Duct Design

Duct Design Example No.(1) By Badran M. Salem

DESIGN OF AIR DUCT DESIGN CONDITIONS:

Inside room conditions Temperature -----------------------------------------------------------------------------------------------------------

22 oC DB

16 o C WB

Outside room conditions Temperature ----------------------------------------------------------------------------------------------------------Temperature of supply air Mass of outside air,mo

34 o C DB 27 oC WB

----------------------------------15 OC ----------------------------------10 % of mS

Mass of recirculated air, mR

------------------------- 90 % of mS

Air Properties From ASHRAE PSYCHROMETRIC CHART, M.E.T.C, page 75 Outside room conditions, @ 32 O C DB & 27 o C WB

1

Duct Design Example No.(1) By Badran M. Salem

WO  0.0199

kgv m3 kJ ;  O  0.898 ; hO  85.6 kgda kg kg

Inside room conditions, @ 22 O C DB & 16 o C WB WO  0.0089

kgv m3 kJ ;  O  0.848 ; hO  45 kgda kg kg

Mass of supply air, mS Qs = mS c pa ( t r - t s ) ; mS =

Qs c pa (t r - t s )

Where: mS = mass of supply air c pa

= specific heat of air, 1.0062 kJ kg °K

tr

= inside room temperature, 22 C

t s = supply air temperature, 15 °C

Therefore, mS 

230.037 kW kg  32.66 sec  kJ    1.0062  22 - 15 O K O   kg K   

Mass of outside air, mO mS = m1 = m4 ; mo + mR

Where: mS = mass of supply air, 32.66 kg s

mR = mass of recirculated air, 90 % of mS

Therefore, mO  mS 1  0.9  32.66 kg sec 1  0.9  3.266 kg sec

2

Duct Design Example No.(1) By Badran M. Salem

Volume flow rate of outside air, Vo VO = mO  O

Where: mO = mass of outside air, 3.266 kg s

 O = specific volume of outside air, 0.898 m 3 kg

Therefore, kg   m3  m3    2.93 VO   3.266   0.898 sec   kg  sec 

Mass of recirculated air, mR mS  mo  mR ; mR  mS

mO

Where: mS = mass of supply air, 32.66 kg s mO = mass of outside air, 3.266 kg sec

Therefore, mR  32.66 kg sec  3.266 kg sec  29.39 kg sec

Volume flow rate of recirculated air, VR V R = m R υi

Where: m R = mass of recirculated air, 29.39 kg s υ i = specific volume of inside air, 0.848 m 3 kg

Therefore, kg   m3  m3    VR   29.39  24.93  0.848 sec   kg  sec 

3

Duct Design Example No.(1) By Badran M. Salem

Temperature of air entering the AHU, t2 From the figure, considering point 2

By temperature balance, mo t O  mR t 4  mS t 2 ; t 2 

Where: mO = mass of outside air, 3.266 kg sec

t O = temperature of outside air, 34 O C

m R = mass of recirculated air, 29.93 kg s t 4 = inside air temperature, 22 o C mS = mass of supply air, 32.66 kg s

Therefore, t2

3.266 kg 









sec 34 O C  29.93 kg sec 22 O C  23.56 O C 32.66 kg sec

Capacity of AHU, QAHU QCDA = mS (h4 - h1 )

Where: mS = mass of supply air, 32.66 kg s

h4 = enthalpy of air entering the AHU, kJ kg h1 = enthalpy of air leaving the AHU, kJ kg

