Flo-=6;T*I
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a
THEN[tl!ODYl\lA[tl|IOS
HIPOLITO B. STA. MARIA
COIVTENTSPreface
vii
Chapter 1 Basic Principles, Concepts and Defrnitions
I
cific Weight, Pressule, Conservation of Mass.
Mass, Werght, Specilc Volume and Density; Spe-
2
Conservation of
Energy
Zg
Potential E_1ergy, Kiletic Energy, Internal Energy, $eat, Work,
Flow Work, Enthalpy, General EnergT Equation.
3 , The Ideal Gas 87Constant, Specific Heats of an tddal
Gas.
4 5 67
Processes of
Ideal Gas
5fdo""sr.
Process, Isentropic Process, polytropic Gas
Isometric Process, Isobaric process, Isothermal
Cycles 81
Camot Cycle, Three-process Cycle.
Internal Combustion Engines ggOtto Cycle, Diesel Cycle, Dual
Combustion Cycle.Gas Compressors
ll5
Single-Stage Con pression, Twestage Compression, Three-Stage
Compression.
8
Brayton Cycle 16l
PREEACEThe purpose of this text is to present a simple yet
rigorous approach to the fundamentals of thermodynamics. The author
expects to help the engineering students in such a way that
learning would be easy and effective, and praetical enough for
workshop practice and understanding.
Chapters 1 and 2 present the development of the first la'ar of
thermodynamics, and energy analysis of ope:r systemsJhapters 3 and
4 give a presentatign of equation of state and involvingideal
gases. The second law of thermodynamics andits applications to
different thermodynamic cycles are discussed in Chapters 5 and 6.
Chapter ? deals with gas compressors andits operation. Chapter 8
develops the Brayton eycle which can be omitted if sufficient time
is not available.;he process
The author is grateful for the comments and suggestions received
from his colleagues at the University of Santo Tomas, Faculty of
Engineering.
The Author
vll
1 I
Basic Ppq"iples, Concepts and Definitions
Thermodynamics is that branch of the physical sciences that
treats of various phenomena of energ-Jr and the related properties
ofmatter, especially of the laws of transformation of heat into
other forrns of energy and vice versa.
Systems of Unitsbody is directly proportional to the
resultantforce acting on and inversely proportional to its
mass.o
Newton's law states that 'the aceeleration of a particular
it
"-
hE, F= D8, m k
k =+F
k is a proportionality constantSystenns of units where k is
unity but not dimensionless: cgs system: I dyne forcre accelerates
1 g mass at1 cm,/s2m./sz
mks system: 1 newton force accelerates
I
I
kg mass at
fps system: 1 lb force accelerates 1 slug mass at
l
Nsz
l--t;]*ldyne I t -* i*l newton [T,,'*-l-'r,0" /777r/7mrV
/7furm,h n77v77v?rrvr 1 cm./s2 _+ 1m/s2
1&,/szt=r,4'cm-cyne.s"
o=t#;p
k=rw
Systems of units where k is not unity:
47If the same word is used for both mass and force in a given
system, k is neither unity nor dimensionless. 1 Ib force
acceierates a I lb mass at 32.L74 fVs2 1 g force accelerates a I g
mass at 980.66 cm/s2 L kg force accelerates a 1 kg mass at 9.8066
m/s2
r=f,a1 poundal = (1 lb_) (1 fVs2)
F is force in poundals
d7mzm'V /72zv7m77k=
f-.,.-f*
,
,0, l- t ,. l-.
t
u
/7V7v77v77v7
[-t u*.
f-,
nr'
#
tr mass in poundsin ftls2
a is acceleration
32.L74 fVsz----+ 980.66 cm"/s2 -------> 9.8066 mlsz
--'-+
rz.tllthP
k = e80.66-*F k = e.80668#.U
l* /7V7V7mV ',0,L fVs2 --------+
[T**
Relation between kilogram force (kgr) and Newton (N)
=r-8. l(= 1 lb"
m
k=1k#Therefore, t
k = e.8066 Ets"
ks .m
1 pound = (1 slug) (1 fvsz); 1 slug
-lrs2
k#
= e.8066
H#
F is force in pounds
S K
1kg"= 9.8066 NRelation between pound psss (lb-) and slug
is -ass in slugs a is acceleration in
fl;/s2
Mass and lVeightThe 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.
