three different values of (UA)h the base temperatures are, respectively, 5.6, 12.3, and 14.5 C and annual Energy requirements are, respectively, 13.6, 50.3, and 89.9 GJ Note: Decreasing the building UA has two effects on building heating loads. It reduces the number of degree-days, and the loss per degree-day.
Solution q˙ = 1500 [W]
σyr = 10.6 [C]
Tset = 19 [C]
Arrays Row 1 2 3 4 5 6 7 8 9 10 11 12
Row 1 2 3 4 5 6 7 8 9 10 11 12
T¯a,i [C] -8 -5 1 9 14 19 22 20 15 11 2 -5
Ni [day] 31 28 31 30 31 30 31 31 30 31 30 31
DegDayAi,2 [C-Day] 630.1 485.3 354.5 125.8 39.11 5.606 1.54 3.951 26.67 87.01 313.8 537.6
σm,i [C] 2.386 2.299 2.125 1.893 1.748 1.603 1.516 1.574 1.719 1.835 2.096 2.299
Tbase,i [C] 5.6 12.3 14.5
DegDayAi,3 [C-Day] 698.6 547.1 421.4 179.6 70.04 12.63 3.709 9.041 50.79 134.4 378.1 605.8
U Ah,i [W/C] 112 224 335
Li [GJ] 13.6 50.3 89.9
DegDayB i,1 [C-Day] 422 297 143 -102 -260 -402 -508 -446 -282 -167 108 329
DegDayAi,1 [C-Day] 426 302 168 26 5 0 0 0 3 14 138 336
DegDayB i,2 [C-Day] 629.4 484.5 350.4 99.11 -52.59 -200.9 -300.6 -238.6 -80.89 40.41 309.1 536.4
189
DegDayB i,3 [C-Day] 698.2 546.6 419.2 165.7 16.19 -134.3 -231.8 -169.8 -14.33 109.2 375.7 605.2
DegDay 0 [i, 1] [C-Day] 422 297 168 26 5 0 0 0 3 14 138 329
DegDay 0 [i, 2] [C-Day] 629.4 484.5 350.4 125.8 39.11 5.606 1.54 3.951 26.67 87.01 309.1 536.4
Row 1 2 3 4 5 6 7 8 9 10 11 12
Row 1 2 3 4 5 6 7 8 9 10 11 12
DegDay 0 [i, 3] [C-Day] 698.2 546.6 419.2 179.6 70.04 12.63 3.709 9.041 50.79 134.4 375.7 605.2
Li,1 [MJ] 4082 2874 1629 252.1 44.75 4.191 0.9999 2.902 26.71 139 1337 3182
Li,2 [MJ] 12181 9377 6782 2435 757 108.5 29.8 76.46 516.1 1684 5982 10381
DegDay i,1 [C-day] 421.8 297 168.3 26.05 4.624 0.4331 0.1033 0.2999 2.76 14.36 138.2 328.8
Li,3 [MJ] 20209 15822 12133 5197 2027 365.6 107.4 261.7 1470 3889 10873 17517
DegDay i,2 [C-day] 629.4 484.5 350.4 125.8 39.11 5.606 1.54 3.951 26.67 87.01 309.1 536.4
hi,1 [-] 1.024 0.872 0.3894 -0.3273 -0.8624 -1.526 -1.942 -1.643 -0.9977 -0.5279 0.3142 0.8287
hi,2 [-] 1.528 1.422 0.9555 0.3186 -0.1743 -0.7628 -1.149 -0.8783 -0.2864 0.1276 0.8976 1.352
DegDay i,3 [C-day] 698.2 546.6 419.2 179.6 70.04 12.63 3.709 9.041 50.79 134.4 375.7 605.2
hi,3 [-] 1.695 1.605 1.143 0.5327 0.05368 -0.51 -0.886 -0.6251 -0.05073 0.3448 1.091 1.525
190
∆T i,1 [C] 13.61 10.61 4.607 -3.393 -8.393 -13.39 -16.39 -14.39 -9.393 -5.393 3.607 10.61
∆T i,2 [C] 20.3 17.3 11.3 3.304 -1.696 -6.696 -9.696 -7.696 -2.696 1.304 10.3 17.3
∆T i,3 [C] 22.52 19.52 13.52 5.522 0.5224 -4.478 -7.478 -5.478 -0.4776 3.522 12.52 19.52
P10 01
Equations 10.1 Energy balance on a system for 1 hour. Knowns Ts = 43 [C] ; Ac = 5 m2 ;
Ls = 5.20;
IT = 18.0/5;
M ass = 400 [kg] ;
U AT ank = 0.81 [W/C] ;
F RT A = 0.76;
Cp = 4190 [J/kg ·C] ;
Ta = 12 [C] ;
F RU L = 4.80 W/m2 ·C ∆T ime = 1 [hr]
(2)
T 0 a = 19 [C]
(3)
Equation 10.1.3 in finite difference form, for fully mixed Tank, in MJ: J M ass · Cp · (Tplus,s − Ts ) = (Qu − Ls − LossT ank ) · 1000000 MJ MJ/hr Qu = Ac · F RT A · IT − F RU L · (Ts − Ta ) · 1 [hr] · 0.0036 J/s
LossT ank
(4)
(5)
MJ/hr = U AT ank · (Ts − T a ) · ∆T ime · 0.0036 J/s 0
(6)
Solution 2 Ac = 5 m IT = 3.6 MJ/m2 Ta = 12 [C]
Cp = 4190 [J/kg-C] LossT ank = 0.07 [MJ] Tplus,s = 46.42 [C]
∆T ime = 1 [hr] Ls = 5.2 [MJ] Ts = 43 [C]
191
(1)
F RT A = 0.76 M ass = 400 [kg] T 0 a = 19 [C]
F RU L = 4.8 W/m2 -C Qu = 11 [MJ] U AT ank = 0.81 [W/C]
P10 02
Equations 10.2
Energy balance on fully mixed Tank over 24 hours
$Tabstops 0.5 1 1.5 2 2.5 Knowns Ac = 30 m2 ; ∆T ime = 1 [hr] ;
FRT auAlpha = 0.78; T 0 a = 20 [C] ;
FRU,L = 7.62 W/m2 ·C ; U AT ank = 11.1 [W/C] ;
M ass = 1500 [kg]
hr = 1 [hr] ;
Cp = 4190 [J/kg ·c]
(1) (2)
Use finite difference form of Equation 10.1.3, in MJ in the parametric table. Note that the Solve Table must start at run 2.: J (Qu − Ls − LossT ank ) · 1000000 MJ (Ts+ − Ts ) M ass · Cp · = (3) ∆T ime hr MJ Qu = max 0, Ac · FRT auAlpha · IT − FRU,L · hr · (Ts − Ta ) · 0.0036 (4) W · hr MJ 0 LossT ank = U AT ank · hr · (Ts − T a ) · 0.0036 (5) J · h/s Ts = Tablevalue(tablerun# − 1, ‘Tplus s’ )
(6)
$DoLast
192
Qu,day = Sumparametric(‘Table 1’ , ‘Q u’ , 2, 25)
(7)
Qloss,day = Sumparametric(‘Table 1’ , ‘Loss tank’ , 2, 25)
(8)
Loadday = Sumparametric(‘Table 1’ , ‘L s’ , 2, 25)
(9)
Tinitial = Tablevalue(1, ‘Tplus s’ )
(10)
Tf inal = Tablevalue(25, ‘Tplus s’ )
(11)
J M ass · Cp · (Tf inal − Tinitial ) = (Qu,day − Qloss,day − Loadday ) · 1000000 + EBE MJ
(12)
$EndDoLast
Table 1 Run 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
T IM E [HR]
Ls [MJ]
IT 2 MJ/m
Ta [C]
Ts [C]
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
12 12 11 11 13 14 18 21 20 20 18 16 14 14 13 18 22 24 18 20 15 11 10 9
0 0 0 0 0 0 0 0 1.09 1.75 2.69 3.78 3.87 3.41 2.77 1.82 1.53 0 0 0 0 0 0 0
-4 -4 -5 -3 -1 0 0 1 0 2 4 10 10 8 8 6 4 4 3 5 6 6 7 6
45.0 42.9 40.9 39.0 37.1 34.9 32.6 29.7 26.3 23.7 24.1 28.6 37.7 46.1 51.4 53.8 51.2 47.5 43.5 40.5 37.2 34.7 32.9 31.2
Tsplus [C] 45.0 42.9 40.9 39.0 37.1 34.9 32.6 29.7 26.3 23.7 24.1 28.6 37.7 46.1 51.4 53.8 51.2 47.5 43.5 40.5 37.2 34.7 32.9 31.2 29.7
193
Qu [MJ]
LossT ank [MJ]
0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 3.9 23.1 46.4 73.1 67.8 48.4 29.1 3.3 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0
1.0 0.9 0.8 0.8 0.7 0.6 0.5 0.4 0.3 0.1 0.2 0.3 0.7 1.0 1.3 1.3 1.2 1.1 0.9 0.8 0.7 0.6 0.5 0.4
P10 03
194
P10 04
Equations 10.4 Calculate collector heat exchanger factor. Knowns = 0.60;
F RU L = 3.75 W/m2 ·C ;
Ac = 50 m2 ;
m ˙ coll = 0.7 [kg/s] ;
m ˙ tank = 0.7 [kg/s]
(1)
Cp,coll = 3350 [J/kg ·C] ;
Cp,tank = 4190 [J/kg ·C]
(2)
C˙ tank = m ˙ tank · Cp,tank ;
C˙ coll = m ˙ coll · Cp,coll
(3)
−1 ˙ Ccoll − 1 · · C˙ min
F 0 R/F R = 1 + Ac · F RU L/C˙ coll
Eqn 10.2.3:
C˙ min = min C˙ tank , C˙ coll
(5)
Solution Ac = 50 m2 Cp,coll = 3350 [J/kg-C] F 0 R/F R = 0.9494
C˙ coll = 2345 [J/s-C] Cp,tank = 4190 [J/kg-C] m ˙ coll = 0.7 [kg/s]
(4)
C˙ min = 2345 [J/s-C] = 0.6 m ˙ tank = 0.7 [kg/s]
195
C˙ tank = 2933 [J/s-C] F RU L = 3.75 W/m2 -C
P10 05
Equations 10.5 Pipe loss factors modifying Qu equation. Knowns C˙ coll = 295 [W/C] ; F Rta = 0.72;
Do = Di + 2 · Insul;
F RU L = 4.65 W/m2 ·C ;
Di = 0.019 [m] ;
L = 20 [m] ;
hconv = 25 W/m2 ·C ;
insul = 0.018 [m]
Ac = 5.2 m2
(1) (2)
kinsul = 0.05 [W/m ·C]
(3)
Ao = π · Do · L;
(4)
ta0 /ta =
1 1 + Upipe · Ao /C˙ coll
U L0 /U L =
Upipe =
Ai = π · Di · L Eqn 10.3.9
(5)
i +Ao 1 − Upipe · Ai /C˙ coll + Upipe · AA c ·F RU L 1 + Upipe · Ao /C˙ coll
Eqn 10.3.10
(6)
1 Do ·
ln (Do /Di ) (2·kinsul )
(7)
+ 1/hconv
Equation 6.7.6 (or 10.1.1) is then modified as follows: Qu = Ac *(FRta*ta’ \ta *GT - FRUL*UL’ \UL*(Ti - Ta ))
Solution Ac = 5.2 m2 Di = 0.019 [m] hconv = 25 W/m2 -C ta0 /ta = 0.9816
Ai = 1.194 m2 Do = 0.055 [m] Insul = 0.018 [m] U L0 /U L = 1.277
Ao = 3.456 m2 F Rta = 0.72 kinsul = 0.05 [W/m-C] Upipe = 1.601 W/m2 -C
196
C˙ coll = 295 [W/C] F RU L = 4.65 W/m2 -C L = 20 [m]
P10 06
Equations 10.6 Air heater simulation with fully mixed liquid storage tank Knowns: TsupplyAir = 35 [C] ; Ac = 80 m2 ;
Tdelivery = 55 [C] ;
F 0 Rta = 0.78;
m ˙ water = 0.056 [kg/s] ;
∆T ime = 1 [hr] ;
F 0 RU L = 4.45 W/m2 ·C ;
m ˙ air = 0.056 [kg/s] ;
M ass = 5000 [kg] ;
Tinitial = 47 [C]
U AT ank = 140 [kJ/hr ·C] ;
Cp,air = 1013 [J/kg ·C] ;
(1)
hr = 1 [hr] (2)
Cp,water = 4190 [J/kg ·C]
(3)
For a perfect heat exchanger the fluid with the minimum capacity will go through the maximum temperature rise. C˙ h = m ˙ water · Cp,water
(4)
C˙ c = m ˙ air · Cp,air C˙ min = min C˙ h , C˙ c
(5)
Qmax
(6)
MJ/hr ˙ = Cmin · (Ts − TsupplyAir ) · hr · 0.0036 W
197
(7)
The required energy to heat the air from 35 to 55 C is: MJ/hr Qreqd = C˙ min · (Tdelivery − TsupplyAir ) · hr · 0.0036 J/s
(8)
The energy removed from storage is then: Ls = min (Qreqd , Qmax )
(9)
The auxiliary energy is then: Aux = max (0, Qreqd − Qmax )
(10)
Note that this control would be implemented by reducing the water mass flow rate. The tank energy balance (Equation 10.9.2) is: (Tplus,s − Ts ) MJ M ass · Cp,water · = (Qu − Ls − LossTank ) · hr · 1 × 10·6 ∆T ime J Ts = Tablevalue(‘Table 1’ , tablerun# − 1, ‘T plus s’ ) MJ/hr Qu = max 0, Ac · F 0 Rta · IT − F 0 RU L · hr · 0.0036 · (T − T ) s a W MJ LossT ank = U AT ank · hr · 0.001 · (Ts − Ta ) kJ
(11) (12) (13) (14)
Energy removed from store to supply load Estore = Sumparametric(‘Table 1’ , ‘L s’ , 2, T ableRun#)
(15)
AuxiliaryEnergy required to meet the load EAux = Sumparametric(‘Table 1’ , ‘Aux’ , 2, T ableRun#)
(16)
Energy supplied to store from solar Esolar = Sumparametric(‘Table 1’ , ‘Q u’ , 2, T ableRun#)
(17)
Energy lost from store to surroundings Eloss = Sumparametric(‘Table 1’ , ‘Loss tank’ , 2, T ableRun#)
(18)
Energy gain (or loss) of tank over 12 hours. MJ Estored = M ass · Cp,water · (Tplus,s − Tinitial ) · 1 × 10·6 J
(19)
Store energy balance to determine Energy Balance Error: EBE = Esolar − Eloss − Estored − Estore
(20)
Energy balance on air stream. Check = Qreqd · 12 − Estore − Eaux
(21)
Solution
198
Aux = 0.00 [MJ] C˙ h = 234.6 [J/s-C] ∆T ime = 1 [hr] Esolar = 821.1 [MJ] F 0 RU L = 4.45 W/m2 -C Ls = 4.08 [MJ] Qmax = 9.353 [MJ] Tdelivery = 55 [C] TsupplyAir = 35 [C]
Ac = 80 m2 C˙ min = 56.73 [J/s-C] EBE = 7.326 × 10−7 [MJ] Estore = 40.97 [MJ] hr = 1 [hr] M ass = 5000 [kg] Qreqd = 4.084 [MJ] Tinitial = 47 [C] U AT ank = 140 [kJ/hr-C]
Check = 1.776 × 10−15 [MJ] Cp,air = 1013 [J/kg-C] EAux = 8.044 [MJ] Estored = 694.2 [MJ] IT = 0.00 MJ/m2 m ˙ air = 0.056 [kg/s] Qu = 0 [MJ] Tplus,s = 80.1 [C]
Table 1 Run 1 2 3 4 5 6 7 8 9 10 11 12 13 Sum
Ta [C] 7 8 9 9 9 12 15 14 12 12 12 11 130
IT MJ/m2
Ts [C]
Qu [MJ]
Ls [MJ]
Aux [MJ]
LossT ank [MJ]
0.00 0.00 1.09 1.75 2.69 3.78 3.87 3.41 2.77 1.82 1.53 0.00 22.71
47 46.62 46.24 46.85 49.38 54.51 62.7 70.79 76.9 80.55 81.13 80.8 743.5
0 0 20.28 60.69 116.1 181.4 180.4 140 89.67 25.71 6.879 0 821.1
2.45 2.37 2.30 2.42 2.94 3.99 4.08 4.08 4.08 4.08 4.08 4.08 40.97
1.63 1.71 1.79 1.66 1.15 0.10 0.00 0.00 0.00 0.00 0.00 0.00 8.04
5.60 5.41 5.21 5.30 5.65 5.95 6.68 7.95 9.09 9.60 9.68 9.77 85.89
199
Tplus,s [C] 47.0 46.6 46.2 46.9 49.4 54.5 62.7 70.8 76.9 80.6 81.1 80.8 80.1 823.6
C˙ c = 56.73 [J/s-C] Cp,water = 4190 [J/kg-C] Eloss = 85.89 [MJ] F 0 Rta = 0.78 LossT ank = 9.77 [MJ] m ˙ water = 0.056 [kg/s] Ta = 11 [C] Ts = 80.8 [C]
P10 07
Equations 10.7 Air heating simulation of Problem 10.6 with mixed liquid tank and a real load HX. Knowns: TsupplyAir = 35 [C] ; Ac = 80 m2 ;
Tdelivery = 55 [C] ;
F 0 Rta = 0.78;
m ˙ water = 0.056 [kg/s] ;
∆T ime = 1 [hr] ;
F 0 RU L = 4.45 W/m2 ·C ;
m ˙ air = 0.056 [kg/s] ;
M ass = 5000 [kg] ;
Tinitial = 47 [C]
U AT ank = 140 [kJ/hr ·C] ;
Cp,air = 1013 [J/kg ·C] ;
(1)
hr = 1 [hr] (2)
Cp,water = 4190 [J/kg ·C]
= 0.58
(3) (4)
For a perfect heat exchanger the fluid with the minimum capacity will go through the maximum temperature rise. C˙ h = m ˙ water · Cp,water
(5)
C˙ c = m ˙ air · Cp,air C˙ min = min C˙ h , C˙ c
(6) (7)
Calculating the new maximum heat transfer in the HX is the only change from Problem 10.6 MJ/hr ˙ Qmax = · Cmin · (Ts − TsupplyAir ) · hr · 0.0036 W The required energy to heat the air from 35 to 55 C is: MJ/hr ˙ Qreqd = Cmin · (Tdelivery − TsupplyAir ) · hr · 0.0036 J/s
(8)
(9)
The energy removed from storage is then: Ls = min (Qreqd , Qmax )
(10)
The auxiliary energy is then: Aux = max (0, Qreqd − Qmax )
(11)
Note that this control would be implemented by reducing the water mass flow rate. The tank energy balance (Equation 10.9.2) is: (Tplus,s − Ts ) ·6 MJ M ass · Cp,water · · hr · 1 × 10 = (Qu − Ls − LossTank ) ∆T ime J
200
(12)
Ts = Tablevalue(‘Table 1’ , tablerun# − 1, ‘T plus s’ ) MJ/hr · (T − T ) Qu = max 0, Ac · F 0 Rta · IT − F 0 RU L · hr · 0.0036 s a W MJ LossT ank = U AT ank · hr · 0.001 · (Ts − Ta ) kJ
(13) (14)
(15)
Energy removed from store to supply load Estore = Sumparametric(‘Table 1’ , ‘L s’ , 2, T ableRun#)
(16)
AuxiliaryEnergy required to meet the load EAux = Sumparametric(‘Table 1’ , ‘Aux’ , 2, T ableRun#)
(17)
Energy supplied to store from solar Esolar = Sumparametric(‘Table 1’ , ‘Q u’ , 2, T ableRun#)
(18)
Energy lost from store to surroundings Eloss = Sumparametric(‘Table 1’ , ‘Loss tank’ , 2, T ableRun#)
(19)
Energy gain (or loss) of tank over 12 hours. Estored
·6 MJ = M ass · Cp,water · (Tplus,s − Tinitial ) · 1 × 10 J
(20)
Store energy balance to determine Energy Balance Error: EBE = Esolar − Eloss − Estored − Estore
(21)
Energy balance on air stream. Check = Qreqd · 12 − Estore − Eaux
(22)
Reducing the heat exchanger effectiveness from 1 to 0.58 increased the Aux from 8.04 to 15.65 MJ and reduced the energy supplied from the store from 40.97 to 33.36 MJ. These are significant chages but general conclusions about the impact over a full year should not be drawn from this one-day simulation. In an annual simulation the expected percentage increase in auxiliary energy is expected to be much smaller.
Solution Aux = 0.00 [MJ] C˙ h = 234.6 [W/C] ∆T ime = 1 [hr] Eloss = 86.21 [MJ] F 0 Rta = 0.78 LossT ank = 9.80 [MJ] m ˙ water = 0.056 [kg/s] Ta = 11 [C] Ts = 81.03 [C]
Ac = 80 m2 C˙ min = 56.73 [W/C] EBE = 7.334 × 10−7 [MJ] Esolar = 818.5 [MJ] F 0 RU L = 4.45 W/m2 -C Ls = 4.08 [MJ] Qmax = 5.452 [MJ] Tdelivery = 55 [C] TsupplyAir = 35 [C]
Check = −3.553 × 10−15 [MJ] Cp,air = 1013 [J/kg-C] = 0.58 Estore = 33.36 [MJ] hr = 1 [hr] M ass = 5000 [kg] Qreqd = 4.084 [MJ] Tinitial = 47 [C] U AT ank = 140 [kJ/hr-C]
201
C˙ c = 56.73 [W/C] Cp,water = 4190 [J/kg-C] EAux = 15.65 [MJ] Estored = 698.9 [MJ] IT = 0 MJ/m2 m ˙ air = 0.056 [kg/s] Qu = 0.00 [MJ] Tplus,s = 80.36 [C]
Table 1 Run 1 2 3 4 5 6 7 8 9 10 11 12 13 Sum
Ta [C] 7 8 9 9 9 12 15 14 12 12 12 11 130
IT MJ/m2
Ts [C]
Qu [MJ]
Ls [MJ]
Aux [MJ]
LossT ank [MJ]
0 0 1.09 1.75 2.69 3.78 3.87 3.41 2.77 1.82 1.53 0 22.71
47.00 46.66 46.34 46.99 49.56 54.73 62.98 71.09 77.18 80.81 81.37 81.03 745.75
0.00 0.00 20.16 60.51 115.88 181.10 180.00 139.61 89.31 25.38 6.57 0.00 818.52
1.42 1.38 1.34 1.42 1.72 2.34 3.31 4.08 4.08 4.08 4.08 4.08 33.36
2.66 2.70 2.74 2.66 2.36 1.75 0.77 0.00 0.00 0.00 0.00 0.00 15.65
5.60 5.41 5.23 5.32 5.68 5.98 6.72 7.99 9.13 9.63 9.71 9.80 86.21
202
Tplus,s [C] 47.00 46.66 46.34 46.99 49.56 54.73 62.98 71.09 77.18 80.81 81.37 81.03 80.36 826.11
P10 08
Equations 10.8 Simulation of water heater with variable load. Knowns ∆T ime = 1 [hr] ; F Rta = 0.73;
M ass = 1000 [kg] ;
F RU L = 5 W/m2 ·C ;
C˙ tank = 1120 [W/C] ;
Tsupply = 11 [C] ;
Ac = 20 m2 ;
Cp = 4190 [J/kg ·C] ; U AT ank = 6.0 [W/C] ; Tinitial = 35 [C] ;
C˙ coll = 1020 [W/C]
Troom = 20 [C] ;
Use F’R \FR to account for the collector-store heat exchanger, Equation 10.2.3. C˙ min = min C˙ T ank , C˙ coll
203
= 0.6
hr = 1 [hr]
(1) (2) (3)
(4)
−1 ˙ Ccoll − 1 · · C˙ min
F 0 R/F R = 1 + Ac · F RU L/C˙ coll
(5)
F 0 Rta = F Rta · F 0 R/F R
(6)
F 0 RU L = F RU L · F 0 R/F R
(7)
Use a form of Equation 10.9.2, in MJ: + Ts − Ts MJ = (Qu − Ls − LossTank ) M ass · Cp · · hr · 1 × 10·6 ∆T ime J MJ/hr 0 0 Qu = max 0 [MJ] , Ac · F Rta · IT − F RU L · 0.0036 · (T − T ) · hr s a W MJ · (Ts − Tsupply ) · hr Ls = m ˙ load · Cp · 1 × 10·6 J MJ/hr LossT ank = U AT ank · 0.0036 · (Ts − Troom ) · hr W
(8)
(9)
(10)
(11)
Ts = Tablevalue(tablerun# − 1, ‘Tplus s’ )
(12)
Find total quantities. Eu = Sumparametric(‘Table 1’ , ‘Q u’ , 2, tableRun#)
(13)
Eloss = Sumparametric(‘Table 1’ , ‘Loss tank’ , 2, tableRun#)
(14)
Eload = Sumparametric(‘Table 1’ , ‘L s’ , 2, tableRun#)
(15)
Tf inal = Ts+
(16)
·6 MJ ∆U = M ass · Cp · (Tf inal − Tinitial ) · 1 × 10 J
(17)
Tank Energy Balance Error EBE = Eu − Eloss − Eload − ∆U
(18)
Solution Ac = 20 m2 Cp = 4190 [J/kg-C] = 0.6 F Rta = 0.73 F 0 R/F R = 0.9387 [-] Ls = 19.99 [MJ] Ta = 20 [C] Ts = 42.81 [C]
C˙ coll = 1020 [W/C] ∆T ime = 1 [hr] Eload = 47.85 [MJ] F RU L = 5 W/m2 -C hr = 1 [hr] M ass = 1000 [kg] Tf inal = 45.08 [C] Tsupply = 11 [C]
C˙ min = 1020 [W/C] ∆U = 42.22 [MJ] Eloss = 1.337 [MJ] F 0 Rta = 0.6852 IT = 2.75 MJ/m2 m ˙ load = 150 [kg/hr] Tinitial = 35 [C] Ts+ = 45.08 [C]
Table 1
204
C˙ tank = 1120 [W/C] EBE = 8.462 × 10−8 [MJ] Eu = 91.41 [MJ] F 0 RU L = 4.693 W/m2 -C LossT ank = 0.49 [MJ] Qu = 29.98 [MJ] Troom = 20 [C] U AT ank = 6 [W/C]
Run 1 2 3 4 5
Ta [C]
IT 2 MJ/m
m ˙ load [kg/hr]
Qu [MJ]
LossT ank [MJ]
Ls [MJ]
Ts [C]
14 17 18 20
0.09 1.75 3.45 2.75
150 150 0 150
0.00 19.14 42.28 29.98
0.32 0.24 0.28 0.49
15.08 12.77 0.00 19.99
35.00 31.32 32.78 42.81
205
Tsplus [C] 35.00 31.32 32.78 42.81 45.08
P10 09
Equations 10.9 Effects of changes in the conditions of Problem 10.8 Each change can be tested using the program of Problem 10.8 but here we wanted answers based on knowledge of solar system performance.
a) Tplus s will be higher, as FR0 will be higher and Qu will be higher. b) Tplus s will be lower, as UL will be higher resulting in higher collector losses and lower Qu. c) Tplus s will be lower, as FR and FR0 will be lower, resulting in lower Qu. d) Tplus s will be higher, as the temperature of operation of the whole system would be higher. The increase would be less than 3 C, since there would be increased thermal losses from the collector and tank.