4

mo t O  mR t 4 mS

Duct Design Example No.(1) By Badran M. Salem

Enthalpy of air entering the AHU, h2 From the figure, considering point 2

By heat balance, mo hO  mR h4  mS h2 ; h2 

mo hO  mR h4 mS

Where: mO = mass of outside air, 3.266 kg sec hO = enthalpy of outside air, 85.6 kJ kg

m R = mass of recirculated air, 29.93 kg s

h4 = enthalpy of recirculated air, 45 kJ kg mS = mass of supply air, 32.66 kg s

Therefore, h4 

3.266 kg sec85.6 kJ

kg   29.93 kg sec45 kJ kg  kJ  49.8 32.66 kg sec kg

Enthalpy of air leaving the AHU, h3 QT  mS h4 - h3  ; h3  h4 -

QT mS

Where: QT = total heat load, 476.228 kWT mS = mass of supply air, 32.66 kg s

h4 = enthalpy of inside air, 45 kJ kg

5

Duct Design Example No.(1) By Badran M. Salem

Therefore, h3  45

kJ 476.228 kWT kJ  30.42 kg 32.66 kg sec kg

Thus, kg  kJ  Q AHU   32.66  632.9 kW or 179.98 TOR 49.8 - 30.42 sec  kg 

Volume of supply air, VS VS = m S υ1

Where: mS = mass of supply air, 32.66 kg s

υ1 = specific volume of supply air, m3 kg

From ASHRAE PSYCHROMETRIC CHART, M.E.T.C, page 75 @ 15 O C DB & h1  30.42 kJ kg

1  0.824

m3 kg

Therefore,





VS  32.66 kg sec 0.824 m 3 kg  26.9

m3 sec

AIR DUCT SIZING DESIGN CONDITIONS: Volume flow rate of supply air ------------------------------ 26.9 m 3 sec Duct dimensions

----------------------------------------------

No. of units

------------------------------------------------ 2

6

W = 4H

Duct Design Example No.(1) By Badran M. Salem

No. of branch take-off

---------------------------------------------

5

No. of outlet per branch take off------------------------------------

4

Method of calculation-----------------------------------------------------Fitting Data Elbow, inner & outer radii ratio --------------------------

0.2

Angle, turns and branches-----------------------------------

90 O

Volume flow rate of supply air per branch take-off, QB QB 

VS 26.9 m 3 sec m3   5.38 Number o Branch Take off 5 sec

Volume flow rate of supply air per outlet, Q O QO 

Volume Flow Rate of Supply Air per Branch Take - off 5.38 m 3 sec m3   1.345 Number of Outlets 4 sec

7

Duct Design Example No.(1) By Badran M. Salem

Duct Sizes For the main duct, from the fan outlet to point O,

From figure 6-2, Pressure drop in straight, circular, sheet metal ducts, REFRIGERATION AND AIRCONDITIONING by Stoecker & Jones, page 107. For a volume flow rate of supply air of, QFO  26.9 m 3 s and a P  1.5 Pa m , Deq. f  1.436 m .

From equation 6-8, REFRIGERATION AND AIRCONDITIONING by Stoecker & Jones, page 108 Deq. f

0.625  ab   1.30 a  b0.25

Where: a = H = height of the rectangular duct, m b = W = width of the rectangular duct, m

Substituting, W  4H to the equation above

8

Duct Design Example No.(1) By Badran M. Salem

Deq. f

0.625  HW   1.30 H  W 0.25

4H   1.30

2 0.625

5H 0.25

0.625  4 H 1.25  1.30 0.25 0.25 5 H

 2.0677 H

Thus, H  0.4836 Deq. f  working equation

For the height, H FO H FO  0.4836 1.436 m  0.694 m

For the width, WFO WFO  4 0.694 m  2.78 m

For the size of duct, AFO AFO  H FO x WFO  0.694 m x 2.78 m  1.93 m2  1930 mm2

Run O to A, two branch take – off supplied

From figure 6-2, Pressure drop in straight, circular, sheet metal ducts, REFRIGERATION AND AIRCONDITIONING by Stoecker & Jones, page 107. For a volume flow rate of supply air of, QOA  26.9 m3 s and a P  1.5 Pa m , Deq. f  1.436 m .