k=1#Therefore,L
k= 32.r74ffi
t*5& = 82.r74ffislug = 32.L74Lb
mFF" k =t=gwhere
AccelerationA unit of force is one that produces unit
acceleration in abody of unit mass.AL or
g a
= acceleration produced by force F* = acceleration produced by
another force F
near the surface of the earth, k and g are numerically
,.r1rr:rl, so are m and
F-
II I
poundal
I
fvs2
:.._l
--)
E
1(Problcms:tion?
l.Whatistheweightofa66-kg-manatstandardcondi-
--'l- J S'= I,l ls-P- =l 0.4e tu.ll+se.o#-l K sBo.b+
lb
.ft
Solution
'L
^"J
= 222.26
g,,,
m=66k9-
I
= 9.8066 m/s2
F"ok Fto.lF' mo=-?-= Bosg_.-
t*tfuflF
"f*J
= 1435.49 g-
?2. The weight of an object is 50 lb. What is its mass at
standard condition?
= (o
ro,r"er
fz.rt- U|nu-r
rtrJ
= 1459.41 g,"
Total
mass
SolutionF, = 5o lbr
= mr + m2 + na + m4 + m5 = 500 + 843.91 +222.26 + 1435.49 +
1459.41 = 446t.07 g^
g= 32.L74ftlsz
(b) Total mass = 446L.0J
g^
= g.EB lb-
* =d-=
FK
Fo
r rb! rf-
lb.rrlft PSo lb_
453.6 9'83
ils]!-o'= 0.306 slug
fztz+14s'j
(t') Total mass
-
32.L74
32.174;ifis
3.Fivemassesinaregionwheretheaceelerationdueto grr"itv i. 30. 5
fVs2 are as follo**t m, is- 500 g of masq rq, y^eighs [oo eim,
weighs 15 poundals; mo weight-g.lli mu is 0'10 slug ;i *]',r. trnuf
iu theiotal mass expressed (a) in grams, 16) in pounds, and (c) in
slugs.Solu,tiong = (30.5 fVsz) (12
that the gravity acceleration at equatorial sea level rr s =
32.088 fpsz and that its variation is - 0.003 fps2 per 1000 (a)
l't, :rscent. Find the height in miles above this point for which
llr:, gravity acceleration becomes 30.504 fps2, (b) the weight of
,r lsivcn man is decreased by \Vo. (c) What is the weight of a 180
I r,,, rn an atop the 29,131-ft, Mt. Everest in Tibet, relative to
this4. NoteI
por
rr
L'?
in/ft)
(2.54 cm/in) = 929'64 cmls2
,\til
tr
tionin acceleration = 30.504 - 32.088=
(;r ) change
*
1.584 fps2
(a) mz =4
F't [roo4frro.uuM=
e2e.64
+
= 843.91
g;
p:; llcight, h = - I lP* fps' = 528,000 ft or 0.003 - -T0008
100 miles
+T(b) F = 0.9b Fg
-tI
.a
Let Fg = weight of the man at sea level
Specifrc Volume, Density and Specifrc WeightThe density p of any
substance is its mass (not weight) per unit volume.
hI
FF= -ag 0.95 F" F" a =g____q
rl=D rvThe specific volume v is the volume of a unit mass.lt
----
a = 0.959 = (0.95) (32.088) = 30.484 fps2
-L 'Fgg = 32.088 fps2t.,
mp
V1
"-
(30.484
- 32'088) fps'z= b34,6z0 ft or tOt.B miles o.oosTS;r _
-TmorrFa
The specificweightTof any substance is the force of gravity on
unit volume. g=
(c)
,vF
8
29.1.31
ft
Since the specific weight is to the local acceleration of
gravity as the density is to the standard acceleration,Tlg= pk,
conversion is easily made;
r_.6 F8
Tk os P='g orY ='fr-
g = 32.088 fps2 m = 1801ba = 32.088 fps'
fTdriil [0'003
rIto
1"1 {}l
-1
At or near the surface of the earth, k and g are numerically
cqual, so are p and yfpsz] = 32'001 fpsz
Problemsr
o
=T-=
ma
tlso lb-l
pz.oor&l
#=179.03 32.174F"1T"
_^ ^^ lbr ,,
1.tion?
What is the specific weight of,water at standard condi.
Stilutiong = 9.8066 m/sz
kg_ P = 1000 n5.
*_pg I- E-
[*,SE**de.8066ffi#
= looo mo
kgF
rydensities (p, = 1500 kg/m3,Pzi^ 500 kg/m3) are poured together
into a 100-L tank, frlling it' If the resulting density of the
mixture is 800 kg/mt, frnd the respective quantities of liquids
used. Also, find the weight of the mixture; Iocal g = 9.675
mps2.