206
P10 10
Equations 10.10 Develop Qu equation for collector subsystem. Work from the collector back to the tank, making a series of corrections on FRta and FRUL. The bo correction can be made any time; it is left to the end. $UnitSystem C J kPa Knowns F avtan = 0.75;
F avU L = 0.75/0.165;
v˙ tank = 0.050 [l/s] ; Ttest = 30 [C] ;
Nseries = 2;
v˙ test = 0.0246 [l/s] ;
Dpipe = 0.012 [m] ;
Pf luid = 200 [kPa] ;
Nparallel = 3;
thick insulation = 0.009 [m] ;
Conc = 50;
v˙ coll = 0.040
Ac = 1.64 m2 ;
[l/s] Nparallel
Ta = 20 [C] ;
= 0.45
Lpipe = 12 [m] ; Ncoll = 6
(1)
F I$ = ‘EG’ (2)
assumed
(3)
so the flow through each collector is:
(4)
kinsulation = 0.043 [W/m ·C]
(5)
Fluid properties - watch units in BRINEPROP. call BRIN EP ROP 2 F I$, Conc, Ttest
Cp,EG , k , µ , P rEG : F reezingP tEG , ρEG , 1000 J EG EG kJ
ρ = ρ (water, T = Ttest , P = pf luid )
! (6) (7)
207
µ = µ (water, T = Ttest , P = Pf luid )
(8)
Cp = cp (water, T = Ttest , P = Pf luid )
(9)
1. Convert from basis of Fav to FR , Section 6.19. m3 m ˙ test = v˙ test · ρ · 0.001 l
(10)
−1 Ac · F avU L F Rtan,1 = F avtan · 1 + 2·m ˙ test · Cp
Ac · F avU L F RU L1 = F avU L · 1 + 2·m ˙ test · Cp
(11)
−1 (12)
2. Apply flow rate correction, Equation 6.20.1: m3 m ˙ use = v˙ coll · ρEG · 0.001 l 0 m ˙ use ·Cp,EG Ac ·F U L · 1 − exp − Ac m ˙ use ·Cp,EG r= Eqn 6.20.2 F RU L1 m ˙ test · Cp Ac 0 F UL = − · ln 1 − F RU L1 · Equation 6.20.4 Ac m ˙ test · Cp F RU L2 = F RU L1 · r;
F Rtan,2 = F Rtan,1 · r
(13)
(14) (15) (16)
3. Two arrays in series: K=
Ac · F RU L2 m ˙ use · Cp,EG
Eqn 10.5.4
(17)
F Rtan,3 = F Rtan,2 · (1 − K/2)
Eqn 10.5.8
(18)
F RU L3 = F RU L2 · (1 − K/2)
Eqn 10.5.9
(19)
4. Pipe heat loss - Section 10.3 Dinsul = Dpipe + 2 · thick insulation U Ao = 2 · π · kinsulation ·
(20)
Lpipe ln (Dinsul /Dpipe )
(21)
U Ao = U Ai
(22)
The pipe heat loss is for all 6 collectors. So the total flow rate and total collector area are used here. −1 U Ai F Rtan,4 = F Rtan,3 · 1 + Eqn 10.3.9 m ˙ use · Nparallel · Cp,EG F RU L4 = F RU L3 ·
1−
U Ai U Ai +U Ao (m ˙ use ·Nparallel ·Cp,EG ) + Ncoll ·Ac ·F RU L3 U Ai 1 + (m˙ use ·Nparallel ·Cp,EG )
208
Eqn 10.3.10
(23)
(24)
5. Collector-heat exchanger correction C˙ coll = Nparallel · m ˙ use · Cp,EG m3 C˙ tank = v˙ tank · ρ · Cp · 0.001 l C˙ min = min C˙ coll , C˙ tank
(25) (26) (27)
−1 ˙ Ncoll · Ac · F RU L4 Ccoll − 1 F R0 /F R = 1 + · C˙ coll · C˙ min
Eqn 10.2.3
(28)
F Rtan,5 = F Rtan,4 · F R0 /F R
(29)
F RU L5 = F RU L4 · F R0 /F R
(30)
6. Incidence angle modifier and Qu equation Kta = 1 - 0.15*(1/cos(θ)-1) Eqn 6.17.10 a = Ncoll · Ac · F Rtan,5
(31)
MJ/hr b = Ncoll · Ac · F RU L5 · 0.0036 W
(32)
The final equation for Qu in MJ/hr (assuming IT is in MJ/m2 -hr) for the complete assembly is: Qu = 4.498*IT *(1 - 0.15(1/cos(θ)-1)) - 0.1138*(Ti - Ta )
Solution a = 4.498 m2 C˙ coll = 142 [J/s-C] Cp,EG = 3349 [J/kg-K] F avtan = 0.75 [-] F Rtan,1 = 0.7237 [-] F Rtan,5 = 0.4571 F RU L4 = 4.413 W/m2 -C K = 0.1458 µ = 0.0007977 [kg/m-s] Ncoll = 6 [-] Pf luid = 200 [kPa] thick insulation = 0.009 [m] U Ao = 3.538 [W/C]
Ac = 1.64 m2 C˙ min = 142 [J/s-C] Dinsul = 0.03 [m] F avU L = 4.545 W/m2 -C F Rtan,2 = 0.6942 F RU L1 = 4.386 W/m2 -C F RU L5 = 3.212 W/m2 -C kEG = 0.3953 [W/m-K] µEG = 0.002831 [Pa-s] Nparallel = 3 [-] r = 0.9593 Ta = 20 [C] v˙ coll = 0.01333 [l/s]
209
b = 0.1138 [MJ/C-hr] C˙ tank = 208.2 [J/s-C] Dpipe = 0.012 [m] F I$ = ‘EG’ F Rtan,3 = 0.6436 [-] F RU L2 = 4.207 W/m2 -C F R0 /F R = 0.728 kinsulation = 0.043 [W/m-C] m ˙ test = 0.02449 [kg/s] Nseries = 2 [-] ρ = 995.7 kg/m3 Ttest = 30 [C] v˙ tank = 0.05 [l/s]
Conc = 50 Cp = 4183 [J/kg-K] = 0.45 F reezingP tEG = −36.03 [C] F Rtan,4 = 0.628 [-] F RU L3 = 3.901 W/m2 -C F 0 U L = 4.547 W/m2 -C Lpipe = 12 [m] m ˙ use = 0.01414 [kg/s] P rEG = 23.98 ρEG = 1060 kg/m3 U Ai = 3.538 [W/C] v˙ test = 0.0246 [l/s]
P10 11
Equations 10.11 Performance of partially shaded collectors Assume that the solar azimuth angle is 0, i.e., that the sun is directly south of the arrays; then the angle of incidence of beam radiation is 5 deg on the unshaded collector. Assume that the combination of diffuse and reflected radiation is isotropic and at the same level as the diffuse on the unshaded collector; the mean angle of incidence is then about 59 deg. Refer to the diagram on p. 39 for dimensions. Use as a basis of calculation a unit width of row, i.e., 2.10 m2 area. Knowns Ac1 = 0.88 m2 ;
Ac2 = 1.22 m2 ;
m ˙ = 0.011 kg/s ·m2 · (Ac1 + Ac2 ) ; IT 1 = 0.88 MJ/m2 ·hr ;
F Rtan = 0.83;
F RU L = 4.07 W/m2 ·C ;
Cp = 4190 [J/kg ·C] ;
IT 2 = 0.88 + 2.95;
Ti = 40 [C] ;
θ1 = 59 [deg] ;
bo = −0.14
θ2 = 5 [deg]
Ta = 14 [C]
(1) (2) (3)
Eqn 10.6.4 MJ/hr Qu = Ac1 · F Rta1 · (1 − K) · IT 1 + Ac2 · F Rta2 · IT 2 − F RU L · (Ac1 · (1 − K) + Ac2 ) · (Ti − Ta ) · 0.0036 (4) W K = Ac2 ·
F RU L m ˙ · Cp
Eqn 10.6.5
(5)
F Rta1 = F Rtan · 1 + bo ·
1 −1 Cos(θ1 ) 1 F Rta2 = F Rtan · 1 + bo · −1 Cos(θ2 )
(6)
(7)
Solution Ac1 = 0.88 m2 F Rta1 = 0.7206 IT 1 = 0.88 MJ/m2 -hr Qu = 3.623 [MJ/hr] Ti = 40 [C]
Ac2 = 1.22 m2 F Rta2 = 0.8296 IT 2 = 3.83 MJ/m2 -hr θ1 = 59 [deg]
bo = −0.14 F Rtan = 0.83 K = 0.0513 θ2 = 5 [deg]
210
Cp = 4190 [J/kg-C] F RU L = 4.07 W/m2 -C m ˙ = 0.0231 [kg/s] Ta = 14 [C]
P10 12
Equations 10.12 Output of collectors with two parts at different orientations Knows Lat = 40 [deg] ;
month = 3;
Slope1 = 20 [deg] ; Ac1 = 50 m2 ;
day = 16;
Slope2 = 70 [deg] ; Ac2 = 50 m2 ;
Cp = 4190 [J/kg ·C] ;
ω = 52.5 [deg] ; Ti = 40 [C] ;
F Rtan = 0.83;
m ˙ = 1.10 [kg/s] ;
γ = 0 [deg]
(1)
Ta = 14 [C]
F RU L = 4.07 W/m2 ·C ;
GrRef = 0.2;
(2) bo = −0.14
G = 580 W/m2
(3) (4)
N = nDay (month, day)
(5)
dec = Dec (N )
(6)
CosZen = Cos(Lat) · Cos(Dec) · Cos(ω) + sin (Lat) · sin (Dec)
Eqn 1.6.5
(7)
CosT heta1 = sin (Dec) · sin (Lat) · Cos(Slope1) − sin (Dec) · Cos(Lat) · sin (Slope1) · Cos(γ) + Cos(Dec) · Cos(Lat) · Cos(Slope1) · Cos(ω) + Cos(Dec) · sin (Lat) · sin (Slope1) · Cos(γ) · Cos(ω) + Cos(Dec) · sin (Slope1) · sin (γ) · sin (ω)
Eqn 1.6.2
(8)
CosT heta2 = sin (Dec) · sin (Lat) · Cos(Slope2) − sin (Dec) · Cos(Lat) · sin (Slope2) · Cos(γ) + Cos(Dec) · Cos(Lat) · Cos(Slope2) · Cos(ω) + Cos(Dec) · sin (Lat) · sin (Slope2) · Cos(γ) · Cos(ω) + Cos(Dec) · sin (Slope2) · sin (γ) · sin (ω) Rb1 = Costheta1/CosZen; Rb2 = CosT heta2/CosZen;
Eqn 1.6.2
T heta1 = arccos (CosT heta1) T heta2 = arccos (CosT heta2)
(9) (10) (11)
It could be assumed that most radiation is beam - it is a very sunny day with kT =0.958 and Go = 604 W/m2, so the diffuse fraction will be 0.165 by Equation 2.10.1. Then GT 1 = G*Rb1 and GT 2 = G*Rb2. This leads to Qu = 42.2 kW. Using the isotropic sky model to calculate radiation on the collectors: Go = Gsc# · (1 + .033 · Cos(360 [deg] · n/365)) · CosZen
211
(12)
Apply the concept of kT to irradiance rather than irradiation: kT = G/Go
(13)
Gd = 580 · 0.165;
Gb = 580 · 0.835
(14)
GT 1 = Gb · Rb1 + Gd ·
1 + Cos(Slope1) 1 − Cos(Slope1) + G · GrRef · 2 2
(15)
GT 2 = Gb · Rb2 + Gd ·
1 + Cos(Slope2) 1 − Cos(Slope2) + G · GrRef · 2 2
(16)
K = Ac2 ·
F RU L m ˙ · Cp
Eqn 10.6.5
F Rta1 = F Rtan · 1 + bo · F Rta2 = F Rtan · 1 + bo ·
(17)
1 −1 Cos(T heta1)
1 −1 Cos(T heta2)
(18)
(19)
Qu = Ac1 · F Rta1 · (1 − K) · GT 1 + Ac2 · F Rta2 · GT 2 − F RU L · (Ac1 · (1 − K) + Ac2 ) · (Ti − Ta )
Solution Ac1 = 50 m2 CosT heta2 = 0.5478 dec = −2.418 [deg] 2 F RU L = 4.07 -C W/m Gb = 484.3 W/m2 GT 2 = 707 W/m2 month = 3 Qu = 40673 [W] Slope2 = 70 [deg] Ti = 40 [C]
Ac2 = 50 m2 CosZen = 0.4388 F Rta1 =0.7376 2 G = 580 W/m Gd = 95.7 W/m2 K = 0.04415 m ˙ = 1.1 [kg/s] Rb1 = 1.27 T heta1 = 56.14 [deg]
bo = −0.14 Cp = 4190 [J/kg-C] F Rta2 = 0.7341 γ = 0 [deg] Go = 605.3 W/m2 kT = 0.9582 [-] N = 75 Rb2 = 1.248 T heta2 = 56.78 [deg]
212
CosT heta1 = 0.5571 day = 16 F Rtan = 0.83 GrRef = 0.2 GT 1 = 711.2 W/m2 Lat = 40 [deg] ω = 52.5 [deg] Slope1 = 20 [deg] Ta = 14 [C]
Eqn 10.6.4 (20)
P11 01
Equations 11.1 Calculation of present worth. Knowns A = 5700 [$] ;
N = 10;
d1 = .08;
d2 = .12
(1)
From Equation 11.4.1 PW1 =
PW2 =
A N
(2)
N
(3)
(1 + d1) A (1 + d2)
Solution A = 5700 [$] P W 1 = 2640 [$]
d1 = 0.08 [-] P W 2 = 1835 [$]
d2 = 0.12 [-]
N = 10 [-]
213
P11 02
Equations 11.2 Calculation of uniform loan payment Knowns M = 20000 [$] ; 1 PWF = · d−i
NL = 15; 1−
i = 0;
1+i 1+d
P eriodP ay = M/P W F
d = 0.085
(1)
NL ! Eqn 11.5.1
(2)
Eqn 11.5.2
(3)
Solution d = 0.085 P eriodP ay = 2408.41 [$]
i=0 P W F = 8.304
M = 20000 [$]
214
NL = 15
P11 03
Equations 11.3 Calculation of Present Worth Factor PWF Knowns d = .095; PWF =
i = .078; 1 · d−i
N = 18
1−
1+i 1+d
(1)
N ! Eqn 11.5.1
(2)
Solution d = 0.095 [-]
i = 0.078 [-]
N = 18 [-]
P W F = 14.44 [-]
215
P11 04
Equations 11.4 Calculation of present worth of a series of costs. Knowns A = 1000;
i = .05;
d1 = .03;
d2 = .07;
N = 10
(1)
Use Equation 11.5.1 with each discount rate: N ! 1 1+i PWF1 = · 1− d1 − i 1 + d1
(2)
PW1 = A · PWF1 1 PWF2 = · d2 − i
(3) 1−
1+i 1 + d2
N ! (4)
PW2 = A · PWF2
(5)
Solution A = 1000 [$] N = 10 [-] P W F 2 = 8.598 [-]
d1 = 0.03 [-] P W 1 = 10603 [$]
d2 = 0.07 [-] P W 2 = 8598 [$]
216
i = 0.05 [-] P W F 1 = 10.6 [-]
P11 05
Equations 11.5 Present worth of a series of triennial payments. Knowns: Glazing cost = 15 $/m2 ; i = 0.07;
Ac = 75 m2 ;
Nyr = 20
d = 0.10
(1) (2)
Two solutions are shown. The PWF factors of Chapter 11 can be used for any time period; e.g., month, year or multiple years. We need to determine the 3-year interest rate from yearly interest (and discount) rates. Note that the rates are not 3 times the yearly rates. The 3-year interest and discount rates are: 3
(3)
(1 + d) = (1 + d3)
3
(4)
N3 = 6
(5)
(1 + i) = (1 + i3)
there are 6 three-year periods in the 20 years.
We can now use the PWF. 3
Cost3 = Glazing cost · Ac · (1 + i)
(6)
P W 1 = Cost3 · P W F (N 3, i3, d3)
(7)
The second method could use either the parametric table or arrays. We have chosen to use arrays. Cost0 = Glazing cost · Ac
(8)
duplicate j = 1, 6
(9)
Y rj = 3 · j
(10)
Costj = costj−1 · (1 + i3)
(11)
PWj =
Costj
(12)
j
(1 + d3)
end
(13)
P W 2 = Sum(P W k , k = 1, 6)
(14)
Solution
217
Ac = 75 m2 Glazing cost = 15 $/m2 Nyr = 20
Cost3 = 1378.17 [$] i = 0.07 P W 1 = 5099.87 [$]
d = 0.1 i3 = 0.225 P W 2 = 5099.87 [$]
Arrays Row 0 1 2 3 4 5 6
Costi [$] 1125.00 1378.17 1688.32 2068.27 2533.72 3103.91 3802.42
Y ri
PWi [$]
3 6 9 12 15 18
1035.44 953.01 877.15 807.32 743.05 683.90
218
d3 = 0.331 N3 = 6
P11 06
Equations 11.6 Loan payments and present worth of payments. Knowns: N = 8;
m = 0.105;
d = 0.095;
Loan = 2700 [$]
(1)
a) Annual payment to lender is given by Equation 11.5.2. The PWF is calculated with i = 0 and d = mortgage interest rate. P W F 1 = P W F (N, 0, m) AnnualP ay = Loan/P W F1
(2) Eqn 11.5.1
(3)
b) The monthly payment calculation is the same except that i = 0.105/12 and N = 8*12 = 96. M onthP ay =
Loan P W F (12 · N, 0, m/12)
Eqn 11.5.2
(4)
c) The present worth of interest payments of part a. is given by Equation 11.5.3. This equation contains 3 PWFs. Note that Nmin =N P W int = Loan · (P W F 2 /P W F 1 + P W F 3 · (m − 1/P W F 1 ))
(5)
P W F 2 = P W F (N, 0, d)
(6)
P W F 3 = P W F (N, m, d)
(7)
Solution AnnualP ay = 515.35 [$] N =8
d = 0.095 P W F 1 = 5.239
Loan = 2700 [$] P W F 2 = 5.433
219
m = 0.105 P W F 3 = 7.544
M onthP ay = 41.69 [$] P W int = 1051.10 [$]
P11 07
Equations 11.7 Present worths of single and series payments. a Present worth of a single cost in the future: Ca = 4500 [$] ;
da = 0.08;
Na = 7;
PWa =
Ca
Eqn 11.4.1
Na
(1 + da )
(1)
b An inflating cost, starting with known cost now: Ab = 700 [$] ;
ib = 0.05;
Nb = 7;
Nb
Cb = Ab · (1 + ib )
Eqn 11.4.4
(2)
c An inflating cost, starting with known cost at end of first period: Ac = 700;
ic = 0.05;
Nc = 7;
Nc −1
Cc = Ac · (1 + ic )
11.4.2
(3)
d Present worth of one of a series of inflating costs, with cost at end of first period known: Nd −1
Ad = 175;
id = 0.05;
dd = 0.08;
Nd = 10;
P W d = Ad ·
(1 + id )
Eqn 11.4.3
Nd
(1 + dd )
(4)
e Present worth of the series of inflating costs: Ae = 175;
ie = 0.05;
de = 0.08;
Ne = 10;
1 · P W F e = Ae · de − ie
1−
1 + ie 1 + de
Ne !! Eqn 11.5.1(5)
Solution Ab = 700.00 [$] Cc = 938.1 [$] id = 0.05 [-] Ne = 10
Ac = 700 [$] da = 0.08 [-] ie = 0.05 [-] P W F e = 1432 [$]
Ad = 175 [$] dd = 0.08 [-] Na = 7 [-] P W a = 2625.71 [$]
220
Ae = 175 [$] de = 0.08 [-] Nb = 7 [-] P W d = 125.75 [$]
Ca = 4500.00 [$] ib = 0.05 [-] Nc = 7 [-]
Cb = 984.97 [$] ic = 0.05 [-] Nd = 10 [-]
P11 08
Equations 11.8 Effects of increasing income tax rates in late years of an economic analysis. An increase in the effective income tax rate at any time in the analysis when interest on a mortgage is being paid, i.e. until period NL , would result in increased income tax savings and thus in increased solar savings. (At the same time, tax payments would go up.) The dollar impact would not be great, as the additional savings would occur late in the period of the analysis when the present worths of the savings are small.
221
P11 09
Equations 11.9 Life cycle savings - by tabular method. Knowns i = 0.10; Load = 124;
d = 0.09;
NL = 5;
F = 0.64;
DownP ayment = 1000;
CF = 12;
iF = 0.07;
Investment = 6000;
T axBracket = 0.42
(1) (2)
a. 10 year analysis and no resale value M ortgage = Investment − DownP ayment interest = OldP rincipal · i AnnualP ay = If year, NL + 1,
(3) (4)
M ortgage P W F (NL , 0, i) year−1
F uelSav = CF · Load · F · (1 + iF )
, 0, 0
Mortgage payments for NL years
CF assumed to be value at end of 1st year
(5) (6)
OldP rincipal = Tablevalue(‘Table 1’ , year, 4)
(7)
P rincipalBal = max (0, OldP rincipal − AnnualP ay + interest)
(8)
T axSav = interest · T axBracket
(9)
SolarSav = F uelSav + T axSav − AnnualP ay P W Sav,A =
(10)
SolarSav year (1 + d)
(11)
b. 5 year analysis and no resale value P W Sav,B = If(year, 6, P W Sav,A , 0, 0)
(12)
222
c. 5 year analysis with resale of $2000 P W Resale =
If(year, 5, 0, 2000, 0) year (1 + d)
(13)
P W Sav,C = P W Sav,B + P W Resale
(14)
The LCS is the sum of the annual LCS, i.e., the sums of the last three columns. These sums are: a: LCS = $2467; b: LCS = -1372; c: LCS = $-72
Solution AnnualP ay = 0.00 F = 0.64 Investment = 6000 NL = 5 P W Sav,A = 739.56 T axBracket = 0.42
CF = 12 F uelSav = 1750.80 iF = 0.07 OldP rincipal = 6.334 × 10−14 P W Sav,B = 0.00 T axSav = 0.00
d = 0.09 i = 0.1 Load = 124 P rincipalBal = 0.00 P W Sav,C = 0.00 year = 10
DownP ayment = 1000 interest = 0.00 M ortgage = 5000 P W Resale = 0 SolarSav = 1750.80
Table 1 Run
year
AnnualP ay
interest
P rincipalBal
F uelSav
T axSav
SolarSav
P W Sav,A
P W Sav,B
P W Sav,C
1 2 3 4 5 6 7 8 9 10 11 Sum
0 1 2 3 4 5 6 7 8 9 10 55
1318.99 1318.99 1318.99 1318.99 1318.99 0.00 0.00 0.00 0.00 0.00 6594.94
500.00 418.10 328.01 228.92 119.91 0.00 0.00 0.00 0.00 0.00 1594.94
5000.00 4181.01 3280.13 2289.15 1199.08 0.00 0.00 0.00 0.00 0.00 0.00 15949.37
952.32 1018.98 1090.31 1166.63 1248.30 1335.68 1429.18 1529.22 1636.26 1750.80 13157.68
210.00 175.60 137.77 96.14 50.36 0.00 0.00 0.00 0.00 0.00 669.87
-156.67 -124.40 -90.91 -56.21 -20.33 1335.68 1429.18 1529.22 1636.26 1750.80 7232.62
-1000.00 -143.73 -104.71 -70.20 -39.82 -13.21 796.42 781.81 767.46 753.38 739.56 2466.96
-1000.00 -143.73 -104.71 -70.20 -39.82 -13.21 0.00 0.00 0.00 0.00 0.00 -1371.67
-1000.00 -143.73 -104.71 -70.20 -39.82 1286.65 0.00 0.00 0.00 0.00 0.00 -71.81
223
P11 10
Equations 11.10 Redo Problem 11.9 using the P1, P2 method. The functions for Equation 11.8.2 and 11.8.3 are used in all three parts of the problem Knowns: m = 0.10; Down = 1/6;
d = 0.09;
NL = 5 [yr] ;
M s = 0;
Investment = 6000 [$] ;
V = 0;
Ne = 10 [yr] ; Rv = 0;
Load = 124;
t = 0;
C = 0;
ND = 1 [yr]
i=0
CF = 12;
(Can not be zero.)(1) (2)
iF = 0.07;
tbar = 0.42;
F = 0.64
(3)
DownP ayment = Down · Investment
(4)
M ortgage = Investment − DownP ayment
(5)
a. 10 year analysis and no resale value F uelSav,A = CF · Load · F
(6)
P 1A = P 1 (Ne , iF , d, 0, 0)
(7)
P 2A = P 2 (m, i, d, Ne , NL , ND , t, tbar, Down, M s, V, Rv, C)
(8)
LCS A = F uelSav,A · P 1A − P 2A · Investment
(9)
Eqn 11.8.1
b. 5 year analysis with no resale value This is the same calculation as in a. but with Ne = 5. P 1B = P 1 (5 [yr] , iF , d, 0, 0)
(10)
P 2B = P 2 (m, i, d, 5 [yr] , NL , ND , t, tbar, Down, M s, V, Rv, C)
(11)
LCS B = CF · Load · F · P 1B − P 2B · Investment
(12)
c. 5 year analysis with resale of $2000 This is the same as b. but with resale term added in to P2. Resale = 1/3
(13)
P 1C = P 1 (5 [yr] , iF , d, 0, 0)
(14)
P 2C = P 2 (m, i, d, 5 [yr] , NL , ND , t, tbar, Down, M s, V, Resale, C)
(15)
LCS C = CF · Load · F · P 1B − P 2C · Investment
(16)
Solution
224
C=0 DownP ayment = 1000 [$] Investment = 6000 [$] LCS C = −71.81 [$] Ms = 0 P 1A = 8.453 P 2B = 0.9305 t=0
CF = 12 [$/GJ] F = 0.64 iF = 0.07 Load = 124 [GJ] ND = 1 [yr] P 1B = 4.422 P 2C = 0.7138 tbar = 0.42
d = 0.09 F uelSav,A = 952.3 [$] LCS A = 2466.96 [$] m = 0.1 Ne = 10 [yr] P 1C = 4.422 Resale = 0.3333 V =0
225
Down = 0.1667 i=0 LCS B = −1371.67 [$] M ortgage = 5000 [$] NL = 5 [yr] P 2A = 0.9305 Rv = 0
P11 11
Equations 11.11 Estimation of uncertainties in LCS in results of Problem 11.9/11.10 Knowns: m = 0.10; Down = 1/6;
d = 0.09;
NL = 5 [yr] ;
M s = 0;
Investment = 6000 [$] ;
V = 0;
Ne = 10 [yr] ; Rv = 0;
Load = 124;
t = 0;
C = 0;
ND = 1 [yr]
i=0
CF = 12;
(Can not be zero.)(1) (2)
iF = 0.07;
tbar = 0.42;
F = 0.64
(3)
DownP ayment = Down · Investment
(4)
M ortgage = Investment − DownP ayment
(5)
Parts a, b, and c refer to the three parts of the previous problem. a. 10 year analysis and no resale value ∆iF = 0.03;
F uelSav = CF · Load · F
∆F = 0.02;
P 1 = P 1 (Ne , iF , d, 0, 0)
(6) (7)
From Equations 11.9.5 and 11.9.13:
P artialP 1,A,wrt,i,F =
P 1 − Ne ·
1+iF 1+d
Ne
1+iF
(8)
d − iF
∆LCS,i,F,A = F uelSav · P artialP 1,A,wrt,i,F · ∆iF
Eqn 11.9.1
(9)
∆LCS,F,A = P 1 · CF · Load · ∆F Eqn 11.9.11 q ∆LCS,P robable,A = ∆2LCS,i,F,A + ∆2LCS,F,A Eqn 11.9.3
(10)
∆LCS,M ax,A = abs (∆LCS,i,F,A ) + abs (∆LCS,F,A )
(12)
(11)
b. 5 year analysis with no resale value. Calculations are similar to part a. Ne,B = 5 [yr]
(13)
P 1B = P 1 (Ne,B , iF , d, 0, 0)
(14)
P artialP 1,B,wrt,i,F =
P 1B − Ne,B ·
1+iF 1+d
Ne,B
1+iF
(15)
d − iF
226
∆LCS,i,F,B = F uelSav · P artialP 1,B,wrt,i,F · ∆iF
(16)
∆LCS,F,B = P 1B · CF · Load · ∆F q ∆LCS,P robable,B = ∆2LCS,i,F,B + ∆2LCS,F,B
(17)
∆LCS,M ax,B = abs (∆LCS,i,F,B ) + abs (∆LCS,F,B )
(19)
(18)
c. 5 year analysis with resale of $2000 Answers are the same as part b since adding resale value only changes P2, and P2 does not enter into the partial derivatives of LCS wrt iF or F.
Solution C=0 ∆iF = 0.03 ∆LCS,i,F,B = 231.76 [$] ∆LCS,P robable,B = 266.52 [$] F uelSav = 952.32 [$] Load = 124 [GJ] ND = 1 [yr] P 1 = 8.453 Rv = 0
CF = 12 [$/GJ] ∆LCS,F,A = 251.55 [$] ∆LCS,M ax,A = 1232.70 [$] Down = 0.1667 i=0 m = 0.1 Ne = 10 [yr] P 1B = 4.422 t=0
227
d = 0.09 ∆LCS,F,B = 131.60 [$] ∆LCS,M ax,B = 363.36 [$] DownP ayment = 1000 [$] Investment = 6000 [$] M ortgage = 5000 [$] Ne,B = 5 [yr] P artialP 1,A,wrt,i,F = 34.34 tbar = 0.42
∆F = 0.02 ∆LCS,i,F,A = 981.15 [$] ∆LCS,P robable,A = 1012.89 [$] F = 0.64 iF = 0.07 Ms = 0 NL = 5 [yr] P artialP 1,B,wrt,i,F = 8.112 V =0
P11 12
Equations 11.12 Calculation of payback times for Problem 11.9/11.10 Knowns DownP ayment = 1000;
Investment = 6000;
a. Fuel costs not discounted iF ln 1 + Investment · F ·Load·C F Np = ln (1 + iF )
Load = 124;
F = 0.64;
Eqn 11.7.3
b. Fuel costs discounted iF −d ln 1 + Investment · F ·Load·C F N pDiscounted = 1+iF ln 1+d
d = 0.09 iF = 0.07
iF = 0.07;
d = 0.09(1)
(2)
Eqn 11.7.6
(3)
Solution CF = 12 [$/GJ] Investment = 6000 [$] N pDiscounted = 7.27
CF = 12;
DownP ayment = 1000 [$] Load = 124 [GJ]
228
F = 0.64 N p = 5.40
P11 13
Equations 11.13 For the system of Problem 11.9/11.10, given an F vs Ac curve, optimize Ac. Knowns: m = 0.10;
d = 0.09;
Down = 1/6; i = 0;
NL = 5 [yr] ;
M s = 0;
Ce = 120;
V = 0;
Ca = 96;
Ne = 10 [yr] ;
t = 0;
C = 0;
ND = 1 [yr] ;
Rv = 0
Load = 124;
Can not be zero(1) (2)
CF = 12;
iF = 0.07;
t¯ = 0.42
(3)
a. 10 year analysis and no resale value P 1A = P 1 (Ne , iF , d, 0, 0)
(4)
F uelSav = CF · Load · F
(5)
P 2A = P 2 (m, i, d, Ne , NL , ND , t, t¯, Down, M s, V, Rv, C)
(6)
Area and F determined from figure with dF/dAc as given by Equation 11.8.5. It is necessary to run EES program twice, once to find dF/dAc and then find LCS with correct F and Ac. dF/dAc,A = P 2A ·
Ca P 1A · CF · Load
Eqn 11.8.5
(7)
From figure with dF/dAc equal to 0.00710, FA and Ac,A are: F = 0.60;
Ac,A = 43
InvestmentA = Ac,A · Ca + Ce
(8) Eqn 11.1.1
(9)
DownP aymentA = Down · InvestmentA
(10)
M ortgage = InvestmentA − DownP aymentA
(11)
LCS A = F uelSav · P 1A − P 2A · InvestmentA
Eqn 11.8.1
(12)
b. 5 year analysis with no resale value P 2B = P 2 (m, i, d, 5 [yr] , NL , ND , t, t¯, Down, M s, V, Rv, C)
(13)
P 1B = P 1 (5 [yr] , iF , d, 0, 0)
(14)
dF/dAc,B = P 2B ·
Ca P 1B · CF · Load
(15)
229
From figure with dF/dAc,B equal to 0.014, FB and Ac,B are: FB = 0.38;
Ac,B = 20
(16)
InvestmentB = Ca · Ac,B + Ce
(17)
LCS B = CF · Load · FB · P 1B − P 2B · InvestmentB
(18)
c. 5 year analysis with resale of $2000 Resale = 1/3
(19)
P 2C = P 2 (m, i, d, 5 [yr] , NL , ND , t, t¯, Down, M s, V, Resale, C)
(20)
P 1C = P 1 (5 [yr] , iF , d, 0, 0)
(21)
dF/dAc,C = P 2C ·
Ca P 1C · CF · Load
(22)
From figure with dF/dAc,C equal to 0.010, F and Ac are: FC = 0.46;
Ac,C = 28
(23)
InvestmentC = Ca · Ac,C + Ce
(24)
LCS C = CF · Load · FC · P 1C − P 2C · InvestmentC
(25)
Solution Ac,A = 43 m2 Ca = 96 $/m2 dF /dAc,A = 0.007102 1/m2 DownP aymentA = 708 [$] FC = 0.46 InvestmentC = 2808 [$] LCS C = 1022 [$] Ms = 0 P 1A = 8.453 P 2B = 0.9305 t=0
Ac,B = 20 m2 Ce = 120 [$] dF /dAc,B = 0.01358 1/m2 F = 0.6 i=0 iF = 0.07 Load = 124 [GJ] ND = 1 [yr] P 1B = 4.422 P 2C = 0.7138 t¯ = 0.42
230
Ac,C = 28 m2 CF = 12 [$/GJ] dF /dAc,C = 0.01041 1/m2 F uelSav = 892.8 [$] InvestmentA = 4248 [$] LCS A = 3594 [$] m = 0.1 Ne = 10 [yr] P 1C = 4.422 Resale = 0.3333 V =0
C=0 d = 0.09 Down = 0.1667 FB = 0.38 InvestmentB = 2040 [$] LCS B = 602.2 [$] M ortgage = 3540 [$] NL = 5 [yr] P 2A = 0.9305 Rv = 0
P11 14
Equations 11.14 Calculation of LCS Knowns: m = 0.09; M s = 250
d = 0.10; [$] ; investment
Down = 0.20;
NL = 10; V = 0;
Ne = 15; Rv = 0;
Investment = 18000;
t = 0; i = 0.06;
F = 0.7;
C = 0;
ND = 1
CF = 8.8;
Any number but zero
iF = 0.10;
t¯r = 0.40
load = 3000 [GJ]
(1) (2)
(3)
DownP ayment = Down · Investment
(4)
M ortgage = Investment − DownP ayment
(5)
Calculate PW of fuel savings: P 1 = P 1 (Ne , iF , d, 0, 0)
(6)
F uelSav = CF · Load · F
(7)
P W F uel,Sav = F uelSav · P 1
(8)
Calculate PW of system nvestment costs: P 2 = P 2 (m, i, d, Ne , NL , ND , t, t¯r, Down, M s, V, Rv, C)
(9)
P W System = P 2 · Investment
(10)
LCS = P W F uel,Sav − P W System
Eqn 11.8.1
(11)
Solution C=0 DownP ayment = 3600 [$] Investment = 18000 [$] m = 0.09 Ne = 15 [yr] P W F uel,Sav = 252000 [$] t¯r = 0.4
CF = 8.8 [$/GJ] F = 0.7 iF = 0.1 M ortgage = 14400 [$] NL = 10 [yr] P W System = 17846 [$] V =0
d = 0.1 F uelSav = 18480 [$] LCS = 234154 [$] M s = 0.01389 [-] P 1 = 13.64 Rv = 0 231
Down = 0.2 i = 0.06 load = 3000 [GJ] ND = 1 [yr] P 2 = 0.9914 t=0
P11 15
Equations 11.15 Calculation of LCS and ROI. Knowns m = 0.09;
d = 0.15;
Down = 0.20; iF = 0.14;
NL = 10 [yr] ;
Investment = 11000;
t¯ = 0.40;
t = 0.02; M s = 0;
C = 0; V = 1;
solar = 110 [GJ]
ND = 1 [yr] Rv = 0.80;
Any number but zero i = 0.06;
CF = 10
(1) (2) (3)
DownP ayment = Down · Investment
(4)
M ortgage = Investment − DownP ayment
(5)
a. 15 year economic analysis Ne,A = 15 [yr]
(6)
P 1A = P 1 (Ne,A , iF , d, 0, 0)
(7)
P W F uel,Sav,A = CF · solar · P 1A
(8)
P 2A = P 2 (m, i, d, Ne,A , NL , ND , t, t¯, Down, M s, V, Rv, C)
(9)
P W System,A = P 2A · Investment
(10)
LCS A = P W F uel,Sav,A − P W System,A
(11)
b. 20 year economic analysis Ne,B = 20 [yr]
(12)
232
P 1B = P 1 (Ne,B , iF , d, 0, 0)
(13)
P W F uel,Sav,B = CF · solar · P 1B
(14)
P 2B = P 2 (m, i, d, Ne,B , NL , ND , t, t¯, Down, M s, V, Rv, C)
(15)
P W System,B = P 2B · Investment
(16)
LCS B = P W F uel,Sav,B − P W System,B
(17)
c. Find ROI for a 15 year economic analysis With EES this is a very simple calculation since it is only necessary to set LCS to zero and let EES find the discount rate which will be the ROI. Without EES or a similar program, it would be necessary to solve the problem for a range of discount rates and choose the one for which LCS is zero. Ne,C = 15 [yr] ;
LCS C = 0 [$]
(18)
P 1C = P 1 (Ne,C , iF , ROI, 0, 0)
(19)
P W F uel,Sav,C = CF · solar · P 1C
(20)
P 2C = P 2 (m, i, ROI, Ne,C , NL , ND , t, t¯, Down, M s, V, Rv, C)
(21)
P W System,C = P 2C · Investment
(22)
LCS C = P W F uel,Sav,C − P W System,C
(23)
Thus the Return on Investment ROI, part c, is 29.4%. If the time of the economic analysis had been 20 years, as used in part b, then the ROI would be 0.3115 or 31.15%.