9

Duct Design Example No.(1) By Badran M. Salem

Therefore, H OA  0.4836 1.436 m  0.694 m WOA  4 0.694 m  2.78 m AOA  H OA x WOA  0.694 m x 2.78 m  1.93 m 2  1930 mm2

Run A to B, two branch take – off supplied

From figure 6-2, Pressure drop in straight, circular, sheet metal ducts, REFRIGERATION AND AIRCONDITIONING by Stoecker & Jones, page 107. For a volume flow rate of supply air of,





QOB  5.38 m3 s 4  21.52 m 3 s P  1.5 Pa m ,

Deq. f  1.306 m

Therefore, H OB  0.4836 1.306 m  0.6316 m WOB  4 0.324 m  2.52 m AOB  H OB x WOB  0.6316 m x 2.52 m  1.59 m 2

11

Duct Design Example No.(1) By Badran M. Salem

Run B to C, 3 Branch take – off supplied,

From figure 6-2, Pressure drop in straight, circular, sheet metal ducts, REFRIGERATION AND AIRCONDITIONING by Stoecker & Jones, page 107. For a volume flow rate of supply air of, QBA  ( 5.38 m3 s )(3)  16.14 m3 s P  1.5 Pa m ,

Deq. f  1.186 m

Therefore, H BA  0.4836 1.186 m  0.5735 m

WBC  4 0.5735 m  2.29 m ABA  H BA x WBA  0.5735 m x 2.29 m  1.316 m 2

11

Duct Design Example No.(1) By Badran M. Salem

Run C to D, 4 branch take – off supplied

From figure 6-2, Pressure drop in straight, circular, sheet metal ducts, REFRIGERATION AND AIRCONDITIONING by Stoecker & Jones, page 107.





QOC  5.38 m 3 s 2  10.76 m3 s & P  1.5 Pa m , Deq. f  0.966 m

Therefore, H OC  0.4836 0.966 m  0.467 m WOC  4 0.467 m  1.868 m AOC  H OC x WOC  0.467 m x1.868 m  0.873 m 2

Run D to E, 5 branch take - off supplied

12

Duct Design Example No.(1) By Badran M. Salem

From figure 6-2, Pressure drop in straight, circular, sheet metal ducts, REFRIGERATION AND AIRCONDITIONING by Stoecker & Jones, page 107. QCD  5.38 m 3 s & P  1.5 Pa m , Deq. f  0.756 m

Therefore, H CD  0.4836 0.756 m  0.3656 m WCD  4 0.3656 m  1.46 m ACD  H CD x WCD  0.3656 m x1.46 m  0.5338 m 2

Branch outlet dimensions Note: Since the number of outlets in each branch take – off are equal for the whole system, only one branch take – off will be considered in computing for the duct dimensions and the values obtained will be replicated in all seven (7) branch outlets.

Considering Branch Take – off A,

13

Duct Design Example No.(1) By Badran M. Salem

Run A – 1, 4 outlets supplied From figure 6-2, Pressure drop in straight, circular, sheet metal ducts, REFRIGERATION AND AIRCONDITIONING by Stoecker & Jones, page 107. For a volume flow rate of supply air of, QA1  5.38 m3 s P  1.5 Pa m ,

Deq. f  0.756 m

Therefore, H A1  0.4836 0.756 m  0.3656 m WA1  4 0.3656 m  1.4624 m

AA1  H A1 x WA1  0.3656 m x 1.4624 m  0.5347 m 2

Run 1 – 2, 3 outlets supplied From figure 6-2, Pressure drop in straight, circular, sheet metal ducts, REFRIGERATION AND AIRCONDITIONING by Stoecker & Jones, page 107. For a volume flow rate of supply air of,





Q12  1.345 m3 s 3  4.035 m3 s

P  1.5 Pa m ,

Deq. f  0.7 m

Therefore, H12  0.4836 0.7 m  0.33852 m W12  4 0.33852 m  1.354 m

A12  H12 x W12  0.33852 m x 1.354 m  0.4584 m 2 14

Duct Design Example No.(1) By Badran M. Salem

Run 2 – 3, 2 outlets supplied From figure 6-2, Pressure drop in straight, circular, sheet metal ducts, REFRIGERATION AND AIRCONDITIONING by Stoecker & Jones, page 107. For a volume flow rate of supply air of,