2. Two Iiquids of different
PressureThe standard reference atmospheric pressure is 760 mm Hg
or 29.92 in. Hg at 32"F, or 1"4.696 psia, or 1 atm.
Measuring Pressure
Solutionmass of mixture, mm = pmvm = (800 kg/m3) (0'100 m3) =
80
1.kgI
By using manometers (a) Absolute pressure is greater than
atmospheric pressure.po
mt+m2=mm
PrVt+PrV,=D1500 Vr + 500
q = 80 V, + V, = 0'100Vt = 0'03mg
(r)Q)
I
p = D Po = 'lt p" = ' I p =
solving equations (1) and (2) simultaneously
absolute pressure atmospheric pressure gage pressure, the
pressure due to the liquid column hPo+Pg
(b) Absolute pressure is less than atmospheric pressure
Ve = 0'07 m3
m, = P,Vr = (1500 kg/m3) (0.03 m3) = 45kg
P=Po-P,The gage reading is called vacuum pressum or the
vacuum.
mr= prY2= (500 kglm3) (0.07weight of mixture,
m3) = 35
kg
re-=x"=@
e.8066*#
=?8.esksr
I
ll"y using pressure gages
rilr,,1||llr rt ng gage
A Jrrt:ssure gage is a device for pressure,
'l'lrin picture shows therrr,vr.rn(.1)t, in one type!, I r r
ofpres-
. l::ll{(',p1i
k
nown as the single-
I
rr
lrr.
r13..
lnlrr, llrrrrrrglr t,lrc thrcnded,
'l'hc f'luid enters the
',,rur.r'lrorr. A$ t.hc prOssur:e
Fig. 1 Pressure Gage
I
ry_increases, the tube with an elliptical section tends to
straighten, the end that is nearest the linkage toward the right.
The linkage causes the sector to rotate. The sector engages a small
pinion gear. The index hand moves with the pinion gear. The whole
mechanism is of course enclosed in a case, and a gpaduated dial,
from which the pressure is read, and is placed under the index
hand.
Solution
pr=*#= FuuS ', kg-'4 ' N.sz
["*S
(30 m)
= b48,680 N/mz or b43.6g pps(gage)
(p=po+p")+Pt
Atmospheric Pressure
,=O,P=Po)
A barometer is used to measure atmospheric pressure.
-P,
V
Absolutet Pressure
(p=p"-pr)
(p=0,Pr=P")Gage Pressurepo
--T--ps
I
P=Po+Pg
_ F" 1V yAhPr=*-A-=:6lP, = Tb,
P.=Y\Where ho = the height of column of liquid supportedby
atmospheric pressure {
=ry'=*
Problem
l)roblems
A 30-m vertical column of fluid (density 1878 kg/ms) islocated
where g = 9.65 mps2. Find the pressure at the base of the
column.
1. A vertical column of water will be supported to what lrcight
by standard atmospheric pressure.
IO
Absolute Pressure
SolutionAt standard condition P=Th
\* = 62'4lblfts Po = 14'7 Psi ;-l T ..-rr lu.z *l lt++'#l , p,,
- L----:n!-!_--!t"! = 33.9 ft t'=
yh"-* h = ho * hr, the - absolute pressure p. height of column
of liquid supported byIf the liquid used in the barometer is
mercury, the atmospheric pressure beconoes,
t;
62.4Y -'- ft3
P"
= THshs = (sp S)H, (T*) (h")
Thespecificgravity(*pg')ofasubstanceistheratioofthe spccifrc
weight of the substance to that of water'
trg.ol
Fz.+
H rL'" i',1
1728Hpo = 0.491h"
sps=Tis 9.5 kg/cm2. The}arometric pressure of the atmosphere is
768mm of Hg. Find the absolute p".*r,r"* in the boiler. (ME Board
Problem - Oct' 1987)
^{
l4
2, The pressure of a boiler
where
ho = column of mercury in inches= 0.491 n-
then, psSolutionPg
h
= 9'5
kg/cm3
ho = 768 mm
Hg
and,l)roblems
p
=0.491 hP-= ln."
At standard conditionT* = 1000
kdmt
po
=
(ynr) (h") = (sp gr) nr(T*) (h")
(13.6)
Fooo
S
to.?68 m)
l. A pressure gage regrsters 40 psig in a region where the
l,irrometer is 14.5 psia. Find the absolute pressure in psia, and
'rr kPa._ 1.04
10.000 'm'
c!*
kgcm-E
Srilutionp = 14.5 + 40 = 54.5 psia
= po
* p, = 1.04 + 9.5 = 10.54#
/Tnvrnh
t-t k-+'r newtona=
/vTTvvmmiV
[ , "[-ft,
,0,
I
m./sz
a=1fUs2
l2
1Tlkgn=
-tE KgJ P.