Solution C=0 DownP ayment = 2200 [$] LCS A = 5622 [$] M ortgage = 8800 [$] Ne,B = 20 [yr] P 1B = 16.03 P 2C = 0.5514 P W System,A = 7884 [$] Rv = 0.8 V =1
CF = 10 [$/GJ] i = 0.06 LCS B = 9057 [$] Ms = 0 Ne,C = 15 [yr] P 1C = 5.514 P W F uel,Sav,A = 13507 [$] P W System,B = 8572 [$] solar = 110 [GJ]
233
d = 0.15 Investment = 11000 [$] LCS C = 0 [$] ND = 1 [yr] NL = 10 [yr] P 2A = 0.7167 P W F uel,Sav,B = 17630 [$] P W System,C = 6065 [$] t = 0.02
Down = 0.2 iF = 0.14 m = 0.09 Ne,A = 15 [yr] P 1A = 12.28 P 2B = 0.7793 P W F uel,Sav,C = 6065 [$] ROI = 0.2944 t¯ = 0.4
P11 16
Equations 11.16 Comparison of the PW of fuel savings and the cost of a system. Knowns: Investment = 1550 [$] ;
Celect = 0.045 [$/kwh] ;
iF = 0.12;
d = 0.07;
load = 19;
Ne = 25 [yr]
F = 0.60
(1) (2)
The question is: is ∆ i.e., (δ=PWF uel,Sav - Investment) positive? $/GJ CF = Celect · 277.777778 $/kwh
(3)
P 1 = P 1 (Ne , iF , d, 0, 0)
(4)
P W F uel,Sav = CF · load · F · P 1
(5)
∆ = P W F uel,Sav − Investment
(6)
Since ∆ is positive, the criterion is met.
Solution Celect = 0.045 [$/kwh] F = 0.6 Ne = 25 [yr]
CF = 12.5 [$/GJ] Investment = 1550 [$] P 1 = 42.64 [-]
d = 0.07 iF = 0.12 P W F uel,Sav = 6077 [$]
234
∆ = 4527 [$] load = 19 [GJ]
P11 17
235
Equations Find economic optimum solar system. Economic data Collector cost per unit area set in parametric table as either 240[$/m2 ] or 150[$/m2 ] CE = 550 [$]
(1)
Invest = Ac · CA + CE
(2)
inf F = 0.052
(3)
dis = 0.045
(4) t¯ = 0.39;
Ne = 20; m = 0.058;
C=0
inf = .020;
Down = 0.2;
(5)
NL = 20;
Ms = 0.050;
ND = 0.0001;
V al = 0.90;
P rT ax = 0.035;
Rv = 0.25;
η = 0.8 CF 1 = 11
(6) (7) (8)
[$/GJ] η
(9)
Thermal data Load = (EDHW
·6 GJ + EHouse ) · 1 × 10 kJ
(10)
EDHW = 2.0645 × 107 [kJ]
(11)
EHouse = 1.42 × 108 [kJ]
(12)
a. LCS Method LCS = LCS (P 1, CF 1 , Load, SolF ract, P 2, Invest)
(13)
P 1 = P 1 (Ne , inf F , dis, t¯, C)
(14)
P 2 = P 2 (m, inf , dis, Ne , NL , ND , P rT ax, t¯, Down, Ms , V al, Rv , C)
(15)
b. Payback Time Method inf F −dis ln Invest · SolF ract·Load·C + 1 F1 NP = 1+infF ln 1+dis
Eqn 11.7.6
(16)
c. Return on Investment Method Return on Investment is the discount rate at which LCS = 0. EES can solve for the discount rate (i.e., ROI) with LCS set to zero for each of the collector areas. The equations are the same as in part a; but P1, P2 have new values and the LCS is 0. 0= LCS(P 10 ,CF 1 ,Load,SolFract,P2’,Invest) P1’ = P1(N,e , infF , ROI, t¯, C) P2’ = P2(m , inf, ROI, Ne , NL , ND , PrTax, t¯, Down, Ms , Val, Rv , C)
236
EES has a problem solving these three equations without very good guesses. Doing a little algebra to eliminate P1’ and P2’ results in:
0 = LCS (P 1 (Ne , inf F , ROI, t¯, C) , CF 1 , Load, SolF ract, P 2 (m, inf , ROI, Ne , NL , ND , P rT ax, t¯, Down, Ms , V al, Rv EES has trouble at zero area (row 1) but all of these values are obvious so run Tables from 2 to 7. d. Cash Flow (PITI) Method This method compares the solar system out-of-pocket first year expenses - for principle, interest, taxes, and insurance (PITI) to the first year fuel savings. P &If irst = (1 − Down) ·
Invest P W F (NL , 0, m)
First year principle and interest
(18)
F uelSav 1 = SolF ract · Load · CF 1
First year fuel savings
(19)
P ropT ax = V al · P rT ax · Invest
First year property taxes
(20)
IncT ax = t¯ · ((1 − Down) · m · Invest + P ropT ax) Ins = Invest · Ms
First year income tax savings
First year insturance
(22)
CashF low = F uelSav 1 + IncT ax − P &If irst − P ropT ax − Ins
First year cash flow
Solution Ac = 75.0 m2 CE = 550 [$] η = 0.8 IncT ax = 358.5 [$] Invest = 11, 800 [$] Ms = 0.05 NP = 10.00 P ropT ax = 371.7 [$] SolF ract = 0.535
C=0 CF 1 = 13.75 [$/GJ] EDHW = 2.064 × 107 [kJ] inf = 0.02 LCS = 2, 107 [$] ND = 0.0001 P &If irst = 809.7 [$] P rT ax = 0.035 t¯ = 0.39
CashF low = −216.46 [$] dis = 0.045 EHouse = 1.420 × 108 [kJ] inf F = 0.052 Load = 162.6 [GJ] Ne = 20 [yr] P 1 = 20.41 ROI = 0.081 V al = 0.9
CA = 150 $/m2 Down = 0.2 F uelSav 1 = 1, 196 [$] Ins = 590 [$] m = 0.058 NL = 20 [yr] P 2 = 1.891 Rv = 0.25
CA=240 Run 1 2 3 4 5 6 7
Ac2 m 0.0 5.0 15.0 30.0 45.0 60.0 75.0
SolF ract
Ac2 m 0.0
SolF ract
0.000 0.063 0.172 0.293 0.389 0.469 0.535
CA 2 $/m 240 240 240 240 240 240 240
Invest [$] 550 1,750 4,150 7,750 11,350 14,950 18,550
LCS [$]
NP
-433 4 -1,280 -3,705 -6,860 -10,654
12.49 10.91 11.91 13.09 14.25 15.43
CA 2 $/m 150
Invest [$]
LCS [$]
CashF low [$] -550.00 -68.65 -112.26 -272.72 -489.08 -741.23 -1,024.69
ROI
-0.008 0.045 0.010 -0.027 -0.060 -0.091
CA=150 Run 1
0.001
(21)
NP
237
CashF low [$]
ROI
(23)
2 3 4 5 6 7
5.0 15.0 30.0 45.0 60.0 75.0
0.063 0.172 0.293 0.389 0.469 0.535
150 150 150 150 150 150
1,300 2,800 5,050 7,300 9,550 11,800
417 2,556 3,824 3,952 3,349 2,107
9.37 7.44 7.87 8.55 9.25 10.00
LCS
Yeas to payback
238
-14.77 49.39 50.58 -4.14 -94.65 -216.46
0.109 0.225 0.194 0.152 0.115 0.081
PITI
ROI
239
P12 01
Equations 12.1
Effect of load timing.
The effect of concentrating the loads to early evening hours will be determined by two opposing factors. The collector will run a bit hotter because there is no energy delivery during collection hours, and essentially all collected energy will have to be stored. But, assuming a fully mixed tank, if energy is removed from store in the early evening, the store will be cooled more quickly and will be at a lower 24 hour average temperature, thus reducing store losses. The net effect should be to increase the solar energy delivered by a small amount.
240
P12 02
Equations 12.2
Effects of design changes on performance of the system of Example 10.9.1
a. Collector losses are very much greater, as UL in Equation 10.2.2 will be substantially increases. This means reduced Qu (possibly to zero for the four hours of the problem), reduced Ts and reduced solar energy delivery to load. b. Doubling the collector area will increase Qu, but will not double it as Ts and thus Ti in Equation 10.2.2 will increase. Increased Ts will lead to greater solar energy delivery to load. c. Applying the load in the evening hours will mean increased Ts and thus Ti because all energy to load must be stored. But, Ts will be brought down more quickly by the concentrated load and the 24 hour average Ts will be lower. The net effect should be to somewhat increase solar energy delivery. d. Stratification of the tank will lead to lower temperatures in the bottom of the tank and lower Ti. On the load side, higher Ts in the top of the tank will lead to higher solar energy delivery. The total effect will be to increase Qu and solar energy delivery.
241
P14 01
Equations 14.1 Month’s absorbed radiation on vertical wall with overhang. ¯b, H ¯ b, H ¯ d , and the effective τ -α products for the beam, The basic equation to be used is 14.6.2. We need to find fibar , Frs , R diffuse and ground-reflected components. Knowns M onth = 4; ρg = 0.2;
¯ = 25.3; H
¯ T = 0.71; K
Slope = 90 [deg] ;
Lat = 35.1 [deg] Ar = 6.0 m2 ; αn = 0.89
(1) Assumed
n = AveDay (month)
(2) (3)
δ = dec (n)
(4)
¯T Hbardif f /Hbar = Hdif f BarHBar Lat, n, K
(5)
¯ dif f = H ¯ · Hbardif f /Hbar H
(6)
SunSetHrAngle0 = min (arccos (− tan (Lat) · tan (δ)), arccos (− tan (Lat − Slope) · tan (δ)))
(7)
SunSetHrAngle = arccos (− tan (Lat) · tan (δ))
(8)
¯ beam = R ¯ BEAM, (n, Lat, Slope) R
(9)
¯T = H ¯ T,LJ, H, ¯ Lat, n, Slope, ρg H
(10)
¯b = H ¯ −H ¯ dif f H
(11)
¯b · R ¯ beam · f¯i · T auAlphabeam + H ¯ dif f · T auAlphadif f · Frs + H ¯ · ρg · T auAlphaGrRef /2 AbsRad = Ar · H
Eqn 14.6.2(12)
From Fig 5.10, θef f,beam = 0.71; at this angle, from Fig. 5.6.1, TauAlpha/TauAlphan = 0.57. From Fig. 5.4.1: θef f,dif f and θef f -GrRef both = 59; at this angle, TauAlpha/TauAlphan = 0.82. from Fig. 5.6.1. The normal transmittace (τ n ) for this glazing system from Figure 5.3.1 is 0.78. For this wall and glazing: τn = 0.78
(13)
T auAlphan = τn · αn
(14)
T auAlphabeam = T auAlphan · 0.57
(15)
T auAlphadif f = T auAlphan · 0.82
(16)
T auAlphaGrRef = 0.57
(17)
242
1
We do not have f¯i for infinite extensions, so will use e = 0.3. Using the charts in Appendix I for w = 1 and w = 4 at Latitudes 35 and 45: At Lat = 35, w = 1: At Lat = 35, w = 4:
f¯i at g = 0.2 is 0.51 and at g = 0.4 is 0.71, so at g = 0.25, f¯i is 0.56 f¯i at g = 0.2 is 0.43 and at g = 0.4 is 0.63, so at g = 0.25, f¯i is 0.48
Using Equation 14.4.6 to interpolate for w = 1.5: f¯i = 0.52: At Lat = 45, w = 1: f¯i at g = 0.2 is 0.72 and at g = 0.4 is 0.90, so at g = 0.25, f¯i is 0.76 At Lat = 45, w = 4: f¯i at g = 0.2 is 0.70 and at g = 0.4 is 0.89, so at g = 0.25, f¯i is 0.75 Thus at w = 1.5, f¯i = 0.76 From Table 14.4.4, at e = 0.3: Frs = 0.44 The following are calculated above: Frs = 0.44;
f¯i = 0.59
(18)
Solution AbsRad = 28.26 [MJ] f¯i = 0.59 ¯ b = 19.2 MJ/m2 H Lat = 35.1 [deg] ¯ beam = 0.3883 R T auAlphabeam = 0.3957 τn = 0.78
αn = 0.89 Frs = 0.44 ¯ dif f = 6.102 MJ/m2 H M onth = 4 Slope = 90 [deg] T auAlphadif f = 0.5692
Ar = 6 m2 Hbardif f /Hbar = 0.2412 ¯ T = 13.05 MJ/m2 H n = 105 SunSetHrAngle = 96.69 [deg] T auAlphaGrRef = 0.57
243
2
δ = 9.415 [deg] ¯ = 25.3 MJ/m2 H ¯ T = 0.71 K ρg = 0.2 SunSetHrAngle0 = 76.35 [deg] T auAlphan = 0.6942
P14 02
Equations 14.2 Calculation of monthly radiation on a direct gain window, with and without overhangs. Knowns Lat = 44.9 [deg] ;
Slope = 90 [deg]
(1)
Frs = 0.43
(2)
dec = Dec (month)
(3)
n = AveDay (month)
(4)
¯ beam = R ¯ BEAM, (n, Lat, Slope) R ¯T Hbardif f /Hbar = Hdif f barHbar Lat, n, K
(5)
(6)
¯T: Rewrite Equation 14.4.4 to give the three components of H ¯T = H ¯ T,beam + H ¯ T,dif f + H ¯ T,GrRef H
(7)
¯ T,beam = H ¯ beam · R ¯ beam · f¯i H
(8)
¯ T,dif f = H ¯ dif f · Frs H
(9)
¯ T,GrRef = H ¯ · GrRef /2 H
(10)
Calculate monthly diffuse and beam: ¯ dif f = H ¯ · Hbardif f /Hbar H
(11)
¯ beam = H ¯ −H ¯ dif f H
(12)
Get f¯i from Appendix I (d) and Frs from Table 14.4.1. Designate the radiation terms for no overhang with double prime, ”: ¯ T 00 = H ¯ T,beam00 + H ¯ T,dif f 00 + H ¯ T,GrRef 00 H
(13)
¯ T,beam00 = H ¯ beam · R ¯ beam H
(14)
¯ T,dif f 00 = H ¯ dif f /2 H
(15)
¯ T,GrRef 00 = H ¯ · GrRef /2 H
(16)
Parametric Table: Table 1
244
1
Run
M onth
¯T K
1 2 3 4 5 6 7 8 9 10 11 12
1 2 3 4 5 6 7 8 9 10 11 12
0.47 0.52 0.48 0.47 0.54 0.55 0.55 0.56 0.49 0.48 0.42 0.47
Run 1 2 3 4 5 6 7 8 9 10 11 12
¯ H T,GrRef MJ/m2 1.93 1.98 1.29 1.61 2.14 2.30 2.26 2.03 1.46 1.05 1.27 1.72
¯ H 2 MJ/m 6.44 9.89 12.86 16.05 21.36 23.04 22.58 20.33 14.59 10.48 6.37 5.74
¯ HT 2 MJ/m 13.86 15.47 11.33 8.76 8.07 7.45 7.74 9.37 10.57 12.84 11.23 13.58
f¯i
GrRef
¯ beam R
1 1 0.94 0.78 0.55 0.4 0.46 0.69 0.88 0.99 1 1
0.6 0.4 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.4 0.6
2.89 1.92 1.13 0.59 0.33 0.24 0.28 0.46 0.88 1.63 2.59 3.26
¯ HT,beam200 MJ/m 10.76 11.91 8.04 5.12 4.25 3.39 3.85 5.90 7.22 10.04 8.65 10.81
¯ HT,dif f200 MJ/m 1.36 1.84 2.88 3.68 4.18 4.41 4.32 3.80 3.20 2.15 1.52 1.21
¯ HT,beam MJ/m2 10.76 11.91 7.56 3.99 2.34 1.36 1.77 4.07 6.36 9.94 8.65 10.81
¯ T,GrRef 00 H MJ/m2 1.93 1.98 1.29 1.61 2.14 2.30 2.26 2.03 1.46 1.05 1.27 1.72
Plot 1
245
2
¯ HT,dif f2 MJ/m 1.17 1.58 2.48 3.17 3.60 3.79 3.71 3.27 2.75 1.85 1.30 1.04
¯ HT 00 2 MJ/m 14.05 15.73 12.21 10.40 10.57 10.11 10.42 11.73 11.88 13.24 11.44 13.75
P14 03
Equations 14.3
Gain (or loss) from a direct-gain south facing window in Example 14.5.1.
Knowns Lat = 43 [deg] ;
Uwindow = 3.2 W/m2 ·C ;
Slope = 90 [deg] ;
Height = 1.25 [m] ; w = W idth/Height;
W idth = 15 [m] ;
P rojection = 0.625 [m] ;
p = P rojection/Height;
Troom = 20 [C] Gap = 0.25 [m] ;
e = Extension/Height;
(1) Extension = 0 [m](2)
g = Gap/Height
(3)
For Feb: month = 2;
¯ = 9.12; H
¯ T = 0.49; K
T¯a = −7;
GrRef = 0.2
(4)
n = AveDay (month)
(5)
dec = dec (n)
(6)
Calculate monthly beam and diffuse: ¯ dif ¯f /H = Hdif f barHbar Lat, n, K ¯T H
(7)
¯ dif f = H ¯ ·H ¯ dif ¯f /H H
(8)
¯ beam = H ¯ −H ¯ dif f H
(9)
¯ beam = R ¯ BEAM, (n, Lat, Slope) R
(10)
¯ i from Appendix I (d): Getf At Lat=45 and w=4,f¯i =0.93, and at w=25,f¯i =0.93 At Lat=35 and w=4,f¯i =0.83, and at w=25,f¯i =0.82 The interpolation is thus linear for Lat=43 and gives: f¯i = 0.91
(11)
Frs from Table 14.4.1. Interpolating between g=0 and g=0.25 at e=0 and p=0.5 gives: Frs = 0.38
(12)
From Example 14.5.1, the room absorptance is 0.96 and the ground reflected and diffuse sky transmittances are about 0.70. Thus: αr = 0.96;
τ¯dif f = 0.70;
τ¯gr = 0.70
(13)
τ¯beam = 0.72
(14)
246
1
From Eqn 14.5.3 ¯ beam · R ¯ beam · f¯i · τ¯beam + H ¯ dif f · Frs · τ¯dif f + H ¯ · GrRef · τ¯gr /2 Absorbedroom = Height · W idth · αr · H MJ Eqn 14.5.4 Lossroom = Height · W idth · Uwindow · 24 [hr] · Troom − T¯a · 0.0036 W · hr
(15) (16)
The net gain is then: N etGainroom = Absorbedroom − Lossroom
(17)
The net is nearly zero, as the monthly daytime gains nearly balance the day-long losses.
Solution Absorbedroom = 143.8 [MJ] f¯i = 0.91 Height = 1.25 [m] ¯ T = 0.49 K N etGainroom = 4 [MJ] τ¯beam = 0.72 Uwindow = 3.2 W/m2 -C
αr = 0.96 Frs = 0.38 ¯ = 9.12 MJ/m2 H Lat = 43 [deg] p = 0.5 τ¯dif f = 0.7 w = 12
dec = −12.95 [deg] g = 0.2 ¯ beam = 5.463 MJ/m2 H Lossroom = 140 [MJ] P rojection = 0.625 [m] τ¯gr = 0.7 W idth = 15 [m]
247
2
e=0 Gap = 0.25 [m] ¯ dif f = 3.657 MJ/m2 H month = 2 ¯ beam = 1.782 R ¯ Ta = −7 [C]
Extension = 0 [m] GrRef = 0.2 ¯ dif f /H ¯ = 0.401 H n = 47 Slope = 90 [deg] Troom = 20 [C]
P14 04
Equations 14.4
Calculation of effect of adding night insulation to the window of Problem 14.3
Knowns RN = 1.6;
Uwindow = 3.2;
Troom = 20;
Height = 1.25;
W idth = 15;
T¯a = −7;
fN = 0.5
(1)
From Problem 14.3, the energy absorbed from the room is 144MJ/day: Absorbedroom = 143.8 [MJ] ¯L = (1 − fN ) · Uwindow + fN · U
Lossroom
(2)
Uwindow 1 + Rn · Uwindow
Eqn 14.5.5
(3)
MJ ¯ ¯ = Height · W idth · UL · 24 [hr] · Troom − Ta · 0.0036 W · hr
Eqn 14.5.4
(4)
The net gain is then: N etGainroom = Absorbedroom − Lossroom
(5)
Solution Absorbedroom = 143.8 [MJ] RN = 1.6 C-m2 /W W idth = 15 [m]
fN = 0.5 T¯a = −7 [C]
Height = 1.25 [m] Troom = 20 [C]
248
1
Lossroom = 81.42 [MJ] ¯L = 1.861 W/m2 -C U
N etGainroom = 62.38 [MJ] Uwindow = 3.2 W/m2 -C
P18 01
Equations 18.1
Calculation of energy flows in a basin type still.
$UnitSystem K Pa kJ Make the following assumptions: the vapor pressure of solution in the basin is the same as that of water at the same temperature; the area of cover is the same as area of the basin; and the condensate and cover are at a single uniform temperature Twg. Knowns: Twb = ConvertTemp(C, K, 60);
(1)
σ = sigma#;
(2)
Tsky
Ta = ConvertTemp(C, K, 24) = ConvertTemp(C, K, 20); hw = 10 W/m2 ·K
Radiation from basin to cover, Equation 18.4.3: 4 4 qrad = 0.9 · σ · Twb − Twg
(3)
Convection from basin to cover, Equation 18.4.6: (The 0.1333 converts from Pa to mm Hg.) mmHg Pwb = P (ST EAM , T = Twb , x = 0) · 0.007500615 Pa mmHg Pwg = P (ST EAM , T = Twg , x = 0) · 0.007500615 Pa 1/3 Pwb − Pwg hc0 = 0.884 · (Twb − Twg ) + Twb · 2016 [mmhg] − Pwb qconv = hc0 · (Twb − Twg )
(4) (5) (6) (7)
Heat transfer from basin to cover by evaporation, Equation 18.4.9: hf g = h (ST EAM , T = Twg , x = 1) − h (ST EAM , T = Twg , x = 0) −7
qevap = 9.15 × 10
· hc0 · (Pwb − Pwg ) · hf g · 1000
(8) (9)
Wind convection: qwind = hw · (Twg − Ta )
(10)
Radiation from cover to sky at 20 C, Equation 3.9.1: 4 4 qradSky = 0.88 · σ · Twg − Tsky
(11)
To check the cover temperature, it is necessary that the energy balance on the cover, Equation 18.4.2, be satisfied. qrad + qconv + qevap = qwind + qradSky
(12)
With EES it is as easy to solve for the cover temperature to see if it is the same as that in the problem statement. T C cover = ConvertTemp(K, C, Twg )
(13)
249
1
Solution hc0 = 2.454 W/m2 -K Pwg = 89.5 [mmHg] qradSky = 171 W/m2 Ta = 297.2 [K]
hf g = 2, 384 [kJ/kg] qconv = 26.22 W/m2 qwind = 253.2 W/m2 Tsky = 293.2 [K]
2 hw = 10 W/m -K 2 qevap = 321.1 W/m σ = 5.670 × 10−8 W/m2 -K4 Twb = 333.2 [K]
250
2
Pwb = 149.5 [mmHg] qrad = 76.83 W/m2 T C cover = 49.32 [C] Twg = 322.5 [K]
P19 01
Equations 19.1
Design of a water heating system to provide 75% of a week’s loads by solar energy.
This is a simulation problem, with the basic approach to compute the solar F for a range of collector areas and select that which gives the desired fraction. The solution shown here was done with EES; TRNSYS or other simulation program could have been used. Many of the functions used in this solution are from problem 6.19. Knowns: Tmin = 60 [C] ;
Store/Ac = 60 kg/m2 ;
UL = 4 W/m2 ·C ; tan = 0.77;
LoadF low = 45 · 4/14;
F 0 = 0.95;
Surf AzAng = 0 [deg] ;
β = 40 [deg] ; hr = 1 [hr] ;
m ˙ c = 50 kg/m2 ·hr ; H/Dtank = 3;
Lat = 40 [deg] ; Ncov = 1;
Tmains = 15 [C] ;
T 0 a = 21;
ρg = 0.2;
251
1
αn = 0.9
(1)
Ref rInd = 1.526
(2)
UL,tank = 1.05 W/m2 ·C
(3)
KL = 0.0125
(4)
ρ = 1000 kg/m3 ;
Cp = 4190 [J/kg ·C]
(5)
the last iteration Ac = 7.9 m2
(6)
The initial tank temperature must be entered into the first row of the Parametric Table and the solution must start in row 2. This is a double trial and error solution. It is necessary to guess both Ac and Tplus tank and see if the solar F is 0.75 and if the final tank temperature is equal to the initial tank temperature. (finding the ‘steady periodic’ solution removes the effects of initial tank temperature, which for a simulation as short as a week can have a significant effect on system performance.) The results are Ac =7.9 m2 and initial Ttank =69.1. This system is probably over designed since it supplies a large F in the middle of winter. It would have excess capacity in the summer. Day = Lookup(‘Boulder week’ , T ableRun#, ‘Day’ )
(7)
hour = Lookup(‘Boulder week’ , T ableRun#, ‘hour’ ) MJ I = Lookup(‘Boulder week’ , T ableRun#, ‘I’ ) · 0.001 kj
(8) (9)
Ta = Lookup(‘Boulder week’ , T ableRun#, ‘T a’ ) HrAng = 15 [deg/hr] · (hour − 12 [hr]) − 7.5 [deg]
(10) Need to use the middle of the hour.
(11)
Estimation of incident and absorbed solar radiation, assuming isotropic sky, Equation 2.15.1 and 5.9.1: IT = IT,LJ, (I, Lat, Day, β, Surf AzAng, ρg , HrAng)
(12)
ST = ST,LJ, (I, Lat, day, β, Surf AzAng, ρg , HrAng, Ncov , KL, Ref rInd, αn )
(13)
Warnings are generated for the 17th hour on January 10. The measured horizontal data is suspect since accurate measurements are difficult to make near sunrise and sunset. Collector useful output: m ˙ c · Cp ·4 1/s x= 2.77778 × 10 · 0 UL · F 1/hr
(14)
FR = F 0 · x · (1 − exp (−1/x)) Eqn 6.7.5 Qu = max 0 [MJ] , Ac · FR · ST − UL · (Ttank − Ta ) · hr · 0.0036
(15) MJ W · hr
Eqn 6.7.6
(16)
For the tank: M assT ank = Ac · Store/Ac V olT ank =
(17)
M assT ank ρ (water, T = 55, p = 200)
(18)
Height = H/Dtank · Dia V olT ank = π · Height · Atank = 2 · π ·
(19)
Dia2 4
(20)
Dia2 + π · Dia · Height 4
(21)
252
2
MJ mC tank = Cp · M assT ank · 1 × 10·6 J QtankLoss
(22)
MJ = UL,tank · Atank · hr · (Ttank − T a ) · 0.0036 W · hr 0
(23)
The load: On = If(hour, 7, 0, 1, 1) · If(hour, 21, 1, 0, 0)
Sets load flow to between 7am and 9pm. ·6 MJ M axLoad = Cp · LoadF low · (Tmin − Tmains ) · hr · 1 × 10 J MJ QLoad = On · min M axLoad, Cp · LoadF low · (Ttank − Tmains ) · hr · 1 × 10·6 J
(24) (25)
(26)
Calculation of new tank temperature - at end of an hour’s time step: Ttank = Tablevalue(‘Table 1’ , T ableRun# − 1, ‘Tplus tank’ ) + Ttank = Ttank +
Qu − QtankLoss − QLoad mCtank
(27)
Eqn 10.9.2
(28)
SumQLoad = Tablevalue(T ABLERU N # − 1, ‘SumQ load’ ) + QLoad F =
(29)
SumQLoad 7 [day] · 14 [1/day] · M axLoad
(30)
The results are Ac =7.9 m2 and initial Ttank =69.1. The Parametric Table shows the first 12 and the last 12 hours of the week.
Parametric Table: Table 1 Run 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
0.69 0.69 0.68 0.68 0.67 0.67 0.65 0.62 0.60 0.58 0.56
plus Ttank [C] 69.1 68.8 68.4 68.1 67.7 67.4 65.8 64.3 62.7 61.2 59.7 58.2
SumQLoad [MJ] 0.00 0.00 0.00 0.00 0.00 0.00 2.42 4.85 7.27 9.70 12.12 14.53
-0.00 0.00 0.00 -0.00 -0.00 0.01 0.02 0.03 0.04 0.05 0.06
0.61 0.70 0.76 0.80 0.81 0.78 0.76
69.4 73.9 76.6 77.0 75.4 73.8 72.2
161.66 164.08 166.51 168.93 171.35 173.78 176.20
0.68 0.69 0.70 0.71 0.72 0.73 0.74
Day
hour [hr]
I 2 MJ/m
Ta [C]
IT 2 MJ/m
ST 2 MJ/m
QLoad [MJ]
Qu [MJ]
QtankLoss [MJ]
8 8 8 8 8 8 8 8 8 8 8
2 3 4 5 6 7 8 9 10 11 12
0.000 0.000 0.000 0.000 0.000 0.000 0.017 0.134 0.331 0.636 0.758
-3.3 -2.8 -2.2 -2.8 -2.8 -2.2 -2.2 -1.1 1.1 2.2 2.8
0.00 0.00 0.00 0.00 0.00 0.00 0.02 0.13 0.31 0.62 0.75
0.00 0.00 0.00 0.00 0.00 0.00 0.01 0.09 0.22 0.45 0.55
0.00 0.00 0.00 0.00 0.00 2.42 2.42 2.42 2.42 2.42 2.41
0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
14 14 14 14 14 14 14
13 14 15 16 17 18 19
1.968 1.733 1.331 0.837 0.096 0.004 0.000
6.7 6.7 7.2 6.7 7.2 3.3 0
3.55 3.23 2.71 2.13 0.25 0.00 0.00
2.84 2.57 2.12 1.57 0.15 0.00 0.00
2.42 2.42 2.42 2.42 2.42 2.42 2.42
14.73 12.15 8.46 4.09 0.00 0.00 0.00
253
3
F [-]
21 22 23 24 25
14 14 14 14 14
20 21 22 23 24
0.000 0.000 0.000 0.000 0.000
3.9 -3.9 -3.9 -6.1 -6.7
0.00 0.00 0.00 0.00 0.00
0.00 0.00 0.00 0.00 0.00
254
4
2.42 0.00 0.00 0.00 0.00
0.00 0.00 0.00 0.00 0.00
0.74 0.71 0.71 0.70 0.70
70.6 70.2 69.8 69.5 69.1
178.63 178.63 178.63 178.63 178.63
0.75 0.75 0.75 0.75 0.75
P19 01a
Equations 19.1
Design of a water heating system to provide 75% of a week’s loads by solar energy.
