3. COIVTENTS
Preface vii
Chapter
1 Basic Principles, Concepts and Defrnitions I
-
Mass, Werght, Specilc Volume and Density; Spe-
cific Weight, Pressule, 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 do""sr.
5 Gas 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
4. 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
;he process involvingideal gases. The second law of thermody-
namics 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.
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
5. 1 Basic Ppq"iples, Concepts
I 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 it
and inversely proportional to its mass.o
"-
hE, F=
m
k =+F
D8,
k
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 kg mass at
I m./sz
fps system: 1 lb force accelerates 1 slug mass at l Nsz
1 cm./s2 _+
t=r,4'cm-
-cyne.s"
1m/s2
o=t#;p
1&,/sz
k=rw
l--t;]*ldyne I t -* i*l newton [T,,'*-l-'r,0"
/777r/7mrV /7furm,h n77v77v?rrvr
Systems of units where k is not unity:
6. 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 [-t u*. f-, nr'
d7mzm'V /72zv7m77 /7V7v77v77v7
32.L74 fVsz----+ 980.66 cm"/s2 -------> 9.8066 mlsz --'-+
k = rz.tllthP k = e80.66-*F k = e.80668#
Relation between kilogram force (kgr) and Newton (N)
k=1k# ks .m
k = e.8066
Ets"
Therefore, t k# = e.8066
H#
1kg"= 9.8066 N
Relation between pound psss (lb-) and slug
k=1# k= 32.r74ffi
Therefore, t*5& = 82.r74ffi
L slug = 32.L74Lb
Acceleration
A unit of force is one that produces unit acceleration in a
body of unit mass.
I poundal
I fvs2 --)
I
I
:.._l
E
r=f,a
1 poundal = (1 lb_) (1 fVs2)
F is force in poundals
# tr mass in pounds
a is acceleration in ftls2
L fVs2 --------+
1 slug = 1 lb" s2
-lr-
mFF"
k =t=g-
[T** l* ',0,
/7V7V7mV
m
.U =r-8.
l(
1 pound = (1 slug) (1 fvsz);
F is force in pounds
S is -ass in slugs
K
a is acceleration in fl;/s2
Mass and lVeight
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.
where g = acceleration produced by force F*
a = acceleration produced by another force F
AL or near the surface of the earth, k and g are numerically
,.r1rr:rl, so are m and F-
7. 1(
Problcms:
l.Whatistheweightofa66-kg-manatstandardcondi-
tion?
Solution
m=66k9- I = 9.8066 m/s2
2. The weight of an object is 50 lb. What is its mass at
standard condition?
Solution
F, = 5o lbr g= 32.L74ftlsz
So lb_
32.L74 ft
P
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.
Solu,tion
g = (30.5 fVsz) (12 in/ft) (2.54 cm/in) = 929'64 cmls2
(a) mz =
F't
=
4
r rf- lb.rrl
Fo
rb!
fztz+14s'j
[roo4frro.uuM
FK
* =d-=
e2e.64 +
= 843.91 g;
lb .ft
S'= I,l ls-P-
K s Bo.b+
F"ok Fto.lF' t*tfufl
mo=-?-= Bosg_.-
--'l- J
=l 0.4e tu.ll+se.o#-l
'L ^"J
F "f*J
? = (o ro,r"er
fz.rt- U|nu-r
rtrJ
= 222.26 g,,,
= 1435.49 g-
= 1459.41 g,"
Total mass = mr + m2 + na + m4 + m5
= 500 + 843.91 +222.26 + 1435.49 + 1459.41
= 446t.07 g^
(b) Total mass = 446L.0J g^
= g.EB lb-
453.6
ils
(t') Total mass -
9'83 ]!-o' = 0.306 slug
32.174;ifis
4. Note that the gravity acceleration at equatorial sea level
rr s = 32.088 fpsz and that its variation is - 0.003 fps2 per 1000
l't, :rscent. Find the height in miles above this point for which (a)
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 I r,,, rn an atop the 29,131-ft, Mt. Everest in Tibet, relative to this
por r r L'?
,til tr tion
(;r ) change in acceleration = 30.504 - 32.088 = * 1.584 fps2
llcight, h =
- I lP* p:; = 528,000 ft or 100 miles
- 0.003 fps'
-T0008
8. +T
(b) F
-t
I
h
I
-L
= 0.9b Fg
.a
Let Fg = weight of the man at sea level
FF-
- = ____q
ag
0.95 F" F"
a =g
a = 0.959 = (0.95) (32.088) = 30.484 fps2
'Fg
g = 32.088 fps2
t., -
(30.484 - 32'088) fps'z=
b34,6z0 ft or tOt.B miles
" - _ o.oosTS;r
-Tmorr
(c) F
a
29.1.31 ft
F8
r_.6
g = 32.088 fps2
m = 1801b- r- -1
Ito 1"1 {}l
a = 32.088 fps' -
fTdriil [0'003 fpsz] = 32'001 fpsz
tlso lb-l pz.oor&l r _^ ^^ ,,
#=179.03 lbr
32.174F"1T"
ma
o =T-=
Specifrc Volume, Density and Specifrc Weight
The density p of any substance is its mass (not weight) per
unit volume.
rl=D
rv
The specific volume v is the volume of a unit mass.
V1
lt
----
mp
The specificweightTof any substance is the force of gravity
on unit volume.
F
g= 8
,v
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;
Tk os
P='g orY ='fr
At or near the surface of the earth, k and g are numerically
cqual, so are p and y
Problems
1. What is the specific weight of,water at standard condi.
tion?
Stilution
g = 9.8066 m/sz P = 1000
kg_
n5.
[*,SE**d
*_pg -
I- E- e.8066ffi#
= looo
kgF
mo
9. ry
2. Two Iiquids of different 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.
Solution
mass of mixture, mm = pmvm = (800 kg/m3) (0'100 m3) = 80 kg
mt+m2=mm
PrVt+PrV,=D-
1500 Vr + 500 q = 80 (r)
V, + V, = 0'100 Q)
solving equations (1) and (2) simultaneously
Vt = 0'03 mg
Ve = 0'07 m3
m, = P,Vr = (1500 kg/m3) (0.03 m3) = 45kg
mr= prY2= (500 kglm3) (0.07 m3) = 35 kg
weight of mixture,
I
I
re-=x"=@ =?8.esksr
e.8066*#
Pressure
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
1. By using manometers
(a) Absolute pressure is greater than atmospheric pres-
sure.
po
p = absolute pressure
D Po = atmospheric pressure
'lt p" = gage pressure, the pres-
I ' sure due to the liquid
column h
p = Po+Pg
(b) Absolute pressure is less than atmospheric pressure
P=Po-P,
The gage reading is called
vacuum pressum or the vacuum.
I ll"y using pressure gages
A Jrrt:ssure gage is a device for
rilr,,1||llr rt ng gage pressure,
'l'lrin picture shows the
rrr,vr.rn(.1)t, in one type ofpres-
!, I r r . l::ll{(', k nown as the single-
I r r lrr. p1i r13.. 'l'hc f'luid enters the
lnlrr, llrrrrrrglr t,lrc thrcnded
,
',,rur.r'lrorr. A$ t.hc prOssur:e
I
Fig. 1 Pressure Gage
10. ry_
increases, the tube with an elliptical section tends to straighten,
the end that is nearest the linkage toward the right. The link-
age 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 gpadu-
ated dial, from which the pressure is read, and is placed under
the index hand.
(p=po+p")
,=O,P=Po)
(p=p"-pr)
(p=0,Pr=P")
Gage Pressure
P=Po+Pg
_ F" 1V yAh-
Pr=*-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
po
I
--T---
ps
+Pt
-P, V
Absolutet Pressure
Solution
FuuS ["*S
pr=*#= (30 m)
= b48,680 N/mz or b43.6g pps(gage)
Atmospheric Pressure
A barometer is used to measure atmospheric pressure.
P.=Y
Where ho = the height of column of liquid supportedby atmos-
pheric pressure {
', kg-'4
' N.sz
l)roblems
1. A vertical column of water will be supported
lrcight by standard atmospheric pressure.
to what
11. Solution
At standard condition
* = 62'4lblfts Po = 14'7 Psi
T ..-rr ;-l
lu.z *l lt++'#l
, p,,
- L----:n!-!_--!t"! = 33.9 ft
t'= t; 62.4Y
-'- ft3
Thespecificgravity(*pg')ofasubstanceistheratioofthe
spccifrc weight of the substance to that of water'
^{
sps=T
2, The pressure of a boiler 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)
Solution
Pg = 9'5 kg/cm3 ho = 768 mm Hg
At standard condition
T* = 1000 kdmt
po = (ynr) (h") = (sp gr) nr(T*)
(h")
Fooo S to.?68 m)
_
10.000 c!*
'm'
kg
cm-E
(13.6)
1.04
l2
= po * p, = 1.04 + 9.5 = 10.54#
a = I m./sz a=1fUs2
Absolute Pressure
P=Th
- yh"-* h = ho * hr, the height of column of liquid supported
by absolute pressure p.
If the liquid used in the barometer is mercury, the atmos-
pheric pressure beconoes,
P" = THshs = (sp S)H, (T*) (h")
trg.ol
Fz.+ H rL'" i',1
1728H
po = 0.491 h" l4
where ho = column of mercury in inches
then, ps = 0.491 n-
h
and, p =0.491 hP-=
ln."
l)roblems
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.
Srilution
p = 14.5 + 40 = 54.5 psia
t-t k-+'r newton [ ,
"[-ft, ,0,
/Tnvrnh /vTTvvmmiV
12. 1T-
lkgn = = 0.06853 slug
= FS][tr'fl =8.28$
F,lbf
a = 3.28 Nsz
t =
ff = (0.06863 slug)
[.za {l= o.zzas tb,
1+
1 newton = 0.2248Ib"
rl4 =
ln'
(1rb)
F**H
114= osgs
mo
1.1b" = 4.4484 newtons
lrr.ut;]
ln-
= 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
-t-
E KgJ
P.
Solution
(a)p =
Pr=
Po * Ps = 14.7 + 80 = 94.7 Psia
ao Ps]L = S.A4atmospheres
r,. t7 Psla
I':t. | --:-
af,m
h"= Z9.tilt".
-1f-
ll
th'
$.. J
lrg = 2o in.
P = 10 psia
= 4.7 psi vacuum
r o"_l
= (4.7 esi)
l:8e5;-s!
=32,407 Ps(gage)
h = 9.92 in. Hg abs
P = 0.491 h
p
=
(0.491) (9.92) = 4.87 psia
p8
ps
(rl)
h = 29.92 + 15 = 44.92 in. Hg abs
P, = 0'491 h,
=[r"H F"!F*'H
= 50,780 Pa(gage)
h =15in.
15
13. .lF
I'empcraturc
1. Derive th. r.l:rtion between degrees Fahrenheit and de-
grees Centigrndo. (FlE Board euestion)
T212.F T
100"c
tl
*uu *r".
1 ,r"" I0".
It follows that,
1Fo=1Po
and
lc.-1K"
2. Show that the specific heat ofa substance in Btu/(lb) (F")
is numerically equal to caV(g)(C").
Solution
t.F -32 _ t"C-0
212 - n
.=
lbb: o
r Btu
(lb) (r")
toF =
toC =
t"C + 32
t.F - 32)
I
o
5(
I
, 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 = t.F + 460, degtees Rankine
TK=t"C+z71,Kelvin
Degrees Fahrenheit ("F) and degrees Centigrade ("C) indi-
cate temperature reading (t). Fahrenheit degrees iFJ) and
Centigrade degress (C") indicate tempertu""
"h"ogu
or differ-
ence (At).
180 Fb = 100 C"
1p"-5g"
9
1 C. =!-1l,"
o
16
- Btu - cal
Ir-IEXD =IG'(E
. Conservation of Mass
'l'lr. law of conservation of mass states rhat mass is inde-
,tr ttr.ltltl.e.
'l'lr,r. rluantity of fluid passing through a given section is
,'r r'n t)y fne lOfmUla
V=Au
-: VAu
III = i__
v v- =Aup
Wltcrc V = volume flow rate
A = cross sectional area ofthe stream
l) :, ilvcrage Speed
rir ,., m:rss llow rutc
t7
14. F7---
Applying the law of consewation of mass'
- - - =-n;
ArDrpr = rtrPz
Problems
1. Two gaseous stre?ms enter a combining tube and leave
as a single mi*trrr". These data apply at the entrance section:
- -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'
Solution
tu'",=il'i,=ffi =4oorps
-[.'9!d=2604+
= --------r6Tt3-
ib
. Aru,
mr = --vr
18
(b)
rh, = rh, + rh, = 26.04+ 16'6? = 42'?1+
t
I
I
=Erf,El a,E4zftz I
4=ff =' *T- T
2. A 10-ft diameter by 15-ft height vertical tank is receiv- I
ing water (p = 62.1 lb/cu ft) at the rate of 300 gpm and is I
discharging through a 6-in ID line with a constant speed of 5 I
: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
rlirrur lr,,w rate enreri", =
[ffi] [rr
r
fi = z4so.
rt,r'* tuwrateleavins=Aup=
? Bd'F.uo*J F +
= ru* S*
15. Mass change = (3658 - 2490.6) (15) = 17,511 lb (decreased)
volume ch^nge = 17'51-l:-!b
= 282 ft'
62.1#
Decrcased in height = ffi# = 3'59 ft
Water level after 15 min. = 7.5 - 3'59 = 3'91 ft
20 2l
Review Problems
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,
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 accelera-
tion 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 (gravita-
l.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
,l-r. The mass of a fluid systenis 0.311 slug, its density is 30
ll,/l'1,:r and g is 31.90 fpsz. Find (a) the specific volume, (b) the
"1,,'r'ific weight, and (c) the total volume.