Q23  1.345 m3 s 2  2.69 m3 s

P  1.5 Pa m ,

Deq. f  0.58 m

Therefore, H 23  0.4836 0.58 m  0.28 m W23  4 0.28 m  1.122 m A23  H 23 x W23  0.28 m x 1.122 m  0.314 m 2

Run 3 – 4, 1 outlets supplied From figure 6-2, Pressure drop in straight, circular, sheet metal ducts, REFRIGERATION AND AIRCONDITIONING by Stoecker & Jones, page 107. For a volume flow rate of supply air of, Q34  1.345 m3 s P  1.5 Pa m ,

Deq. f  0.43 m

Therefore, H 34  0.4836 0.43 m  0.208 m W34  4 0.208 m  0.832 m A34  H 34 x W34  0.208 m x 0.832 m  0.173 m 2

15

Duct Design Example No.(1) By Badran M. Salem

Air Velocity, υ Q = A υ; υ =

Q A

Where: υ = velocity of air, m s Q = volume

of supply air, m 3 s

A = area of the air duct, m 2

Branch take - off From fan outlet to branch take – off O,  FO  FA

QF 0 26.9 m3 sec    13.937 m sec AF 0 1.93 m 2

Where: QFO = 26.9 m 3 s

AFO = 1.93 m 2

Run A-B,  AB  AB 

Q AB 21.52 m 3 sec   13.53 m sec AAB 1.59 m 2

Where: Q AB = 21.52 m 3 s

16

Duct Design Example No.(1) By Badran M. Salem

AAB = 1.59 m 2

Run B -C,  BC  BC 

QBC 16.14 m 3 sec   12.26 m sec ABC 1.316 m 2

Where: QBC = 16.14 m 3 s ABC = 1.316 m 2

Run C - D,  CD  CD 

QCD 10.76 m 3 sec   12.325 m sec ACD 0.873 m 2

Where: QCD = 10.76 m 3 s ACD = 0.873 m 2

Run D - E, υ DE  DE 

QDE 5.38 m 3 sec   10.078 m sec ADE 0.5338 m 2

Where: Q DE = 5.38 m 3 s

ADE = 0.5338 m 2

17

Duct Design Example No.(1) By Badran M. Salem

Branch Outlets Note: Since the number of outlets in each branch take – off are equal for the whole system, only one branch take – off will be considered in computing for the duct velocities and the values obtained will be replicated in all seven (7) branch outlets.

Run A - 1, υ A1  A1 

Q A1 5.38 m 3 sec   10.06 m sec AA1 0.5347 m 2

Where: Q A1 = 5.38 m 3 s AA1 = 0.5347 m 2

Run 1 - 2, υ12 12

Q12 4.035 m 3 sec    8.802 m sec A12 0.4584 m 2

Where: Q12 = 4.035 m 3 s A12 = 0.4584 m 2

18

Duct Design Example No.(1) By Badran M. Salem

Run 2 - 3, υ23  23 

Q23 2.69 m 3 sec   8.57 m sec A23 0.314 m 2

Where: Q23 = 2.69 m 3 s A23 = 0.314 m 2

Run 3 - 4, υ34  34 

Q34 1.345 m 3 sec   7.77 m sec A34 0.173 m 2

Where: Q34 = 1.345 m 3 s A34 = 0.173 m 2

Determination of Pressure Drops Considering the Run FOA A1 A2 A3 A4

Run FO, Straight Duct 19

Duct Design Example No.(1) By Badran M. Salem

PFO  LFO P m  4.6387 m 1.5 Pa m  6.958 Pa

Run O, Turn From figure 6 – 8, Pressure losses in rectangular elbows, REFRIGERATION and AIRCONDITONING by: Stoecker & Jones, page 113 For a width to height ratio (W/H) equal to 4, inner and outer radius ratio of 0.20 2 ΔPTOB (vOB )2 (ρair ) = 0.25

