thermodynamics 1 by hipolito sta. maria (optimized)
DESCRIPTION
solutionTRANSCRIPT
Flo--=6;T*I
Z2i
!
a
THEN[tl!ODYl\lA[tl|IOS
HIPOLITO B. STA. MARIA
COIVTENTSPreface viiChapter
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 EnergTEquation.
3 , The Ideal Gas 87
Constant, Specific Heats of an tddal Gas.
4 Processes of Ideal Gas 5fIsometric 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
PREEACE
The purpose of this text is to present a simple yet rigorousapproach to the fundamentals of thermodynamics. The authorexpects to help the engineering students in such a way thatlearning would be easy and effective, and praetical enough forworkshop practice and understanding.
Chapters 1 and 2 present the development of the first la'arof thermodynamics, and energy analysis of ope:r systemsJhapters 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 cyclesare discussed in Chapters 5 and 6. Chapter ? deals with gascompressors andits operation. Chapter 8 develops the Braytoneycle which can be omitted if sufficient time is not available.
The author is grateful for the comments and suggestionsreceived from his colleagues at the University of Santo Tomas,Faculty of Engineering.
The Author
vll
1 Basic Ppq"iples, ConceptsI and Definitions
Thermodynamics is that branch of the physical sciencesthat treats of various phenomena of energ-Jr and the relatedproperties ofmatter, especially of the laws of transformation ofheat into other forrns of energy and vice versa.
Systems of Units
Newton's law states that 'the aceeleration of a particularbody is directly proportional to the resultantforce acting on itand inversely proportional to its mass.o
"- hE, F=m k =+FD8,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,/s2mks system: 1 newton force accelerates I kg mass at
I m./szfps system: 1 lb force accelerates 1 slug mass at l Nsz
1 cm./s2 _+
t=r,4'cm--cyne.s"
1m/s2
o=t#;p1&,/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:
47
If the same word is used for both mass and force in a givensystem, k is neither unity nor dimensionless.
1 Ib force acceierates a I lb mass at 32.L74 fVs21 g force accelerates a I g mass at 980.66 cm/s2L 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 .mk = 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 abody of unit mass.
I poundal
I fvs2 --)II
:.._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 slugsK
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-
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 atstandard condition?
Solution
F, = 5o lbr g= 32.L74ftlsz
So lb_32.L74 ft
P
3.Fivemassesinaregionwheretheaceelerationduetogrr"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) inpounds, 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.rrlFo rb!
fztz+14s'j
[roo4frro.uuM
FK* =d-=
e2e.64 += 843.91 g;
lb .ftS'= I,l ls-P-K s
Bo.b+
F"ok Fto.lF' t*tfuflmo=-?-= 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 slug32.174;ifis
4. Note that the gravity acceleration at equatorial sea levelrr 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 thispor 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
+T(b) F-t
I
h
I
-L
= 0.9b Fg.a
Let Fg = weight of the man at sea level
FF-- = ____qag0.95 F" F"
a =g
a = 0.959 = (0.95) (32.088) = 30.484 fps2
'Fgg = 32.088 fps2
t., - (30.484 - 32'088) fps'z= b34,6z0 ft or tOt.B miles" - _ o.oosTS;r
-Tmorr
(c) Fa
29.1.31 ft
F8r_.6g = 32.088 fps2m = 1801b- r- -1
Ito 1"1 {}la = 32.088 fps' - fTdriil [0'003 fpsz] = 32'001 fpsz
tlso lb-l pz.oor&l r _^ ^^ ,,#=179.03 lbr
32.174F"1T"
mao =T-=
Specifrc Volume, Density and Specifrc Weight
The density p of any substance is its mass (not weight) perunit volume.
rl=Drv
The specific volume v is the volume of a unit mass.
V1lt
----mp
The specificweightTof any substance is the force of gravityon unit volume.
Fg= 8,vSince 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 osP='g orY ='fr
At or near the surface of the earth, k and g are numericallycqual, so are p and y
Problems
1. What is the specific weight of,water at standard condi.tion?
Stilutiong = 9.8066 m/sz P = 1000
kg_n5.
[*,SE**d*_pg -I- E- e.8066ffi#= looo kgF
mo
ry2. Two Iiquids of different densities (p, = 1500 kg/m3,Pzi^
500 kg/m3) are poured together into a 100-L tank, frlling it' Ifthe resulting density of the mixture is 800 kg/mt, frnd the
respective quantities of liquids used. Also, find the weight ofthe 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.esksre.8066*#
Pressure
The standard reference atmospheric pressure is 760 mmHg 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 hp = Po+Pg
(b) Absolute pressure is less than atmospheric pressure
P=Po-P,
The gage reading is calledvacuum pressum or the vacuum.
I ll"y using pressure gages
A Jrrt:ssure gage is a device forrilr,,1||llr rt ng gage pressure,
'l'lrin picture shows therrr,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 thelnlrr, llrrrrrrglr t,lrc thrcnded, ',,rur.r'lrorr. A$ t.hc prOssur:e
I
Fig. 1 Pressure Gage
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 smallpinion gear. The index hand moves with the pinion gear. Thewhole mechanism is of course enclosed in a case, and a gpadu-
ated dial, from which the pressure is read, and is placed underthe 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) islocated where g = 9.65 mps2. Find the pressure at the base of thecolumn.
IO
po
I--T---ps
L
+Pt
-P, V
Absolutet Pressure
Solution
FuuS ["*Spr=*#= (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 supportedlrcight by standard atmospheric pressure.
to what
Solution
At standard condition
\* = 62'4lblfts Po = 14'7 Psi
T ..-rr ;-llu.z *l lt++'#l
, p,, - L----:n!-!_--!t"! = 33.9 ftt'= t; 62.4Y-'- ft3
Thespecificgravity(*pg')ofasubstanceistheratioofthespccifrc weight of the substance to that of water'
^{sps=T
2, The pressure of a boiler is 9.5 kg/cm2. The}arometricpressure 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'kgcm-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 supportedby 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 thel,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
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. Hgabs), 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.A4atmospheresr,. t7 PslaI':t. | --:-
af,m
h"= Z9.tilt". -1f-llth'$.. 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
.lFI'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").
Solutiont.F -32 _ t"C-0212 - n .=
lbb: o
r Btu(lb) (r")toF =
toC =
t"C + 32
t.F - 32)
Io
5(I
, Absolute temperature is the temperature measured fromabsolute zero.
Absolute zero temperature is the temperature at which allmolecular motion ceases.
Absolute temperature will be denoted by T, thus
TbR = t.F + 460, degtees RankineTK=t"C+z71,Kelvin
Degrees Fahrenheit ("F) and degrees Centigrade ("C) indi-cate temperature reading (t). Fahrenheit degrees iFJ) andCentigrade degress (C") indicate tempertu""
"h"ogu or differ-
ence (At).
180 Fb = 100 C"
1p"-5g"9
1 C. =!-1l,"o
16
- Btu - calIr-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
-: VAuIII = i__v v- =Aup
Wltcrc V = volume flow rate
A = cross sectional area ofthe stream
l) :, ilvcrage Speed
rir ,., m:rss llow rutc
t7
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/ftgAt exit, u.. j 350 fPs, v, = 7 ftaAb'
Find (a) the speed u, at section 2, i- 'dft) the flow anii area at the exit section'
Solution\
tu'",=il'i,=ffi =4oorps
-[.'9!d=2604+= --------r6Tt3-ib
. Aru,mr = --vr
Ir 18
(b)
rh, = rh, + rh, = 26.04+ 16'6? = 42'?1+
\tII
=Erf,El a,E4zftz I4=ff =' *T-T
2. A 10-ft diameter by 15-ft height vertical tank is receiv- Iing water (p = 62.1 lb/cu ft) at the rate of 300 gpm and is Idischarging 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
r\lirrur lr,,w rate enreri", =
[ffi] [rr r fi = z4so.\
r\t,r'* tuwrateleavins=Aup= ? Bd'F.uo*J F +
= ru* S*
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 andin 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 externalvertical force of4 lb. Ifthe local gravity acceleration is g = 30.5fps2 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 forr,rrch foot of elevation.
Ans. 20,0001b-; 627.62 slugs; 19,850lbr
4. A lunar excursion module (LEM) weights 150[r kg, onr.rrrth where g = 9.75 mps2. What will be its weight on therrrrrface 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 30ll,/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 ft3llb; (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
the ground level, the barometer read 30.150 in. F,Ig absolute;topside it read 28.607 in. Hg absolute. Assume that the averageatmosphdric air density was 0.075 lb/ft3 and estimate theheight of the building.
Ans. 1455 ft
8. A vacuum gauge mounted on a condenser reads 0.66 mHg.What is the absolute pressure in the condenser in kPa whenthe 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 cmHg 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 twosections 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 specifrcweiglit is 61.5 lb/ft3 (g = 31.95 fpsz), frnd (a) the mass flow ratein lb/min, and (b) and total time required to fill a verticalcylinder 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 energydue 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 calledkinetic energy.