Solution= 0.06853 slug
(a)p =Pr=
Po
* Ps = 14.7 + 80 = 94.7 Psia
ao
Ps]Laf,m
1+
= FS][tr'fl =8.28$F,lbf
Psla r,. t7 --:I':t. |
= S.A4atmospheres
h = 9.92 in. Hg absa = 3.28 Nszlrg = 2o
in.P = 0.491
t = ff
h"= Z9.tilt".= (0.06863 slug)
[.za {l=
o.zzas tb,
$..
ll -1fth'J
h
p
=
(0.491) (9.92) = 4.87 psia
1
newton = 0.2248Ib"1.1b"
= 4.4484 newtons
p8 = 4.7
psi vacuum
(1rb)
rl4 ln'114=
=
F**H lrr.ut;]lnP
ps = (4.7
esi)
r
o"_l
l:8e5;-s!
= 10 psia
=32,407 Ps(gage)
osgs\ mo
(rl)
h =15in.= 375,780 Pa or 375.78 kPa
h = 29.92 + 15 = 44.92 in. Hg abs
Given the barometric pressure of L4.7 psia (2g.g2 in. Hg abs),
make these conversions: (a) 80 psig to psia and to atmosphere, (b)
20 in. Hg vacuum to in. Hg abg and to psia, (c) 10 psia to psi
vacuum and to Pa, (d) 15 in. Hg gage to psia, to torrs, and to pa.
(1 atmosphere = 760 torrs)
2.
P, = 0'491 h,
=[r"H F"!F*'H= 50,780 Pa(gage)15
t4
.lFI'empcraturc
1. Derive th. r.l:rtion between degrees Fahrenheit and degrees
Centigrndo. (FlE Board euestion)100"c
It follows that,1Fo=1Poand
T212.F
*uu1212
tl
T
lc.-1K"2. Show that the specific heat ofa substance in Btu/(lb)
(F") is numerically equal to caV(g)(C").Solution
,r""-
*r". I0".lbb:o
t.F -32 .= t"C-0 _
n
toF =
Io
rt"C + 32
Btu(lb) (r")
toC =
I
5( t.F
-
32)
- Btu =IG'(E Ir-IEXD - cal .,tr ttr.ltltl.e.'l'lr,r.
, Absolute temperature is the temperature measured from absolute
zero. Absolute zero temperature is the temperature at which all
molecular motion ceases. Absolute temperature will be denoted by T,
thusTK=t"C+z71,KelvinTbR =
Conservation of Massmass is inde-
'l'lr. law of conservation of mass states rhat
t.F + 460, degtees Rankine
,'r\ r'n t)y fne lOfmUla
rluantity of fluid passing through a given section is
V=AuIII = i__
Degrees Fahrenheit ("F) and degrees Centigrade ("C) indicate
temperature reading (t). Fahrenheit degrees iFJ) and Centigrade
degress (C") indicate tempertu"" or differ"h"ogu ence (At). 180 Fb
= 100 C"
-: VAu =Aupv v-
Wltcrc V = volume flow rateA = cross sectional area ofthe
stream
1p"-5g"9
l) :, ilvcrage Speedrir ,., m:rss llow rutc
1 C. =!-1l,"o
16
t7
F7--Applying the law of consewation of mass'
\t - - \ArDrpr =
I
=-n;
=Erf,El 4=ff =' *T-
a,E4zftz
I I
T
\rtrPz
2. A 10-ft diameter by 15-ft height vertical tank is receiving
water (p = 62.1 lb/cu ft) at the rate of 300 gpm and is discharging
through a 6-in ID line with a constant speed of 5
I I I
ProblemsTwo gaseous stre?ms enter a combining tube and leave
section: as a single mi*trrr". These data apply at the entrance
-fot 6rr" gur, A'r= 75 in,z, o, = 590 fps,-vt] 10 ft3llb For the
other gas, A, = 59^i1''.:T, = 16'67 }b/s P" = 0.12lb/ftg At exit,
u.. j 350 fPs, v, = 7 ftaAb' Find (a) the speed u, at section 2, i-
'd ft) the flow anii area at the exit section'
1.
