Thermodynamics 1

Flo-=6;T* I

Z2 i !

a

THEN[tl!ODYl\lA[tl|IOS

HIPOLITO B. STA. MARIA

COIVTENTS vii

Preface

Chapter 1 Basic Principles, Concepts and Defrnitions

I

Mass, Werght, Specilc Volume and Density; Spe-

- Weight, Pressule, cific Conservation of Mass.

2

Conservation of

Energy

Zg

Potential E_1ergy, Kiletic Energy, Internal Energy, $eat, Work, Flow Work, Enthalpy, General EnergT Equation.

3 , The Ideal Gas 87 Constant, Specific Heats of an tddal Gas.

4

Processes of

Ideal Gas

5f

Isometric Process, Isobaric process, Isothermal

Process, Isentropic Process, polytropic

5

Gas

do""sr.

Cycles 81

Camot Cycle, Three-process Cycle.

6

Internal Combustion Engines gg Otto Cycle, Diesel Cycle, Dual Combustion Cycle.

7

Gas Compressors

ll5

Single-Stage Con pression, Twestage Compression, Three-Stage Compression.

8

Brayton Cycle 16l

PREEACE The purpose of this text is to present a simple yet rigorous approach to the fundamentals of thermodynamics. The author expects to help the engineering students in such a way that learning would be easy and effective, and praetical enough for workshop practice and understanding.

Chapters 1 and 2 present the development of the first la'ar of thermodynamics, and energy analysis of ope:r systems Jhapters 3 and 4 give a presentatign of equation of state and involvingideal gases. The second law of thermodynamics andits applications to different thermodynamic cycles are discussed in Chapters 5 and 6. Chapter ? deals with gas compressors andits operation. Chapter 8 develops the Brayton eycle which can be omitted if sufficient time is not available. ;he process

The author is grateful for the comments and suggestions received from his colleagues at the University of Santo Tomas, Faculty of Engineering.

The Author

vll

1 I

Basic Ppq"iples, Concepts and Definitions

Thermodynamics is that branch of the physical sciences that treats of various phenomena of energ-Jr and the related properties ofmatter, especially of the laws of transformation of heat into other forrns of energy and vice versa.

Systems of Units Newton's law states that 'the aceeleration of a particular

body is directly proportional to the resultantforce acting on and inversely proportional to its mass.o

"-

hE, F= D8, m k

it

k =+F

k is a proportionality constant Systenns of units where k is unity but not dimensionless: cgs system: I dyne forcre accelerates 1 g mass at 1 cm,/s2

mks system: 1 newton force accelerates

I

I

kg mass at

m./sz

fps system: 1 lb force accelerates 1 slug mass at

l

Nsz

l--t;]*ldyne I t -* i*l newton [T,,'*-l-'r,0" /777r/7mrV /7furm,h n77v77v?rrvr 1 cm./s2 _+ 1m/s2 1&,/sz t=r,4'cm-cyne.s"

o=t#;p

Systems of units where k is not unity:

k=rw

47 If the same word is used for both mass and force in a given system, k is neither unity nor dimensionless. 1 Ib force acceierates a I lb mass at 32.L74 fVs2 1 g force accelerates a I g mass at 980.66 cm/s2 L kg force accelerates a 1 kg mass at 9.8066 m/s2

f-.,.-f*

,0, l- t ,. l-.

,

t

u

d7mzm'V /72zv7m77

[-t u*.

f-,

rz.tllthP

1 poundal = (1 lb_) (1 fVs2)

F is force in poundals

a is acceleration

[T**

l* /7V7V7mV ',0,

k = e80.66-*F k = e.80668#

L fVs2 --------+ .U

m

=r-8. l(

ks .m

k=1k#

k = e.8066 Ets"

k#

= e.8066

1 pound = (1 slug) (1 fvsz); 1 slug

H#

is -ass in slugs a is acceleration in

k= 32.r74ffi

t*5& = 82.r74ffi L

The mass of a body is the absolute quantity of matter in it. The weight o,f a body means the force of gravity F, on the lrody.

Acceleration A unit of force is one that produces unit acceleration in a body of unit mass.

:.._l

E

poundal

I

fl;/s2

mFF" k =t=g-

slug = 32.L74Lb

I

s2

Mass and lVeight

where

I I

-lr-

S K

Relation between pound psss (lb-) and slug

k=1#

= 1 lb"

F is force in pounds

1kg"= 9.8066 N

Therefore,

in ftls2

--'-+

Relation between kilogram force (kgr) and Newton (N)

Therefore, t

tr mass in pounds

#

nr'

/7V7v77v77v7

32.L74 fVsz----+ 980.66 cm"/s2 -------> 9.8066 mlsz

k=

r=f,a

fvs2

--)

AL or

g a

= acceleration produced by force F* = acceleration produced by another force F

near the surface of the earth, k and g are numerically

,.r1rr:rl, so are m and

F-

1( Problcms:

lb

tion?

I

m=66k9-

F"ok Fto.lF' mo=-?-= Bosg_.-

2. The weight of an object is 50 lb. What is its mass at standard condition?

= (o

Total

ro,r"er

mass

Solution

r rb!rf-

FK

* =d-=

Fo

32.L74

ft P

So lb_

3.Fivemassesinaregionwheretheaceelerationdueto grr"itv i. 30. 5 fVs2 are as follo**t m, is- 500 g of masq rq, y^eighs [oo eim, weighs 15 poundals; mo weight-g.lli mu is 0'10 slug ;i *]',r. trnuf iu theiotal mass expressed (a) in grams, 16) in pounds, and (c) in slugs.

(t') Total mass

in/ft)

(2.54 cm/in) = 929'64 cmls2

-

=

e2e.64

+

= 843.91

g;

"f*J

rtrJ

= 1435.49 g-

= 1459.41 g,"

9'83

g^

= g.EB lb-

ils ]!-o'

32.174;ifis

= 0.306 slug

that the gravity acceleration at equatorial sea level rr s = 32.088 fpsz and that its variation is - 0.003 fps2 per 1000 (a) l't, :rscent. Find the height in miles above this point for which llr:, gravity acceleration becomes 30.504 fps2, (b) the weight of ,r lsivcn man is decreased by \Vo. (c) What is the weight of a 180 I r,,, rn an atop the 29,131-ft, Mt. Everest in Tibet, relative to this I

,\til

rr

L'?

tr

tion

(;r ) change

F't [roo4frro.uuM

F

g,,,

4. Note

por

Solu,tion g = (30.5 fVsz) (12

= 222.26

= mr + m2 + na + m4 + m5 = 500 + 843.91 +222.26 + 1435.49 + 1459.41 = 446t.07 g^

453.6

lb.rrl

fztz+14s'j

fz.rt- U|nu-r

(b) Total mass = 446L.0J

g= 32.L74ftlsz

F, = 5o lbr

^"J

t*tfufl

= 9.8066 m/s2

?

4

'L

Bo.b+

Solution

(a) mz =

.ft

--'l- J ls-PI,l S'= tu.ll+se.o#-l 0.4e =l K s

l.Whatistheweightofa66-kg-manatstandardcondi-

in acceleration = 30.504 - 32.088

p:; = 528,000 ft or llcight, h = - I lP* fps' 0.003 - -T0008

=

*

1.584 fps2

100 miles

+T (b) F = 0.9b Fg

-t

Specifrc Volume, Density and Specifrc Weight

Let Fg = weight of the man at sea level

.a

FF= -ag 0.95 F" F" a =g

The density p of any substance is its mass (not weight) per unit volume.

____q

I

h I

rl=D rv

a = 0.959 = (0.95) (32.088) = 30.484 fps2

-L 'Fg

The specific volume v is the volume of a unit mass.

g = 32.088 fps2

" --

t.,

lt

(30.484

- 32'088) fps'z= b34,6z0 ft or tOt.B miles o.oosTS;r _

-Tmorr

,vF

g=

a

29.1.31

Tk orY ='fr os P='g

r_.6 F8

g = 32.088 fps2 m = 1801ba = 32.088 fps'

o

=T-=

8

Since the specific weight is to the local acceleration of gravity as the density is to the standard acceleration,Tlg= pk, conversion is easily made;

ft

ma

V1

mp

The specificweightTof any substance is the force of gravity on unit volume.

F

(c)

----

-

rIto

1"1 {}l

fTdriil [0'003

tlso lb-l

At or near the surface of the earth, k and g are numerically cqual, so are p and y

-1

fpsz] = 32'001 fpsz

pz.oor&l

#=179.03 32.174F"1T"

Problems r

_^ ^^ lbr ,,

1.

