thermosolutions chapter12
TRANSCRIPT
12-1
Chapter 12 THERMODYNAMIC PROPERTY RELATIONS
Partial Derivatives and Associated Relations
12-1C
yx zzdzdyydxx
12-2C For functions that depend on one variable, they are identical. For functions that depend on two or more variable, the partial differential represents the change in the function with one of the variables as theother variables are held constant. The ordinary differential for such functions represents the total change as a result of differential changes in all variables.
12-3C (a) ( x)y = dx ; (b) ( z) y dz; and (c) dz = ( z)x + ( z) y
12-4C Only when ( z/ x) y = 0. That is, when z does not depend on y and thus z = z(x).
12-5C It indicates that z does not depend on y. That is, z = z(x).
12-6C Yes.
12-7C Yes.
y
( z)x
( z)y
dx
dz
x +dx
y + dy
z
dyx
y
x
12-2
12-8 Air at a specified temperature and specific volume is considered. The changes in pressurecorresponding to a certain increase of different properties are to be determined.Assumptions Air is an ideal gasProperties The gas constant of air is R = 0.287 kPa·m3/kg·K (Table A-1). Analysis An ideal gas equation can be expressed as P = RT/v. Noting that R is a constant and P = P(T,v),
2v
v
vvv
dTRdTRdv
T
PdTTPdP
(a) The change in T can be expressed as dT T = 400 0.01 = 4.0 K. At v = constant,
kPa276.1/kgm0.90
K)K)(4.0/kgmkPa(0.2873
3
vvdTR
dP
(b) The change in v can be expressed as dv v = 0.90 0.01 = 0.009 m3/kg. At T = constant,
kPa276.123
33
2 /kg)m(0.90/kg)mK)(0.009K)(400/kgmkPa(0.287
v
vdTRdP T
(c) When both v and T increases by 1%, the change in P becomes0)276.1(276.1)()( TdPdPdP v
Thus the changes in T and v balance each other.
12-9 Helium at a specified temperature and specific volume is considered. The changes in pressure corresponding to a certain increase of different properties are to be determined.Assumptions Helium is an ideal gasProperties The gas constant of helium is R = 2.0769 kPa·m3/kg·K (Table A-1). Analysis An ideal gas equation can be expressed as P = RT/v. Noting that R is a constant and P = P(T,v ),
2v
v
vv
vv
dTRdTRd
T
PdTTPdP
(a) The change in T can be expressed as dT T = 400 0.01 = 4.0 K. At v = constant,
kPa2319./kgm0.90
K)K)(4.0/kgmkPa(2.07693
3
vvdTR
dP
(b) The change in v can be expressed as dv v = 0.90 0.01 = 0.009 m3/kg. At T = constant,
kPa2319./kg)m(0.90
)mK)(0.009K)(400/kgmkPa(2.076923
33
2v
vdTRdP T
(c) When both v and T increases by 1%, the change in P becomes0)231.9(231.9)()( TdPdPdP v
Thus the changes in T and v balance each other.
12-3
12-10 It is to be proven for an ideal gas that the P = constant lines on a T-v diagram are straight lines and that the high pressure lines are steeper than the low-pressure lines. Analysis (a) For an ideal gas Pv = RT or T = Pv/R. Taking the partial derivative of T with respect to vholding P constant yields
RPT
PvT
P = const
v
which remains constant at P = constant. Thus the derivative( T/ v)P, which represents the slope of the P = const. lines on a T-v diagram, remains constant. That is, the P = const. lines are straight lines on a T-v diagram.(b) The slope of the P = const. lines on a T-v diagram is equal toP/R, which is proportional to P. Therefore, the high pressure linesare steeper than low pressure lines on the T-v diagram.
12-11 A relation is to be derived for the slope of the v = constant lines on a T-P diagram for a gas that obeys the van der Waals equation of state.Analysis The van der Waals equation of state can be expressed as
baPR
T vv 2
1
Taking the derivative of T with respect to P holding v constant,
Rbb
RPT v
vv
011
which is the slope of the v = constant lines on a T-P diagram.
12-4
12-12 Nitrogen gas at a specified state is considered. The cp and cv of the nitrogen are to be determinedusing Table A-18, and to be compared to the values listed in Table A-2b.Analysis The cp and cv of ideal gases depends on temperature only, and are expressed as cp(T) = dh(T)/dTand cv(T) = du(T)/dT. Approximating the differentials as differences about 400 K, the cp and cv values aredetermined to be
KkJ/kg1.045
K390)(410kJ/kg.011,347)/28(11,932
K390410K390K410
)()()K400(K400K400
hh
TTh
dTTdhc
TTp cp
h
T(Compare: Table A-2b at 400 K cp = 1.044 kJ/kg·K)
KkJ/kg0.748K390)(410
kJ/kg08,104)/28.(8,523
K390410K390K410
)()()K400(K400K400
uu
TTu
dTTduc
TTv
(Compare: Table A-2b at 400 K cv = 0.747 kJ/kg·K)
12-13E Nitrogen gas at a specified state is considered. The cp and cv of the nitrogen are to be determinedusing Table A-18E, and to be compared to the values listed in Table A-2Eb.Analysis The cp and cv of ideal gases depends on temperature only, and are expressed as cp(T) = dh(T)/dTand cv(T) = du(T)/dT. Approximating the differentials as differences about 600 R, the cp and cv values are determined to be
RBtu/lbm0.249R580)(620
Btu/lbm8.04,028.7)/2(4,307.1
R580620R580R620
)()()R600(R600R600
hh
TTh
dTTdhc
TTp
(Compare: Table A-2Eb at 600 R cp = 0.248 Btu/lbm·R )
RBtu/lbm0.178R580)(620
Btu/lbm8.02,876.9)/2(3,075.9
R580620R580R620
)()()R600(R600R600
uu
TTu
dTTduc
TTv
(Compare: Table A-2Eb at 600 R cv = 0.178 Btu/lbm·R)
12-5
12-14 The state of an ideal gas is altered slightly. The change in the specific volume of the gas is to be determined using differential relations and the ideal-gas relation at each state. Assumptions The gas is air and air is an ideal gas. Properties The gas constant of air is R = 0.287 kPa·m3/kg·K (Table A-1). Analysis (a) The changes in T and P can be expressed as
kPa4kPa100)96(K4K400)(404
PdPTdT
The ideal gas relation Pv = RT can be expressed as v = RT/P. Note that R is a constant and v = v (T, P).Applying the total differential relation and using average values for T and P,
/kgm0.0598 3/kg)m(0.04805/kg)m(0.0117
kPa)(98kPa)4K)((402
kPa98K4
K)/kgmkPa(0.287
33
23
2PdPTR
PdTR
dPP
dTT
dTP
vvv
(b) Using the ideal gas relation at each state,
/kgm1.2078kPa96
K)K)(404/kgmkPa(0.287
/kgm1.1480kPa100
K)K)(400/kgmkPa(0.287
33
2
22
33
1
11
PRT
PRT
v
v
Thus,
/kgm0.0598 31480.12078.112 vvv
The two results are identical.
12-6
12-15 Using the equation of state P(v-a) = RT, the cyclic relation, and the reciprocity relation at constant vare to be verified. Analysis (a) This equation of state involves three variables P, v, and T. Any two of these can be taken as the independent variables, with the remaining one being the dependent variable. Replacing x, y, and z byP, v, and T, the cyclic relation can be expressed as
1v
vv P
TT
P
PT
where
Ra
PT
RaPT
PR
Ta
PRT
aP
aRTP
aRTP
P
T
vv
vv
vvvv
v
)(
2
Substituting,
1R
aPR
aP
PT
TP
PT
vv
vv v
which is the desired result.(b) The reciprocity rule for this gas at v = constant can be expressed as
aR
TP
aRTP
Ra
PT
RaPT
PTTP
vv
vv
v
v
vv
)()/(
1
We observe that the first differential is the inverse of the second one. Thus the proof is complete.
