Flo=6;T* I
Z2 i !
a
THEN[tl!ODYl\lA[tlIOS
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, Threeprocess Cycle.
6
Internal Combustion Engines gg Otto Cycle, Diesel Cycle, Dual Combustion Cycle.
7
Gas Compressors
ll5
SingleStage Con pression, Twestage Compression, ThreeStage 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 energJr 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
lt;]*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'cmcyne.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
=r8. 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 weightg.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 29,131ft, 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 Unur
(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 lsPI,l S'= tu.ll+se.o#l 0.4e =l K s
l.Whatistheweightofa66kgmanatstandardcondi
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 lbl
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 100L 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=PoP,
kg
The gage reading is called vacuum pressum or the vacuum.
weight of mixture,
re=x"
[email protected]
e.8066*#
=?8.esksr
I
ll"y using pressure gages
A Jrrt:ssure gage is a device for pressure,
rilr,,1llr 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
Tps
P=Po+Pg
_ F" 1V yAhPr=*A=:6l P, = Tb,
=ry'=*
Problem
A 30m 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 cmE
Srilution p = 14.5 + 40 = 54.5 psia
= po
* p, = 1.04 + 9.5 = 10.54#
tt 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"C0 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 IrIEXD =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 10ft diameter by 15ft height vertical tank is receiving water (p = 62.1 lb/cu ft) at the rate of 300 gpm and is discharging through a 6in 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
=

17'51l:!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 gramforce 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 50,000ft elevation. 'l'he acceleration of gravity g decreases by 3.33 x 106 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 ,lr. The mass of a fluid
ll,/l'1,:r
{;.
A cylindrical drum (2ft diameter, 3ft 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, =
[email protected] 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=4K,=fttoiui) 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 ancl 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)dLpdv
Piston At ea = .zl
I
'"**F
which is the area under the curve ef 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 1ef2.
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*
tl
W,,
II,
2
Wl"+ U" + W
Au=25Btu/lb Q=0
Energy Diagtam
d=l"P+ak+lwr+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'+Ql;..?J*ril*
fi, W,,
,lf = Offiimi=le.e?r+b l',v,
llr V.l *
26
ffL
E*'ii,lEHl (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 100hp 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
Erul 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 lr 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 ^ kl '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
asfu
2. For a certain gas R =320 Jll (0.1714) (T,
306'1 K
tz =
33.1"C
AT =

289)
constantvorum,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 V0.06ms
,/
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 lhp stirring motor
l 'l
(2685.6) = 1835.6
lrrrddle work. The system.coStains 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,T2Tr)
T, =

AU = mc" (AT)
= 427.34kg
a = Llruooealtl9 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" coR= = 0.b29W=0.40ss#S (n\
AU=
2. A perfect eas a If 120 kJ *" \1s value of R = 319 .2 Jlkg.lfurrrtt lt
r.2G.
iaggJfi;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"=eApAK W"=Q
(AP0,4K=0) ft)
'i!:{t

From pV = C, pdV + Vdp
Fo'{
,!'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,pC1rr
IsentroPic Process An isentropic process is a
=
k1
W"
reversible adiabatic process' A reversible adiabatic is one
,2
rz
,2
= lpdv=J CV+dV= C { VndV t'Itl
Integrating and simplifing,
w
lk
n
pvn=9
lk
'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
1L
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 lk
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= 15822) =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 2T^'Lpil I
t"= 68
\z
r._T r.666
=
7001__{q_l Lsool
pVk= 6
\
1.6661 k1 'lk
\
=
828 kPa 7L + 273 = i44 l( 138 kPa
211.8'R
248'7"F (il)
kr
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
AKAh=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
rlP. l:€
lp.
I I
t_^ tl
I
Nonflow work It,
where n is any constant. I
(paV = PrY,  P,V, " ,'ln
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
1n
Ic nc +Rl (r2r,) = *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 heatengine 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 l2, and the piston moves from tr' to 2'. Next, the t:ylinder is removed from thehot 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 23 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 B4 occrut while tlrc piston moves from 3'to 4'. Finally, the insulator I is again lllnced over the head and the isentropic cor.npression 41 ret,urns the substance toits initial condition, as the piston moves ftom 4'to 1'.
W=Wrr+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 l2nm1 (1,, = T3 (S4 Ss), area B4mnB 83
E BE :E H" ff.8 xie a; E ilg t Frg
tlal.{ lll
d
Nl
tl*
H
tr
E<
fyr.
E I
a{
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tr
I
tr
Pl>
d
;; tl
l*t'
'
rc. f,
E il
dd
:@
G
T
lrll
iF
E
I
+
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
Fi
F
F:
E
t' Y
t'r>
I
E{
Efr E
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'l>
Ni
f1
i$ !g; f; fl: .i;lr Eils
+
t<
F{
g i$e
fld]
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I
S
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q?l'Hric'$ r d r H F t.E* g g*THE €
+
cl*
>t> 5 a
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i
rY
H.jj ., +) ll (n (E ()(H
tl
;E>gi :=:
F
B
P.