Ans. (a) 0.0333 ft3Ab; (b) 29.75 lb/ft3; (c) 0.3335 ft3
{;. 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
! r
"t, ir l llrxrr to tris ofl'icc atop the Sears Towcr in Chicago. On
16. 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.
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 abso-
lute, 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
10. 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
11. 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
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
22
23
Consenration of Energy
Gravitational Potential Energy (P)
The gravitational potential energ:y of a body is its energy
due to its position or elevation.
p=Fsz=ry
AP
=
P, - P, =
ff@r- zr)
AP = change in potential energy
Datum.plane
Kinetic EnergT (K)
The energy or stored capacity for performing work pos'
Hrls$ed by a moving body, by virtue of its momentum is called
kinetic energy.
K=#
nK=4-K,=fttoi-ui)
AK = change in kinetic energy
17. qT
Internal EnergY (U' u)
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'
u = speci{ic internal energy (unit mass) Au = tlz - ul
fJ = mu = total internal energy (m mass) AU = Uz - Ur
Work (W)
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
Piston
At ea = .zl
'"**F I
Cylinder ---. Final Position of Piston The work done as the
piston moves from e to f is
dW=F,d*=(pA)dL-pdv
which is the area under the
curve e-f on the pV plane.
Therefore, the total work
done as the pistonmoves from
lto2is
w =Jlndv
which is the area under the
curve 1-e-f-2.
nV
Fig. 2 woRK ot 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
Flow lVork (Wr)
Flow work or flow energry is work done in pushing a fluid
across a boundary, usually into or out of
"
uy*L-.
l"ig. 3 FIow Worh"
lVr=Fi=pAL
Wr=PV
AW,=Wr,-Wrr=pr%-FrV,
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 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
r I t A r'lrr.se d' system is one in which mass does not cross its
l,or r ntlaries.
' ' r . r | (
'r,t'n
system 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,nlyeCl-
i l,, f u:,1 l;rw ol'l.lrr:r'modynarnics states that one fornt
:::i:':. , !tttt . ltt. (..,ttIt('t.l((1. i.n.l.O U.nOthCf.
ls
oI
13orr nrll lr'_
;1=Area of Sur.face
18. SteadY Flow EnergY Equation
Characteristics of steady flow system'
- 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'
Fig. 4 Energy Diagram of a Steady Flow System
Energy Entering System = Energy Leaving System
P, + K, + Wr, + U, + Q = Pa* t-l Wl"+ U" + W
d=l"P+ak+l-wr+aU+W
(SteadY Flow Energy Equation)
EnthalPY (H, h)
Enthalpy is a composite property applicable to all fluids
and is defined bY
h=u+pv and H=mh=U+PV
The steady flow energy equation becomes
+K'+H'+Q-l;..?J*ril*
26
Problems
t. During a steady flow process, the pressure of the work-
ing 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
pr = 200 peia p, = 20 psia
o, = 200 fps rlr = 1000 fps
vr=lfts/lb vc=8 ffnb
Au=-25Btu/lb Q=0
Energy Diagtam
,F, + K, + W' + U, + A,=Pr+ 4 + W* + U, + W
llrrnis I lb-
Kl
W,,
II,
2
lr"3 ]
fi, ,lf =
Offiimi=le.e?r+b
E*'ii,lE-Hl
W,, l',v,
= o.8o
ffL
= sz,o2
Bfi
(20) (r44) (8)
= 2e.6rff
n
2
llr V.l -*
778
27
19. -T'r--
Kr+Wrr=Iq+W,r+Au+W
0.8 + 3?.02 = 19.9? + 29.61 -25 + W
w = 13.24
ff,0t,
t-
lr
L-
s24ffi["*il
2. Steam is supplied to afully loaded 100-hp turbine bt 200
priu *itft
"r = 116'bT nl"/lb,"t, ::'1U
ftsAb and u'.=^19'0 fp*'
Exhaust is at r prl" *ilrt * J ozs Btunb, Y,=-29!
ft3Ab and
"
-=
rioo fps. tne heat loss from the steam in the turbin is L0
glJu. il;;ipor""tiur enersy change and determine (a) the
*o"t p"" tU steam and (b) the steam flnw rate in lb/h'
w:
Solution
p, = 200 psia
p, - l Psia
u, = 400 fPs
= 3,12 hp
42.4(mi#)hp)
u, = L163.3 Btunb v, = 2'65 ftsnb
u" = 925 Btunb vr= 294 fts/l.b
Q = -10 Btu/lb
u, = 1100 fps
W=t00hp
2B
/r+Kr+ Wr, + Ur + Q=/r+ Iq + Wo + U, + W
(a) Basis f lb'?n'
K,=S= ,Cffio,, =3'20ff!
,q =*=
Wr, = PrVr =
(1100)2 = Z+.t7 BJu
(z',) (32.174) (778) rb-
(200) (144) (2.65)
= 98.lC #E
779 --'-- lb_
wrz= 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
Fl{
w= 251ff
r Eru-l
(roo hp) P544lrr) trro) r
---
251 Btu
E;
(b) Steam flow = = 1014
+
:t. An air compressor (an open system ) receives 272kgper
r r 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.
29
20. Solution
r4
wo
u2
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
y'r*Kr+ W., + U, + Q=/r+ 4 + Wo + U, + W
Basis 1kB-
f2
3
€9.29
t-
I 68e.
t_
+Q=
383 =
I
I
Pe
F
+G
-1
-4.
lvr =
'zYz=
t*1
q4-
:p
=p
w.
Pzv
vflr
594
1.
W,,
wn
2.582 +
,![I
'm1.l
- kli
'o mz
AK+'
' 0.896 1+W
w
6.24
'l
; l=
'J
mil
m
w
6.2,
ol
b-l
,0il
,Ia-
005
u2-
;16-
).026
r
t0.00
L
z* uz
3.516
F
lI
wlz
i+3
KS
;1
+
+
= 2.583 kJ&e
= 3.51.6lnl/kg
i
E
I
II
{ = - 10.g6H
t- kr-l l- _ ke_l
w - j_- to.se6gJ Vzztry)
[I = - 2954*
4. 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?
30
EnergY Diagrom
Solution
= 2270 k'elmin
= 0.1524 m
= 82,740Pa
= 1000 kg/mg
= 0.1016 m
r 275,800 Pa
2270160
= 4.667 mls
(1ooo) (o.oo81o7)
ilgxrcd at exit, D, =
llnHis 1 kg-
K, =;ik=
fr
dr
Pr
p
q
Pz
1
Area at entrance, A, = t (0.1524F = 0.01824 mz
Area at exit, Ao =ftO.rOro)2, = 0.00810? mg
2270k9^
H1r,r,if at entrance, Dr = U* = # =2.074m1s
- Pr-l
[oootrl P'0t824 {
m
m
Q.orni]'
Fffi
N.m
= 2.151q;
K =D? =
(4.667Y
= to.gg T.-
'" -DE- (zxit ks-
,, ., -l.
t21o*'
= 82.24+,.rts
l)'vr =E =;;E- = oL''* kgm
C
w
3l
21. Kr+Wrr=Iq+{Io+W
2.L5L + 82.74 = 10.89.+ 275.8 + W
[-,'"ffiE*H
kI
W = -4b8.1ffn
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
zr=3m
h. = 2?85 E
' IKg
ur=33'3fl
l&I
Q = -O.29 s
zz= 0m
4=2512H
u, = 100*
'
fi = 0:4#
32 88
Basis 1kg,
Pr=?=
Kl=
.4_
m-
a-
fs.eooof'(B m)
'E-E
.TT.F
= 0.0294 4
ks
2
fm_l
L33.3g:l
= 0.55a4 P
Ks
q=;i =1#f
-o.zey
;F- = {).6eo5H
Pr+Kr+hr+Q=%+4+ L+W
Pr+Kr+hr+Q=4++W
,hI
= s.o00E
0,0!fg4 + 0.5544 + 2785+ ({.690b) = b.000 + 2bt2 + W
W = ZG?.gg
*_
w= l:^- ^ hrl I- k;
,
roT.eEl 19.42fl
W = 112.52 kW
22. T
1. Assuming that there are no heat effects and no fric'
tionaleffects,nnatnekineticenerg]andspeedofaS220.lb
;;d**; iiiar, 778 ft,from rest. starr wfth the steady flow
a;;til, deleting energy terms which are inelevant'
Ans. 224 fPs l
. - ? ,:"?l:..
' 2. A reciproc"ti"e di"pressor draws in 500 cubic feet per
mir'rte of air whose density is 0.0?9 lb/cu ft and discharges it
*iiit
"
au"sity of 0.304lUcu ft' At the suction' p, = LS.psia; at
ait"ftt"g", Pz = 80 psia' The increase in the specific-internal
enerm/ is gAS Btudb anrl the heat transferred from the air by
;ft ; ri et"nU. Determine the work on lhe air in Btu/min
u"a irittp. Neglect change in kinetic energy'
Ans. 56.25 hP
3. Steem enters a turbine with an,enthalpy of 1292B,h,1|b
*dl;;;;;h an enrhalpy of 1098 Btu/tb. The transferred
Review Problems
heat is 13 Btu/lb. what is the work in Btrlmin and in hp for a
flow of 2 lb/sec?
Ans. 512.3 hP
4. A thermodynamic steady flow system receives^4'56
p"" Li" "i" n"ii where n1 1JBQ0 T?.Y'= 0'0ll-8:-]1
i.-= tii J", aod ,r, = 17.16 k nte' The fluid leaves the sys
ui u t"""aary wheie Pz = 551'6 kPa, v, = 0'193 m3/kg' o, =
;;ffi %. ="sz.eo uttfite DurFs pasiage through tbe sv
inu nnid receives 3,000 J/s of heat. Determine the work'
Ans. -486 kJ/min
5. Air flows steadily at the rate of 0'5 kg/s through qn
compressor, entering at 7 mls speed, 100 kPa pressure l
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
it4
-6..
In a steady flow apparatus, L3b lc.I of work is done by
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.
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?
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)
lfl-r
23. 3 The rdeal Gas
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.
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,1 1e | ;1 1, l . | | rr. r r, wi th nny change of state, the pressure will vary
,f i* e' | !r' 1ri lli,' lrllsll utC te mpe ratUfe.
V* l or V=9
pp
,tt
24. -7
E
I
-t
P-T or P=CT
Equation of State or Characteristic
Perfect Gas
Combining Boyle's and Charles' Iawg,
fr=c or
t=+,
+=ry =c,aconstant
pV
T =mR
pV = mRT
pv =RT
(unit mass)
where = absolute pressure
= volume
= specific volume
= maSS
= absolute temperature
= specific gas constant or simply gas constant
English units
SI units kg
Problems
1. A drum 6 in. in diameter and 40 in. long
acetylene at250 psia and 90"F. After some ofthe acetylene
3rl
}F
N
;t
p
V
v
m
T
R
ft3 lb_
R
T
oR
K
V
m3
Equation of a
ml=
(25cD $44) (0.6545)
= 0.7218Ib
(59.35) (550)
i
m EltTr
Vr=
:l 'l'lrc volume of a 6 x 12-ft tank is 339.3 cu ft. It contains
sir rrl '.1(X) psig and 85"F. How many l-cu ft drums can bc fillcd
l' rru 1rrr1.f :rnd 80'F if it is assumed that the air temperasturtt
irr llrr' lrrrrh remains at 85"F? The drurns have been silting
€*,iurrl rrr l.hu atmosphere which is at 14.7 psia anrl [t0"1"
Pa
;t 1)
used, the pressure was 200 psia and the temperature was 85oF,
(a) What proportion of the acetylene was used? (b) What
volume would the used acetylene occufiy at L4.7 psia and fl0'F?
R for acetylene is 59.35 ft.lb/lb."R.