This is a simulation problem, with the basic approach to compute the solar F for a range of collector areas and select that which gives the desired fraction. The solution shown here was done with EES; TRNSYS or other simulation program could have been used. Many of the functions used in this solution are from problem 6.19. Knowns: Tmin = 60 [C] ;
Store/Ac = 60 kg/m2 ;
UL = 4 W/m2 ·C ; tan = 0.77;
LoadF low = 45 · 4/14;
F 0 = 0.95;
Surf AzAng = 0 [deg] ;
β = 40 [deg] ; hr = 1 [hr] ;
m ˙ c = 50 kg/m2 ·hr ; H/Dtank = 3;
Lat = 40 [deg] ; Ncov = 1;
Tmains = 15 [C] ;
T 0 a = 21;
ρg = 0.2;
255
1
αn = 0.9
(1)
Ref rInd = 1.526
(2)
UL,tank = 1.05 W/m2 ·C
(3)
KL = 0.0125
(4)
ρ = 1000 kg/m3 ;
Cp = 4190 [J/kg ·C]
(5)
the last iteration Ac = 7.9 m2
(6)
The initial tank temperature must be entered into the first row of the Parametric Table and the solution must start in row 2. This is a double trial and error solution. It is necessary to guess both Ac and Tplus tank and see if the solar F is 0.75 and if the final tank temperature is equal to the initial tank temperature. (finding the ‘steady periodic’ solution removes the effects of initial tank temperature, which for a simulation as short as a week can have a significant effect on system performance.) The results are Ac =7.9 m2 and initial Ttank =69.1. This system is probably over designed since it supplies a large F in the middle of winter. It would have excess capacity in the summer. Day = Lookup(‘Boulder week’ , T ableRun#, ‘Day’ )
(7)
hour = Lookup(‘Boulder week’ , T ableRun#, ‘hour’ ) MJ I = Lookup(‘Boulder week’ , T ableRun#, ‘I’ ) · 0.001 kj
(8) (9)
Ta = Lookup(‘Boulder week’ , T ableRun#, ‘T a’ ) HrAng = 15 [deg/hr] · (hour − 12 [hr]) − 7.5 [deg]
(10) Need to use the middle of the hour.
(11)
Estimation of incident and absorbed solar radiation, assuming isotropic sky, Equation 2.15.1 and 5.9.1: IT = IT,LJ, (I, Lat, Day, β, Surf AzAng, ρg , HrAng)
(12)
ST = ST,LJ, (I, Lat, day, β, Surf AzAng, ρg , HrAng, Ncov , KL, Ref rInd, αn )
(13)
Warnings are generated for the last hour of sunlight on January 10. The measured horizontal data is suspect since the hourly value is almost equal to that of the preceding hour. Accurate measurements are difficult to make near sunrise and sunset. Collector useful output: m ˙ c · Cp ·4 1/s x= · 2.77778 × 10 0 UL · F 1/hr
(14)
FR = F 0 · x · (1 − exp (−1/x)) Eqn 6.7.5 Qu = max 0 [MJ] , Ac · FR · ST − UL · (Ttank − Ta ) · hr · 0.0036
(15) MJ W · hr
Eqn 6.7.6
(16)
For the tank: M assT ank = Ac · Store/Ac V olT ank =
(17)
M assT ank ρ (water, T = 55, p = 200)
(18)
Height = H/Dtank · Dia V olT ank = π · Height ·
(19)
Dia2 4
(20)
256
2
Dia2 + π · Dia · Height 4 ·6 MJ = Cp · M assT ank · 1 × 10 J
Atank = 2 · π ·
(21)
mC tank
(22)
QtankLoss = UL,tank · Atank · hr · (Ttank − T 0 a ) · 0.0036
MJ W · hr
(23)
The load: On = If(hour, 7, 0, 1, 1) · If(hour, 21, 1, 0, 0)
Sets load flow to between 7am and 9pm. ·6 MJ M axLoad = Cp · LoadF low · (Tmin − Tmains ) · hr · 1 × 10 J ·6 MJ QLoad = On · min M axLoad, Cp · LoadF low · (Ttank − Tmains ) · hr · 1 × 10 J
(24) (25)
(26)
Calculation of new tank temperature - at end of an hour’s time step: Ttank = Tablevalue(‘Table 1’ , T ableRun# − 1, ‘Tplus tank’ ) + Ttank = Ttank +
Qu − QtankLoss − QLoad mCtank
(27)
Eqn 10.9.2
(28)
SumQLoad = Tablevalue(T ABLERU N # − 1, ‘SumQ load’ ) + QLoad F =
(29)
SumQLoad 7 [day] · 14 [1/day] · M axLoad
(30)
The results are Ac =7.9 m2 and initial Ttank =69.1.
Solution αn = 0.9 Cp = 4, 190 [J/kg-C] F 0 = 0.95 hr = 1 [hr] IT = 0.00 MJ/m2 M assT ank = 474 [kg] Ncov = 1 Qu = 0.00 [MJ] Store/Ac = 60 kg/m2 tan = 0.77 Ttank = 69.49 [C] UL,tank = 1.05 W/m2 -C
Ac = 7.9 m2 Day = 14 FR = 0.9196 HrAng = 172.5 [deg] KL = 0.0125 M axLoad = 2.424 [MJ] On = 0 [-] Ref rInd = 1.526 SumQLoad = 178.63 [MJ] Ta = −6.7 [C] T 0 a = 21 [C] V olT ank = 0.4809 m3
Atank = 3.811 m2 Dia = 0.5888 [m] Height = 1.766 [m] H/Dtank = 3 Lat = 40 [deg] mC tank = 1.986 [MJ/C] QLoad = 0.00 [MJ] ρ = 1, 000 kg/m3 Surf AzAng = 0 [deg] Tmains = 15 [C] + Ttank = 69.1 [C] x = 15.31
β = 40 [deg] F = 0.75 [-] hour = 24[hr] I = 0.000 MJ/m2 LoadF low = 12.86 [kg/hr] m ˙ c = 50 kg/m2 -hr QtankLoss = 0.70 [MJ] ρg = 0.2 ST = 0.00 MJ/m2 Tmin = 60 [C] UL = 4 W/m2 -C
Parametric Table: Table 1 Run
Day
hour [hr]
I 2 MJ/m
Ta [C]
IT 2 MJ/m
ST 2 MJ/m
QLoad [MJ]
Qu [MJ]
QtankLoss [MJ]
1 2
8
2
0.000
-3.3
0.00
0.00
0.00
0.00
0.69
257
3
plus Ttank [C] 69.1 68.8
SumQLoad [MJ] 0.00 0.00
F [-] -0.00
3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56
8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10
3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 1 2 3 4 5 6 7 8
0.000 0.000 0.000 0.000 0.000 0.017 0.134 0.331 0.636 0.758 1.105 1.252 0.641 0.167 0.046 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.004 0.071 0.155 0.343 0.402 1.185 1.009 0.796 0.389 0.134 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.033
-2.8 -2.2 -2.8 -2.8 -2.2 -2.2 -1.1 1.1 2.2 2.8 2.8 3.8 3.3 2.2 0.6 -0.6 -1.1 -1.7 -1.7 -2.2 -2.2 -2.2 -2.8 -3.3 -3.3 -3.3 -3.9 -3.9 -3.9 -3.9 -3.9 -3.3 -2.8 -2.2 -2.2 -1.3 -0.6 -0.6 -2.2 -2.8 -3.3 -5.6 -6.7 -7.8 -8.3 -8.3 -9.4 -10 -8.9 -10.6 -8.3 -8.3 -10 -8.9
0.00 0.00 0.00 0.00 0.00 0.02 0.13 0.31 0.62 0.75 1.36 1.98 0.72 0.16 0.06 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.07 0.14 0.32 0.37 1.55 1.28 1.09 0.49 1.27 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.04
0.00 0.00 0.00 0.00 0.00 0.01 0.09 0.22 0.45 0.55 1.04 1.55 0.54 0.11 0.04 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.05 0.10 0.23 0.27 1.19 0.98 0.83 0.36 0.72 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.02
258
4
0.00 0.00 0.00 0.00 2.42 2.42 2.42 2.42 2.42 2.41 2.33 2.30 2.38 2.30 2.22 2.15 2.07 2.01 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 1.83 1.76 1.71 1.65 1.60 1.54 1.49 1.55 1.57 1.55 1.50 1.46 1.42 1.37 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 1.25 1.21
0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 1.78 5.66 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 3.97 2.41 1.32 0.00 0.52 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
0.69 0.68 0.68 0.67 0.67 0.65 0.62 0.60 0.58 0.56 0.54 0.53 0.55 0.53 0.51 0.49 0.47 0.45 0.43 0.43 0.43 0.42 0.42 0.42 0.41 0.41 0.41 0.40 0.40 0.39 0.37 0.35 0.34 0.33 0.31 0.33 0.33 0.33 0.31 0.31 0.29 0.28 0.27 0.27 0.26 0.26 0.26 0.26 0.26 0.25 0.25 0.25 0.25 0.24
68.4 68.1 67.7 67.4 65.8 64.3 62.7 61.2 59.7 58.2 57.7 59.1 57.6 56.2 54.8 53.5 52.2 51.0 50.8 50.5 50.3 50.1 49.9 49.7 49.5 49.3 49.1 48.9 47.8 46.7 45.6 44.6 43.6 42.7 43.8 44.1 43.8 42.8 42.2 41.3 40.4 39.6 39.5 39.3 39.2 39.1 38.9 38.8 38.7 38.5 38.4 38.3 37.5 36.8
0.00 0.00 0.00 0.00 2.42 4.85 7.27 9.70 12.12 14.53 16.86 19.16 21.53 23.83 26.05 28.19 30.27 32.27 32.27 32.27 32.27 32.27 32.27 32.27 32.27 32.27 32.27 32.27 34.10 35.86 37.57 39.22 40.81 42.36 43.85 45.40 46.96 48.51 50.01 51.48 52.89 54.26 54.26 54.26 54.26 54.26 54.26 54.26 54.26 54.26 54.26 54.26 55.52 56.73
0.00 0.00 -0.00 -0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.14 0.14 0.14 0.14 0.14 0.14 0.14 0.14 0.14 0.14 0.14 0.14 0.15 0.16 0.17 0.17 0.18 0.18 0.19 0.20 0.20 0.21 0.22 0.22 0.23 0.23 0.23 0.23 0.23 0.23 0.23 0.23 0.23 0.23 0.23 0.23 0.24
57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110
10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 12 12 12 12 12 12 12 12 12 12 12 12 12 12
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 1 2 3 4 5 6 7 8 9 10 11 12 13 14
0.419 1.047 1.570 1.805 1.872 1.733 1.352 0.775 0.205 0.004 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.004 0.042 0.092 0.138 0.163 0.138 0.096 0.084 0.042 0.004 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.017 0.071 0.180 0.247 0.331 0.389 0.477
-7.2 -5 -2.2 -1.1 2.2 4.4 6.1 6.7 6.1 3.3 0.6 0.6 0 0.6 1.7 0.6 -1.7 -2.2 -2.2 -2.8 -4.4 -5 -5.6 -6.1 -5.6 -5.6 -5.6 -5.6 -5 -3.9 -4.4 -3.9 -5 -5.6 -6.7 -7.8 -9.4 -8.9 -9.4 -11.1 -11.7 -12.8 -15.6 -16.7 -16.7 -16.1 -17.2 -17.8 -13.3 -11.1 -7.8 -5.6 -2.2 -0.6
0.57 1.90 2.94 3.29 3.42 3.28 2.81 2.04 2.02 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.04 0.08 0.13 0.15 0.13 0.09 0.08 0.04 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.02 0.07 0.17 0.23 0.30 0.36 0.44
0.41 1.48 2.34 2.64 2.74 2.61 2.20 1.50 1.15 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.03 0.06 0.09 0.11 0.09 0.06 0.06 0.03 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.01 0.05 0.12 0.17 0.22 0.26 0.32
259
5
1.17 1.14 1.28 1.58 1.92 2.26 2.42 2.42 2.42 2.42 2.42 2.42 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 2.35 2.28 2.20 2.13 2.06 1.99 1.92 1.86 1.80 1.74 1.68 1.62 1.57 1.52 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 1.39 1.34 1.30 1.26 1.22 1.18 1.14 1.10
0.00 6.47 12.71 14.41 14.87 13.51 10.11 4.72 2.01 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
0.23 0.22 0.25 0.34 0.43 0.52 0.60 0.65 0.66 0.65 0.63 0.61 0.58 0.58 0.58 0.57 0.57 0.56 0.56 0.56 0.55 0.55 0.54 0.52 0.50 0.48 0.46 0.45 0.43 0.41 0.39 0.38 0.36 0.35 0.33 0.32 0.31 0.30 0.30 0.30 0.30 0.30 0.29 0.29 0.29 0.29 0.28 0.27 0.26 0.25 0.24 0.23 0.22 0.21
36.1 38.7 44.3 50.6 56.9 62.3 65.9 66.7 66.2 64.6 63.1 61.6 61.3 61.0 60.7 60.4 60.1 59.8 59.5 59.3 59.0 58.7 57.2 55.8 54.5 53.2 51.9 50.7 49.5 48.3 47.2 46.2 45.1 44.2 43.2 42.3 42.1 42.0 41.8 41.7 41.5 41.4 41.2 41.1 40.9 40.8 39.9 39.1 38.3 37.6 36.8 36.1 35.5 34.8
57.91 59.04 60.32 61.89 63.81 66.07 68.49 70.92 73.34 75.77 78.19 80.61 80.61 80.61 80.61 80.61 80.61 80.61 80.61 80.61 80.61 80.61 82.97 85.24 87.44 89.57 91.63 93.61 95.54 97.39 99.19 100.93 102.61 104.23 105.80 107.32 107.32 107.32 107.32 107.32 107.32 107.32 107.32 107.32 107.32 107.32 108.71 110.05 111.35 112.61 113.83 115.00 116.14 117.24
0.24 0.25 0.25 0.26 0.27 0.28 0.29 0.30 0.31 0.32 0.33 0.34 0.34 0.34 0.34 0.34 0.34 0.34 0.34 0.34 0.34 0.34 0.35 0.36 0.37 0.38 0.39 0.39 0.40 0.41 0.42 0.42 0.43 0.44 0.45 0.45 0.45 0.45 0.45 0.45 0.45 0.45 0.45 0.45 0.45 0.45 0.46 0.46 0.47 0.47 0.48 0.48 0.49 0.49
111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164
12 12 12 12 12 12 12 12 12 12 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14
15 16 17 18 19 20 21 22 23 24 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
0.532 0.461 0.033 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.017 0.314 0.724 1.809 2.299 1.926 1.750 1.340 0.703 0.059 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.038 0.452 1.110 1.608 1.884 1.968 1.733 1.331 0.837 0.096 0.004 0.000 0.000
2.8 -0.6 -1.7 -4.4 -7.8 -7.8 -8.9 -10.6 -12.8 -11.7 -10.6 -10.6 -10 -11.1 -10.6 -9.4 -7.2 -10.6 -8.3 -1.7 1.7 3.3 5.6 7.2 8.3 8.9 6.7 4.4 1.1 0 -2.2 2.8 1.7 1.7 -0.6 -1.1 -0.6 -3.9 -1.7 -2.8 -2.8 -5 -5 -1.7 2.8 3.8 6.7 6.7 7.2 6.7 7.2 3.3 0 3.9
0.53 0.68 0.03 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.02 0.32 0.87 3.38 4.16 3.48 3.27 2.74 1.71 0.07 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.04 0.62 2.04 2.96 3.39 3.55 3.23 2.71 2.13 0.25 0.00 0.00 0.00
0.39 0.49 0.02 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.01 0.24 0.66 2.70 3.34 2.79 2.61 2.15 1.26 0.05 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.03 0.46 1.59 2.36 2.72 2.84 2.57 2.12 1.57 0.15 0.00 0.00 0.00
260
6
1.07 1.03 1.00 0.97 0.94 0.91 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.84 0.81 0.79 0.76 0.78 1.21 1.73 2.11 2.42 2.42 2.42 2.42 2.42 2.38 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 2.16 2.09 2.02 1.95 2.05 2.31 2.42 2.42 2.42 2.42 2.42 2.42 2.42 2.42
0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 1.53 16.70 20.66 15.96 14.06 10.23 3.43 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 6.05 11.89 14.14 14.73 12.15 8.46 4.09 0.00 0.00 0.00 0.00
0.20 0.19 0.18 0.17 0.16 0.16 0.15 0.15 0.15 0.14 0.14 0.14 0.14 0.14 0.14 0.14 0.14 0.13 0.12 0.12 0.12 0.24 0.38 0.48 0.56 0.61 0.62 0.59 0.57 0.55 0.53 0.52 0.52 0.52 0.51 0.51 0.51 0.50 0.50 0.49 0.49 0.47 0.45 0.44 0.46 0.53 0.61 0.70 0.76 0.80 0.81 0.78 0.76 0.74
34.2 33.5 32.9 32.4 31.8 31.3 31.2 31.1 31.1 31.0 30.9 30.8 30.8 30.7 30.6 30.6 30.1 29.6 29.1 29.5 37.4 47.1 54.1 59.9 63.5 63.7 62.2 60.7 59.1 57.7 57.4 57.1 56.9 56.6 56.4 56.1 55.8 55.6 55.3 55.1 53.8 52.5 51.2 53.1 57.8 63.5 69.4 73.9 76.6 77.0 75.4 73.8 72.2 70.6
118.31 119.34 120.34 121.31 122.24 123.15 123.15 123.15 123.15 123.15 123.15 123.15 123.15 123.15 123.15 123.15 123.99 124.80 125.59 126.35 127.13 128.33 130.06 132.17 134.58 137.01 139.43 141.86 144.28 146.66 146.66 146.66 146.66 146.66 146.66 146.66 146.66 146.66 146.66 146.66 148.82 150.91 152.93 154.88 156.93 159.23 161.66 164.08 166.51 168.93 171.35 173.78 176.20 178.63
0.50 0.50 0.51 0.51 0.51 0.52 0.52 0.52 0.52 0.52 0.52 0.52 0.52 0.52 0.52 0.52 0.52 0.53 0.53 0.53 0.54 0.54 0.55 0.56 0.57 0.58 0.59 0.60 0.61 0.62 0.62 0.62 0.62 0.62 0.62 0.62 0.62 0.62 0.62 0.62 0.63 0.64 0.64 0.65 0.66 0.67 0.68 0.69 0.70 0.71 0.72 0.73 0.74 0.75
165 166 167 168
14 14 14 14
21 22 23 24
0.000 0.000 0.000 0.000
-3.9 -3.9 -6.1 -6.7
0.00 0.00 0.00 0.00
0.00 0.00 0.00 0.00
261
7
0.00 0.00 0.00 0.00
0.00 0.00 0.00 0.00
0.71 0.71 0.70 0.70
70.2 69.8 69.5 69.1
178.63 178.63 178.63 178.63
0.75 0.75 0.75 0.75
P19 02
Equations 19.2
Add a collector-store HX to problem 19.1.
Knowns: Tmin = 60 [C] ;
Store/Ac = 60 kg/m2 ;
UL = 4 W/m2 ·C ; tan = 0.77;
LoadF low = 45 · 4/14;
F 0 = 0.95;
Surf AzAng = 0 [deg] ; ρ = 1000 kg/m3 ;
β = 40 [deg] ; hr = 1 [hr] ;
m ˙ c = 50 kg/m2 ·hr ; H/Dtank = 3;
Lat = 40 [deg] ; Ncov = 1;
Tmains = 15 [C] ;
T 0 a = 21;
ρg = 0.2;
αn = 0.9
(1)
Ref rInd = 1.526
(2)
UL,tank = 1.05 W/m2 ·C
(3)
KL = 0.0125
Cp = 4190 [J/kg ·C]
(4) (5)
the last iteration Ac = 8.8 m2
(6)
Tinitial = Tablevalue(1, ‘Tplus tank’ )
(7)
The initial tank temperature must be entered into the first row of the Parametric Table and the solution must start in row 2. This is a double trial and error solution. It is necessary to guess both Ac and Tplus tank and see if the solar F is 0.75 and if the final tank temperature is equal to the initial tank temperature. (finding the ‘steady periodic’ solution removes the effects of initial tank temperature, which for a simulation as short as a week can have a significant effect on system performance.) The results are Ac =7.9 m2 and initial Ttank =69.1. This system is probably over designed since it supplies a large F in the middle of winter. It would have excess capacity in the summer. Day = Lookup(‘Boulder week’ , T ableRun#, ‘Day’ )
(8)
hour = Lookup(‘Boulder week’ , T ableRun#, ‘hour’ ) MJ I = Lookup(‘Boulder week’ , T ableRun#, ‘I’ ) · 0.001 kj
(9)
Ta = Lookup(‘Boulder week’ , T ableRun#, ‘T a’ ) HrAng = 15 [deg/hr] · (hour − 12 [hr]) − 7.5 [deg]
(10) (11)
Need to use the middle of the hour.
(12)
Estimation of incident and absorbed solar radiation, assuming isotropic sky, Equation 2.15.1 and 5.9.1: IT = IT,LJ, (I, Lat, Day, β, Surf AzAng, ρg , HrAng)
(13)
ST = ST,LJ, (I, Lat, day, β, Surf AzAng, ρg , HrAng, Ncov , KL, Ref rInd, αn )
(14)
262
1
Warnings are generated for the 17th hour on January 10. The measured horizontal data is suspect since accurate measurements are difficult to make near sunrise and sunset. Collector useful output: m ˙ c · Cp ·4 1/s 2.77778 × 10 · x= 0 UL · F 1/hr FR = F 0 · x · (1 − exp (−1/x))
(15)
Eqn 6.7.5
(16)
The following four equations were added and the fifth changed from problem 19.1 1/s ˙ = Ac · m AC ˙ c · 2.77778 × 10·4 1/hr
(17)
F RU L = FR · UL
(18)
HX = 0.35
(19)
˙ cp , AC, ˙ cp , HX F 0 R = FR · F R0 /F R Ac , F RU L, AC,
Qu = max 0 [MJ] , Ac · F R · ST − UL · (Ttank − Ta ) · hr · 0.0036
0
(20) MJ W · hr
Eqn 6.7.6
(21)
For the tank: M assT ank = Ac · Store/Ac V olT ank =
(22)
M assT ank ρ (water, T = 55, p = 200)
(23)
Height = H/Dtank · Dia V olT ank = π · Height ·
(24)
Dia2 4
(25)
Dia2 + π · Dia · Height 4 MJ = Cp · M assT ank · 1 × 10·6 J
Atank = 2 · π ·
(26)
mC tank
(27)
QtankLoss
MJ = UL,tank · Atank · hr · (Ttank − T a ) · 0.0036 W · hr 0
(28)
The load: On = If(hour, 7, 0, 1, 1) · If(hour, 21, 1, 0, 0)
Sets load flow to between 7am and 9pm. ·6 MJ M axLoad = Cp · LoadF low · (Tmin − Tmains ) · hr · 1 × 10 J MJ QLoad = On · min M axLoad, Cp · LoadF low · (Ttank − Tmains ) · hr · 1 × 10·6 J
(29) (30)
(31)
Calculation of new tank temperature - at end of an hour’s time step: Ttank = Tablevalue(‘Table 1’ , T ableRun# − 1, ‘Tplus tank’ )
263
2
(32)
+ Ttank = Ttank +
Qu − QtankLoss − QLoad mCtank
Eqn 10.9.2
(33)
SumQLoad = Tablevalue(T ABLERU N # − 1, ‘SumQ load’ ) + QLoad F =
(34)
SumQLoad 7 [day] · 14 [1/day] · M axLoad
(35)
The results are a) epsilonHX =0.78 Ac =7.9 m2 and initial Ttank =69.1. a) epsilonHX =0.35 Ac =8.7 m2 and initial Ttank =66.8. When the HX is large the impact is small but for the small HX there is an 11% incresas in collector area.
Solution ˙ = 0.1222 [kg/s] AC β = 40 [deg] HX = 0.35 FR = 0.9196 hr = 1 [hr] IT = 0.00 MJ/m2 M assT ank = 528 [kg] Ncov = 1 Qu = 0.00 [MJ] Store/Ac = 60 kg/m2 tan = 0.77 Tmin = 60 [C] UL = 4 W/m2 -C
Ac = 8.8 m2 Day = 14 F RU L = 3.679 W/m2 -C Height = 1.831 [m] H/Dtank = 3 Lat = 40 [deg] mC tank = 2.212 [MJ/C] QLoad = 0.00 [MJ] ρ = 1, 000 kg/m3 Surf AzAng = 0 [deg] Tinitial = 66.8 [C] T 0 a = 21 [C] V olT ank = 0.5356 m3
αn = 0.9 Cp = 4, 190 [J/kg-C] F = 0.75 [-] F 0 R = 0.823 HrAng = 172.5 [deg] KL = 0.0125 M axLoad = 2.424 [MJ] On = 0 [-] Ref rInd = 1.526 SumQLoad = 177.59 [MJ] Ta = −6.7 [C] Ttank = 67.17 [C] UL,tank = 1.05 W/m2 -C
264
3
Atank = 4.096 m2 Dia = 0.6103 [m] F 0 = 0.95 [-] hour = 24[hr] I = 0.000 MJ/m2 LoadF low = 12.86 [kg/hr] m ˙ c = 50 kg/m2 -hr QtankLoss = 0.71 [MJ] ρg = 0.2 ST = 0.00 MJ/m2 Tmains = 15 [C] + Ttank = 66.8 [C] x = 15.31
P20 01
Equations 20.1
Calculation of a months performance of a liquid-based heating system with f-Chart.
Knowns: U Ahouse = 145 [W/C] ; T¯a = −8;
Days = 31 [day] ;
F 0 RU L = 3.80; a
Lhw = 2.06 · 109
DegDays = 828 [C ·day] ; Ac = 22;
F 0 Rtan = 0.79;
¯ T = 10.7 · 106 ; H
¯ ta/ta n = .96
Vstore = 1650 [l]
(1) (2) (3)
Load calculation with space heating estimated with Equation 9.3.6:
Lhtg = U Ahouse · DegDays · 24 [hr/day] · 3600 [s/hr]
(4)
Load = Lhtg + Lhw
(5)
b
Calculation of X and Y by Equations 20.2.3 and 20.2.4:
The store size is 1650/22 = 75 liter/m2, so no X correction is required. The load heat exchanger parameter is 2.00, so no Y correction is needed. s 86400 day X = F 0 RU L · 100 [C] − T¯a · Days · Ac · (6) Load ¯ ¯ Y = F 0 Rtan · ta/ta n · HT · Days · Ac /Load c
(7)
Calculation of solar fraction with Equation 20.3.1 (or Figure 20.3.1):
fc = 1.029 · Y − .065 · X − .245 · Y 2 + .0018 · X 2 + .0215 · Y 3
265
1
(8)
d
Recalculation of X and fraction for reduced store size:
Correction for non-standard storage, Equation 20.3.2: Storeactual = 75/2
(9) −.25
Xd = X · (Storeactual /75)
(10)
fd = 1.029 · Y − .065 · Xd − .245 · Y 2 + .0018 · Xd2 + .0215 · Y 3
(11)
e
Recalculation of Y and fraction for reduced load heat exchanger size:
Correction for non-standard load heat exchanger, Equation 20.3.3: LHX parameter = 0.5
(12)
Y e/Y = .39 + .65 · exp (−.139/LHX parameter )
(13)
Ye = Y · Y e/Y
(14)
fe = 1.029 · Ye − .065 · Xd − .245 · Ye2 + .0018 · Xd2 + .0215 · Ye3
(15)
Solution Ac = 22 m2 fd = 0.2707 ¯ T = 1.070 × 107 J/m2 -day H Lhw = 2.060 × 109 [J] U Ahouse = 145 [W/C] Y = 0.4451 [-]
Days = 31 [day] fe = 0.2269 LHX parameter = 0.5 Storeactual = 37.5 Vstore = 1, 650 [l] Y e/Y = 0.8822
DegDays = 828 [C-day] F 0 Rtan = 0.79 [-] Load = 1.243 × 1010 [J] ¯ ta/ta n = 0.96 X = 1.945 [-] Ye = 0.3927
266
2
fc = 0.2918 F 0 RU L = 3.8 W/m2 -c Lhtg = 1.037 × 1010 [J] T¯a = −8 [C] Xd = 2.313
P20 02
Equations 20.2
Liquid solar system f-Chart calculations for one month.
Knowns: U Ahouse = 350 [W/C] ; Daysmonth = 30;
DD = 553 [C ·day] ;
T¯ = −2 [C] ;
Ac = 100 m2
F 0 RU L = 7.0 W/m2 ·C
Storeactual = 180 l/m2 F 0 Rtan = 0.78;
¯ ta/ta n = 0.95;
(1) (2) (3)
¯ T = 15 MJ/m2 H
Calculation of month’s loads by degree-day method, Equation 9.3.6: s Load = U Ahouse · DD · 86400 day
(4)
(5)
Calculation of X, assuming standard storage capacity, with Equation 20.2.3: ∆t = Daysmonth · 24 [hr] · 3600 [s/hr]
(6)
X = F 0 RU L · 100 [C] − T¯ · ∆t · Ac /Load
(7)
Correction for non-standard storage, Equation 20.3.2: −.25 Storeactual Xc/X = 75 [l/m2 ]
(8)
267
1
a) value of X Xa = X · Xc/X
(9)
b) value of Y using Equation 20.2.4: J ¯ ¯ 1000000 · Daysmonth · Ac /Load Yb = F 0 Rtan · ta/ta · H · n T MJ
(10)
c) Solar fraction for Xa and Yb fc = 1.029 · Yb − .065 · Xa − .245 · Yb2 + .0018 · Xa2 + .0215 · Yb3
(11)
d) Solar fraction with 1/2 the area both X and Y are reduced by 1/2. 2
2
3
fd = 1.029 · Yb /2 − .065 · Xa /2 − .245 · (Yb /2) + .0018 · (Xa /2) + .0215 · (Yb /2)
Solution Ac = 100 m2 fc = 0.81 ¯ T = 15 MJ/m2 H T¯ = −2 [C] Xa = 8.891
Daysmonth = 30 [-] fd = 0.55 Load = 1.672 × 1010 [J] U Ahouse = 350 [W/C] Yb = 1.994
DD = 553 [C-day] F 0 Rtan = 0.78 Storeactual = 180 l/m2 X = 11.07 [-]
268
2
∆t = 2.592 × 106 [s] F 0 RU L = 7 W/m2 -C ¯ ta/ta n = 0.95 Xc/X = 0.8034
(12)
P20 03
Equations 20.3
a. Calculation of area required for an air system to meet 50% of January load.