12-7
The Maxwell Relations
12-16 The validity of the last Maxwell relation for refrigerant-134a at a specified state is to be verified.Analysis We do not have exact analytical property relations for refrigerant-134a, and thus we need toreplace the differential quantities in the last Maxwell relation with the corresponding finite quantities. Using property values from the tables about the specified state,
K/kgm101.0095K/kgm101.005
C60)(100/kgm0.018404)(0.022442
kPa1000)(1400KkJ/kg1.0458)(1.0056
C60100kPa10001400
3434
3?
kPa1200
C60C100?
C80
kPa1000kPa1400
kPa1200
?
C80
?
PT
PT
PT
ss
TPs
TPs
vv
v
v
since kJ kPa·m³, and K C for temperature differences. Thus the last Maxwell relation is satisfied.
12-8
12-17 EES Problem 12-16 is reconsidered. The validity of the last Maxwell relation for refrigerant 134a atthe specified state is to be verified. Analysis The problem is solved using EES, and the solution is given below.
"Input Data:"
T=80 [C] P=1200 [kPa] P_increment = 200 [kPa] T_increment = 20 [C] P[2]=P+P_incrementP[1]=P-P_incrementT[2]=T+T_incrementT[1]=T-T_increment
DELTAP = P[2]-P[1] DELTAT = T[2]-T[1]
v[1]=volume(R134a,T=T[1],P=P)v[2]=volume(R134a,T=T[2],P=P)s[1]=entropy(R134a,T=T,P=P[1])s[2]=entropy(R134a,T=T,P=P[2])
DELTAs=s[2] - s[1]DELTAv=v[2] - v[1]
"The partial derivatives in the last Maxwell relation (Eq. 11-19) is associated with the Gibbsfunction and are approximated by the ratio of ordinary differentials:"
LeftSide =DELTAs/DELTAP*Convert(kJ,m^3-kPa) "[m^3/kg-K]" "at T = Const."RightSide=-DELTAv/DELTAT "[m^3/kg-K]" "at P = Const."
SOLUTION
DELTAP=400 [kPa] DELTAs=-0.04026 [kJ/kg-K] DELTAT=40 [C] DELTAv=0.004038 [m^3/kg] LeftSide=-0.0001007 [m^3/kg-K] P=1200 [kPa] P[1]=1000 [kPa] P[2]=1400 [kPa] P_increment=200 [kPa]
RightSide=-0.000101 [m^3/kg-K] s[1]=1.046 [kJ/kg-K] s[2]=1.006 [kJ/kg-K] T=80 [C] T[1]=60 [C] T[2]=100 [C] T_increment=20 [C] v[1]=0.0184 [m^3/kg] v[2]=0.02244 [m^3/kg]
12-9
12-18E The validity of the last Maxwell relation for steam at a specified state is to be verified. Analysis We do not have exact analytical property relations for steam, and thus we need to replace the differential quantities in the last Maxwell relation with the corresponding finite quantities. Using property values from the tables about the specified state,
R/lbmft101.635R/lbmft101.639
F00)7(900/lbmft.6507)1(1.9777
psia0)35(450RBtu/lbm1.7009)(1.6706
F007900psia035450
3333
3?
psia400
F700F900?
F800
psia350psia450
psia400
?
F800
?
PT
PT
PT
ss
TPs
TPs
vv
v
v
since 1 Btu 5.4039 psia·ft3, and R F for temperature differences. Thus the fourth Maxwell relation issatisfied.
12-19 Using the Maxwell relations, a relation for ( s/ P)T for a gas whose equation of state is P(v-b) = RTis to be obtained.
Analysis This equation of state can be expressed as bP
RTv . Then,
PR
T P
v
From the fourth Maxwell relation,
PR
PT TPs v
12-20 Using the Maxwell relations, a relation for ( s/ v)T for a gas whose equation of state is (P-a/v2)(v-b)= RT is to be obtained.
Analysis This equation of state can be expressed as 2vv
ab
RTP . Then,
bR
TP
vv
From the third Maxwell relation,
bR
vv vTPs
T
12-10
12-21 Using the Maxwell relations and the ideal-gas equation of state, a relation for ( s/ v)T for an ideal gas is to be obtained.
Analysis The ideal gas equation of state can be expressed as v
RTP . Then,
vv
RTP
From the third Maxwell relation,
vv v
RTPs
T
The Clapeyron Equation
12-22C It enables us to determine the enthalpy of vaporization from hfg at a given temperature from the P,v, T data alone.
12-23C It is exact.
12-24C It is assumed that vfg vg RT/P, and hfg constant for small temperature intervals.
12-25 Using the Clapeyron equation, the enthalpy of vaporization of refrigerant-134a at a specifiedtemperature is to be estimated and to be compared to the tabulated data. Analysis From the Clapeyron equation,
kJ/kg163.00K4
kPa.68)963(1072.8/kg)m0.00087202K)(0.01995273.15(40
C83C42)(
)(
3
C83@satC42@satC04@
C04sat,C04@
sat
PPT
TPT
dTdPTh
fg
fg
fgfg
vv
vv
v
The tabulated value of hfg at 40°C is 163.00 kJ/kg.
12-11
12-26 EES Problem 12-25 is reconsidered. The enthalpy of vaporization of refrigerant 134-a as a functionof temperature over the temperature range -20 to 80°C by using the Clapeyron equation and the refrigerant134-a data in EES is to be plotted.Analysis The problem is solved using EES, and the results are tabulated and plotted below.
"Input Data:"
T=30 [C] T_increment = 5 [C]
-20 0 20 40 60 800.05
0.1
0.15
0.2
0.25
0.3
0.35
T [C]
PercentError[%]
T[2]=T+T_incrementT[1]=T-T_incrementP[1] = pressure(R134a,T=T[1],x=0) P[2] = pressure(R134a,T=T[2],x=0) DELTAP = P[2]-P[1] DELTAT = T[2]-T[1]
v_f=volume(R134a,T=T,x=0)v_g=volume(R134a,T=T,x=1)h_f=enthalpy(R134a,T=T,x=0)h_g=enthalpy(R134a,T=T,x=1)
h_fg=h_g - h_f v_fg=v_g - v_f
"The Clapeyron equation (Eq. 12-22) provides a means to calculate the enthalpy of vaporization,h_fg at a given temperature by determining the slope of the saturation curve on a P-T diagramand the specific volume of the saturated liquid and satruated vapor at the temperature."
h_fg_Clapeyron=(T+273.2)*v_fg*DELTAP/DELTAT*Convert(m^3-kPa,kJ)PercentError=ABS(h_fg_Clapeyron-h_fg)/h_fg*Convert(, %) "[%]"
hfg
[kJ/kg]hfg,Clapeyron
[kJ/kg]PercentError
[%] T
[C]212.91 213.68 0.3593 -20205.96 206.56 0.2895 -10198.60 199.05 0.2283 0190.73 191.07 0.1776 10182.27 182.52 0.1394 20173.08 173.28 0.1154 30163.00 163.18 0.1057 40151.79 151.96 0.1081 50139.10 139.26 0.1166 60124.32 124.47 0.1195 70106.35 106.45 0.09821 80
12-12
12-27 Using the Clapeyron equation, the enthalpy of vaporization of steam at a specified pressure is to beestimated and to be compared to the tabulated data.Analysis From the Clapeyron equation,
kJ/kg2161.1C130.58)(136.27
kPa50/kg)m0.001073K)(0.60582273.15(133.52
kPa75)2(325)(
)(
3
kPa@275satkPa253@satkPa003@kPa300@sat
kPa300sat,kPa300@
sat
TTT
TPT
dTdPTh
fg
fg
fgfg
vv
vv
v
The tabulated value of hfg at 300 kPa is 2163.5 kJ/kg.