:j:rlil"ffJrr;,'frh'iisfilTil;1lo' Irs,I
Solution
\
I l=:-:_-_*--l -l-, I F'--=- -:-1J tiu'e""" =-f, (10)2 = 78.54
ftz
f___ _
_]=
t__
tu'",=il'i,=ffi(b)
=4oorpsr\lirrur lr,,w
rate enreri", =
[ffi]=
[rr
r
fi
= z4so.\
mr
.
= --vr
Aru,
-[.'9!d=2604+ --------r6Tt3=
r\t,r'* tuwrateleavins=Aup=
ibrh, = rh, + rh,18
? Bd'F.uo*J F +ru* S*
= 26.04+ 16'6? = 42'?1+
Mass change = (3658
-
2490.6) (15) = 17,511 lb (decreased)
Review Problems1. What is the mass in grams and the weight in
dynes and in gram-force of 12 oz of salt? Local gis 9.65 m/s2 1 lb-
= 16 oz. Ans. 340.2 g-;328,300 dynes; 334.8 g,
volume ch^nge
=
17'51-l:-!b
62.1#
=
282
ft'
Decrcased in height
=
ffi#-
= 3'59
ftft
Water level after 15 min. = 7.5
3'59 = 3'91
2. A mass of 0"10 slug in space is subjected to an external
vertical force of4 lb. Ifthe local gravity acceleration is g = 30.5
fps2 andiffriction effects are neglected, determine the
acceleration of the mass if the external vertical force is acting
(a) upward and (b) downward Ans. (a) 9.5 fps2; (b) 70.5 fps'?The
mass of a given airplane at sea level (g = 32.1 fps2) is 10 tons.
Find its mass in lb, slugs, and kg and its (gravital.ional) weight
in lb when it is travelling at a 50,000-ft elevation. 'l'he
acceleration of gravity g decreases by 3.33 x 10-6 fpsz for r,rrch
foot of elevation. Ans. 20,0001b-; 627.62 slugs; 19,850lbr
3.
4. A lunar excursion module (LEM) weights 150[r kg, on r.rrrth
where g = 9.75 mps2. What will be its weight on the rrrrrface of
the moon where B. = 1.70 mpsz. On the surface of the ,noon, what
will be the force in kg, and in newtons required to ',,'ttlerate
the module at 10 mps2? Ans. 261.5 kg; 1538.5 kgr; 15,087 Nsystenis
0.311 slug, its density is 30 g is 31.90 fpsz. Find (a) the
specific volume, (b) the and (c) the total volume. "1,,'r'ific
weight, and Ans. (a) 0.0333 ft3Ab; (b) 29.75 lb/ft3; (c) 0.3335
ft3,l-r. The mass of a fluidll,/l'1,:r
A cylindrical drum (2-ft diameter, 3-ft height) is filled *'rllr
:r tluid whose density is 40lb/ft3. Determine (a) the total
,,,lrrrno of fluid, (b) its total mass in pounds and slugs, (c) its
,'1r'r'rlit: volume, and(d) its specific weight where g = 31.90
fps2. Ans. (a) 9.43 ft'; (b) 377.21b; 11.72 slugs; (c) 0.025 ft3l
lb; (d) 39.661b/ft3. A wuathcrman carried an aneroid barometer from
the ! "t, l llrxrr to tris ofl'icc atop the Sears Towcr in Chicago.
Onr
{;.
'i
ir
20
2l
atmosphdric air density was 0.075 lb/ft3 and estimate the height
of the building.
the ground level, the barometer read 30.150 in. F,Ig absolute;
topside it read 28.607 in. Hg absolute. Assume that the average
Ans. 1455 ft
Consenration of Energy
A vacuum gauge mounted on a condenser reads 0.66 m Hg.What is
the absolute pressure in the condenser in kPa when the atmospheric
pressure is 101.3 kPa? Ans. 13.28 kPa
8.
Gravitational Potential Energy (P)The gravitational potential
energ:y of a body is its energy due to its position or
elevation.
Convert the following readings of pressure to kPa absolute,
assuming that the barometer reads 760 mm ltrg: (a) 90 cm Hg gage;
(b) 40 cm Hgvacuum; (c) 100 psiS; (d) 8 in. Hg vpcuum, and (e) 76
in. Hg gage. Ans. (a) 221..24 kPa; (b) 48 kPa; (c) ?90.83 kPa; (d)
74.219 kPa; (e) 358.591 kPa
9.
p=Fsz=ryAP =
A fluid moves in a steady flow manner between two sections in a
flow line. At section 1:A, =10 fLz,Dr= 100 fpm, v, = 4 ft3/lb. At
section 2: Ar- 2ft2, pz = 0.201b/f13. Calculate (a) the mass
flow'rate and (b) the speed at section 2. Ans. (a) 15,000lb/h; (b)
10.42 fps10.