What is the specific weight of,water at standard condi.

tion?

Stilution g = 9.8066 m/sz

*_pg I- E--

kg_ P = 1000 n5.

[*,SE**d e.8066ffi#

kgF

= looo mo

ry Pressure

densities (p, = 1500 kg/m3,Pzi^ 500 kg/m3) are poured together into a 100-L tank, frlling it' If the resulting density of the mixture is 800 kg/mt, frnd the respective quantities of liquids used. Also, find the weight of the mixture; Iocal g = 9.675 mps2.

2. Two Iiquids of different

The standard reference atmospheric pressure is 760 mm Hg or 29.92 in. Hg at 32"F, or 1"4.696 psia, or 1 atm.

Measuring Pressure

Solution

1.

mass of mixture, mm = pmvm = (800 kg/m3) (0'100 m3) = 80

kg

By using manometers

I

(a) Absolute pressure is greater than atmospheric pressure.

mt+m2=mm

po

PrVt+PrV,=D-

q = 80 V, + V, = 0'100

1500 Vr + 500

p = Po = D 'lt p" = ' I

I

(r) Q)

solving equations (1) and (2) simultaneously

Vt = 0'03

p =

absolute pressure atmospheric pressure gage pressure, the pressure due to the liquid column h Po+Pg

mg

(b) Absolute pressure is less than atmospheric pressure

Ve = 0'07 m3

m, = P,Vr = (1500 kg/m3) (0.03 m3) = 45kg

mr= prY2= (500 kglm3) (0.07

m3) = 35

P=Po-P,

kg

The gage reading is called vacuum pressum or the vacuum.

weight of mixture,

re-=x"=@

e.8066*#

=?8.esksr

I

ll"y using pressure gages

A Jrrt:ssure gage is a device for pressure,

rilr,,1||llr rt ng gage

'l'lrin picture shows the rrr,vr.rn(.1)t, in one type !, I r r

I

rr

. l::ll{(',

lrr.

p1i

r13..

k

ofpres-

nown as the single-

'l'hc f'luid enters the

lnlrr, llrrrrrrglr t,lrc thrcnded ,

',,rur.r'lrorr. A$ t.hc prOssur:e

Fig. 1 Pressure Gage

I

ry_ increases, the tube with an elliptical section tends to straighten, the end that is nearest the linkage toward the right. The linkage causes the sector to rotate. The sector engages a small pinion gear. The index hand moves with the pinion gear. The whole mechanism is of course enclosed in a case, and a gpaduated dial, from which the pressure is read, and is placed under the index hand.

Solution

["*S

pr=*#= FuuS ', kg-'4 ' N.sz

(30 m)

= b48,680 N/mz or b43.6g pps(gage)

(p=po+p") +Pt

Atmospheric Pressure

,=O,P=Po)

-P,

A barometer is used to measure atmospheric pressure.

V

(p=p"-pr)

Absolutet Pressure

(p=0,Pr=P") Gage Pressure po

I

--T--ps

P=Po+Pg

_ F" 1V yAhPr=*-A-=:6l P, = Tb,

=ry'=*

Problem

A 30-m vertical column of fluid (density 1878 kg/ms) is located where g = 9.65 mps2. Find the pressure at the base of the column.

IO

P.=Y\ Where ho = the height of column of liquid supportedby atmospheric pressure {

l)roblems

1. A vertical column of water will be supported to what lrcight by standard atmospheric pressure.

Absolute Pressure

Solution P=Th

At standard condition

yh"-* h = ho * hr, the height of column of liquid supported -by absolute pressure p.

\* = 62'4lblfts Po = 14'7 Psi ;-l T ..-rr lu.z *l lt++'#l ,t'= p,, - L----:n!-!_--!t"! = 33.9 ft

t;

If the liquid used in the barometer is mercury, the atmospheric pressure beconoes, = THshs = (sp S)H, (T*) (h")

P"

62.4Y -'- ft3

trg.ol

Thespecificgravity(*pg')ofasubstanceistheratioofthe spccifrc weight of the substance to that of water'

Fz.+

H rL'" i',1

1728H

^{

sps=T po = 0.491 is 9.5 kg/cm2. The}arometric pressure of the atmosphere is 768mm of Hg. Find the absolute p".*r,r"* in the boiler. (ME Board Problem - Oct' 1987)

2, The pressure of a boiler

where

then, ps = 9'5

kg/cm3

ho = 768 mm

Hg

l4

ho = column of mercury in inches

Solution Pg

h"

and,

p

= 0.491 n-

h

=0.491 hP-= ln."

At standard condition T* = 1000

po

=

l)roblems

kdmt

l. A pressure gage regrsters 40 psig in a region where the l,irrometer is 14.5 psia. Find the absolute pressure in psia, and 'rr kPa.

(ynr) (h") = (sp gr) nr(T*) (h")

(13.6)

Fooo

S

10.000 'm'

to.?68 m)

c!*

_ 1.04

kg cm-E

Srilution p = 14.5 + 40 = 54.5 psia

= po

* p, = 1.04 + 9.5 = 10.54#

t-t k-+'r newton

/Tnvrnh

a=

l2

[ , "[-ft,

,0,

/vTTvvmmiV

I

m./sz

a=1fUs2

1Tlkgn

1+

=

-tE KgJ P.

Solution = 0.06853 slug

(a)p = Pr=

= FS][tr'fl =8.28$

Po

* Ps = 14.7 + 80 = 94.7 Psia

ao

Ps]L

t7 Psla r,. I':t. | --:-

= S.A4atmospheres

af,m

F,lbf h = 9.92 in. Hg abs a = 3.28 Nsz

t = ff

lrg = 2o

in. P = 0.491

h"= Z9.tilt". = (0.06863 slug)

1

[.za {l=

llth' -1f-

o.zzas tb,

$..

p

=

h

(0.491) (9.92) = 4.87 psia

J

newton = 0.2248Ib" 1.1b"

p8 = 4.7

= 4.4484 newtons

(1rb)

rl4 ln' 114=

=

ps = (4.7

F**H lrr.ut;] ln-

osgs\ mo

P

= 10 psia

(rl)

h =15in.

psi vacuum

esi)

r

o"_l

l:8e5;-s!

=32,407 Ps(gage)

h = 29.92 + 15 = 44.92 in. Hg abs

= 375,780 Pa or 375.78 kPa

2.

Given the barometric pressure of L4.7 psia (2g.g2 in. Hg abs), make these conversions: (a) 80 psig to psia and to atmosphere, (b) 20 in. Hg vacuum to in. Hg abg and to psia, (c) 10 psia to psi vacuum and to Pa, (d) 15 in. Hg gage to psia, to torrs, and to pa. (1 atmosphere = 760 torrs)

t4

P, = 0'491 h,

=[r"H F"!F*'H = 50,780 Pa(gage)

15

.lF I'empcraturc

It follows that,

1. Derive th. r.l:rtion between degrees Fahrenheit and degrees Centigrndo. (FlE Board euestion)

1Fo=1Po and

100"c

T212.F

T

*uu

*r". I0".

tl

1

,r""

lc.-1K"

t.F -32 _.= t"C-0 212

-

n

toF =

toC =

lbb:

Solution

o

r

I

t"C + 32

o

5( t.F

I

-

t.F + 460, degtees Rankine

TK=t"C+z71,Kelvin

Degrees Fahrenheit ("F) and degrees Centigrade ("C) indicate temperature reading (t). Fahrenheit degrees iFJ) and Centigrade degress (C") indicate tempertu"" or differ"h"ogu ence (At). 180 Fb = 100 C"

1p"-5g" 9

1 C. =!-1l," o

Btu (lb) (r")

Btu - cal -Ir-IEXD =IG'(E

32)

, Absolute temperature is the temperature measured from absolute zero. Absolute zero temperature is the temperature at which all molecular motion ceases. Absolute temperature will be denoted by T, thus TbR =

2. Show that the specific heat ofa substance in Btu/(lb) (F") is numerically equal to caV(g)(C").

.

