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.<'rxrling water circulated around the cylindercanffis away
.l:ul:t .f/kg of air. Thc change in kinetic energ"y is 896 J&g
rr{ n.nso. Sketch an enerry diagram. Compute the work.
rr
W=t00hp
/r+Kr+
2B
Wr, + Ur +
Q=/r+ Iq + Wo + U, + W
29
Solution
P, = 99.29 kPa
v, = 0.026 m3/kg
u, = L594 J/kg
Q = -4383 Jlkg
h = 272 kg/min
Pz = 689.5 kPa
vz = 0.0051 m3/hg
uz= 6241J/kg
AK = 896 J&g
r4
wo
u2
Solution
fr =
2270 k'elmin
0.1524 m
=
= 82,740Pa
p
1000 kg/mg
q == 0.1016 m
275,800 Pa
Pz r
dr
Pr
C
1
EnergY Diagrom
y'r*Kr+
W., + U, +
Q=/r+ 4
+ Wo + U, + W
Basis 1kB-
W,,
:p lvr =
wn =pPzv
'zYz=
Area at entrance, A, =
,![I
IIP€9.29
e 'm1.l F
).026
t-
-
I 68e.'o mz
t_
F
2270k9^
'J
rt0.00
mil
,Ia005 ,0il
m
L
lI
kl\i
'l; l== 2.583 kJ&e
ol
KS
b-l
;1
H1r,r,if
-
at entrance, Dr =
= 3.51.6lnl/kg
ilgxrcd at exit, D, =
q4--4. 383 = 0.896i+333.516
;16-+ 6.2,
6.24 1+W
2.582 + 1. 594
'
llnHis 1
w
t- to.se6gJ
- j_Vzztry)
\[I = -
kr-l l- _
K, =;ik=
ke_l
4.
I
II
IL
30
m
m
P'0t824
{
=2.074m1s
2270160
= 4.667 mls
(1ooo) (o.oo81o7)
Q.orni]'
Fffi
N.m
=
2.151q;
(4.667Y
K
= (zxit= to.gg T.-DE'" =D?
ks-
2954*
A centifugal pump operating under steady flow condi'
tions delive rs 2,270 t glmin of water from an initial pressure of
82,740Patoa final p"essore of 2?5,800 Pa. The diameter of the
inlet pipe to the pump is .15.24 cm and the diameter of the
ilischaree prpe is 10.16 cm. What is the work?
U*
Pr-l= #
[oootrl
kg-
\{ = - 10.g6H
E
(0.1524F = 0.01824 mz
Area at exit, Ao =ftO.rOro)2, = 0.00810? mg
u2-+ w
+Q= AK+'wlzf2z* uz
vflr
\w.
t*1 -1 +G
i
t
w
,,
.,
l)'vr
t21o*'
-l.
=E =;;E-
82.24+,.rts
= oL''*
kgm
3l
Basis 1kg,
Pr=?= fs.eooof'(B
fm_l
2.L5L + 82.74 = 10.89.+ 275.8 + W
Kl=
W=
5. Aturbine operates under steadyflow conditions, recei
iag steqm at the following state: pnessure 1200 kPa,
tue 188"C, enthalpy 2785kJ/kg, speed 33.3 m/s and elevati
3 m. The steam leaves the turbine at the following
pressure 20 kPa, enthalpy 25L2 klkg, speed 100 m/s
elevation 0 m. Heat is lost to the surioundings at the rate of 0.
hVs. If, the rate of steam flow throughthe turbine is 0.42
what is the power output of the turbine in kW?
Solution
.4_ L33.3g:l
-o.zey
;F-
=
zz= 0m
E
IKg
ur=33'3fl
ks
P
= 0.55a4 K
s
,hI
=
s.o00E
{).6eo5H
Pr+Kr+hr+Q=%+4+ L+W
Pr+Kr+hr+Q=4+\+W
0,0!fg4 + 0.5544
+ 2785+ ({.690b) = b.000 + 2bt2 + W
W = ZG?.gg
*_
w=, l:^-
zr=3m
h. = 2?85
'
a-
4
2
m-
q=;i =1#f
kI
-4b8.1ffn
= 0.0294
.TT.F
'E-E
Kr+Wrr=Iq+\{Io+W
[-,'"ffiE*H
m)
^
hrl I- k;
roT.eEl 19.42fl
W = 112.52 kW
4=2512H
u, =
100*
'
l&I
Q = -O.29 s
32
fi = 0:4#
88
T
Review Problems
1.
fric'
Assuming that there are no heat effects and no
tionaleffects,nnatnekineticenerg]andspeedofaS220.lb
starr wfth the steady flow
;;d**; iiiar, 778 ft,from rest.which
are inelevant'
deleting energy terms
a;;til,
fPs l
. - ? ,:"?l:..
Ans. 224
The discharge conditions are 0.62 ms/kg,-100 kpa, and 270
m/s. The total heat loss between the inlet and discharge ie g
kJlkg of fluid. In flowing through this apparatus, does the
specific internal energy increase or decrease, and by how
rnuch?
' 2. A reciproc"ti"e di"pressor draws in 500 cubic feet per
ft and discharges it
mir'rte of air whose density is 0.0?9 lb/cu
*iiit au"sity of 0.304lUcu ft' At the suction' p, = LS.psia; at
"
in the specific-internal
ait"ftt"g",
Pz = 80 psia' The increase
the air by
enerm/ is gAS Btudb anrl the heat transferred from
ri et"nU. Determine the work on lhe air in Btu/min
u"a irittp. Neglect change in kinetic energy'
Ans. 56.25 hP
;
;ft
In a steady flow apparatus, L3b lc.I of work is done by
-6..
each kgof fluid. The specific volume of the fluid, p""*s.r"*, und
speed at the inlet are 0.37 mslkg, G00 kpa, and 16 m/s. The inlet
is 32 m above the floor, and the discharge pipe is at floor level.
Ans. -20.01 kJ/kg
7. Steam enters a turbine stage with an enthalpy of 862g
k.l/hg at 70 m/s and leaves the same stage with an entharpy of
:ltl46 kr&g and a velocity of L2a n/s. calculate the work done
l,y the steam.
Ans. 776.8 kJ&e (ME Board Problem - Oct. 1996)
Steem enters a turbine with an,enthalpy of 1292B,h,1|b
an enrhalpy of 1098 Btu/tb. The transferred
hp for a
heat is 13 Btu/lb. what is the work in Btrlmin and in
flow of 2 lb/sec?
Ans. 512.3 hP
3.
*dl;;;;;h
A thermodynamic steady flow system receives^4'56
n"ii where n1 1JBQ0 T?.Y'= 0'0ll-8:-]1
p"" Li"
J", aod ,r, = 17.16 k nte' The fluid leaves the sys
i.-=
ui u t"""aary wheie Pz = 551'6 kPa, v, = 0'193 m3/kg' o, =
DurFs pasiage through tbe sv
%. ="sz.eo uttfite
inu nnid receives 3,000 J/s of heat. Determine the work'
Ans. -486 kJ/min
4.
tii "i"
;;ffi
5. Air flows steadily at the rate of 0'5 kg/s through qn
compressor, entering at 7 mls speed, 100 kPa pressure
0.95 m3/kg specific volume, and leaving at 5 m/s, 700 kPa,
0.1"9 m34rg. The internal energy of the air leaving is 90
greater t[an that of the air entering. Cooling water in
io*p""rror jackets absorbs heat from the air at the rate of
kW. Compute the work in kW.
Ans. -122 kW
l
it4
lfl-r
The rdeal Gas
3
An ideal,gas is ideal ronly in the sense that it conforns rc
llrc simple perfect gas laws.
Boyle's Law
lf' the temperature of a given quantity of gas is held
,,rr'l,irnt, the volume of the gas varies inversely with the
rrl*rolute pressure during a change of state.
l or V=9
V* pp
pV=C or prV, =prYz
Charles'Law
r I r lf' thc pressure on a particular quantity of gas is held
,,,*irt;rrrl., t,hon, with any change of state, the volume will vary
rlirr.r tly :rrr lhc absolute temperature.
V,."1
or
V=CT
v (:
' or L-IL
'r'
q=q
r,:r
ll
tlrr.volurnc of a particular quantity of gas is held
r r, wi th nny change of state, the pressure
.
will vary
,f i* e' | !r' 1ri lli,' lrllsll utC te mpe ratUfe.
, r,1 1e |
;1 1,
l
| | rr.
,tt
-7
P-T
or
or
fr=c
used, the pressure was 200 psia and the temperature was 85oF,
P=CT
(a) What proportion of the acetylene was used? (b) What
volume would the used acetylene occufiy at L4.7 psia and fl0'F?
R for acetylene is 59.35 ft.lb/lb."R.
t=+,
Equation of State or Characteristic Equation
Perfect Gas
of a
Solution
(a) Let
Combining Boyle's and Charles' Iawg,
+=ry =c,aconstant
pV
T
frr
=
oz =
Be =
Pr =
Pz = 200 Psia
Tz =85oF+460=5451R
=mR
volume of
m
T
R
= absolute pressure
= volume
= specific volume
= maSS
= absolute temperature
= specific gas constant or simply gas constant
}F
ffiffi
= 0.6545 cu
RT,
T
V
English units
dr,r* =
(25cD $44)
ml= PrV, =
(59.35)
pv =RT
(unit mass)
V
v
rllass of acetylene used
250 Psia
Tr =90oF+460=550'R
pV = mRT
where p
initialiy in the drum
ltrass of acetylene left in the drum
rlrBss of acetylene
ft3
lb_
oR
m3
kg
K
R
mz =
ms
(0.6545)
(550)
ft
= 0.7218Ib
o-E= (200,)=911)!9.6j45)
ifq'=
- ml
(bgsb)
-"'""-- lb
(54b) = 0.b828
mz= 0.72L8 - 0.5828 = 0.1390Ib
Acetylene used =
#i = 3+#
= 0'1e26 or re'26vo
tlr) p, = 14.7 psia
'f.=80oF+460=540oR
i
SI units
N
;t
Vr=
m
EltTr
Pa
(b=e.Bit t5+01
\roils$l
' (r4.7)
= 2.'0b fr3
(L44\
:l
E
Problems
1. A drum 6 in. in diameter and 40 in. long
acetylene at250 psia and 90"F. After some ofthe acetylene
I
-t
3rl
'l'lrc volume of a 6 x 12-ft tank is 339.3 cu ft. It contains
psig and 85"F. How many l-cu ft drums can bc fillcd
l' rru 1rrr1.f :rnd 80'F if it is assumed that the air temperasturtt
irr llrr' lrrrrh remains at 85"F? The drurns have been silting
€*,iurrl rrr l.hu atmosphere which is at 14.7 psia anrl [t0"1"
sir rrl
'.1(X)
;t 1)
r
Solution
Solution
Let
Dr = IIlBss of air initially in the tank
Dz = rnoss of air lelt in the tank
Ds = mas$ of air initially in the dmm
ha = rnsss of air in the drum after filling
20,000 kg
l
I
P,Vr
RT,
-z=
IDo
R"S
RT,
of air displaced by
the balloon
EH" = mass of Helium
v = volume of the balloon
= "mass
+
ms
(2L4.7)(r44) (33J.3)
= _*-(SmtGaSI
= 360.9Ib.
-=
(64;3),(l*t)=(?gie'3)
(53.34)
(545t-
mass of air that can be used = 860.9
- 10g.? = 252.2Ib.
mass of air put in each drum = 0.323b
Numberof drums filled
"p=
E__
T,=21.t +273=294.iK
p-V
101.32bV
'nu=ili" = tffirl
-
0.0?gb = 0.25Ib
= 1009
3. It is planned to lift and move logs from almost inaccessible forest qery by means of balloons. Helium at atmospheric
pressure (101-.325 kPa) and temperature 21.1oC is to be used
in the balloons. What 6inims6 balloon diameter (assumo
spherical shape) will be required for a gross lifting force of 20
metric tons?
=l'2oolvkg
f','t lltt'heliUm
&r" = 2,077.67
P11"
=o'3235rb
2#
J
= 287.08
P, = 101,325 Pa
= 108.? lb
p.v. (t4.7) (r44) (1)
m3 = 'ff = 'GBiJAIGadf
= o'0735 lb
"'o= Sf =ttf#[]l*}$i
Itor the air
R
For the drums
40
mr
EH.