E H
C!
r
Et t;
€! trE o9l
c;
Eg .!e @Y
r,r
'.+
il
;q. Fti
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EE
S1rEDI^
d h l^.[ui  .i d €'l rn* v 
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i
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i i illl
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il
flt
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:
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
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H i"*l*
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8E: E , ,, E tif E E fi * o'* i 6B
i {S i! F'i'rJ
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'23'4'l
r;s;;r o"=
W
e=
Sr)
TrT,
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.
TtTt
T,
Work from the pV plane.
W = IW
f =
mRTrln
= Wr_, + Wr, + Wrn + Wr,
w = p,v,l"
V.V3
1;
Q* = mRTrt"
i
t. &+: :J,+ p,v,rnf,.&tJ{.
From process 23,
T3 lv, l*'
T
=Lv'J
Mean Effective Pressure (p_ or mep)
P=W VD
From process 41,
T,
lfJ 11 lv,J.' but Tn = Ts and Tr =T2
  = therefore,l V"kr
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 trro L,+) g
:(D ttll
€L
tl
\ro
lA oilS u; vlJ/l*

 ll
ll
'tirt'r r. dl ct
ll
Eil € ld EI
'F lF,
Els
LJ
N
EI
'5
lcotFI ilu?
r.'ilco c\
.+
lv oilX rO l: "l
o2cd c6o
EE
ll
+r
€;9 'i{r tv ()r F{') a''+) ,kq) t{ls li r Ei o' ) a,(;r' \4,),^: ;,> la \ li,i(a.7442)(950 886.9)
131 IGI
= QoQ*=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= ffiv'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 aryJt: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 bringtle 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
= QoQ*=131 L32.2=L.2kJ
 if r#iFosgfl
=12okw 1)

,V
wnere
Otto Cycle The Otto
cYcle
is the ideal prototype'of sparkignition
engines.
the isentrcpic compression ratio "* =vr.,
Derivation of the form ,la for e Process l"2:
5_ Tr trll
LVol
T, = Tr"oot
'
(2)
Process B4:
FiS. 14. Airstandard Otto CYcle
Air.standardcyglemeansthatairaloneistheworking medium. 12: isentroPic comPression 2'3: constant volume addition of heat 34: isentmPic exPansion 41: constant volume rejection of heat
& Iv;l*'' =F T= Lr*J
Ll (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)
TnT,
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 'rwoStage, No pressure Drop
v _Fig.,23. Conventional Cards,. TwoStage, with pressure Diop
kw The figures aboveshow the bvents ofthe conventional cards of a twostage 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 t2 occurs and the gas is The discharged gas passes through the $yharc.ei interc*te" cooled by circulating water
G
through the
Ii"*r
"*ir".
interc*t." ".rd"is i"U"r. Co"uu"tio'Jfi,"it i, t:f7
entering the
Pr = P'
el rrpcvrindeTiu.ir,?,ug*g;;^iil:;tt*mi*""1i$
gas leaving the intercool:l assumed that the
'* Hft *u*kil*t=P**T'*'**fr r must reexpand FE 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 twostage
1\ 2nm'Rr,f1P,$ ;1 ='+Pfel* 1n 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*ori"gr.oilpr"rrion when the intercooler cools
initial 6 temPerature. Fi"h trr" displacement 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 * ""*"ffis 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 
NmtztilJ
(i,l.BBi lTga.BqtY/
11.4
=
l
(1.a)11s3.a2) (11.33) 11.4 L\

 1416 # mln Tqtal work
 3327# ot 55.45 kw

(c)n"=L+cc
o" 28.6 kW
(2) (23.6) = 47.2 kW l&1
+
LP'l
tr"=1+cc
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'[];
[email protected]
**=+#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 cclfil = = 0.9178
tfffi#ffi#P
;,
r, = r, r42
[*{*
767 oR
=
= (bbo)
t#.f
b50)
(rh*) (c,") (At*) = er.z
678 Btu
rh*=414l 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 lp,
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
ThreeStage Compression nrh'RT, \i/. "LP =  ln l7gt+ _ il
L\prl
]
(1.34) (e0) (53.34) (550) 1t52.effi
[email protected]\ra.si jl
'1
=

b2G5
Total work,
(0
mrn = L24.2hp
fr
IP cylinder
LP cylinder
B!t'
Fig. 24. ThreeStage 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
2P,
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"arlThreestage, No pressure Dr'p
,,1'T,r,
nm'Rro Zluf+il =,,p'or. f&\.'l = 1*L\&i If+l+ l=In[\d/ ln l\Pr/ l
11
Pr P" P,
Pn
=F, = P,
I
P, = (PrPr)
2
P, = (P,Po)
2
(1)
I
(2)
ft)T3Tr=BgbK
Solving equations (1) and (2) simultaneously,
p,=\/ir'p,
and p,
nl
T /&\  =,ruc /3ss'olff/ = 411 K 'z = Trrgf/ "no (ib3;)
=t6trJ
1n [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 threestage 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
ln 7&.* l\P,)"1.J
,., _ 3nm'RT, vY

(3) (1.32) (1) (0.28708) (305) 11.31
=

L.IZJ l/353.6\ r'3211 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 doubleacting 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 L2in., single'cylinder, doubleacting 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 out"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