Conservation of Mass

'l'lr. law of conservation of mass states rhat

,tr ttr.ltltl.e. 'l'lr,r.

mass is inde-

rluantity of fluid passing through a given section is

,'r\ r'n t)y fne lOfmUla

V=Au

-: VAu =Aup

III = i__

v v-

Wltcrc V = volume flow rate A = cross sectional area ofthe stream

l) :, ilvcrage Speed rir ,., m:rss llow rutc

16

t7

F7---

\t

I

Applying the law of consewation of mass'

- - \ArDrpr =

=-n;

=' *T4=ff =Erf,El

a,E4zftz

T

2. A 10-ft diameter by 15-ft height vertical tank is receiving water (p = 62.1 lb/cu ft) at the rate of 300 gpm and is discharging through a 6-in ID line with a constant speed of 5

\rtrPz

I I

I I I

Problems Two gaseous stre?ms enter a combining tube and leave single mi*trrr". These data apply at the entrance section: as a -fot 6rr" gur, A'r= 75 in,z, o, = 590 fps,-vt] 10 ft3llb For the other gas, A, = 59^i1''.:T, = 16'67 }b/s P" = 0.12lb/ftg At exit, u.. j 350 fPs, v, = 7 ftaAb' Find (a) the speed u, at section 2, i- 'd ft) the flow anii area at the exit section'

1.

Solution

:j:rlil"ffJrr;,'frh'iisfilTil;1lo' I I

rs,

\

f___ _

_]=

t__

I l=:-:_-_*--l -l-, I F'--=- -:-1J tiu' e""" =-f, (10)2 = 78.54 ftz

tu'",=il'i,=ffi

=4oorps r\lirrur lr,,w

(b)

.

mr

Aru,

= --vr

-[.'9!d=2604+ --------r6Tt3=

rate enreri", =

[ffi]

r\t,r'* tuwrateleavins=Aup=

ib rh, = rh, + rh, 18

= 26.04+ 16'6? = 42'?1+

=

[rr

r

fi

= z4so.\

? Bd'F.uo*J F + ru* S*

Mass change = (3658

volume ch^nge

=

-

17'51-l:-!b

Decrcased in height

62.1#

=

Review Problems

2490.6) (15) = 17,511 lb (decreased)

=

282

= 3'59

ffi#

Water level after 15 min. = 7.5

-

1. What is the mass in grams and the weight in dynes and in gram-force of 12 oz of salt? Local gis 9.65 m/s2 1 lb- = 16 oz. Ans. 340.2 g-;328,300 dynes; 334.8 g,

ft'

ft

3'59 = 3'91

ft

2. A mass of 0"10 slug in space is subjected to an external vertical force of4 lb. Ifthe local gravity acceleration is g = 30.5 fps2 andiffriction effects are neglected, determine the acceleration of the mass if the external vertical force is acting (a) upward and (b) downward Ans. (a) 9.5 fps2; (b) 70.5 fps'? 3.

The mass of a given airplane at sea level (g = 32.1 fps2) is 10 tons. Find its mass in lb, slugs, and kg and its (gravital.ional) weight in lb when it is travelling at a 50,000-ft elevation. 'l'he acceleration of gravity g decreases by 3.33 x 10-6 fpsz for r,rrch foot of elevation. Ans. 20,0001b-; 627.62 slugs; 19,850lbr

4. A lunar excursion module (LEM) weights 150[r kg, on r.rrrth where g = 9.75 mps2. What will be its weight on the rrrrrface of the moon where B. = 1.70 mpsz. On the surface of the ,noon, what will be the force in kg, and in newtons required to ',,'ttlerate the module at 10 mps2? Ans. 261.5 kg; 1538.5 kgr; 15,087 N systenis 0.311 slug, its density is 30 g and is 31.90 fpsz. Find (a) the specific volume, (b) the (c) the total volume. "1,,'r'ific weight, and Ans. (a) 0.0333 ft3Ab; (b) 29.75 lb/ft3; (c) 0.3335 ft3 ,l-r. The mass of a fluid

ll,/l'1,:r

{;.

A cylindrical drum (2-ft diameter, 3-ft height) is filled *'rllr :r tluid whose density is 40lb/ft3. Determine (a) the total ,,,lrrrno of fluid, (b) its total mass in pounds and slugs, (c) its ,'1r'r'rlit: volume, and(d) its specific weight where g = 31.90 fps2. Ans. (a) 9.43 ft'; (b) 377.21b; 11.72 slugs; (c) 0.025 ft3l lb; (d) 39.661b/ft3.

'i

A wuathcrman carried an aneroid barometer from the ! "t, l llrxrr to tris ofl'icc atop the Sears Towcr in Chicago. On r

20

ir

2l

the ground level, the barometer read 30.150 in. F,Ig absolute; topside it read 28.607 in. Hg absolute. Assume that the average

atmosphdric air density was 0.075 lb/ft3 and estimate the height of the building.

Consenration of Energy

Ans. 1455 ft

8.

A vacuum gauge mounted on a condenser reads 0.66 m Hg.What is the absolute pressure in the condenser in kPa when the atmospheric pressure is 101.3 kPa? Ans. 13.28 kPa

9.

Convert the following readings of pressure to kPa absolute, assuming that the barometer reads 760 mm ltrg: (a) 90 cm Hg gage; (b) 40 cm Hgvacuum; (c) 100 psiS; (d) 8 in. Hg vpcuum, and (e) 76 in. Hg gage. Ans. (a) 221..24 kPa; (b) 48 kPa; (c) ?90.83 kPa; (d) 74.219 kPa; (e) 358.591 kPa

Gravitational Potential Energy (P) The gravitational potential energ:y of a body is its energy due to its position or elevation.

p=Fsz=ry

A fluid moves in a steady flow manner between two sections in a flow line. At section 1:A, =10 fLz,Dr= 100 fpm, v, = 4 ft3/lb. At section 2: Ar- 2ft2, pz = 0.201b/f13. Calculate (a) the mass flow'rate and (b) the speed at section 2. Ans. (a) 15,000lb/h; (b) 10.42 fps

AP

10.

=

P,

-

P, =

ff@r- zr)

AP = change in potential energy

Datum.plane

If a pump

discharges 75 gpm of water whose specifrc weiglit is 61.5 lb/ft3 (g = 31.95 fpsz), frnd (a) the mass flow rate 11.

in lb/min, and (b) and total time required to fill a vertical cylinder tank 10 ft, in diameter and 12 ft high. Ans. (a) 621.2lblmin, (b) 93.97 min

Kinetic EnergT (K) The energy or stored capacity for performing work pos' by a moving body, by virtue of its momentum is called kinetic energy.

Hrls$ed

K=# nK=4-K,=fttoi-ui) AK = change in kinetic energy

22

23

qT Internal EnergY (U' u)

Flow lVork (Wr)

Internal energy is energy stored within a body or substance by virtue of the r"ti.rity an-cl configuration of its molecules and ol thu vibration of the atoms within the molecules'

Flow work or flow energry is work done in pushing a fluid across a boundary, usually into or out of uy*L-.

u = speci{ic internal energy (unit

mass)

Au = tlz

- ul

fJ = mu = total internal energy (m mass) AU = Uz

-

"

13orr

nrll

lr'_

lVr=Fi=pAL

;1=Area of Sur.face

Wr=PV

Ur

Work (W)

l"ig. 3

FIow Worh"

work is the product of the displacement of the body and the component of the force in the direction of the displacement. w,r.k is energy in transition; that is, it exists only when a force is "moving through a distance."

Work of a Nonflow SYstem Cylinder

---.

The work done as the piston moves from e to f is

Final Position of Piston

dW=F,d*=(pA)dL-pdv

Piston At ea = .zl

I

'"**F

which is the area under the curve e-f on the pV plane. Therefore, the total work done as the pistonmoves from

lto2is

AW, = change in llow work

Ideat (e)

lleal is energ'y in transit (on the move) from one booy or '::1"11.1'ry1 to another solely because of a temperature difference I'r'l wr:err the bodies or

nV Fig. 2

woRK

ot

which is the area under the curve 1-e-f-2.

EXPANSIoN.