K=#
nK=4-K,=fttoi-ui)
AK = change in kinetic energy
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
PistonAt ea = .zl
'"**F I
Cylinder ---. Final Position of Piston The work done as thepiston moves from e to f is
dW=F,d*=(pA)dL-pdv
which is the area under thecurve e-f on the pV plane.Therefore, the total workdone as the pistonmoves fromlto2is
w =Jlndv
which is the area under thecurve 1-e-f-2.nV
Fig. 2 woRK ot EXPANSIoN.
The area und.er the curue of the prrcess on the pV plnnercpresents 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 fluidacross 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 differenceI'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 itsl,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
"t,lt
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 sYstem3. 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 incneasesfrom 200 to 1000 fps, the internal energy ofthe opeh system de.creases 25 Btu/lb, and the specific volume increases ftom I to8 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-HlW,, l',v,
= o.8o ffL
= sz,o2 Bfi
(20) (r44) (8)= 2e.6rff
n
2
llr V.l -*778
27
-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-lrL-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 hp42.4(mi#)hp)
u, = L163.3 Btunb v, = 2'65 ftsnb
u" = 925 Btunb vr= 294 fts/l.b
Q = -10 Btu/lbu, = 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 #E779 --'-- 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 272kgperr r r r l of air at 99.29 kPa and a specific volume of 0.026 m3/kg. Thenr r" llrws steady through the compressor and is discharged atfrrllf 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&grr{ n.nso. Sketch an enerry diagram. Compute the work.
29
Solution
r4
wo
u2
P, = 99.29 kPav, = 0.026 m3/kgu, = L594 J/kgQ = -4383 Jlkgh = 272 kg/min
Pz = 689.5 kPavz = 0.0051 m3/hguz= 6241J/kgAK = 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 =
IIPe
F+G-1
-4.
lvr =
'zYz=
t*1q4-
:p
=p
w.
Pzv
vflr
594
\
1.
W,,
wn
2.582 +
,![I'm1.l
- kl\i'o mz
AK+'
' 0.896 1+W
w
6.24
'l; l='J
milmw
6.2,
olb-l
,0il,Ia-005
u2-
;16-
).026
rt0.00L
z* uz
3.516
F
lIwlz
i+3
KS
;1
+
+
= 2.583 kJ&e
= 3.51.6lnl/kg
i
E
IIIIL
\{ = - 10.g6H
t- kr-l l- _ ke_lw - 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 mr 275,800 Pa
2270160= 4.667 mls
(1ooo) (o.oo81o7)ilgxrcd at exit, D, =
llnHis 1 kg-
K, =;ik=
frdrPrpqPz
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+,.rtsl)'vr =E =;;E- = oL''* kgm
C
w
3l
Kr+Wrr=Iq+\{Io+W
2.L5L + 82.74 = 10.89.+ 275.8 + W
[-,'"ffiE*HkI
W = -4b8.1ffn
5. Aturbine operates under steadyflow conditions, receiiag steqm at the following state: pnessure 1200 kPa,tue 188"C, enthalpy 2785kJ/kg, speed 33.3 m/s and elevati3 m. The steam leaves the turbine at the followingpressure 20 kPa, enthalpy 25L2 klkg, speed 100 m/selevation 0 m. Heat is lost to the surioundings at the rate of 0.hVs. If, the rate of steam flow throughthe turbine is 0.42what is the power output of the turbine in kW?
Solution
zr=3m
h. = 2?85 E' IKg
ur=33'3fl
l&IQ = -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_lL33.3g:l
= 0.55a4 PKs
q=;i =1#f
-o.zey
;F- = {).6eo5H
Pr+Kr+hr+Q=%+4+ L+WPr+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
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:-]1i.-= 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 90greater t[an that of the air entering. Cooling water inio*p""rror jackets absorbs heat from the air at the rate ofkW. Compute the work in kW.
Ans. -122 kW
it4
-6.. In a steady flow apparatus, L3b lc.I of work is done byeach kgof fluid. The specific volume of the fluid, p""*s.r"*, undspeed at the inlet are 0.37 mslkg, G00 kpa, and 16 m/s. The inletis 32 m above the floor, and the discharge pipe is at floor level.The discharge conditions are 0.62 ms/kg,-100 kpa, and 270m/s. The total heat loss between the inlet and discharge ie gkJlkg of fluid. In flowing through this apparatus, does thespecific internal energy increase or decrease, and by howrnuch?
Ans. -20.01 kJ/kg
7. Steam enters a turbine stage with an enthalpy of 862gk.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 donel,y the steam.
Ans. 776.8 kJ&e (ME Board Problem - Oct. 1996)
lfl-r
3 The rdeal Gas
An ideal,gas is ideal ronly in the sense that it conforns rcllrc 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 therrl*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 varyrlirr.r tly :rrr lhc absolute temperature.
V,."1 or V=CT
v (: or L-IL'r' ' q=qr,: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=9pp
,tt
-7
E
I
-t
P-T or P=CT
Equation of State or CharacteristicPerfect Gas
Combining Boyle's and Charles' Iawg,
fr=c or t=+,
+=ry =c,aconstant
pVT =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. longacetylene at250 psia and 90"F. After some ofthe acetylene
3rl
}F
N;t
pVvmTR
ft3 lb_
RT
oR
K
V
m3
Equation of a
ml=(25cD $44) (0.6545)
= 0.7218Ib(59.35) (550)
im EltTr
Vr=
:l 'l'lrc volume of a 6 x 12-ft tank is 339.3 cu ft. It containssir rrl '.1(X) psig and 85"F. How many l-cu ft drums can bc fillcdl' rru 1rrr1.f :rnd 80'F if it is assumed that the air temperasturttirr llrr' lrrrrh remains at 85"F? The drurns have been silting€*,iurrl rrr l.hu atmosphere which is at 14.7 psia anrl [t0"1"
Pa
;t 1)
used, the pressure was 200 psia and the temperature was 85oF,(a) What proportion of the acetylene was used? (b) Whatvolume would the used acetylene occufiy at L4.7 psia and fl0'F?R for acetylene is 59.35 ft.lb/lb."R.
Solution
(a) Let frr = rlrBss of acetylene initialiy in the drumoz = ltrass of acetylene left in the drumBe = rllass of acetylene usedPr = 250 PsiaTr =90oF+460=550'RPz = 200 PsiaTz =85oF+460=5451R
volume of dr,r* = ffiffi = 0.6545 cu ft
PrV, =RT,
o-E= (200,)=911)!9.6j45) = 0.b828 lbmz = ifq'= (bgsb) (54b) "'""-- --
ms - ml mz= 0.72L8 - 0.5828 = 0.1390Ib
Acetylene used = #i = 3+# = 0'1e26 or re'26vo
tlr) p, = 14.7 psia
'f.=80oF+460=540oR
\roils$l (b=e.Bit t5+01 = 2.'0b fr3' (r4.7) (L44\
rSolution
Let Dr = IIlBss of air initially in the tankDz = rnoss of air lelt in the tankDs = mas$ of air initially in the dmmha = rnsss of air in the drum after filling
Pr = 200 + 14.7 = 214.7 psia p, = 14.? psiaTr = 85 + 460 = 545.R T, = 80 + aOO = b40RPz = 50 + 14.7 = 64.7 psiaTr=8S+460=545oR
For the tank
[l=
Po = 50 + 14.7 = 64.7 psiaTn=80+460=540R
P,VrRT,
(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 atmosphericpressure (101-.325 kPa) and temperature 21.1oC is to be usedin the balloons. What 6inims6 balloon diameter (assumospherical shape) will be required for a gross lifting force of 20metric tons?
40
Solution
20,000 kgl
Itor the air
= "mass of air displaced bythe balloon
= mass of Helium= volume of the balloon
I€t mr
EH"v
JR = 287.08 E__P, = 101,325 PaT,=21.t +273=294.iK
p-V 101.32bV'nu=ili" = tffirl =l'2oolvkg
f','t lltt'heliUm
&r" = 2,077.67 #RP11" = 101,325Pa
T""=21.1 +278=Zg4.lK
,,, _ Pn.v _101,325 Vrrrrl,,= ffiT"" = qOngZffim =0.1658Vkg
fr,=DH,+20,000
1.200f V =0.1658V +,20,000V = l9,BB7 mJ
1
.l rf = 19,337
.l
r = 16.6b md - 2(16.65) = 3B.B m
I
IEH.
+ms
4l
G{
4. TVo vessels A and B of different sizes are connected bya pipe with a valve. Vessel A contains L42L of air at2,767.92kPa, 93.33oC. Vessel B, of unknown volume, contains air at68.95 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
L
(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 quantityof heat required to change the temperature of unit masethrough one degree.
In dimensional form,
c__*
In differential quantities,
c^e= ;ffif or dQ=mcdT
nrr<l for a particular masg m,
a=* !'.arI
(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
^uIVolume I
( lorrstant I
, ---la,
-y'r
a
., tEiil
tIt-.