Where: 3

ρair = density of supply air, kg / m

vOB = velocity of air in run O, 13.937 m / sec

From table 6 – 2, Viscosity and Density of dry air at standard atmospheric pressure, REFRIGERATION and AIRCONDITONING by: Stoecker & Jones, page 106 Temperature Viscosity  , Density , O C kg / m3  Pa - s 10 17.708 1.2467 20 18.178 1.2041 Since the temperature of the supply air is at 15 0C, by interpolation, @ 15 C = 1.2254 kg / m3 Therefore, POT

2  vOB   air  0.25

2

13.937 m  0.25

Run OA, Straight Duct POA  LOA P m 10.4360 m 1.5 Pa m  15.654 Pa

Branch take – off A, Turn

21



sec 1.2254 kg m3 2 2



 29.75 Pa

Duct Design Example No.(1) By Badran M. Salem

From figure 6 – 8, Pressure losses in rectangular elbows, REFRIGERATION and AIRCONDITONING by: Stoecker & Jones, page 113 For a width to height ratio (W/H) equal to 4, inner and outer radius ratio of 0.20 2 PAT

v AT 2  air 

 0.25

Where: 3

ρair = density of supply air, 1.2254 kg / m

v AT = velocity of air in run OA, 13.937 m / sec

Therefore, PAT  0.25

vOB 2  air 2

 0.25

13.937 m

2

Run AA1, Straight Duct PAA1  LAA1 P m  2.1703 m 1.5 Pa m  3.255 Pa

Run AA2, Straight Duct PAA2  LAA2 P m  5.7749 m 1.5 Pa m  8.6623 Pa

Run AA3, Straight Duct PAA3  LAA3 P m  9.613 m 1.5 Pa m  14.4195 Pa

Run AA4, Straight Duct PAA4  LAA4 P m 12.4634 m 1.5 Pa m  18.6951 Pa

21



sec 1.2254 kg m3 2



 29.75 Pa

Duct Design Example No.(1) By Badran M. Salem

Therefore, PFOA 4  PFO  POT  POA  PAT  PAA1  PAA2  PAA3  PAA4

 6.958  29.75  15.654  29.75  3.225  8.6623  14.4195  18.6951 Pa PFOBA 7  127.1439 Pa

Considering the Run FOBB1B 2 B3 B 4

Run FO, Straight Duct PFO  LFO P m  4.6387 m 1.5 Pa m  6.958 Pa

Run O, Turn From figure 6 – 8, Pressure losses in rectangular elbows, REFRIGERATION and AIRCONDITONING by: Stoecker & Jones, page 113 For a width to height ratio (W/H) equal to 4, inner and outer radius ratio of 0.29 2 PTOB

vTO 2  air 

 0.25

Where: 3

ρair = density of supply air, 1.2254 kg / m

22

Duct Design Example No.(1) By Badran M. Salem

vTO = velocity of air in run OB, 13.937 m / sec

Therefore, PTO  0.25

vTO 2  air 2

 0.25

13.937 m



sec 1.2254 kg m3 2 2



 29.75 Pa

Branch take – off A, Duct Straight Through From Equation 6 – 16, REFRIGERATION and AIRONDITIONING by Stoecker & Jones, page 114 PDST A

2    VOA   air 0.41  VOA  

PDST A

2  13.53 m sec  1.2254 0.41 

 

2

2

V AB   

2

13.53   13.937 

2

PDST A  0.038 Pa

Run OB, Turn From figure 6 – 8, Pressure losses in rectangular elbows, REFRIGERATION and AIRCONDITONING by: Stoecker & Jones, page 113 For a width to height ratio (W/H) equal to 4, inner and outer radius ratio of 0.29 2 PTOB