Solution
(a) Let frr = rlrBss of acetylene initialiy in the drum
oz = ltrass of acetylene left in the drum
Be = rllass of acetylene used
Pr = 250 Psia
Tr =90oF+460=550'R
Pz = 200 Psia
Tz =85oF+460=5451R
volume of dr,r* =
ffiffi = 0.6545 cu ft
PrV,
=
RT,
o-E= (200,)=911)!9.6j45)
= 0.b828 lb
mz = ifq'= (bgsb) (54b) "'""-- --
ms - ml mz= 0.72L8 - 0.5828 = 0.1390Ib
Acetylene used = #i = 3+# = 0'1e26 or re'26vo
tlr) p, = 14.7 psia
'f.=80oF+460=540oR
roils$l (b=e.Bit t5+01
= 2.'0b fr3
' (r4.7) (L44
25. r
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
Pr = 200 + 14.7 = 214.7 psia p, = 14.? psia
Tr = 85 + 460 = 545.R T, = 80 + aOO = b40R
Pz = 50 + 14.7 = 64.7 psia
Tr=8S+460=545oR
For the tank
[l=
Po = 50 + 14.7 = 64.7 psia
Tn=80+460=540R
P,Vr
RT,
(2L4.7)(r44) (33J.3)
= 360.9Ib.
= _*-(SmtGaSI
IDo = R"S =
(64;3),(l*t)=(?gie'3)
= 108.? lb
-z RT, - (53.34) (545t-
mass of air that can be used = 860.9 - 10g.? = 252.2Ib.
For the drums
p.v. (t4.7) (r44) (1)
m3 = 'ff =
'GBiJAIGadf
= o'0735 lb
"'o= Sf =ttf#[]l*}$i =o'3235rb
mass of air put in each drum = 0.323b - 0.0?gb = 0.25Ib
Numberof drums filled
"p=
2# = 1009
3. It is planned to lift and move logs from almost inacces-
sible 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?
40
Solution
20,000 kg
l
Itor the air
= "mass of air displaced by
the balloon
= mass of Helium
= volume of the balloon
I€t mr
EH"
v
J
R = 287.08
E__
P, = 101,325 Pa
T,=21.t +273=294.iK
p-V 101.32bV
'nu=ili" = tffirl =l'2oolvkg
f','t lltt'heliUm
&r" = 2,077.67
#R
P11" = 101,325Pa
T""=21.1 +278=Zg4.lK
,,, _ Pn.v _101,325 V
rrrrl,,=
ffiT"" = qOngZffim =0.1658Vkg
fr,=DH,+20,000
1.200f V =0.1658V +,20,000
V = l9,BB7 mJ
1
.l rf = 19,337
.l
r = 16.6b m
d - 2(16.65) = 3B.B m
I
I
EH.
+
ms
4l
26. G{
4. TVo vessels A and B of different sizes are connected by
a pipe 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 prcper-
ties have been determined, it is found that p- = 1378.96 kPa,
t- = 43.33'C. What is the volume of vessel B?
Solution
For vessel A
Po= 2,767.92 kPa
Yn= L4?liters
TA = 93'33 + 273= 366'33 K
For vessel B
Ps = 68'95 kPa
TB =4.44+273=277.44K
For the mixture
P- = 1378.96 kPa
T- = 43.33 + 273 = 316.33 K
III,,,=IIIO*IIIU
p-v* p^V^
* bY!
RT_ RTn RTu
(13?8.e6)V
^ (2767.s2) (yLD , 68.e5 VB
4.36 V- = 1072.9 + 0.25 Vu
V-=142+Vn
(1)
(2)
42 4:l
solving equations L and 2 simultaneously
Vs = 110.4 liters
Specifrc Heat
_ _The specific heat of a substance is defined as the quantity
of heat required to change the temperature of unit mase
through one degree.
In dimensional form,
c__*
In differential quantities,
c^
e= ;ffif or dQ=mcdT
nrr<l for a particular masg m,
a=* !'.ar
I
(The specific heat equation)
ll llrr: mean or instantaneous value of specific heat is used,
Q = mc !'u, = mc (T, - T,)
l-
(constant specific heat)
I'orrnltnt Volume Specifrc Heat (c,)
Q"=aU
Qu = mcu (T2
- Tr)
I
^uI
Volume I
( lorrstant
I
, ---l
a,
27. -y'r
a
., t
Eiil
t
It
Constant Pressure Specifrc Heat (co)
Qn
Qn
Qn
mco (T, -Tr)
AU+W=AU+
al
pdv
-l
= AU+p(%-Vr)
= Ur-ur+pz%-prV,
Q, = I{-H'=AH
Ratio of Specific lleats
c
k=d:>r
Internal Energy of an Ideal Gas
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,
AIJ = rtrc" (T2 _ Tr)
whether the volume remains constant or not.
Enthalpy of an Ideal Gas
The change of enthalpy of an ideal gas is given by
formula,
AH = ECo (Tz - T1)
whether the pressure remains constant or not.
g
44
Relation Between cn and c,
Fromh =u+pvandpv=RT
dh = d11+ RdT
codT = c"dT+RdT
co -c,+R
c" =Eh
^ -B
'p -k-l
lfroblems
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?
Htilution
""',, =
* = #ig = su.4z**" oro.aotffi
,. -% = T3# = 0.868#"
ll,r V lScuft, p=75psia T=80+460=b40o3
pV _ (75) (1114) (rb)
-= ffi= -6ffi =11'631b
r r I tf n,c" (T, _ Tr)
'ilr I t.63 (0.3685) (T, _ 540)
4{t
28. E T
Tz = 547"R
Pz = Pr (Tuftr) = 75 (5471540) = ?6 Psia
2. For a certain gas R =320 Jll<g. K and c, = 0.84 kJlkg. K"
(a) Find co 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.
Solutlon
(a) cp = c" + R = 0.84 + 0.32 = 1.16
f# + t ='t.3st
(b)r-
=
pr[.
=
(6901909[!.133)
= 488.6 K
'r - mR - (5) (320)
AU = rnc, (T, - T1) = 5 (0.84) (828 - 488.6)
= 1425.51r.I
AH = trrcn (Ts - T1) = 5(1.16) (828 - 488.6)
= 1968.5 k I
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= F
kI
IFF
k= &+1=
cY
-2
or As =JF
I
46
where:dQ = heat transferred at the temperature T
AS = total change ofentropy
as--fu
as = -lftl ; mc hr _&
T1
(constant specific heat)
'l'emperature-Entropy Coordinates
I lt lrr.r Enerry Relations
dQ = TdS
a2
Q = jTds
I
'The area under the curve
ofthe process on the TS plane
represents the quantity of
heat transfered during the
process."
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
29. -{
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.
48
Review Problems
1. An automobile tire is inflated to g2 psig pressurs at
60"F. Alter being driven the temperature rise to zb"F. Deter-
mine 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 tenpera-
lrlrj"" compressed to a volume of 1 fts at a temperaiuoe or
?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
"og"d
to
70"F and 300 psia, respectively. Heat was applied 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
::]u,:
uo that_the pressure inside the tire never will exceed 240
ll'^ (sage). 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"""
relrrr.rrH to 28"C.
ArrH (a) 0.006,1kS; ft) 192.48 kpa (gage)
{i A 6-m3 tank contains helium at 400 K and is evacuated
F,nr rrl,mospheric pressure to a pressure of 240 mm Hg
te, urrrn. I)etermine (a) mass of helium remaining in the tank;
f 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?
49
30. Ans. (a) 0.01925 ke; ft) 0.7L23 ks; (c) 1.886 kPa
7 . 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
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
9. A reservoir contains 2.83 cu m of carbon monoxide
6895 kPa and 23.6"C. An evacuated tank is filled from I
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
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".
Ans. 0.451 m3
10. A gas 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,
(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= 1'
(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
sure?
Ans. (a) A.7214 and 0.994 kJ/kg.R"; (b> 2'M7
(c) 43.27"C, 545.75 kPa
50
4 Processes of Ideal Gases
Constant Volume process
An isometric process is a reversible constant volume proc-
.gs- A constant volume process may be reversible or irreiers-
rlrle.
2T
I
I
I
I
'l
T_
Pz
Tt Pz
;fr- =-
It Pr
Hl
F-_sz
Fig. 5. Isometric Process
(;r) Relation between p and T.
(b) Nonflow work.
,'2
W.=JpdV=0
5l
31. (c) The change of internal energy'
6{J = rtr'c" (T2 - Tr)
(d) The heat transfened'
Q = Itrc' (Tz - Tr)
(e) The change of enthalPY'
6tl = mco (T2 - T1)
(0 The change of entroPY'
lS = mc"h
ft
(g) Reversible steady flow constant volume'
ta) ( =16+AK+AWr+W"+AP
W"=-(AWr+AK+AP)
W"=-AWr=V(Pr-Pr)
(AP=0'AK-0)
/2
&)- lVdP=W"+lK
-l
-V(Pz-Pr)=W"+AK
v(Pr-Pr)=W"+AK
v(Pr-P')=w"
166 = 0)
(h) Ireversible nonflow constant volume process'
Q=AU+W"
53
For reversible nonflow, Wn = 0'
For irreversible nonflow, Wo + 0'
W = nonflow work
!d = steadY flow work
l': oblemg
l.TencuftofairatS00psiaand400.Fiscooledtol40"F
*t <.onstant rroto*". Wnat are (a) the final pressure, (b) the
w o rh, (c) the change of internal energy' ( d) the' tralsferred heat'
i,,, ,.r," .frurrg" of
"oittatpy,
ana (0 ihe change of entropy?
Hululion
ll
V
Pr
Tr
T2
i0 cu ft
300 psia
400+ 460= 860'R
140+460=600"R
V
I
I
2
v
llr)
Ir I
t z--
+=
Ag#q = 2oe psia
W=0
"' = S'= l##li6?#) =g'4?tb
,lI= mC"(Tr-Tr)
. (s.4L7) (0.1?14) (600 - 860)
-420 Btu
r,tr (,f mc" (T, - Tr) = -420 Btu
32. (e) AH = mcn (T, - Tr)
= (9.417) (0.24) (600 - 860)
= -588 Btu
(0
os = -...1o $
' lr
= (e.4tz) (0.1?14) t" 333
= -0.581H
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
Solution
k = 1.25
R = 377 Jlke.k
m = 1.36 kg
Q = 105.5 kJ
Pr = 551.6 kPa
Pz = L655 kPa
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
ofentropy.
2
/
/
/
t-r4
lrlr
Tr=60+273= 333K
(a) ,p _ T,p,
=
gPS652 = 999 K
'2 Pr DOI.O
(b)" - R = 377 =1b0g-J==
vv - k-l - 7.25-1- kg.K"
AU= mc, (T, - Tr)
= (1.36) (1.508) (999 - 333)
= 1366 kJ
W"=Q-AU=105.5-1366
= -1260.5 kJ
(")
ls = mculn
q99
= (1.36) (1.508) l" i=g
l"
Tr
=2.2ffiY
:t. 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
vnl11111o 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
1u r ;rlrrre after l0 minutes. (ME Board Problem - April ll)f't4 )
lit,l rr lion
= 101"3 kPa
= 16 + 27:f . ',tt{lf l(
z rl z
ll/Pr
ll/
ll/r,
I l','
L
Vg
33. c, = 0.1?14
#. = 0.1714# = 0.r7r4ffi
Q = (50 persons) (150 kcaVperson.hour) = 7500 kcal/h
volume of room = (L2) (10) (3) = 360 m3
volume of air, V = 360 - (0.2) (50) = 350 m3
mass of air, m = -4 = ,(191,31(l5ol
. RT, (0.28708) (289)
a = l-ruooealt-l9 hl = rzsok.ul
L h llliO I
a = mc,T2-Tr)
1250 = (427.34> (0.1714) (T, - 289)
T, = 306'1 K
tz = 33.1"C
4. A l-hp stirring motor 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) (
Solution
-l
'l
I
l
c.
I
Vs
Irreversible Constant Volume Process
a = (-850 kJ/h) (1 h) = -€50 kJ
56
/
W= (-1 hp) (h) =r(-lhp) (0.74G kWhp) (h) (8600 n/lr r
= 427.34kg
= -2685.6 k I
a = AU+W
AU = Q - W = -850 - (-2685.6) = 1835.6 kI
AU = mc" (AT)
AT = -AU. =
rffi5.6 kJ
= 19.3 C"
DC" (22.7 kS) @.t87 kJ/kg.C")
5. A closed constant-vorum,e system receives r0.5 lr.I of
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) f #e nnat
k.mperature is 400 K. Gn Board problem _ April lg, l"ggg)
Solution
:1 = 0.6SgS kJ(kc) (K)
lt = 2b9.90 J(ks) (K)
p, = _344
kPa Tr = 278 K
V-0.06ms Tz=400X
2T
I
I
t
I
1
,lr
vs
p,v _ (344) (0.06) - _
q =
id:t500n?s) = 0'2857 ke
mc" (T, - Tr)
Q.2857) (0.6595) (400 - 278)
22.99 kJ
AU+W
22.99 + (*r0.5)
t2.49 kJ
,/
fr7
34. Isobaric Process
An isobaric process is an internally reversible prccess of
substance during which the pressure remains constant.
Fig.6. Isohric Process
(a) Relation between V and T.