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
ck=d:>r
Internal Energy of an Ideal Gas
Joule's law states that "the change of internal energy of anideal 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 byformula,
AH = ECo (Tz - T1)
whether the pressure remains constant or not.
g
44
Relation Between cn and c,
Fromh =u+pvandpv=RTdh = 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 gagworrld occupy a volume of l5 cu ft dt ZS psia and gO"F? (c) lfgOlll.rr are transferred to this gas at constant volume in (b), whatnrr. 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
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 nonflow oonstant pressure process from V, = 1.133 m3 and Pr = 690kPa 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 whichconstant if no heat enters or leaves the substance, while itwork or alters its volume, but which increases or diminishould a small amount of heat enter or leave.
The change of entropy of a substance receiving (or deliing) heatis defined by
dS= F
kIIFF
k= &+1=cY
-2or As =JF
I
46
where:dQ = heat transferred at the temperature TAS = 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
a2Q = jTds
I
'The area under the curveofthe process on the TS planerepresents the quantity ofheat transfered during theprocess."
12
-)VdP=W+AKI
(Reversiblesteadyflow,AP= 0)
"The area behind thecurve ofthe process on the pVplanes represents the workofa steady flow process whenAK * 0, or it represents AKwhen W' = 0."
47
-{
Any process that can be made to go in the reverse directionby aninfinitesimal change in the conditions is called a nrersibleprocess.
Any process that is not reversible is irreversible.
48
Review Problems
1. An automobile tire is inflated to g2 psig pressurs at60"F. Alter being driven the temperature rise to zb"F. Deter-mine the final gage pressure assuming the volume remainaconstant.
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 gasw^s discharged and the temperature ind p""**" .t
"og"d to70"F and 300 psia, respectively. Heat was applied and theI.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 Hgalr"lut'e and 18oc undergoes a proc€ss uotit ttre pr?ssunelp.rmes 760 mm Hg absolute andihe temperature 0"c. whattr 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 240ll'^ (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 tirexrrrlrrins 0.11 kg of air. Determine (a) the mass of air escapingeer 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 evacuatedF,nr rrl,mospheric pressure to a pressure of 240 mm Hgte, 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
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 thepressure 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 ofhydrogen, other conditions remaining the same, what load can
itlift? (c) Helium is nearly twice as heavy as hydrogen. Does ithave half the lifting force? R for hydrogen is 766.54 and forhelium 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 surrouby zrir at 60"F and29.92in Hg abs. (a) If the balloon is filled
reservoir to a pressure of 3497 kPa and a temperatureLz.4}C,while tfe pressure in the reservoir decreases to 62
kPa and the temperature to 18.3"C. What is the volume oftank? 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 inby 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 ofgas 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 atvolume 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.
2TII
II
'l
T_Pz
Tt Pz;fr- =-It Pr
HlF-_szFig. 5. Isometric Process
(;r) Relation between p and T.
(b) Nonflow work.
,'2W.=JpdV=0
5l
(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
llVPrTrT2
i0 cu ft300 psia400+ 460= 860'R140+460=600"RV
II
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
(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 ak = 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) theofentropy.
2//
/
t-r4lrlr
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 = mculnq99
= (1.36) (1.508) l" i=gl"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 ilirrllill ,l rneters. The room is completely sealed off and insulrtl'r'rll,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/Prll/ll/r,I l','LVg
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.ulL 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 contai22.7 kg of water. The stirring action is applied for I hourthe tank loses 850 kJ/h of heat. Calculate the rise inture of the tank after I hour, assuming that the processat constant volume and that c" for water is 4.187 kJ/(kg) (
Solution
-l'lI
l
c.
IVs
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 kIAU = 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 oflrrrddle 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 nnatk.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 KV-0.06ms Tz=400X
2TI
I
tI
1
,\lr
vsp,v _ (344) (0.06) - _q = id:t500n?s) = 0'2857 ke
mc" (T, - Tr)
Q.2857) (0.6595) (400 - 278)22.99 kJAU+W22.99 + (*r0.5)t2.49 kJ
,/
fr7
Isobaric Process
An isobaric process is an internally reversible prccess ofsubstance during which the pressure remains constant.
Fig.6. Isohric Process
(a) Relation between V and T.
Tz Vz
Tr=vi(b) Nonflow work.
t2W" {,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.lVlh."R, expands from b cu ft and g0"F to 15 cu ft while thetrrcsgutre remains constant at lb.b psia. Compute (a) T", (b) AH,(r') AU and (d) AS. (e) For an internally reversible'nonflowf r'ocess, what is the work?
Solution
T
l __>_2 p = 15.5 psiaV, = 5cuft% = l5cuftT, = 80+460=540"R
vc
,^)'r', =1:,= g+lP =r620R
'r'\,, = ffi i##ffif) =o.2r48rb
51)
2/
/
,/
= 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$;(1620 _ b4o)
= 94 Btu
(d) os = mcorn ftI
= (0.2148) (0.52e) h ffi= 0.1249 Btu
oR
(e) \=
=
p(% - v,)
(r5.5) (144) (15 - 5)778
28.7 Btu
2. A perfect eas \1s a value of R = 319 .2 Jlkg.lfurrrtt ltr.2G. If 120 kJ *" iaggJ-fi;ik; of this gas ar c''r.rlrrrrl,fiTre):f: jli.i?Ttlmrnlm{:m1t,'i,i,t?,,,,,
Solution
kmRaTr
= 1.26= 2.27 kg= 319.2 J&g.K= f20 kW= 32.2 + ZZg - BO5.Z K
f{_kg.Ku
(c) cv =
(a) co = * -(1.2gxo.a1e2)= t.b46e
a = mco (T, - T,)
r20 = (2.27) (r.b469) (T, _ g05.2)
Ta = s39.4 K
(b) aH= mco (T2 _ Tr) = l20 kI
h=ffit$ =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-
=
,i I
-Fr Isothermal process
isothermal process is an internally reversible constanttemperature 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 orrrr 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 ai,,,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,,lt8tb80 psiar.t + 14.7 = 1.9.? 1lsi1
- pdv-v-/2j oouI
T1*--__r.__2 m
pl
Pr
r Tlt pV,=[ I,'ul\l
\\.2 I
-LV
'i!:{tF-o'-{
62
2. During a reversible process there are abstracted 317kJ/s from 1.134 kg/s of a certain gas while the temperatureremains constant at 26.7'C. For this gas, cD = 2.232 and c"1.713 kJ/kg.K. The initial pressure is 586 kPa. Fornonflow and steady flow (AP = 0, AK = 0) process, determine (
Vr,% and pr, (b) the work and Q, (c) AS and AH.
Solution
-317 kJ/s1.134 ks/s586 kPa26.7 +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/spr 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
=$
v2q
V,In "r-vl
%qv,
TRPrPrp,
:l Air flows steadily through an engine at constant tem_u'r rrl,'re,4_09 K.Find the workperkilogram ifthe exitpressurei,',, r r r' l.hird the inlet pressure and the inlet pressure is zoz kpa.Arrarrrro that the kinetic and potential energy variation is111'plrplible. (EE Board Problem - April lggS)
tlnlttlitttt
tT\ l)V=C
\'\2
V
R't'I -_.(9,?87_q8) gog)l), 207
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$u,td q =+' we have
(c) Relation between T and p.
k-112 [p,l r-q = LP-'l
2. Nonflow work.
Fromp\A=C,p-C1r-r,2 rz ,2
W" = lpdv=J CV+dV= C { V-ndVt'Itl
Integrating and simplifing,
1.
w-n l-k l-k
'fhe change of internal energy.
AIJ = ncu (T2 - Tr)
'l'he heat transferred.
Q=0
'l'hc change of enthalpy.
AI{ = mcp (Tz * Tl)
'l'lrr: change of entropy.
ns=0
I iI r.rrrly flow isentropic.
,,,r(c,.AP+AK+AH+W"
wo,,_-AP_AK_AH
W. -AH
r \l' O, Al( = 0)
- k'l
T, lvt I
T, = LqJ
pvn=9
.pv=Q\ tJl
I
(i(; 67
T-
k-1l?r -'l-k-
T -T lr2 I-2- ^'Lpil
t"= -248'7"F68
.2(b)- lVdp=W"+AKt' 1-LLetC=pIVorV=Cpk
'.2.1
- lVap =!C pk dpt'
Integrating and simPlifYing,
- fiao - k (P'v' - P'v') = r. f'navt' ' l-k i
(a) \ = v, H$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)Vand tr, (b) AU and AH, (c)JpdV, (d) -5vdp, (e) Q and AS. Whais the work (f) if the process is nonflow, (g) if the process isteady flow with AK = 10 Btu?
Solution
Pr = 300 PsiaPz= 0.3 +'l'4.7 = 15 psiaV, = 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=0As-- 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 kPa7L + 273 = i|44 l(138 kPa
I
\ pVk= 6
\\z
(il)
k-r
-ktnlT"=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 reversibleduring 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.*.1L
r-rn r--Le lP. IRa - l:-€- I'r', -lp. II 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
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 -;lcn = 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"rAKI
- fvao = {&t:!& = ,2 .