The area und.er the curue of the prrcess on the pV plnne rcpresents the work d'one during a nonflow reuersible process. Work done by the system is positive (outflow of energy) Work dnne on the system is negatiue (inflow of energy) 24

systems"

u{-_. ,{,.-.

t) is poslfiue when heat is added to the body or system. (l is negatiue when heat is rejected by the body or system. Classificati.on of Systems rI

' w =Jlndv

AW,=Wr,-Wrr=pr%-FrV,

'

t

A

r

l,or r ntlaries. .\ r | ( system

r'lrr.se

d' system is one in which mass does not cross its

'r,t'n

is one in which mass crosses its bounda-

Cnnservation of Energy

|1,, l.riv ol r:orrservation of energy states Lhat energy

:. r.ti, I r r'rtlr.tl ttttt't/t,St,nlyeCli l,, f u:,1 l;rw ol'l.lrr:r'modynarnics states :::i:':. , !tttt \. ltt. (..,ttIt('t.l((1. i.n.l.O U.nOthCf.

ls

that one fornt oI

SteadY Flow EnergY Equation

of steady flow system' Characteristics -

i. There is neither accumulation nor diminution of mass within the sYstem' 2. There is neitier accumulation nor diminution of energy within the sYstem 3. The state of"the working substance at any point'in the system remains constant'

Problems

t. During a steady flow process, the pressure of the working substance drops from 200 to 20 psia, the speed incneases from 200 to 1000 fps, the internal energy ofthe opeh system de. creases 25 Btu/lb, and the specific volume increases ftom I to 8 ftsnb. No heat is transferred. Sketch an energy diagram. Determine the work per lb. Is it done on or by the substance? Determine the work in hp for 10lb per *io. (t hp = 42.4Btu/ min). Solution

peia p, = 20 psia o, = 200 fps rlr = 1000 fps vc=8 ffnb n vr=lfts/lb pr = 200

Kl Fig. 4 Energy Diagram of a Steady Flow System Energy Entering System = Energy Leaving System

P, + K, + Wr, + U, + Q = Pa*

t-l

W,,

II,

2

Wl"+ U" + W

Au=-25Btu/lb Q=0

Energy Diagtam

d=l"P+ak+l-wr+aU+W

,F, +

(SteadY Flow Energy Equation)

llrrnis

K, + W' + U,

+

A,=Pr+

4

+ W* + U, + W

I lb2

lr"3 ]

EnthalPY (H, h)

= o.8o

fluids Enthalpy is a composite property applicable to all and is defined bY

h=u+pv and H=mh=U+PV The steady flow energy equation becomes

+K'+H'+Q-l;..?J*ril*

fi, W,,

,lf = Offiimi=le.e?r+b l',v,

llr V.l -*

26

ffL

E*'ii,lE-Hl (20) (r44) (8) = 778

= sz,o2

Bfi

2e.6rff 27

-T'r-(a) Basis

Kr+Wrr=Iq+W,r+Au+W 0.8 + 3?.02 = 19.9? + 29.61

w = 13.24

-25

f

lb'?n'

K,=S= ,Cffio,,

+W

ff,0t,

,q =*=

t-

lr s24ffi["*il

w:

L-

Wr, = PrVr =

= 3,12 hp

turbine bt 200 2. Steam is supplied to afully loaded 100-hp u'.=^19'0 fp*' and ftsAb priu *itft = 116'bT nl"/lb,"t, ::'1U "r at r prl" *ilrt * J ozs Btunb, Y,=-29! ft3Ab and is Exhaust -= turbin is L0 fps.

(1100)2 = Z+.t7 BJu (778) rb-

(32.174)

(200) (144) (2.65)

779

tne heat loss from the steam in the

(a) the "glJu. enersy change and determine *o"t p"" tU steam and (b) the steam flnw rate in lb/h'

PzYz=A+#@=s+'z+ff K, + Wr, + ur + Q- IL + Wo + u, + W

;t.20+ 98.10 + 1163.3 + (-10) =24.L7 + 54.42 + 925 + W

il;;ipor""tiur

w=

Solution p, = 200 psia

p,

-l

Psia

u, = 400 fPs

--'-- #E

= 98.lC lb_

42.4(mi#)hp) wrz=

rioo

(z',)

=3'20ff!

(roo

u, = L163.3 Btunb v, = 2'65 ftsnb u" = 925 Btunb

vr= 294

u, = 1100 fps

Q = -10 Btu/lb

(b) Steam flow =

r

Eru-l hp) P544lrr) trro)

251 Btu

---

fts/l.b

Fl{ 251ff

E;

r

= 1014

+

:t.

An air compressor (an open system ) receives 272kgper r r l of air at 99.29 kPa and a specific volume of 0.026 m3/kg. The nr r" llrws steady through the compressor and is discharged at frrllf l-r kPa and 0.0051 mslkg"The initial internal enerry of the ,r r rrr | 594 Jlkg; at discharge, the internal energy is 6241ilkg. 'l'lrr.r

., t Eiil

co

-B ^ -k-l 'p

Ratio of Specific lleats

AH = ECo (Tz

and c,

Fromh =u+pvandpv=RT dh = d11+ RdT

Qn

g

cn

r

I tf

'ilr

=11'631b

n,c" (T, _ Tr) I

t.63 (0.3685) (T, _ 540) 4{t

E

T where:dQ = heat transferred at the temperature T AS = total change ofentropy

Tz = 547"R Pz

= Pr (Tuftr) = 75 (5471540) = ?6 Psia

as--fu

2. For a certain gas R =320 Jll (0.1714) (T,

306'1 K

tz =

33.1"C

AT =

-

289)

constant-vorum,e system receives r0.5 lr.I of

Gn

Board problem _ April

f

lg,

l"ggg)

Solution

2T I

t

p,v

/ Vs

Irreversible Constant Volume Process (-850 kJ/h) (1 h) = -€50 kJ

Tr = 278 K Tz=400X

vs q

I

p, = _344 kPa V-0.06ms

,/

1

,\lr

a=

lt

I

c.

l

:1 = 0.6SgS kJ(kc) (K) = 2b9.90 J(ks) (K)

I

Solution

56

(22.7 kS) @.t87 kJ/kg.C") = 19.3 C"

k.mperature is 400 K.

is applied to a tank contai 22.7 kg of water. The stirring action is applied for I hour the tank loses 850 kJ/h of heat. Calculate the rise in ture of the tank after I hour, assuming that the process at constant volume and that c" for water is 4.187 kJ/(kg) (

I

kI

rffi5.6 kJ

-AU. DC"=

5. A closed

4. A l-hp stirring motor

-l 'l

(-2685.6) = 1835.6

lrrrddle work. The system.coSt-ains o*yg"r, at B44kpa, 2?g K, rr.d occupies 0.0G cu m. Find the t eat (gain or loss) #e nnat

mc,T2-Tr)

T, =

-

AU = mc" (AT)

= 427.34kg

a = Ll-ruooealt-l9 h llliO hl I = rzsok.ul a =

-2685.6 k I

a = AU+W

volume of room = (L2) (10) (3) = 360 m3 volume of air, V = 360

r

_ =

(344)

(0.06) = 0'2857 ke _

id:t500n?s)

mc" (T,

-

Tr)

Q.2857) (0.6595) (400 22.99 kJ

-

278)

AU+W 22.99 + (*r0.5)

t2.49 kJ fr7

(g) Steady flow isobaric.

Isobaric Process

-

(a)Q=AP+AK+AH+W'

An isobaric process is an internally reversible prccess of substance during which the pressure remains constant.

W =-(AK+Ap) W" =

(AP = 3;

N\ \s\:i\

(b)

-

.2 JVdp = W + aK I

0=W"+AK W" =

Fig.6. Isohric Process

(a)

Relation between V and T.

Tz

Vz

(b) Nonflow work. t2

{,ndV

= F(V2

AIJ = rDC" (T2

-

Vr)

-

(c) The change of internal

Q = mcn (T,

from b cu

to 15 cu ft while the at lb.b psia. Compute (a) T", (b) AH, (r') AU and (d) AS. (e) For an internally reversible'nonflow f r'ocess, what is the work? Solution T

mcohfr

2

l

__>_2

/ ,/

Tr)

p= V, = %= T, =

15.5 psia

5cuft

l5cuft

80+460=540"R

vc

ofenthalpy.

-

ft and

/

Tr)

(f) The change ofentropy.

58

A certain gas, with c, = 0.b29 Btu/lb.R" and R 96.2 ft.lV .lh."R, l. expands = g0"F

-Tr)

AH = rlc, (T,

aS =

l'roblems

energ:y.

(d) The heat transferred.

(e) The change

-aK

trrcsgutre remains constant

Tr=vi W"

-aK

,^)'r',

=1:,=

'r'\,,

=

g+lP

=r620R

.

ffi i##ffif)

=o.2r48rb

51)

= = =

mce(Tz

_ Tr)

(0.2148) (0.529) (1620_ 540) 122.7

Btu

(c' c" co-R= = 0.b29-W=0.40ss#S (n\

AU=

2. A perfect eas a If 120 kJ *" \1s value of R = 319 .2 Jlkg.lfurrrtt lt

r.2G.

iaggJ-fi;ik;

Solution

mc, (T2 _ Tr)

= = = a = Tr =

= (0.214s) (0.40$;(1620 _ b4o) = 94 Btu (d)

os

=

mcorn

ftI

= (0.2148) (0.52e) h =

of this gas ar

c''r.rlrrrrl ,fiTre):f: jli.i?Ttlmrnlm{:m1t,'i,i,t?,,,,,

ffi

(a) co =

Btu

*

k

1.26

m R

2.27 kg 319.2 J&g.K

f20 kW 32.2 + ZZg

-(1.2gxo.a1e2)= t.b46e

a = mco (T, - T,)

0.1249 oR

-

BO5.Z

f{_

kg.Ku

r20 = (2.27) (r.b469) (T, _ g05.2) (e)

p(%

\=

- v,)

(r5.5) (144) (15 778 =

28.7 Btu

Ta

-

5)

(b)

=

aH=

(c) cv

=

s39.4

K

mco (T2 _ Tr) =

l20

h=ffit$

=r.22??#h

kI

AU- mc, (T, - Tr) =

(d)

(2.27) (r.2277)(33e.4 _ 305.2) 95.3 kJ

-ITl =mR(Tr*T,) ^LP, --tri] ' plg,_

W = p(%- V,) =

= =

(2.22) (0.8192) (Js9.4 _ g0s.z) Z4.Zg kJ

K

-Fr

Isothermal process

G) Steady flow isothermal.

isothermal process is an internally reversible constant temperature process of a substance.