For the tank
[l=
I€t
I
+ 14.7 = 214.7 psia p, = 14.? psia
Tr = 85 + 460 = 545.R
T, = 80 + aOO = b40R
Pz = 50 + 14.7 = 64.7 psia
Po = 50 + 14.7 = 64.7 psia
Tr=8S+460=545oR
Tn=80+460=540R
Pr = 200
#R
= 101,325Pa
T""=21.1 +278=Zg4.lK
,,,
_
rrrrl,,=
Pn.v = _101,325 V
ffiT"" qOngZffim
=0.1658Vkg
fr,=DH,+20,000
V =0.1658V +,20,000
V = l9,BB7 mJ
.l rf = 19,337
.l
1.200f
1
r = 16.6b m
d
-
2(16.65) = 3B.B m
4l
G{
4.
TVo vessels A and B of different sizes are connected by
with a valve. Vessel A contains L42L of air at2,767.92
kPa, 93.33oC. Vessel B, of unknown volume, contains air at
68.95 kPa,4.44"C. The valve is opened and, when the prcperties have been determined, it is found that p- = 1378.96 kPa,
t- = 43.33'C. What is the volume of vessel B?
solving equations
a pipe
L and 2 simultaneously
Vs = 110.4 liters
Specifrc Heat
specific heat of a substance is defined as the quantity
_ _The
of heat required to change the temperature of unit mase
through one degree.
In dimensional form,
Solution
For vessel A
c__*
Po= 2,767.92 kPa
In differential quantities,
Yn= L4?liters
TA = 93'33
+ 273= 366'33 K
c^
e= ;ffif
For vessel B
nrr<l
for a particular masg m,
Ps = 68'95 kPa
TB
a=* !'.ar
I
=4.44+273=277.44K
(The specific heat equation)
For the mixture
ll llrr: mean or instantaneous value of specific heat is used,
P- = 1378.96 kPa
T-
= 43.33 + 273 = 316.33
Q=
K
4.36
p^V^
RTn*
(2767.s2)
V- = 1072.9
!'u,
l-
= mc (T,
- T,)
I'orrnltnt Volume Specifrc Heat (c,)
bY!
RTu
(yLD , 68.e5 VB
+ 0.25 Vu
V-=142+Vn
mc
(constant specific heat)
III,,,=IIIO*IIIU
p-v*
RT_
(13?8.e6)V
^
or dQ=mcdT
(1)
^uI
Volume
(
lorrstant
I
Q"=aU
I
Qu = mcu (T2
I
(2)
- Tr)
, ---l
a,
42
4:l
-y'r
Relation Between
Constant Pressure Specifrc Heat (co)
Qn
mco (T,
Qn
AU+W=AU+
-Tr)
al
pdv
-l\
codT = c"dT+RdT
= AU+p(%-Vr)
= Ur-ur+pz%-prV,
Q, = I{-H'=AH
Joule's law states that "the change of internal energy of an
ideal gas is a function of only the temperature change." There.
fore, AU is given by the formula,
=Eh
Htilution
AIJ = rtrc" (T2 _ Tr)
""',,
whether the volume remains constant or not.
=
*
=
#ig
,. -% = T3#
Enthalpy of an Ideal Gas
The change of enthalpy of an ideal gas is given by
formula,
-
ll,r
su.4z**" oro.aotffi
0.868#"
pV _ (75)
(1114) (rb)
-6ffi
-= ffi=
r
whether the pressure remains constant or not.
=
=
V lScuft, p=75psia T=80+460=b40o3
T1)
a
44
c"
1. For a certain ideal gas R = 2b.8 {t.lb b..R and k - f.09
(r) What are the values of co and c,? (b) What
mass of this gag
worrld occupy a volume of l5 cu ft dt ZS psia and gO"F? (c) lfgO
lll.rr are transferred to this gas at constant volume in (b), what
nrr. the resulting temperatur,e and pressure?
Internal Energy of an Ideal Gas
tIt
-c,+R
lfroblems
c
k=d:>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<g. K and c, = 0.84 kJlkg. K"
and k. (b) If 5 kg of this gas u4dergo a reversible non
flow oonstant pressure process from V, = 1.133 m3 and Pr = 690
kPa to a etate where tc = 555"C, find AU and AH.
(a) Find
co
as =
cp
;
mc hr _&
T1
(constant specific heat)
Solutlon
(a)
-lftl
= c" + R = 0.84 + 0.32 = 1.16
k= &+1=
cY
f#
+
kI
IFF
'l'emperature-Entropy Coordinates
dQ = TdS
t ='t.3st
a2
Q
(b)r-
'r
= jTds
I
pr[.
(6901909[!.133)
= 488.6 K
=
-= mR - (5) (320)
AU = rnc, (T,
-
T1) = 5 (0.84) (828
-
'The area under the curve
ofthe process on the TS plane
488.6)
represents the quantity of
heat transfered during the
= 1425.51r.I
AH = trrcn (Ts
-
process."
T1) = 5(1.16) (828
-
488.6)
= 1968.5 k I
I lt
lrr.r Enerry Relations
Entnopy (S, s)
Entropy is that property of a substance which
constant if no heat enters or leaves the substance, while it
work or alters its volume, but which increases or dimini
should a small amount of heat enter or leave.
The change of entropy of a substance receiving (or deli
ing) heatis defined by
dS=
46
F
-2
or As
=JF
I
12
-)VdP=W+AK
I
(Reversiblesteadyflow,AP= 0)
"The area behind the
curve ofthe process on the pV
planes represents the work
ofa steady flow process when
AK * 0, or it represents AK
when W' = 0."
47
-{
Any process that can be made to go in the reverse direction
by aninfinitesimal change in the conditions is called a nrersible
process.
Any process that is not reversible is irreversible.
Review Problems
1. An automobile tire is inflated to g2 psig pressurs at
60"F. Alter being driven the temperature rise to
zb"F. Determine the final gage pressure assuming the volume remaina
constant.
Ans. 84.29 psig (EE Board problem)
2. If 100 fts ofatJnospheric air at zero Fahrenheit tenperacompressed to a volume of 1 fts at a temperaiuoe or
lrlrj""
?00oF, what will be the pressure of the air in psi? Ans. 2109 psia (EE Board problem)
3. A 10-ft3 tank co-ntains gas at a pressure of b00 psia,
l.rnperature of 8b"F and a weight of 2b pounds. A part
oithe gas
w^s discharged and the temperature ind p""**"
.t
to
70"F and 300 psia, respectively. Heat was applied "og"d
and the
I.rnperature was back to 8b"F. Find the nnd weight.
volume,
nrrrl pressure of the gas.
Ans. 1b.48 lb; 10 fts;808.b psia (EE Board problem)
4. Four hundred cubic centimeters of a gas at ?40 mm Hg
alr"lut'e and 18oc undergoes a proc€ss uotit ttre pr?ssune
lp.rmes 760 mm Hg absolute andihe temperature 0"c.
what
tr l,hc final volume of the gas?
Ans. 36b cc (EE Board problem)
fi.
A motorist equips his automobile tires
with a relief-tlpe
that_the pressure inside the tire never will exceed 240
::]u,:(sage).
ll'^
He starts
1tlp wilh a pressru€ of 200 kpa (gage)
e.rrrl rr uemperature of 2B"c in the tires. During
the long drive,
lf*r l.mperature of the air in the tires reaches-g8"c. nich
tire
xrrrlrrins 0.11 kg of air. Determine (a) the mass
of air escaping
eer lr l.ire, (b)
lhe pressure of the tire when tfre tempe""t"""
uo
relrrr.rrH to 28"C.
ArrH (a) 0.006,1kS;
{i
ft)
192.48 kpa (gage)
A 6-m3 tank contains helium at 400 K and is
F,nr rrl,mospheric pressure to a pressure of 240evacuated
mm Hg
te,
urrrn. I)etermine (a) mass of helium remaining
in the tank;
kf rrrrrHs of helium pumped out, (c) tfre tempei*ui"
of tfr"
l€*r'rrrr^g helium falls to 10"C. What is the pi*u*rr"" in kpa?
f
48
49
Ans. (a) 0.01925 ke;
ft)
0.7L23 ks; (c) 1.886 kPa
.
An automobile tire contains 3730 cu in. of air at 32 psig
and 80"F. (a) What mass of air is in the tire? ft) In operation,
the air temperature increases to 145''c .If the tire is inflexible,
what is the resulting percentage increase in gage pressure?
(c) What mass of the 145"F air must be bled off to reduce the
pressure back to its original value?
Ans. (a) 0.5041 Ib; (b) 17'53Vo; (c) 0'0542lb
7
4
Processes of Ideal Gases
-
8.
A spherical balloon is 40 f,t in diameter and surrou
by zrir at 60"F and29.92in Hg abs. (a) If the balloon is filled
hydrogen at a temperature of 70"F and atmospheric pressure'
what iotal load can it lift? (b) If it contains helium instead of
hydrogen, other conditions remaining the same, what load can
itlift? (c) Helium is nearly twice as heavy as hydrogen. Does it
have half the lifting force? R for hydrogen is 766.54 and for
helium is 386.04 ft.lb/lb."R.
Ans. (a) 2381 lb; (b) 2209 lb
A reservoir contains 2.83 cu m of carbon monoxide
6895 kPa and 23.6"C. An evacuated tank is filled from I
reservoir to a pressure of 3497 kPa and a temperature
Lz.4}C,while tfe pressure in the reservoir decreases to 62
kPa and the temperature to 18.3"C. What is the volume of
tank? R for CO is 296'.92 J/kg.K".
9.
Constant Volume process
An isometric process is a reversible constant volume proc.gs- A constant volume process may be reversible or irreiersrlrle.
2T
I
T_
I
I
Pz
I
'l
Hl
Ans. 0.451 m3
F-_sz
initially at 15 psia and 2 cu ft undergoes a
to 90 psia and 0.60 cu ft, during which the enthalpy in
by 15.5 Btu; c" =2.44Btunb. R". Determine (a) AU, (b) cn,
10. A gas
Fig. 5. Isometric Process
(c) R.
Ans. (a) 11.06 Btu; (b) 3.42 Btunb.R'; (c) 762.4ft.lVlb.
11. For a certain gas, R = 0.277 kJ/kg.Kandk=
(a) What are the value of co and c,? ft) What mass of
gas would occupy a volurire 6t O.+ZS cu m at517.l'l kPa
26.7'C? (c) If 31.65 kJ are transferred to this gas at
volume in (b), what are the resulting temperature and
1'
sure?
Ans. (a) A.7214 and 0.994 kJ/kg.R"; (b> 2'M7
(;r) Relation between p and T.
Tt
Pz
;fr- =It
Pr
(b) Nonflow work.