Tz Vz
Tr=vi
(b) Nonflow work.
t2
W"
{,ndV = F(V2 - Vr)
(c) The change of internal energ:y.
AIJ = rDC" (T2 - Tr)
The heat transferred.
Q = mcn (T, -Tr)
The change ofenthalpy.
AH = rlc, (T, - Tr)
The change ofentropy. .
(d)
(e)
aS = mcohfr
N
s:i
58
(f)
-
(g) Steady flow isobaric.
(a)Q=AP+AK+AH+W'
W =-(AK+Ap)
W" = -aK
(AP = 3;
.2
(b) - JVdp = W + aK
I
0=W"+AK
W" = -aK
l'roblems
. l. A certain gas, with c, = 0.b29 Btu/lb.R" and R = 96.2 ft.lV
lh."R, expands from b cu ft and g0"F to 15 cu ft while the
trrcsgutre remains constant 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
l __>_2
p = 15.5 psia
V, = 5cuft
% = l5cuft
T, = 80+460=540"R
vc
,^)'r', =1:,= g+lP =r620R
'r',, =
ffi i##ffif) =o.2r48rb
51)
2
/
/
,/
35. = mce(Tz _ Tr)
= (0.2148) (0.529) (1620_ 540)
= 122.7 Btu
(n
(c' c" = co-R= 0.b29-W=0.40ss#S
AU= mc, (T2 _ Tr)
= (0.214s) (0.40$;(1620 _ b4o)
= 94 Btu
(d)
os = mcorn
ftI
= (0.2148) (0.52e) h
ffi
= 0.1249
Btu
oR
(e) =
=
p(% - v,)
(r5.5) (144) (15 - 5)
778
28.7 Btu
2. A perfect eas
1s a value of R = 319 .2 Jlkg.lfurrrtt lt
r.2G. If 120 kJ *" iaggJ-fi;ik; of this gas ar c''r.rlrrrrl
,fiTre):f: jli.i?Ttlmrnlm{:m1t,'i,i,t?,,,,,
Solution
k
m
R
a
Tr
= 1.26
= 2.27 kg
= 319.2 J&g.K
= f20 kW
= 32.2 + ZZg - BO5.Z K
f{_
kg.Ku
(c) cv =
(a) co = * -(1.2gxo.a1e2)= t.b46e
a = mco (T, - T,)
r20 = (2.27) (r.b469) (T, _ g05.2)
Ta = s39.4 K
(b) aH= mco (T2 _ Tr) = l20 kI
h=ffit$ =r.22??#h
mc, (T, - Tr)
(2.27) (r.2277)(33e.4 _ 305.2)
95.3 kJ
(d) W = p(%- V,) = plg,_ -ITl
' ^LP, --tri] =mR(Tr*T,)
= (2.22) (0.8192) (Js9.4 _ g0s.z)
= Z4.Zg kJ
AU-
=
36. -Fr Isothermal process
isothermal process is an internally reversible constant
temperature process of a substance.
Fig. Z. Isothermal process
(a) Retation between p and V.
PrVr = Pz%
ft) Nonflow work.
f2 )2
w" = Jpav=l$Y= Cln5= n,v,rr *
r { v vr ' v,
(c) The change of internal energy.
AU=9
(d) The heat transfenred.
Q= N + W" = p,Vrln *= -nrrn&
(e) The change of enthalpy.
Y
r Pz
AH=9
(f) The change of entropy.
n
^s=+-mRrn$j
G) Steady flow isothermal.
(a)Q = Ap+AK+AH+W
w"=e-Ap-AK
W"=Q
(AP-0,4K=0)
.2
ft) - JVdp = W + aK
From pV = C, pdV + Vdp -_ 0, dp =
-,!'uoo=-l;,i #l =
I
P'1n -w
W"=W"
(AK = 6;
f 'r'olrlcms
I l)uring an isothermal process at ggoF, the pressurc orr
rr tt, .t''ir drops fr.om g0 p.i" tol gsic. For
"rilJ"lr",,
*r,,r.11;i[lls process,
_d:,tennile fal lfru ipaV and the work of a
i,,,i1ll1v1y process, (b) the-_ JVdp;ndllie *o"k of a steady llow
f 'r , !,
'.,,:, rluring which AK = 0, ("i e, iai aU il;fi,;liii oS.
88+460=54fi,,lt
8tb
80 psia
r.t
+ 14.7 = 1.9.? 1lsi1
- pdv
-v-
/2
j oou
I
T
1*--__r.__2 m
pl
Pr
r Tl
t pV,=[ I
,'ul
l
.2
I
-L
V
'i!:{t
F-o'-{
62
37. 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.
Solution
-317 kJ/s
1.134 ks/s
586 kPa
26.7 +273=299.7
(a)
lndv = p,V,tnV' = mRT r" *
Vr Pz
= tltt#ftQ t" f# = 42L.2Btu
W,= jOaV=42l.2Btu'
(b) - jvap = p,V,ln .f, = 42L.2Btu
(c) a = ryt *W"= 421.28tu
(d) AU=0
AH=0
(e) m= 3=W=0.2686#
vs
(a) R - cp c, = 2.232 - 1.713 = 0.5L9 kl/kg.K
i. = _*xTl= (1.134) (0:5_U)) (299.7)
= 0.301 m3/s
pr 586
a=.
fi=
Pr=
,n
64
= 0.5547 m,t/kg
(;5
Q = Prvrlo
=
Uft = m#oO =-r.80
-1.80
= € = 0.1653
= (0.1653) (0.30r) = 0.0498 m3/s
- P,t, - (b86) (0.
-T- o:oa#l) =3542kPa
(b) Since AP = 6 and AK = 0, W" = lV" = e = -B1Z kJ/s
(t)ns=
+= # =-1.ob8kJ/r(.s
AH=0
= 400K
= 282.08 kJ(ke) (K)
= 2O7 kPa
=$
v2
q
V,
In
"r-
vl
%
q
v,
T
R
Pr
Pr
p,
:l Air flows steadily through an engine at constant tem_
u'r rrl,'re,4_09 K.Find the workperkilogram ifthe exitpressure
i,',, r r 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)
tlnlttlitttt
tT
l)V=C
'2
V
R't'I -_.(9,?87_q8) gog)
l), 207
38. W = prvrl"
t=nrvr1nfl
= (20?) (0.5547) ln 3
= 126.1 kJ
IsentroPic Process
An isentropic process is a reversible adiabatic process'
Adiabatic simply *"t"t-"theat' A reversible adiabatic is one
of constant entroPY'
Fig. 8. IsentroPic Process
Relation among P, V, and T'
(a) Relation between P and V'
P'VI=PrVb=C
(b) Relation between T and V'
From p,VT = pr$u,td q =+' we have
(c) Relation between T and p.
k-1
12 [p,l r-
q =
LP-'l
2. Nonflow work.
FrompA=C,p-C1r-r
,2 rz ,2
W" = lpdv=J CV+dV= C { V-ndV
t'Itl
Integrating and simplifing,
1.
w-
n
l-k l-k
'fhe change of internal energy.
AIJ = ncu (T2 - Tr)
'l'he heat transferred.
Q=0
'l'hc change of enthalpy.
AI{ = mcp (Tz * Tl)
'l'lrr: change of entropy.
ns=0
I iI r.rrrly flow isentropic.
,,,r(c,.AP+AK+AH+W"
wo,,_-AP_AK_AH
W. -AH
r l' O, Al( = 0)
-
k'l
T, lvt I
T, =
LqJ
pvn=9
.pv=Q
tJl
I
(i(; 67
39. T-
k-1
l?r -'l-k-
T -T lr2 I
-2- ^'Lpil
t"= -248'7"F
68
.2
(b)- lVdp=W"+AK
t'
1-L
LetC=pIVorV=Cpk
'.2.1
- lVap =!C pk dp
t'
Integrating and simPlifYing,
- fiao - k (P'v' - P'v') = r. f'nav
t' ' l-k i
(a) = v,
H$t= 1oo[,!9f
1'666
= 608.4 rtg
1.666-1
r.-_T r.666
= 7001__{q_l = 211.8'R
Lsool
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?
Solution
Pr = 300 Psia
Pz= 0.3 +'l'4.7 = 15 psia
V, = 100 cu ft.
T, = 240+46A=700'R
s
I
E
(lr) _ p,V, (800) (t44)(100)
m = ftfr=-6f6ffi =l5'eelb
AII = ms, (f, * Tr) = (1b.99) (1.241) (211.8 _70{)= _9698 tstu
AL.I = mc, (T, - Tr) = (15.99) (0.74b) (211.S
- 200) = _5822 Btu
tt')6av
= &!;f,J' =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
W" = 9636 31rt
'.', An adiabatic expansion of air occurs through a nor,zlt,
h
"rrr ll28 kPa and ?1oc to 1Bg kpa. The initial kinetlc energy i"
..'11lr1lible. For an isentropic expansion, compute the spcr:if i.
.r,lrnnr), temperature and speed at the exit section.
titi rr lion
828 kPa
7L + 273 = i|44 l(
138 kPa
I
pVk= 6
z
(il)
40. k-r
-k
tnl
T"=T, ll2l
- 'Lpil
tz= -67oC
", =
#,
_ (0.287q8X344)
= 0.1193 m'/ks
lI
ve = vr [g'l. = 0.1198 lHgl'n = 0.429m'/ks
- -
LprJ 11381
Ah = cp (T, * Tr) = 1.0062 (20G
- 344) = -188.9 kJ/kg
A =&*aK+Ah+/"
AK--Ah=136,900J/kg
AK=4-^r=*
D2r= (2k)(AK) = zf r ffil 1rg,966S ) = 277,800 m
1Jz
= 527.1m/s
Polytropic Process
A polytropic procebs is an internaliy reversible
during which
pV" = C and prVl = prVl = p,I"
where n is any constant.
Fig. 9. Polytropic Process
Itelation among p, V, and T
(a) Relation between p and V.
P,vi = Prvi
(b) Relation between T and V.
To /-vJ "-t
T =1q1.
t, t li.elation between T and p.
*.1
L
r-
rn r--
Le lP. I
Ra - l:-€- I
'r', -lp. I
I t_^ t--l
Nonflow work
r.4_l
-.-1.4
= 344lHgl = 206 K
18281
70
75yty:;
'iivr2i
;,,
it> I
't.h^
., // i,
22Q..,
'Zzt
(paV = PrY, - P,V, - mR (T, - T,)
" ,'- l-n
'l'hc change of internal energy
AIJ = mcu (T, - T1)
It,
I
41. 4. The heat transferred
a = AU+W-
= mc" (T2 - T,) + mR-(T, - Tr)
1-n
Ic -nc +Rl
= *Lffj (r2-r,)
[c - nTl
= - lffl (r'?-rr)
= ,n." f-!- "-j (T, _ T,)
Lr - I}_l
a = mc. (T, - Tr)
l'-t -;l
cn = cu
lfrl , the polytropic specific heat
The change of enthalpy
AH = mcp (T2
- Tr)
The c.hange of entropy
AS=mc ln It
"T,
Steady flow polytropic
(a)Q=AP+AK+AH+
w"=Q_AP_AK_AH
w = Q_AH
(AP=0,aK=g;
D.
7.
72
'/3',
(b)- Juao=W"rAK
I
- fvao = {&t:!& = ,2 .
, T_n-- -n JPdv
I'rohlems
l.^ 3X"^1u:
polytropic process, t0Ib of an ideal gas, whose
It 40 ft.lbnb.R and co = o.-zs etju.&1;;;;;;il l#;;
lrlr;r and 40'F to 120 psla
p __:_ _vwrv.r!, luau6,cs suate Irom zu
ra and 340"F. Determine (a) n, (f;4g urr4
;ll,l !ilil,-(11'9:l"ljf dY, (? - il{t (g) rf the pi"*,, i ,iuuav
f l,'rv <luring which AK= 0, whaf is w"i]wuut i. axirw" J;it1
Vlr;rI is the work fo, u
"o"n*-p."i"rrZ
s
Se ilution
l', ilO psia
ffn 120 psia
l" ,10 + 460 = 500"R
l'" it4o + 460 = g00"R
n_l
=T'
Tr
n-l
liio J_ _ g00
:ro I - b00
m = 10lb
R=40**
cp = o.2b
#
l),
l),
tr I
tl
ln6=ln1.6
rr- l 0.4700
=-
rr 1.7918
n = l.Bbo
42. (b) c, - cp R = 0.25 - #= 0.1986
m
AIJ = DCu (T2 - Tr)
= (10) (0.1986) (800 - 5oo)
= 595.8 Btu
AH = mcp (T2 - T1)
= (10) (0.25) (800 - 500)
= 750 Btu
(c) k = 5= ^9'^4 =r.25s
q 0.1e86
?= (10) (0'0541) r"ffi= 0'2543+#
AS = -c" lt d,
(d)Q = mc"(Tr-Tr)
= (10) (0'0541) (800 - 500)
L62.3 Btu
(e)Jnav- eE+*L)-ffi
= -433.3 Btu
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) as ionflow, (b) as a stleady flow
l)rocesg where AP = 0, AK = g.