(a)Q = Ap+AK+AH+W

w"=e-Ap-AK W"=Q

(AP-0,4K=0) ft)

'i!:{t

-

From pV = C, pdV + Vdp

F-o'-{

-,!'uoo=-l;,i I

Fig. Z. Isothermal process

P'\1n

(a) Retation between p and V.

W"=W"

ft) Nonflow work.

(AK = 6;

f2

r

-_

0, dp = -

#l

pdv

-v/2

=

j oou I

-w

PrVr = Pz%

)2

Cln5= n,v,rr * vr ' v, {v

w" = Jpav=l$Y=

(c) The change of internal energy. AU=9 (d) The heat transfenred. Q= N + W" = p,Vrln (e) The change of enthalpy.

AH=9 (f) The change of entropy.

n

^s=+-mRrn$j 62

.2

JVdp = W + aK

'r'olrlcms

I

l)uring an isothermal process ggoF, at the pressurc orr drops fr.om g0 p.i" tol For gsic. *r,,r.11;i[lls process, lfru ipaV and the work of a i,,,i1ll1v1y process, (b)_d:,tennile fal the-_ JVdp;ndllie *o"k of a steady llow f 'r , !, '.,,:, rluring which AK = 0, ("i e, iai aU oS. rr tt,

.t''ir

"rilJ"lr",,

il;fi,;liii

*=r -nrrn& Y

f

Pz

Tl r t pV,=[ ,'ul I

\l

\\.2 I

-L V

1*--__r.__2

T m pl Pr

88+460=54fi,,lt 8tb 80 psia + 14.7 = 1.9.? 1lsi1

r.t

(a)

mRT r" * lndv = p,V,tnV' Pz Vr=

= tltt#ftQ

t"

f#

= 42L.2Btu

V,

In vl = "r-

W,= jOaV=42l.2Btu' jvap

(d)

= p,V,ln

.f,

= 42L.2Btu

v,

=

m#oO =-r.80 = 0.1653

= (0.1653) (0.30r) = 0.0498 m3/s

P,t, - (b86) (0. --To:oa#l) =3542kPa

AU=0 AH=0

(e) m=

Uft

q

% = €-1.80 q

(c) a = ryt *W"= 421.28tu (b)

v2

Q = Prvrlo

(b) Since AP = 6 and AK 0, W" lV" = = = e = -B1Z kJ/s

(t)ns=

3=W=0.2686#

+=

Solution

=-1.ob8kJ/r(.s

AH=0

2. During a reversible process there are abstracted 317 kJ/s from 1.134 kg/s of a certain gas while the temperature remains constant at 26.7'C. For this gas, cD = 2.232 and c" 1.713 kJ/kg.K. The initial pressure is 586 kPa. For nonflow and steady flow (AP = 0, AK = 0) process, determine ( Vr,% and pr, (b) the work and Q, (c) AS and AH.

#

:l

Air flows steadily through an engine at constant tem_ K.Find the workperkilogram ifthe exitpressure i,',, r' l.hird the inlet pressure and the inlet pressure is zoz kpa. Arrarrrro that the kinetic and potential energy variation is 111'plrplible. (EE Board Problem - April lggS) u'r rrl,'re,4_09 r

r

tlnlttlitttt

a=. fi= Pr= ,n

-317 kJ/s

1.134 ks/s 586 kPa 26.7 +273=299.7

vs (a)

c, = 2.232 - 1.713 = 0.5L9 kl/kg.K (1.134) (0:5_U)) (299.7) = 0.301 m3/s = _*xTl= pr 586

T R

\

Pr

'\2

= = =

400K 282.08 kJ(ke) (K) 2O7

kPa

Pr

p, =$ V

R - cp

\i. 64

tT \ l)V=C

R't'I

l),

-_.(9,?87_q8)

207

gog)

= 0.5547 m,t/kg (;5

(c) Relation between T and p.

W = prvrl" t=nrvr1nfl

= =

(20?) (0.5547)

ln

12

3

q

126.1 kJ

2.

Adiabatic simply *"t"t-"theat' of constant entroPY'

[p,l rLP-'l

Nonflow work.

Fromp\A=C,p-C1r-r

IsentroPic Process An isentropic process is a

=

k-1

W"

reversible adiabatic process' A reversible adiabatic is one

,2

rz

,2

= lpdv=J CV+dV= C { V-ndV t'Itl

Integrating and simplifing,

w-

l-k

n

pvn=9

l-k

'fhe change of internal energy.

.pv=Q tJl

AIJ = ncu (T2 - Tr)

\

I

'l'he heat transferred.

Q=0 'l'hc change of enthalpy.

Fig. 8. IsentroPic Process 1.

AI{ = mcp (Tz * Tl)

Relation among P, V, and T'

'l'lrr: change of entropy.

(a) Relation between P and V'

ns=0

P'VI=PrVb=C

I iI

(b) Relation between T and V' From p,VT = pr$u,td

T,= lvt-

T, (i(;

q =+'

r.rrrly flow isentropic.

,,,r(c,.AP+AK+AH+W" we have

wo,,_-AP_AK_AH

k'l

W. -AH

I

LqJ

r

\l'

O,

Al( = 0) 67

T-

E

(lr) _ p,V, (800) (t44)(100) m=

.2

(b)- lVdp=W"+AK t'

ftfr=-6f6ffi

1-L

LetC=pIVorV=Cpk

AII

'.2.1

- t'lVap =!C pk dp

AL.I =

Integrating and simPlifYing,

- t'fiao' -

= ms,

k (P'v'

- P'v') r. = f'nav l-k

i

Problems

1. From a state defined by 300 psia, 100 cu ft and 240" helium undergoes andisentropic process to 0.3 psig. Find (a)V and tr, (b) AU and AH, (c)JpdV, (d) -5vdp, (e) Q and AS. Wha is the work (f) if the process is nonflow, (g) if the process i steady flow with AK = 10 Btu?

(f, * Tr) = (1b.99) (1.241) (211.8 _70{)= _9698 tstu

mc, (T,

tt')6av

=

=l5'eelb

-

Tr) = (15.99) (0.74b) (211.S

&!;f,J'

- 200) = _5822 Btu

=ffi

= b822 Btu

rrlt *!Vdp = kjpdV = (1.606) (b822)= 9698 Btu

lr,)a=0 As-- 0 rlr a = AU+W" W"= -AU= 1-5822) =b822 Btu Irir

JVdp = W" + AK

1Xj9g=W"+10

Solution

W" = 9636 31rt

Pr

=

300 Psia +'l'4.7 = 15 psia 0.3 Pz= V, = 100 cu ft.

T, = 240+46A=700'R

h

'.', An adiabatic expansion of air occurs through a nor,zlt,

"rrr ll28 kPa and ?1oc to 1Bg kpa. The

initial kinetlc energy i" For an isentropic expansion, compute the spcr:if i. .r,lrnnr), temperature and speed at the exit section.

..'11lr1lible.

titi rr lion s

I (a)

\

= v,

1'666

H$t= 1oo[,!9f

I

= 608.4 rtg

l?r

T lr2 -2--T^'Lpil I

t"= 68

\z

r.-_T r.666

=

7001__{q_l Lsool

pVk= 6

\

1.666-1 k-1 -'l-k-

\

=

828 kPa 7L + 273 = i|44 l( 138 kPa

211.8'R

-248'7"F (il)

k-r

ll2l 'Lpil

T"=T, -

r.4_l

-k

tnl

-.-1.4

=

344lHgl 18281

= 206 K

;,,\

it>\ 't.h^I

tz= -67oC

",

=

#,

_ (0.287q8X344) = 0.1193 m'/ks

., // i,

22Q..,

'Zzt

75yty:; 'iivr2i

lI

-

ve = vr -

Ah =

= 0.429m'/ks [g'l. = 0.1198 lHgl'n LprJ 11381

cp

(T,

* Tr) = 1.0062 (20G -

Fig. 9. Polytropic Process

344) = -188.9 kJ/kg

A =&*aK+Ah+/"

Itelation among p, V, and T

AK--Ah=136,900J/kg

(a) Relation between p and V.