12-28 The hfg and sfg of steam at a specified temperature are to be calculated using the Clapeyron equationand to be compared to the tabulated data. Analysis From the Clapeyron equation,
kgkJ82206 /.K10
kPa.18)169(232.23/kg)m0.001060K)(0.89133273.15(120
C115C125)(
)(
3
C115@satC125@satC120@
C120sat,C120@
sat
PPT
TPT
dTdPTh
fg
fg
fgfg
vv
vv
v
Also,
Kkg/kJ6131.5K273.15)(120
kJ/kg2206.8T
hs fg
fg
The tabulated values at 120 C are hfg = 2202.1 kJ/kg and sfg = 5.6013 kJ/kg·K.
12-13
12-29E [Also solved by EES on enclosed CD] The hfg of refrigerant-134a at a specified temperature is to be calculated using the Clapeyron equation and Clapeyron-Clausius equation and to be compared to thetabulated data.Analysis (a) From the Clapeyron equation,
error)(0.2%/lbmftpsia444.0
R20psia49.776)(72.152
/lbm)ft0.01270R)(0.79136459.67(50
F40F60)(
)(
3
3
F40@satF60@satF50@
F50sat,F50@
sat
Btu/lbm82.16
PPT
TPT
dTdPTh
fg
fg
fgfg
vv
vv
v
since 1 Btu = 5.4039 psia·ft3.(b) From the Clapeyron-Clausius equation,
error)(14.4%
R67.459601
R67.459401
RBtu/lbm0.01946psia776.49psia72.152
ln
11lnsat21sat1
2
Btu/lbm93.80fg
fg
fg
h
h
TTRh
PP
The tabulated value of hfg at 50 F is 82.00 Btu/lbm.
12-14
12-30 EES The enthalpy of vaporization of steam as a function of temperature using Clapeyron equationand steam data in EES is to be plotted.Analysis The enthalpy of vaporization is determined using Clapeyron equation from
TPTh fgfg vClapeyron,
At 100ºC, for an increment of 5ºC, we obtain
/kgm6710.1001043.06720.1
/kgm6720.1
/kgm001043.0kPa29.3661.8490.120
C1095105kPa90.120
kPa61.84C1055100
C955100
3
3C100@
3C100@
12
12
C105@sat2
C95@sat1
increment2
increment1
fgfg
g
f
PPPTTT
PPPP
TTTTTT
vvv
v
v
Substituting,
kJ/kg2262.8K10kPa36.29/kg)mK)(1.671015.273100( 3
Clapeyron, TPTh fgfg v
The enthalpy of vaporization from steam table is
/kgm2256.4 3C100@fgh
The percent error in using Clapeyron equation is
0.28%1004.2256
4.22568.2262orPercentErr
We repeat the analysis over the temperature range 10 to 200ºC using EES. Below, the copy of EES solution is provided:
"Input Data:""T=100" "[C]"T_increment = 5"[C]"T[2]=T+T_increment"[C]"T[1]=T-T_increment"[C]"P[1] = pressure(Steam_iapws,T=T[1],x=0)"[kPa]"P[2] = pressure(Steam_iapws,T=T[2],x=0)"[kPa]"DELTAP = P[2]-P[1]"[kPa]"DELTAT = T[2]-T[1]"[C]"
v_f=volume(Steam_iapws,T=T,x=0)"[m^3/kg]"v_g=volume(Steam_iapws,T=T,x=1)"[m^3/kg]"h_f=enthalpy(Steam_iapws,T=T,x=0)"[kJ/kg]"h_g=enthalpy(Steam_iapws,T=T,x=1)"[kJ/kg]"
h_fg=h_g - h_f"[kJ/kg-K]"v_fg=v_g - v_f"[m^3/kg]"
12-15
"The Clapeyron equation (Eq. 11-22) provides a means to calculate the enthalpy of vaporization,h_fg at a given temperature by determining the slope of the saturation curve on a P-T diagramand the specific volume of the saturated liquid and satruated vapor at the temperature."
h_fg_Clapeyron=(T+273.15)*v_fg*DELTAP/DELTAT*Convert(m^3-kPa,kJ)"[kJ/kg]"PercentError=ABS(h_fg_Clapeyron-h_fg)/h_fg*100"[%]"
hfg
[kJ/kg]hfg,Clapeyron
[kJ/kg]PercentErro
r[%]
T[C]
2477.20 2508.09 1.247 102429.82 2451.09 0.8756 302381.95 2396.69 0.6188 502333.04 2343.47 0.4469 702282.51 2290.07 0.3311 902229.68 2235.25 0.25 1102173.73 2177.86 0.1903 1302113.77 2116.84 0.1454 1502014.17 2016.15 0.09829 1801899.67 1900.98 0.06915 2101765.50 1766.38 0.05015 240
0 50 100 150 200 2501700
1800
1900
2000
2100
2200
2300
2400
2500
2600
T [C]
hfg
[kJ/kg]
hfg calculated by EES
hfg calculated by Clapeyron equation
12-16
12-31 The sublimation pressure of water at -30ºC is to be determined using Clapeyron-Clasius equationand the triple point data of water.Analysis The sublimation pressure may be determined using Clapeyron-Clasius equation from
211
CCsub, 11lnTTR
hP
P ig
where the triple point properties of water are P1 = 0.6117 kPa and T1 = 0.01ºC = 273.16 K (first line inTable A-4). Also, the enthalpy of sublimation of water at -30ºC is determined from Table A-8 to be 2838.4kJ/kg. Substituting,
kPa0.03799CCsub,
CCsub,
211
CCsub,
K15.273301
K16.2731
kJ/kg.K0.4615kJ/kg4.2838
kPa6117.0ln
11ln
P
P
TTRh
PP ig
The sublimation pressure of water at -30ºC is given in Table A-8 to be 0.03802 kPa. Then, the error involved in using Clapeyron-Clasius equation becomes
0.08%10003802.0
03799.003802.0orPercentErr
12-17
General Relations for du, dh, ds, cv, and cp
12-32C Yes, through the relation
PT
p
TT
Pc
2
2v
12-33 It is to be shown that the enthalpy of an ideal gas is a function of temperature only and that for anincompressible substance it also depends on pressure. Analysis The change in enthalpy is expressed as
dPT
TdTcdhP
Pv
v
For an ideal gas v = RT/P. Then,
Thus,dTcdh
PRT
TT
p
P0vvv
vv
To complete the proof we need to show that cp is not a function of P either. This is done with the help of the relation
PT
p
TT
Pc
2
2v
For an ideal gas,
Thus,
0
0)/(and2
2
T
P
PPP
Pc
TPR
TPR
Tvv
Therefore we conclude that the enthalpy of an ideal gas is a function of temperature only.For an incompressible substance v = constant and thus v/ T = 0. Then,
dPdTcdh p v
Therefore we conclude that the enthalpy of an incompressible substance is a function of temperature andpressure.