, T_n-- -n JPdv
I'rohlems
l.^ 3X"^1u: polytropic process, t0Ib of an ideal gas, whoseIt 40 ft.lbnb.R and co = o.-zs etju.&1;;;;;;il l#;;
lrlr;r and 40'F to 120 pslap __:_ _vwrv.r!, luau6,cs suate Irom zura and 340"F. Determine (a) n, (f;4g urr4
;ll,l !ilil,-(11'9:l"ljf dY, (? - il{t (g) rf the pi"*,, i ,iuuavf 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
tlln6=ln1.6
rr- l 0.4700=-rr 1.7918
n = l.Bbo
(b) c, - cp R = 0.25 - #= 0.1986 mAIJ = 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.25sq 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 gastction, frld pr, ry, e;lS (a) as ionflow, (b) as a stleady flowl)rocesg where AP = 0, AK = g.
Solution
(h) W" = JpdV = -433.3 Btu
Pr = 103.4 kPa fi=4\gs
Tr = 60 +273 = 333 K T, =227 +Z7B = b00K
trr ) Nonflow
*#,o, = o, [+..| = (10s.4)F$$] = r184.e kpaL 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,
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 isentropicpf9c9s9^fr9m gb.B psig and a volume of 0.6 {tr to a final volumeof 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- = 1.01HAg.X expands isentropically from lbt? kFa, ie8"t t" gOkPa. 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 1occurs 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 clby 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, -86.140.0141Btu/"R; (c) 34.4f Btu,44.78 Btu; (d) gBtu; (e) 34.41Btu; (0 44.?B Btu
11. The work required to compress a gas reversibly acconing 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 = 1R = 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, rejectslu,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; *a4 ' 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 arv'l.t'm experiencer, u.ly.*,ber of processes that Jventuailynrlrrrn the fluid to its initial state.
Cycle lVork and Thermal Effrciency
(1. QA = heat added
Qn = heat rejected
W - net work
ftl
Available energy is that part of the heat that was convertedinto mechanical work.
Unavailable energy is the remainder of the heat that hadbe rejected into the receiver (sink).
The Second Law of Thermodynamics
AII energy receiued as heat by a heat-engine cycle cannotconuerted 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 ofthedone by the individual processes.
W= LW
W=Wr-r+Wr"r+W'n+..
The Carnot Cycle
The Carnot cycle is the most efficient cycle conceiThere are otherideal cycles as effr-cient as the Carnot cycle; butnone more so, such a perfect cycleforms a standard ofcomparisonfor actual engines and actual cy-cles and also for other less effi-sient ideal cycles, permitting as
to judge how much room theremight be for improvement.
H'
m
n82
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 cylindorhead, the only place where heat may enter or leave the sub-gtance (system) is placed in contact with the sounoe of heat orhot body which has a constant temperature Tr. Heat flows fromthe hot body into the substance in the cylinCler isothermally,l)rocess l-2, and the piston moves from tr' to 2'. Next, thet:ylinder is removed from the-hot body and the insulator I ieplaced over the head of the cylinder, so that no heat may bel,ransfemed in or out. As a result, any further process isndiabatic. The isentrppic change 2-3 now occurs and the pistonmoves from 2' to 3'. When the piston reaches the end of thesl.roke 3', the insulator I is removed and the cylinder head isplaced in contact with the receiver or sink, which remains at aronstant temperature T". Heat then flows from the substancet,rr the sink, and the isothermal compression B-4 occrut whiletlrc piston moves from 3'to 4'. Finally, the insulator I is againlllnced over the head and the isentropic cor.npression 4-1 re-t,urns the substance toits initial condition, as the piston movesftom 4'to 1'.
VmFig. 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
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;;rTr-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 fV.-V3
Qn = mRTrln 1; = -mRTrln iFrom 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-rtherefore,l V" | =
LqIthen, &
v,
=T2
%=-vr
84ft5
Q* = -mRTrt" t\[ = A^ - a- = mRTrtnt mRTrh t
- Tr) mRl" +-v(Tr - Ts) mR ln f
w-
g=
(Tt
wa;
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-=WVD
Vp = displacement volume, the volume swept by the pistonrr one stroke.
Mean effective pressure is the average constant pressurel,ir:rt, acting through one stroke, will do on the piston the network of a single cycle.
Ratio of Expansion, Ratio of Compr.ession
I,)xpansion ratio = -., vglute,3t
the end of expansiql
volumeattheffiili
Problems
Isothermal exPansion "atio = t
,:- VLIsentroPic exPansion ratro = 1;
Overall exPansion 'utio = h
Compression ratio = EH*#v
lsothermal comPression ratio = #
Y^-Isentropic compression ratio' rr. = 1;
VOverall 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 :fl'ffi:S J'";n#J'#* ;bd n* ilu^*
11:, : 11 :'-' i;1'":ff31lX"-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 rlrrrineiil'6::?.i' 6Ji";fi:ie :, g*: :** $""#"ffi:T ffil,3lll,?*,$*;1il* h[fi; ft;;rr iutio or"*pansion, and (h)
the mean effective Pressure'
Solution2lb400 psia960'R199.7 Psia530'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$ffit#,=8.b61 na
--*p, [:t:^ = 11ee.7'
l-sso-l
L.aJ 'b-dmRT" (2) (53.34) (530)-Ti =-(24,s7) ( lll*4)
= 24.57 psia
= 15.72 f13
v, [q = (1b.?2)F-ffi = 2.84e rtg
= 7.849 ftg
(b) ^s,-,
= mRln t= Q.%19 h*ffi = o"oeoz{fi
(c) Qo = Tr (AS) = (960) (0.0952) = 91.43 Btu
(d) QR - -T, (AS) = {530) (0.0952) = - 50.46 Btu
(e) W = Qn - Qn = 91.4g -50.46 = 40.97 Btu
(o " = l[=4sa^ fl'43- = o'4481 ot 44'8Lvo
(8)I*oth""-al expansion ratio = * =ffi =,
i./ :;
)!cf; l-dH
J_co
-@eqqct?F
rlil
lF-l
"ql'. lro
-lH
rlqlj
qolH,_
(ol ,
I 116l
@l
'A
IA
'IIvtyJ
-t,'@
la tllI
lOrt
:̂t{F
'lrorO
I-
coA.:v
svvllil"F
l€I cld
NlO
IrlatA
, I I
r
r I
lJlrl-*
+,-()lt
tl.ol
E
claaIoca-qotl
Hla
l-lvfl6Itl
#
,--E
! a EF
vvosoo{r6lIO
ti€rO
rt€19;t=---r
6It=I
l*l
tl
{.rr;rRlR
I
t'll c,a
*{
0r
fi|r)o?<
ltlrn
c\roO(aotl
OJ
oio,tl
iloFIE
-l lolol'€lC
'l'o!i{rotl
ilillllttl
ddt' dE *
I
-l3lvallt?<
{ lvst3-l-,:R
IE-lutl
>1+
dltrllgle."le''*.
flallq
il'r*;,R
lAl
Ef1lN
E{
c..i
c.Hd
so15tEE
rtE
fl a
[E! E
s E
i gtE
fi;JE
(0h I
E'E
:E;
3 ?€:3+
HE
' 3 T
B.
B*E
E,
H-ti-L$*
EiiE
}if
Fil" * dF
" H g e iA
E*q
" ,r T
"" r {
H €H
,FE
do
o'' d le
o g i 'E
Hd i'
G
o of;E
-EA
ES
8
M X
'aronoJt-O
<r
C- C
Q rO
Er
Ro,farb3ou)
MI
to9€€ bor.EF
Bsarr:<F
aoO/?
t:Eg
-ob0d€Egh$E9ii.ab6I.iE
;Otssd.5x a_ d<
EE
ts
9.X
B..E
E $
.dagcOqlr:F{
<r
lloq"olQll
l<t
--r ^lF
rg{l!o
colY\Itr
FIA
rril\ f.ll
'li :l\
il c-lF
i.d. d
('JI>
"b 3l-l+
t9{,
rt el\
l^lrOr
.X
lF{
l€lvIcElrrlllr|
5 t>
I 'E
Bl ,-
| (
ti'IpI,lxrlIO
AlaB
l;lH
rlIO
)lA
dlv.EIVIIIIIIIII
illlll
drlF
*/
Solution
v
Pr=rF11 -Q*=
827.41,Pa677 +273= 950K- 132.2 kJ
Qe = (m) (c") (T3 - Tr) = (0'1382) (-0'6808) (540 - 939'9)
= 37.63 Btu
en= mRr.rn{=,Wt"*hn
= -27.82FJttt
'!{ = Qo - Q* = 37.63 -27.82 = 9'81 Btu
o -A sz = _o.osrdlgAs.ir:fl=-bao
w (9.8!X179)- = B.lb psip-=ql172= ffi-v'LvEe'
2. T\vo 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'
1X)1) |
cp = c, + R = 0.7442+ 0.2969 = 1.0411
r c_ 1.0411o={=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
'
Review Problemsl.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
inwhict the respective iolumes at the four corners of the cycle,
rt"*frtg ;tlnetUegittning of the isothermal expansion' arg Vriib. 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 interms -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
kPa92
t
L
6 fnternal Combustion Engines
Internal combustion-engine'is a heat engine deriving itspower from the energy liberated by the exploJion oi" *l*trr"of some hydrocarbon, in gur*o.r, or vaporized form, withatmospheric air.