AK=4-^r=* D2r=

1Jz

(2k)(AK) = zf r

P,vi = Prvi

ffil

(b) Relation between T and V.

1rg,966S ) = 277,800 m

To

= 527.1m/s

T t,

t

/-vJ "-t

=1q1.

li.elation between T and p. *.1

L

Polytropic Process

rn Le

Ra 'r',

A polytropic procebs is an internaliy reversible during which pV" = C and prVl = prVl = p,I"

I

r-lP. l:-€-

-lp.

I I

t_^ t--l

I

Nonflow work It,

where n is any constant. I

(paV = PrY, - P,V, " ,'l-n

mR (T,

'l'hc change of internal energy AIJ = mcu (T,

70

r-

-

T1)

-

T,)

4.

The heat transferred

a= =

(b)-

AU+Wmc" (T2

- T,) +

mR-(T,

-

- ,fvao = {&t:!& = T_n-- -n

Tr)

,2 . JPdv

1-n

Ic -nc +Rl (r2-r,) = *Lffj

I'rohlems

=

- lffl

=

,n." f-!- "-j (T, _ T,) Lr - I}_l

polytropic process, t0Ib of an ideal gas, whose l.^ 3X"^1u: and 40 ft.lbnb.R cop = o.-zs __:_ _vwrv.r!, luau6,cs suate Irom zu lrlr;r and 40'F to 120 psla ra and 340"F. Determine (a) n, (f;4g urr4 dY, (? (g) rf the pi"*,, ,iuuav ;ll,l !ilil,-(11'9:l"ljf l,'rv F{

0(|)

E:r

.; bb

r-.

N

0o

Available energy is that part of the heat that was converted into mechanical work. Unavailable energy is the remainder of the heat that had be rejected into the receiver (sink).

The Second Law of Thermodynamics AII energy receiued as heat by a heat-engine cycle cannot conuerted into mechanical work.

Work of a Cycle

(a)W=IQ W=Qo+(-Qn)

(Algebraic sum)

W=Qo-

(Arithmetic difference)

Q*

(b) The net work of a cycle is the algebraic sum ofthe done by the individual processes.

W= LW

Operation of the Carnot Engine A cylinder C contains m mass of a substance. The cylindor head, the only place where heat may enter or leave the subgtance (system) is placed in contact with the sounoe of heat or hot body which has a constant temperature Tr. Heat flows from the hot body into the substance in the cylinCler isothermally, l)rocess l-2, and the piston moves from tr' to 2'. Next, the t:ylinder is removed from the-hot body and the insulator I ie placed over the head of the cylinder, so that no heat may be l,ransfemed in or out. As a result, any further process is ndiabatic. The isentrppic change 2-3 now occurs and the piston moves from 2' to 3'. When the piston reaches the end of the sl.roke 3', the insulator I is removed and the cylinder head is placed in contact with the receiver or sink, which remains at a ronstant temperature T". Heat then flows from the substance t,rr the sink, and the isothermal compression B-4 occrut while tlrc piston moves from 3'to 4'. Finally, the insulator I is again lllnced over the head and the isentropic cor.npression 4-1 ret,urns the substance toits initial condition, as the piston moves ftom 4'to 1'.

W=Wr-r+Wr"r+W'n+.. The Carnot Cycle The Carnot cycle is the most efficient cycle concei There are otherideal cycles as effrcient as the Carnot cycle; but none more so, such a perfect cycle forms a standard ofcomparison for actual engines and actual cycles and also for other less effisient ideal cycles, permitting as to judge how much room there might be for improvement.

H' m

Fig. 11. The Carnot Cycle 82

n

Vm Fig. 12 Canrot Cycle

Anulysis of the Carnot Cycle (ln

= Tl (S2 - Sr), area l-2-n-m-1 (1,, = T3 (S4 Ss), area B-4-m-n-B 83

E BE :E H" ff.8 xie a; E ilg -t Frg

tlal.{ lll

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fyr.

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tr

Pl>

d

;; t-l

l-*|t'

'

rc. f,

E il

dd

:@

G

T

lrll

iF

E

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+

E*l * tA B llts 9 EJI tB EA .L;

il

EJ

F lcr"

tl

tl

>^t

dlo tlt

tl

"€rl -g

B

Of{

E

ot

gd

Ei

I

tr +f; q fi € HI

d

+

t'l* pll HB' .9+ .:

iij

Er

6l

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F

F:

E

t' Y

t'r>

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E{

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f1

i$ !g; f; fl: .i;lr Eils

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t<

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fld]

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S

jE ,Ll

q?l'Hric'$ r d r H F t.E* g g*THE €

+

cl*

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i

rY

H.jj ., +) ll (n (E ()(H

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;E>gi :=:

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Et t;

€! trE o9l

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il

;q. Fti

I

EE

S1rEDI^

d h l^.[ui | .i d €'l rn* v |

* (.l

i|

rr)* FfllF. lg_ X E| il rr

i i illl

F

^l-^*

dl _ tlFi

Flo' dl il

il

flt

E

:

irt , ?

€.*!; i-''- tt rl j 'q

i

i

o --,dfi rEi sis :N r rr E >1s Ei T c j EF n [' E

t

dll

.e

$E Et

Aoa

Gi ; ;

H- i"*l*

$ H *..-l,.* H E=ld iE.'E EEA E : s

8E: E , ,, E tif E E fi * o'* i 6B

i {S i! F'i-'r---J

E

rr

trlt ''

s

E

= cc

-TB (Ss

w-

-

S. ) = *Tr (S2 - Sr)

Qn

-

Q* = Tr (Sz - Sr) - Ts (S2 - Sr)

(Tl

-

Ts) (S2

(Tr

-

S1), arca

T3) (Sz

L'2-3'4'l

r;s;;r o"=

W

e=

Sr)

Tr-T,

e = ---Erl The thermalefficiencye is definedas the fractionoftheheat cycle that is converted into work ; supplied to a thermodynamic

Work from the TS Plane Q^

=

mRTrfn

Qn

=

mRTrln

t

\[ = A^ - a- = mRTrtnt

w-

(Tt

g=

w a;

g=

-

mRl"

Tr) (Tr

-

mRTrh

t

+-v

Ts) mR

ln fvl

kL ,V,

mRT.

Tt-Tt

-T,

Work from the pV plane.

W = IW

f =

-mRTrln

= Wr_, + Wr-, + Wr-n + Wr-,

w = p,v,l"

V.-V3

1;

Q* = -mRTrt"

i

t. &+: :J,+ p,v,rnf,.&tJ{.

From process 2-3,

T3 l-v, l*-'

T

=Lv'J

Mean Effective Pressure (p_ or mep)

P-=W VD

From process 4-1,

T,

-lfJ 11 -l-v,J.-' but Tn = Ts and Tr =T2

- | = therefore,l V"-k-r

LqI

then,

84

&

% =v, vr

Vp = displacement volume, the volume swept by the piston rr one stroke. Mean effective pressure is the average constant pressure l,ir:rt, acting through one stroke, will do on the piston the net work of a single cycle.