12-18
12-34 General expressions for u, h, and s for a gas that obeys the van der Waals equation of state foran isothermal process are to be derived.Analysis (a) For an isothermal process dT = 0 and the general relation for u reduces to
2
1
2
1
2
112
v
v v
v
v vv vv dP
TPTdP
TPTdTcuuu
T
T
The van der Waals equation of state can be expressed as
Thus,
22
2
vvvv
vvv
v
v
aab
RTb
RTPTPT
bR
TPa
bRTP
Substituting,
212
112
1 vvv
v
v
vadau
(b) The enthalpy change h is related to u through the relation
1122 vv PPuh
where
vvv
va
bRTP
Thus,
211
1
2
21122
11vvv
v
v
vvv a
bbRTPP
Substituting,
bbRTah
1
1
2
2
21
112v
v
v
v
vv
(c) For an isothermal process dT = 0 and the general relation for s reduces to
2
1
2
1
2
112
v
v v
v
v v
v vv dTPd
TPdT
Tc
sssT
T
Substituting ( P/ T)v = R/(v - b),
bb
Rdb
Rs1
2ln2
1 v
vv
v
v
v
12-19
12-35 General expressions for u, h, and s for a gas whose equation of state is P(v-a) = RT for an isothermal process are to be derived.Analysis (a) A relation for u is obtained from the general relation
2
1
2
112
v
v vv vdP
TPTdTcuuu
T
T
The equation of state for the specified gas can be expressed as
Thus,
0PPPa
RTPTPT
aR
TP
aRTP
v
vv
v
v
Substituting,2
1
T
TdTcu v
(b) A relation for h is obtained from the general relation
2
1
2
112
P
P P
T
TP dP
TvTdTchhh v
The equation of state for the specified gas can be expressed as
Thus,
aaPRT
TvT
PR
Ta
PRT
P
P
)(vvvv
vv
Substituting,
122
1
2
1
2
1
PPadTcdPadTchT
Tp
P
P
T
Tp
(c) A relation for s is obtained from the general relation
2
1
2
112
P
P P
T
T
p dPT
dTTc
sss v
Substituting ( v/ T)P = R/T,
1
2ln2
1
2
1
2
1 PP
RdTTc
dPPRdT
Tc
sT
T
pP
P P
T
T
p
For an isothermal process dT = 0 and these relations reduce to
1
212 lnand,,0
PP
RsPPahu
12-20
12-36 General expressions for ( u/ P)T and ( h/ v)T in terms of P, v, and T only are to be derived.Analysis The general relation for du is
vv
v dPTPTdTcdu
Differentiating each term in this equation with respect to P at T = constant yields
TTTT PP
PTPT
PP
TPT
Pu vvv
vv
0
Using the properties P, T, v, the cyclic relation can be expressed as
PTTP TPTP
PT
TP vvv
v vv
1
Substituting, we get
TPT PP
TT
Pu vv
The general relation for dh is
dPT
TdTcdhP
pv
v
Differentiating each term in this equation with respect to v at T = constant yields
TPTTPT
PT
TPPT
Thv
vv
vv
vv
v0
Using the properties v, T, P, the cyclic relation can be expressed as
vv vv
vv
PTP
TP
PT
T TPTP1
Substituting, we get
vvv
v PTTPh
TT
12-21
12-37 Expressions for the specific heat difference cp-cv for three substances are to be derived. Analysis The general relation for the specific heat difference cp - cv is
TPp
PT
Tccv
vv
2
(a) For an ideal gas Pv = RT. Then,
vvvv
vv
PRTPRTP
PR
TPRT
T
P
2
Substituting,
RRPTR
PRPTcc p vvv
2
(b) For a van der Waals gas RTbaP vv 2
. Then,
bT
Rba
aPR
baR
TbaPR
TP
vv
vv
vvv
vv
3
232
2
1211
Inverting,
bT
RbaT P
vv
vv
32
1
Also,322
2vvvvv
ab
RTPab
RTPT
Substituting,
32
2
3
22
1vv
vv
vva
bRT
bT
Rba
Tcc p
(c) For an incompressible substance v = constant and thus ( v / T)P = 0. Therefore, 0vcc p
12-22
12-38 The specific heat difference cp-cv for liquid water at 15 MPa and 80 C is to be estimated.Analysis The specific heat difference cp - cv is given as
TPp
PT
Tccv
vv
2
Approximating differentials by differences about the specified state,
Kkg/kJ0.3214K/kgmkPa3114.0
/kgm0.0010244)(0.0010199kPa10,000
K40/kgm0.0010105)(0.0010361
)K15.353(
MPa)10(20C)60100(
)K15.27380(
3
3
23
C80MPa10MPa20
2
MPa15
C60C100
C80
2
MPa15
TP
TPp
PT
Tcc
vv
vv
vv
v
12-39E The specific heat difference cp-cv for liquid water at 1000 psia and 150 F is to be estimated.Analysis The specific heat difference cp - cv is given as
TPp
PT
Tccv
vv
2
Approximating differentials by differences about the specified state,
)ftpsia5.4039Btu(1R/lbmftpsia3081.0
/lbmft0.016317)(0.016267psia1000
R50/lbmft0.016177)(0.016427
R)609.67(
psia500)(1500F)125175(
)R67.459150(
33
3
23
F150psia500psia1500
2
psia1000
F125F175
F150
2
psia1000
RBtu/lbm0.0570
TP
TPp
PT
Tcc
vv
vv
vv
v
12-23
12-40 Relations for the volume expansivity and the isothermal compressibility for an ideal gas and for a gas whose equation of state is P(v-a) = RT are to be obtained. Analysis The volume expansivity and isothermal compressibility are expressed as
TP PTv
vv
v1and1
(a) For an ideal gas v = RT/P. Thus,
P11
T11
2 PPPRT
P
PR
PR
T
T
P
vv
vv
vv
(b) For a gas whose equation of state is v = RT/P + a,
vvv
vvv
vv
Pa
Pa
Pa
PRT
P
aPRTR
PR
PR
T
T
P
1
1
2
12-41 The volume expansivity and the isothermal compressibility of refrigerant-134a at 200 kPa and 30 C are to be estimated.Analysis The volume expansivity and isothermal compressibility are expressed as
TP PTv
vv
v1and1
Approximating differentials by differences about the specified state,
1K003810.K20
/kgm0.11418)(0.12322/kgm0.11874
1
C)2040(11
3
3
kPa200
C20C40
kPa200 PPTvv
vv
v
and
1kPa004820.kPa60
/kgm0.13248)(0.09812/kgm0.11874
1
kPa180)(24011
3
3
C30
kPa180kPa240
C30 TTPvv
vv
v
12-24
The Joule-Thomson Coefficient
12-42C It represents the variation of temperature with pressure during a throttling process.
12-43C The line that passes through the peak points of the constant enthalpy lines on a T-P diagram is called the inversion line. The maximum inversion temperature is the highest temperature a fluid can be cooled by throttling.
12-44C No. The temperature may even increase as a result of throttling.
12-45C Yes.
12-46C No. Helium is an ideal gas and h = h(T) for ideal gases. Therefore, the temperature of an ideal gas remains constant during a throttling (h = constant) process.
12-47 The equation of state of a gas is given to be P(v-a) = RT. It is to be determined if it is possible to cool this gas by throttling. Analysis The equation of state of this gas can be expressed as
PR
Ta
PRT
P
vv
Substituting into the Joule-Thomson coefficient relation,
0111
pppPp caa
cPRT
cTT
cvvv
vv
Therefore, this gas cannot be cooled by throttling since is always a negative quantity.