,'2
W.=JpdV=0
(c) 43.27"C, 545.75 kPa
50
5l
For reversible nonflow, Wn = 0'
For irreversible nonflow, Wo + 0'
W = nonflow work
!d = steadY flow work
(c) The change of internal energy'
6{J = rtr'c" (T2
-
Tr)
(d) The heat transfened'
l': oblemg
(Tz - Tr)
Q = Itrc'
(e) The change of enthalPY'
6tl = mco (T2 -
(0
T1)
d) the' tralsferred
rh, (c) the change of internal energy'
ana (0 ihe change of entropy?
i,,, ,.r," .frurrg" of
wo
"oittatpy,
The change of entroPY'
lS = mc"h
l.TencuftofairatS00psiaand400.Fiscooledtol40"F
(b) the
*t <.onstant rroto*". Wnat are (a) the( final pressure, heat'
ft
Hululion
ll
I
I
volume'
(g) Reversible steady flow constant
ta)
(
2
=16+AK+AWr+W"+AP
v
W"=-(AWr+AK+AP)
W"=-AWr=V(Pr-Pr)
t z-- +=
(AP=0'AK-0)
/2
&)- -llVdP=W"+lK
-V(Pz-Pr)=W"+AK
llr)
W=0
Ir
"' = S'=
I
v(Pr-Pr)=W"+AK
,\lI=
.
v(Pr-P')=w"
166 = 0)
volume process'
(h) Ireversible nonflow constant
V
Ag#q
i0 cu ft
300 psia
V
Pr
400+ 460= 860'R
140+460=600"R
Tr
T2
= 2oe psia
l##li6?#)
=g'4?tb
mC"(Tr-Tr)
(s.4L7) (0.1?14) (600
-
860)
-420 Btu
r,tr
(,f
mc" (T,
-
Tr) = -420 Btu
Q=AU+W"
53
(e) AH =
=
=
(0
mcn
(T,
-
Tr=60+273=
Tr)
(9.417) (0.24) (600
-
(a) ,p _ T,p, gPS652
= DOI.O
'2
Pr
860)
-588 Btu
= 999 K
(b)"vv - R
377 =1b0g-J==
=
kg.K"
7.25-1k-l
$
os = -...1o
'
lr
=
333K
AU=
(e.4tz) (0.1?14)
t"
=
=
333
= -0.581H
mc, (T,
-
Tr)
(1.36) (1.508) (999
-
333)
1366 kJ
W"=Q-AU=105.5-1366
=
2. There are 1.36 kg of gas, for which R= 377 J/kg'k a
k = 1.25, that undergo a nonflow constant volume process
pr = 551.6 kPa and t, = 6OC to p, = 1655 kPa. During the proc
tlie gas is internally stirred and there are also added 105'5
of heat. Determine (a) tr, (b) the workinput and (c) the
(")
ls
-1260.5 kJ
l"
= mculn Tr = (1.36) (1.508)
q99
l" i=g
=2.2ffiY
ofentropy.
:t.
Solution
2
//
/
k =
R=
m=
1.25
of 0.2 cubiC meter. The room has an initial presstrrc ol'
lo t tt hPa and temperature of 16"c. calculate the roortt lcrrr
ll)f't4 )
1u r ;rlrrre after l0 minutes. (ME Board Problem - April
vnl11111o
377 Jlke.k
1.36 kg
Q = 105.5 kJ
Pr = 551.6 kPa
Pz
A group of 50 persons attended a secret meeting irr rr
,,u,rrr which is 12 meters wide by 10 meters long and a ce ilirrll
ill ,l rneters. The room is completely sealed off and insulrtl'r'rl
l,lirr.lr Jrerson gives off 150 kcal per hour of heat and occultit'r, rr
= L655 kPa
lit,l
rr
lion
z rl
ll/Pr
ll/
ll/r,
I l','
L
z
= 101"3 kPa
= 16 + 27:f
.
',tt{lf l(
Vg
lrlr
t-r4
c, = 0.1?14
W= (-1 hp) (h) =r(-lhp) (0.74G kWhp) (h) (8600 n/lr
= 0.1714# = 0.r7r4ffi
#.
=
Q = (50 persons) (150 kcaVperson.hour) = 7500 kcal/h
-
AU = Q - W = -850
(0.2) (50) = 350 m3
-4 =(0.28708)
. = RT,
,(191,31(l5ol
(289)
mass of air, m
1250 = (427.34> (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 <luring which AK= 0, whaf is
w"i]wuut i. axirw"s
\Vlr;rI is the work fo, u
mc. (T,
Se
[c -
a=
cn =
D.
Juao=W"rAK
I
cu
nTl
-
(r'?-rr)
Tr)
-;l
lfrl
l'-t
, the polytropic specific heat
The change of enthalpy
AH = mcp (T2
-
It
etju.&1;;;;;;il l#;;
il{t
i
J;it1
f
"o"n*-p."i"rrZ
ilution
l',
ilO psia
ffn
120
l"
,10
l'"
it4o + 460 = g00"R
psia
R=40**
+ 460 = 500"R
Tr)
The c.hange of entropy
m = 10lb
cp
= o.2b
#
n_l
ln It
"T,
AS=mc
7. Steady
l),
=T'
Tr
l),
flow polytropic
(a)Q=AP+AK+AH+\
w"=Q_AP_AK_AH
w = Q_AH
(AP=0,aK=g;
n-l
liio
:ro
tr
J_ _
-
I
g00
b00
ln6=ln1.6
I
tl
l 0.4700
rr =-1.7918
rr-
n = l.Bbo
72
'/3',
(b) c,
- cp
R = 0.25
AIJ = DCu (T2
-
-
#=
0.1986
(h) W" = JpdV = -433.3 Btu
m
Tr)
(800
= (10) (0.1986)
-
5oo)
= 595.8 Btu
AH
(c)
=
mcp (T2
=
=
(10) (0.25) (800
\,
- T1)
-
500)
2. Compress 4 kg/s of COrgas polytropically (pVr.z = C)
{ro3 pr = 103.4
!lu,-t, = 60oC to-tr- zzT.C.Assumingideal gas
tction, frld pr,
ry, e;lS (a)g.as ionflow, (b) as a stleady flow
l)rocesg where AP = 0, AK =
750 Btu
k = 5=
^9'^4 =r.25s
q 0.1e86
Solution
Pr = 103.4 kPa
AS
?= (10) (0'0541) r"ffi=
= -c" lt d,
0'2543+#
Tr = 60 +273 = 333
trr
(d)Q
=
=
mc"(Tr-Tr)
(10) (0'0541) (800
-
*#,
500)
eE+*L)-ffi
=
-433.3 Btu
(0 -JVap = nJRdV = (1'356) (-433'3) = -587'6 Btu
T, =227 +Z7B = b00K
) Nonflow
o, =
o,
L62.3 Btu
(e)Jnav-
K
fi=4\gs
[+..|
L rl
=
(10s.4)F$$]
w = ,hR %u
=
Lgo'-l
= r184.e kpa
__,4),0,1T16):900
- 33o
KJ
-631.13
c =c ll-d
" "Ll-ul
;-s
=ro.osorffi;]
(g) W" = -fVdP = -58?.6 Btu
AK
74
=
-JVap = -587"6 Btu
=
-0.2887
[]*
IT,
TIIF'
=,
7. If 10 kg/min of air are compressedisothermally from p,
kPa *{Vr.= 7.G5 ms/min to p, = 620 kpa, find tie worh,
96
ofentropy and the heat for (a) nonflow process and
a
steady
flow proce-s-s_with or = lb m/s and u,
.b)
='60 Js.
Ans. (a) -tBZ0 kJ/min, _b. gbo kJK.min;iU)_f
386.9kJ
:he change
min
8.
5
Gas Cycles
One pound of an ideal gas undergoes an isentropic
pf9c9s9^fr9m gb.B psig and a volume of 0.6 {tr
to a final volume
of 3.6 ft3. If c^ = 0.1,^2{3nd c, - 0.098 Btunb.R,
----'--' what
a.eia)
\
'!-asw *rv
(b) pr, (c) AH'and (d) W.
t'
Ans. (a) -2€.r"F; (b) 10.09 psia; (c) _21.96
(d) 16.48 Btu
9. A certain ideal gas whose R = 22g.6 J/kg.K and c- = 1.01
HAg.X expands isentropically from lbt? kFa, ie8"t t" gO
kPa. For454 glsof this gas determine, (a)W",
fljV'i.iAU
(s) AH.
Ans. (a) 21.9 kJ/s;(b) 0.0649b m'/s; (d) 80.18
kJ/s
-
10. A polytropic process ofair from lbO psia, 800.F, and 1
occurs to p, = 20 psia in accordance with pVt.g - C. Determir
ft) lU, AH and AS, (c) JpaV and JVap. 1
9) t,
-%,Compute the heat from the polytropic splcific heat and cl
by the equation Q = AU + fpdV. (e) Fina tne nonflow work
(f) the steady flow work for AK 0.
=
Ans. (a) 17.4"F, 4.71t ft3; (b) -2b.8f Btu, -86.14
0.0141Btu/"R; (c) 34.4f Btu,44.78 Btu; (d) g
Btu; (e) 34.41Btu; (0 44.?B Btu
*d
Fleat engine or thermal engine is a closed
system (no mass
.r'osses its boundaries) that exchanges
only heai
-
11. The work required to compress a gas reversibly accon
ing to p[r'ao = C is 67,790 J, if there is no flow. Detennine A
3"d Q if the gas is (a) air, (b) methane.For methane, k = 1
R = 518.45 J/kg.K, c, = 1.6lg7, co= Z.lB77 kJ/kg.K'Ans.(aiso.gi KI, -ro.esokl;ruiog.bo kJ, 4.zgkJ
-
rts surrounding and
-"rr. *itr,
""a
that operates in cyclls.
Illements of a thermodinemic heat engine
with a fluid as
I lrr. working substance:
a working substance, matter that receives
heat, rejects
.lu,rrl,I and
does work;
2. a source of heat (also called a hot body, a heat reservoir,
,r'.;ust source), from which the working zubstancei*.*iuuc
lrlrr [;
3. a heat sink (also called a receiver, a cold body, just
or
rrrrk), to which the working substance can
reject rr""i; *a
4 ' an engine, wherein the
working substa'nce
*""r.
lr. lurve work done on it.
"rr"h"
A thermodynamic cycle occurs when the
working fluid of a
rv'l.t'm experiencer, u.ly.*,ber of processes that
Jventuaily
nrlrrrn the fluid to its initial state.
Cycle lVork and Thermal Effrciency
(1.
QA
=
heat added
Qn
=
heat rejected
W-
net work
ftl
c€
@
iiiiiit;iiiiiffiii
s €g:Er:
r-{
AA
iA
o7T
crJ
EqY
FHiEsEll+igi€$s€gEi
C)(!d
rigg
g6dc!
f{an
I
I(Da'
\rll
ON!
tcna
9vv
+i .l
OrrF
o
.r
ID
igiiiilitig;l;giiil,
€3
fd
g.H
E*
c).q
o+)
arl
EE
*.e
+)
cd
+)
+!
'"li
E
s
r/q)
():i
.Fl
q)
R
.f>
q)
:i
()
q)
t{
EB
C!
tr
(|)
e
F'S
3F
Oq)
ri
E ;:*
T F.H
dHVJ
X
>4Xrr .rV
t ig
'e'5
€
FE Hiq
F?-9
x.9€;
-{j tri:.;
(E'l rd
i3
i€
gHEA
< €IP g'
dv
!t
g{,hIi
{
.ts
G
i$
tistP-v
d':
d
E
TS
o
5d
O ':i$
€) xe
(n
FF
h;
Poc
:'$
Xts
o
S!)
=R
<5
.F
c)
(.)
al
I
tr
EH
i
Utr
?
c)
()
-+t
fi
o
t
B
.+a€
I
-c)
lrY
Y>
f
F'
ililtl
'^@@
n
-+)
r'>)
a-Q
tt
-o)
()
HE:*Ei+
E tr
(.)
e;A
t^
q)
E5-E
-il
-. ll a.C);i.-
'Ji(/J
Ttr{rF
Ns
F,t
ti
r\cE
d
F{
r)p
NN r/
fi,EE$H$3
Jrq
Y;;
I
Itrorm
YbFl
*d'\o)
FVE a\
c)
\v
FFB
€
q)
.FV
+tA
he
td
-?G
\J
.o2
€r,
orR
ra Q)
.71
a
(!
r-r
+l
F-i
lrl------------)
9,9
-rH
c)
.\t/i€
l\t
-Ll
€tEl
Oq)
b0
SF
4_
T--+E
q)
p
o
I
ae
a5
C)
F3:
Ea+{
: sQ
Ic!Y
+-
EgJ
Cl
,15
!-
.F,
rl
G'
R
Eaiq)
+)
.a
ll
.s
5€gF*tE
FH
Eo
HHa
ii;
uiiE
f?BEHtbx'i
c6*'_d.Fc!o
3f;eS:tti€
FKETf;E'sSf
b; H F F ;i 1f,
f, 3 E.H"E€ E s
E-
(J
+>
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
d
Nl
tl*
H
tr
E<
fyr.