Solution
(h) W" = JpdV = -433.3 Btu
Pr = 103.4 kPa
fi=4g
s
Tr = 60 +273 = 333 K T, =227 +Z7B = b00K
trr ) Nonflow
*#,
o, = o,
[+..| = (10s.4)F$$] = r184.e kpa
L rl Lgo'-l
w = ,hR %u __,4),0,1T16):900 - 33o
= -631.13
c =c ll-d
" "Ll-ul
KJ
;-
s
(0 -JVap = nJRdV = (1'356) (-433'3) = -587'6 Btu
(g) W" = -fVdP = -58?.6 Btu
AK = -JVap = -587"6 Btu
,
=ro.osorffi;]
= -0.2887 []*
74 IT,
43. TIIF'
7. If 10 kg/min of air are compressedisothermally from p,
=, 96 kPa *{Vr.= 7.G5 ms/min to p, = 620 kpa, find tie worh,
:he change ofentropy and the heat for (a) nonflow process and
.b) a steady flow proce-s-s_with or = lb m/s and u, ='60 Js.
Ans. (a)
-tBZ0 kJ/min, _b. gbo kJK.min;iU)_f 386.9kJ
min
8. 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) t'
(b) pr, (c) AH'and (d) W.
----'--' '!-asw *rv
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
9) t, *d -%,-
ft) lU, AH and AS, (c) JpaV and - JVap. 1
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
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
5 Gas Cycles
Fleat engine or thermal engine is a closed system (no mass
.r'osses its boundaries) that exchanges only heai
""a -"rr. *itr,
rts surrounding and that operates in cyclls.
Illements of a thermodinemic heat engine with a fluid as
I lrr. working substance:
. I a working substance, matter that receives heat, rejects
lu,rrl, 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, or just
rrrrk), to which the working substance can reject rr""i; *a
4 ' an engine, wherein the working substa'nce
"rr"h"
*""r.
lr. lurve work done on it.
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
44. 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)
W=Qo- Q*
(b) The net work of a cycle is the algebraic sum ofthe
done by the individual processes.
W= LW
W=Wr-r+Wr"r+W'n+..
The Carnot Cycle
The Carnot cycle is the most efficient cycle concei
There are otherideal cycles as effr-
cient as the Carnot cycle; but
none more so, such a perfect cycle
forms a standard ofcomparison
for actual engines and actual cy-
cles and also for other less effi-
sient ideal cycles, permitting as
to judge how much room there
might be for improvement.
H'
m
n
82
Fig. 11. The Carnot Cycle
83
(Algebraic sum)
(Arithmetic difference)
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 sub-
gtance (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 re-
t,urns the substance toits initial condition, as the piston moves
ftom 4'to 1'.
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
45. w-
e=
-TB (Ss - S. ) = *Tr (S2 - Sr)
Qn - Q* = Tr (Sz - Sr) - Ts (S2 - Sr)
(Tl - Ts) (S2 - S1), arca L'2-3'4'l
W (Tr - T3) (Sz - Sr)
o"= - r;s;;r
Tr-T,
e = ---E-
rl
The thermalefficiencye is definedas the fractionoftheheat
; supplied to a thermodynamic cycle that is converted into work
Work from the TS Plane
Q^ = mRTrfn
f
V.-V3
Qn = mRTrln 1; = -mRTrln i
From process 2-3,
T3 l-v, l*-'
T =Lv'J
From process 4-1,
T,
-l-v,J.-'
11 -lfJ
but Tn = Ts and Tr
- -k-r
therefore,l V" | =
LqI
then, &
v,
=T2
%
=-vr
84
ft5
Q* = -mRTrt"
t
[ = A^ - a- = mRTrtnt mRTrh
t
- Tr) mRl"
+-v
(Tr - Ts) mR ln f
w-
g=
(Tt
w
a;
vl
mRT. k L
,V,
g=
Work from the pV plane.
W = IW = Wr_, + Wr-, + Wr-n + Wr-,
Tt-Tt
-T,
w = p,v,l"
t. &+: :J,+
p,v,rnf,.&tJ{.
Mean Effective Pressure (p_ or mep)
P-=W
VD
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
the
end of expansiql
volumeattheffiili
46. Problems
Isothermal exPansion
"atio
= t
,:- VL
IsentroPic exPansion ratro = 1;
Overall exPansion
'utio
=
h
Compression ratio = EH*#
v
lsothermal comPression ratio = #
Y^-
Isentropic compression ratio' rr. = 1;
V
Overall comPression tutio = t
The isentropic compression ratio rn is the compression ratio
most commonlY used'
1. A Carnot power cvcle operates on 2 lb
{l*j::?*ii?
ri*i, # "ttffi 'ffi Ho"n b;:. n""q;l"^:l :l: *,".':t'H :f
l'ffi:S J'";n#J'#* ;bd n* ilu^* 11:, : 11 :'-' i;1'":ff31
lX"-lff ffilT.f#H; ff;;il; ?qif
'
vorume at the end
-.";^- rlt nS durine an isothermal proce
?xpallsrv[ rD rvu ]'"'b' - - - ,
an isothermal process,
isothermal compression, G) lP $"t"q: - ^r ^-aanoinn rlrrrine
iil'6::?.i' 6Ji";fi:ie :, g*: :** $""#"ffi:T ffil,3
lll,?*,$*;1il* h[fi; ft;;rr iutio or"*pansion, and (h)
the mean effective Pressure'
Solution
2lb
400 psia
960'R
199.7 Psia
530'R
m=
Pr=
Tr=
Pz=
Tr=
of
ft(;
87
Point 1:
vr
Point 2:
%
Point 3:
Pg=
%=
Point 4:
v4=
(a)
naRT. (2) (53.34) (960)
= -E- = -I4OOXI4;JI= = L.778 ft,3
=
+=ti$ffit#,=8.b61 na
--*
p, [:t:^ = 11ee.7'
l-sso-l
L.aJ 'b-d
mRT" (2) (53.34) (530)
-Ti =-(24,s7) ( lll*4)
= 24.57 psia
= 15.72 f13
v,
[q = (1b.?2)F-ffi = 2.84e rtg
= 7.849 ftg
(b)
^s,-,
= mRln
t=
Q.%19 h*ffi = o"oeoz{fi
(c) Qo = Tr (AS) = (960) (0.0952) = 91.43 Btu
(d) QR - -T, (AS) = {530) (0.0952) = - 50.46 Btu
(e) W = Qn - Qn = 91.4g -50.46 = 40.97 Btu
(o
" = l[=4s
a^ fl'43- = o'4481 ot 44'8Lvo
(8)I*oth""-al
expansion ratio = * =ffi =,
47. i./ :;
)!c
f; l-d
HJ_
co
-@
eqq
ct?
Frl
il lF-l
"ql
'. lro
-lH rlq
lj qolH
,_ (ol ,
I 116l @l '
AIA 'II
vtyJ
-t,'
@la tll
I lOrt
^
:t{
F'lro
rO I-
coA
.:v
sv
vll
il"
Fl€
I cld
NlOIr
latA, I I r
r I lJlrl
-* +
,-()
lt tl
.ol E
cl
a
a
Io
ca
-q
o
tl
Hla
l-lvfl
6
Itl
#
,--
E! a EF
vv
osoo{r
6lIOti€rO
rt
€
19
;
t=---r
6It=
I l*l
tl
{.r
r;
rRlR I
t'
ll c,a
*{
0r
fi
|r)
o?
<l
tl
rn cr
oO
(a
o
tl
OJ
oi
o,
tl
ilo
FIE
-l
lolol
'€lC'l'
o
!i{
ro
tl
ilillllttl
ddt' dE
*
I
-l
3lv
allt?
<{ lv
st3
-l-,:
RIE
-lu
tl
>1+
dltr
ll
g
le."le''*.
fl
a
ll
q
il'r*;
,RlAl
Ef
1l
N
E{
c..i
c
.H
d
s
o
15
t
E
E
rtE fl a
[E! E
s Ei gt
E fi;JE
(0
h I E'E:E
; 3 ?€:3+
HE
' 3 TB. B*EE,
H-ti-L$* EiiE}
if Fil" * dF" H g e iAE*q
" ,r T "" r { H €H,FEd
o o'' d le o g i 'EHd i'
G o of;E-EAES8
M X'a
ronoJ
t-O<r
C- CQ rO
Er
R
o
,fa
rb
3
o
u)
MI
to9
€€ bo
r.E
FB
sa
rr:<
Fa
o
O/?
t:
Eg
-o
b0d
€E
gh
$E
9ii
.ab
6I.i
E;O
tssd
.5x a_ d
<EE ts
9.X B
..EE
$
.d
a
g
cO
q
lr:
F{
<r ll
oq
"olQ
ll l<t
--r ^lFr
g{l!o colY
Itr FIA
rril f.ll
'li :l
il c-lFi
.d. d ('JI
>"b 3l-l
+t9{
, rt el
l^lrO
r .X lF{
l€lv
IcE
lrrll
lr
| 5 t>
I 'E Bl ,-
| ( ti'
IpI,
lxrl
IOA
laBl;
lHrl
IO)
lAd
lv
.E
IV
I
I
I
I
I
I
I
I
illlll
drlF
48. Solution
v
Pr=
rF
11 -
Q*=
827.41,Pa
677 +273= 950K
- 132.2 kJ
Qe = (m) (c") (T3 - Tr) = (0'1382) (-0'6808) (540 - 939'9)
= 37.63 Btu
en= mRr.rn{=,Wt"*h
n
= -27.82FJttt
'!{ = Qo - Q* = 37.63 -27.82 = 9'81 Btu
o -A sz =
_o.osrdlg
As.ir:fl=-bao
w (9.8!X179)- = B.lb psi
p-=ql172= ffi-v'LvEe'
2. Tvo and a half kg of an ideal gas with- R = 2963 Jfte)
(K) and c" =6i++i r'"lltr'?Xrc11i a-ryJt:y"" 9f 127
't kPa and a
temperatrfe
"f
b6Fc *J*t 132.2 kJ of heat at constant pres'
sure. The e""1;it""-"d;a*a "tto"ails
to nJis = C to a point
where a constant volume p"ot"tt wil bring-tle e: back to its
original ttateS;t"rttil; er;q' *d the poier in kW for 100 Hz'
1) |
cp = c, + R = 0.7442+ 0.2969 = 1.0411
r c_ 1.0411
o={=yiffi,=1'3ee
Point 1:
v. = -IT, -
(2.5) (q396e) (e50)
= 0.8522 m3
'' - & 827.4
Point 2:
Qn = mco (T, - Tr)
-132.2 = (2.5) (1.0411) (T, - 9b0)
Tz = 899'2 X
% = u,F,] = (0.8b22)ffi21 = 0.8066 mg
Point 3:
r, = r,
H]"''
= rsro.rlffi"u-' = 880.e K
Qo = mco (T, - Tr) + mcv (Tr - T3)
Qn = (2.5X-{.4435X886.9
- S99.2) + (2.5>(a.7442)(950- 886.9)
Qn = 131 IGI
'![
= Qo-Q*=131 -L32.2=-L.2kJ
w - if
r#iFosgfl =-12okw
KI
EAIF
49. Review Problems
l.ThbworkingsubstanceforaCarnotcycleis8lbofair.
The volume at the feginning of isothermal expansion is.9 cu ft
*a tn" pressure is 360 psia. The ratio of expansion during the
uaaiuo" of heat is 2 and the temperature of the cold body is
;0"F, Fi;J (a) Qe, o) QR, (c) vr, (d) pr, (e) vn, (0 pn, (g) P-,, (h) the
ratio of u*purrsion duffng the isenlropic process' and (i) the
overall ratio of comPression.
Ans. @) gia.a, Btu; (b) -209.1 Btu; (c) 63.57 99.ft;
(d)
25.(/-p*iu; t"> ef.Zg cu ft; (f) 51.28 psia; (g) 13'59 psia; (h) 3"53;
(8) 7.06
2. Gaseous nitrogen actuates a Carnot power
-cycle
in
whict the respective iolumes at the four corners of the cycle,
rt"*frtg ;tlnetUegittning of the isothermal expansion' arg Vri
ib. iit i; v, = 1 4.bI L, v
"Z
zza.r+!, *1 Yr : r57'7 3 L
Jhc
cvcle
receives zi.r t<.1 of it"it. Determine (a) the work and (b) the
mean effective Pressure.
Ans. (a) 14.05 kJ; (b) &'91kPa
3. show that the thermal efficiency of the carnct cycle in
terms
-of
the isentropic compression ratio rk is glven
bv . 1.
g=l-
L-l
rk
4. Two and one'halfpounds of air actuate a cyclecomposed
of the following pro"u*t"*t polytropic compressiol Y' urith n =
1.5; constant pressure 2-3-; constant volume 3-1' The known
au1,a *", p, = i0 p.iu, t, = 160'F, Q* = -1682 Btu' Determine (a)
i^
""a
t- iul th;;;;k'of the cvcle'using the pV plane' in Btu;
(J) Q^, (ai tne thermal efficiency, and (e) p-' . -: -.