Knowns: U A = 400 [W/C] ; F Rtan = 0.6;
Store/Ac = 0.25;
F ractAlow = 10;
F RU L = 4.0 W/m2 ·C ;
ta/tan = 0.96;
Slope = 50 [deg] ;
DHW = 2.15/31
Tref = 100 [C]
For the month of January - data from App G and Example 2.19.1: ¯ T = 11.9 MJ/m2 ·day ; month = 1; H T¯a = −8 [C] ; DD = 828 [C ·day]
(1) (2)
821
(3)
nDays = N umDay (M onth)
(4)
∆t = 3600 [s/hr] · 24 [hr/day] · nDays
(5)
Determine the Load, X and Y. MJ/day Load = DD · U A · 0.0864 + DHW · nDays Eqn 9.3.6 J/s ·6 MJ × 10 1 j X = F RU L · Tref − T¯a · ∆t · Ac · Eqn 20.2.3 Load ¯ T · Ac · nDays/Load Eqn 20.2.4 Y = F Rtan · ta/ta · H n
(6)
(7) (8)
Equation 20.4.1 represents the air f-Chart, and must be used in the range of values shown in Figure 20.4.1. f = 1.040 · Y − 0.065 · X − 0.159 · Y 2 + 0.00187 · X 2 − 0.0095 · Y 3
(9)
Set f=0.50 and let EES solve for Ac : f=0.50 f = 0.5
(10)
Solution
269
1
Ac = 113.8 m2 f = 0.5 ¯ T = 11.9 MJ/m2 -day H Slope = 50 [deg] Tref = 100 [C]
DD = 828 [C-day] F ractAlow = 10 Load = 28, 618 [MJ] Store/Ac = 0.25 U A = 400 [W/C]
∆t = 2.678 × 106 [s] F Rtan = 0.6 month = 1 ta/tan = 0.96 X = 4.601 [-]
270
2
DHW = 0.06935 [MJ/day] F RU L = 4 W/m2 -C nDays = 31 [day] T¯a = −8 [C] Y = 0.8449 [-]
P20 03a
Equations 20.3
a. Calculation of area required for an air system to meet 50% of a January load.
Knowns: U A = 400 [W/C] ; F Rtan = 0.6;
Store/Ac = 0.25;
F ractAlow = 10;
F RU L = 4.0 W/m2 ·C ;
ta/tan = 0.96;
Slope = 50 [deg] ;
DHW = 2150
Tref = 100 [C]
[MJ] 31 [day]
(1) (2)
For the month of January - data from App G and Example 2.19.1: ¯ T = 13.7 MJ/m2 ·day ; month = 1; H T¯a = −8 [C] ; DD = 821 [C ·day]
(3)
nDays = N umDay (M onth)
(4)
∆t = 3600 [s/hr] · 24 [hr/day] · nDays
(5)
Determine the Load, X and Y. MJ/day House = DD · U A · 0.0864 J/s
(6)
HW = DHW · nDays MJ/day Load = DD · U A · 0.0864 + DHW · nDays Eqn 9.3.6 J/s ·6 MJ × 10 1 j X = F RU L · Tref − T¯a · ∆t · Ac · Eqn 20.2.3 Load ¯ T · Ac · nDays/Load Eqn 20.2.4 Y = F Rtan · ta/ta · H n
(7) (8)
(9) (10)
Equation 20.4.1 represents the air f-Chart, and must be used in the range of values shown in Figure 20.4.1. f = 1.040 · Y − 0.065 · X − 0.159 · Y 2 + 0.00187 · X 2 − 0.0095 · Y 3
(11)
Set f=0.50 and let EES solve for Ac : f = 0.5
(12)
271
1
Solution Ac = 98.36 m2 f = 0.5 House = 28, 374 [MJ] month = 1 ta/tan = 0.96 X = 3.728 [-]
DD = 821 [C-day] F ractAlow = 10 HW = 2, 150 [MJ] nDays = 31 [day] T¯a = −8 [C] Y = 0.7883 [-]
∆t = 2.678 × 106 [s] F Rtan = 0.6 ¯ T = 13.7 MJ/m2 -day H Slope = 50 [deg] Tref = 100 [C]
272
2
DHW = 69.35 [MJ/day] F RU L = 4 W/m2 -C Load = 30, 524 [MJ] Store/Ac = 0.25 U A = 400 [W/C]
P20 03b
Equations 20.3
b. Determine the annual solar fraction for a collector area of 98.36 m2.
$Arrays On Knowns: Ac = 98.36 m2
(1)
U A = 400 [W/C] ;
F Rtan = 0.6;
Store/Ac = 0.25;
F ractAlow = 10;
F RU L = 4.0 W/m2 ·C ;
ta/tan = 0.96;
Slope = 50 [deg] ;
DHW = 2150
[MJ] 31 [day]
(2)
Tref = 100 [C]
(3)
¯ T,1..12 = [13.7, 17.2, 15.8, 14.7, 16.6, 16.5, 16.8, 17.5, 15.6, 15.2, 11.4, 12.7] H
(4)
DD1..12 = [821, 656, 537, 280, 136, 30, 1, 25, 103, 232, 479, 699]
(5)
T¯a,1..12 = [-8, -5, 1, 9, 14, 19, 22, 20, 15, 11, 2, -5]
(6)
duplicate i = 1, 12
(7)
Weather data from Appendix G and Example 2.19.1
monthi = i
(8)
∆ti = 3600 [s/hr] · 24 [hr/day] · nDaysi
(9)
nDaysi = N umDay (M onthi )
(10)
Determine the Load, X and Y. MJ/day Housei = DDi · U A · 0.0864 J/s
(11)
273
1
HW i = DHW · nDaysi MJ/day + DHW · nDaysi Eqn 9.3.6 Loadi = DDi · U A · 0.0864 J/s ·6 MJ × 10 1 j Eqn 20.2.3 Xi = min 15, F RU L · Tref − T¯a,i · ∆ti · Ac · Loadi
Xi ¯ T,i · Ac · nDaysi + 1.2, F Rtan · ta/tan · H Yi = min 1.5 · 16 Loadi
(12) (13)
(14)
Eqn 20.2.4
(15)
Equation 20.4.1 represents the air f-Chart, and must be used in the range of values shown in Figure 20.4.1. With the limits shown the solar fraction never exceeds 0.91 but this high solar fraction is for months with very small loads so the annual solar fractions is not significantly impacted. fi = 1.040 · Yi − 0.065 · Xi − 0.159 · Yi2 + 0.00187 · Xi2 − 0.0095 · Yi3
(16)
end
(17)
The annual solar fraction is the load weighted average monthly solar fraction. F =
Sum(fi · Loadi , i = 1, 12) Sum(Loadi , i = 1, 12)
(18)
Solution Ac = 98.36 m2 F Rtan = 0.6 ta/tan = 0.96
DHW = 69.35 [MJ/day] F RU L = 4 W/m2 -C Tref = 100 [C]
F = 0.699 Slope = 50 [deg] U A = 400 [W/C]
F ractAlow = 10 Store/Ac = 0.25
Arrays Table: Main Row
monthi
1 2 3 4 5 6 7 8 9 10 11 12
1 2 3 4 5 6 7 8 9 10 11 12
nDaysi [day] 31 28 31 30 31 30 31 31 30 31 30 31
DDi [C-day] 821 656 537 280 136 30 1 25 103 232 479 699
¯ T,i H MJ/m2 -day 13.7 17.2 15.8 14.7 16.6 16.5 16.8 17.5 15.6 15.2 11.4 12.7
∆ti [s] 2.678E6 2.419E6 2.678E6 2.592E6 2.678E6 2.592E6 2.678E6 2.678E6 2.592E6 2.678E6 2.592E6 2.678E6
274
2
Housei [MJ] 28,374 22,671 18,559 9,677 4,700 1,037 35 864 3,560 8,018 16,554 24,157
HW i [MJ] 2,150 1,942 2,150 2,081 2,150 2,081 2,150 2,150 2,081 2,150 2,081 2,150
Loadi [MJ] 30,524 24,613 20,709 11,757 6,850 3,117 2,185 3,014 5,640 10,168 18,635 26,307
T¯a,i [C] -8 -5 1 9 14 19 22 20 15 11 2 -5
Xi [-] 3.73 4.06 5.04 7.89 13.23 15.00 15.00 15.00 15.00 9.22 5.36 4.21
Yi [-] 0.79 1.11 1.34 1.94 2.44 2.61 2.61 2.61 2.61 2.06 1.04 0.85
fi [-] 0.50 0.71 0.81 0.95 0.92 0.91 0.91 0.91 0.91 0.95 0.60 0.52
P20 04
Equations 20.4
Calculation of annual performance for an air system.
$UnitSystem SI C J kg KPa Knowns: Lat = 43 [deg] ; F Rtan = 0.76;
Slope = 60 [deg] ;
Ac = 40 m2 ;
azim = 0 [deg]
(1)
F RU L = 4.85 W/m2 ·C
(2)
V olF lowAir = 12 l/m2 ·s ;
V olF lowAirStd = 10 l/m2 ·s ;
nCov = 2;
Ref rInd = 1.526;
KL = 0.0125;
StorageV olStd = 0.25 m3 /m2 ;
αn = .9;
StorageV ol = 0.15 m3 /m2 Tref = 100 [C]
good quality glass
U Ah = 300 [W/C]
(3) (4) (5)
Weather data from Appendix G and Problem 2.18 DD1..12 = [821, 656, 537, 280, 136, 30, 1, 25, 103, 232, 479, 699]
(6)
T¯a,1..12 = [-8, -5, 1, 9, 14, 19, 22, 20, 15, 11, 2, -5]
(7)
¯ 1 = 6.44; H
¯ 2 = 9.89; H
¯ 7 = 22.58; H
¯ 3 = 12.86; H
¯ 8 = 20.33; H
¯ 4 = 16.05; H
¯ 9 = 14.59; H
¯ 5 = 21.36; H
¯ 10 = 10.48; H
ρg,1 = 0.7;
ρg,2 = 0.7;
ρg,3 = 0.4;
ρg,4 = 0.2;
ρg,7 = 0.2;
ρg,8 = 0.2;
ρg,9 = 0.2;
ρg,10 = 0.2;
¯ 6 = 23.04 H
¯ 11 = 6.37; H
ρg,5 = 0.2; ρg,11 = 0.4;
¯ 12 = 5.74 H
ρg,6 = 0.2 ρg,12 = 0.7
(8) (9) (10) (11)
It is first necessary to estimate monthly average incident and absorbed radiation on sloped surfaces. These two quantities are then used to calcualte the monthly average τ *α product. This solution uses the Liu and Jordan mode Eqn 5.10.2 but other assumptions could be made.. Day = 1 [day]
(12)
τn = 0.83
(13)
Figure 5.3.1
tan = 1.01 · τn · αn
Eqn 5.5.2
(14)
275
1
duplicate i = 1, 12
(15)
ni = AveDay (i)
(16)
δi = dec (ni ) Cos θ¯b,i = Cos(lat − slope)·Cos(δi )·Cos(2.5 · 15)+sin (lat − slope)·sin (δi ) ¯ i , Lat, ni , Slope, ρg,i , θ¯b,i , N cov, KL, Ref rInd, αn S¯T,i = S¯T,LJ, H ¯ T,i = H ¯ T,LJ, H ¯ i , Lat, ni , Slope, ρg,i H
(17) Eqn 1.6.7a at 2.5 hrs from noon(18)
S¯T,i tabar/tan,i · tan = ¯ HT,i
(19) (20) (21)
end
(22)
F r0 /F r = 1
No collector operating corrections.
(23)
System corrections on storage size and air flow. 0.28
Xc/X = (V olF lowAir/V olF lowAirStd)
−0.3
· (StorageV ol/StorageV olStd)
duplicate i = 1, 12
(25)
N domi = N umDay (i) s ∆ti = Day · N domi · 86400 day s Loadi = U Ah · DDi · 86400 day
(26) (27)
(28)
X and Y are not needed as they are also calculated internally inside the function for solar fraction. Xi = Xc/X · XX F rU l, F R0 F R, Tref , T¯a,i , ∆ti , Ac , Loadi J 0 ¯ , N domi , Ac , Loadi Yi = Y Y F rtan , F R /F R, tabartan,i , HT,i · 1000000 MJ Fi = SolF ract0 A F rU l, F R0 /F R, Tref , T¯a,i , ∆ti , Ac , Loadi , F rtan , tabartan,i , J ¯ HT,i · 1000000 , N dom , StorageV ol, V olF lowAir i MJ end Fannual =
(24)
(29) (30)
(31) (32)
Sum(Fi · Loadi , i = 1, 12) Sum(Loadi , i = 1, 12)
(33)
Solution αn = 0.9 F rtan = 0.76 KL = 0.0125 Slope = 60 [deg] tan = 0.7545 V olF lowAirStd = 10 l/m2 -s
azim = 0 [deg] F rU l = 4.85 W/m2 -C Lat = 43 [deg] StorageV ol = 0.15 m3 /m2 Tref = 100 [C] Xc/X = 1.227 276
2
Ac = 40 m2 F R0 /F R = 1 N cov = 2 StorageV olStd = 0.25 m3 /m2 U Ah = 300 [W/C]
Day = 1 [day] Fannual = 0.54 Ref rInd = 1.526 τn = 0.83 V olF lowAir = 12 l/m2 -s
Arrays Table: Main
1 2 3 4 5 6 7 8 9 10 11 12
DDi [C-Day] 821 656 537 280 136 30 1 25 103 232 479 699
δi [deg] -20.9 -13.0 -2.4 9.4 18.8 23.1 21.2 13.5 2.2 -9.6 -18.9 -23.0
Row
N domi
ρg,i
1 2 3 4 5 6 7 8 9 10 11 12
31 28 31 30 31 30 31 31 30 31 30 31
0.7 0.7 0.4 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.4 0.7
Row
∆ti [s] 2.68E6 2.42E6 2.68E6 2.59E6 2.68E6 2.59E6 2.68E6 2.68E6 2.59E6 2.68E6 2.59E6 2.68E6 ¯ ST,i 2 MJ/m 9.661 12.07 10.93 10.01 11.06 10.85 11.08 11.87 10.8 10.74 8.243 9.292
Fi [-] 0.35 0.52 0.60 0.86 0.87 1.00 1.00 1.00 0.88 0.98 0.46 0.38
¯ Hi 2 MJ/m 6.44 9.89 12.86 16.05 21.36 23.04 22.58 20.33 14.59 10.48 6.37 5.74
tabar\tan,i 0.93 0.93 0.92 0.90 0.88 0.87 0.88 0.90 0.92 0.94 0.94 0.94
¯ HT,i 2 MJ/m 13.71 17.25 15.79 14.69 16.58 16.53 16.76 17.47 15.58 15.18 11.67 13.15
θ¯b,i [deg] 35.6 36.4 39.6 45.5 51.4 54.3 53.0 47.9 41.7 37.2 35.7 35.7
T¯a,i [C] -8 -5 1 9 14 19 22 20 15 11 2 -5
277
3
Loadi [J] 2.13E10 1.70E10 1.39E10 7.26E9 3.53E9 7.78E8 2.59E7 6.48E8 2.67E9 6.01E9 1.24E10 1.81E10
Xi [-] 3.235 3.555 4.533 7.734 15.55 64.25 1,918 78.69 19.64 9.433 4.869 3.694
ni 17 47 75 105 135 162 198 228 258 288 318 344
Yi [-] 0.57 0.80 0.98 1.67 3.92 16.86 534.12 22.87 4.89 2.23 0.80 0.64
P20 05
Equations 20.5
Calculation of annual solar fraction for a combined water and space heating load.
It is first necessary to estimate monthly average radiation on sloped surfaces. The radiation calculations are the same as Problems 2.18 and 20.4. Knowns: Ac = 50 m2 ;
Lat = 43 [deg] ;
Slope = 60 [deg] ;
F 0 Rtan = 0.72;
F 0 RU L = 3.5 W/m2 ·C ;
azim = 0 [deg]
LoadDHW = 1.9 [GJ]
V olF lowAir = 12 l/m2 ·s ;
V olF lowAirStd = 10 l/m2 ·s ;
nCov = 2;
Ref rInd = 1.526;
KL = 0.0125;
U Ah = 300 [W/C] ;
(1)
αn = .9;
(2)
StorageV ol = 5000 [l] Tref = 100 [C]
good quality glass
StorageV olStd = Ac · 75 l/m2
(3) (4) (5)
Weather data from Appendix G and Problem 20.4 DD1..12 = [821, 656, 537, 280, 136, 30, 1, 25, 103, 232, 479, 699]
(6)
T¯a,1..12 = [-8, -5, 1, 9, 14, 19, 22, 20, 15, 11, 2, -5]
(7)
¯ 1 = 6.44; H
¯ 2 = 9.89; H
¯ 7 = 22.58; H
¯ 8 = 20.33; H
¯ 3 = 12.86; H
¯ 4 = 16.05; H
¯ 9 = 14.59; H
¯ 5 = 21.36; H
¯ 10 = 10.48; H
ρg,1 = 0.7;
ρg,2 = 0.7;
ρg,3 = 0.4;
ρg,4 = 0.2;
ρg,7 = 0.2;
ρg,8 = 0.2;
ρg,9 = 0.2;
ρg,10 = 0.2;
¯ 6 = 23.04 H
¯ 11 = 6.37; H
ρg,5 = 0.2; ρg,11 = 0.4;
¯ 12 = 5.74 H
ρg,6 = 0.2 ρg,12 = 0.7
(8) (9) (10) (11)
It is first necessary to estimate monthly average incident and absorbed radiation on sloped surfaces. These two quantities are then used to calcualte the monthly average τ *α product. This solution uses the Liu and Jordan mode Eqn 5.10.2 but other assumptions could be made.. Day = 1 [day]
(12)
278
1
τn = 0.83
Figure 5.3.1
tan = 1.01 · τn · αn
(13)
Eqn 5.5.2
(14)
duplicate i = 1, 12
(15)
ni = AveDay (i)
(16)
δi = dec (ni ) Cos θ¯b,i = Cos(lat − slope)·Cos(δi )·Cos(2.5 · 15)+sin (lat − slope)·sin (δi ) ¯ i , Lat, ni , Slope, ρg,i , θ¯b,i , N cov, KL, Ref rInd, αn S¯T,i = S¯T,LJ, H ¯ T,i = H ¯ T,LJ, H ¯ i , Lat, ni , Slope, ρg,i H
(17) Eqn 1.6.7a at 2.5 hrs from noon(18)
S¯T,i tabar/tan,i · tan = ¯ HT,i
(20) (21)
end F r0 /F r = 1
(19)
(22) Collector operating corrections contained in F‘Rtan and F‘RUL.
(23)
System corrections on storage size and air flow. −0.25
Xc/X = (StorageV ol/StorageV olStd)
(24)
load heat exchanger: =1
(25)
· Cmin /U Ah = 2
(26)
StoreCap = StorageV ol/Ac
(27)
duplicate i = 1, 12
(28)
N domi = N umDay (i) s ∆ti = Day · N domi · 86400 day s 9 J 1 × 10 + Load · Loadi = U Ah · DDi · 86400 DHW day GJ X and Y are not needed as they are also calculated internally inside the function for solar fraction. Xi = Xc/X · XX F 0 RU L, F R0 F R, Tref , T¯a,i , ∆ti , Ac , Loadi ¯ T,i · 1000000 J , N domi , Ac , Loadi Yi = Y Y F 0 Rtan , F R0 /F R, tabartan,i , H MJ Fi = SolF ract0 L F 0 RU l, F R0 /F R, Tref , T¯a,i , ∆ti , Ac , Loadi , F 0 Rtan , tabartan,i , J ¯ HT,i · 1000000 , N domi , StoreCap, , Cmin , U Ah MJ end
(29) (30) (31)
(32) (33)
(34) (35)
Find the annual solar fraction Sum(Fi · Loadi , i = 1, 12) Fannual = Sum(Loadi , i = 1, 12)
(36)
Solution
279
2
αn = 0.9 Day = 1 [day] F 0 Rtan = 0.72 LoadDHW = 1.9 [GJ] StorageV ol = 5, 000 [l] tan = 0.7545 V olF lowAirStd = 10 l/m2 -s
azim = 0 [deg] =1 F 0 RU l = 3.5 W/m2 -C N cov = 2 StorageV olStd = 3, 750 [l] Tref = 100 [C] Xc/X = 0.9306
Ac = 50 m2 F R0 /F R = 1 KL = 0.0125 Ref rInd = 1.526 StoreCap = 100 l/m2 U Ah = 300 [W/C]
Cmin = 600 [W/C] Fannual = 0.63 [-] Lat = 43 [deg] Slope = 60 [deg] τn = 0.83 V olF lowAir = 12 l/m2 -s
Arrays Table: Main Row 1 2 3 4 5 6 7 8 9 10 11 12
Row 1 2 3 4 5 6 7 8 9 10 11 12
DDi [C-day] 821 656 537 280 136 30 1 25 103 232 479 699 θ¯b,i [deg] 35.6 36.4 39.6 45.5 51.4 54.3 53.0 47.9 41.7 37.2 35.7 35.7
δi [deg] -20.92 -12.95 -2.42 9.41 18.79 23.09 21.18 13.46 2.22 -9.60 -18.91 -23.05
T¯a,i [C] -8 -5 1 9 14 19 22 20 15 11 2 -5
∆ti [s] 2.678E6 2.419E6 2.678E6 2.592E6 2.678E6 2.592E6 2.678E6 2.678E6 2.592E6 2.678E6 2.592E6 2.678E6
¯ Hi 2 MJ/m 6.44 9.89 12.86 16.05 21.36 23.04 22.58 20.33 14.59 10.48 6.37 5.74
Loadi [J] 2.318E10 1.890E10 1.581E10 9.158E9 5.425E9 2.678E9 1.926E9 2.548E9 4.570E9 7.913E9 1.431E10 2.002E10
ni
N domi
ρg,i
Xi
Yi
17 47 75 105 135 162 198 228 258 288 318 344
31 28 31 30 31 30 31 31 30 31 30 31
0.7 0.7 0.4 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.4 0.7
2.032 2.188 2.73 4.195 6.915 12.77 17.67 13.7 7.852 4.906 2.89 2.288
0.6165 0.8529 1.022 1.565 3.015 5.8 8.513 6.889 3.384 2.007 0.8243 0.6866
¯ HT,i 2 MJ/m 13.71 17.25 15.79 14.69 16.58 16.53 16.76 17.47 15.58 15.18 11.67 13.15
280
3
¯ ST,i 2 MJ/m 9.66 12.07 10.93 10.01 11.06 10.85 11.08 11.87 10.80 10.74 8.24 9.29
Fi [-] 0.420 0.577 0.652 0.849 1.000 1.000 1.000 1.000 1.000 0.975 0.519 0.457
tabar\tan,i 0.93 0.93 0.92 0.90 0.88 0.87 0.88 0.90 0.92 0.94 0.94 0.94
P20 06
Equations 20.6
Calculation using the bin method of energy requirements to heat a building with a heat pump.
$SumRow on Since there is logic required in the heat pump performance, a Procedure is used. procedure HP (Load, hours, Tbin , Delmax , W ork max : Qdel , Qwork , Qaux )
(1)
Three cases must be considered: (1) At low temperatures the COP is less than 1: use all auxiliary. (2) COP>1 and Qdel,maxQload : HP runs only part of the time and auxiliary is zero. COP := Delmax /W ork max
(2)
If(COP > 1) then
(3)
goto 2
Case 1 Qwork := 0
(4)
Qaux := Load · hours
(5)
Qdel := 0
(6)
goto 4
(7)
Case 2 2 : IF (Delmax>Load ) then
goto 3
(8)
Qwork := W ork max · hours
(9)
Qdel := Delmax · hours
(10)
Qaux := max (0, (Load − Delmax ) · hours)
(11)
goto 4
(12)
281
1
Case 3 3 : F ractionT ime := Load/Delmax
(13)
Qwork := F ractionT ime · hours · W ork max
(14)
Qdel := Load · hours
(15)
Qaux := 0
(16)
4 : end HP
(17)
The heat pump work and delivered energy as a function of ambient temperature are placed in a lookup table. Column 1 is the temperature, 2 is the work and 3 is the delivered energy. Knowns: U A = 300 [W/C] ;
Tbal = 18.3 [C]
(18)
hr = 1
(19)
A parametric table is made with the bins and associated hours as knowns. The various energy quanities are also in the parametric table. Tbin = Lookup(‘Bins’ , T ableRun#, ‘T bin’ )
(20)
Hours = Lookup(‘Bins’ , T ableRun#, ‘hours’ )
(21)
The Load in each bin is: kW Load = max 0, U A · (Tbal − Tbin ) · 0.001 W
(22)
Qload = Load · hours
(23)
The maximum HP input and output at T=Tbin : Delmax = Lookup(‘HP’ , Lookuprow(‘HP’ , 1, Tbin ), ‘Q del’ )
(24)
W ork max = Lookup(‘HP’ , Lookuprow(‘HP’ , 1, Tbin ), ‘Work’ )
(25)
call HP (Load, hours, Tbin , Delmax , W ork max : Qdel , Qwork , Qaux )
(26)
Qpurchased = Qwork + Qaux
(27)
The month’s auxiliary energy requirement is 2545 kWh, or 9.2 GJ; the work is 1525 kWh, or 5.5 GJ
Solution Variables in Main program Delmax = 7.67 [kW] Qaux = 0 [kWh] Qwork = 0 [kWh] W ork max = 3.2 [kW]
Hours = 0 [Hr] Qdel = 0 [kWh] Tbal = 18.3 [C]
hr = 1 Qload = 0 [kW-Hr] Tbin = 8 [C]
282
2
Load = 3.09 [kW] Qpurchased = 0 [kWh] U A = 300 [W/C]
Parametric Table: Table 1 Run 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 Sum
Tbin Delmax [C] [kW] -26 1.25 -24 1.25 -22 1.25 -20 1.25 -18 1.70 -16 2.14 -14 2.58 -12 3.03 -10 3.48 -8 3.92 -6 4.37 -4 4.81 -2 5.26 0 5.70 2 6.50 4 6.89 6 7.28 8 7.67 8 7.67 -154 77.97
Hours [Hr] 5.5 4 11.5 12 33.5 34 40 44 43 77 76 92 110 103.5 46.5 8.5 3 0 0 744
Load [kW] 13.29 12.69 12.09 11.49 10.89 10.29 9.69 9.09 8.49 7.89 7.29 6.69 6.09 5.49 4.89 4.29 3.69 3.09 3.09 150.5
Qdel [kWh] 0 0 0 0 57 73 103 133 149 302 332 443 578 568 227 36 11 0 0 3,013
Qaux [kWh] 73 51 139 138 308 277 284 267 216 306 222 173 92 0 0 0 0 0 0 2,545
Lookup Table: Bins Row 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
Tbin [C] -26 -24 -22 -20 -18 -16 -14 -12 -10 -8 -6 -4 -2 0 2 4 6 8
Hours [Hr] 5.5 4.0 11.5 12.0 33.5 34.0 40.0 44.0 43.0 77.0 76.0 92.0 110.0 103.5 46.5 8.5 3.0 0.0
Lookup Table: HP Row
Tbin [C]
W ork [kW]
Qdel [kW]
283
3
Qpurchased [kWh] 73 51 139 138 362 335 356 350 302 467 389 385 356 249 98 15 5 0 0 4,070
Qwork [kWh] 0 0 0 0 54 58 72 84 86 162 167 212 264 249 98 15 5 0 0 1,525
Qload [kW-Hr] 73 51 139 138 365 350 388 400 365 608 554 615 670 568 227 36 11 0 0 5,558
W ork max [kW] 1.5 1.5 1.5 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.8 2.9 3.0 3.2 3.2 41.6
1 2 3 4 5 6
-100 -20 0 2 20 100
1.5 1.5 2.5 2.8 3.9 3.9
1.25 1.25 5.7 6.5 10 10
284
4
P20 07
Equations 20.7
Add a solar heating system in parallel with the heat pump of Problem 20.6; calculate month’s performance.
Knowns: FSolar = 0.30
(1)
Data from Problem 20.6 - all in kWh: W ork = 1525 [kWh] ; F W 0 = W ork/Load
Aux = 2545 [kWh] ;
Del = 3013 [kWh]
Work fraction for stand alone HP
F W = F W 0 · (1 − FSolar )
(5)
Non-pruchased fraction for stand alone HP
F AU X = (1 − F AT M 0 − F W 0 ) · (1 − FSolar ) F AU X = AuxW ithSolar /Load
(4)
Find WorkW ith Solar
Load − Aux − W ork Load
(6)
Equation 20.7.3, aux fraction with solar
(7)
Find AuxW ithSolar
(8)
Solution Aux = 2, 545 [kWh] F AU X = 0.3205 [-] Load = 5, 558 [kWh]
(2) (3)
Equation 20.7.2, work fraction with solar
F W = W ork W ithSolar /Load F AT M 0 =
Load = 5558 [kWh] ;
AuxW ithSolar = 1, 781 [kWh] F W = 0.1921 [-] W ork = 1, 525 [kWh]
Del = 3, 013 [kWh] F W 0 = 0.2744 [-] W ork W ithSolar = 1, 067 [kWh]
285
1
F AT M 0 = 0.2677 [-] FSolar = 0.3
P21 01
Equations 21.1 Calculation of a month’s performance of a heating system using generalized utilizability charts. Knowns: lat = 43;
month = 1;
slope = 60 [deg] ;
F RU L = 4.85 W/m2 ·C ; From problem 20.04: ¯ = 6.44 MJ/m2 ; H
F Rtan = 0.76;
ρg = 0.7;
Tin = 20 [C] ;
Surf AzAng = 0
Ac = 40 m2
T¯a = −8 [C] ;
Load = 21.3 [GJ]
(1) (2)
(3)
HrAng = (T ime − 12 [hr]) · 15 [deg/hr]
(4)
n = AveDay (month) 284 + n δ = 23.45 [deg] · sin 360 · 365
(5) (6)
ωs = (arccos (− tan (lat) · tan (δ)))
(7)
a = 0.409 + 0.5016 · sin (ωs − 60 [deg])
(8)
b = 0.6609 − 0.4767 · sin (ωs − 60 [deg])
(9)
¯ T = (a + b · Cos(HrAng)) · K ¯T Lk
(10)
¯ o = HZero, (n, Lat) H
(11)
286
1
¯ =K ¯T · H ¯o H
(12)
Rb = RBeam, (Lat, n, HrAng, Slope, Surf AzAng) ¯T Hd/Hbar = HDIF F BARHBAR Lat, n, K
(13)
rd = rd, (HrAng, Lat, n)
(15)
rt = rt, (HrAng, Lat, n) ¯T · H ¯ o · (rt − Hd/Hbar · rd ) · Rb + HdHbar · rd · 1 + Cos(Slope) + ρg · rt · 1 − Cos(Slope) I¯T = K 2 2
(16)
¯h = R
(14)
I¯T ¯ rt · H
(18)
Xm = 1.85 + 0.169 ·
¯T ¯h R Cos(slope) Lk − 0.0696 · − 0.981 · 2 2 2 ¯ ¯ Cos(δ) Lk Lk T T
(19)
Xc = I¯T,c /I¯T
(20)
Xm − 1 2 − Xm q 2 2 φ = abs abs (g) − g + (1 + 2 · g) · (1 − Xc /Xm ) g=
(21) (22)
hr = 1 [hr]
I¯T,c =
(17)
(23)
F RU L · Tin − T¯a · hr · 0.0036
MJ/hr J/s
(24)
F Rtan · tabar/ta
The hourly energy is then: QBar/Ac = F Rtan · tabarta · I¯T · φ
(25)
The following statement sums the QBar \Ac to get Sum for the day: Sum = Sumparametric(‘Table 1’ , ‘QBarA c’ , 1, T ableRun#)
(26)
The fraction by solar is: f = 31 · 2 · Ac · Sum ·
0.001
GJ MJ
(27)
Load
Problem 20.04 estimated the solar fraction for January as 35%. Here the estimate is also 35%.