Spark.Ignition (SI) or Gasoline Engine
Infoh ttrcb
Fig. lB. Four-stroke Cycle Gasoline EngineA cycla beginr wilh the intoke slroke or fhe pirlon move3 down the cylinder ond drows in o fuet.oirmixlure' 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 fhoi".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 thepiston, two revolutions, are required to complet" u.y.l".'
Erh06l
Ittr.u!t lkol.Comprarrlcn Strol.
9:i
The Otto cYcle
engines.
Otto Cycle
is the ideal prototype'of spark-ignition
FiS. 14. Air-standard Otto CYcle
Air.standardcyglemeansthatairaloneistheworkingmedium.
1-2: isentroPic comPression2'3: constant volume addition of heat
3-4: isentmPic exPansion4-1: 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=ffie = r-#+F (1)
'rr - rz
e = 1-+rl
94
,Vwnere "* =vr., the isentrcpic compression ratio
Derivation of the form ,la for e
Process l"-2:
t-rl-lLVol
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
96
& I-v;l*'' FT= Lr*J = tIL-l
T, = Tn"*(3)
Substituting equations (2 ) and
a - , Tn-T,'-E4rffie = 1_n+
-t
IVorh from the pVplane
lrr
where s = p€r cent clearance
% = clearance volumeVn = dsplacement volume
Ideal standard of comparison
Cold-air standard, k = 1.4Hot-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 k3. 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 80,000miles of motoring.
Problems
1. An Otto cycle operates on 0.1 lb/s of air from 13 psia and13trF at the beginning of compression. The temperasture atthe 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/so.o1.313 psia130 + 460 =5000"R
m=^kk=Pr=Tr=Ts=
L96
o, = t [+J=
(2ee8)H= 66.r psia
(a) Point t:
v, = s"- (0.1x€-.94)l_5eo)
= 1.oar $Point 2:
fV *rt
p, = prLfrJ = P, (r*)h = (13) (5.5;r.e = 119.2 psia
l.l
= \ (r*)h-r = (590) (b.b;r.s-r = ggB.9.R
tz = 523.9,F
v- = li = l'6=81
= o.Bob6 &i-z-t 5.8 s
Point B:
%=%=0.3056ts
elTr=Tt
Point 4:
l-ti : r'rr. = 4Li-J
tr = 2538"tr'
=(boo)m"' = 2998"R
97
(h)c= R - 53.34 =0.22t- Btu
\u'f cv = L11 = (zzgfitm = v'o'c'o l6.R"
Qo = rhc" (T, - Tr) = (0.1) (0.2285) (5000 - 983.9)
Qn = sr.zz ntrts
Qp = rhcu (T, - Tn) = (0.1) (0.2285) (590 - 2998)
Qn = -55'03 Btus
W = Qo - Q* = 91.77 - 55.03 ; 36.75 ry
o =W =3!'75=0.4005 ot4A.O1Vo
W'=(36.?5 BtuX60+)'smrn
=52hp
c'= E*=m =o'8444*k
-=*+ =ffi=o'o43e6lce
"* =f= tdi%to =,,(a) Point 2:
' v, 0.0!' "' =T= # = o'003455 m3
T, = Tr"*t't = (805) {ll;t't-t = 6g9 K
tlPz = Pr{ = (101.8) (tt; t'e
= 2blg lipa
Point 3:
Q^ = mc" (T, - Tr)
12.6 = (0.04396) (O.UU)(TB - 689)
Tg = 1028 X
Ps = r,ltJ= (2518) t8rfl = BZbzkpa
Point 4:
n'*t#ftnr
2. The conditions at the beginning of compression in anOtto engine operating on hot-air standard with k ='1.34, are101.3 kPa,0.038 m3 and lz'C.The clearanceisL0%oand 12.6hIare added per cycle. Determine (a) V' T*P* T3, Ps, Tn atd p.'(b) W, (c) e, and (d) p-.
Solution
1.t4.1
P, = 101.3 kPaV, = 0'038 mg
Ti=32"C +273 =306
t =t{W"'=r&l'],r*r{*J
n, =n,ffi:r,91]ruzuaftl'
=455K
h.
= 16l kPa
(b) Qn = mc" (T1- T1) = (0'04396) (0'8444) (305 - 455)
Q* = -5'57 kJ
W = Qn * Qn = L2'6-5'5? = ?'03 kJ
, \ - W - 7.99-= 0.558 or 55.87o(c) e=q= 12S-
(d) p. =#" = #T,=
out of the cYlinder'
Compression-Ignition or Diesel Engine
ComF'trlon Sftok' ?ow'r Stlol'
Fig. 15. Four-stroke Cycle Diesel Engine
A cycle begins with the intake stroke when the piston moves
downanddraws"ilffi iil;t;*:: j*:t:-::f".1fit:11
burns explosrvery' rra$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 kPao55s - oso3455
Crh!urt Sitol'
ComF.trlon Sftok'ln|!l. Sl.ok.
loo
(")
Fig.
(b)
16. Air-standard Diesel CYcle
1-2: isentropic comPression2-3: constant-pressure addition of heat3-4: isentropic expansion4-1: 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
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-lT" lv, Iq =Lv^,l
T, = Tr"* k-l (5)
ft={;
Tn=Trrnk-l H
. 1 f-t"*-rle=r-,r-rlq:11l
L r02
t=F;-'=m-'=*'
Tr = Trr"k (7)
Substituting equations (5), (6), and (7) in equation (4)'
T.t"ni.--e=1-m\f-'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
trFTgreater than 1, because r" is always greater than l. Thus, lirr rrparticularcompression 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 actualDiesel engine with a compression ratio of lb is mo"e efficierrtthan 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 anafor 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 = 13.5L = 1.84p, = 14 PsiaTr=140+460=600'Ry, =lcuft
Io;l
53.34 =OrOtUffi(078) (1.34 - 1)
cn = kc" = (1 .34) (0'2016) = 0'2702 ffi"p,V, _ (14) (144]jp
= o.68o rb* = alf = (b&lr+,1 (buu)
(a) Point 2:
v, 1
V, =# =1fS = 0'0741 ft3x
T, = Tr#-1 = (600) (13.5)1 31-t = 1454oR
tz = 994oF
pz = prrr.k = (14) (13.5I'34 = 457.9 psia
Point 3:
% = V, + 0:06VD = % + 0.06 (Vl -V2)
% = 0.0741 + (0.06) (1 - 0.0?41) = 0.1'297 ftc
- r\il 0.L297r, = Trl_C = G454) i,^g?A
t, = 2085'F
Point 4:
rn = r, l_sf''LvrI
tr = 811oF
= (2545) lli2gfl '''n-' = 12?1"R
L1J
Rc, =FIf =
= 2545"R
r-v-r r oo., IgJZgZl'''no. = n,lt'J = (45't.e) [-T.] = 29.7 psia = 0.2143 mg
10s
(e) w_ft''lnin-l
i ..
(b) QA = DCo (T3 - Tr) = (0.063) (0.2702) (2545 - iaga)
Qe = 18.57 Btu
Qn = mc" (T, - T.) = (0.063) (0.2016) (600 -72i:l1r)
Qn = 8.52 Btu
(c) W= QA- Qn = 18.57 -8.52= 10.05 Btu
(d) e = W = f0.05 = 0.54L2 or 54.L2Eo
a^ 18.57
P- = (10.05) (778) = 58.64 psi
min.hp
2. There are supplied 317 kJ/cycle to an ideal Dieselengine operating on227 g air: p, = 9?.91 kPa, t, = 48.9oC. At theend ofcompression, pz = 3930 kPa. Deteruineia) ro, (b) c, (c) r",(d) W, (e) e, and (f) p-.
Solutionm = 0.227 kg.P, = 97.91 kPaTr = 48.9 + 273 = 321.g KPz = 3930 kPaQo= gf7 kJ/cycle
\------(\\
\
l)oint 1:
mRT.v --r'l- ll.:l
(l -.0:,0741) (144)
[""ir*f fo*42.4 lltu '= 287 hp
* (0.227) (0.28708) (32r.e)97.9r
4
I
ry-
Point 2:
lo;l+'IJil
,,=*b{'=(rrrr)
(a)
(b)
(c)
" =vr=0.2143_14'* -V--o.oi^re -'1+cf,=--*c
1r 1+cI4t =- c
c = 0.0769 or 7.69Vo
v^ 0.0383 tf = -!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 mgv, = 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 kIW = Qo - QR = 317 - fg6.g = lg0.l kJ
(e) e = P= lao.t = 0.b6g1 or 56.glvoQA 317
1fl P- =g= =.w = l0o.l _= 9ob kpavD vr_% o-zr+s:00rog
DuaI Combustion Engine
In modern compression ignition engines the pressure is notconstant during the.combristio" p"o"ess but varies in themanners illustrated in the ng"*.-ili;*J il ffi* ol" *combustion can be conside*dt";il;ach a constant-vorumeprocess, and the late burning, u *;rilunt-pressure process.