'-' - '
Arrr. (a) itzo'R,4485'R; (b) 384'4'Btu; (c) 2067 Btu;
(d) 18.60%; (e) 106.8 Psi
5. Athree-process cycle of anideal gas'.forwhi*.htr= 1'0et
*aI." = 0.804 lr,yl*e.K', tl-tTlt"Fiby an isentropic compres-
sion 1-2 from rog.a"kpa, 27 "C 1060g. 1 IiPa. A cbnstant volume
p"".*t Z-S and a *-ftti*t 3:l 11ll
n= L'Zcomplete the cvcle'
Circulation ir
"
rtiuiv raL of o.go5 kg/s, compute (a) Qa, ft) W'
(c) e, and (d) p-.
Ans. (a) 41.4 k'ys; &) - 10 kJ/s; @ 24'157o; (d) 19'81
kPa
92
t
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
Infoh ttrcb
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
i".ilTrili?ii;lte
piston rhen,o"1, ,p the cytinde-gJ", prrhrg rhe'burneJ for", ori!"rins *,o
The four-stmkg cycre is one wherein four strokes of the
piston, two revolutions, are required to complet" u.y.l".'
Erh06l
Ittr.u!t lkol.
Comprarrlcn Strol.
9:i
50. The Otto cYcle
engines.
Otto Cycle
is the ideal prototype'of spark-ignition
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
Analysis of the Otto CYcle
Qe = mc" (T, - Tr)
Qn = mc, (T, - Tn) = -mc" (Tn- Tr)
{ = Qn - Q* ' BC" (Ts - Tr) - BC' (T4 - Tr)
e=fr=ffi
e = r-#+F (1)
'rr - rz
e = 1-+
rl
94
,V
wnere
"* =vr., the isentrcpic compression ratio
Derivation of the form ,la for e
Process l"-2:
t-rl-l
LVol
T, = Tr"oo-t
Process B-4:
5_
Tr-
' (2)
(3) in equation (t)
W = IW =
Pr%'- 9rV, * O,? - O, -%
Clearance volume, per cent'clearance
"*=f=q;r=Hg6
".
_l+c
*c
& I-v;l*'' F
T= Lr*J = tI
L-l
T, = Tn"*
(3)
Substituting equations (2 ) and
a - , Tn-T,
'-E4rffi
e = 1_n+
-t
IVorh from the pVplane
51. lrr
where s = p€r cent clearance
% = clearance volume
Vn = dsplacement volume
Ideal standard of comparison
Cold-air standard, k = 1.4
Hot-air standard, k < 1.4
The thermal elficiency of the theoretical Otto cycle is
1. Increased by increase in r*
2. Increased by increase in k
3. Independent of the heat added
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; hot-
air 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
0.1 lb/s
o.o
1.3
13 psia
130 + 460 =
5000"R
m=
^k
k=
Pr=
Tr=
Ts=
96
o, = t
[+J=
(2ee8)H= 66.r psia
(a) Point t:
v, =
s"-
(0.1x€-.94)l_5eo)
= 1.oar
$
Point 2:
fV
*rt
p, = prLfrJ = P, (r*)h
= (13) (5.5;r.e
= 119.2 psia
l.l
= (r*)h-r
= (590) (b.b;r.s-r = ggB.9.R
tz = 523.9,F
v- =
li =
l'6=81
= o.Bob6 &i
-z-t 5.8 s
Point B:
%=%=0.3056t
s
el
Tr=Tt
Point 4:
l-ti : r'r
r. = 4Li-J
tr = 2538"tr'
=(boo)m"' = 2998"R
52. (h)c= R - 53.34 =0.22t-
Btu
u'f cv =
L11 = (zzgfitm = v'o'c'o l6.R"
Qo = rhc" (T, - Tr) = (0.1) (0.2285) (5000 - 983.9)
Qn = sr.zz ntrt
s
Qp = rhcu (T, - Tn) = (0.1) (0.2285) (590 - 2998)
Qn = -55'03 Btu
s
W = Qo - Q* = 91.77 - 55.03 ; 36.75
ry
o =W =3!'75=0.4005 ot4A.O1Vo
W'=
(36.?5 BtuX60+)
'smrn
=52hp
c'= E*=m =o'8444*k
-=*+ =ffi=o'o43e6lce
"*
=f= tdi%to
=,,
(a) Point 2:
' v, 0.0!
' "' =T= # = o'003455 m3
T, = Tr"*t't = (805) {ll;t't-t = 6g9 K
tl
Pz = Pr{ = (101.8) (tt; t'e
= 2blg lipa
Point 3:
Q^ = mc" (T, - Tr)
12.6 = (0.04396) (O.UU)(TB
- 689)
Tg = 1028 X
Ps = r,ltJ= (2518)
t8rfl = BZbzkpa
Point 4:
n'*t#ftnr
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-.
Solution
1.t4.1
P, = 101.3 kPa
V, = 0'038 mg
Ti=32"C +273 =306
t =t{W"'=r&l'],r*r{*J
n, =n,ffi:r,91]ruzuaftl'
=455K
= 16l kPa
53. (b) Qn = mc" (T1- T1) = (0'04396) (0'8444) (305 - 455)
Q* = -5'57 kJ
W = Qn * Qn = L2'6-5'5? = ?'03 kJ
, - W - 7.99-= 0.558 or 55.87o
(c) e=q= 12S-
(d) p.
=#" =
#T,=
out of the cYlinder'
Compression-Ignition
or Diesel Engine
ComF'trlon Sftok' ?ow'r Stlol'
Fig. 15. Four-stroke Cycle Diesel Engine
A cycle begins with the intake stroke when the piston moves
downanddraws"ilffi iil;t;*:: j*:t:-::f".1fit:11
burns explosrvery' rra$tru Prvuuvv- -"
,k". During the exhaust
ffi;"tfit"oo do*o ror the Power strt
*t*k", the piston #"; ;;Jt; ""d
forces the burned gases
down and draws'"t1':-:ini'".-""ussion stroke' the tem''
n*J,ffi"; ll: 1l Htr :H3:j!rye?tio"' when o' is
iniected into the
'U'" "tU"a"1
it -i*"t *iift ttt" hot air and
burnsexplosivelv'e;'";;;;'"'*:Jg1*;if ::f,1f'l
12.6
= 364.7 kPa
o55s - oso3455
Crh!urt Sitol'
ComF.trlon Sftok'
ln|!l. Sl.ok.
(")
Fig.
(b)
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
Analysis of the Diesel CYcle
Qn = mcn (Ts - T2)
Q* I -." (T, - Tn) - -DC, Tn - Tr)
W = Qe - QR = mcn (T, -T, ) -DC" (T1 - Tr)
"=frW
. T.-T
e = 1- Fd:fJ (4)
€=1-
where
"*
=F the comPression ratio
""
= +, the cutoffratio
l0l
54. Point 3 is called the cutoffPoint.
Derivation of the fornula for e
Process 1-2:
Process 2-3:
=f"
Ts = Trrrk'tr. (6)
Process 3-4:
'- *k-l
T" lv, I
q =Lv^,l
T, = Tr"*
k-l (5)
ft={;
Tn=Trrnk-l
H
. 1 f-t"*-rl
e=r-,r-rlq:11l
r02
t=F;-'=m-'=*'
Tr = Trr"k (7)
Substituting equations (5), (6), and (7) in equation (4)'
T.t"ni.--
e=1-mf-'r--ffii)
-The
efficiency ofthe Diesel cycle differs from that of.th* ( )r,r.r,
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.
Relation among rLr r.r and r" (expansion ratio)
L
%
t-
e -L
-%
' =f"f"
Problems
"
1' A Diesel cycie operates with a compression ratio of l3.b
and 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.
Solution
rk- t =[+][q
'.,'4^ rn = 13.5
L = 1.84
p, = 14 Psia
Tr=140+460=600'R
y, =lcuft
Io;l
56. ry-
Point 2:
lo;l+'
IJil
,,=*b{'=(rrrr)
(a)
(b)
(c)
" =vr=0.2143_14
'* -V--o.oi^re -'
1+c
f,=--
*c
1r 1+c
I4t =- c
c = 0.0769 or 7.69Vo
v^ 0.0383 t
f = -!iL =--:-::= - 2.50
-c v, 0.0153
u, = urffl
Tr=T,
Point 3:
Qn = mco (Ts - T2)
3r7 = Q.227) (1.0062) (T3 - 924.4)
T, = 2312I(
I m | ).olb3) lW1= o.oB8B mg
v, = vr,if i=
((
L-2) P24A
Point 4:
B*?H" = 1161k
1
1.1
= (0-2143)
ffi 0.0153 m3
=(821.e) Hfl1f = sz4lK
106
(d) QB - &c, (T, - Tr) = (0.227)(0.2186) (B zt.g -tt6t )
Qn = -136.9 kI
W = Qo - QR = 317 - fg6.g = lg0.l kJ
(e) e = P= lao.t = 0.b6g1 or 56.glvo
QA 317
1fl P- =g=
=.w = l0o.l _= 9ob kpa
vD vr_% o-zr+s:00rog
DuaI Combustion Engine
In modern compression ignition engines the pressure is not
constant during the.combristio" p"o"ess but varies in the
manners illustrated in the ng"*.-ili;*J il ffi* ol" *
combustion can be conside*dt";il;ach a constant-vorume
process, and the late burning, u *;rilunt-pressure process.
Fig. tZ. Air_Standard Dual Cycle
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)
57. Q* = me, (T1 - T6) = -mc" (Tr - Tr)
W = Qe - Qn = mc" ( - Tr) + mco (T1 - Ts) - DC" (T6 - Tr)
mc" (T, - Tr) + mc, (T, - Tr) - mc, ('t'o - T,)
mc, (T, - Tr) + mco (Ta - T, )
g='W=
QA
e=l- (8).
where
""
=S, the pressure ratio during the consant volume
o P, ' poii"" of co-U"stio"
v
rr =titr, the compression ratio
,2
r
r. =#, the cutoffratio
' Y3
Th'b thernal,efficiency of this cycle lies between that of the
ideal Otto qnd the ideal Diesel.
Derivation of the formula for e
Proccss 1-2:
- -k-1
T" lv,l
q=LrJ /
T" = Trr*I'r
Process 2-3:
t=#="
T, = Trrrk-t rn (10)
r0g
Procesg B-4:
tn /
4a v t
il= f,=""
^g t g
Tn = Trrr t'lr;{" , (lt)
Tu = Trr*'t-l ror.
Tu= Tpor"r
or.
too"otuting equatirins (9), (10), (11), and (12) in equation
(r2)
€=l-
o=l-
Problems
. L. At the *tllpg d:op-p."*rsion in an ideal dual com-
bustion cvcle, the w.orki"ng n"ia-ir i ru
"irri"i-iijT#" ""a
99:F..
The compre*io.l
:.ilI i il"- p"*rru* at the end of tlre
constant volume addrtion or n*ullrito ;;i""#;;#;
"*
added 100 Btu uA* th;,;il;;ilpor*,ro expansion. Find
(a) ro, (b) r", (c) the percentage cfearence, (d) e, and 1e) p_.
l-T
*L
58. Solution
Point 1:
m = llbair
p., = 14.1 psia
T, = 80+460=540oR
pa = 470 psia
rk= 9
Qr-n = 100 Btu
Point 2:
v. 14.186
%=t=-t-= 1.576ft3
rir-'l k-l
Tr= T, l+ I = (540) (9) ''n-' = 1300R
L'rJ
l-v,l*
l, = n,l_if = (14.1) (9) 1'4
= 305.6 psia
Point 3:
u,=-3l'-=%#ffi#=la186rt3
Tr=T,[pJ
LF;J
Point 4:
Qr-n = (m) (co) (T. - Tr)
100 = (1) (0.24) (T4 - 1999)
Tn = 24J.6"R
v. = v,R] = o.b?o)
f+f
il0
= 1.905 ftg
I'
4
A'
,-/i
'/ -""
,2'
ill
*'
Point 5:
t, = t l+ln.'=
(rnru)
E&1" = 1082"R
L_'I-J
(a) r^ =g= +!y = L.54
P Pz 305.6
(b) r =t= !g!tg = L.Zr
" v, 1.576
(c)r.-1+c
*c
9=1+c
c = 0.125 or ]'Z.EVo
(d) QA = Q-, + Qr.n = (m) (e") (T, - Tr) + 1oo
= (1) (0.1?14) (1999 - lB00) + 100 = 219.8 Btu
Qn = (mXc"XT,
- Tu) = (1X0.1714X840- 1082) = -92.9 Btu
^ W 219.8-e2q
" =Q;= --fts-=:: = o'5773 0r 57 '73Vo
w (126.e) (778
P*=V,-% = ffi =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 t p",?g.g"c.