Solution a = 0.4885 f = 0.3517 [-] Hd/Hbar = 0.4091 ¯ o = 13.36 MJ/m2 H lat = 43 [deg] n = 17 ρg = 0.7 rt = 0.055 tabar/ta = 0.94 Xc = 0.78 [-]
Ac = 40 m2 F Rtan = 0.76 hr = 1 [hr] I¯T = 0.88 MJ/m2 ¯ T = 0.4071 Lk ωs = 69.12 [deg] Rb = 3.76 slope = 60 [deg] T ime = 15.5 [hr] Xm = 3.709
b = 0.5853 F RU L = 4.85 W/m2 -C HrAng = 52.5 [deg] I¯T,c = 0.6843 MJ/m2 Load = 21.3 [GJ] φ = 0.51 ¯ h = 2.477 R Sum = 3.021 MJ/m2 287 [C] T¯a = −8 2
δ = −20.92 [deg] g = −1.585 ¯ = 6.44 MJ/m2 H ¯ T = 0.4819 K month = 1 QBar/Ac = 0.32 MJ/m2 rd = 0.065 Surf AzAng = 0 [deg] Tin = 20 [C]
Parametric Table: T able 1 Run 1 2 3 4
T ime [hr] 12.5 13.5 14.5 15.5
rt
rd
tabar\ta
0.176 0.152 0.108 0.055
0.165 0.147 0.113 0.065
0.96 0.96 0.95 0.94
¯ IT 2 MJ/m 2.12 1.87 1.43 0.88
Xc [-] 0.32 0.36 0.47 0.78
288
3
Rb
φ
2.28 2.40 2.72 3.76
0.72 0.69 0.62 0.51
QBar\A c MJ/m2 1.11 0.94 0.64 0.32
P21 03
Equations 21.3
Calculation of a month’s collected energy with generalized utilizability method.
Knowns: F RU L = 4.2 W/m2 ·C ; Slope = 55 [deg] ;
F Rta = 0.75;
Dec = −20.9;
Lat = 40 [deg] ;
¯ T = 0.5 K
¯ o = 15.06 MJ/m2 H
Surf AzAng = 0 [deg] ;
month = 1;
(1) (2)
Ta = 0 [C] ;
Ti = 50 [C] ;
ρg = 0.2
Assumed
n = AveDay (month)
(3) (4)
Find rt and rd from Equations 2.13.2 and 1.13.4 ω = 15 [deg/hr] · Hour
Hour is set in the parametric table. It is hours from solar noon.
ωs = arccos (− tan (Lat) · tan (Dec))
(5) (6)
¯ T =0.5 and ω s < 81.4: From Figure 2.12.2 with K ¯ = 0.39 H¯d /H
(7)
Find hourly radiation on tilted surface from Equation 2.23.4: Rb = RBeam, (Lat, n, ω, Slope, Surf AzAng)
(8)
rt = rt, (ω, Lat, n)
(9)
rd = rd, (ω, Lat, n) ¯ · rd · Rb + H¯d H ¯o · K ¯ T · rt − H¯d /H ¯ · rd · 1 + Cos(Slope) + ρg · rt · 1 − Cos(Slope) I¯T = H 2 2
(10) (11)
Find critical level with Equation 21.1.7: hr = 1 [hr]
(12)
289
1
Xc = F RU L · (Ti − Ta ) · hr ·
0.0036
MJ/hr W
(13)
FRta · ¯IT
¯ b from Equation 2.19.3: Find R ¯b = R ¯ BEAM, (n, Lat, Slope) R
(14)
¯ b , find φ from Figure 2.23.1(c). Then find the hourly useful energy from Equation 21.1.5: With critical level and R Qu/Ac = F Rta · I¯T · φ
(15)
The monthly output is two times sum of the five hourly values times the number of days in the month: SumQu = 2 · N umDay (month) · Sumparametric(‘Table 1’ , ‘Q uAc’ , 1, T ableRun#)
Solution Dec = −20.9 [deg] hr = 1 [hr] ¯ T = 0.5 K ω = 67.5 [deg] ρg = 0.2 rt = 0.01072 Ta = 0 [C]
F Rta = 0.75 ¯ o = 15.06 MJ/m2 H Lat = 40 [deg] ωs = 71.31 [deg] Rb = 9.87 Slope = 55 [deg] Ti = 50 [C]
F RU L = 4.2 W/m2 -C ¯ = 0.39 H¯d /H month = 1 φ = 0.00 [-] ¯ b = 2.472 R SumQu = 142.4 MJ/m2 Xc = 2.486 [-]
Hour = 4.5 [hr] I¯T = 0.41 MJ/m2 n = 17 Qu /Ac = 0.00 MJ/m2 rd = 0.0148 Surf AzAng = 0 [deg]
Parametric Table: Table 1 Run 1 2 3 4 5
hour [hr] 0.5 1.5 2.5 3.5 4.5
ω [deg] 7.5 22.5 37.5 52.5 67.5
Rb 2.06 2.15 2.39 3.11 9.87
I¯T MJ/m2 2.11 1.87 1.45 0.92 0.41
Xc [-] 0.477 0.538 0.696 1.091 2.486
Qu \Ac2 MJ/m 0.92 0.77 0.47 0.14 0.00
φ [-] 0.58 0.55 0.43 0.20 0.00
290
2
SumQ2u MJ/m 56.95 104.9 133.8 142.4 142.4
(16)
P21 04
Equations 21.4
Calculation of a month’s collected energy with daily utilizability method.
Knowns: F RU L = 4.2 W/m2 ·C ; Slope = 55 [deg] ;
F Rta = 0.75;
Dec = −20.9 [deg] ;
Lat = 40 [deg] ;
¯ T = 0.5 K
¯ o = 15.06 MJ/m2 H
Surf AzAng = 0 [deg] ;
month = 1;
(1) (2)
Ta = 0 [C] ;
Ti = 50 [C] ;
ρg = 0.2
Assumed
n = AveDay (month)
(3) (4)
Find rt and rd at noon from Equations 2.13.2 and 1.13.4: noon = 0 [deg]
(5)
rt,noon = rt, (noon, lat, n)
(6)
rd,noon = rd, (noon, lat, n)
(7)
¯ T =0.5 and ω s < 81.4: From Figure 2.12.2 with K ¯T Hdbar/Hbar = HDIF F BARHBAR Lat, n, K
(8)
¯ b from Equation 2.19.3: Find R ¯b = R ¯ BEAM, (n, Lat, Slope) R
(9)
Find radiation on tilted surface from Equation 2.19.1: 1 + Cos(Slope) 1 − Cos(Slope) ¯ ¯ ¯ ¯ HT = Ho · KT · (1 − Hdbar/Hbar) · Rb + HdbarHbar · + ρg · 2 2 ¯ ¯ = HT R ¯o · K ¯T H
(10)
(11)
Find Rb,noon from Equation 2.24.2: Rb,noon = RBeam, (Lat, n, 0 [deg] , Slope, Surf AzAng)
(12)
From Figure 2.11.2 with ω s < 81.4 ¯T Hd/H = Hdif f H Lat, n, K
(13)
Dif = rd,noon · Hd/H/rt,noon
(14)
291
1
Rnoon = (1 − Dif ) · Rb,noon + Dif ·
1 + Cos(Slope) 1 − Cos(Slope) + ρg · 2 2
(15)
Find critical level with Equation 21.1.7: hr = 1 [hr]
(16)
Xc = F RU L · (Ti − Ta ) · hr ·
0.0036
MJ W·hr
(17)
¯o · K ¯T FRta · rt,noon · Rnoon · H
¯ N oon =1.31: Find φ from Figure 2.24.2(c) with R/R ¯ noon RBar/RN oon = R/R
(18)
φ = 0.40
(19)
The monthly output is found from Equation 21.2.3: ndom = numDay (month)
(20)
¯T · φ SumQ/Ac = ndom · F Rta · H
(21)
Solution Dec = −20.9 [deg] Hdbar/Hbar = 0.3911 ¯ T = 13.97 MJ/m2 H n = 17 RBar/RN oon = 1.305 Rb,noon = 2.046 Slope = 55 [deg] Ti = 50 [C]
Dif = 0.5293 Hd/H = 0.5688 ¯ T = 0.5 K noon = 0 [deg] ρg = 0.2 rd,noon = 0.1622 SumQ/Ac = 129.9 MJ/m2 Xc = 0.5398
F Rta = 0.75 hr = 1 [hr] Lat = 40 [deg] ndom = 31 ¯ = 1.856 R Rnoon = 1.422 Surf AzAng = 0 [deg]
292
2
F RU L = 4.2 W/m2-C ¯ o = 15.06 MJ/m2 H month = 1 φ = 0.4 ¯ b = 2.472 R rt,noon = 0.1744 Ta = 0 [C]
P21 05
Equations 21.5 Calculation of July solar contribution to operation of a n absorption air conditioner, neglecting storage losses. Use method of Section 21.3 Knowns: Ac = 25 m2 ;
COP = 0.7;
F RU L = 2.63;
Storage = 4180;
Lat = 25.8 [deg] ;
Load = 5.1
T¯a = 28 [C] ;
Tmin = 70 [C] ;
Slope = 25 [deg] ;
F Rtan = 0.72;
Dec = 21.2; month = 7; ρg = 0.2 assumed ¯ = 20.01 MJ/m2 ; ¯ T = 0.5; H K Surf AzAng = 0 [deg]
[GJ] ; COP
(1) (2) (3)
(4)
n = AveDay (month)
(5)
numDay = N umDay (month)
(6)
Find rt and rd at noon from Equations 2.13.2 and 1.13.4: noon = 0 [deg]
(7)
rt,noon = rt, (noon, lat, n)
(8)
rd,noon = rd, (noon, lat, n)
(9)
¯ T =0.5 and ω s < 81.4: From Figure 2.12.2 with K ¯T Hdbar/Hbar = HDIF F BARHBAR Lat, n, K
(10)
¯ b from Equation 2.19.3: Find R ¯b = R ¯ BEAM, (n, Lat, Slope) R
(11)
Find radiation on tilted surface from Equation 2.19.1: ¯=H ¯ T /H ¯ R ¯T = H ¯ · (1 − Hdbar/Hbar) · R ¯ b + HdbarHbar · 1 + Cos(Slope) + ρg · 1 − Cos(Slope) H 2 2
293
1
(12) (13)
Find Rb,noon from Equation 2.24.2: Rb,noon = RBeam, (Lat, n, 0 [deg] , Slope, Surf AzAng)
(14)
From Figure 2.11.2 with ω s < 81.4 ¯T Hd/H = Hdif f H Lat, n, K
(15)
Find Rnoon from Equation 2.24.2: Dif = rd,noon · Hd/H/rt,noon
(16) 1 + Cos(Slope) 1 − Cos(Slope) + ρg · 2 2
Rnoon = (1 − Dif ) · Rb,noon + Dif ·
(17)
Find average TauAlpha from Equation 5.10.4. The equivalent beam radiation angle is 50 from Figure 5.10.1(a). The equiva¯ lent diffuse and ground reflected angles are 57 and 76 from Figure 5.4.1 From Figure 5.6.1 the ratios of (TauAlpha)/(T auAlpha) for these three angles are 0.92, 0.84 and 0.32. ¯b beam = 0.92 · (1 − Hdbar/Hbar) · R dif f use = 0.84 · Hdbar/Hbar · ground = 0.32 · ρg ·
(18)
1 + Cos(Slope) 2
(19)
1 − Cos(Slope) 2
(20)
¯ ¯ ta/ta n · R = beam + dif f use + ground
(21)
¯ = F Rtan · ta/ta ¯ F Rta n
(22)
Find minimum critical level with Equation 21.3.2: hr = 1 [hr]
XcM in
(23)
MJ |0.0036 W·hr | F RU L · Tmin − T¯a · hr · ¯ FRta = ¯ rt,noon · Rnoon · H
(24)
¯ N oon =0.95 and XcM in =0.26 or from φ¯function.: Find φmax from Figure 2.24.2(c) with R/R ¯ N oon = R/R ¯ noon R/R Tmin − T¯a IT c = F RU L · ¯ F Rta
(25) (26)
¯ n, Lat, Slope, ρg , IT c φ¯max = φ¯ H,
(27)
Find X’ and Y: ∆t = 3600 [s/hr] · 24 [hr] · N umDay
0
X = Ac · F RU L · 100 [C] · ∆t ·
1 × 10·9
(28)
GJ J
Load
0.001 GJ MJ ¯ ·H ¯ T · N umDay · Y = Ac · F Rta Load
Eqn 21.3.4 Eqn 21.3.3
294
2
(29) (30)
Find the solar fraction, f, from Equation 21.3.5: Storage/Ac = Storage · 4.190
[kJ/kg] Ac
(31)
f = Y · φ¯max − 0.015 · (exp (3.85 · f ) − 1) · (1 − exp (−0.15 · X )) · 0
!0.77 kJ/m2 350 Storage/Ac
Solution Ac = 25 m2 ∆t = 2.678 × 106 [s] ¯ = 0.6319 F Rta Hdbar/Hbar = 0.4291 ¯ T = 18.12 MJ/m2 H Load = 7.286 [GJ] numDay = 31 ¯ N oon = 0.9466 R/R Rnoon = 0.9569 Storage/Ac = 700.6 kJ/m2 Tmin = 70 [C]
beam = 0.4484 Dif = 0.56 F Rtan = 0.72 Hd/H = 0.6083 IT c = 174.8 W/m2 month = 7 φ¯max = 0.6458 ¯ b = 0.8538 R rt,noon = 0.1287 Surf AzAng = 0 [deg] XcM in = 0.2554
COP = 0.7 dif f use = 0.3436 F RU L = 2.63 W/m2 -C hr = 1 [hr] ¯ T = 0.5 K n = 198 ρg = 0.2 Rb,noon = 0.9404 Slope = 25 [deg] ¯ ta/ta n = 0.8777 X 0 = 2.417 [-]
295
3
Dec = 21.2 [deg] f = 0.74 ground = 0.002998 ¯ = 20.01 MJ/m2 H Lat = 25.8 [deg] noon = 0 [deg] ¯ = 0.9058 R rd,noon = 0.1185 Storage = 4, 180 [kg] T¯a = 28 [C] Y = 1.218 [-]
(32)
P21 06b
Solution Variables in Main program Ac = 25 m2 ¯ = 0.632 F Rta ¯ KT = 0.5 month = 7 φ¯max = 0.625 Storage/Ac = 700.6 kJ/m2 Tmin = 70 [C] X 0 = 1.904 [-]
COP = 0.7 F RU L = 2.63 W/m2 -C Load = 9.25 [GJ] numDay = 31 Qst = 0.8089 [GJ] T¯a = 28 [C] U Ast = 7 [W/C] Y = 0.9584 [-]
296
1
∆t = 2.678 × 106 [s] fT L = 0.5808 Loadhouse = 5.1 [GJ] P arta $ = ‘False’ ¯ N oon = 0.947 R/R ¯ Ts = 71.14 [C] Xc = 0.2689
f = 0.4678 ¯ T = 18.1 MJ/m2 H Loado = 8.441 [GJ] φ¯ = 0.606 Storage = 4, 180 [kg] Tmax = 72.29 [C] XcM in = 0.255
P22 01
Equations 22.1
Estimation of a month’s auxiliary energy requirement for a direct-gain building.
In this problem we start with meteorological data from Appendix G,work through the radiation calcuLations, and apply unutilizability ideas to building performance calcuLations. function P hiBar (X, R, K)
Equation 2.24.4:
(1)
a = 2.943 − (9.271 − 4.031 · K) · K
(2)
b = −4.345 + K · (8.853 − 3.602 · K)
(3)
c = 0.170 − K · (0.306 − 2.936 · K)
(4)
P hiBar= exp ((a + b · R) · (X · (1 + c · X)))
(5)
end φ¯
(6)
function DirectGain (X, Y, P hiBar)
Equations 22.3.13
0.652
(7)
P = (1 − exp (−0.294 · Y ))
(8)
DirectGain= min (1, P · X + (1 − P ) · (3.082 − 3.142 · P hiBar) · (1 − exp (−0.329 · X)))
(9)
end DirectGain
(10)
function ACP S (X, Y, P hiBarc , P hiBaru )
Equations 22.5.10:
.652
(11)
P = (1 − exp (−.294 · Y ))
(12)
ACP S = min (P · X · P hiBarc + (1 − P ) · (3.082 − 3.142 · P hiBaru ) · (1 − exp (−.329 · X)) , 1)
(13)
end ACP S
(14)
Knowns:
297
1
Slope = 90 [deg] ;
Azimuth = 0 [deg] ;
¯ = 9.75 MJ/m2 ; H
T¯a = 6 [C] ;
Uwindow = 3.45 W/m2 ·C ; U Ahouse = 200 [W/C] ;
DD = 346;
Lat = 35 [deg] ;
Tbase = 18.3 [C]
ρg = 0.4
(16)
Awindow = 16.1 m2
mchouse = 20.2 [MJ/C] ;
(15)
(17) ∆T b = 5 [C]
(18)
Assume the following: KL = 0.0125; month = 2
Ref rInd = 1.526;
N cov = 2;
αn = .97
(19)
All 12 months could be evaluatedby using a parametric table with Month=tablerun#.
(20)
n = AveDay (M onth)
(21)
DaysinM onth = N umDay (M onth)
(22)
dec = Dec (n)
(23)
¯ o = HZero, (n, Lat) H
(24)
¯ T = H/ ¯ H ¯o K
(25)
Find incident and absorbed radiation on window. ¯ Lat, n, Slope, ρg ¯T = H ¯ T,LJ, H, H
(26)
¯=H ¯ T /H ¯ R
(27)
Use the 2.5 hour approximation of section 5.10 for the effective beam incidence angle. Cos(IncAngEf f Beam) = cosIncAng (Lat, n, 37.5 [deg] , Slope, Azimuth) ¯ Lat, n, Slope, ρg , IncAngEf f Beam, N cov, KL, Ref rInd, αn S¯T = S¯T,LJ, H, ¯T ¯ = S¯T /H ta
(28)
(29) (30)
Utilizability calcuLations: MJ Load = U Ahouse · DD · 24 [hr/day] · 0.0036 W · hr
Itc = U Ahouse ·
Tbase − T¯a ¯ Awindow · ta
(31)
Eqn 22.3.5
(32)
ωnoon = 0 [deg]
(33)
rt,n = rt, (ωnoon , lat, n)
(34)
rd,n = rd, (ωnoon , lat, n)
(35)
¯T Hd/H = HDIF F H Lat, n, K
(36)
298
2
Rb,n = RBeam, (Lat, n, ωnoon , Slope, Azimuth)
Eqn 2.24.2
Rn = (1 − (rd,n /rt,n ) · Hd/H) · Rb,n + (rd,n /rt,n ) · HdH ·
1 + Cos(Slope) 1 − Cos(Slope) + ρg · 2 2
¯ RnORbar = Rn /R ¯ c = Itc · 1 [hr] · X
LaZ
MJ W·hr
Eqn 2.24.3
¯ rt,n · Rn · H
(40)
(41)
For a zero thermal capacity building: = Load − 1 − φ¯ · Awindow · S¯T · DaysinM onth
FzeroCap =
Eqn 2.24.2(38) (39)
0.0036
¯ c , RnORbar , K ¯T φ¯ = P hiBar X a
(37)
Load − LaZ Load
Eqn 22.3.8
Eqn 22.3
(42) (43)
The fraction of absorbed energy that reduces auxiliary energy demand is: Load − LaZ Fusef ul,solar = ¯ ST · Awindow · DaysInM onth b
(44)
For a building with finite thermal capacity with a temperature swing of 5C:
¯ T · DaysinM onth · Awindow · ta/Load ¯ X=H QD = Awindow · S¯T · DaysinM onth · φ¯ Y = mchouse · ∆T b · DaysinM onth/QD FF initeCap = DirectGain X, Y, φ¯
Eqn 22.3.9
Eqn 22.3.7
(45) (46)
Eqn 22.3.12
(47) (48)
LaDeltaT 5 = Load · (1 − FF initeCap )
(49)
c
For a building with finite thermal capacity with a temperature swing of 10C: FDoubleDeltaT = DirectGain X, 2 · Y, φ¯
(50)
LaDeltaT 10 = Load · (1 − FDoubleDeltaT )
(51)
Solution Variables in Main program αn = 0.97 DD = 346 [C-day] FF initeCap = 0.4851 ¯ = 9.75 MJ/m2 H Itc = 212.9 W/m2 Load = 5, 979 [MJ] mchouse = 20.2 [MJ/C] ωnoon = 0 [deg] ρg = 0.4 Rn = 0.898 S¯T = 7.945 MJ/m2 U Ahouse = 200 [W/C] Y = 1.961
Awindow = 16.1 m2 ∆Tb = 5 [C] FzeroCap = 0.3579 ¯ T = 11.07 MJ/m2 H ¯ T = 0.4179 K LaDeltaT 5 = 3, 079 [MJ] n = 47 QD = 1, 442 [MJ] Rb,n = 1.109 rt,n = 0.1559 T¯a = 6 [C] X = 0.5991
Azimuth = 0 [deg] dec = −12.95 [deg] Fusef ul,solar= 0.5974 ¯ o = 23.33 MJ/m2 H KL = 0.0125 LaDeltaT 10 = 2, 756 [MJ] month = 2 φ¯ = 0.4026 ¯ = 1.136 R RnORbar = 0.7908 ¯ = 0.7176 ta Uwindow = 3.45 W/m2 -C 299
3
DaysinM onth = 28 FDoubleDeltaT = 0.539 Hd/H = 0.7279 IncAngEf f Beam = 51.16 [deg] Lat = 35 [deg] LaZ = 3, 839 [MJ] N cov = 2 Ref rInd = 1.526 rd,n = 0.1445 Slope = 90 [deg] Tbase = 18.3 [C] ¯ c = 0.5616 X
P22 02
Equations 22.2
Estimation of a month’s auxiliary energy requirements for an ACPS system.
function P hiBar (X, R, K)
Equation 2.24.4:
(1)
a = 2.943 − (9.271 − 4.031 · K) · K
(2)
b = −4.345 + K · (8.853 − 3.602 · K)
(3)
c = 0.170 − K · (0.306 − 2.936 · K)
(4)
P hiBar= exp ((a + b · R) · (X · (1 + c · X)))
(5)
end φ¯
(6)
function ACP S (X, Y, P hiBarc , P hiBaru )
Equations 22.5.10:
(7)
.652
P = (1 − exp (−.294 · Y ))
(8)
ACP S = min (P · X · P hiBarc + (1 − P ) · (3.082 − 3.142 · P hiBaru ) · (1 − exp (−.329 · X)) , 1)
(9)
end ACP S
(10)
Data for Jan from Example 22.5.1 and other data are: ¯ = 6.63 MJ/m2 ; H
¯ = 1.764; R
U Ah = 175 [W/C] ;
∆T b = 5 [C] ;
αn = 0.93;
Lat = 40 [deg] ;
Rn = 1.30;
rt,n = 0.174;
F RT An = 0.62;
Ta = −3 [C] ;
DD = 657 [C ·Day] ;
S¯ = 9.50 MJ/m2
Tbase = 18.3 [C]
Slope = 55 [deg] ; hr = 1 [hr] ;
F RU L = 3.1 W/m2 ·C ;
(12)
Azimuth = 0 [deg] ;
¯ T = 0.436; K tan = 0.846
300
1
(11)
¯ = 0.81; ta
T¯i = 21 [C] ;
ρg = .3
DaysinM onth = 31
(13) (14) (15)
¯ = F RT An · ta/ta ¯ F Rta n
(16)
FR = F RT An /tan
(17)
¯ Rn/Rbar = Rn /R
(18)
Critical radiation levels and utilizabilities: T¯i − Ta 2 IT cc = max F RU L · , 0 W/m ¯ F Rta 0.0036 MJ w·hr Xcc = IT cc · hr · ¯ rt,n · Rn · H ¯T P hiBarc = P hiBar Xcc , Rn/Rbar , K
Eqn 22.5.2
(19) (20) (21)
U Ah · (Tbase − Ta ) + Area · F RU L · T¯i − Ta ¯ Area · F Rta 0.0036 MJ w·hr Xcd = IT cd · hr · ¯ rt,n · Rn · H ¯T P hiBard = P hiBar Xcd , Rn/Rbar , K
IT cd = max
! 2 , 0 W/m
Eqn 22.5.4
P hiBaru = 1 − P hiBarc + P hiBard Eqn 22.5.8
(24)
(26)
Load and dumped energy for January: QD = Area · FR · S¯ · DaysinM onth · P hiBard Eqn 22.5.5 MJ Load = 24 [hr/day] · U Ah · DD · 0.0036 W · hr Eqn 22.5.9
(27) (28) (29)
f = ACP S (X, Y, P hiBarc , P hiBaru ) GJ LAux = Load · (1 − f ) · 0.001 Eqn 22.5.11 MJ a
(23)
(25)
X = S¯ · DaysinM onth · Area · FR /Load
Y = Capbldg · ∆T b · DaysinM onth/QD
(22)
(30) (31)
For the building with heavy’ construction
The January auxiliary decreased from 7.2 GJ in Example 22.5.1 to 7.1 GJ when heavy construction is assumed. b
With collector area doubled and medium construction
The January auxiliary decreased from 7.2 GJ in Example 22.5.1 to 6.0 GJ when the collector area is doubled. c
With doubled collector area and heavy construction
The January auxiliary decreased from 7.2 to 5.1 GJ on doubling the collector area and with heavier construction. d
Advisability of the changes
The large increase in building mass and collector area do not make large changes in the January auxiliary energy. The percentage change on an annual basis will certainly be even smaller. Thus these two changes probably would not be cost effective. The larger building capacity will result in smaller interior temperature swings and thus more comfortable conditions.
301
2
Solution αn = 0.93 DaysinM onth = 31 ¯ = 0.5936 F Rta hr = 1 [hr] ¯ T = 0.436 K P hiBarc = 0.6545 ρg = 0.3 rt,n = 0.174 tan = 0.846 U Ah = 175 [W/C] Y = 3.473
Area = 40 m2 DD = 657 [C-Day] F RT An = 0.62 ¯ = 6.63 MJ/m2 H Lat = 40 [deg] P hiBard = 0.321 ¯ = 1.764 R Slope = 55 [deg] Ta = −3 [C] X = 0.8691
Azimuth = 0 [deg] ∆Tb = 5 [C] F RU L = 3.1 W/m2-C IT cc = 125.3 W/m2 Load = 9, 934 [MJ] P hiBaru = 0.6665 Rn = 1.3 S¯ = 9.5 MJ/m2 T¯i = 21 [C] Xcc = 0.3009 [-]
Capbldg = 62.1 [MJ/C] f = 0.49 FR = 0.7329 IT cd = 282.3 W/m2 LAux = 5.1 [GJ] QD = 2, 771 [MJ] Rn /Rbar = 0.737 ¯ = 0.81 ta Tbase = 18.3 [C] Xcd = 0.6777 [-]
Parametric Table: Table 1 Run
P art$
1 2 3
a b c
Capbldg [MJ/C] 62.1 22.9 62.1
Area 2 m 20 40 40
P hiBard
P hiBaru
0.1287 0.321 0.321
0.4742 0.6665 0.6665
302
3
QD [MJ] 555.5 2,771 2,771
Y
X
17.33 1.281 3.473
0.4345 0.8691 0.8691
LAux [GJ] 7.1 6.0 5.1
f 0.28 0.40 0.49
P22 03
Equations 22.3
CalcuLation of month’s gain of a perfectly controlled direct gain system.
function P hiBar (X, R, K)
Equation 2.24.4:
(1)
a = 2.943 − (9.271 − 4.031 · K) · K
(2)
b = −4.345 + K · (8.853 − 3.602 · K)
(3)
c = 0.170 − K · (0.306 − 2.936 · K)
(4)
P hiBar= exp ((a + b · R) · (X · (1 + c · X)))
(5)
end φ¯
(6)
Knowns: Lat = 40 [deg] ;
Slope = 90 [deg] ;
Ref Index = 1.526; T¯i = 20 [C] ; θ¯b = 57 [deg] ;
Nglazings = 2;
T¯a = 3 [C] ;
Azimuth = 0 [deg] ;
KL = 0.0370;
Uw = 3.2 W/m2 ·C ;
M onth = 3;
αroom = .96;
Area = 15 [m] · 1.25 [m]
θ¯d = 59 [deg]
From Example 14.5.1 ¯ = 19.93 MJ/m2 ; H
¯ d = 6.26; H
(7) ρg = 0.2
(8) (9) (10)
¯ b = 13.67; H
f¯i = 0.87;
Frs = 0.42
(11)
The monthly utilizability method can be used to estimate the maximum useful gain for the month if we assume that the effects of shading and transmission are constant throughout the month. With this assumption we can use monthly utilizbility ¯ If hourly shading algorithms were readily available, then the and the the monthly average absorbed solar radiation, S. solution to this problem would be improved by using hourly utilizability. With a window loss coefficient of Uw , the solar energy per unit area absorbed inside of the building must be at least equal to Uw *(T¯i -T¯a ) or there will be a net loss. Thus the critical level must be: IT c = Uw · T¯i − T¯a (12) ¯ Calculate H nDay = AveDay (month)
(13)
Dec = Dec (N day)
(14)
303
1
¯b = R ¯ BEAM, (nDay, Lat, Slope) R
(15)
Ho = HZero, (nDay, Lat)
(16)
¯T = H ¯b · R ¯ b · f¯i + H ¯ d · Frs + H ¯ · ρg · 1 − Cos(Slope) H 2 ¯ ¯ ¯ R = HT /H
(17) (18)
¯ CalcuLate S¯ and T auAlpha: τBeam = T ransIncAng Nglazings , θ¯b , KL, Ref Index τdif f = T ransIncAng Nglazings , θ¯d , KL, Ref Index τgrnd = T ransIncAng Nglazings , θ¯d , KL, Ref Index 1 − Cos(Slope) ¯ ¯ ¯ ¯ ¯ ¯ S = αroom · fi · Hb · Rb · τBeam + Hd · τDif f · Frs + H · ρg · τGrnd · 2 ¯ ¯ H ¯T T auAlpha = S/
(19) (20) (21) (22) (23)
Noon values on an average day: HrAng n = 0 [deg]
(24)
¯ T = H/H ¯ o K
(25)
Rb,n = RBeam, (Lat, nDay, HrAng n , Slope, Azimuth)
(26)
rd,n = rd, (HrAng n , Lat, nDay)
(27)
rt,n = rt, (HrAng n , Lat, nDay) ¯T Hd/H = Hdif f H Lat, nDay, K
(28) (29)
Rn = (1 − rd,n · Hd/H/rt,n ) · Rb,n + rd,n · HdH/rt,n ·
1 − Cos(Slope) 1 + Cos(Slope) + ρg · 2 2
¯ = Rn /R ¯ Rn/R
(30) (31)
Utilizability calcuLations: hr = 1 [hr] ¯ c = IT c · hr · X
(32) 0.0036
MJ W·hr
Eqn 2.24.3 except that Sbar is used in place of Rn *HBar
¯ rt,n · S
¯ c , Rn/R ¯ , K ¯T φ¯ = P hiBar X
(33) (34)
The maximum useful energy gain is given by: QuM ax = Area · S¯ · φ¯
(35)
An alternate solution is given below in which calcuLations are based upon radiation incident upon an unshaded vertical window. The critical level is increased by the monthly average effects of shading, transmittance and room absorptance. The two methods give nearly the same results. ¯ T =H ¯ b *R ¯ b +H ¯ d /2+H*ρ ¯ g *(1-cos(Slope))/2 = 18.2 MJ/m2 H Monthly average shading, transmittance and absorptance: ¯ ¯ tBar,noshade = 0.60 f T auAlpha= S/H ¯ ¯ ¯ IT c =Uw *(T i -T a )/f T auAlpha = 90.8 W ¯ XBarc =IT c,Alternate *3600/(rtn *Rn *H*1e6 =0.120 ¯ ¯ ¯ ¯ P hialternate =PhiBar( X c ,Rn \R,K T ) = 0.82 ¯ ¯ T *φ¯*f T auAlpha QuM ax,alternate =Area*H = 168 MJ/day
304
2
Solution Variables in Main program αroom = 0.96 f¯i = 0.87 HrAng n = 0 [deg] ¯ T = 15.99 MJ/m2 H ¯ T = 0.7263 K Nglazings = 2 ρg = 0.2 rd,n = 0.1336 Slope = 90 [deg] τdif f = 0.7019 T¯a = 3 [C]
Area = 18.75 m2 Frs = 0.42 ¯ = 19.93 MJ/m2 H Ho = 27.44 MJ/m2 Lat = 40 [deg] φ¯ = 0.8086 ¯ = 0.8024 R Rn = 0.9468 S¯ = 10.93 MJ/m2 τgrnd = 0.7019 T¯i = 20 [C]
Azimuth = 0 [deg] Hd/H = 0.175 ¯ b = 13.67 MJ/m2 H IT c = 54.4 W/m2 M onth = 3 QuM ax = 165.8 [MJ] ¯ b = 0.956 R ¯ = 1.18 Rn /R ¯ T auAlpha = 0.6838 θ¯b = 57 [deg] Uw = 3.2 W/m2 -C
305
3
Dec = −2.418 [deg] hr = 1 [hr] ¯ d = 6.26 MJ/m2 H KL = 0.037 nDay = 75 Ref Index = 1.526 Rb,n = 0.9137 rt,n = 0.1445 τBeam = 0.7164 θ¯d = 59 [deg] ¯ c = 0.1239 X
P23 01
Equations 23.1 Calculation of power from known array conditions and constants Knowns: IL = 13.6 [A]
(1)
Io = .008 [A]
(2)
a = 23.6 [V]
(3)
Rs = 0.9 [ohm]
(4)
Rload = 20 [ohm]
(5)
Cell Equations for I-V and P-V: V + I · Rs I = IL − Io · exp −1 a P =I ·V
Eqn 23.2.1
(6)
Eqn 23.2.2
(7)
V = I · Rload
(8)
Solution a = 23.6 [V] Io = 0.008 [A] Rs = 0.9 [Ω]
I = 7.496 [A] P = 1124 [W] V = 149.9 [V]
IL = 13.6 [A] Rload = 20 [Ω]
306
1
P23 02
Equations 23.2
Calculation of maximum power from known array conditions and constants
$UnitSystem J K Pa Knowns: Isc,ref = 13.6 [A] ;
Voc,ref = 186.6 [V] ;
Vmp,ref = 131.4 [V] ;
Imp,ref = 11.26 [A] ;
GT,ref = 1000 W/m2 ; GT = 800 W/m2 ;
k = k#;
µV oc = −0.716 [V/K] ;
µIsc = 0.02 [A/K]
Ns = 300
(2)
Tc,ref = 298 [K]
(3)
Tc = 303 [K]
q = e#;
(1)
(4)
C = 0.0002677 [1/K] ;
Eg,ref
·19 J = 1.12 [eV] · 1.60207 × 10 eV
(5)
The following reference parameters were found from the EES program FINDREFPAR Rsh,ref = 82.71 [ohm]
(6)
aref = 8.1022 [V]
(7)
IL,ref = 14.09 [A]
(8)
Io,ref = 1.177 × 10−9 [A]
(9)
Rs,ref = 2.995 [ohm]
(10)
Temperature and radiation dependent parameters are: a/aref = Tc /Tc,ref
Eqn 23.2.8
(11)
IL = (GT /GT,ref ) · (IL,ref + µIsc · (Tc − Tc,ref ))
Eqn 23.2.9
3
Io = Io,ref · (Tc /Tc,ref ) · exp ((1/k) · (Eg,ref /Tc,ref − Eg /Tc )) Eg = Eg,ref · (1 − C · (Tc − Tc,ref )) Rsh /Rsh,ref = GT,ref /GT
Eqn 23.2.10b
(12) Eqn 23.2.10a
(13) (14)
Eqn 23.2.11
(15)
307
1
Rs = Rs,ref
Eqn 23.2.12
(16)
Cell Equations for I-V and P-V: V + I · Rs V + I · Rs I = IL − Io · exp −1 − a Rsh P =I ·V
Eqn 23.2.1
(17)
Eqn 23.2.2
(18)
The easiest way to find the maximum power with EES is to use the Min/Max solution method to maximize P with respect to V. Allow V to vary between 0 and 150 V.