Fig. tZ. Air_Standard Dual Cycle
l-2: isentropic compression2-B: constant_volume addition of heat3-4: constant-pressure addition of heat4-b: isentroplc expansion5-1: constant-volume rejection of heat
Analysis of Dual Combustion Cycle
Qo = mc, (T, - Tr) + mcp (T. _ fr)
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 volumeo P, ' poii"" of co-U"stio"
vrr =titr, the compression ratio
,2
\rr. =#, the cutoffratio' Y3
Th'b thernal,efficiency of this cycle lies between that of theideal Otto qnd the ideal Diesel.
Derivation of the formula for e
Proccss 1-2:
- -k-1T" lv,lq=LrJ /
T" = Trr*I'r
Process 2-3:
t=#="T, = Trrrk-t rn (10)
r0g
Procesg B-4:
tn \/4a v til= 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#" ""a99:F.. The compre*io.l :.ilI i il"- p"*rru* at the end of tlreconstant 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
Solution
Point 1:
m = llbairp., = 14.1 psiaT, = 80+460=540oRpa = 470 psiark= 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-' = 1300RL'rJ
l-v,l*l, = n,l_if = (14.1) (9) 1'4 = 305.6 psia
Point 3:
u,=-3l'-=%#ffi#=la186rt3
Tr=T,[pJLF;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+fil0
= 1.905 ftg
I'4A',-/i
'/ -"",2'ill
*'
Point 5:
t, = t l+ln.'= (rnru) E&1" = 1082"R
L_'I-J
(a) r^ =g= +!y = L.54P 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 airP, = 96.53 kPaT, = 43.3 + 273 = 816.3 Krp = l'5r" = 1'60rr = ll
W = Qr - Q* = 474-L95.7 = 278'3 kJ(a)- -1+crk-
c
1+c11 =-;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 m3vr =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.Pap, = 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-lrn = t'L+l= (1238'1) (1"6) = Le81 K
- I-vln', = (1e81) Bm''n-' = e16.2 K,, = r.LirJ
l-m-l .6 kpapu = 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
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
DischargeValve
IntakeValye
Di5charge
Compressbn
vl 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,lva ves being closed, and the gas is compressed along the curvet-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 asl'lrose with no clearance, except that since trre piston J* ,rotlirrce.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
Preferred Compression Curves
The work necessary todrive the compresor decreasesas the value of n decreases.Polytropic compression andvalues of n less than k arebrought about by circulatingcooling water.
Comparison of work forIsothermal and for IsentropicCompression.
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 deliversit 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 K105 kPa630 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'630105,
-l
T_
ffi
f- r-r
I rp,t-Llp;jE#
= (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, II-'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'nF'ql
l-p] #LEJ
t.67-
y1 rszI
I
I
c- l'oq = 0.6168
= rycv =f = L.6z kg.k"
c = c [-t -rr-l= 0.6168 I r.oz - r.el = 4.41G8 kIn "Lr*l [-T:T-.a J [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"[*-lLP?-i
=(105)(6)rn tffi= - 1129 kJ/min
2. A centrifugal comprcssor handles 300 crr ft per ninuteof 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_rr,=r, L*J'=(E4o) L#t= =64r.eeR
co = c" k-l =(o.lzr4) fILls v U-qJ - ' Ll-tful =-o.oaze ffiAH = 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+Ww = Q-aK-aH
= - 96.4 _ tZ.Z_ bBg.B
= - fl47.gBtu/min or _ lb.2g hp
Volunetric Effidiency
Conventional volumetric effciency = ffin,=$=kX"VDVD
Displacement volume Vo is the volume swept by the face ofl,he piston in one etroke.
ffi
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 pistonL = length of strokeN = number of cycle completed per minuteN = (n) (1) (number of cylinders), for
single'acting compressorsN = (n) (2) (number of cylinders), for
double-acting compressorsn = 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 inone revolution.
Fie. 20. Single-acting Compressor
Connecting rod
then,D"=1+c-c [+-]tLP'J
, Pision rinRs
7l ,''"on.
-J/
Wrist pin'
nk pin
,-- Crank
! Crankshaft
Crosshead
Crosshead guard
Lt 722Fig. 21. Double-acting Compressor
(b)n, =1+c*c
1
Free AirFree air is air at normal atmospheric conditions irr nparticular 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"cUo !Z! ̂ H*. ggrypression.ana urp""Jio" .r" p"fyt""pil .itf,n = 1.30. Find (a) the work, (b) the hialre5ected, and (c) the boreand stroke for I"b0 rpm and UD = f .gO.
Solution
V;PrP2Tr
= 20 m3/min.= 100 kPa= 725 Wa= 37+273=Bl0K
e=\Von = lbO rpmIID = 1.39
(a) W =T#[A* -_l
l-pJ F
l!-,J
l2:l
= -5023 Ymrn
PVt's - "
t
= 1 + o.ob - (o.ob) lzzq-l fi
"'Llo0l
= 0.9205
n.' 20 = z+.ss 4vo=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 = $.7442)Fffi#:l = 4'4b'#
6r-, = rhrc" (T, - Tr)
= (27.83) (-0.2456) (489.7 - 310)
= {ZZs Ymrn
(c) vo ={nrlN =tD',(1.3 D) (r50) (2) (2)- 612.6 O'#
24.38= 612.6 D3
D = 0.3414 m or 34.14 cm
L = (1.30) (34.14) = 44.38 cm
2:. A single.acting air compressor operates at- 150 rpm withinitial condilion of air at 97.9 kPa and 27"c and discharges the
air at 3?9 kPa to a cylindrical tank. The bore and stroke are 355
t24
I
p-T 3ld 381 mm, respectivery with a percentage cre'r'rrr.o 'f5?o, rf su'oundins air ar* it r00 kFa ""a
zi-.c *hJio tt,,,compression and expansion processes are pVr.s _ C. Dutor,r,,,,u(a) Freg air capacity in mtZs. iU) power of the **pr"rro" i" f, W(ME Board hoblem - Oct. 19S6)
SolutianP, = 100 kPaT =293K
(a) n" = 1 + c-. [#J* = I + 0.0b-(0.0b)
m]*=0.e0e4vD =-tDpLN =f, {o.essft0.Bsl) (r50) = s.osz -'
V;= (n,) (Vo) = (0.9094) (b.6bz) = 5.r+a #o. = vr
F,]hl = (b r44) t+rtffi = 4.els#or 0.082$
(b)w =T#'tre,J*:,]
= (1.3) (97.9) (b.
C= f%o-T P=$SSmm+. L = 381mm
It n=150rpm
Pr = 97.9 kPaTr=300K
;
trill
;1- 1.3
t26
= - 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 is0.25 cu ft per cycle measured at discharge pressure. If thecompression is isentropic, frnd (a) piston displacement percycle, 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,3lcYcleT, = 85+460=545oR
= 900"R
KJmrn
(a)r,=r,H* =(545) [q*
l-ar lfih47l
[r;ttLP'-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
IW=" - -IF-to,/
_1J = '?i#iffifiea- [ia,z/ 'l
= 96hp
4' A single-acting compressor has a volumetic effici'rt,yof 87vo and operates at 500ipm. Il trk"r in air at 100 kpa nrrrll[CA! esc\argel 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"riliilalSolution
= 100 kPa= 600kPa= 6 ms/min= 3O+273=B0BK
= 21.58 m3/min
Pr
fzV2Tr
_Jlooo l'''Lrooj
ra) vi =o,Fno = (6)
v^ =&=?rs" q, o.87
24.8 T'_ mrn
UOO stlgkesmln
= 24.8 Itmrn
= 0.M9G ,-Lstroke
It",,7
(h) w= ++lZ+r+ -r-k l\p,/
_@ffi@Kml*_!= -sosa.g Ymln
n =_li{_= bob3.g
= 208.8 kparn 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
qTr
14.7 psia90.3 + L4.7 = 105 psia500 ft3/min70+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 rlgNew 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 airas in (a)
vD at 6000 ft = displacement volume to handle the samemass 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
Compressor EfficiencY
ideal workIn 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 systemis
indicated work of compressor"-'- indicated work of driving engine
B. Compression. EfEciencY
Adiabatic compression effieiency is
adiabatic ideal workS--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)
130I it I
Adiabatic overall efficiency is
,, .. = adiabAtic ideal workoc%
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.,tdmean adiabatic compressi; "ffi";;;
Problems
1. A twocylinl":f:gl:__actils air compressor is direcilycoupled to an electric motor *rrrririg at 1000 rpm.Other data are as follows:Size of each cylinder, lbO mm x 200 mmClearance
"?\-9, f OZ.of JirpfacementExponent (n) for both comp.e5ri""
""J *-expansionprocess, 1.6Airconstant,k= t.{Air molecular mass, 29
J
no, = (n-) (n") = Sltpolvtmpic ideal worli
Calculate:(a) The volume rate of air delivery in terms of standard
air for a delivery pressure of 8 times ambient pressureunder ambient conditions of 300 K and 1 bar.