Itet ro - 1.5,,r..= 1.!-0, an{ r* = 11. Determine (a) the percentage
('lea.rance, (b) p, V, and T at each corner of the cycle, tc) e-n,
(d) s, an6 (e) p-.
Solution
'f-.
t,:
m = 0.454kgof air
P, = 96.53 kPa
T, = 43.3 + 273 = 816.3 K
rp = l'5
r" = 1'60
rr = ll
59. W = Qr - Q* = 474-L95.7 = 278'3 kJ
(a)- -1+c
rk-
c
1+c
11 =-;
g = 0'10 or IUVo
mRT, (0.454) (0.28?08) (316'3)
= e.427r ms
(b)Vr=-p;=re
*, -
vt- o.42t]-
= o.oB88B m3
vr =T;-= --11
l-v-lr'-r
,, = t,FJ*-'= T, ("n) *-'= (316'3) (11)'n-' = 8254K
I-vlF = pr(roy = (96.b3) (11) ''n =2770'81.Pa
p, = n,
ft'1
ps = (Pz) ("n) = (2??0'8) (t'5) = 4156'2 kPa
,, = r,fog = (82b.4)
ffi =
'288.1
K
Vn = (Vr) (r.) = (0'03883) (t'60) = 0'06213 m3'
l-ri-l
rn = t'L+l= (1238'1) (1"6) = Le81 K
- I-vln', = (1e81)
Bm''n-' = e16.2 K
,, = r.LirJ
l-m-l .6 kpa
pu = p,l+l= (e6.53) e1g.? =27s
'L'
d 316'3
(c) Qe - (m) (c") (T, - Tr) * (m) (cn) (T4 - T3)
=
(0.454X0.?186X1238'1 - 825'4) + (0'454X1'0062X1981-1238' l)
= 474kJ
(d) QR = (m)(c"XT, -Tu) = (0'454X0'?186X316'3 - 916'2) = 195'?
w 278.3
"=6o= 474 =
w
(e,p_=Vr5,=
0.5871 or 58.7lVo
278.3
= 716.8 kPa
o.427L - 0.03883
I t:l
60. Review Problems
1. An ideal Otto engine, operating on the hot'air standard
with k = 1.34, h^t ;;;;;;tfi ratiJof 5' At the beginning of
;;;;t;;;irt" uor"-"is 6 cu ft' the pressure is 13'?5 psia and
the temperature i. fOO"f' Ouring the constant'volume heat-
t"g, il;'Bl" ^t"
uaJJp"t cvcle' ritta (u) c' (b) T" (c) p" (d) e'
and (e) p-.
Ans,
mrn.
(a) 257o; (b) 5209"R; (c) 639'4 psia; (d) 42'14Vo;
(e) 161.2 Psi
2. An ideal Otto cycle engine 'lrrtlnll%o
clearance operates
on 0.227 kg/s of
"ii
i"Lx" !tut". is 100'58 kPa' 37'7oC' The
energy released d;l;;;*bustion is 110 kJ/s' For hot-air
standard with k = isi,-"o-pute (a) p' V' and T at each corner'
(b) W, (c) e, and (d) P-'
a";.*Ai;.idig *'r., o 029?qm'hl:9:* t"1t':f:
i.pili ;6. + x, zazo.t r<P a, 5s2,1K 19 1'71 kPa;
(b) 52'7 kJis; (c) 47 '9LVo;(d) 301'1 kPa
3. In an ideal Diesel engine compression is from 14'7 psia'
80"F, 1.43 cu ft to 5d0;tt* i"hi" tu
Btu/cvcle are added as heat'
Make computatio,', f* cold-air standard and find (a) T" V2' T3'
v3, Ta, and pn, ft) w;i;;""Jp-' and (d) the hp for 300 cvcles/
Ans. (a) t4?9"R,0'1152 ft3' 2113:l' 0'1&6 ft3' 890'I
zi.ipui^;(Ujg'Z gt"; (e) 60'637o' 39'9 psi; (d) 68'
hp
4. For an ideal Diesel eycle with the overall value of k =
1.33. r,- = 15, r. =2.l,Pr=
gi'g kPa' find P2 and p,"'
' ^Anr. 35-89 kPa, 602 kPa
5. State 1 for a dual combustion engine is pJ = 1 atm
t, Joo.g;Cfrn = 18; a! th9
"i*{*::"Y?L::t:*",?fr
;t;J,"o;;J;ilp*til" i' zogr kPa'-r" = 1'5' tsase on l kg/
;ilil;i-;r 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.*
114
f.p"pZO.g K; (c) 803.5 kJ; (d) 57'a3%; (e) 900
-l
7 ""s
Compressors
Operation of Compressor
Discharge
Valve
Intake
Valye
Di5charge
Compressbn
v
l rtruenlionol Diogrom without Clearance. Conuenttonal Diagram witn Clearance.
Fig. 18. Fig. t9
Figure 18 shows a conventional indicator card for a com-
pressor 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"
"e1,r*
ui"nr.", u,l
va ves being closed, and the gas is compressed 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
l'lrose 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
Il5
61. Preferred Compression Curves
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.
Comparison of work for
Isothermal and for Isentropic
Compression.
Heat Rejected
The heat rejected during compression 1-2 is,
Qr-, = mrcr, (T, - Tr)
Problems
1. 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
Solution
vf=
Tr=
Pr=
Pz=
6 m3/min.
27 +273 = 300 K
105 kPa
630 kPa
118
Tr=T, = 300 = 500.5 K
il9
r^, p,V,, (lob) (6)
"'=f;4= (ozmtGoo
(a) Isentropic compression
= 5.722 kg/min
w-
=
(1.67) (105) (6)
1-1.67
Another solution:
= - 1652 kJ/min
= - 1474 kJ/min
"'-t
I
T
'630
105,
-l
T_
ffi
f- r-r
I rp,t-
Llp;j
E#
= (300) = 615.6 K
e
(T2 - Tr)
= - (5.122) (1.03) (615.6 - 300) = - 1665 kJ/min
(b) Polytropic compression
T2
w
-
k-l
-,r l&l*
- ^t lP, I
I-'J
= -AH = -fi'c
w =+Fffi.,1* I
= (1.a) (rOs) (0)
1-1.4
Another solution
f- 1.4-l _.1
l1f3gl
" -11
_ l.,l-l
iogol'n
F'ql
l-p] #
LEJ
t.67-
y1 rsz
I
62. I
I
c- l'oq
= 0.6168 =
ry
cv =f = L.6z kg.k"
c = c [-t -rr-l= 0.6168 I r.oz - r.el = 4.41G8 kI
n "Lr*l [-T:T-.aJ [sF
if = -afr* 6=-th'co(Tr-Tr)+ fi'co(Tz-Tr)
= -(5.L22) (1.03) (600.5 - 3oo)
+ (5.L22)(-0.4163) (500.6 - S00)
=-1486kl/min
(c) Isothermal compression
w = p,tf r"[*-l
LP?-i
=(105)(6)rn tffi
= - 1129 kJ/min
2. 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?
Solution
f;= 300 ctu
Pr = 14.7 Psia
Pz = 30 Psia
Tr=80+460=540R
u, = 35 fps
u, = 170 Ss
rh' =
off = !#}.r-ffi = 2*.o'rb/min
n r* _r.32_r
r,=r,
L*J'=(E4o) L#t= =64r.eeR
co = c" k-l =(o.lzr4) fILl
s v
U-qJ - ' Ll-tful =-o.oaze ffi
AH = drbp (Tz
- Tr)
= (22.05) (0.24)(641.9
- 540) = bB9.B Btu/min
a -nqTr-T,)
= (22.05) (- 0.0429) (er.g _ 540) = * 96.4 Btu/min
o* =
m'* u'J
6 =#*ar(+ali+W
w = Q-aK-aH
= - 96.4 _ tZ.Z_ bBg.B
= - fl47.gBtu/min or _ lb.2g hp
Volunetric Effidiency
Conventional volumetric effciency =
ffi
n,=$=kX
"VDVD
Displacement volume Vo is the volume swept by the face of
l,he piston in one etroke.
ffi
63. l4r
The clearance ratio or per cent cleararrce, c =
t,
If the compression process is isentropic, let n = k.
vo ={ortN
where:
D = diameter of piston
L = length of stroke
N = number of cycle completed per minute
N = (n) (1) (number of cylinders), for
single'acting compressors
N = (n) (2) (number of cylinders), for
double-acting compressors
n = compressor speed, revolution per min., rpm
A single-acting compressor makes one complete cycle in one
revolution.
A double-acting compressor makes two complete cycles in
one revolution.
Fie. 20. Single-acting Compressor
Connecting rod
then,D"=1+c-c [+-]t
LP'J
, Pision rinRs
7l ,''"on.
-J/
Wrist pin'
nk pin
,-- Crank
! Crankshaft
Crosshead
Crosshead guard
L
t 722
Fig. 21. Double-acting Compressor
(b)n, =1+c*c
1
Free Air
Free air is air at normal atmospheric conditions irr n
particular geographical location.
Problens
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
!Z! ^H*. ggrypression.ana urp""Jio" .r" p"fyt""pil .itf,
n = 1.30. Find (a) the work, (b) the hialre5ected, and (c) the bore
and stroke for I"b0 rpm and UD = f .gO.
Solution
V;
Pr
P2
Tr
= 20 m3/min.
= 100 kPa
= 725 Wa
= 37+273=Bl0K
e=Vo
n = lbO rpm
IID = 1.39
(a) W =T#[A* -_l
l-pJ F
l!-,J
l2:l
= -5023 Y
mrn
PVt's -
"
64. = 1 + o.ob - (o.ob) lzzq-l
fi
"'Llo0l
= 0.9205
n.' 20 = z+.ss 4
vo=n'.,=o8Do5=
t, = Vo * V, = Vo + cVu = Vo (1 + c)
= (24.38) (1 + 0'05) = 25'60 4
*,=*=#Hffi=27.''*t
rn
- r I-Cl+ : (s'o) Fz{lst = 48s.7 K
,, = t,
l_n, |
: !,rvl
[ool
o, = ."ffi = $.7442)Fffi#:l = 4'4b'#
6r-, = rhrc" (T, - Tr)
= (27.83) (-0.2456) (489.7 - 310)
= {ZZs Y
mrn
(c) vo ={nrlN =tD',(1.3 D) (r50) (2) (2)- 612.6 O'#
24.38= 612.6 D3
D = 0.3414 m or 34.14 cm
L = (1.30) (34.14) = 44.38 cm
2:. A single.acting air compressor operates at- 150 rpm with
initial condilion of air at 97.9 kPa and 27"c and discharges the
air at 3?9 kPa to a cylindrical tank. The bore and stroke are 355
I
p-T 3ld 381 mm, respectivery with a percentage cre'r'rrr.o
'f
5?o, rf su'oundins air ar* it r00 kFa
""a
zi-.c *hJio tt,,,
compression and expansion processes are pVr.s _ C. Dutor,r,,,,u
(a) Freg air capacity in mtZs. iU) power of the **pr"rro" i" f, W
(ME Board hoblem - Oct. 19S6)
Solutian
P, = 100 kPa
T =293K
(a) n" = 1 + c-.
[#J* = I + 0.0b-(0.0b)
m]*=0.e0e4
vD =-tDpLN =f, {o.essft0.Bsl) (r50) = s.osz -'
V;= (n,) (Vo) = (0.9094) (b.6bz) = 5.r+a
#
o. = vr
F,]hl
= (b r44)
t+rtffi = 4.els#or 0.082$
(b)w =T#'tre,J*:,]
= (1.3) (97.9) (b.
C= f%o
-T P=$SSmm
+. L = 381mm
It
n=150rpm
Pr = 97.9 kPa
Tr=300K
;
t
ri
ll
;
1- 1.3
t26
65. = - 800.3 or 13.34 kW
3. A single-acting air compressor with a clearance of 6Vo
takes in air at atmospheric pressure and a temperature of 85oF,
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)
Solution
Pr = 14.7 Psia
,Pz = 85 Psia
% = 0.25 ft,3lcYcle
T, = 85+460=545oR
= 900"R
KJ
mrn
(a)r,=r,H* =(545)
[q*
l-ar lfi
h47l
[r;tt
LP'-J
-' = #,, =
(%### = o.o68z4 ib/cycte
v"
+ =3f;3# = 1'o3o n3lcYcre
tbl V; = (0.8754) (750) = 656'6 ft3lmin
126
ni RT. (0.06374X53.34X545)
v,=ff=ffi =o'87i4ftvcYcle
D"=L+c-c -1+0.06-(0.06) = 0.8499
I
W=
" - -IF-to,/
_1J
= '?i#iffifiea- [ia,z/ 'l
= 96hp
4' A single-acting compressor has a volumetic effici'rt,y
of 87vo and operates at 500ipm. Il trk"r in air at 100 kpa nrrrl
l[CA! escargel jr ar 600 kpa. iil ai, rraodted is o .i * p,,.
mrn measured at discharge condition. If the comii#io' i,
isentropic, find (a) piston d;;pi;;;;;t per stroke in cu m, and
(b) mean effective pressure in kpa.