E
I
a{
>r>
tr
I
tr
Pl>
d
;;
t-l
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
,+
'l>
Ni
f1
i$ !g;
f; fl: .i;lr
Eils
+
t<
F{
g i$e
fld]
>tr
-
I
S
jE
,Ll
q?l'Hric'$
r
d
r
H
F
t.E*
g g*THE €
+
cl*
>t>
5
a
E
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
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<i
0lsv
F---=iJ
CDla
ll
ol
a
El
HIH
i]8
El+
?IE
ElE.T
.+
s
H
H
+Gt
+t
H
_
(D::
+)r
€rr
9Fhr
'9;
bo?
a=
E
a>.=v
't
9i
.sL
&FrF
cl
li d{
ot H- iic oFE<
a,dfr
o).;
;
Br d)++r
ir't
9p .e +)C
Xc|
#C
;i
H
a
CI
E.
q{i
E
qnc!
Sr
+>i oi
+i(
6l
lfi'
'e
ci
€r
+i,q(
66
. gDr
!8.
a
q){(,)
I
i€gEfi
2'da
lt ll ll ll
(
+)i
a.
xt
E
X
c)
ot
dddd
r+) ,+t+
bo:
'Gl tct
0
Frt
Itr
Cd
E
E
:
* I,#e H
E I;El'H
a
B
$c
e
I
H
r E
H.
F
E
ep
g
E
F
Fa
II ar rH H €:
:G
;
fiE
i()
E
A
:
I E q i A E E T t?<
€ E E L E 5 t sg
IlE'9oe;vo
I
H
ai
l{
a
=
a
tr
o
)q
J+a
iro
+)
Pco g
(J
x@ I
€€) ,e
qX:f, i6
trbr
E
'n
lr \c)(l) (t) a)
5G
3.q
+>
o2cd
c6o
EE
ll
+r
€;9 'i{r tv ()r F{')
a''+) ,k-q)|
t{ls li r
Ei o'
) a,(;r' \4,),^:
;,>
la \ li,i<i hoi
JL
r(D
lp )o,tt)g
()) r-'. ).Yr
)s€
i$
h
ia:
c)?b-\ (nl
tJ ,1
i."
t-r
0)
I
uir
wU) >CE
tbo
;(n
FIro aa io)di
c€
t'6
P^
AQ
ag
E
E
€)
t{
I
+
b-
E
I
€
$
iiH
""1"
.,gifi
f
'r>1; #" rd ti; ".1t
.l 5 ry
i]"llgE
I :
E i
I
;
i{
vl
.dl
l-88
I
H
3
;q:" do
f'c,ttnYrrj
t''
o E
,, t' 2 idF
3
O ggOs
l>.]>"r
HI
tl
t
"Y?x?qiq
E 3 3 t's;d *
nd
o)
Fld
:.
of dtq
t
\
v
Ft
ei a3.j
nls $ s
tl
,^I
c'd
'^
co
l<1
rt)l<{
NIF
c'OfF{
t-
gY)
il
.N
o)
!r
I
I
r*'l *
c.i
t-
o)
^ir-{
6l lv
vl
al
C\l
I
oi
;l\
i^,ilo
irlo)
ca
,
EEiEE
j ?dY S >*'.
d
ll
rIolcD
^lFi
$
lCA l.^
o
c\
Cral(O
5l$
!r l-
'^EHH
ddor
*)
_l-
Vl+
,:l-{
{i-€
+
C\
!l+
6t
H ie
lltt
,--
AtsIII
hH
H
J
Y
la
oR
)tr-gE tcd g;i do
\o oH*mlll
,huHV
tr ^tr a;t
<q)
(nq)
ed-o)9 ,-r<
.a€5
Ji5 HX
Cg 'H
ts\r
da
ed
E*9XXh.ECD994
€
-)z
Point
Isothermal exPansion "atio =
IsentroPic
t
naRT. (2) (53.34) (960)
vr = -E-
,:- VL
exPansion ratro = 1;
% =
of
EH*#
lsothermal comPression ratio
#
on 2 lb
A Carnot power cvcle operates
'ffi Ho"n b;:. n""q;l"^:l {l*j::?*ii?
:l: *,".':t'H :f
;bd n* ilu^*
l'ffi:S J'";n#J'#*ff;;il;
?qif
-
ffilT.f#H;
na
l-sso-l
= 24.57 psia
(2) (53.34) (530)
=-(24,s7) ( lll*4) = 15.72
f13
Point 4:
ratio
ratio rn is the compression
The isentropic compression
most commonlY used'
# "ttffi
[:t:^ = 11ee.7''b-d
L.aJ
mRT"
V
1.
--*
%= -Ti
Overall comPression tutio = \t
ri*i,
+=ti$ffit#,=8.b61
Pg= p,
rr. Y^Isentropic compression ratio' = 1;
Problems
-I4OOXI4;JI= = L.778 ft,3
Point 3:
v
=
=
Point 2:
Overall exPansion 'utio = h
Compression ratio =
1:
11:,
: 11 :'-'
at
vorume
[q = (1b.?2)F-ffi =
v4=
v,
(a)
= 7.849 ftg
\
(b)
i;1'":ff31
the end
^s,-,
=
mRln
t=
Q.%19
2.84e rtg
h*ffi = o"oeoz{fi
lX"-lff
?xpallsrv[ rD rvu ]'"'b' - ' isothermal process,
proce
nS durine , an
-.";^- rlt
G) lP
isothermal compression,
$"t"q: - ^r ^-aanoinn rlrrrine
(c) Qo = Tr (AS) = (960) (0.0952) = 91.43 Btu
lll,?*,$*;1il*
(d) QR
g*: iutio or"*pansion,
(h)
iil'6::?.i' 6Ji";fi:ie
ffil,3
$""#"ffi:T and
h[fi;:, ft;;rr :**
the mean effective Pressure'
(AS) =
{530) (0.0952) = - 50.46 Btu
(e) W = Qn - Qn = 91.4g -50.46 = 40.97 Btu
Solution
m= 2lb
Pr=
Tr=
400 psia
Pz=
199.7 Psia
Tr=
ft(;
- -T,
960'R
530'R
(o
o'4481 ot
" l[=4s
a^ fl'43- =
=
(8)I*oth""-al
44'8Lvo
expansion ratio =
* =ffi =,
87
,-E!vv a EF
osoo{r
6lIOti€rO
ilillllttl
ddt'
dE
i./
f;
HJ_l-d
cl
*
rn
c\r
OJ
oi
fi
o,
|r)
tl
<l
a
a
Io
oO
co
-@
eqq
ct?
Frl
il"ql lF-l
ca
-q
o?
o
tl
(a
tl
o
tl
'.
ilo
I
FIE
3lv
allt?
<{ lv
st3
lolol
'€lC'l'
-l
o
rt
€
19
-l ;
t=---r
6It=
I l*l
RIE
-lu
il'r*;
s
o
ll
,RlAl
g
Ef
15
t
E
c..i
c
.H
E
a
g
MI
to9
q
r.E
F{
sa
rr:<
Fa
lr:
<r
oq
"olQ
ll
ll
l<t
--r
^lFr
g{l!o
colY
FIA
\Itr
rril\
f.ll
.d.
>"b
3l-l
+t9{
'liil d :l\
c-lFi
('JI
rt
el\
€€
lxrl
IOA
laBl;
lHrl
IO)
lAd
lv
.E
IV
I
II
I
II
I
I
I
illlll
drlF
t:
Eg
-o
b0d
€E
gh
5 t>,'E
( Blti' 6I.i
$E
9ii
.ab
E;O
tssd
.5x a_
<EE
9.X
..EE
X'a
ronoJ
t-O<r
C- CQ rO
o
O/?
|
lF{
M
FB
.X
I|IpI,
E{
r;
rRlR
t'll
I
a
fl
*{
c,a
ll
Er
d
R
o
ts
,fa
rb
B
3
o
u)
$
:^t{
I-
il"
NlOIr
le."le''*.
'
-t,'
coA
Fl€
#
,
sv.:v
vll
tl
{.r
'II
F'lro
rO
I
cld
II
rlatA,I lJlrl
-*
,-()
r
+
lt
tl
.ol
E
q
0r
rtE
bo
rl€lv
IcE
lrrll
lr
N
d
cO
,l^lrO
1l
@la
I lOrt tll
I
tl
tl
>1+
dltr
AIA
vtyJ
6
tl
qolH
(ol
,_
I 116l @l
l-lvfl
ro
lro
rlq
-lH
lj
Hla
!i{
-l-,:
.d
:;
)!c
sE
(0
h
;HE 3
3
TB.
'
fl a
[E! E
Ei gt
fi;JE
I E'E:E
?€:3+
B*EE,
H-ti-L$*
EiiE}
g
if * H e iAE*q
Fil" dF"
,r T "" r { iH €H,FEd
'EHd i'
"o o''
d le o g
G
o
of;E-EAES8
a
d
ro
l\l
blnb0
J4lI l,l
co
co
|
HE
i-
ad{F
EXEaa
c\
IIE
.<lr
.j
tl
OJ
o)
c!
o
+
c\
.rtr
.S
t-
o
tl
E
+
I
tl
g)A
-q
o)
I
."
E
,dil,n' n.r)
smi!
FF
fr
n
u
H
TII sy I
o *_ ,q_,,dlj
3x T
s
A
cr
'
lx
m.
S
a
El$;glc; f $1 *#
31.
ffi
li,H5;'g
TBt,
f; sl$ J T
6r
.Fi
rr
lti
co
1-g99va"-"A;L_3j
;iu?l
iE" d3
oj'* sYl
Jr
!l$
1
c)
$
n
t
X
,, e-i r't"
r i
llr {!i
:-
_l € t' €
Ep:Td
ll
+r.d
g E- >' E #
gE
c
E
,i
h '#:ts
d
d i d F
F
eBia
Ri EsfiE
T*
uu;',{E
lg
I1 L*[J.ah
+l
1q
l@odFl?ii
Im'
rI
F
x r-----r
,
Er{
ftr:4l$gEiil$GJ
:GrH{
tr
ld
'.].o El*
,f
,
io r r i r t T 6.j ? ? 5 t ? ; T
cr
a
$
n
31fr
| 1
l * + 6
g
I f,, i4,F n llxg€1
I
B
_&
i}=F
ffiggig
iffiglg s
X
ro
ct)
tl
dp!
-S
-q
c-q c\r
+ti
t-E-=
c\l C-'.'
00(^o I
lt ll
ll
de.
o
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)
|
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<.1 of
mean effective Pressure.
Ans. (a) 14.05 kJ; (b) &'91kPa
2.
6
fnternal Combustion Engines
Internal combustion-engine'is a heat engine
deriving its
power from the energy liberated by
the exploJion oi" *l*trr"
of some hydrocarbon, in gur*o.r, or vaporized form,
with
atmospheric air.
Spark.Ignition (SI) or Gasoline Engine
:
Erh06l
the thermal efficiency of the carnct cycle in
3. show
-of thatisentropic
compression ratio rk is glven
the
terms
bvg=l- .
1.
L-l
rk
Two and one'halfpounds of air actuate a cyclecomposed
urith n =
of the following pro"u*t"*t polytropic compressiol Y'
known
The
3-1'
1.5; constant pressure 2-3-; constant volume
Btu' Determine (a)
au1,a *", p, = i0 p.iu, t, = 160'F, Q* = -1682
plane' in Btu;
i^
iul th;;;;k'of the cvcle'using the pV
(e)
(J) ""a
tne thermal efficiency, and
. -: -.
'-' Q^,' (ai
Arrr. (a) itzo'R,4485'R; (b) 384'4'Btu; (c) 2067 Btu;
4.
t-
p-'
-
Infoh ttrcb
Comprarrlcn Strol.
Ittr.u!t lkol.
Fig. lB. Four-stroke Cycle Gasoline Engine
A cycla beginr wilh the intoke slroke or fhe pirlon move3
down the cylinder ond drows in o fuet.oir
mixlure' Next, the pisron compresse3 rhe mitture
whire rnoving up ri,. iyiiJ"r.-iiri.'i"o
or n.
comprersion ttroke. fhe spork prug ignites rhe mixrure.
Br:rning gq!es puth ,he pirton down for fho
piston rhen,o"1, ,p the cytinde-gJ", prrhrg
rhe'burneJ
ori!"rins
i".ilTrili?ii;lte
for",
*,o
(d) 18.60%; (e) 106.8 Psi
1'0et
5. Athree-process cycle of anideal gas'.forwhi*.htr=
compresisentropic
an
*aI." = 0.804 lr,yl*e.K', tl-tTlt"FibyIiPa.