Solution a = 8.238 [V] Eg,ref = 1.794 × 10−19 [J] IL = 11.35 [A] Io,ref = 1.177 × 10−9 [A] µV oc = −0.716 [V/K] Rs = 2.995 [Ω] Tc = 303 [K] Voc,ref = 186.6 [V]
aref = 8.102 [V] GT = 800 W/m2 IL,ref = 14.09 [A] Isc,ref = 13.6 [A] Ns = 300 Rsh = 103.4 [Ω] Tc,ref = 298 [K]
C = 0.0002677 [1/K] GT,ref = 1, 000 W/m2 Imp,ref = 11.26 [A] k = 1.381 × 10−23 [J/K] P = 1, 206 [W] Rsh,ref = 82.71 [Ω] V = 131.5 [V]
308
2
Eg = 1.792 × 10−19 [J] I = 9.168 [A] Io = 2.691 × 10−9 [A] µIsc = 0.02 [A/K] q = 1.602 × 10−19 [Coulomb] Rs,ref = 2.995 [Ω] Vmp,ref = 131.4 [V]
P23 03
Equations 23.3
Calculation of best series/parallel connections for resistive loads of 4, 7, and 12 ohms.
Knowns from Example 23.2.1 IL = 4.549;
Io = 4.782 × 10−9 ;
Rs = 0.567;
I-V for an individual module: V + I · Rs V + I · Rs I = IL − Io · exp −1 − a Rsh
a = 0.9352;
Rsh = 52.4
Eqn 23.2.1
(1)
(2)
I-V for various series and parallel combinations: (Ms is # modules in series in each string, and Mp is # strings in parallel: V 2 = Ms · V
(3)
I2 = M p · I
(4)
V 2 = I2 · Rload
(5)
P = I2 · V 2
(6)
Imax
Vmax + Imax · Rs Vmax + Imax · Rs −1 − = IL − Io · exp a Rsh
Eqn 23.2.1
(7)
Imax,2 = Imax · M p
(8)
Vmax,2 = Vmax · M s
(9)
Vmax,2 = Imax,2 · Rmax
(10)
Pmax = Imax,2 · Vmax,2
(11)
For Rload=4, the combination of Ms=2 & Mp=4 is best but Ms=4 and Mp=2 is essentially the same but Ms=3 & Mp=2 is a close second with only 6 cells. For Rload=7, Ms=4 & Mp=2 is best. For Rload=12, Ms=4 and Mp=2 is best.
309
1
Parametric Table: Table 1 Run
Ms
Mp
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
1 2 3 2 4 1 2 3 2 4 1 2 3 2 4
1 2 2 4 2 1 2 2 4 2 1 2 2 4 2
Rload [Ω] 4 4 4 4 4 7 7 7 7 7 12 12 12 12 12
P [W] 56.6 226.6 292.1 300.9 300.8 41.6 166.4 321.5 188.7 456.0 27.0 107.9 225.7 115.6 370.6
310
2
P23 04
Equations 23.4
Calculation of cell operating temperature and NOCT.
Knowns Ta = 35; ηnotc = 0;
UL = 25 W/m2 ·C ;
GT = 1100;
τ = 0.94;
α = 0.92
GT N OCT = 800 W/m2
TaN OCT = 20 [C] ;
(1) (2)
a. Calculate cell operating temperature with Equation 23.3.3: The efficiency is unknown, but probably ranges between 5 and 15%. This fixes minimum and maximum possible temperatures: ηmin = .05;
ηmax = 0.15
(3)
T cM in = Ta + (GT · τ · α/UL ) · 1 −
ηmax τ ·α
Eqn 23.3.3
(4)
ηmin T cM ax = Ta + (GT · τ · α/UL ) · 1 − τ ·α
(5)
b. Calculate the cell temperature at NOCT conditions with Equation 23.3.2. By definition, the cell efficiency at NOCT conditions is zero. ηnotc TcN OCT = TaN OCT + (GT N OCT · τ · α/UL ) · 1 − Eqn 23.3.2 (6) τ ·α
Solution α = 0.92 GT = 1, 100 W/m2 T cM in = 66.45 [C] UL = 25 W/m2 -C
ηmax = 0.15 GT N OCT = 800 W/m2 Ta = 35 [C]
ηmin = 0.05 τ = 0.94 TaN OCT = 20 [C]
311
1
ηnotc = 0 T cM ax = 70.85 [C] TcN OCT = 47.67 [C]
P23 05
Equations 23.5
Power calculation from data at reference conditions.
$Units J K Knowns: Isc = 1.2 [A] ;
GT,ref = 1000 W/m2 ;
µIsc = .0005 [A/K] ; N s = 24;
µV oc = −.05 [V/K] ;
Voc = 15 [V] ;
GT = 750 W/m2 ;
Vmp = 12.5 [V] ;
Tc,ref = 298 [K] ;
Tc = 288 [K] ;
Imp = 1 [A]
Eg,ref = 1.12 [eV]
RLoad = 10 [ohm]
(1) (2) (3)
Use the EES program 23.RefProp to find the reference parameters. Rsh,ref = 78.78 [ohm]
(4)
aref = 0.601 [V]
(5)
IL,ref = 1.21 [A]
(6)
Io,ref = 1.464 × 10−11 [A]
(7)
Rs,ref = 0.596 [ohm]
(8)
Calculate parameters at operating conditions: a/aref = Tc /Tc,ref
Eqn 23.2.8
(9)
IL = (GT /GT,ref ) · (IL,ref + µIsc · (Tc − Tc,ref )) 3
Io /Io,ref = (Tc /Tc,ref ) · exp
1.60207 × 10·19 k#
Eg /Eg,ref = 1 − C · (Tc − Tc,ref )
Eqn 23.2.9 !
J eV
(10)
· (Eg,ref /Tc,ref − Eg /Tc )
Eqn 23.2.10b
Eqn 23.2.10a
(11) (12)
C = 0.0002677
(13)
Rsh /Rsh,ref = GT,ref /GT
(14)
312
1
Rs = Rs,ref
Eqn 23.2.12
(15)
a Find Power with Rload Cell Equations for I-V and P-V: V + I · Rs V + I · Rs −1 − I = IL − Io · exp a Rsh P =I ·V
Eqn 23.2.1
(16)
Eqn 23.2.2
(17)
V = I · RLoad
(18)
b Find Rmax that maximizes the Power. The EES Min/Max option was used to find the value of Rmax that maximizes the power. Vmax + Imax · Rs Vmax + Imax · Rs −1 − Eqn 23.2.1 Imax = IL − Io · exp a Rsh Pmax = Imax · Vmax
Eqn 23.2.2
(19) (20)
Vmax = Imax · Rmax
(21)
Solution a = 0.5808 [V] Eg,ref = 1.12 [eV] IL = 0.9038 [A] Io = 2.576 × 10−12 [A] µV oc = −0.05 [V/K]
aref = 0.601 [V] GT = 750 W/m2 IL,ref = 1.21 [A] Io,ref = 1.464 × 10−11 [A] N s = 24
C = 0.0002677 [1/K] GT,ref = 1, 000 W/m2 Imax = 0.7467 [A] Isc = 1.2 [A] P = 6.739 [W]
Eg = 1.123 [eV] I = 0.8209 [A] Imp = 1 [A] µIsc = 0.0005 [A/K] Pmax = 9.706 [W]
RLoad = 10 [Ω] Rsh,ref = 78.78 [Ω] V = 8.209 [V]
Rmax = 17.41 [Ω] Rs,ref = 0.596 [Ω] Vmax = 13.00 [V]
Rs = 0.596 [Ω] Tc = 288 [K] Vmp = 12.5 [V]
Rsh = 105 [Ω] Tc,ref = 298 [K] Voc = 15 [V]
313
2
P23 06
Equations 23.6
Calculation of power at a new set of conditions from data at measured conditions.
Data: Measured conditions: GT = 800 W/m2 ; Ta = ConvertTemp(C, K, 30); GT,ref = 1000 W/m2 ;
P1 = 12.2 [W] ;
Area = 0.1 m2
(1)
T1 = ConvertTemp(C, K, 18.6)
(2)
Ef f Equip = 0.92;
P2 = 10.0 [W] ;
T2 = ConvertTemp(C, K, 59.2); Tref,N OCT
TN OCT = ConvertTemp(C, K, 40) = ConvertTemp(C, K, 20); GT,N OCT = 800 W/m2 ;
(3) ta = 0.9
Use NOCT Conditions for finding cell loss coefficient: ta UL = GT,N OCT · Eqn 23.3.2 TN OCT − Tref,N OCT
(4)
(5)
Determine the array efficiencies at the two given temperatures: η1 =
P1 Area · GT,ref · Ef f Equip
(6)
η2 =
P2 Area · GT,ref · Ef f Equip
(7)
Determine the temperature coefficient of power from : η2 − η1 µP mp = T2 − T1
(8)
Use Equationn 23.3.3 to find cell temperature at the new operating radiation level and ambient temperature. UL is assumed to remain constant. Tc − Ta = (GT /GT,noct ) · (1 − η/ta) (9) TN OCT − Tref,noct Choose one of the measured conditions (e.g. condition one) as a reference and use Equation 23.2.16 to find array efficiency as a function of cell temperature. η − η1 = µP mp · (Tc − T1 )
Eqn 23.2.16
(10)
Use definition of maximum power point efficiency to find power output of system. P = Area · GT · Ef f Equip · η
(11)
Solution
314
1
Area = 0.1 m2 η2 = 0.1087 µP mp = −0.000589 [1/K] ta = 0.9 [-] Tc = 320.6 [K]
Ef f Equip = 0.92 GT = 800 W/m2 P = 8.51 [W] T1 = 291.8 [K] TN OCT = 313.2 [K]
η = 0.1156 GT,N OCT = 800 W/m2 P1 = 12.2 [W] T2 = 332.4 [K] Tref,N OCT = 293.2 [K]
315
2
η1 = 0.1326 GT,ref = 1, 000 W/m2 P2 = 10 [W] Ta = 303.2 [K] UL = 36 W/m2 -K
P23 07
Equations 23.7
A computer program to do Example 23.7.1
Without EES and the SETP functions this is a long problem which starts with the month’s average daily radiation and works through the PV array characteristics and output. Only the March results are presented. The first sets of calculations give the absorbed radiation hour by hour through a day estimated from monthly average daily radiation. ¯ Rh, Slope, Dec function φClark XCrit, K, Eqn 2.23.5 (1) ¯ Cos(Slope) K Rh − 0.981 · Xm = 1.85 + 0.169 · ¯ 2 − 0.0696 · 2 ¯2 K K (Cos(Dec)) Xm − 1 2 − Xm q 2 p = abs abs (g) − g 2 + (1 + 2 · g) · (1 − XCrit/Xm) g=
φClark = (1 − Step(XCrit − Xm)) · p
(2)
(3) (4) (5)
end
(6)
function Zi(Rb , slope, kT , I0, IT , ρ)
Eqn 23.7.6
(7)
cosB = Cos(slope)
(8) 2
a1 = Rb2 + ρ · (1 − cosB) · Rb +
(ρ · (1 − cosB)) 4
(9)
b1 = −.1551 + .9226 · kT
(10)
a2 = Rb · (1 + cosB − 2 · Rb ) + ρ · (1 + cosB − 2 · Rb ) ·
1 − cosB 2
b2 = .1456 + .0544 · ln (kT ) a3 =
1 − cosB − Rb 2
(11) (12)
2 (13)
b3 = (.2769 − .3184 · kT ) · kT
(14) 2
Zi = (a1 · b1 + a2 · b2 + a3 · b3) · (I0/IT )
(15)
end
(16)
Knowns: TN OCT = 46 [C] ;
ηref = 0.104;
ηmppt = 0.9;
Ac = 4 · 0.427 m2
316
1
(17)
Bc = 1200 [W ·h] ;
ηbatt = 0.8;
Lat = 40 [deg] ;
slope = 35 [deg] ;
µM P = −.00050 [1/C] ; ta = 0.9
ρg = 0.2;
hr = 1 [hr]
Surf AzAng = 0 [deg] ;
Tref = 25 [C] ;
TN OCT,ref = 20 [C] ;
(18) M onth = 3;
¯ = 13.2 MJ/m2 H
GN OCT = 800 W/m2
Assumed but does not matter as it cancels out.
(19) (20) (21)
N day = AveDay (month)
(22)
HrAng = (time − 12 [hr]) · 15 [deg/hr]
(23)
dec = dec (N day)
(24)
rt = rt, (HrAng, Lat, N day)
(25)
rd = rd, (HrAng, Lat, N day)
(26)
Ho = HZero, (N day, Lat)
(27)
¯ T = H/H ¯ o K
(28)
Rb = RBeam, (Lat, N day, HrAng, Slope, Surf AzAng)
(29)
¯ I = rt · H
(30)
¯T Hd/H = HDIF F BARHBAR Lat, N day, K ¯ · (rt − Hd/H · rd ) · Rb + HdH · rd · 1 + Cos(slope) + ρg · rt · 1 − Cos(slope) I¯T = H 2 2
(31) (32)
We could have used the following call to find I¯T and eliminated the previous 9 lines but some of the intermediate results are used below. ¯ Lat, Month, HrAng, Slope, SurfAzAng, ρg ) I¯T =I¯T,( H, Rh = I¯T /I
(33)
¯ T · rt /rd k¯ = K
(34)
I¯o = GZero, (N day, Lat, HrAng) ¯ I¯o , I¯T , ρg Z = Zi Rb , Slope, k,
(35)
UL · (TN OCT − TN OCT,ref ) = ta · GN OCT Eqn 23.3.2 ¯T · ta · (1 − ηref ) · Z T − T µ · I a ref M P η¯array = ηmppt · ηref · 1 + µM P · + ηmppt hr · ηmppt · UL · 0.0036 MJ/hr J/s
(37)
¯ ¯ Ei = η¯array · Ac · IT /hr · 277.777778
Xc =
¯ i · 0.0036 L
MJ/hr W
Ac · η¯array · I¯T
(36)
W MJ/hr
Eqn 23.7.4
(38)
(39)
· hr Eqn 23.7.7
(40)
317
2
¯ Rh , Slope, dec φ = φClark Xc , k,
(41)
¯ex,i = E ¯i · φ E
(42)
Eqn 23.7.8
¯L,i = E ¯i − E ¯ex,i E
Eqn 23.7.9
(43)
Find the Sum row in the Parametric table. ¯ex = Sumparametric(‘Table 1’ , ‘E bar ex i’ , 1, T ableRun#) E
(44)
¯L = Sumparametric(‘Table 1’ , ‘E bar L i’ , 1, T ableRun#) E
(45)
¯ = Sumparametric(‘Table 1’ , ‘L bar i’ , 1, T ableRun#) + 0.001 L ¯L /L ¯ fo = E
The 0.001 is to prevent divide by zero
Eqn 23.3.12
(47)
¯ex /L ¯ do = ηbatt · E Bc ∆f max = min 1 − fo , ¯ L · hr P = 1.315 − 0.1059 ·
(46)
(48) (49)
¯ · hr fo · L ¯T − 0.1847/K Bc
(50)
x = do + ∆f max p x − x2 − 4 · P · do · ∆f max ∆f s = 2·P
(51)
fmonth = fo + ∆f s
(53)
(52)
$ShowWindow Solution
Solution Variables in Main program Ac = 1.708 m2 Bc = 1, 200 [W-h] ∆fmax = 0.2844 [-] ∆fs = 0.2306 η¯array = 0.094 [-] ηbatt = 0.8 ¯ex = 252 [W] ηref = 0.104 E ¯ ¯L = 429 [W] Ei = 7 [W] E fmonth = 0.9461 fo = 0.7156 Hd/H = 0.4476 hr = 1 [hr] ¯ = 13.2 MJ/m2 H H MJ/m2 o = 27.44 I¯o = 0.3613 MJ/m2 I¯T = 0.15 MJ/m2 ¯ T = 0.481 K Lat = 40 [deg] ¯ i = 0 [W] L M onth = 3 N day = 75 P = 0.8931 ρg = 0.2 Rb = 1.734 Rh = 1.226 rt = 0.009238 Surf AzAng = 0 [deg] ta = 0.9 Ta = 9 [C] TN OCT = 46 [C] Tref = 25 [C] UL = 27.69 W/m2 -C Xc = 0.00 [-] Z = 2.24 [-]
dec = −2.418 [deg] do = 0.3361 ηmppt = 0.9 ¯ex,i = 7 [W] E ¯L,i = 0 [W] E GN OCT = 800 W/m2 HrAng = 82.5 [deg] I = 0.1219 MJ/m2 k¯ = 0.3374 ¯ = 600 [W] L µM P = −0.0005 [1/C] φ = 1.000 [--] rd = 0.01317 slope = 35 [deg] time = 17.5 [hr] TN OCT,ref = 20 [C] x = 0.6205
318
3
Parametric Table: Table 1 Run 1 2 3 4 5 6 7 8 9 10 11 12
time [hr] 6.5 7.5 8.5 9.5 10.5 11.5 12.5 13.5 14.5 15.5 16.5 17.5
Ta [C] 4 5 5 6 8 10 11 12 12 11 10 9
¯ IT 2 MJ/m 0.15 0.60 1.11 1.61 2.01 2.23 2.23 2.01 1.61 1.11 0.60 0.15
Z [-] 2.24 1.92 1.74 1.61 1.53 1.49 1.49 1.53 1.61 1.74 1.92 2.24
η¯array [-] 0.095 0.094 0.094 0.093 0.093 0.093 0.093 0.093 0.093 0.094 0.094 0.094
¯i E [W] 7 27 49 71 89 98 98 89 71 49 27 7
¯i L [W] 0 0 0 100 100 100 100 100 100 0 0 0
319
4
Xc [-] 0.00 0.00 0.00 1.40 1.13 1.02 1.02 1.13 1.41 0.00 0.00 0.00
φ [--] 1.000 1.000 1.000 0.090 0.175 0.220 0.219 0.174 0.089 1.000 1.000 1.000
¯ex,i E [W] 7 27 49 6 16 22 22 15 6 49 27 7
¯L,i E [W] 0 0 0 65 73 77 77 73 65 0 0 0
P24 01
P24_01_D.jpg
Equations Show Weibull distribution with k=2 is identical to Rayleigh distribution. This can be done analytically. The Weibull probablity distribution Eqn 24.2.7 with k=2 equals P(U)=(2/c)*(U/c)*EXP(-(U/c)2 ) Line before Fig 24.2.6 ¯ 2 =c2 *Γ ( 1+1/2)=c2 *π/4 U
320
1
Solve for c2 , ¯ 2 /π c2 =4*U Substitute c2 into the Weibull probability distribution results in ¯ 2 )*Exp(-(π/4)*(U/U ¯ )2 ) P(U)=(π/2)*(U/U which is Eqn 24.2.11, the Rayleigh distribution In the following we show both the Rayleigh and Weibull (with k=2) distributions on the same plot. k=2
(1)
¯ = c · Γ (1 + 1/k) U
(2)
¯2 · σ2 = U
(3)
Eqn 24.2.9 ! Γ (1 + 2/k) Eqn 24.2.10 2 −1 Γ (1 + 1/k) k−1
PW ei = (k/c) · (U/c) PRay = (π/2) ·
U ¯ U2
k · exp − (U/c)
(4)
¯ 2 · exp − (π/4) · U/U
(5)
Solution c = 16.93 σ = 7.841
k=2 U = 10
PRay = 0.0492 ¯ = 15 U
PW ei = 0.0492
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2
P24 02
Equations Find solutions to two simultaneous nonlinear equations. ¯ = 9 [m/s] U
(1)
σ = 3.5 [m/s]
(2)
¯ = c · Γ (1 + 1/k) U
(3)
2
2
σ =u ¯ ·
Γ (1 + 2/k) 2
Γ (1 + 1/k)
! −1
(4)
Solution c = 10.11 [m/s] σ = 3.5 [m/s]
k = 2.782 ¯ = 9 [m/s] U
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1
P24 03
Equations Determine the wind speed range for a range in power. $UnitSystem SI kPa C Knowns: ¯min = 7 [m/s] ; U
¯max = 7.5 [m/s] ; U
p = 101.3 [kPa] ;
T = 20 [C]
Standard conditions
ρ¯ = ρ (air, T = T, P = p)
(1) (2)
For Rayleigh distribution the parameter k is equal to 2. Then the parameter c can then be found from Eqn 24.2.9: ¯min = cmin · Γ (1.5) U
(3)
¯max = cmax · Γ (1.5) U
(4)
The power per unit area for the two average velocities is found from Eqn 24.2.13. 3 cmax Γ (2.5) P/Amax = ρ¯ · · 2 2 3 cmin Γ (2.5) P/Amin = ρ¯ · · 2 2
(5)
(6)
The power range in the Wind Atlas for these two wind speeds is 400 to 500 W/m2. This analysis gives 394 to 485 W/m2 which is pobably OK in a wind atlas.
Solution cmax = 8.463 [m/s] P/Amin = 197.2 W/m2 ¯min = 7 [m/s] U
cmin = 7.899 [m/s] ρ¯ = 1.204 kg/m3
p = 101.3 [kPa] T = 20 [C] 323
1
P/Amax = 242.5 W/m2 ¯max = 7.5 [m/s] U
P24 05
Equations Find average power knowing only average wind speed and standard deviation. ¯ , σ : c, k SubprogramF indCK U
(1)
¯ = c · γ (1 + 1/k) U
(2)
γ (1 + 2/k)
¯2 · σ2 = U
2
γ (1 + 1/k)
! −1
(3)
end CK
(4)
function Cp, (U )
(5)
If((U < 8) or (U > 22)) then
(6)
Cp,=0
(7)
else
(8)
If U < 18 then h i h i 2 3 h i (m/s) (m/s) [(m/s)] -1 − 2506 + 309.38 − 15.09 + 0.2602 (m/s) · U Cp,=6936.4 3 2 U U U else
(10)
(11) h
Cp,=1388
3
(m/s)
i (12)
U3
endif
(13)
endif end
(9)
(14)
function
(15)
Knowns η = 0.89;
A = 300 m2 ;
ρ = 1.225
(16)
¯ high =4.75[m/s]; U ¯ low =3.61[m/s];σ high =2.6[m/s]; σ low =2.20[m/s]:. A Parametric Table is defined From Figure 24.2.3, U with a range of average wind speeds and standard deviations. Find Weibull c and k ¯ , σ : c, k call F indCK U
(17)
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1
k−1
p = (k/c) · (U/c)
k · exp − (U/c)
Weibull probability density function
Integrate from 8 to 22 Z 22 [m/s] A·ρ·η Cp, (U ) · U 3 · p dU P¯ = · 2 8 [m/s]
kW · 0.001 W
(18)
(19)
¯ for various values of σ that the maximum power occurs when both the average and It is clear from the graph of P¯ vs U standard deviation of the velocity are at a maximu. Probably the best year is then 1988 or 1990. The worst year is at a low velocity and low standard deviation. The worst year is 1980.
Parametric Table: Pbar vs Ubar Run 1 2 3 4 5 6 7 8 9 10 11 12
¯ U [m/s] 3.5 4.125 4.75 3.5 4.125 4.75 3.5 4.125 4.75 3.5 4.125 4.75
σ [m/s] 2.2 2.2 2.2 2.3 2.3 2.3 2.4 2.4 2.4 2.5 2.5 2.5
c [m/s] 3.910 4.652 5.362 3.893 4.645 5.364 3.872 4.635 5.363 3.848 4.623 5.359
k [-] 1.632 1.956 2.289 1.554 1.862 2.178 1.484 1.776 2.077 1.420 1.698 1.985
P¯ [kW] 1.55 1.92 2.68 2.00 2.44 3.31 2.50 3.01 3.99 3.04 3.64 4.73
Pbar vs Ubar
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2
Solar Energy A semester course for engineering seniors and graduate students
Text: Solar Engineering of Thermal Processes - Third Edition, 2006, Second (or subsequent) Printing, by Duffie and Beckman Assignments: Noted as sections of the text. Supplemental notes will be provided for several assignments. Problems: There will be 12 weekly problem sets (called A through L) plus a semester project. You will be assigned to work with a new partner on each problem set; you and your partner will turn in one solution. The problem sets will be evaluated and the grades will contribute to the final grade. Most of these problems are best done with EES, using the EES/SETP functions that are provided. Note that some of the problems require qualitative statements or interpretations; these are important parts of the assignments and should not be neglected. Problem sets are due on the day indicated and must be turned in at the end of class. Problem due dates will be used to discuss the solution and, if time permits, start on the next day’s material. Most of the problem sets constitute an interrelated series of calculations, with the results of one set providing information for subsequent sets. The first problem set, A, will be an introduction to the course and will be a prediction of performance of a solar process. You will run a simulation program (which will be provided, and which represents a practical solar heating system operating in a specific location) and prepare a brief summary and interpretation of the results. Subsequent problem sets (with one or two exceptions) will treat in some detail parts of the simulation: Calculation of solar radiation absorbed by the collector, Calculation of performance of collector and other components in the system, Calculation of system performance. Calculation of system economics. There will be a design project that will integrate various aspects of the course. It Project: will replace the final examination. A report will be required and will be due on final examination day.
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Solar Energy Technology Day 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44
Topic Introduction: Collector, Storage Energy Balances; Systems Simulation of a Solar Energy System Solar Radiation: Definitions, Angles, Solar Geometry, Shading, Extraterrestrial Radiation Engineering Equation Solver (EES) Measurements, Radiation Data & Processing Radiation on Sloped Surfaces Utilizability Concepts; Heat Transfer Topics Heat Transfer Topics; Spectral Radiation Properties Spectral Radiation Properties; Transmittance Transmitted Radiation; Absorbed Radiation Flat Plate Collectors; Loss Coefficients F, F', F", FR, QU Review EXAM I Critical Radiation; Mean Temperatures; Heat Capacity Effects; Liquid and Air Heaters Collector Characterizations and Tests, Practical Considerations Concentrating and Evacuated Collectors Energy Storage Solar Process Loads; System Analysis System Analysis and Simulation Photovoltaics; Cell Energy Balances More photovoltaics Design Methods: f-Charts; Utilizability Basic and Life Cycle Economics P1, P2 Method; Optimizations; Uncertainties Wind Power More wind Power Project discussion Review EXAM II Passive Processes Passive and Hybrid Systems Other Applications Closure
Assignment (Chapter 10) Chapter 1 Set A 2.1-2.6, 2.9-2.13 2.14-2.21 Set B 2.22-2.25, 3.1-3.10 3.11-3.17, 4.1-4.6 Set C 4.7-4.13, 5.1-5.2 5.3 –5.12, PV Notes Set D 6.1-6.4 6.5-6.7 Set E 6.8-6.15 Set F 6.16-6.24 Chapter 7 Set G Chapter 8 Ch. 9, 10.1-10.7 Set H 10.8-10.10 Chapter 23 Set I Chapters 20 & 21 Set J 11.1-11.7 11.8-11.10 Set K Chapter 24 Set L
Ch 14, 22.1-22.2 22.3 - 22.5 Ch 15, 17, 18
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Problem Set A A solar heating system for a residence is to be located in Madison, WI, and is to supply space and hot water heat for a residence. A schematic diagram of the system is shown below. Your task is to calculate how much energy this system will deliver to meet the building heating and water heating loads of the house for a typical year.
The system has the following major components: The collector is a flat-plate liquid heater with one glass cover, similar to those shown in Figures 6.1.1 and 6.3.1. The collectors are mounted flush on the building (in a manner similar to that in Figure 13.2.5b and c). The total roof area suitable for collectors is 75 m2. The collectors are facing south at a slope of 60 degrees from the horizontal. The collector parameter “intercept efficiency” is essentially the optical efficiency of the collector and is 0.8. The “negative of the first order loss coefficient” is essentially an overall thermal loss coefficient from the collector fluid to the ambient and is 3.50 W/m2C. The “negative of the second order loss coefficient” indicates how much the thermal losses deviate from being linear with temperature. For this simulation assume a value of 0.013 W/m2C. The “First order IAM from ASHRAE test” is related to the change in optical efficiency as a function of the angle of the beam solar radiation. The default value of 0.1 is typical of modern collectors. These parameters are built into the “Average performing flat-plate (0.8, 3.50, 0.013)” when you choose to “Select solar collector from a list”.