(ME Board p"otrem :ep"riliilal
Solution
= 100 kPa
= 600kPa
= 6 ms/min
= 3O+273=B0BK
= 21.58 m3/min
Pr
fz
V2
Tr
_J
looo l'''
Lrooj
ra) vi =o,Fno = (6)
v^ =&=?rs
" q, o.87
24.8 T'
_ mrn
UOO
stlgkes
mln
= 24.8 It
mrn
= 0.M9G ,-L
stroke
I
t",,7
(h) w= ++lZ+r+ -
r-k lp,/
66. _@ffi@Kml*_!
= -sosa.g Y
mln
n =_li{_=
bob3.g
= 208.8 kpa
rn vD 24.9
6. A compressor is to be designed ntith 64o clearane tn
handle 500 cfin of air at L4.7 pcia and 70pF, the state at the
beginning of compression stroke. The compression is isentropic
to 90.3 peig.
(a) What displaoement in cfu is neessary?
iU) f 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)?
Solution
Pr
q
Tr
14.7 psia
90.3 + L4.7 = 105 psia
500 ft3/min
70+460=530"R
1+c-"[fl*
*', lr0ilfr
=I+0.60-(0.0_.1;14.fl
y-=Yt==5=ry== =orgq-
'o- o" - 0.91b6
- min.
= 0.8156
l2{,
(b) Barosetric pressure at 6000 ft = 1r.?g psia or 23.gg in rlg
New intake pressure, pr* = ll.Zg psia
New discharge pressur€, pz* = g0.B + ll.Zg = 102.0g psia
New volumetric efliciency, r
DvN = 1 + o.o6 -(0.06)
ffiff"o = o.77e|
New capacity, Vi* = @.7795)(6tB) = 472.8 fr:
mln
Percentage decreased in cqpacity =
5010:j[r?.8
= 4.44Vo
(c) pr = 14.7 psia
Vi = 500 cfu
Tr = 530'R
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 4*
Vo at 6000 ft =
ffi= 800.4 g
R, at 6000 ft = 11.78 psia
T, at 6000 ft = 530"R
67. Compressor EfficiencY
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 compression effieiency is
adiabatic ideal work
S-
-c - indicated work of compressor
c.
Isothermal compression efficiency is
- - isothermal ideal work
't - indicated work of compressor
Polytropic compression effrciency is
oolvtropic ideal work
"p = indicated work of compressor
Overall Effrciency
Overall elficiency is
no = (mechanical efficiency) (compression efficiency)
130
Adiabatic overall efficiency is
,, .. =
adiabAtic ideal work
oc%
Isothermal overdll efficiency is
o^, - isotherlpel ideal *"*
or%
Polyhbpic overall efficiency is
Indicated workjs the work done in the cylinder.
Brake work or sh"n *o"r.lr tn" i"* delivered at the shaft.
Adiabatic compressio"
"E"i"i.r r, ,t
"
compression effr-
ciencycommonryused.c;p;;i;;;ffi
"tr;;y;h";;;*,wo.,td
mean adiabatic compressi;
"ffi";;;
Problems
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
"?-9, f OZ.of Jirpfacement
Exponent (n) for both comp.e5ri""
""J
*-expansion
process, 1.6
Airconstant,k= t.{
Air molecular mass, 29
J
no, = (n-) (n") = Sltpolvtmpic
ideal worli
68. 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
pr = lbar=100kPa
o
(a) vo =tryLN ={to.rso)'?(0.200x2x1000) = ?.06e
#
Vl= rr"Vo = (0.?332X7.069) = 5.183#ot 0.0864
(b)w=T#R)* -l
Pz= g
Pr
I
tr, = I * . -.pf = 1 + 0.10 - (0.10X8)t = 0.?332
Lru
[t,*-t]
m3
S
(1.6) (100) (0.0864)
1-1.6
27.2L
= 27.ZlkW
Shaft power = ffi = 33.59 kW
(53.34) (545) = 2o.ss lP
mln
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'
9,u^ _":"d
dischargin g.it at 62' p; i;thu .91 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 pJ;i;;e, if atmospheric condi.
tions are 82'F and r+.2 psia, ?tiil t";;fiffi;i"l ,r,,
indicated work of the-.o-p."rror iitit" compression e-fficiency
is 87Vo, and (d) the ideal *ort .
Solution
P" = 14.7 psia
T = 82"F+460=542"R
Pr = 14.5 psia
Tr=85oF+460=b4b.R
(a)n"=1+c-c
Vf = (o,) (V;) = (0.8924) (214.g)= 248.8 cfm
9 -
(v/ (P,) (r")
= 84!€I(14.s ) (542)
'" - --In"Jnt- - -liz:7t6 = 240'6 cfm
(b) ir = Vn * % = Vo + cVo = Vo(l + c)
= (274.9) (1 + 0.0bb) = 290.02 cfm
Q!.5) O44) (2s0.02)
= r.ob5 - o.obb
m]* = 0.8e2,4
r -L
lP, I'
l&i
vD =4'-D'?LN =
t H' frq
(1b0x2) = 274.e crm
. P,V,
m, = 1i1; =
69. r, = r,
[t]" = 545EH
51
= ?88"R
co = c"
F=; = (o.tz14)
ftfrfl= - o'3025ffi
= (20.83) (-0.03025) (788 - 545)
.,-o'' Btu
= - IDO.I ::::'
mtn
(c) iV,"",, =
(1.4) (14.5) (144) (245.3)
t7g) * -r
=@lrr+.rt )
= - 1185
BtP o" -27.97 hP
mln
adiabatic ideal wor!
,n.i@
Indicated work =H#= 32:15 hP
(d)w =ryreil{-']
(1.34) (14.5) ]44)(245.3) lTsz-t'*g - il
=@lr+si ]
= - 1157
Blu or - 27 -29 hP
mln
.br
k4{&fiq* - rl
r-K Lpr/ -J
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
"t"
polyt"opic wt&
n = 1.31. Determine (a) the con-uentional volumetricefliclency,
9ltlt. heat rejected, (c) the work; and (d)the XW inpui by tfd
driving motor for an overall adiabatic elficiency of ittir.-
Solution
D'=
fr'=
Pr=
Pz=
Tr=
L = 0.3556 m
8.48 kg/min
101.35 kPa
310.27 kPa
26.7 + 273 = 2gg.7 K
(a) v, =fD,tN =t0.Bbb6), (0.sbr6i (100) (z) = 2.068
#
vf=+=W=6.bu#
o" =*- W^ = 0.9227 or gl.z7vo
" vo 2.068
- -*:!.
(b) 12 = r,l+4- = eee.7) t!-lq€fl+#= Beg.b K
- Lrrl L101.351
." =.,p3J = (0.6beb)
H$H = -0.1808 kJ(ks) (K)
F;r+
l-F;l
135
OYt'a
* C
4)-
F_V:
'vo '
D"=l+c-c
70. I-gro.2?l#
0.9227=1+c-cl
Ll0L5il-
c = 0.0573 or 5.737o
r rrt3
V, = Vo (1 + c) = (?.063) (1 + 0.0573) = 7.468 -*
mrn
,- p,v, (101.35 tn aRR kg
'l',=ffi=idffiffi=e.717 ;ff
Q,-, = rhrcn (T, - Tr) = (9.717) (-0.1808) (390.5 -ZggJ)
I-r
= _159.5 ^1
mln
+
l(tl -rl
(c) W= nth'RT,
T.n
= (131) (8.48) (0.25ee) (zss.7> [7 srO.ZztttJil
-
.'l
Tllolsb/ -:l
= -846.1 Y o" -14.1 kW
mln
(d)w,"*=qPR)*-!
(1.3eb) (8.48) (0.25ee) (2ss'7) [121s.2711fH
= l!0135i
= -..309.b
g- or -14.49 kw
' mrn
adiabatic ideal work
'^oc - brake work
' 14'49
= 20.41 kW
DraKe wofK = 0J1
work input by the driving motor = 20.41 hW
Multistage Compression
Multistagingis simply the compression of the gas in two or
more cylinders in place of a singffitinaer como"Jrro". l, iu
usedin reciprocatingcompressors in order to(l) save power, (2)
limit the gas discharge temperaru"q ;;?JiilililJ;;:r"""
differential per cylinder.
4 ------r -
IIP cyUnder
IiS ?2. Conventional Cards,
'rwo-Stage,
No pressure Drop
v
_Fig.,23. Conventional Cards,.
Two-Stage, with pressure Diop
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 G pry"Vai; Ii"*r in. Compression t-2
occurs and the gas is
$yharc.ei
"*ir-".
The discharged gas
passes through the interc*te"
".rd"is
cooled by circulating
water through the interc*t." i"U"r. Co"uu"tio'Jfi,"it i,
t:f7
rvater in water out
71. assumed that the gas leaving the intercool:l el entering the
rrpcvrindeTiu.ir,?,u-g*g;;^iil:;tt*mi*""1i$
Hft *u*kil*t=P**T'*'-**fr '*
fromtheGuuv^'--- r Iearance and must reexpand F-E
; each cylinder because ot c
iirp tvu'ii"'i"*a pe (LP cvlinder)'
!f = W of the loLPlessure cylinder + W of the high
Pressure cYhnoer
= l#,Kkl*-1.#[ft]*-tr
Itis common practice to adjust ll:.o*tution
of multistage
compressor, *o tr'uiipii#;;*y:f works are donejn the
cvlinders,
"
p"u"""Jiil"t "^"'otf
ti":*imum work tbr com-
pressine . gi*'u" q;"iG oru *: :liiiT:H:ftff#Til:
#T- = i- ;d of P, = Pr =.P*' weltave
l;,,h toitrat of the HP stage' or
#trf,*{=+[tlt'i
p,= yTF*'-
where: P, = intermediate pressure for minimum work
since the work of eachcvlila"iillh" sane' tlre t?lawork
for the two-stage #;;;tJtwice the workin each cvlinder' or
2nm'Rr,f-1P,$ ;1 ='+Pfel* -1
w= "iffiLft,? _J - 1-n l9'/ r
A pressure drop in the intercooler could be spread on each
"ide
oi this ideal value'
I
i,
i
I
Pressure droP
Pr=P,*--T--
Pr = P' -- l'rtrHFllrlr tllrtll
Heat Tlansferred in Intercoolor
The heat rejected in the intercooler is'
Qt" = m'cn (T, - T')
where m' is the mass of gas passing through the intercoolor
i Jro tfr" mass clrawnin byifrgif .ili"der and delivered bv tho
HP cylinder).
Problems
l.Therearecompressedl'1'33m3/minofairfrom26'7"C'
L03.42kPa to 821.36 kPa' All clearance are 8Vo'
(a) Find the isentropic power and piston displacement
required for a single stage cornpresslon'
--=ft)-u*ing
the,"-, a""t , nnd the minimum ideal work for
t*o-ri"gr.oilpr"rrion when the intercooler cools the air to the
initial temPerature.
---6
Fi"h trr" di-splacement of each cylinder for the condi-
tions of part (b).
:
ial liow much heat is exchanged in the intercooler?
(e) For * ""*"ff-
*p'"ttiin efficiency of 78Vo' what
driving motor outPut is required?
Solution
vf=
Pr=
Pz=
rT
rl -
11.33 m3/min
103.42 kPa
827.36 kPa
26.7 + 273 = 299.7 K
139
72. r =IilFR)*
l
_(1.4) (108.42) (i,l.BBi lTga.BqtY/ -il
1-1.4 - N-mtz-t -J
= - 3327# ot -55.45 kw
tr"=1+c-c
' lezz'361.r
=1+0.08-(0.08)h1ffi1
11.33 _r^*o *t
mffi -'"'"Y min
tr.
vo=#=
(b)
p
Pr
Pa
103.42 kPa
827,36 kPa
p,=y'[];=@ 292.52kPa
**=+#F)* I
(1.a)11s3.a2) (11.33) ftzgz.szttft;l
L 1o&42l - |
1-1.4
- 1416 # o" -28.6 kW
mln
Tqtal work - (2) (23.6) = -47.2 kW
(c)n"=L+c--c +
=1+0.08-(0.08)
l-&1
LP'l
= 0.9119
vnrp=#=## =12.42#
*' =
n#, =,+ffiffi$?, = 18.62
#
,l-= -,BT€ -
(13.62) (0.2q2q81j299.7)
= 4.006 T3
'3 - Pa 292.52 mln
r/ V; 4006 rn3
vnur =;jf = ffig = 4.393;fr
(d) Qrc = th'cn (Ts - Tr)
(13.62) (1.0062) (299.7403.4) = _ 1427
(e) Outpur of driving motor =!7:? = 60.5 kW
: 0.79
l&I-
min
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 tem-
perature of the cooling water increase by iA F". Find (a) the
volume offree air, (b) tlie discharge pressure ofthe low pr*rruro
t4l