A cbnstant volume
t
sion 1-2 from rog.a"kpa, 27 "C 1060g. 1
3:l 11ll n= L'Zcomplete the cvcle'
p"".*t Z-S and a
(a) Qa, ft) W'
Circulation ir rtiuiv raL of o.go5 kg/s, compute
"
(c) e, and (d) p-.
Ans. (a) 41.4 k'ys; &) - 10 kJ/s; @\ 24'157o; (d) 19'81
kPa
92
The four-stmkg cycre is one wherein four strokes
of the
piston, two revolutions, are required to complet"
u.y.l".'
*-ftti*t
9:i
lrr
where
s
(a) Point
= p€r cent clearance
% = clearance volume
Vn = dsplacement volume
v,
=
t:
s"fV
Cold-air standard, k = 1.4
Hot-air standard, k < 1.4
$
*rt
p, = prLfrJ
= P, (r*)h = (13) (5.5;r.e = 119.2 psia
The thermal elficiency of the theoretical Otto cycle is
Tr=Tt
Increased by increase in r*
Increased by increase in k
Independent of the heat added
l.l
el
=
\
(r*)h-r
= (590) (b.b;r.s-r = ggB.9.R
tz = 523.9,F
The average family car has a compression ratio of about 9:1.
The economical life of the average car is 8 years or 80,000
miles of motoring.
Problems
1. An Otto cycle operates on 0.1 lb/s of air from 13 psia and
13trF at the beginning of compression. The temperasture at
the end of combustion is 5000oR; compression ratio is 5.5; hotair standard, k = 1..3. (a) Find V' p2, t s, ps, V3, tn, and pr. (b-)
Compute Qn, Qj,'W, e, and the corresponding hp.
Solution
m=
^k
k=
Pr=
Tr=
Ts=
96
= 1.oar
Point 2:
Ideal standard of comparison
1.
2.
3.
(0.1x€-.94)l_5eo)
0.1 lb/s
o.o
1.3
13 psia
130 + 460 =
5000"R
li
l'6=81
v-z-t
= 5.8 = o.Bob6 &i
=
s
Point B:
%=%=0.3056ts
Point 4:
l-ti : r'r
r. = 4Li-J
=(boo)m"'
tr = 2538"tr'
o, =
t
[+J=
(2ee8)H=
66.r psia
= 2998"R
Btu
R =53.34 ==0.22t(h)c=
v'o'c'o l6.R"
\u'f cv =
L11 (zzgfitm
Qo = rhc"
Qn =
(T,
-
Tr) = (0.1) (0.2285) (5000
-
c'=
s
Qn =
-
Tn) = (0.1) (0.2285) (590
-
Q*
=
91.77
=f= tdi%to =,,
(a) Point 2:
s
-
"*
2998)
-55'03 Btu
W = Qo
-
55.03
o =W =3!'75=0.4005
;
36.75
ry
v, 0.0!
'' "' =T=
#
W'=
Pz =
BtuX60+)
'smrn
n'*t#ftnr
= o'003455 m3
T, = Tr"*t't = (805)
ot4A.O1Vo
tl
(36.?5
=o'8444*k
-=*+ =ffi=o'o43e6lce
983.9)
sr.zz ntrt
Qp = rhcu (T,
E*=m
Pr{
{ll;t't-t
= (101.8)
(tt;
= 6g9 K
t'e
= 2blg lipa
Point 3:
=52hp
Q^ = mc" (T,
2. The conditions at the beginning of compression in an
Otto engine operating on hot-air standard with k ='1.34, are
101.3 kPa,0.038 m3 and lz'C.The clearanceisL0%oand 12.6hI
are added per cycle. Determine (a) V' T*P* T3, Ps, Tn atd p.'
(b) W, (c) e, and (d) p-.
-
Tr)
12.6 = (0.04396) (O.UU)(TB
Tg
-
689)
= 1028 X
Ps =
Solution
r,ltJ=
(2518)
t8rfl
= BZbzkpa
Point 4:
t =t{W"'=r&l'],r*r{*J
P, = 101.3 kPa
V, = 0'038 mg
Ti=32"C +273 =306
n,
=n,ffi:r,91]ruzuaftl'
1.t4.1
=455K
= 16l kPa
(0'8444) (305
(b) Qn = mc" (T1- T1) = (0'04396)
-
455)
Q* = -5'57 kJ
W = Qn
,\
(c)
(d)
*
Qn
kJ
= L2'6-5'5? = ?'03
- W - 7.99-= 0.558 or 55.87o
e=q=
12Sp.
12.6
=#" = #T,= o55s -
oso3455
(b)
(")
= 364.7 kPa
Fig. 16. Air-standard Diesel CYcle
1-2: isentropic comPression
2-3: constant-pressure addition of heat
3-4: isentropic expansion
4-1: constant-volume rejection of heat
or Diesel Engine
Compression-Ignition
Analysis of the Diesel CYcle
Qn = mcn (Ts
-
T2)
I -."
-
Tn)
Q*
ln|!l.
Sl.ok.
ComF'trlon
ComF.trlon
Sftok'
?ow'r Stlol'
Crh!urt Sitol'
Diesel Engine
Fig. 15. Four-stroke Cycle
piston moves
intake stroke when the
the
with
begins
cycle
A
iil;t;*:: j*:t:-::f".1fit:11
and draws'"t1':-:ini'".-""ussion
downanddraws"ilffi
down
stroke' the tem''
when o' is
:H3:j!rye?tio"'
Htr
1l
ll:
n*J,ffi";
*iift ttt" hot air and
it
-i*"t
iniected into the
"tU"a"1
'U'"
rra$tru
-"
Prvuuvvburnsexplosivelv'e;'";;;;'"'*:Jg1*;if
burns explosrvery'
During the exhaust
do*o ror the Power strt,k". the burned gases
forces
*t*k", the piston #"; ;;Jt; ""d
out of the cYlinder'
ffi;"tfit"oo
W = Qe -
(T,
-
QR = mcn
-DC, Tn - Tr)
(T,
-T,
)
-DC"
"=frW
e=
.1- T.-T
Fd:fJ
(T1
-
Tr)
(4)
€=1-
::f,1f'l
where
"*
""
=F
=
+,
the comPression ratio
the cutoffratio
l0l
Point 3 is called the cutoffPoint.
Derivation of the fornula for
e
efficiency ofthe Diesel cycle differs from that of.th* ( )r,r.r,
-The
cycle by the bracketed factor".o'1 . This factor i*iit*,,vu
trFT
greater than 1, because r" is always greater
than l. Thus, lirr rr
particularcompression ratio rn, the otto cycle
is more efficiont.
However, since the Diesel eigirr" compresses
air only, thr,
compression ratio is higher than in an otto engine.
An actual
Diesel engine with a compression ratio of lb is
mo"e efficierrt
than an actual otto engine with a compression ratio of
9.
Process 1-2:
'- *k-l
T"=Lv^,l
lv,
q
k-l
T, = Tr"*
I
(5)
Relation among rLr r.r and r" (expansion ratio)
Process 2-3:
t-
e
ft={;
=f"
rk-
Ts = Trrrk'tr.
(6)
L% -L
-%
t =[+][q
' \=f"f"
Process 3-4:
Problems
t=F;-'=m-'=*'
Tn=Trrnk-l
Tr = Trr"k
H
(7)
1'
A Diesel cycie operates with a compression ratio
of l3.b
with a outoffoccuring at 6vo of the stroke. state 1 is defined
!f ta psia and 14OF. Foithe hot-air standard with t< = f .ga ana
for an initial I cu ft, comp-ute (a) tz, p2,,.Uz,tsn
%, po, ,rrl-tn, {b)
Q*, (c) w, (d) g uttd p-. (e) For aratlof"ciic,riauon irrooo.r-,
compute the horsepower.
and"
Solution
Substituting equations (5), (6), and (7) in equation (4)'
T.t"ni.-e=1-m\f-'r--ffii)
'.,'4^
rn = 13.5
L
= 1.84
p, = 14 Psia
Tr=140+460=600'R
y, =lcuft
. 1 f-t"*-rl
e=r-,r-rlq:11l
r02
Io;l
i
c,
53.34
= (078) (1.34 1)
-
R
=FIf
cn =
.34) (0'2016)
kc" =
(1
p,V,
_ (14)
* = alf
=
=OrOtUffi
DCo (T3
Qe = 18.57
0'2702
ffi"
Qn = 8.52
(144]jp
= (b&lr+,1 (buu) = o.68o rb
-
-
v,
1
Btu
(c) W= QA- Qn = 18.57 -8.52= 10.05 Btu
= 0'0741 ft3
T, = Tr#-1 = (600)
- iaga)
T.) = (0.063) (0.2016) (600 -72i:l1r)
(d) e = W = f0.05 = 0.54L2 or
=#x =1fS
(13.5)1 31-t = 1454oR
tz = 994oF
Point 3:
54.L2Eo
a^ 18.57
P- = (10.05) (778) = 58.64 psi
(l
(e)
pz = prrr.k = (14) (13.5I'34 = 457.9 psia
w_
-.0:,0741) (144)
[""ir*f
fo*
42.4 lltu
ft''l
nin-l'= 287
hp
min.hp
2.
(Vl -V2)
% = V, + 0:06VD = % + 0.06
There are supplied 317 kJ/cycle to an ideal Diesel
engine operating on227 g air: p, = 9?.91 kPa, t, = 48.9oC. At the
end ofcompression, pz = 3930 kPa. Deteruineia) ro, (b) c, (c) r",
ftc
(1
% = 0.0741 + (0.06) - 0.0?41) = 0.1'297
(d) W, (e) e, and (f)
0.L297
- r\il = G454) i,^g?A
r, = Trl_C
Solution
= 2545"R
\------(
\\
t, = 2085'F
4
\
Point 4:
rn =
Tr) = (0.063) (0.2702) (2545
Btu
Qn = mc" (T,
(a) Point 2:
V,
(b) QA =
..
r, l_sf''
LvrI
= (2545)
lli2gfl
L1J
l)oint
o. =
r
n,lt'J
oo., IgJZgZl'''n = 29.7 psia
= (45't.e)
[-T.]
m = 0.227 kg.
P, = 97.91 kPa
Tr = 48.9 + 273 = 321.g K
Pz = 3930 kPa
Qo= gf7 kJ/cycle
I
'''n-' = 12?1"R
tr = 811oF
r-v-r
p-.
1:
v --r
'l-
mRT.
ll
.:l
* (0.227) (0.28708) (32r.e) = 0.2143 mg
97.9r
10s
ryPoint 2:
(d) QB - &c, (T,
1
Qn = -136.9
1.1
u, =
urffl
= (0-2143)
0.0153 m3
ffi
W = Qo
(e) e =
Tr=T, lo;l+'=(821.e)
IJil
Hfl1f
= sz4lK
Point 3:
kI
fg6.g = lg0.l kJ
P= lao.t = 0.b6g1 or 56.glvo
QA 317
1fl P- =g=
l0o.l _=
vD =.w
vr_%=o-zr+s:00rog
9ob kpa
- 924.4)
In modern compression ignition
engines
constant during the.combristio" p"o"ess the pressure is not
manners illustrated in the ng"*.-ili;*J but varies in the
((
i= ).olb3)
L-2)
lW1=
o.oB8B mg
P24A
combustion can be conside*dt";il;ach
a constant-vorume
process, and the late burning,
u *;rilunt-pressure process.
Point 4:
,,=*b{'=(rrrr) B*?H"
(a)
= 1161k
-'
-V--o.oi^re
"'* =vr=0.2143_14
(b) f,=-*c
Fig. tZ. Air_Standard Dual Cycle
1+c
1+c
1r
I4t
=-
c
(c)
106
c
= 0.0769 or 7.69Vo
0.0383 t
f-c= v^
-!iL =--:-::= - 2.50
v,
0.0153
)
il ffi* ol" *
T, = 2312I(
Im|
vr,if
-tt6t
DuaI Combustion Engine
3r7 = Q.227) (1.0062) (T3
=
Tr) = (0.227)(0.2186) (B zt.g
- QR = 317 -
Qn = mco (Ts - T2)
v,
-
l-2: isentropic compression
2-B: constant_volume addition
of heat
3-4: constant-pressure addition
of heat
4-b: isentroplc expansion
5-1: constant-volume rejection
of heat
Analysis of Dual Combustion Cycle
Qo = mc, (T,
-
Tr) + mcp (T. _
fr)
Q* = me, (T1
W = Qe
g='W=
-
T6) =
-mc" (Tr
-
- Qn = mc" (\ - Tr) + mco (T1 - Ts) - DC" (T6 - Tr)
mc" (T,
\QA
- Tr) + mc, (T, - Tr) - mc, ('t'o
mc, (T, - Tr) + mco (Ta - T, )
e=l-
where
Procesg B-4:
Tr)
-
\/v t
^g
t
il= f,=""
T,)
Tn =
g
Trrr
t'lr;{"
, (lt)
(8).