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The collector-heat exchanger isolates the anti-freeze solution in the collector loop from the water storage tank loop. The effectiveness is assumed to be 0.6. The main storage tank is an insulated tank with a loss coefficient of 1.0 W/m2-C that is sized in proportion to the collector area. Typical values range from 30 to 100 liters/m2. The default case uses 75 liters/m2. The auxiliary energy maintains the upper portion of the tank to 60C. The domestic hot water (DHW) subsystem consists of a heating coil (heat exchanger) in the main storage tank. Mains water enters this heat exchanger and is heated to the delivery temperature of 45C. The hot water load is 60 liters per person per day for a family of five. The mains temperature is assumed to be constant at 10C. If solar energy heats the domestic hot water above 45C (as it will in the summer) then a bypass system (not shown) takes mains water and mixes it with the too-hot water to deliver water at 45C. The auxiliary energy supply maintains the upper section of the tank at 60C. The Auxiliary heat exchanger is located in the top of the tank and is of sufficient capacity that it can supply all of the domestic hot water and house heating energy needs if necessary. The space heating system can meet the entire space heating load. Sixty degree or higher water is removed from the top of the tank, circulated through a water-to-air load heat exchanger (LHX) and returned to the tank. The flow rate of the water is controlled to exactly meet the house heating load. The LHX is large enough to meet the entire house load. The building has an overall loss coefficient, including infiltration, (UA) of 350 W/C. Details of how systems are controlled and related matters will be dealt with as the course progresses. The Problem Using the simulation program provided (CombiSys.zip on the course homepage), simulate the operation of this system in Madison, WI for the entire year using the default parameter set. Perform an energy balance on the main solar tank for the entire year. (Energy in – Energy out – Energy Stored = Error) The error is due to numerical tolerances in solving the equations. Express the error as a percentage of the delivered solar energy, 100*Error/(Solar Energy in) = %error. In addition, you can change several of the design parameters of the system. These include: The collector area, which cannot exceed 75 m2. Storage unit size normally varies in proportion to the collector area; ratios of 50, 75 and 100 liters/ m2 can be assumed. The collector slope can conceivably vary between 30º and 75º. You are to write a brief report that is intended to inform a group of contractors and architects about the performance of the system and the effects of changes in the design on system performance. Use plots or diagrams to illustrate your results. To reduce the number of runs, you can investigate storage size only for the 60º slope. The usual performance figure is the solar fraction, F, defined as the ratio of the solar contribution to the load divided by the load. Note that the solar contribution to the load is NOT ESol due to losses from the tank.
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Estimate how much the home owner can afford to pay for the solar equipment if the auxiliary energy is (a) natural gas and (b) electricity. There are many other design parameters that for this problem you cannot change or do not need to change. These include the heating load of the building and the characteristics of the collector. We will look at the impact of other design parameters during the semester. Installing TRNSYS on your computer: Information on installing TRNSYS (pronounced Transys) follows. Installing: 1. Download the file CombiSys.zip from the course homepage and place it in the c:\temp directory. 2. Right click on the file and select to install into the directory c:\temp. You MUST install in this directory. 3. You should end up with a new directory called c:\temp\combisys on your computer. Running TRNSYS 1. In the c:\temp\combisys directory there is a program called TRNSED (pronounced Transed). Double click it to run. 2. Click “OK” and a dialog box will come up asking you what file you would like to open. Choose CombiSys.trd and a window describing the system in the problem statement will open. The second file CombiSys_defaults.trd is identical to CombiSys.trd and is provided so you can restore the default settings. 3. The various system options should be self-explanatory. Help is available by ckicking on the question marks. You can run the system by either pressing the F8 key or by choosing “Calculate” from the TRNSYS menu. Comments and suggestions: It is suggested that you first simulate the system operation for one week. The computing time will be very small, and you can use this weeklong operation to get accustomed to the program and what it does. Quantities like solar energy collected, energy lost from the tank, auxiliary energy and various temperatures are computed as a function of time. Energy rates are integrated to give monthly energy quantities. Examining the On-Line plots You may manipulate the on-line plots in a variety of ways. The right mouse button will start and stop the simulation. After the simulation is complete select NO to the question “Exit on-line Plotter”. With the plot on the screen, click on the various plot identifiers at the top of the plot – the individual plots should disappear and reappear. Click and drag the mouse over part of the plot for a blow-up of a region. Click near the top or bottom on
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either the right or left had axis numbers to change the scale. If the simulation is more than one page you may move back and forth in time with the tabs at the bottom. There are two tabs at the bottom for looking at either instantaneous values or integrated values. When finished, go to menu item Calculations and choose Exit. Examining the Output Once the simulation has completed and you have returned to TRNSED, you will find an output file, COMBISYS.OUT, under the “Windows” menu item at the top of the screen. The output is a text file that you can copy and paste into Excel. The values printed in the output file are as follows: a. b. c. d. e. f. g. h. i.
UTank: change in internal energy of the storage tank from the start of each month ESol: the integrated energy transferred across the collector loop heat exchanger EAux: integrated auxiliary energy added to the space to meet heating load requirements ELossTank: the integrated energy loss from the tank (assumed to be in an unheated area of the house) EMains: the integrated energy entering the tank with the water from the mains EDHW: the integrated energy leaving the tank with the DHW EHouse: integrated energy losses from the house EIncSol: the integrated solar radiation incident on the collectors EHorSol: the integrated solar radiation incident on a horizontal surface
Values are printed for each month. If the simulation ends within a month, a value will be printed for the completed portion of the last month. You will need to add up the monthly values to obtain yearly values. Copy the file and paste into EXCEL (or other spreadsheet program) to do your calculations. Annual information of this type, as will be seen later, is essential information in determining the economics of the application. If you look at “View Simulation Results”, you will find a summary of the performance for the total time of the simulation.
PROBLEM SET B In Problem Set A, the performance of a solar heating system in Madison, WI was explored. In this and subsequent sets, you will look at a series of aspects of this heating simulation. This problem is concerned with calculations on basic aspects of solar radiation. Set C will use the concepts in this set to deal with calculation of radiation on sloped surfaces starting with measured data on a horizontal surface. Subsequent sets will treat absorbed radiation, collector energy balances, other component performances, and systems calculations. At the end of this series, we will have explored the basis of all of the aspects of Set A, and you will have a picture of how to calculate long-term system performance. Use EES and the EES/SETP function (available on the course home page) for your solution. It is necessary to put in the time to learn to use EES and the EES/SETP functions otherwise future problems will take MUCH longer. A few hours extra on this problem will be well spent.
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For February 16 (the average day for February from table 1.6.1) in Madison, WI: 1. 2. 3. 4. 5.
6. 7. 8.
9.
What is the day length? At 10:30 AM solar time, what is the local clock time? At what time does sunrise occur (both clock time and solar time)? For this day, what is the declination? Is it significantly different at sunrise and sunset? How does the angle of incidence of beam radiation (Theta) vary through the day for a surface sloped 60 degrees to the south? Make a parametric table with time_solar, time_clock and Theta. Make a plot of theta versus solar time (use the midpoint of the hour). Add R_b to your table of part 5 and show a plot of Rb vs. solar time. Add Go to your table of part 6 and show a plot of Go [W/m2] versus solar time. What is the extraterrestrial radiation on a horizontal surface for this day in W-hr/m2? Calculate using the appropriate SETP/EES function and also add up the values of Go (times one hour) from part 7. Note that in EES you can have a final row in a parametric table that contains the sum of the values in the column – see Options\Preferences\Options\Include a sum row in the parametric table. How much of the extraterrestrial solar radiation is in a wavelength range of 0.640 µm to 1.100 µm? Express the result both as a fraction of the total and as energy rate in W/m2.
Problem Set C For the same location and date as in Set B (Madison, WI, south facing surfaces, February 16, longitude=89.4º the hour-by-hour radiation on a horizontal surface, I, the normal beam radiation, Ibn, and the ambient temperature, Tamb, are given in the table. The time is clock time (not solar time) in the central time zone and represents the integrated values for the previous hour (e.g., for the row at time 9 the values of I, Ibn and Tamb are the averages over the time from 8am to 9am). Due to measurement “problems” it may be necessary to set the radiation values for hours that include sunrise or sunset to zero. To transfer the following information into EES, select the entire table, including the column header and units and then copy. In EES open a “Lookup Table” with 4 columns and 24 rows. In the upper left-hand corner of the table select Paste Special and then select to paste column names, units and data. time hr 1 2 3 4 5 6 7 8 9 10 11
I 2
W/m 0 0 0 0 0 0 0 31 201 390 503
Ibn 2 W/m 0 0 0 0 0 0 0 55 674 813 883
Tamb ºC -3 -3.1 -3.3 -3.1 -3 -2.8 -4.1 -5.4 -6.7 -5.9 -5.2
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12 13 14 15 16 17 18 19 20 21 22 23 24
585 597 415 213 198 69 17 0 0 0 0 0 0
913 917 360 40 287 47 42 0 0 0 0 0 0
-4.4 -5 -5.5 -6.1 -7.6 -9.1 -10.6 -11.3 -12.1 -12.8 -13.5 -14.3 -15
For this day, for a slope of 60 degrees to the south and for a ground reflectance of 0.4: 1.
2.
3.
4.
5. 6.
Plot kT , Io [W/m2] and I [W/m2] as functions of hour (use the midpoint of the clock-time hour). Comment on any unusual aspects of the plots. Use the EES parametric table and the appropriate SETP/EES functions. Using the data in the table, plot IT [W/m2] and the beam, diffuse and ground-reflected components of IT as a function of clock time. Use the Liu and Jordan method. Why can’t we use the SETP/EES procedure I_T_LJ_p_ to do this calculation? Estimate Ib and Id from I and use your estimated value of Ib and Id to estimate IT using the L & J method. How do the values compare to those obtained in question 2? Do not use the function I_T_LJ_ to do the calculations. For slopes of 30, 45, 60 and 75 degrees, plot IT [W/m2] vs. time. Use the L&J method and the function I_T_LJ_ to do the calculations. If the selection of slope is based on maximizing incident radiation for the day (as we will see, this is not necessarily the correct basis), what slope would you select for this day? (Note: under Options\Preferences you can have EES sum a parametric table column). How might your answer to this question change if a summer’s day had been used in the calculations? What are H and Ho [MJ/m2] and KT for this location and day? What is H T [MJ/m2] for the month of February for this location using the data in Appendix G and a slope of 60 degrees. Use both the Liu and Jordan method and the K-T method. Compare your answers with the value calculated by TRNSYS in set A. Explain why they are not the same.
Problem Set D A collector that is to be used on the Madison, WI heating system has radiation properties of the absorber plate as given in Figure 4.8.2, curve C. (See the course homepage for 'Figure 4.8.2 Curve C as a lookup file'. The collector is expected to deliver most of its output when the absorber surface temperature is about 325 K. The data in the table ranges from 0.4071 micrometers to 25 micrometers, which is not sufficient for the calculations of this problem. Extend the data down to 0.01 micrometers by adding a point
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at 0.01 with the same reflectance value as at 0.4071. Extend to 100 micrometers in the same manner. 1.
What is the absorptance for solar radiation assuming the sun is a blackbody at 5777K? Write an EES function called Planck_(lambda, T) that will evaluate the blackbody energy at a specified wavelength and temperature (Equation 3.4.1). Use the EES integral function and your Planck function to integrate Equation 4.5.1 where epsilon and epsilon_lambda are replaced by rho and rho_lambda. Comment on the differences when you integrate from 0.01 to 100 micrometers versus integrating from 0.4071 to 25 micrometers.
2.
Repeat problem 1 but assume the solar radiation is distributed as in Table 2.6.1 (i.e., air mass 2). Use the Insert\modify Array command from the Edit menu to insert the wavelength bands from Table 2.6.1. Find an average rho in each band and use Equation 4.6.5 to evaluate the solar absorptance.
3.
What is the emittance for long-wave radiation at 325 K? Comment on the error associated with extrapolating the data.
4.
This surface is proposed for use in other collectors that will operate at different temperatures. Plot the emittance versus surface temperature for a range of temperatures from 300 K to 800 K. Use the EES curve fit option under the menu Plots and find a 2nd order polynomial (in degrees Kelvin) that fits your results.
Problem Set E The surface to be used as the absorber plate in collector has a normal incidence absorptance of 0.85. The collector has a single glass cover of low iron content and a KL of 0.121 per sheet. Use the same day in your calculations as was used in Set C. Do not use the beam radiation values given in the weather data (the SETP functions for S uses total radiation and estimates the beam and diffuse breakup). Use the HDKR method in your analysis so we all use the same algorithms. 1. Plot S and IT for slopes of 30, 45, 60, and 75 degrees as a function of hour angle (four graphs). What are the general trends? 2. Plot (τα)ave for the four slopes as a function hour angle. Can you draw any general conclusions? 3. It is proposed that the collector design is to be modified by the addition of a second cover. For a slope of 45 degrees compare the absorbed radiation for this day for the two cover collector with that of the one cover collector. What can you speculate about the utility of second cover? 4. There is also the possibility of using a flat-black absorber surface which has an absorptance at normal incidence of 0.96 and an emittance of 0.86. What effect would use of this surface have on absorbed radiation for the 45 degree slope?
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Problem Set F A program fragment is given below that describes a solar collector for use in Madison WI. You can copy and paste the fragment into EES. Your job is to complete the program and answer the following questions: 1. For February 16th in Madison, WI for the hour 11 to 12, with a water inlet temperature of 35C, compute the hourly output of the collector (MJ/hr) and the outlet fluid temperature. If your program is working correctly, ηcoll = 0.487. You will be using this program for Sets G and H so you want to get it working correctly. To show that your program is working correctly, print values of: ηcoll , S, F`, F``, FR, UL, T_plate, T_bar_bar_fluid and Qu. 2. The heat transfer coefficient inside the tubes is given as 175 W/m2-K. Use this value in your solution to parts 1 and 2. Calculate the value of the heat transfer coefficient inside the tubes and compare it to the 175 W/ m2-K. Use the Lookup table 'PropyleneGlycol.LKT' from the course homepage \Computer Stuff and read it into your EES program with Tables\Open Lookup Table. This table contains the properties of 50-50 propylene glycol and water from 244.3K to 394.3K. The temperatures in the table are in K. You can easily extract information from the lookup table using the linear interpolate function, Interpolate1. For example, to extract the viscosity at the mean plate temperature (assumed to be the same as the tube wall temperature) use: mu_glycol_wall=Interpolate1('PropyleneGlycol', 'T', 'mu', T=ConvertTemp(C, K, T_bar_plate)) 3. The wind heat transfer coefficient has been assumed to be equal to 10 W/m2-K. Use the 'Uncertainty Propagation' under the 'Calculate’ menu to estimate the impact on not knowing the wind coefficient better than plus or minus 5 W/m2-K. 4. The plate emittance is assumed to be 0.14. The following equation representing the plate emittance as a function of plate temperature was obtained from the solution to Problem set D4: εplate= - 0.00443111 + 0.000345085[1/K]*T_K + 2.61859E-07[1/K^2]*T_K^2 where T_K is the mean plate temperature in K. Check to see if it will be necessary to account for this variation in the plate emittance in evaluating annual performance. The collector in the Madison, WI application has the following characteristics: (If you download the Word copy of this problem you can copy the following from Word and paste into EES.) "
"!Data" Ins_back=0.07 [m] k_ins=0.0245 [W/m-K]
"back insulation thickness" "polyurethane rigid foam"
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U_edge=0.098 W/m^2-K] "edge loss coefficient" Space=0.028 [m] "plate to cover spacing" T_in=35 [C] "also check using 60 and 85" slope=45 [deg] Long=89.4 [Deg] Long_std=90 [deg] Lat=43.1 [deg] Month=2 Day=16 n=nDay_(month, day) SurfAzAng=0 [deg] GrRef=0.4 hour=1 [hr] zero=0 Tubes=8 "number of tubes in parallel - neglect the impact of the headers" Sp_tube=0.115 [m] "distance between tube centers" L=2.5 [m] "length in the flow direction" W=Tubes*Sp_tube "width of collector" A_c=L*W Fluid$='PG' "Propylene Glycol / Water" Conc=50 [%] "Glycol/Water concentration" alpha_n=0.881 "solar absorptance of absorber plate – from problem D4" Dia_Tube=0.008 [m] "use for both inner and outer diameter" Thick_plate=0.0005 [m] k_plate=385 [W/m^2-K] "copper" Flow=0.025 [l/s] "flow through the collector, divided evenly among the tubes" KL = 0.121 "Glass KL product" Refrind=1.526 "Glass refractive index" C_b=1e6 [W/m-K] "bond conductance" h_fi=200 [W/m^2-K] “heat transfer coefficient inside the tubes" N_cov=1 "Number of covers" h_w=10 [W/m^2-K] "with a plus or minus 5 W/m^2-K uncertainty" "!Get weather data from lookup table, ‘Data’." Row=12 HrAng=Lookup('Data', Row, 'HrAng') T_sky=T_amb I=max(0.01[W-hr/m^2], lookup('Data', Row, 'I'))*Convert(W-hr/m^2, MJ/m^2) $Ifnot parametrictable T_amb=lookup('Data', Row, 'T_amb') $EndIf $If parametricTable='Fixed epsilon' epsilon_plate=0.14 "IR emittance of absorber plate" $Else epsilon_plate=-0.00443111 + 0.000345085[1/K]*T_K + 2.61859E-07[1/K^2]*T_K^2
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$EndIf
"Find average plate and fluid temperatures – make guess to get started – later replace by appropriate functions" T_bar_plate= T_bar_Fluid= "Find U_top and U_L" U_top= U_L= "Find F`, collector efficiency factor" F`= "Find F_R and F`` " F_R= F``= "Find absorbed solar using HDKR method" S= "Find useful energy gain" Q_u= "Find the outlet temperature" "Find the collector efficiency" “Find the actual inside tube heat transfer coefficient. Need to find properties, Reynolds, Prandtl and Nusselt numbers. “
Problem Set G This problem simulates a collector test. Modify your program from set F to generate plots of efficiency versus (T_in-T_amb)/I_T and K_tau_alpha versus (1/cos(theta)-1). Assume the tests are carried out in Madison, WI on February 16 but that the day is clear with kt equal to 0.7 at solar noon and the ambient temperature remains constant at 12.1 C. All tests are completed at noon. In reality it takes time to change the test settings and to let the system come to a new equilibrium condition. Use a range of inlet temperatures from just a little above T_amb (e.g., 12.2C) to 100C. The collector slope is such that the incident beam radiation is normal to the collector surface when the surface azimuth angle is zero. Your program should include the effects of the plate emittance as a function of plate temperature and the effects of the inside-thetubes heat transfer coefficient as a function of flow rate and the mean fluid and mean plate temperatures. You should set up a parametric table with each row representing one test point for
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the DELTA T over I plot. In a second parametric table vary the azimuth angle from 0 to some value such that theta goes from 0 degrees to 60 degrees in order to generate the plot of Kta versus (1/Cos(theta)-1).
1. What is the collector slope? 2. Plot collector efficiency versus DELTAT/I and plot the ratio FR(τα)/ FR(τα)n versus (1/Cos(theta)-1). 3. Use the Curve fit option in the Plot menu to determine FR(τα)n, FRUL and bo. Also determine the parameters of equation 6.17.7. Do you expect that FR(τα)n found using the two different models will be the same - explain.
Problem Set H 1. Use your program from Set G to evaluate the impact of flow rate on collector performance. Plot efficiency versus Delta T over I for the conditions of set F and for double and half of the flow rate. Comment on the use of the flow rate correction factor (Section 6.20) in evaluating the impact of varying flow rate on the collector parameters FR(τα)n and FRUL. 2. In Europe the test data is presented in terms of (T_ave-T_amb) rather than (T_in-T_amb) where T_ave = (T_in + T_out)/2. Make a plot showing both eta vs (T_ave - T_amb)/I and eta vs (T_in - T_amb)/I and comment on the use of Equations 6.19.2 and 6.19.3 for converting data from one presentation method to the other. 3. Use your program to investigate the impact of test-day conditions on collector parameters (e.g. kT of 0.5 instead of 0.7 and Tamb=25 instead of 12.1). The following two questions were not used but would make an excellent follow-on problem. 4. There is always an uncertainty in any measured variable. Estimate the resulting uncertainty in determining the linear collector parameters (FR(τα)n and FRUL) when the uncertainty in all temperature measurements is ±0.2C, in flow rate is measurements is ±3% and in solar radiation measurements is ±5% 5. Show uncertainty bars on a plot of efficiency vs DELTA T over I for the nominal conditions (i.e., the conditions of Set F). Note that EES can plot uncertainty bars.
Problem Set I For Madison, WI on February 16, (Latitude 43.1º, Longitude=89.4º), the hour-by-hour radiation on a horizontal surface, I and the ambient temperature, Tamb, are given in the table of problem set
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B. The time is clock time at the local standard meridian (90º). Due to sunrise and sunset “problems” set the hours 8 (i.e., 7-8) and 18 (i.e., 17-18) to zero radiation. Consider a solar system with a collector, a fully mixed storage tank and a house heating load. The losses from the house can be modeled as an overall loss coefficient (UAhouse) times the difference in the house temperature and the ambient temperature. The house has average internal gains due to people and electrical use of 1500W. The storage tank is located outside. The tank is sized with the rule-of-thumb of 75 liters per square meter of collector. Assume that the solar energy and ambient temperature recorded at the end of an hour was constant for that hour. The characteristics of the various components are as follows: "!Data" Lat=43.1 Day=16 Month=2 FRta=0.7 FRUL=4 slope=50 SurfAzAng=0 GrRef=0.4 A_c=35 b_o=-0.14 Vtank\A_c=75 h_tank=2 U_tank=0.30 UA_house=300 T_house=23 G_house=1500 hour=1 DELTAtau=1
[deg]
[W/m^2-K] [deg] [deg] [m^2] [l/m^2] “tank volume per unit of collector area" [m] “tank height" [W/m^2-C] “U-value of tank insulation." [W/C] "loss coefficient of the house" [C] "set temperature of house" [W] "internal gains of house due to people, electrical use, etc." [hr] [hr]
1.
Determine the “steady-periodic” solution for this one day. That is, find the initial tank temperature at midnight such that when the system is operated for the 24 hour period the tank temperature returns to the initial temperature.
2.
Plot the steady-periodic tank temperature as a function of time.
3.
Plot the collector useful gain, Qu,, the tank loss, Qtank loss and the heating demand of the house, Qhouse for the 24 hour period all in MJ/hr.
4.
Discuss your results.
5.
How would you have to change your program if the collector area were 10 m2?
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You may wish to use the EES MAX function. The MAX function returns the maximum of two expressions. So, with the EES expression Z=MAX(0, X-Y), Z will never go below zero no matter the values of X and Y. Use a parametric table with 25 entries where the first entry is the initial condition. Use Euler integration with a one hour time step. You will need to start your “table solution” in row 2 since row 1 should contain the initial conditions. To extract information out of a completed row (i.e., the temperature of the tank at the start of an hour is in the row above the one you are currently working on) use: T_tank=TableValue(‘TableName’, TableRun#-1, ‘T|Plus_tank’) where T|plus_tank is the previously calculated tank temperature (note that EES will format T|Plus_tank as T+tank).
Problem Set J The plot shows current and voltage for a silicon PV module operating at 1000 W/m2 at two cell temperatures, 25 and 50ºC. Determine values for Io, IL, Rs, a µIsc and µVoc. The NOCT for this module is 51ºC. The module area is 3.914 m2. The number of cells in series string is 110 and the number of series strings in parallel is 60. Determine the maximum power point current, voltage and power for this module when the ambient temperature is 10ºC and the incident radiation is 900 W/m2. 10
500
Tcell=25 °C
9
400
8
Tcell = 50 °C
6
300
5 4
200
3 2
100
1 0 0
10
20
30
40
50
Voltage [volts]
Reading (very carefully) from the plots: V_oc_ref=63.2 "[Volt]" V_oc_50=57.6 "[Volt]"
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60
0 70
Power [W]
Current [amps]
7
I_sc_ref=9.0 I_sc_50=9.3 I_mp=8.4 V_mp=51.4
"[Amp]" "[Amp]" "[Amp]" "[Volt]"
Problem Set K In Problem Set A, the annual thermal performance of a solar heating system was determined, and F vs Ac curves were generated for three ratios of storage capacity to collector area. In this set, we will use these results in an economic study of this solar application. Use the F vs Ac data from Problem Set A for 75 liters per square meter of collector. Ac 0 5 7.5 15 30 45 60 75
F 0 9.1 13.3 24.6 41.5 54.0 63.6 71.1
As noted in Chapter 11, there are a number of “figures of merit” that are favored by or used by those designing and evaluating solar energy applications. The criterion of economic viability that we recommend is Life Cycle Solar Savings. In this problem you are to consider LCS and three additional criteria. The four criteria are: a. Life Cycle Savings b. Payback time B (Section 11.3) with fuel costs discounted. c. Return on Investment. d. Cash Flow, considering principle and interest payments, payments for taxes and insurance, and fuel purchase savings, all in the first year. This is sometimes called the PITI (Principle, Interest, Taxes and Insurance) method. Economic parameters are as follows (with rates in percent per year): Mortgage interest rate = 5.8 % General inflation rate = 2.0 % Period of analysis = 20 years Term of loan = 20 years Property tax rate = 3.5 % Effective income tax rate = 39 % Down payment = 0.2 of investment Ratio year 1 miscellaneous costs to investment =0.050 Ratio of assessed valuation to investment = 0.90 Ratio of resale value to investment = 0.25
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C=0 (A non income producing system) First year fuel cost = $11/GJ (what you pay for fuel as it enters the furnace) Furnace efficiency = 0.8 Energy cost inflation rate = 5.2 % Discount rate = 4.5 % Area dependent cost = $ 150m2 Fixed cost = $ 550 The loads (from problem set A) are 2.0645*107 kJ for the domestic hot water and 1.2475*108 kJ for the house heating. 1. The object is to find economic optimum collector areas for the 75 liters/m2 systems (if optima exist) using each the four methods and compare them. How do the results using these four criteria compare? Would you expect them to be different? If so, why? 2. Repeat the process, assuming that the collector unit area costs are $240/m2 instead of $150/m2. Are the results very much different? 3. If you have an EES program to do this problem, you can easily explore the effects of other economic parameters. What might be interesting variables to investigate?
Semester Project The following is an example of a system simulation problem given as a semester project. This requires the integration and application to systems of many of the ideas used in short problems in this appendix. The project is open-ended in the sense that the student must decide what variables will be treated and how the results will be presented. Appropriate meteorological data must be available. There are many variations on problems of this type; this one includes some complications that can be omitted if a shorter project is wanted. INDUSTRIAL AIR HEATER SIMULATION The Application For an industrial operation, a continuous supply of heated air is required at a rate of 1.40 kg/s, at a temperature of at least 65 C. The air supply is at a constant temperature of 22 C. The location is Albuquerque, NM. A liquid solar heating system has been proposed to supply part of the needed energy. The major system components are: collector, collector heat exchanger, storage tank, load heat exchanger, and auxiliary heat exchanger. They are arranged as shown in the diagram. You are assigned the task of evaluating this system and reporting to your management on the technical and economic feasibility of the system. To do this, you are to write a simulation program that will allow you to do year-long simulations of the system, and allow you to explore the effects of several critical variables.. The results of the thermal analysis will
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then be used as the basis for an economic analysis. In the proposed process the air enters the heating section at 22 C, is heated first by solar energy, and is then heated in the auxiliary heat exchanger. It is to leave the heater section at a temperature no lower than 65 C. The auxiliary heat exchanger has the capacity to provide all of the needed energy if necessary. (Thus its cost is common to all systems and does not enter into economic comparisons of systems.) Freeze protection is to be provided by use of an antifreeze solution in the collector loop. You should consider what, if anything, should be done about boiling. The Equipment Some of the component design parameters can be fixed, based on past experience and on equipment that is available. For purposes of these simulations, assume the following: Collector: FR (τα)n = 0.81 and FRUL = 4.45 W/m2C at mɺ /Ac = 0.015 kg/m2s, bo = –0.12, γ = 0, β = latitude and Ac is variable Collector heat exchanger: The UA of the exchanger and thus also its effectiveness εc can be specified at any level. Storage: A wide range of storage tank sizes is available. Based on past experience it is recommended that the ratio of storage tank capacity to collector area be kept at 100 l/m2. Assume that the tanks are cylindrical with a ratio of height to diameter of 1. Assume a loss coefficient for the tanks of 5.5 W/m2C. Load heat exchanger: Cross flow water-to-air heat exchangers are available in a wide range of sizes. It is recommended that the flow rate on the water-side be fixed at 0.30 kg/s. The heat capacity of air under the conditions of this operation can be taken as 1013 J/kgC. Use the effectiveness of this exchanger εL to describe its operation. Other Components: The other components, such as piping, ductwork, pumps, controls, etc., are standard. Any assumptions about them should be clearly stared. The Thermal Analysis Some thought about this problem will lead you to the following: There are three design variables that are or may be important in determining thermal performance and thus economic feasibility. They are the collector area Ac, the collector heat exchanger effectiveness εc, and the load heat exchanger effectiveness εL. You should explore the effects of these design variables on: LA, the auxiliary energy required Ls, the solar energy delivered across the load heat exchanger QL, the energy losses from the storage tank. It is also of interest to explore process dynamics, i.e., the short-term variations of energy flows and temperatures. It is suggested that for a time period of your choice (a period which would be important in process design) you consider what happens with time to the following variables: T, the tank temperature TAs, the temperature of the air leaving the load heat exchanger TAo, the temperature of the air leaving the auxiliary heat exchanger
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Ls the rate of solar energy delivery to the load Qu, the rate of solar energy delivery to the tank IT, the radiation incident on the collector LA, the rate of delivery of auxiliary energy to the load. You will have to make assumptions in order to write the simulation program. In particular, you will have to decide how you wish to model the storage tank. Considering it to be a fully-mixed tank leads to the simplest program, and has the advantage of producing conservative results. You will need some guidance in selecting collector area ranges to explore. This can be obtained by use of the design methods described in Chapters 20 and 21. (These methods provide a first estimate only; they are not a substitute for the simulations.) As a first suggestion, you should cover a range of at least 100 to 400 m2 (but this may not be an adequate range). Also note that you can calculate the total minimum amount of energy required to heat the air from 22 to 65 C, you can estimate roughly what the annual efficiency of the collector might be, and thus you can get a rough first estimate of the solar contribution of a system. The Economic Analysis Based on the thermal analysis, an economic analysis is to be prepared. This does not need to be done in detail, but should be sufficient to indicate to your management what the economic feasibility of this project might be. In order to do this, assume the following economic parameters: Cost of collectors = $ 250/m2 Cost of storage = $ 0.50/liter Fixed costs = $ 20,000 Cost now of delivering auxiliary energy = $ 10/GJ Cost of heat exchangers = $ 5 per unit of W/C Your Procedure Prepare a simulation program in the language of your choice. You will need two meteorological data files, one a one-day file of hourly data for debugging your program and a second a TMY3 (Typical Meteorological Year) file of hourly data which you will use for the simulations. Your Report An executive summary should be included in your report. It should be one or two pages in length and should contain the essential facts of what you did and what conclusions you reached. It is intended for managers who are not familiar with the details of solar processes but who have strong engineering backgrounds. The technical report itself should follow standard technical report format. It should include such items as: the basis of the calculations, assumptions made, tables or charts of data, analysis of your results, notes of what might be done about practical problems such as boiling and freezing, and
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whatever else may be appropriate. It should be the kind of report you would prepare for your supervisor who has some familiarity with the technology.
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