12-48 Relations for the Joule-Thompson coefficient and the inversion temperature for a gas whose equation of state is (P+a/v2) v = RT are to be obtained. Analysis The equation of state of this gas can be expressed as
222322212
v
vvv
v
vv
vvv
v
RaRTT
RaaP
Ra
RTaP
RT
P
Substituting into the Joule-Thomson coefficient relation,
)2(2
211 2
vv
vv
vv
vRTac
aaRT
RTcT
Tc ppPp
The temperature at = 0 is the inversion temperature,
00)2(
2v
vvRTac
a
p
Thus the line of v = 0 is the inversion line. Since it is not physically possible to have v = 0, this gas does not have an inversion line.
12-25
12-49 The Joule-Thompson coefficient of steam at two states is to be estimated.Analysis (a) The enthalpy of steam at 3 MPa and 300 C is h = 2994.3 kJ/kg. Approximating differentials by differences about the specified state, the Joule-Thomson coefficient is expressed as
kJ/kg3.2994hh PT
PT
Considering a throttling process from 3.5 MPa to 2.5 MPa at h = 2994.3 kJ/kg, the Joule-Thomsoncoefficient is determined to be
C/MPa12.3MPa)5.2(3.5
C)2943.306(MPa)5.2(3.5
kJ/kg3.2994
MPa2.5MPa3.5
h
TT
(b) The enthalpy of steam at 6 MPa and 500 C is h = 3423.1 kJ/kg. Approximating differentials bydifferences about the specified state, the Joule-Thomson coefficient is expressed as
kJ/kg1.3423hh PT
PT
Considering a throttling process from 7.0 MPa to 5.0 MPa at h = 3423.1 kJ/kg, the Joule-Thomsoncoefficient is determined to be
C/MPa4.9MPa5.0)(7.0
C)1.4958.504(MPa5.0)(7.0
kJ/kg1.3423
MPa5.0MPa7.0
h
TT
12-50E [Also solved by EES on enclosed CD] The Joule-Thompson coefficient of nitrogen at two states is to be estimated.Analysis (a) The enthalpy of nitrogen at 200 psia and 500 R is, from EES, h = -10.564 Btu/lbm. Note that in EES, by default, the reference state for specific enthalpy and entropy is 0 at 25ºC (77ºF) and 1 atm.Approximating differentials by differences about the specified state, the Joule-Thomson coefficient is expressed as
Btu/lbm564.10hh PT
PT
Considering a throttling process from 210 psia to 190 psia at h = -10.564 Btu/lbm, the Joule-Thomsoncoefficient is determined to be
R/psia0.0297psia)210(190
R500.296)(499.703psia210)(190
Btu/lbm564.10
psia210psia190
h
TT
(b) The enthalpy of nitrogen at 2000 psia and 400 R is, from EES, h = -55.321 Btu/lbm. Approximatingdifferentials by differences about the specified state, the Joule-Thomson coefficient is expressed as
Btu/lbm321.55hh PT
PT
Considering a throttling process from 2010 psia to 1990 psia at h = -55.321 Btu/lbm, the Joule-Thomsoncoefficient is determined to be
R/psia0.0213psia2010)(1990
R400.213)-(399.786psia2010)(1990
Btu/lbm321.55
psia2001psia1999
h
TT
12-26
12-51E EES Problem 12-50E is reconsidered. The Joule-Thompson coefficient for nitrogen over the pressure range 100 to 1500 psia at the enthalpy values 100, 175, and 225 Btu/lbm is to be plotted. Analysis The problem is solved using EES, and the results are tabulated and plotted below.
Gas$ = 'Nitrogen' {P_ref=200 [psia] T_ref=500 [R] P= P_ref}h=50 [Btu/lbm] {h=enthalpy(Gas$, T=T_ref, P=P_ref)}dP = 10 [psia] T = temperature(Gas$, P=P, h=h) P[1] = P + dP P[2] = P - dP T[1] = temperature(Gas$, P=P[1], h=h) T[2] = temperature(Gas$, P=P[2], h=h) Mu = DELTAT/DELTAP "Approximate the differential by differences about the state at h=const."DELTAT=T[2]-T[1]DELTAP=P[2]-P[1]
h = 100 Btu/lbmP [psia] μ [R/psia]
100 0.003675275 0.003277450 0.002899625 0.00254800 0.002198975 0.001871
1150 0.0015581325 0.0012581500 0.0009699
0 200 400 600 800 1000 1200 1400 1600-0.006
-0.005
-0.004
-0.003
-0.002
-0.001
0
0.001
0.002
0.003
0.004
P [psia]
[R/psia]
h = 100 Btu/lbm
h = 175 Btu/lbm
h = 225 Btu/lbm
12-27
12-52 The Joule-Thompson coefficient of refrigerant-134a at a specified state is to be estimated.Analysis The enthalpy of refrigerant-134a at 0.7 MPa and T = 50 C is h = 288.53 kJ/kg. Approximatingdifferentials by differences about the specified state, the Joule-Thomson coefficient is expressed as
kJ/kg53.288hh PT
PT
Considering a throttling process from 0.8 MPa to 0.6 MPa at h = 288.53 kJ/kg, the Joule-Thomsoncoefficient is determined to be
C/MPa18.1MPa0.6)(0.8
C.19)48(51.81MPa0.6)(0.8
kJ/kg53.288
MPa0.6MPa0.8
h
TT
12-53 Steam is throttled slightly from 1 MPa and 300 C. It is to be determined if the temperature of the steam will increase, decrease, or remain the same during this process. Analysis The enthalpy of steam at 1 MPa and T = 300 C is h = 3051.6 kJ/kg. Now consider a throttling process from this state to 0.8 MPa, which is the next lowest pressure listed in the tables. The temperatureof the steam at the end of this throttling process will be
C52.297kJ/kg6.3051
MPa8.02T
hP
Therefore, the temperature will decrease.
12-28
The h, u, and s of Real Gases
12-54C It is the variation of enthalpy with pressure at a fixed temperature.
12-55C As PR approaches zero, the gas approaches ideal gas behavior. As a result, the deviation from ideal gas behavior diminishes.
12-56C So that a single chart can be used for all gases instead of a single particular gas.