=S,' the pressure ratio during the consant volume
poii"" of co-U"stio"
v the
rr
ratio
""o P,
compression
=titr,
,2
r.'
tn
4a
\r
=#,
Y3
the cutoffratio
thernal,efficiency of this cycle lies between that of the
ideal Otto qnd the ideal Diesel.
Th'b
Derivation of the formula for e
Proccss 1-2:
T"
Tu =
Trr*'t-l ror.
Tu=
Tpor"r
(r2)
too"otuting equatirins (9), (10), (11),
and (12) in equation
or.
l-T
€=l-
*L
o=lProblems
the *tllpg
.bustion
L. At
cvcle, the
-lv,l-k-1
d:op-p."*rsion in an ideal dual
com-
n"ia-ir i ru
"irri"i-iijT#"
at the end of ""a
:.ilI i il"- p"*rru*
tlre
w.orki"ng
q=LrJ /
The compre*io.l
99:F..
constant
volume addrtion or n*ullrito
added 100 Btu
th;,;il;;ilpor*,ro expansion. Find
(a) ro, (b) r", (c) the percentage
cfearence, (d) e, and 1e) p_.
uA*
T" = Trr*I'r
Process 2-3:
;;i""#;;#;
"*
t=#="
T, = Trrrk-t
rn
(10)
r0g
*'
Point 5:
Solution
m = llbair
p., = 14.1 psia
T, = 80+460=540oR
pa = 470 psia
rk= 9
Qr-n
t,
= 100 Btu
r" =t=
v,
!g!tg = L.Zr
1.576
(c)r.-1+c
*c
u,=-3l'-=%#ffi#=la186rt3
9=1+c
c
Point 2:
v.
14.186
%=t=-t-=
rir-'l
Tr= T,
l+
I
L'rJ
l-v,l*
= 0.125 or ]'Z.EVo
(d) QA =
1.576ft3
Q-, + Qr.n = (m)
= (540) (9) ''n-' = 1300R
= (14.1) (9)
1'4
Qn = (mXc"XT,
w
P*=V,-%
Tr=T, [pJ
-
Tr) + 1oo
-
- lB00) + 100 = 219.8 Btu
Tu) = (1X0.1714X840- 1082)
=
-92.9 Btu
^ W 219.8-e2q
" =Q;= --fts-=:: = o'5773 0r 57 '73Vo
= 305.6 psia
Point 3:
= ffi (126.e) (778\
=54.3?psi
2. An ideal dual c'ombustion cycre operates on 4b4 g of air.
At the beginning ofcomp_ression, the airis at g6.b3 p",?g.g"c.
t
Itet ro - 1.5,,r..= 1.!-0, an{ r* 11. Determine (a)
the percentage
=
LF;J
Point 4:
(T.
Qr-n = (m) (co)
(e") (T,
= (1) (0.1?14) (1999
k-l
l, = n,l_if
-
100 = (1) (0.24) (T4
('lea.rance, (b) p,
(d) s, an6 (e) p-.
Tr)
-
Tn = 24J.6"R
= v,R] = o.b?o)
f+f
V, and T at each corner of the cycle, tc) e-n,
Solution
1999)
I'
il0
= 1082"R
+!y = L.54
=g=
Pz 305.6
1:
v.
t l+ln.'= (rnru)
L_'I-J
E&1"
(a) r^
P
(b)
Point
=
= 1.905 ftg
'f-\.
t,\:
,-/i
4
A'
'/ -""
,2'
m = 0.454kgof air
P, = 96.53 kPa
T, = 43.3 + 273 = 816.3 K
rp = l'5
r" = 1'60
rr = ll
ill
W = Qr
(a)-rk--1+c
w
c
11
"=6o=
1+c
=-;
g = 0'10
or
(0.28?08) (316'3)
vt- o.42t]*,
vr -=T;-= --11
l-v-lr'-r
,, = t,FJ*-'=
I-vlF
ft'1
278.3
474
= 0.5871
or 58.7lVo
278.3
= 716.8 kPa
o.427L - 0.03883
= e.427r ms
= o.oB88B m3
8254K
T, ("n) *-'= (316'3) (11)'n-' =
(96.b3) (11) ''n =2770'81.Pa
= pr(roy =
ps = (Pz) ("n) = (2??0'8)
(t'5) = 4156'2 kPa
,, = r,fog
ffi
Vn = (Vr)
Q* = 474-L95.7 = 278'3 kJ
w
(e,p_=Vr5,=
IUVo
mRT, (0.454)
(b)Vr=-p;=re
p, = n,
-
= (82b.4)
=
K
'288.1
(r.) = (0'03883) (t'60) = 0'06213 m3'
l-ri-l
K
rn = t'L+l= (1238'1) (1"6) = Le81
- I-vln',
,, = r.LirJ
pu =
= (1e81)
= e16.2
K
l-m-l (e6.53) e1g.?
=27s .6 kpa
p,l+l=
'L'
d
316'3
(c) Qe - (m) (c") (T,
=
Bm''n-'
-
Tr)
(0.454X0.?186X1238'1
-
* (m) (cn) (T4 825'4)
+
T3)
(0'454X1'0062X1981-1238'
l)
= 474kJ
(d) QR = (m)(c"XT,
-Tu)
= (0'454X0'?186X316'3
- 916'2) = 195'?
I t:l
-l
Review Problems
the hot'air standard
An ideal Otto engine, operating on
5'
ratiJof At the beginning of
with k = 1.34, h^t
and
uor"-"is 6 cu ft' the pressure is 13'?5 psiaheatconstant'volume
the temperature i. fOO"f' Ouring the (u) (b) T" (c) p" (d) e'
c'
ritta
cvcle'
1.
;;;;;;tfi
;;;;t;;;irt"
uaJJp"t
t"g, il;'Bl"
^t"
and (e) p-.
Ans,
7 ""s Compressors
(a) 257o; (b) 5209"R; (c) 639'4 psia; (d) 42'14Vo;
(e) 161.2 Psi
operates
2. An ideal Otto cycle engine 'lrrtlnll%o clearance
The
i"Lx" !tut". is 100'58 kPa' 37'7oC'
on 0.227 kg/s of
is 110 kJ/s' For hot-air
energy released d;l;;;*bustion
(a) p' V' and T at each corner'
standard with k = isi,-"o-pute
(b) W, (c) e, and (d) P-'
t"1t':f:
*'r., o
kPa;
;6. + x, zazo.t r<P a, 5s2,1K 19 1'71
kPa
(b) 52'7 kJis; (c) 47 '9LVo;(d) 301'1
"ii
a";.*Ai;.idig
i.pili
029?qm'hl:9:*
Operation of Compressor
Discharge
Di5charge
Valve
Intake
Valye
is from 14'7 psia'
In an ideal Diesel engine compression
Btu/cvcle are added as heat'
80"F, 1.43 cu ft to 5d0;tt* i"hi" tu
and find (a) T" V2' T3'
Make computatio,', f* cold-air standard
w;i;;""Jp-' and (d) the hp for 300 cvcles/
3.
v3, Ta, and pn, ft)
mrn.
ft3' 2113:l' 0'1&6 ft3' 890'I
Ans. (a) t4?9"R,0'1152
psi; (d) 68'
zi.ipui^;(Ujg'Z gt"; (e) 60'637o' 39'9
Compressbn
l rtruenlionol Diogrom without Clearance.
Conuenttonal Diagram witn
Fig.
hp
overall value of k =
For an ideal Diesel eycle with the
gi'g kPa' find P2 and p,"'
1.33.' r,- = 15, r. =2.l,Pr=
^Anr. 35-89 kPa, 602 kPa
4.
is pJ = 1 atm
State 1 for a dual combustion engine
t, Joo.g;Cfrn = 18; a! th9
i' zogr kPa'-r" = 1'5' tsase on l kg/
standard with k = 1-31,.deiermine 1")!l-^P:1
;;;;i;;.""ce, (b) p, v, andr at each corner point on the
(c) W, (d) e, and (e) P-'
ilJ.*-a);.EEq";&) 0.e443 m, Q'!szjo^3i *9q
;;4.; n, i ilio.zK, 0.0?869 Ti' ?^19e;3.*
(e) 900
f.p"pZO.g K; (c) 803.5 kJ; (d) 57'a3%;
5.
;t;J,"o;;J;ilp*til" "i*{*::"Y?L::t:*",?fr
;ilil;i-;r
114
18.
v
Clearance.
Fig. t9
Figure 18 shows a conventional indicator card for
a compressor without clearance. As the piston
starts the stroke 4-r,
the inlet valve opens and gas is drawn into
the cylinder arong
[he line 4.'1. A-t point 1, th; piston starts
ttr"
ui"nr.", u,l
va ves being closed, and the gas is compressed"e1,r*
along the curve
t-2. Atz,the discharge valve opens und th";;pGfigas
is
<lclivered to the receiver.
The events of the d"iagr"m with clearance are
the same as
with no clearance, except that since trre piston J* ,rot
lirrce.all the gas from the cylirrdu" at the pr"rrrrr"-o.,
tfr*
rcmsifilg gas must re-expand to the intake p"urr".*,
irL*r,
it 4, before intake starts again. without
clearance, th* ioi r-o
l'lrose
Il5
r^, p,V,,
Preferred Compression Curves
"'=f;4=
The work necessary to
drive the compresor decreases
as the value of n decreases.
Polytropic compression and
values of n less than k are
brought about by circulating
cooling water.
f-
w- E#
=
r-r
"'-t
I rp,t-
I
T
t.67-
T_
'630y1 rsz
ffi105,
=
I
- 1652 kJ/min
Another solution:
k-l
The heat rejected during compression 1-2 is,
-
Tr)
l&l*
T2 -,r
- ^t lP,
I-'J
w
= -AH =
Problems
=
A rotary compressor receives 6 m3/min. of a gas (R = 410
J/kg.K,c- = 1.03 kJ/kg.K, k = 1.67) at 105 kPa, 27"C and delivers
it at 630 LPa. Find the work if compression is (a) isentropic' (b)
polytropic with pvt'r = C, and isothermal
1.
-
Tr=
Pr=
Pz=
6 m3/min.
27 +273 = 300 K
105 kPa
630 kPa
= (300)
-fi'c e (T2 -
= 615.6
K
Tr)
(5.122) (1.03) (615.6
=+Fffi.,1*
=
vf=
I
-
300) =
-
1665 kJ/min
(b) Polytropic compression
w
Solution
(1.a) (rOs) (0)
1-1.4
I
1.4-l
f-
l1f3gl
_.1
" -11
=
-
1474
kJ/min
Another solution
Tr=T,
l-p] #
LEJ
118
-l
Llp;j
(1.67) (105) (6)
1-1.67
Heat Rejected
Qr-, = mrcr, (T,
= 5.722 kg/min
(a) Isentropic compression
Comparison of work for
Isothermal and for Isentropic
Compression.