Ratio of Expansion, Ratio of Compr.ession I,)xpansion ratio

vglute,3t

end of expansiql -., the = volumeattheffiili ft5

tco

giF

HEc{

6 aa

e

*F, !od tr-ro L,+) g

:(D ttll

€L

tl

\ro

lA oilS u; vlJ/l*

-

- ll

ll

'tir---t'---r r. dl- ct

ll

Eil € ld EI

'F lF,

Els

L-J

N

EI

'5

lcotFI i-lu?

r.'ilco c\

.+

lv oilX rO l-: "l

o2cd c6o

EE

ll

+r

€;9 'i{r tv ()r F{') a''+) ,k-q)| t{ls li r Ei o' ) a,(;r' \4,),^: ;,> la \ li,i(a.7442)(950- 886.9)

131 IGI

= Qo-Q*=131 -L32.2=-L.2kJ

- if r#iFosgfl

=-12okw 1)

|

Qe

= =

en=

(m) (c") (T3 - Tr) = (0'1382) (-0'6808) (540

-

939'9)

c_ 1.0411 ro={=yiffi,=1'3ee

37.63 Btu

mRr.rn{=,Wt"*h

Point

n = '!{ =

-

1:

v. '' = -

-27.82FJttt Qo

c, + R = 0.7442+ 0.2969 = 1.0411 KI EAIF

cp =

Q* = 37.63 -27.82 = 9'81 Btu

o -A sz As.ir:fl=-bao

-IT, - (2.5) (q396e) (e50) = 0.8522 m3 827.4 &

Point 2: Qn = mco (T,

_o.osrdlg =

-

Tr)

-132.2 = (2.5) (1.0411) (T,

(9.8!X179)w p-=ql172= ffi-v'LvEe'

= B.lb psi

Tz = 899'2

gas with- R = 2963 Jfte) 2. T\vo and a half kg of an ideal 't kPa and a a-ryJt:y"" (K) and c" =6i++i r'"lltr'?Xrc11i kJ of heat9fat127constant pres' temperatrfe b6Fc *J*t 132.2 point C to a

to nJis = "f sure. The e""1;it""-"d;a*a "tto"ails back to its bring-tle wil where a constant volume p"ot"tt poier in kW for 100 Hz' original ttateS;t"rttil; er;q' *d the

e:

%=

- 9b0)

X

u,F,]

=

(0.8b22)ffi21 = 0.8066 mg

=

rsro.rlffi"u-'

Point 3:

r, = r,

H]"''

= 880.e K

Solution

Qo = mco (T,

-

Tr) + mcv (Tr

Qn = (2.5X-{.4435X886.9

v Pr= rF 11 -

Q*=

827.41,Pa 677 +273= 950K - 132.2 kJ

Qn

'![

w

=

-

-

T3)

S99.2) + (2.5>(a.7442)(950- 886.9)

131 IGI

= Qo-Q*=131 -L32.2=-L.2kJ

- if r#iFosgfl

=-12okw 1)

|

,V

wnere

Otto Cycle The Otto

cYcle

is the ideal prototype'of spark-ignition

engines.

the isentrcpic compression ratio "* =vr.,

Derivation of the form ,la for e Process l"-2:

5_ Tr- t-rl-l

LVol

T, = Tr"oo-t

'

(2)

Process B-4:

FiS. 14. Air-standard Otto CYcle

Air.standardcyglemeansthatairaloneistheworking medium. 1-2: isentroPic comPression 2'3: constant volume addition of heat 3-4: isentmPic exPansion 4-1: constant volume rejection of heat

& I-v;l*'' =F T= Lr*J

L-l (3) T, = Tn"* Substituting equations (2 ) and (3) in equation

'-E4rffi

a - ,

= Qn = \{ =

mc" (T,

- Tr) mc, (T, - Tn) = -mc" (Tn- Tr) (T4 Qn - Q* ' BC" (Ts - Tr) - BC'

e=fr=ffi r-#+F

e = 'rr - rz e = 1-+ rl

94

(1)

Tn-T,

e = 1_n+ -t

Analysis of the Otto CYcle Qe

tI

IVorh from the pVplane Tr)

W=

IW = Pr%'- 9rV, * O,? - -% O,

Clearance volume, per cent'clearance

"*=f=q;r=Hg6 _l+c

".*c

(t)

Review Problems l.ThbworkingsubstanceforaCarnotcycleis8lbofair.

feginning of isothermal expansion is.9 cu ft during the *a tn" pressure is 360 psia. The ratio of expansion is uaaiuo" of heat is 2 and the temperature of the cold body (g) P-,, (h) the (0 ;0"F, Fi;J (a) Qe, o) QR, (c) vr, (d) pr, (e) vn, pn, and (i) the process' ratio of u*purrsion duffng the isenlropic overall ratio of comPression. (d) Ans. @) gia.a, Btu; (b) -209.1 Btu; (c) 63.57 99.ft; (h) 3"53; 25.(/-p*iu; t"> ef.Zg cu ft; (f) 51.28 psia; (g) 13'59 psia; The volume at the

(8) 7.06

in Gaseous nitrogen actuates a Carnot power -cycle whict the respective iolumes at the four corners of the cycle, Vri rt"*frtg ;tlnetUegittning of the isothermal expansion' arg cvcle L 3 zza.r+!, *1 Yr r57'7 ib. iit i; v, = 1 4.bI L, v Jhc "Z Determine (a) the work and (b) the it"it. receives zi.r t [7

Tl\lolsb/

=

srO.ZztttJil

.'l -:l

Y o" -14.1 kW -846.1 mln

(d)w,"*=qPR)*-! [121s.2711fH =(1.3eb) (8.48) (0.25ee) (2ss'7) l!0135i

g-

= -..309.b mrn or -14.49

' '^oc

-

adiabatic ideal work brake work

'=

DraKe wofK

14'49

20.41 kW 0J1 =

Cards, IiS ?2. Conventional 'rwo-Stage, No pressure Drop

v _Fig.,23. Conventional Cards,. Two-Stage, with pressure Diop

kw The figures abov-e-show the bvents ofthe conventional cards of a two-stage machin", *itl ifr* nigh pressure (Hp; srpe.posed on the low pressure (Lp). suition il th; ilp.ji"a*" begins at A and pry"Vai; in. Compression t-2 occurs and the gas is The discharged gas passes through the $yharc.ei interc*te" cooled by circulating water

G

through the

Ii"*r

"*ir-".

interc*t." ".rd"is i"U"r. Co"uu"tio'Jfi,"it i, t:f7

entering the

Pr = P'

el rrpcvrindeTiu.ir,?,u-g*g;;^iil:;tt*mi*""1i$

gas leaving the intercool:l assumed that the

'* Hft *u*kil*t=P**T'*'-**fr r must reexpand F-E fromtheGuuv^'--ot Iearance and ;

=

high cylinder + W of the

l#,Kkl*-1.#[ft]*-tr

of multistage adjust ll:.o*tution donejn the are Itis common practice to works compressor, *o ti":*imum work tbr comp"u"""Jiil"t cvlinders, "^"'otf oru " pressine . gi*'u" of P, = Pr =.P*' weltave #T- = il;,,h toitrat of the HP stage' or

;d

tr'uiipii#;;*y:f q;"iG *: :liiiT:H:ftff#Til:

#trf,*{=+[tlt'i p,=

i, i I

The heat rejected in the intercooler is'

yTF*'-

work pressure for minimum intermediate where: P, = sane' tlre t?la\work of eachcvlila"iillh" work cvlinder' or the since the workin each #;;;tJtwice for the two-stage

-1\ 2nm'Rr,f-1P,$ ;1 ='+Pfel* 1-n l9'/ r w= "iffiLft,? _J in A pressure drop "ide

Qt" = m'cn (T,

cvlinder)'

W of the loLPlessure Pressure cYhnoer

!\f =

I

Heat Tlansferred in Intercoolor

c each cylinder because pe (LP

iirp tvu'ii"'i"*a

tllrtll -- l'rtrHFllrlr

- T')

the intercoolor where m' is the mass of gas passing through byifrgif .ili"der and delivered bv tho i Jro tfr" mass clrawnin HP cylinder). Problems l.Therearecompressedl'1'33m3/minofairfrom26'7"C'

are 8Vo' L03.42kPa to 821.36 kPa' All clearance piston displacement and power (a) Find the isentropic for a single stage cornpresslon' required --=ft)-u*ing work for the,"-, a""t , nnd the minimum ideal air to the the t*o-ri"gr.oilpr"rrion when the intercooler cools

initial ---6 temPerature. Fi"h trr" di-splacement of each cylinder for the condi-

tions : of part (b). ial liow much heat is exchanged in the intercooler? *p'"ttiin efficiency of 78Vo' what (e) For * ""*"ff-is required? driving motor outPut

Solution

vf= Pr=

Pz= rT rl

-

11.33 m3/min 103.42 kPa 827.36 kPa 26.7 + 273 = 299.7 K

be spread on each the intercooler could

oi this ideal value' Pressure droP

Pr=P,*--T--

139

r =IilFR)* _(1.4) (108.42)

ftzgz.szttft;l -

N-mtz-t-il-J

(i,l.BBi lTga.BqtY/

1-1.4

=

l

(1.a)11s3.a2) (11.33) 1-1.4 L\

-

- 1416 # mln Tqtal work

- 3327# ot -55.45 kw

-

(c)n"=L+c--c

o" -28.6 kW

(2) (23.6) = -47.2 kW l-&1

+

LP'l

tr"=1+c-c

1o&42l |

=1+0.08-(0.08)