12-57 The enthalpy of nitrogen at 175 K and 8 MPa is to be determined using data from the ideal-gas nitrogen table and the generalized enthalpy departure chart. Analysis (a) From the ideal gas table of nitrogen (Table A-18) we read
kg/kmol)013.28(kJ/kmol8.50832NMh kJ/kg181.48
N2175 K 8 MPa
at the specified temperature. This value involves 44.4% error. (b) The enthalpy departure of nitrogen at the specified state is determined from the generalized chart to be
and 6.1)(
360.239.38
387.12.126
175
cr
,ideal
cr
cr
TRhh
Z
PPP
TTT
u
PTh
R
R
Thus,
or,
)error%1.3(kg/kmol28.013kJ/kmol3405.0
kJ/kmol0.34052.126314.86.18.5083crideal
kJ/kg121.6Mhh
TRZhh uh
12-29
12-58E The enthalpy of nitrogen at 400 R and 2000 psia is to be determined using data from the ideal-gasnitrogen table and the generalized enthalpy departure chart.Analysis (a) From the ideal gas table of nitrogen (Table A-18E) we read
)lbm/lbmol28(Btu/lbmol0.27772NMh Btu/lbm99.18
N2400 R
2000 psia
at the specified temperature. This value involves 44.2% error. (b) The enthalpy departure of nitrogen at the specified state is determined from the generalized chart to be (Fig. A-29)
and 18.1)(
065.4492
2000
761.11.227
400,ideal
cr
cr
cru
PTh
R
R
TRhh
Z
PPP
TTT
Thus,
or,
error)(54.9%lbm/lbmol28
Btu/lbmol2244.8
Btu/lbmol8.22441.227986.118.10.2777crideal
Btu/lbm80.17Mhh
TRZhh uh
12-59 The errors involved in the enthalpy and internal energy of CO2 at 350 K and 10 MPa if it is assumedto be an ideal gas are to be determined.Analysis (a) The enthalpy departure of CO2 at the specified state is determined from the generalized chart to be (Fig. A-29)
and 5.1)(
353.139.7
10
151.12.304
350
cr
,ideal
cr
cr
TRhh
Z
PPP
TTT
u
PTh
R
R
CO2350 K
10 MPaThus,
and,
%.557,7
557,7351,11)(Error
kJ/kmol557,72.304314.85.1351,11
,ideal
crideal
250h
hh
TRZhh
PT
uh
(b) At the calculated TR and PR the compressibility factor is determined from the compressibility chart tobe Z = 0.65. Then using the definition of enthalpy, the internal energy is determined to be
and,
%9.48666,5
666,5439,8Error
kJ/kmol666,5350314.865.07557
ideal
uuu
TZRhPhu uv
12-30
12-60 The enthalpy and entropy changes of nitrogen during a process are to be determined assuming idealgas behavior and using generalized charts.Analysis (a) Using data from the ideal gas property table of nitrogen (Table A-18),
and
K/.6
12ln314.8289.183562.193ln)(
/537,69306)(
1
212ideal12
ideal1,ideal2,ideal12
kmolkJ5104
kmolkJ2769
PP
Rssss
hhhh
u
(b) The enthalpy and entropy departures of nitrogen at the specified states are determined from the generalized charts to be (Figs. A-29, A-30)
25.0and6.0770.1
39.36
783.12.126
225
11
cr
11
cr
11
sh
R
R
ZZ
PP
P
TT
T
and
15.0and4.0540.2
39.312
536.22.126
320
22
cr
22
cr
22
sh
R
R
ZZ
PP
P
TT
T
Substituting,
Kkmol/kJ341.5
kmol/kJ2979
510.415.025.0314.8)()(
27694.06.02.126314.8)()(
ideal122112
ideal122112
ssZZRss
hhZZTRhh
ssu
hhcru
12-31
12-61 The enthalpy and entropy changes of CO2 during a process are to be determined assuming ideal gas behavior and using generalized charts.
280 K 12 MPa
CO2250 K 7 MPa
Analysis (a) Using data from the ideal gas propertytable of CO2 (Table A-20),
Thus,
KkJ/kg0.0100
kJ/kg24.32
kg/kmol44kJ/kmol0.442)(
)(
kg/kmol44kJ/kmol1,070)(
)(
Kkmol/kJ442.07
12ln314.8337.207376.211ln)(
kmol/kJ070,1627,7697,8)(
ideal12ideal12
ideal12ideal12
1
212ideal12
ideal1,ideal2,ideal12
Mss
ss
Mhh
hh
PP
Rssss
hhhh
u
(b) The enthalpy and entropy departures of CO2 at the specified states are determined from the generalizedcharts to be (Figs. A-29, A-30)
and
2.4and0.5624.1
39.712
920.02.304
280
3.5and5.5947.0
39.77
822.02.304
250
22
cr
22
cr
22
11
cr
11
cr
11
sh
R
R
sh
R
R
ZZ
PP
P
TT
T
ZZ
PP
P
TT
T
Thus,
KkJ/kg0.198
kg/kJ05.53
010.02.43.51889.0)()(
32.240.55.52.3041889.0)()(
ideal122112
ideal1221cr12
ssZZRss
hhZZRThh
ss
hh
12-32
12-62 Methane is compressed adiabatically by a steady-flow compressor. The required power input to thecompressor is to be determined using the generalized charts.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis The steady-flow energy balance equation for this compressor can be expressed as
12inC,
21inC,
outin
(steady)0systemoutin 0
hhmW
hmhmW
EE
EEE
10 MPa 110 C
CH4 = 0.55 kg/s m·The enthalpy departures of CH4 at the specified states are determined
from the generalized charts to be (Fig. A-29)
and
50.0155.2
64.410
00.21.191
383
21.0431.0
64.42
376.11.191
263
2
cr
22
cr
22
1
cr
11
cr
11
h
R
R
h
R
R
Z
PP
P
TT
T
Z
PP
P
TT
T2 MPa -10 C
Thus,
kg/kJ7.241101102537.250.021.01.1915182.0)()( ideal1221cr12 hhZZRThh hh
Substituting,
kW133kJ/kg241.7kg/s0.55inC,W
12-33
12-63 [Also solved by EES on enclosed CD] Propane is compressed isothermally by a piston-cylinderdevice. The work done and the heat transfer are to be determined using the generalized charts.Assumptions 1 The compression process is quasi-equilibrium. 2 Kinetic and potential energy changes are negligible.Analysis (a) The enthalpy departure and the compressibility factors of propane at the initial and the finalstates are determined from the generalized charts to be (Figs. A-29, A-15)
and
50.0and8.1939.0
26.44
008.1370373
92.0and28.0235.0
26.41
008.1370373
22
cr
22
cr
22
11
cr
11
cr
11
ZZ
PP
P
TT
T
ZZ
PP
P
TT
T
h
R
R
h
R
R
Q
Propane1 MPa 100 C
Treating propane as a real gas with Zavg = (Z1+Z2)/2 = (0.92 + 0.50)/2 = 0.71, constantavg CRTZZRTPv
Then the boundary work becomes
kJ/kg99.640.9210.50lnK373KkJ/kg0.188571.0
ln//
lnln21
12ave
11
22avg
1
22
1
2
1in, PZ
PZRTZ
PRTZPRTZ
RTZCdCdPwb v
vv
vv
Also,
kg/kJ5.7637392.03735.01885.0106)()(
kg/kJ10608.128.03701885.0)()(
11221212
ideal122112
TZTZRhhuu
hhZZRThh hhcr
Then the heat transfer for this process is determined from the closed system energy balance to be
kJ/kg176.1kg/kJ1.1766.995.76 outin,12in
12in,in
systemoutin
qwuuquuuwq
EEE
b
b
12-34
12-64 EES Problem 12-63 is reconsidered. This problem is to be extended to compare the solutions basedon the ideal gas assumption, generalized chart data and real fluid (EES) data. Also, the solution is to beextended to carbon dioxide, nitrogen and methane.Analysis The problem is solved using EES, and the solution is given below.