(lob) (6)
(ozmtGoo
_ l.,l-l
= 300
iogol'n
F'ql
= 500.5 K
il9
cv
c- l'oq = 0.6168 ry
=f = L.6z
kg.k"
rh' =
off
=
cn= c [-t -rr-l= 0.6168 I r.oz - r.el = 4.41G8 kI
"Lr*l [-T:T-.a J
[sF
r,=r,
I
\if =
I
=
-afr* 6=-th'co(Tr-Tr)+ fi'co(Tz-Tr)
-(5.L22) (1.03) (600.5
-
-
L*J'=(E4o)
L#t=
=64r.eeR
=
Ll-tful =-o.oaze ffi
-
540) = bB9.B Btu/min
a -nqTr-T,)
= (22.05) (- 0.0429)
w = p,tf r"[*-l
LP?-i
-
_r.32_r
= (22.05) (0.24)(641.9
(c) Isothermal compression
=
n r*
AH = drbp (Tz Tr)
-
S00)
=-1486kl/min
=(105)(6)rn
!#}.r-ffi = 2*.o'rb/min
s c"v U-qJ
k-l -=(o.lzr4)' fILl
co
3oo)
+ (5.L22)(-0.4163) (500.6
=
o*
tffi
=
m'*
(er.g
_ 540) * 96.4 Btu/min
=
u'J
ffi
1129 kJ/min
6 =#*ar(+ali+W
A centrifugal comprcssor handles 300 crr ft per ninute
of air at t4.7 psia and 80"F. The air is compressed to 80-psia.
The initial speed is 35 fps and the final speed is 1?0 fps. If the
compressionis polytropic with n = 1.32, what is the work?
2.
w = Q-aK-aH
Solution
f;=
300
ctu
Pr = 14.7 Psia
Pz = 30 Psia
- 96.4 _ tZ.Z_ bBg.B
=
-
fl47.gBtu/min or _ lb.2g hp
Volunetric Effidiency
Conventional volumetric effciency
=
ffi
n,=$=kX
"VDVD
Tr=80+460=540R
u, = 35 fps
u, = 170 Ss
=
l,he
Displacement volume Vo
is the volume swept by the
face of
piston in one etroke.
l4r
1
The clearance ratio or per cent cleararrce, c =
then,D"=1+c-c
Free
t,
[+-]t
LP'J
Free air is air at normal atmospheric
conditions irr
particular geographical location.
r' j
twin-cylinder, double-acting compressor
with
a crear=
ance of ,vo handles 20 ms/min. of
nitiogen from roo i.i", az"c
Uo
ggrypression.ana urp""Jio" .r" p"fyt""pil
!Z!
n = 1.30. Find (a) the work, (b) the
hialre5ected, and (c) the bore
={ortN
.itf,
^H*.
and stroke for I"b0 rpm and UD f
= .gO.
where:
D
L
N
N
=
=
=
=
N=
n =
diameter of piston
length of stroke
number of cycle completed per minute
(n) (1) (number of cylinders), for
single'acting compressors
(n) (2) (number of cylinders), for
double-acting compressors
compressor speed, revolution per min., rpm
Solution
PVt's
-
V;
Pr
"
P2
Tr
=
=
=
=
20 m3/min.
100 kPa
725 Wa
37+273=Bl0K
e=\Vo
n = lbO rpm
IID = 1.39
A single-acting compressor makes one complete cycle in one
revolution.
A double-acting compressor makes two complete cycles in
one revolution.
(a) W
=T#[A*
-_l
Fie. 20. Single-acting Compressor
,
Pision rinRs
7l ,''"on.
Connecting rod
nk pin
,-- Crank
! Crankshaft
-J/
Crosshead
Wrist pin'
t
L
Crosshead guard
Y
= -5023
mrn
(b)n, =1+c*c l-pJ
l!-,J
F
Fig. 21. Double-acting Compressor
722
n
Problens
If the compression process is isentropic, let n = k.
vo
Air
l2:l
I
-
= 1 + o.ob
(o.ob) lzzq-l
p-T 3ld 381 mm, respectivery with a percentage cre'r'rrr.o
5?o, rf su'oundins air ar* it r00 kFa
zi-.c *hJio
fi
"'Llo0l
= 0.9205
n.'
20
vo=n'.,=o8Do5= = z+.ss
t,
'f
tt,,,
compression and expansion processes are pVr.s
""a _ C. Dutor,r,,,,u
(a) Freg air capacity in mtZs.
iU) power of the **pr"rro" i" f, W
(ME Board hoblem Oct. 19S6)
-
4
Solutian
P, = 100 kPa
T =293K
Vo (1 + c)
= Vo * V, = Vo + cVu =
25'60
= (24.38) (1 + 0'05) =
C=
-T
4
+.
*,=*=#Hffi=27.''*t
It
f%o
P=$SSmm
L = 381mm
n=150rpm
Pr = 97.9 kPa
Tr=300K
rn
,,
r I-Cl+ : (s'o)
\!,rvl Fz{lst =
[ool
48s.7
."ffi
=
- t,
=
o, =
l_n,
|
6r-, = rhrc" (T,
=
$.7442)Fffi#:l
-
K
4'4b'#
(a) n" = 1 +
[#J*
=
I + 0.0b-(0.0b)
Tr)
= (27.83) (-0.2456) (489.7
-
vD =-tDpLN
310)
Y
= {ZZsmrn
(c) vo
c-.
={nrlN =tD',(1.3 D) (r50) (2) (2)-
=f, {o.essft0.Bsl) (r50) = s.osz
V;= (n,) (Vo) = (0.9094) (b.6bz) = 5.r+a
612.6
O'#
o.
=
vr
m]*=0.e0e4
F,]hl
= (b r44)
t+rtffi
=
-'
#
4.els#or 0.082$
24.38= 612.6 D3
(b)w
D = 0.3414 m or 34.14 cm
L = (1.30) (34.14) = 44.38 cm
;
t
ri
2:. A single.acting air compressor operates at- 150 rpm with
initial condilion of air at 97.9 kPa and 27"c and discharges the
air at 3?9 kPa to a cylindrical tank. The bore and stroke are 355
=T#'tre,J*:,]
ll
;
=
(1.3) (97.9) (b.
1- 1.3
t26
I
=
-
KJ
or 13.34 kW
mrn
800.3
W=
"-
=
3.
A single-acting air compressor with a clearance of 6Vo
in
air at atmospheric pressure and a temperature of 85oF,
takes
and discharges it at a pressure of 85 psia. The air handled is
0.25 cu ft per cycle measured at discharge pressure. If the
compression is isentropic, frnd (a) piston displacement per
cycle, and (b) air hp of compressor if rpm is 750.
(ME Board Problem - March 1978)
-IF-to,/
_1J
'?i#iffifiea- [ia,z/
=
4' A single-acting compressor has a volumetic
effici'rt,y
500ipm. Il trk"r in air at 100 kpa nrrrl
esc\argel jr ar 600 kpa.
ai, rraodted is o .i * p,,.
l[CA!
mrn
measured at discharge condition.
If the comii#io' i,
isentropic,
of 87vo and operates at
iil
d;;pi;;;;;t
find (a) piston
(b) mean effective pressure
in kpa.
(ME Board p"otrem
=
=
%=
T, =
Pr
,Pz
Solution
14.7 Psia
85 Psia
0.25 ft,3lcYcle
85+460=545oR
Pr
fz
V2
Tr
(a)r,=r,H*
-' = #,, =
=(545)
[q*
(%###
D"=L+c-c
[r;tt
LP'-J
ra)
= o.o68z4 ib/cycte
lfi
h47l
=o,Fno
+
" q,
(h)
126
l'''
Lrooj
100 kPa
600kPa
6 ms/min
3O+273=B0BK
= 21.58 m3/min
It
24.8
o.87 =
mrn
T'
_ 24.8mrn
= 0.8499
= 1'o3o n3lcYcre
tbl V; = (0.8754) (750) = 656'6 ft3lmin
_J
looo
(6)
=
v^ =&=?rs
=o'87i4ftvcYcle
-1+0.06-(0.06) l-ar
vi
UOO
=3f;3#
=
=
=
=
= 900"R
ni RT. (0.06374X53.34X545)
v,=ff=ffi
v"
per stroke in cu m, and
:ep"riliilal
Solution
'l
96hp
,-L
0.M9G
stlgkes =
stroke
mln
w= ++lZ+r+
r-k l\p,/ -
I
t",,7
_@ffi@Kml*_!
(b) Barosetric pressure at 6000
ft = 1r.?g psia or 23.gg in rlg
New intake pressure, pr*
=
Y
= -sosa.g mln
ll.Zg psia
New discharge pressur€, pz* g0.B + ll.Zg
=
= 102.0g psia
bob3.g 208.8 kpa
n
rn=_li{_=
vD 24.9 =
New volumetric
DvN
tn
the
at
state
the
handle 500 cfin of air at
beginning of compression stroke. The compression is isentropic
efliciency,
= 1 + o.o6 -(0.06)
6. A compressor is to be designed ntith 64o clearane
L4.7 pcia and 70pF,
to 90.3 peig.
(a) What displaoement in cfu is neessary?
iU) tU" co*presso"is used at an altitude of 6000 ft and if
the initial temperature and dischargp pressure remain the
same as given in (a), by what percentage is the capacity of the
@mpressor reduced?
(c) WUat snouldbe the displacement ofacumpressor at the
altitude of 6000 ft to handle the sa-e mass of air as in (a)?
f
r
ffiff"o
New capacity, Vi* = @.7795)(6tB)
o.77e|
fr:
= 472.8mln
Percentage decreased in cqpacity 5010:j[r?.8
=
= 4.44Vo
(c) pr = 14.7 psia
R, at 6000
Vi = 500 cfu
ft = 11.78 psia
T, at 6000 ft = 530"R
Tr = 530'R
Solution
q
14.7 psia
90.3 + L4.7 = 105 psia
500 ft3/min
Tr
70+460=530"R
Pr
V, at 6000 ft = capacity to handle the same mass
of air
as in (a)
vD at 6000 ft = displacement volume to handle
the same
mass of air as in (a)
-,=#,=
Vl at 6000 ft = q+{H00) = ozs.g
1+c-"[fl*
*', lr0ilfr
=I+0.60-(0.0_.1;14.fl
y-=Yt==5=ry==
-
'o- o"
=
0.91b6
Vo at 6000
ft =
ffi=
800.4
4*
g
= 0.8156
-=orgqmin.
l2{,
Compressor EfficiencY
Adiabatic overall efficiency
is
ideal work
In general, effrciencY = actual work
^{. Mechnnical EffrciencY
The mechanical elficiency of a compressor is
- indica@
n*
If the compressor is driven by a steam or internal combus'
tion engine, the meehanical efficiency ofthe compressor system
is
"-'-
indicated work of compressor
indicated work of driving engine
B. Compression. EfEciencY
adiabAtic ideal work
,,oc%
.. =
Isothermal overdll efficiency
is
- isotherlpel ideal *"*
or%
o^,
Polyhbpic overall efficiency
is
no, =
ideal worli
(n-) (n") =
Sltpolvtmpic
Adiabatic compression effieiency is
S-
-c
-
adiabatic ideal work
indicated work of compressor
Isothermal compression efficiency is
isothermal ideal work
-'t -- indicated
work of compressor
Polytropic compression effrciency is
oolvtropic ideal work
= indicated work of compressor
"p
c. Overall Effrciency
Overall elficiency is
no = (mechanical efficiency) (compression efficiency)
130
Indicated workjs the work
done in the cylinder.
Brake work or sh"n *o"r.lr
tn"
delivered at the shaft.
Adiabatic compressio"
r,
ciencycommonryused.c;p;;i;;;ffi ,t compression effr-
i"*
"E"i"i.r
mean adiabatic compressi;
"ffi";;;
"
"tr;;y;h";;;*,wo.,td
Problems
J
1. A twocylinl":f:gl:__actils air
compressor is direcily
coupled to an electric motor
*rrrririg at 1000 rpm.
Other data are as follows:
Size of each cylinder, lbO
mm x 200 mm
Clearance
f OZ.of Jirpfacement
"?\-9,
Exponent (n)
for both comp.e5ri""
process, 1.6
Airconstant,k= t.{
Air molecular mass, 29
""J
*-expansion
Calculate:
(a) The volume rate of air delivery in terms of standard
air for a delivery pressure of 8 times ambient pressure
under ambient conditions of 300 K and 1 bar.