P,
-
P, =
ff@r- zr)
AP = change in potential energy
Datum.plane
in 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
discharges 75 gpm of water whose specifrc weiglit is 61.5 lb/ft3
(g = 31.95 fpsz), frnd (a) the mass flow rate11.
If a pump
Kinetic EnergT (K)Hrls$ed
The energy or stored capacity for performing work pos' by a
moving body, by virtue of its momentum is called kinetic
energy.
K=#nK=4-K,=fttoi-ui)AK = change in kinetic energy
22
23
qTInternal 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
Flow lVork (Wr)Flow work or flow energry is work done in pushing
a fluid across a boundary, usually into or out of uy*L-.
"
mass)
Au = tlz
- ulUr
13orr
nrll
lr'_
;1=Area of Sur.face
lVr=Fi=pALWr=PV
fJ = mu = total internal energy (m mass) AU = Uz
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."
l"ig. 3
FIow Worh"
AW,=Wr,-Wrr=pr%-FrV,AW, = change in llow work
Ideat (e)
Work of a Nonflow SYstemCylinder
---.
Final Position of Piston
The work done as the piston moves from e to f is
dW=F,d*=(pA)dL-pdv
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{-_.,{,.-.
Piston At ea = .zl
'"**F
I
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 SystemsrI
which is the area under the curve e-f on the pV plane.
Therefore, the total work done as the pistonmoves from
lto2is
tr
A
r'lrr.se
'w =JlndvnVFig. 2woRK
'
l,or r ntlaries. .\ r | ( system
d' system is one in which mass does not cross itsis one in which
mass crosses its bounda-
'r,t'n
which is the area under the curve 1-e-f-2.
Cnnservation of Energy:. r.ti, I r r'rtlr.tl
ttttt't/t,St,nlyeCli l,, f u:,1 l;rw ol'l.lrr:r'modynarnics states
:::i:':. , !tttt \. ltt. (..,ttIt('t.l((1. i.n.l.O U.nOthCf.
ot
EXPANSIoN.
|1,, l.riv ol r:orrservation of energy states Lhat energy
ls
The area und.er the curue of the prrcess on the pV plnne
rcpresents the work d'one during a nonflow reuersible process. Work
done by the system is positive (outflow of energy) Work dnne on the
system is negatiue (inflow of energy)24
that one fornt oI
SteadY Flow EnergY Equation
Problems
Characteristics of steady flow system' -
i. There is neither accumulation nor diminution of mass within
the sYstem' 2. There is neitier accumulation nor diminution of
energy within the sYstem 3. The state of"the working substance at
any point'in the system remains constant'
t. During a steady flow process, the pressure of the working
substance drops from 200 to 20 psia, the speed incneases from 200
to 1000 fps, the internal energy ofthe opeh system de. creases 25
Btu/lb, and the specific volume increases ftom I to 8 ftsnb. No
heat is transferred. Sketch an energy diagram. Determine the work
per lb. Is it done on or by the substance? Determine the work in hp
for 10lb per *io. (t hp = 42.4Btu/ min).Solution
Kl Fig. 4 Energy Diagram of a Steady Flow SystemEnergy Entering
System = Energy Leaving SystemW,,
peia p, = 20 psia o, = 200 fps rlr = 1000 fps vc=8 ffnb n
vr=lfts/lbpr = 2002
II,Energy Diagtam,F, +
Au=-25Btu/lb Q=0
P, + K, + Wr, + U, + Q = Pa*
d=l"P+ak+l-wr+aU+W
t-l
Wl"+ U" + W
K, + W' + U,
+
(SteadY Flow Energy Equation)
A,=Pr+2
4
+ W* + U, + W
llrrnis
I lb-
EnthalPY (H, h)fluids Enthalpy is a composite property
applicable to alland is defined bY
lr"3 ]fi,W,,
= o.8o
ffL
h=u+pv and H=mh=U+PVThe steady flow energy equation becomes
,lf = Offiimi=le.e?r+bl',v,
E*'ii,lE-Hl(20) (r44) (8) = 778
+K'+H'+Q-l;..?J*ril*
= sz,o2
Bfi
llr V.l -*
2e.6rff27
26
-T'r-Kr+Wrr=Iq+W,r+Au+W0.8 + 3?.02 = 19.9? + 29.61
(a) Basis+W
f
lb'?n'
-25
K,=S= ,Cffio,,,q =*== 3,12 hp(z',)
=3'20ff!