vTOB 2  air 

 0.25

Where: 3

ρair = density of supply air, 1.2254 kg / m

vTOB = velocity of air in run OB, 13.53 m / sec

Therefore, PTOB

2  vTOB   air  0.25

2

13.53 m  0.25

23



sec 1.2254 kg m3 2 2



 28.04 Pa

Duct Design Example No.(1) By Badran M. Salem

Run OB, Straight Duct POB  LOB P m 18.136 m 1.5 Pa m  27.204 Pa

Run BB1, Straight Duct PBB1  LBB1 P m  2.3628 m 1.5 Pa m  3.5442 Pa

Run BB2, Straight Duct PBB 2  LBB 2 P m  5.9674 m 1.5 Pa m  8.9511 Pa

Run BB3, Straight Duct PBB3  LBB3 P m  9.8055 m 1.5 Pa m  14.708 Pa

Run BB4, Straight Duct PBB 4  LBB 4 P m 12.6559 m 1.5 Pa m  18.9838 Pa

Therefore,

24

Duct Design Example No.(1) By Badran M. Salem

PFOBB 1B 2 B3B 4  PFO  PTO  PDSTA  PTOB  POB  PBB1  PBB 2  PBB 3  PBB 4

 6.958  29.75  0.038  28.04  27.204  3.5442  8.9511  14.708  18.9838 Pa

PFOBA 7  138.1771 Pa

Considering the Run FOCC1C 2C 3C 4

Run FO, Straight Duct PFO  LFO P m  4.6387 m 1.5 Pa m  6.958 Pa

Run O, Turn From figure 6 – 8, Pressure losses in rectangular elbows, REFRIGERATION and AIRCONDITONING by: Stoecker & Jones, page 113 For a width to height ratio (W/H) equal to 4, inner and outer radius ratio of 0.29 2 PTOB

vTO 2  air 

 0.25

Where: 3

ρair = density of supply air, 1.2254 kg / m

vTO = velocity of air in run OB, 13.937 m / sec

25

Duct Design Example No.(1) By Badran M. Salem

Therefore, PTO  0.25

vTO 2  air 2

 0.25

13.937 m



sec 1.2254 kg m3 2 2



 29.75 Pa

Branch take – off A, Duct Straight Through From Equation 6 – 16, REFRIGERATION and AIRONDITIONING by Stoecker & Jones, page 114 PDST A

V 2  air 0.41  VOA   OA

PDST A

13.53 m sec2 1.2254 0.41 

 

2

2

V AB   

2

13.53   13.937 

2

PDST A  0.038 Pa

Branch take – off B, Duct Straight Through From Equation 6 – 16, REFRIGERATION and AIRONDITIONING by Stoecker & Jones, page 114 PDSTB

V 2  air 0.41  VOB   BC

PDSTB

2 2  12.26 m sec  1.2254  12.26  0.4 1 

PDSTB

2

2  0.03245 Pa

 

2

VBC   

 13.53 

Run OC, Turn From figure 6 – 8, Pressure losses in rectangular elbows, REFRIGERATION and AIRCONDITONING by: Stoecker & Jones, page 113 For a width to height ratio (W/H) equal to 4, inner and outer radius ratio of 0.29

26

Duct Design Example No.(1) By Badran M. Salem

2 PTOC

vTOC 2  air 

 0.25

Where: 3

ρair = density of supply air, 1.2254 kg / m

vTOC = velocity of air in run OB, 12.26 m / sec

Therefore, PTOC  0.25

vTOC 2  air 2

 0.25

12.26 m



sec 1.2254 kg m3 2 2

Run OC, Straight Duct POC  LOC P m  25.836 m 1.5 Pa m  38.754 Pa

Run CC1, Straight Duct PCC1  LCC1 P m  2.4153 m 1.5 Pa m  3.623 Pa

Run CC2, Straight Duct PCC2  LCC2 P m  6.0199 m 1.5 Pa m  9.02985 Pa

Run CC3, Straight Duct PCC3  LCC3 P m  9.858 m 1.5 Pa m  14.787 Pa

Run CC4, Straight Duct PCC4  LCC4 P m 12.7084 m 1.5 Pa m  19.0626 Pa

27



 23.0234 Pa

Duct Design Example No.(1) By Badran M. Salem

Therefore,

PFOCC1C 2C 3C 4  PFO  PTO  PDSTA  PDSTB  PTOC  POC  PCC1  PCC2  PCC3    6.958  29.75  0.038  0.03245  23.0234  38.754  3.623   Pa     9.02985  14.787  19.0626  PFOCC1C 2C 3C 4  145.0583 Pa