(b) Shaft power required if the mechanical efficiency is
81%. (ME Board Problem - April 1984)
Solution
2. A 12 x 14_in., dollle-acting air compresor
with 6.6*"
clearance operates at lS0 ,p*,
ari*ing air at l'.'pnin en'
dischargin g.it at 62' p; i;thu .91
9,u^ _":"d
n"".rion an d ex pH I r,
sron
processes are polytropic
with n = l.Bi. Determini i"l tfru
volume of free air irirnarea
if atmospheric condi.
tions are 82'F and r+.2 psia,
,r,,
indicated work of the-.o-p."rror
iitit" compression e-fficiency
is 87Vo, and (d) the ideal *ort .
pJ;i;;e,
?tiil t";;fiffi;i"l
Solution
pr = lbar=100kPa
Pz=
g
Pr
P" =
T =
Pr =
=tryLN
={to.rso)'?(0.200x2x1000) = ?.06e
#
I
tr, =
I * . -.pf
Vl=
rr"Vo = (0.?332X7.069) =
Lru
= 1 + 0.10
(b)w=T#R)*
-
0.0864
l&i
m3
-l
(1.6) (100) (0.0864)
1-1.6
S
vD =4'-D'?LN =
= r.ob5
- o.obb
m]*
= 0.8e2,4
t H' frq (1b0x2) = 274.e crm
Vf = (o,) (V;) = (0.8924) (214.g)=
248.8 cfm
(v/
(P,) (r")
(542)
9'" - --In"Jnt-= 84!€I(14.s
-liz:7t6) = 240'6 cfm
[t,*-t]
27.2L
Shaft power =
= 33.59 kW
ffi
r
(a)n"=1+c-c lP,-LI'
(0.10X8)t = 0.?332
5.183#ot
14.5 psia
Tr=85oF+460=b4b.R
o
(a) vo
14.7 psia
82"F+460=542"R
= 27.ZlkW
(b) \ir = Vn
* % = Vo + cVo = Vo(l
+ c)
= (274.9) (1 + 0.0bb) = 290.02 cfm
. P,V, = Q!.5) O44) (2s0.02)
lP
m, =
= 2o.ss
(53.34) (545)
1i1;
mln
r, = r,
[t]"
co = c"
= 545EH
F=;
= (o.tz14)
\51
= ?88"R
ftfrfl=
-
o'3025ffi
3. There are compressed g.4g kg/min of oxygen by a g!,0€
x E5. 5 6-cm, double -actin g, motor d"irre' co-p"essor
oporetlnf
at L00 rpm: These data apply: Fr = 101.9b kpa, t, Z$.ZA
inE
=
p,'310.27kPa. compression and expansion
polyt"opic wt&
n = 1.31. Determine (a) the con-uentional volumetricefliclency,
"t"
heat rejected, (c) the work; and (d)the XW inpui by tfd
9ltlt.
driving motor for an overall adiabatic elficiency of ittir.Solution
(788
= (20.83) (-0.03025)
-
545)
.,-o'' Btu
= - IDO.I ::::'
mtn
OYt'a
*
C
D'=
.br
k4{&fiq* - rl
(c) iV,"",, =
r-K
L\pr/
4)F_V:
'vo '
-J
fr'=
Pr=
Pz=
Tr=
L = 0.3556 m
8.48 kg/min
101.35 kPa
310.27 kPa
26.7
+
273
=
2gg.7
K
(1.4) (14.5) (144) (245.3) t7g) * -r
=@lrr+.rt
)
=
- 1185
BtP
mln
adiabatic
,n.i@
=H#=
(14.5) ]44)(245.3) lTsz-t'*g
=@l\r+si
Blu or - 27 -29 hP
= - 1157 mln
#
0.9227 or gl.z7vo
vo W^
"=*2.068 =
32:15 hP
o"
- -*:!.
(b) 12 = r,l+4= eee.7)
=ryreil{-']
(1.34)
v, =fD,tN =t0.Bbb6), (0.sbr6i (100) (z) = 2.068
vf=+=W=6.bu#
ideal wor!
Indicated work
(d)w
(a)
o" -27.97 hP
-
- il
."
Lrrl
=.,p3J
t!-lq€fl+#=
Beg.b
K
L101.351
= (0.6beb)
]
H$H
= -0.1808
kJ(ks) (K)
D"=l+c-c F;r+
l-F;l
135
0.9227=1+c-cl
work input by the driving motor
= 20.41 hW
I-gro.2?l#
Ll0L5il-
Multistage Compression
c = 0.0573 or 5.737o
r
Multistagingis simply the compression
more cylinders in place of a singffitinaerof the gas in two or
como"Jrro". l, iu
usedin reciprocatingcompressors
in order to(l) save power, (2)
limit the gas discharge temperaru"q
differential per cylinder. 4 ------r -
rrt3
V, = Vo (1 + c) = (?.063) (1 + 0.0573) = 7.468 -*
mrn
,- p,v,
(101.35\ tn
aRR\
'l',=ffi=idffiffi=e.717
;;?JiilililJ;;:r"""
kg
;ff
rvater
Q,-, = rhrcn (T,
=
-
Tr) = (9.717) (-0.1808) (390.5 -ZggJ)
in
water out
_159.5 I-r
^1
mln
(c) W= nth'RT,
T.n
+
l(tl -rl
IIP cyUnder
= (131) (8.48) (0.25ee) (zss.7> [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<pa.If the overail
adiabatic efficiency is 6rvo, d"t"r-i.r"
the indicated horsepower of the directly connected
driving steam engine.
Ans. 91.89 hp
14.73 ms/min of
A twin-cylinder, double-acting, compressor with a clear-
in oxygen at 450 kPa, 17"C and discharges it
at 1800 kPa. The mass flow rate is 20 kg/min, compression and
expansion are polytropic with a = 1.25. Find (a) the work, (b) the
heat transferred, and (c) the bore and stroke for 100 rpm and
llD = 1.20.
Ans. (a) -40.23 kW;(b) -829 kJ/min/ (c) 2L.71x25'76
5. From a testjf an.air compressor driven direcily
I'
7. Methane is compressed in a two-stage, double_acting
compressor which is electricaily driven
at rbb rpm. The row
pressure cylinder (3_0. E x Bb, b cm)
receive, O. S6
pe r-mirrute
of air at 96.b3 kpa,4B.B"C, *Jtfr"
""
x
!20..3 35.5 cm) discharges til" -"th* e at 7t7.06 kpa. The
isothermal overall efficiencyi szq,%-.inanu
and the kwoutput
of the raotor.
;
hish;;;iJ.r]ioa""
" Ans. 8O.02Vo,90.g6Vo
+
i,
i'i
A tw-ostage compressor with a clearance
of *Voreceives
14 psia and 8E"F and dcrivers ii
pri".
The comp,ressions
_
"i1io
1.g0,
and
inter_
cooler cools the air Td-polyt*pi;;th
" the worL, (bjthe
Uact to ar"i.. ri"Jfrl
li"
rruut
transferred in the various processes,
;;;il,f#;^.irer"_
i.ith"
gtage m achine, (d) the correspondiog percentage
s avin g for the
two-stage_machine, and (f) tle
water to be circulated
through the intercooler if its t"*p"i"l.rre
rise is 15 F".
Ans. (a)-17_1.0 hp; (b) -Soz.S Bru/min;
l.l _igie stu/
min; (d) _196 hp; (e) t2.4EVo; (ft igo lb/mi;
^^ ,9:
80lumin
of air at
-asiif
8
The Brayton Cycle
Operation of a Simple Gas T\rrbine power plant
Combustor
To
'*'/
Generator
lr?:
Compressor
Turbine
fi:-r-:i-::::i::a
1.,:".:':::Sinki.,'
r...: r:: i : :r't..i: ... .:l
F------
J
Open Cycle
Q*
Closed Cycle
Fig. 26 Diagrammatic Layout of Gas Turbine Units
Air continuously enters the compressor 1. After
compression, it enters the combustors, som'e
of it going u"o,rrra tfru
outside of the comhrrstion chamber proper
and the remainder
fulnish]1* oxygen for burning the fieljwhich ir-.orrti*orrrfv
injected into the combustioniha-ber.
Because of their temrise, the gases expand and enter the
ryTlure
3' After expansion through the turbine, the turbine in state
exhaust t. ilrt:
atmosphere is in some condition 4. In
an ordinary powor'r:r.t.
arrangement, the work of the turbine W,
is g"*i ,,,r,,,,gt, t,.,
drive the compressor W
delive.
U.ut
,,
*,rrL W,, i,,',t,.iu,,,
_and
say, a generator or proptlllrlr;
W, __ W,, I W,. Arr ,,*..il,,,,l H()r(.(,
of power is needed to si lrrt :r
liirH l.rrr.l'r'rrrr. rrrril..
Derivation of the formula for e
Process 1-2:
T =H"=FJT
T2 =
Fig.27. ,Air-standard Brayton (Joule) cycle
L-2
.
Tr"an.t
(2)
rok-t = ro Y
(3)
isentroPic conPression
2-3: constant-pressure addition of heat
3-4: isentroPicexPansion
4-tr; constant-pressure rejectionofheat
t=Fl*=
Analysis of the BraYton CYcle
Qo = mco (Tr
(Tr
- TJ
- T4) = *nrco (T4 -
Q*
-
S
= Q^ - Q* = mco'(Ts - TJ
DCo
-
Ta = Tn"*tt
Tr)
mco (Tn
- T,)
e = W = mc.(Tr-Tr)-mco(Tn-Tr)
q@
e=
+,-+
(1)
1_ rg t2
-
e-1- -11 =1-trL
=
v
+,
the comPression ratio
rppr= -P, I ttre pressure ratio
k
l-p
LF
(4).
Substituttuig equations (2) and (4)
in
(1).
€ =11
.:=
r-r
r
f
1-#I
"J
Total compressor work, W"=&
-AH
W.= -mco(T,
"o*
where rk
L-l
Total turbine work,
W,
={-AH
W, - -mc, (T. -
\)
W, = mco (T, - T,)
W--_
.,].
Net work, W or W" = W,
-
Point 3:
W"
Vg
= v2
tt.]
L
= e.7z)
Problems:
t--*
v=v
l&l
.4 ,.LpJ
ll: ""'
Pr
T1
T3
p4
r.K
= 40,000 cfm
= L5 Psia
= 550'R
= 1900'R
= 15 psia
=5
=S
v2
n
Point
1:
v1
=*,t = --T5rilx550t
v,
--=
IiI
4o.ooo
2945
=
v2
Pz
T2
I
ll4
vl
13.58
= -=
rkD= --=- =
Turbine work, W,
= Prf**-t =
= Trr*k-r =
Heat added,
(550X5)1:a 1=
-
co
=
(0.24)(1900
W,
142'8 Psia
1047"R
(T,
- W" =
-
eesoR
119.8
Btu/lb
T.)
2L6.5
=
c, (T, -
=
(0.24) (1900
= 216.b Btu/lb
-
99S)
-
119.8 = g7.2 Btu/lb
=
6751 hp
Tr)
-1047)
= +Qo= V2204.7 =
p.
=
=
Q
13.58 fta^b
(15X5)"
^
l#l'o'=
124:Ll
= W_Q945)
42.4
f
2.72fbs/,b
r
W = -{o (Tz - Tr)
2945lb/min
Point 2:
= (leoo)
= - (0.24) (L047 - 550) = -
Net work, WB
(15x144X40,000) _
\^rvrlLlbI
=@.94)ll+Z8lt"=
z4.Tfttnb
i+lL'rJ
Compressor work,
I
-
f-,, -lt -l
r.= r,
v1
4.s4 rtsnb
Point 4:
an air-standard BrayThe intake of the compressor of
and 90oF' The compression
ton cycle is 40,000;;;it;sia
turbine inletis 1440"F'
ratio, rr = 5 andth-;;;;;'i""3t11"
is 15 psia Dgterminl
The exit pressure oiiftJi""tine
pressure'
efficiencv and the mean effective
1.
;;;;,;#;al
f+#fl =
=
0.4748
2A4.7 Btu/lb
or
47.48?o
=
g = =-lu- = (97.2) (778)
vD v, - % 6+7 _ zlz) tt,t,n
=
23.89 psi
_
2' There are required 2288 kwnet from
r,rrr'rrrr* rrrrrl
lirr'prrmpi.g of crude oil from thc Nrrth Arrrrrkrrrr
'11:rs ,.rr,1*, .i*
'|'irt'(!r'$ thc compressor scction at gg"n?l-r kPr, ltzH ti, rrr* lr*ee
IFE