w = 13.24t-
ff,0t,
(32.174)
(1100)2 = Z+.t7 BJu (778) rb779= 98.lC lb_
w:
L-
lr s24ffi["*il42.4(mi#)hp)
Wr, = PrVr =
(200) (144) (2.65)
--'-- #E
wrz=
PzYz=A+#@=s+'z+ffK, + Wr, + ur + Q- IL + Wo + u, + W
2. Steam is supplied to afully loaded 100-hp turbine bt 200
ftsAb and u'.=^19'0 fp*' priu *itft = 116'bT nl"/lb,"t, ::'1U ft3Ab
and "r Exhaust is at r prl" *ilrt * J ozs Btunb, Y,=-29! -= turbin
is L0(a) the " enersy change and determine glJu. *o"t p"" tU steam
and (b) the steam flnw rate in lb/h'
rioo
il;;ipor""tiur
fps.
tne heat loss from the steam in the
;t.20+ 98.10 + 1163.3 + (-10) =24.L7 + 54.42 + 925 + W
w=(roo(b) Steam flow =
Fl{ 251ff
Solutionp, = 200 psia u, = L163.3 Btunb v, = 2'65 ftsnb u" = 925
Btunb
Eru-l hp) P544lrr)trro)r
r
251 Btu
p,
-l
Psia
vr= 294
fts/l.b
--:t.rr
E;
= 1014
+
u, = 400 fPs
u, = 1100 fps
Q = -10 Btu/lb
W=t00hp
An air compressor (an open system ) receives 272kgper r r l of
air at 99.29 kPa and a specific volume of 0.026 m3/kg. The nr r"
llrws steady through the compressor and is discharged at frrllf l-r
kPa and 0.0051 mslkg"The initial internal enerry of the ,r r rrr |
594 Jlkg; at discharge, the internal energy is 6241ilkg. 'l'lrr.
2'M7
W.=JpdV=05l
,'2
(c) 43.27"C, 545.75 kPa50
(c) The change of internal energy'6{J = rtr'c" (T2
-
Tr)
For reversible nonflow, Wn = 0' For irreversible nonflow, Wo +
0' W = nonflow work !d = steadY flow work
(d) The heat transfened'(Tz - Tr) Q = Itrc'
l': oblemg
(e) The change of enthalPY'
l.TencuftofairatS00psiaand400.Fiscooledtol40"F (b) the *t 4Xrr
.rV
F3: Ea+{ : sQG .ts i$ tistP-v d': d
Ii g{,h
ae a5
tElq)I
Utr
SFo t B/i .\t l\t -Ll td
?
b0
ra .71 Q) a
r, orRc)F,t
he(!
c)
YbFl *d'\o) FVE a\Jrq
.FV +tA
.o2
.+a
-?G \J
9,9
r-r +l F-i lrl------------)
4_ T--+EI
r)p
NN r/q)
Itrorm
FE Hi F?-9 q
) FF (n xeIY c! h; Poc:'$
E TS o ':i$ 5d OXts oS!)
Y;;lrY
'Ji(/J Ttr{rF Ns(.)
E5-E -il -. ll a.C);i.E o
fi,EE$H$3
()
Y>F'
-c)f
\v
t^n
e;A
ililtl '^@@
FFB
tt
ti cE r\d
ii; uiiE f?BEHtbx'ic6*'_d.Fc!o
5gF*tE
HE:*Ei+ E trFH
(J+>F{
0(|)
a HH
E:r
=R t> 5 a
i
B
Of{
"rl -gI
tl
q?l'Hric'$ r F d H t.E* r g g*THE EE H
tr HI q +f; fi;E>gi :=:H.jj ., +) ll (n (E ()(H rY P.
i !g; .i;lr f; fl: Eils $jE,Ll
g i$e
E BE :E H" ff.8 a; E xie ilg -t Frg
S
E
F
-
C!