(b) Shaft power required if the mechanical efficiency is81%. (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= gPr
I
tr, = I * . -.pf = 1 + 0.10 - (0.10X8)t = 0.?332Lru
[t,*-t]
m3S
(1.6) (100) (0.0864)1-1.6
27.2L
= 27.ZlkW
Shaft power = ffi = 33.59 kW
(53.34) (545) = 2o.ss lPmln
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 tfruvolume 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-fficiencyis 87Vo, and (d) the ideal *ort .
Solution
P" = 14.7 psiaT = 82"F+460=542"RPr = 14.5 psiaTr=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 -LlP, I'l&i
vD =4'-D'?LN = t H' frq (1b0x2) = 274.e crm
. P,V,m, = 1i1; =
l:t:r
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=@l\r+si ]= - 1157 Blu or - 27 -29 hP
mln
.brk4{&fiq* - rlr-K L\pr/ -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 oporetlnfat L00 rpm: These data apply: Fr = 101.9b kpa, t, = Z$.ZA inEp,'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 tfddriving motor for an overall adiabatic elficiency of ittir.-
Solution
D'=fr'=Pr=Pz=Tr=
L = 0.3556 m8.48 kg/min101.35 kPa310.27 kPa26.7 + 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
I-gro.2?l#0.9227=1+c-clLl0L5il-
c = 0.0573 or 5.737o
r rrt3V, = Vo (1 + c) = (?.063) (1 + 0.0573) = 7.468 -*mrn
,- p,v, (101.35\ tn aRR\ kg'l',=ffi=idffiffi=e.717 ;ffQ,-, = 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
Tl\lolsb/ -:l
= -846.1 Y o" -14.1 kWmln
(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 kWDraKe wofK = 0J1
work input by the driving motor = 20.41 hW
Multistage Compression
Multistagingis simply the compression of the gas in two ormore cylinders in place of a singffitinaer como"Jrro". l, iuusedin 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 cardsof 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-2occurs and the gas is $yharc.ei "*ir-". The discharged gaspasses through the interc*te" ".rd"is
cooled by circulatingwater through the interc*t." i"U"r. Co"uu"tio'Jfi,"it i,
t:f7
rvater in water out
assumed that the gas leaving the intercool:l el entering the
rrpcvrindeTiu.ir,?,u-g*g;;^iil:;tt*mi*""1i$
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]*-trItis 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'ip,= yTF*'-
where: P, = intermediate pressure for minimum work
since the work of eachcvlila"iillh" sane' tlre t?la\work
for the two-stage #;;;tJtwice the workin each cvlinder' or
2nm'Rr,f-1P,$ ;1 ='+Pfel* -1\w= "iffiLft,? _J - 1-n l9'/ rA pressure drop in the intercooler could be spread on each
"ide oi this ideal value'
I
i,i
I
Pressure droPPr=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=rTrl -
11.33 m3/min103.42 kPa827.36 kPa26.7 + 273 = 299.7 K
139
r =IilFR)* l_(1.4) (108.42) (i,l.BBi lTga.BqtY/ -il1-1.4 - 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 *tmffi -'"'"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;lL\ 1o&42l - |1-1.4
- 1416 # o" -28.6 kWmln
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 rn3vnur =;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 gnpsia'. $e lormal barometer is 29. g in. Hg and the tempern t rr rois 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., thospeed is 210 rpm and pVt.er = C during compregeion undexpansion. The clearance is E% for both cylinders. Ths tem-perature of the cooling water increase by iA F". Find (a) thevolume offree air, (b) tlie discharge pressure ofthe low pr*rruro
t4l
cylinder for minimum work, (c) the tempprature at dischargefrom both low pressure and high pressure cylinders, (d) themass of cooling water to be circulated about each cylinder andthrough the.intercooler, (e) the work, and (f) if, for the lowpressure cylinder, IJD = 0.68 and if both cylinders have thesam: stroke, what should be the cylinder dimentions?
Solution
(d) c, = *-ffi = (0.1?r4)Htf = -{.0302 dhi
Low pressure cylinder
r-*D" = I + c-clfil = I + 0.0b{0.0b)
vn =*= ;€g = rB98 cfm' ru 0.9173
V, = VD (1 + c) = (1393) (1 + 0.0b) = L46Z cfm
;, =$1f= tra,'f*ggxpzr = 1oB rb/min
Q"z = frrc" (T, - Tr) = (10S) (-0.0302 ) (767 - b50)
= -675 Btu/min
Heat to water = Heat from air
(rh*) (c,") (At*) = er.z
678 Bturh*=----414-- =37.5l f Btu\ (18F")
\6F/High pressure cylinder
.ll"a* = BZ.E #Intercooler
Q," = rir,co (\ - Tr)
= (90) (0.24) (bbo - 767) =+osz Bltmln
Fzslt[a s]
= 0.9178
mpo
To
PrTrPr
90lb/min(29.8) (0.491) = 14.63 Psia80+460=540oRL4.3 psia90+460=550oR185 psia
(a) Vr= 6'RTr _ (90) (bB.B4) (bb0) = 1Zg2 cfm, Pr (143t(144) -'
v" = ffi|$) = tfffi#ffi#P
= 12Bo crm
(b) p- = ilFm, =J043) (185t= 51.4 psia
pz= 5!.4+&= 52.9 psia
(c) ps = 51.4 -9= +g.gpsia
,n - ,rr l-p, | +
- ,,trR^\ [Bzd #' - ,t, = t,b:l = (550) Lffi]
''"" = 767 oR
r, = r, [*{* = (bbo) t#.f
'#' = 7G7'u,
lbmrn
l4lj
l" Vi*:--l
r42
mass ofcooling wate" = {y
= 260.4
(e) Low pressure cylinder
\i/. = nrh'RT, l7gt+ _ il"LP - l-n L\prl ]
(1.34) (e0) (53.34) (550) 1-t52.effi '1=@l\ra.si -jl
= - b2G5 B!t'
= -L24.2hpmrn
Total work, fr = (2) (-124.2) = *248.4 hp
(0 Low pressure cylinder
y^ =3.D2LN =!pe (0.68 D) (210) (2) = 224.3 D3 cfmD44
224.3 D3 = 1398
D = 1.84 ft or 22.08 in.
L = (1.84) (0.68) = L.25 ft or 15.01 in.
High pressure cylinde
v _ fi'Rr3 _ (eg) !5giq1)
(?50) = 36?.4 cfm
": p, (49.9) (144)
\r -i; - gal'D - n-, Ufr* = 400'5 cfm
V^ =ID2LN =3D2 (t.zb) (210) (z) = 4Lz.g D2 cfmu44
lbmin
4L2.3 D2 = 400.5
D = 0.986 ft or 11.88 in.
L = 15.01 in.
Three-Stage Compression
LP cylinder IP cylinder
Fig. 24. Three-Stage Compression
pV=CpV"=C
py Condltlone for nlnlnum rorkPV"=C 1) wr,p = wrp - %p
e)TS =T3 =Tl2-P,
I
(1)
(2)
-PV" = C
Fig. 25. conventiorrut cu"arlThree-stage, No pressure Dr'p
,,1'T,r, If+l+- il =,,p'or. f&\.'-l = nm'Rro Zluf+- 11l-n l\Pr/ l=-I-n-[\d/ --1*-L\&i
-l
Pr P" Pn
P, =F, = P,
IP, = (PrPr) 2
I
P, = (P,Po) 2
Solving equations (1) and (2) simultaneously,
p,=\/ir'p, and p, =t6trJ
3nm'Rr, [gf#-il1-n l\P'r lProblem
Air is compressed from 103.4 kPa and 32"C to 4136 kPa bya three-stage compresor with value of n = 1.32. Determine (a)
the work per kg of air and (b) the heat rejected in the intercool-ers.
Solution
p
lke103.4 kPa4136 kPa32"C + 273 = 305 K
mPrPoTr
(a) p, = (p,,pu)*= fioa.aX (4136j#= 353.6 kPa
,., _ 3nm'RT, 7&.* -1
vY- l-n l\P,)"-1.J
(3) (1.32) (1) (0.28708) (305)- 1-1.31
= - 376.2 kJ
L.IZJl/353.6\ r'32-11It- | r I
l]103.4/ _1
ft)T3-Tr=BgbK
n-l
T = T /&\ -- "no
/3ss'olff/'z = rrgf/ =,ruc (ib3;) = 411 K
Heat rejected in the first intercooler,
Qrc= m'co (\ - Tr)
= (1) (1.0062) (305 - 4rr) = -106.2 kITotal heat rejectred = (Z) (_t06.7) = _218.4 kJ
t47
IT il
',t
i
Review Problems
1. A reciprocating compressor handles 1000 cfm of airmeasured at intake where P, = 14 psia and t, = 80"F. The
discharge pressure is 84 psia. Cdlculate the workifthe process
of compression is (a) isothermal, (b) polytropic with n * L.25,
and (c) isentropic.Ans. (a) -109.5 hp; (b) -131.7 hp; (c) - 143 hp
2. A twin-cylinder, double-acting, compressor with a clear-ance of \Vo draws in oxygen at 450 kPa, 17"C and discharges itat 1800 kPa. The mass flow rate is 20 kg/min, compression andexpansion are polytropic with a = 1.25. Find (a) the work, (b) theheat transferred, and (c) the bore and stroke for 100 rpm andllD = 1.20.