Considering the Run FODD1D 2 D3 D 4

Run FO, Straight Duct PFO  LFO P m  4.6387 m 1.5 Pa m  6.958 Pa

Run O, Turn From figure 6 – 8, Pressure losses in rectangular elbows, REFRIGERATION and AIRCONDITONING by: Stoecker & Jones, page 113 For a width to height ratio (W/H) equal to 4, inner and outer radius ratio of 0.29 2 PTOB

vTO 2  air 

 0.25

Where:

28

Duct Design Example No.(1) By Badran M. Salem 3

ρair = density of supply air, 1.2254 kg / m

vTO = velocity of air in run OB, 13.937 m / sec

Therefore, PTO

2  vTO   air  0.25

2

13.937 m  0.25



sec 1.2254 kg m3 2 2

Branch take – off A, Duct Straight Through From Equation 6 – 16, REFRIGERATION and AIRONDITIONING by Stoecker & Jones, page 114 PDST A

V 2  air 0.41  VOA   OA

PDST A

13.53 m sec2 1.2254 0.41 

 

2

2

V AB   

2

13.53   13.937 

2

PDST A  0.038 Pa

Branch take – off B, Duct Straight Through From Equation 6 – 16, REFRIGERATION and AIRONDITIONING by Stoecker & Jones, page 114 PDSTB

2    VBC   air 0.41  VOB  

PDSTB

2 2  12.26 m sec  1.2254  12.26  0.4 1 

PDSTB

2

2  0.03245 Pa

 

2

VBC   

 13.53 

Branch take – off C, Duct Straight Through From Equation 6 – 16, REFRIGERATION and AIRONDITIONING by Stoecker & Jones, page 114 29



 29.75 Pa

Duct Design Example No.(1) By Badran M. Salem

PDSTC

2    VOC   air 0.41  VOC  

PDSTC

2 2  12.325 m sec  1.2254  12.325  0.4 1 

PDSTC

 

2

2

VBC   

2  0.001046 Pa

 12.26 

Run OD, Turn From figure 6 – 8, Pressure losses in rectangular elbows, REFRIGERATION and AIRCONDITONING by: Stoecker & Jones, page 113 For a width to height ratio (W/H) equal to 4, inner and outer radius ratio of 0.29 2 PTOD

vTOD 2  air 

 0.25

Where: 3

ρair = density of supply air, 1.2254 kg / m

vTOD = velocity of air in run OD, 12.325 m / sec

Therefore, PTOD

2  vTOD   air  0.25

2

12.325 m  0.25



sec 1.2254 kg m3 2 2

Run OD, Straight Duct POD  LOD P m  33.536 m 1.5 Pa m  50.304 Pa

Run DD1, Straight Duct PDD1  LDD1 P m 1.9931 m 1.5 Pa m  2.9896 Pa

31



 23.268 Pa

Duct Design Example No.(1) By Badran M. Salem

Run DD2, Straight Duct PDD2  LDD2 P m  5.904 m 1.5 Pa m  8.856 Pa

Run DD3, Straight Duct PDD3  LDD3 P m  8.6386 m 1.5 Pa m  12.9579 Pa

Run DD4, Straight Duct PDD4  LDD4 P m 11.1831 m 1.5 Pa m  16.7746 Pa

Therefore, PFODD1D 2 D3D 4  PFO  PTO  PDSTA  PDSTB  PDSTD  PTOD POD  PDD1  PDD 2 

P

 6.958  29.75  0.038  0.03245  0.001046  23.268  50.304   Pa     2.9896  8.856  12.9579  16.7746   151.932 Pa 2 3 4