Procedure INFO(Name$, T[1] : Fluid$, T_critical, p_critical)If Name$='Propane' then
T_critical=370 ; p_critical=4620 ; Fluid$='C3H8'; goto 10 endifIf Name$='Methane' then
T_critical=191.1 ; p_critical=4640 ; Fluid$='CH4'; goto 10 endifIf Name$='Nitrogen' then
T_critical=126.2 ; p_critical=3390 ; Fluid$='N2'; goto 10 endifIf Name$='Oxygen' then
T_critical=154.8 ; p_critical=5080 ; Fluid$='O2'; goto 10 endifIf Name$='CarbonDioxide' then
T_critical=304.2 ; p_critical=7390 ; Fluid$='CO2' ; goto 10 endifIf Name$='n-Butane' then
T_critical=425.2 ; p_critical=3800 ; Fluid$='C4H10' ; goto 10 endif
10: If T[1]<=T_critical thenCALL ERROR('The supplied temperature must be greater than the critical temperature for the fluid. A value of XXXF1 K was supplied',T[1])endif
end
{"Data from the Diagram Window"T[1]=100+273.15p[1]=1000p[2]=4000Name$='Propane'Fluid$='C3H8' }
Call INFO(Name$, T[1] : Fluid$, T_critical, p_critical)R_u=8.314M=molarmass(Fluid$)R=R_u/M
"****** IDEAL GAS SOLUTION ******""State 1"h_ideal[1]=enthalpy(Fluid$, T=T[1]) "Enthalpy of ideal gas"s_ideal[1]=entropy(Fluid$, T=T[1], p=p[1]) "Entropy of ideal gas"u_ideal[1]=h_ideal[1]-R*T[1] "Internal energy of ideal gas""State 2"h_ideal[2]=enthalpy(Fluid$, T=T[2]) "Enthalpy of ideal gas"s_ideal[2]=entropy(Fluid$, T=T[2], p=p[2]) "Entropy of ideal gas"u_ideal[2]=h_ideal[2]-R*T[2] "Internal energy of ideal gas"
12-35
"Work is the integral of p dv, which can be done analytically."w_ideal=R*T[1]*Ln(p[1]/p[2])
"First Law - note that u_ideal[2] is equal to u_ideal[1]"q_ideal-w_ideal=u_ideal[2]-u_ideal[1]
"Entropy change"DELTAs_ideal=s_ideal[2]-s_ideal[1]
"***** COMPRESSABILITY CHART SOLUTION ******""State 1"Tr[1]=T[1]/T_criticalpr[1]=p[1]/p_criticalZ[1]=COMPRESS(Tr[1], Pr[1]) DELTAh[1]=ENTHDEP(Tr[1], Pr[1])*R*T_critical "Enthalpy departure"h[1]=h_ideal[1]-DELTAh[1] "Enthalpy of real gas using charts"u[1]=h[1]-Z[1]*R*T[1]
"Internal energy of gas using charts"DELTAs[1]=ENTRDEP(Tr[1], Pr[1])*R "Entropy departure"s[1]=s_ideal[1]-DELTAs[1] "Entropy of real gas using charts""State 2"T[2]=T[1]Tr[2]=Tr[1]pr[2]=p[2]/p_criticalZ[2]=COMPRESS(Tr[2], Pr[2]) DELTAh[2]=ENTHDEP(Tr[2], Pr[2])*R*T_critical "Enthalpy departure"DELTAs[2]=ENTRDEP(Tr[2], Pr[2])*R "Entropy departure"h[2]=h_ideal[2]-DELTAh[2] "Enthalpy of real gas using charts"s[2]=s_ideal[2]-DELTAs[2] "Entropy of real gas using charts"u[2]=h[2]-Z[2]*R*T[2] "Internal energy of gas using charts"
"Work using charts - note use of EES integral function to evaluate the integral of p dv."w_chart=Integral(p,v,v[1],v[2])"We need an equation to relate p and v in the above INTEGRAL function. "p*v=COMPRESS(Tr[2],p/p_critical)*R*T[1] "To specify relationship between p and v""Find the limits of integration"p[1]*v[1]=Z[1]*R*T[1] "to get v[1], the lower bound"p[2]*v[2]=Z[2]*R*T[2] "to get v[2], the upper bound"
"First Law - note that u[2] is not equal to u[1]"q_chart-w_chart=u[2]-u[1]
"Entropy Change"DELTAs_chart=s[2]-s[1]
"***** SOLUTION USING EES BUILT-IN PROPERTY DATA *****""At state 1"u_ees[1]=intEnergy(Name$,T=T[1],p=p[1])s_ees[1]=entropy(Name$,T=T[1],p=p[1])"At state 2"u_ees[2]=IntEnergy(Name$,T=T[2],p=p[2])s_ees[2]=entropy(Name$,T=T[2],p=p[2])
"Work using EES built-in properties- note use of EES Integral funcion to evaluate the integral of pdv."w_ees=integral(p_ees, v_ees, v_ees[1],v_ees[2]) "The following equation relates p and v in the above INTEGRAL"
12-36
p_ees=pressure(Name$,T=T[1], v=v_ees) "To specify relationship between p and v"
"Find the limits of integration"v_ees[1]=volume(Name$, T=T[1],p=p[1]) "to get lower bound"v_ees[2]=volume(Name$, T=T[2],p=p[2]) "to get upper bound"
"First law - note that u_ees[2] is not equal to u_ees[1]"q_ees-w_ees=u_ees[2]-u_ees[1]
"Entropy change"DELTAs_ees=s_ees[2]-s_ees[1]
"Note: In all three solutions to this problem we could have calculated the heat transfer by q/T=DELTA_s since T is constant. Then the first law could have been used to find the work.The use of integral of p dv to find the work is a more fundamental approach and can be used if T is not constant."
SOLUTION
DELTAh[1]=16.48 [kJ/kg] DELTAh[2]=91.96 [kJ/kg] DELTAs[1]=0.03029 [kJ/kg-K] DELTAs[2]=0.1851 [kJ/kg-K] DELTAs_chart=-0.4162 [kJ/kg-K] DELTAs_ees=-0.4711 [kJ/kg-K] DELTAs_ideal=-0.2614 [kJ/kg-K] Fluid$='C3H8'h[1]=-2232 [kJ/kg] h[2]=-2308 [kJ/kg] h_ideal[1]=-2216 [kJ/kg] h_ideal[2]=-2216 [kJ/kg] M=44.1Name$='Propane'p=4000p[1]=1000 [kPa] p[2]=4000 [kPa] pr[1]=0.2165pr[2]=0.8658p_critical=4620 [kPa] p_ees=4000q_chart=-155.3 [kJ/kg] q_ees=-175.8 [kJ/kg] q_ideal=-97.54 [kJ/kg] R=0.1885 [kJ/kg-K] R_u=8.314 [kJ/mole-K]s[1]=6.073 [kJ/kg-K]
s[2]=5.657 [kJ/kg-K] s_ees[1]=2.797 [kJ/kg-K] s_ees[2]=2.326 [kJ/kg-K] s_ideal[1]=6.103 [kJ/kg-K] s_ideal[2]=5.842 [kJ/kg-K] T[1]=373.2 [K] T[2]=373.2 [K] Tr[1]=1.009Tr[2]=1.009T_critical=370 [K] u[1]=-2298 [kJ/kg] u[2]=-2351 [kJ/kg] u_ees[1]=688.4 [kJ/kg] u_ees[2]=617.1 [kJ/kg] u_ideal[1]=-2286 [kJ/kg] u_ideal[2]=-2286 [kJ/kg] v=0.01074v[1]=0.06506 [m^3/kg] v[2]=0.01074 [m^3/kg] v_ees=0.009426v_ees[1]=0.0646 [m^3/kg] v_ees[2]=0.009426 [m^3/kg] w_chart=-101.9 [kJ/kg] w_ees=-104.5 [kJ/kg] w_ideal=-97.54 [kJ/kg] Z[1]=0.9246Z[2]=0.6104
12-37