Ans. (a) -40.23 kW;(b) -829 kJ/min/ (c) 2L.71x25'76cm
3. A double-acting compressor with c = 7Vo draws 40 lb perminute of air atl4.7 psia and 80"F and discharges it at 90 psia.
Compression and expansion are polytropic with n = 1.28. Find(a) the work, (b) the heat rejected, and (c) the bore and strokefor 90 rpm and UID = L.25.
Ans. (a) 77.68 hp;(b) -1057 Btu/min; (c) 18.96 x23.70in'
4. A 14 x L2-in., single'cylinder, double-acting air com-pressor wit}'5.5Vo clearance operates atL25 rpm. The suctionpressure and temperature arc14 psia and t00oF, respectively.The discharge pressure is 42 psia. Compression and expansionprocesses are polytropic, with n - 1.30. Determine (a) thevolumetric effrciency, (b) the mass and volume at suctionconditions handled each minute, (c) the work, (d) the heatrejected, (e) the indicated air. hp developed if the polytropiccompression efficiency is 75Vo, and (f) the compression effr-ciency.
Ans. (a)92.7Vo;(b) 247.8 cfm,L6.72lblmin; (c) -18.93hp; (d) -175.7 Btu/min; (e) -25.24 hp; (f) 77.42Vo
5. From a testjf an.air compressor driven direcily by art"uq engine, the following data and resurts ** o-ut"i,r.,a,capacity, 800 cfm; suction it t+.2 psia; disch;d;;; iio pri,,;indicated work of lhe compressor,'i5S frp; indicated work ol.steam engine, IZ2 hp^aCal..rlute (u) tt u.";p;;i""im.i"n.yand (b) the overall efficiency.
Ans. (a) 90,06Vo; (b) Bt.t6qo
6. An air compressor with a clearance of 4Vo compresses14.73 ms/min of airfrom gz kpa, z7ic to 462r<pa.If the overailadiabatic efficiency is 6rvo, d"t"r-i.r" the indicated horse-power of the directly connected driving steam engine.Ans. 91.89 hp
7. Methane is compressed in a two-stage, double_actingcompressor which is electricaily driven at rbb rpm. The rowpressure cylinder (3_0. E x Bb, b cm) receive, O. S6 ""
; pe r-mirruteof air at 96.b3 kpa,4B.B"C, *Jtfr" hish;;;iJ.r]ioa""!20..3 x 35.5 cm) discharges til" -"th* e at 7t7.06 kpa. Theisothermal overall efficiencyi szq,%-.inanu and the kwoutputof the raotor.
" Ans. 8O.02Vo,90.g6Vo
^^ ,9: A tw-ostage compressor with a clearance of *Voreceives80lumin of air at 14 psia and 8E"F and dcrivers ii "i1io
pri".The comp,ressions Td-polyt*pi;;th " _ 1.g0, and the inter_cooler cools the air Uact to ar"i.. ri"Jfrl the worL, (bj li" rruuttransferred in the various processes, i.ith" ;;;il,f#;^.irer"_
gtage m achine, (d) the correspondiog percentage s avin g for thetwo-stage_machine, and (f) tle -asiif water to be circulatedthrough the intercooler if its t"*p"i"l.rre rise is 15 F".Ans. (a)-17_1.0 hp; (b) -Soz.S Bru/min; l.l _igie stu/min; (d) _196 hp; (e) t2.4EVo; (ft igo lb/mi;
l,l!f
;
I
I
ft
I'
+
i,i'i
l4 f]
8 The Brayton Cycle
Operation of a Simple Gas T\rrbine power plant
Combustor
Compressor Turbine
ToGenerator'*'/
lr?:
fi:-r-:i-::::i::a1.,:".:':::Sinki.,' F------ J
r...: r:: i : :r't..i: ... .:l
Open Cycle
Q*Closed Cycle
Fig. 26 Diagrammatic Layout of Gas Turbine Units
Air continuously enters the compressor 1. After compres-sion, it enters the combustors, som'e of it going u"o,rrra tfruoutside of the comhrrstion chamber proper and the remainderfulnish]1* oxygen for burning the fieljwhich ir-.orrti*orrrfvinjected into the combustioniha-ber. Because of their tem-ryTlure rise, the gases expand and enter the turbine in state3' After expansion through the turbine, the exhaust t. ilrt:atmosphere is in some condition 4. In an ordinary powor'r:r.t.arrangement, the work of the turbine W, is g"*i ,,,r,,,,gt, t,.,drive the compressor W
_and delive. U.ut ,, *,rrL W,, i,,',t,.iu,,,say, a generator or proptlllrlr; W, __ W,, I W,. Arr ,,*..il,,,,l H()r(.(,of power is needed to si lrrt :r liirH l.rrr.l'r'rrrr. rrrril..
lIl
Fig.27. ,Air-standard Brayton (Joule) cycle
. L-2 isentroPic conPression2-3: constant-pressure addition of heat
3-4: isentroPicexPansion4-tr; constant-pressure rejectionofheat
Analysis of the BraYton CYcle
Qo = mco (Tr - TJ
Q* - DCo (Tr - T4) = *nrco (T4 - Tr)
S = Q^ - Q* = mco'(Ts - TJ - mco (Tn - T,)
e = W = mc.(Tr-Tr)-mco(Tn-Tr)q@e = 1_ +,-+ (1)
rg - t2
11e-1- - =1-t-rL
"o*
vwhere rk = +, the comPression ratio
r = -P, I ttre pressure ratioppr
Derivation of the formula for e
Process 1-2:
T =H"=FJTT2 = Tr"an.t
rok-t = ro Y(2)
(3)
l-pLFt=Fl*=
1.:=r r-r
f
L-lk
Ta = Tn"*tt (4).
Substituttuig equations (2) and (4) in (1).
€ =1-
I1-#"J
Total compressor work, W"=& -AHW.= -mco(T,
W, ={-AHW, - -mc, (T. - \)W, = mco (T, - T,)
Total turbine work,
W--_]..,
Net work, W or W" = W, - W"
Problems:
1. The intake of the compressor of an air-standard Bray-
ton cycle is 40,000;;;it;sia and 90oF' The compression
ratio, rr = 5 andth-;;;;;'i""3t11" turbine inletis 1440"F'
The exit pressure oiiftJi""tine is 15 psia Dgterminl ll: ""';;;;,;#;al efficiencv and the mean effective pressure'
v1PrT1T3p4
r.K
= 40,000 cfm
= L5 Psia= 550'R= 1900'R= 15 psia
= S =5v2
n =*,t =
v, 4o.ooo--=- IiI 2945
vl 13.58= -= = --=-rkD
(15x144X40,000) _--T5rilx550t 2945lb/minf
Point 1:
v1
Point 2:
v2
Pz
T2
I ll4
= Prf**-t =
= Trr*k-r =
= 13.58 fta^b
= 2.72fbs/,b
(15X5)" = 142'8 Psia
(550X5)1:a 1= 1047"R
Point 3:
Vg = v2 tt.] = e.7z) f+#fl = 4.s4 rtsnbL
Point 4:
t--* - I
v=v l&l [email protected])ll+Z8lt"= z4.Tfttnb.4 ,.LpJ \^rvrlLlbI
f-,, -lt -l r ^r.= r, i+l- = (leoo) l#l'o'= eesoRL'rJ 124:Ll
Compressor work, W = -{o (Tz - Tr)
= - (0.24) (L047 - 550) = - 119.8 Btu/lb
Turbine work, W, - co (T, - T.)
= (0.24)(1900 - 99S) = 216.b Btu/lb
Net work, WB = W, - W" = 2L6.5 - 119.8 = g7.2 Btu/lb
= W_Q945) = 6751 hp42.4
Heat added, Q = c, (T, - Tr)
= (0.24) (1900 -1047) = 2A4.7 Btu/lb
= + = V2- = 0.4748 or 47.48?oQo 204.7
= g = =-lu- = (97.2) (778) _vD v, - % 6+7 _ zlz) tt,t,n
= 23.89 psi
p.
2' There are required 2288 kwnet from '11:rs
r,rrr'rrrr* rrrrrllirr'prrmpi.g of crude oil from thc Nrrth Arrrrrkrrrr ,.rr,1*, .i*'|'irt'(!r'$ thc compressor scction at gg"n?l-r kPr, ltzH ti, rrr* lr*ee
IFE