B
r
Et t;c;
! trEo9l
Aoa
;q.Fti
'.+S1r EDI^t
r-,r
Eg .!e @Yil
I
$E Et* (.l
.e
flt?
d h l^.[ui | .i rn* d 'l v |
EE[' E
d ll
i|
i iilll
rr)* Ff lg_ X llF. E- il | rril
^l-^*
Flo' dl
dl _ tlFiil
o --,dfi r rEi Ei E >1s sis : T rr j c N EF n
irt ,i
E
i
.*!;j 'q i-''- tt rlH- i"*l*Es
:
{Si i F! 'i-'r---J''
Gi ; ;6B
F
8E: E , ,, E tiE E o'* i fi f *
$ H *..-l,.* H E=ld i EE A E E.'E :srr
trlt
E
=
cc
-TB (Ss
-
S. ) = *Tr (S2 - Sr)
Q* = -mRTrt"
we=
Qn
-
Q* = Tr (Sz - Sr) - Ts (S2 - Sr)Ts) (S2
tmRTrh
(Tl
-
S1), arca
L'2-3'4'l
\[ = A^ - a- = mRTrtnt
t
W
- r;s;;r o"= Tr-T,
(Tr
T3) (Sz
Sr)
wg=
(Tt
-
Tr)(Tr
mRl"
+-v
e = ---ErlThe thermalefficiencye is definedas the
fractionoftheheat cycle that is converted into work ; supplied to a
thermodynamic
w a;
-
Ts) mR
ln f vl
mRT.
k ,V,L
g=
Tt-Tt
-T,
Work from the TS Plane Q^
Work from the pV plane.
==
mRTrfn
f1;V.-V3=
W = IW-mRTrln
= Wr_, + Wr-, + Wr-n + Wr-,
Qn
mRTrln
w = p,v,l"
i
t. &+: :J,+ p,v,rnf,.&tJ{.
From process 2-3,
T
T3 l-v, l*-'=Lv'J
Mean Effective Pressure (p_ or mep)
P-=WFrom process 4-1,VD
-lfJ 11 -l-v,J.-'but Tn = Ts and Tr =T2
T,
Vp = displacement volume, the volume swept by the piston rr one
stroke. Mean effective pressure is the average constant pressure
l,ir:rt, acting through one stroke, will do on the piston the net
work of a single cycle.
- -k-r therefore,l V" | =
LqI&
Ratio of Expansion, Ratio of Compr.essionI,)xpansion ratio
then,
% =-vr v,
end of -., the = volumeattheffiili expansiqlft5
vglute,3t
84
giF6aa
tco
*F, !od tr-ro L,+) g H
hH
e*)c\ll
Y
AtsIIIo
HEc{NIFFt :. t
+ C\tl
Vl+ ,:l-{!r lca
6t H ie
:(D ttll
\ ro,I I
d!rtl
o)co
!l+_l-
tr.'ilco c\F---=iJ
L^lFi $ lCA l.^^ir-{ 6l vl lvll
5l$
rIolcD
Cral(O
-
gY)
tll
l.]>"r.++tH i{
EHH '^ ddor
{i-
lltt ,--
c.i
.N
ilc'd
'^
'5
a
f ;q:" do 'c,ttnYrrj t'' o E ,, t' 2 idF 3 O ggOssH
"Y?x?qiq * E 3 3 t's;da= bo? a>.=v E 't9ili d{
+
H
$
+Gt
vl .dl
E E * I,#e H : E I;El'H B a $c r E H e I H. F E ep r H H ; A Fa
I a r g :E E :F fiE i() :G I E q I iE A E E T sg t?< I E LE5t
lE'9oe;voEI
iiH ""1" f .,gifi 'r>1; #" rd ti; ".1t 5 ry .l i]"llgE E i I
:'e
HIH i]8 El+ ?IE ElE.T_
El
rrH
(D:: +)r
o).;
aE.
Br d)+ ir't .e +r 9p Xc| +)C #C ;iq{i
E< ot H- iic oF dfr a,
9F hr
'9;
;
CI
E 6l
qn Src! +>i oi+i(66
a 2'd(
igEfilt ll ll llll
.sL &FrFcl
lfi'
r +i,q(ICd
ci
+)i a.
xtot
X
+r
c)
E
dddd
E
H
;9 'i{r tv ()r F{ ') a' ,k | ls t{ li r '+) E o' a, \4, -q) i
)la (;r' ),^: ;,> i
gDr a.{(,)!8. q)
r+) + bo: ,+t 'Gl tct 0 Frt Itr
ail{
a = atr
o c6o 2cdE)
o iro Pco g )q I J+a +) (J x@ qX i6 trbr :f, (t) a) ) ,e 'n 5G E
lr \c) (l) 3.q
la
EE
t{
l-88 II
)trtcd g;i do \o gE *mlll HV ^tr a;t oH ,hu tr (a.7442)(950-
886.9)
Qn
=
131 IGI
'