= 0.9119

vnrp=#=## =12.42# ' lezz'361.r =1+0.08-(0.08)h1ffi1

tr. vo=#=

11.33

mffi

*' = n#, =,+ffiffi$?,

*t _r^*o -'"'"Y

min

,l-=- -,BT€ '3 Pa

V; r/ vnur

(b)

=;jf

p Pr

103.42 kPa

Pa

827,36 kPa

=

= 18.62

#

(13.62) (0.2q2q81j299.7)

292.52

=

4.006 T3

mln

rn3 4006 4.393;fr =

ffig

(d) Qrc = th'cn (Ts Tr) (13.62) (1.0062) (299.7403.4) _ 1427 l&I=

min

(e) Outpur of driving motor =!7:? = 60.5 kW : 0.79

p,=y'[];=@

**=+#F)*

292.52kPa

I

lb/min of air from l4.B psia and gb,r to a final pressurer tf I gn psia'. $e lormal barometer is 29. g in. Hg and the tempern t rr ro is 80"F. The pressure drop in the intercooler is B paiand th, temperature of the air at the exit of the intercooler is g0,,1., tho speed is 210 rpm and pVt.er = C during compregeion und expansion. The clearance is E% for both cylinders. Ths temperature of the cooling water increase by iA F". Find (a) the volume offree air, (b) tlie discharge pressure ofthe low pr*rruro

t4l

cylinder for minimum work, (c) the tempprature at discharge from both low pressure and high pressure cylinders, (d) the mass of cooling water to be circulated about each cylinder and through the.intercooler, (e) the work, and (f) if, for the low pressure cylinder, IJD = 0.68 and if both cylinders have the sam: stroke, what should be the cylinder dimentions?

(d) c, =

m

90lb/min

po

(29.8) (0.491) = 14.63 Psia

To

80+460=540oR

Pr

l"

L4.3 psia

Tr

90+460=550oR

Pr

185 psia

ffi|$)

+

r-* I + 0.0b{0.0b) Fzslt c-clfil = = 0.9178

tfffi#ffi#P

;,

r, = r, r42

[*{*

767 oR

=

= (bbo)

t#.f

b50)

(rh*) (c,") (At*) = er.z

678 Btu

rh*=----414-l f Btu\ (18F") \6F/

=37.5 lb mrn

= BZ.E

#

Intercooler

,,trR^\ [Bzd #' ''"" =- , t,b:l+ - (550) Lffi] |

-

High pressure cylinder

= 51.4 -9= +g.gpsia

,rr l-p,

Tr) = (10S) (-0.0302 ) (767

-675 Btu/min

.ll"

=

-

= 1oB rb/min

Heat to water = Heat from air

a*

,n t,

= rB98 cfm

=$1f= tra,'f*ggxpzr

= 12Bo crm

pz= 5!.4+&= 52.9 psia ps

s]

V, = VD (1 + c) = (1393) (1 + 0.0b) = L46Z cfm

=

(b) p- = ilFm, =J043) (185t= 51.4 psia

(c)

[a

ru ;€g 0.9173 ' =*=

Pr

=

I

vn

6'RTr _ (90) (bB.B4) (bb0) 1Zg2 cfm (a) Vr= = , (143t(144) -'

v" =

=

Q"z = frrc" (T,

Vi*:--l

dhi

Low pressure cylinder D"

Solution

*-ffi = (0.1?r4)Htf = -{.0302

'#'

=

7G7'u,

Q," = rir,co (\

- Tr)

= (90) (0.24) (bbo

- 767) =+osz Blt mln l4lj

mass ofcooling wate" =

{y

4L2.3 D2 = 400.5

lb

D = 0.986 ft or 11.88 in.

= 260.4 min

L = 15.01 in.

(e) Low pressure cylinder

Three-Stage Compression nrh'RT, \i/. "LP = - l-n l7gt+ _ il

L\prl

]

(1.34) (e0) (53.34) (550) 1-t52.effi =@l\ra.si -jl

'1

=

-

b2G5

Total work,

(0

mrn = -L24.2hp

fr

IP cylinder

LP cylinder

B!t'

Fig. 24. Three-Stage Compression

= (2) (-124.2) = *248.4 hp

pV=C pV"=C

Low pressure cylinder

py

y^D44 =3.D2LN =!pe (0.68 D) (210) (2) = 224.3 D3 cfm

Condltlone

nlnlnum rork 1) wr,p = wrp %p

PV"=C

2-P,

224.3 D3 = 1398

-PV" =

I

D = 1.84 ft or 22.08 in.

L = (1.84) (0.68) = L.25 ft or 15.01 in. High pressure cylinde

v ":

(eg) (?50) _ fi'Rr3 _ !5giq1) = 36?.4 cfm (144) (49.9) p,

\r'D --i;n-,- gal Ufr*

= 400'5 cfm

V^u44 =ID2LN =3D2 (t.zb) (210) (z) = 4Lz.g D2 cfm

for

C

e)TS =T3 =Tl

Fig. 25. conventiorrut cu"arlThree-stage, No pressure Dr'p

,,1'T,r,

nm'Rro Zluf+il =,,p'or. f&\.'-l = --1*-L\&i If+l+- l=-I-n-[\d/ l-n l\Pr/ -l

11

Pr P" P,

Pn

=F, = P,

I

P, = (PrPr)

2

P, = (P,Po)

2

(1)

I

(2)

ft)T3-Tr=BgbK

Solving equations (1) and (2) simultaneously,

p,=\/ir'p,

and p,

n-l

T /&\ -- =,ruc /3ss'olff/ = 411 K 'z = Trrgf/ "no (ib3;)

=t6trJ

1-n [gf#-il l\P'r l

3nm'Rr,

Heat rejected in the first intercooler,

Qrc=

Problem Air is compressed from 103.4 kPa and 32"C to 4136 kPa by a three-stage compresor with value of n = 1.32. Determine (a) the work per kg of air and (b) the heat rejected in the intercool-

=

m'co

(\ - Tr)

(1) (1.0062) (305

- 4rr) = -106.2 kI

Total heat rejectred = (Z) (_t06.7) _218.4 kJ =

ers.

Solution p

m

lke

Pr

103.4 kPa 4136 kPa 32"C + 273 = 305 K

Po

Tr

(a) p, = (p,,pu)*= fioa.aX (4136j#= 353.6 kPa -1

l-n 7&.* l\P,)"-1.J

,., _ 3nm'RT, vY-

-

(3) (1.32) (1) (0.28708) (305) 1-1.31

=

-

L.IZJ l/353.6\ r'32-11 It-

|

l]103.4/

r

I

_1

376.2 kJ

t47

IT

il

Review Problems ,t

'

i

handles 1000 cfm of air psia and t, = 80"F. The 14 measured at intake where P, = discharge pressure is 84 psia. Cdlculate the workifthe process * of compression is (a) isothermal, (b) polytropic with n L.25, and (c) isentropic. Ans. (a) -109.5 hp; (b) -131.7 hp; (c) - 143 hp

1. A reciprocating compressor

2.

;

I

I f

t

cm A double-acting compressor with c = 7Vo draws 40 lb per minute of air atl4.7 psia and 80"F and discharges it at 90 psia. Compression and expansion are polytropic with n = 1.28. Find (a) the work, (b) the heat rejected, and (c) the bore and stroke for 90 rpm and UID = L.25. Ans. (a) 77.68 hp;(b) -1057 Btu/min; (c) 18.96 x23.70

in'

4. A 14 x L2-in., single'cylinder, double-acting air compressor wit}'5.5Vo clearance operates atL25 rpm. The suction pressure and temperature arc14 psia and t00oF, respectively. The discharge pressure is 42 psia. Compression and expansion processes are polytropic, with n - 1.30. Determine (a) the volumetric effrciency, (b) the mass and volume at suction conditions handled each minute, (c) the work, (d) the heat rejected, (e) the indicated air. hp developed if the polytropic compression efficiency is 75Vo, and (f) the compression effrciency.

Ans.

l4 f]

(a)92.7Vo;(b) 247.8 cfm,L6.72lblmin; (c) -18.93 hp; (d) -175.7 Btu/min; (e) -25.24 hp; (f) 77.42Vo

**

6.

ance of \Vo draws

3.

by a engine, the following data and resurts o-ut"i,r.,a, capacity, 800 cfm; suction it t+.2 psia; disch;d;;; iio pri,,; indicated work of compressor,'i5S frp; indicated work ol. lhe steam engine, IZ2 hp^aCal..rlute (u) tt u.";p;;i""im.i"n.y and (b) the overall efficiency. Ans. (a) 90,06Vo; (b) Bt.t6qo

rt"uq

An air compressor with a clearance of 4Vo compresses airfrom gz kpa, z7ic to 462r