FODD1D D D

Considering the Run FOEE1E 2 E 3 E 4

31

Duct Design Example No.(1) By Badran M. Salem

Run FO, Straight Duct PFO  LFO P m  4.6387 m 1.5 Pa m  6.958 Pa

Run O, Turn From figure 6 – 8, Pressure losses in rectangular elbows, REFRIGERATION and AIRCONDITONING by: Stoecker & Jones, page 113 For a width to height ratio (W/H) equal to 4, inner and outer radius ratio of 0.29 2 PTOB

vTO 2  air 

 0.25

Where: 3

ρair = density of supply air, 1.2254 kg / m

vTO = velocity of air in run OB, 13.937 m / sec

Therefore, PTO  0.25

vTO 2  air 2

 0.25

13.937 m



sec 1.2254 kg m3 2 2

Branch take – off A, Duct Straight Through From Equation 6 – 16, REFRIGERATION and AIRONDITIONING by Stoecker & Jones, page 114

32



 29.75 Pa

Duct Design Example No.(1) By Badran M. Salem

PDST A

2    VOA   air 0.41  VOA  

PDST A

2  13.53 m sec  1.2254 0.41 

 

2

2

V AB   

2

13.53   13.937 

2

PDST A  0.038 Pa

Branch take – off B, Duct Straight Through From Equation 6 – 16, REFRIGERATION and AIRONDITIONING by Stoecker & Jones, page 114 PDSTB

2    VOB   air 0.41  VOB  

PDSTB

2 2  12.26 m sec  1.2254  12.26  0.4 1 

PDSTB

2

 

2

VBC   

2  0.03245 Pa

 13.53 

Branch take – off C, Duct Straight Through From Equation 6 – 16, REFRIGERATION and AIRONDITIONING by Stoecker & Jones, page 114 PDSTC

2    VOC   air 0.41  VOC  

PDSTC

2 2  12.325 m sec  1.2254  12.325  0.4 1 

PDSTC

2

2  0.001046 Pa

 

2

VCD   

 12.26 

Branch take – off D, Duct Straight Through

33

Duct Design Example No.(1) By Badran M. Salem

From Equation 6 – 16, REFRIGERATION and AIRONDITIONING by Stoecker & Jones, page 114 PDSTD

2    VOD   air 0.41  VOD  

 

2

PDSTD 

2

VDE 

10.078 m sec2 1.2254 0.41 10.078  2  

2

 12.325 

PDSTD  0.827 Pa

Run OE, Turn From figure 6 – 8, Pressure losses in rectangular elbows, REFRIGERATION and AIRCONDITONING by: Stoecker & Jones, page 113 For a width to height ratio (W/H) equal to 4, inner and outer radius ratio of 0.29 2 PTOE

vTOE 2  air 

 0.25

Where: 3

ρair = density of supply air, 1.2254 kg / m

vTOE = velocity of air in run OD, 10.078 m / sec

Therefore, PTOE  0.25

vTOE 2  air 2

 0.25

 10.078 m

2

Run OE, Straight Duct POE  LOE P m  41.236 m 1.5 Pa m  61.854 Pa

34



sec 1.2254 kg m3 2



 15.557 Pa

Duct Design Example No.(1) By Badran M. Salem

Run EE1, Straight Duct PEE1  LEE1 P m  2.1971 m 1.5 Pa m  3.295 Pa

Run EE2, Straight Duct PEE 2  LEE 2 P m  5.5437 m 1.5 Pa m  8.316 Pa

Run EE3, Straight Duct PEE 3  LEE 3 P m  8.4626 m 1.5 Pa m  12.6939 Pa

Run EE4, Straight Duct PEE 4  LEE 4 P m 11.20005 m 1.5 Pa m  16.8 Pa

Therefore,

PFOEE 1E 2 E 3 E 4  PFO  PTO  PDSTA  PDSTB  PDSTC  PDSTD  PTOE POE  PEE 1  PEE 2  PEE 3    6.958  29.75  0.038  0.03245  0.001046  0.827  15.557  61.854   Pa    3 . 295  8 . 316  12 . 6939  16 . 8   P 1 2 3 4  156.122 Pa FOEE E E E

35