12-65E Propane is compressed isothermally by a piston-cylinder device. The work done and the heattransfer are to be determined using the generalized charts.Assumptions 1The compression process is quasi-equilibrium. 2 Kinetic and potential energy changes are negligible. 3 The device is well-insulated and thus heat transfer is negligibleAnalysis (a) The enthalpy departure and the compressibility factors of propane at the initial and the finalstates are determined from the generalized charts to be (Figs. A-29, A-15)
and
22.0and2.4297.1
617800
991.09.665
660
88.0and37.0324.0
617200
991.09.665
660
22
cr
22
cr
22
11
cr
11
cr
11
ZZ
PP
P
TT
T
ZZ
PP
P
TT
T
h
R
R
h
R
R
Q
Propane200 psia 200 F
Treating propane as a real gas with Zavg = (Z1+Z2)/2 = (0.88 + 0.22)/2 = 0.55, constantavg CRTZZRTPv
Then the boundary work becomes
Btu/lbm45.380088.020022.0lnR660RBtu/lbm0.0450455.0
ln//
lnln21
12avg
11
22avg
1
22
1
2
1in, PZ
PZRTZ
PRTZPRTZ
RTZCdCdPwb v
vv
vv
Also,
lbm/Btu3.9566088.066022.004504.09.114)()(
lbm/Btu9.11402.437.09.66504504.0)()(
11221212
0ideal122112
TZTZRhhuu
hhZZRThh hhcr
Then the heat transfer for this process is determined from the closed system energy balance equation to be
lbmBtu6140 /.Btu/lbm6.1403.453.95 outin,12in
12in,in
systemoutin
qwuuquuuwq
EEE
b
b
12-38
12-66 Propane is compressed isothermally by a piston-cylinder device. The exergy destruction associated with this process is to be determined.Assumptions 1The compression process is quasi-equilibrium. 2 Kinetic and potential energy changes are negligible.Properties The gas constant of propane is R = 0.1885 kJ/kg.K (Table A-1). Analysis The exergy destruction is determined from its definition where the entropy generation is determined from an entropy balance on the contents of the cylinder. It gives
x Tdestroyed gen0s
)(=)(surr
out12gen12gen
surr,
out
systemgenoutin
Tq
sssssmSTQ
SSSS
b
where
KkJ/kg261.014ln1885.00lnln)(
)()(
1
20
1
2ideal12
ideal122112sys
PP
RTT
css
ssZZRsss
p
ss
and
5.1939.0
26.44
008.1370373
21.0235.0
26.41
008.1370373
2
cr
22
cr
22
1
cr
11
cr
11
s
R
R
s
R
R
Z
PP
P
TT
T
Z
PP
P
TT
T
Thus,KkJ/kg.5040261.05.121.01885.0)()( ideal122112sys ssZZRsss ss
and
kgkJ423 /.KkJ/kgK303
kJ/kg176.10.504K303)(
surr
out120gen0destroyed T
qssTsTx
12-39
12-67 Carbon dioxide passes through an adiabatic nozzle. The exit velocity is to be determined using thegeneralized enthalpy departure chart. Assumptions 1Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3The nozzle is adiabatic and thus heat transfer is negligibleProperties The gas constant of CO2 is 0.1889 kJ/kg.K (Table A-1). Analysis The steady-flow energy balanceequation for this nozzle can be expressed as
P2 = 2 MPaT2 = 350 K
CO2
)(2
)2/()2/(
0
212
222
0211
outin
(steady)0systemoutin
hhV
VhVh
EE
EEE P1 = 8 MPaT1 = 450 K
The enthalpy departures of CO2 at the specified states are determined from the generalized enthalpydeparture chart to be
and
20.027.0
39.72
15.12.304
350
55.008.1
39.78
48.12.304
450
2
cr
22
cr
22
1
cr
11
cr
11
h
R
R
h
R
R
Z
PP
P
TT
T
Z
PP
P
TT
T
Thus,
kJ/kg8.7344/483,15351,112.055.02.3041889.0)()( ideal1221cr12 hhZZRThh hh
Substituting,
m/s384kJ/kg1
/sm1000kJ/kg73.82
22
2V
12-40
12-68 EES Problem 12-67 is reconsidered. the exit velocity to the nozzle assuming ideal gas behavior, thegeneralized chart data, and EES data for carbon dioxide are to be compared.Analysis The problem is solved using EES, and the results are given below.
Procedure INFO(Name$: Fluid$, T_critical, p_critical) If Name$='CarbonDioxide' then
T_critical=304.2 ; p_critical=7390 ; Fluid$='CO2'endif
END
T[1]=450 [K] P[1]=8000 [kPa] P[2]=2000 [kPa] T[2]=350 [K] Name$='CarbonDioxide'
Call INFO(Name$: Fluid$, T_critical, P_critical) R_u=8.314M=molarmass(Fluid$)R=R_u/M
"****** IDEAL GAS SOLUTION ******""State 1 nd 2"h_ideal[1]=enthalpy(Fluid$, T=T[1]) "Enthalpy of ideal gas"h_ideal[2]=enthalpy(Fluid$, T=T[2]) "Enthalpy of ideal gas""Exit velocity:"V_2_ideal=SQRT(2*(h_ideal[1]-h_ideal[2])*convert(kJ/kg,m^2/s^2))"[m/s]"
"***** COMPRESSABILITY CHART SOLUTION ******""State 1"Tr[1]=T[1]/T_criticalPr[1]=P[1]/P_criticalDELTAh[1]=ENTHDEP(Tr[1], Pr[1])*R*T_critical "Enthalpy departure"h[1]=h_ideal[1]-DELTAh[1] "Enthalpy of real gas using charts"
"State 2"Tr[2]=T[2]/T_criticalPr[2]=P[2]/P_criticalDELTAh[2]=ENTHDEP(Tr[2], Pr[2])*R*T_critical "Enthalpy departure"h[2]=h_ideal[2]-DELTAh[2] "Enthalpy of real gas using charts"
"Exit velocity:"V_2_EnthDep=SQRT(2*(h[1]-h[2])*convert(kJ/kg,m^2/s^2)) "[m/s]"
"***** SOLUTION USING EES BUILT-IN PROPERTY DATA *****""At state 1 and 2"h_ees[1]=enthalpy(Name$,T=T[1],P=P[1])h_ees[2]=enthalpy(Name$,T=T[2],P=P[2])
"Exit velocity:"V_2_ees=SQRT(2*(h_ees[1]-h_ees[2])*convert(kJ/kg,m^2/s^2))"[m/s]"
12-41
SOLUTIONDELTAh[1]=34.19 [kJ/kg] DELTAh[2]=13.51 [kJ/kg] Fluid$='CO2'h[1]=-8837 [kJ/kg] h[2]=-8910 [kJ/kg] h_ees[1]=106.4 [kJ/kg] h_ees[2]=31.38 [kJ/kg] h_ideal[1]=-8803 [kJ/kg] h_ideal[2]=-8897 [kJ/kg] M=44.01 [kg/kmol]Name$='CarbonDioxide'P[1]=8000 [kPa] P[2]=2000 [kPa]
Pr[1]=1.083Pr[2]=0.2706P_critical=7390 [kPa] R=0.1889 [kJ/kg-K] R_u=8.314 [kJ/kmol-K]T[1]=450 [K] T[2]=350 [K] Tr[1]=1.479Tr[2]=1.151T_critical=304.2 [K] V_2_ees=387.4 [m/s] V_2_EnthDep=382.3 [m/s] V_2_ideal=433 [m/s]
12-69 A paddle-wheel placed in a well-insulated rigid tank containing oxygen is turned on. The final pressure in the tank and the paddle-wheel work done during this process are to be determined.Assumptions 1The tank is well-insulated and thus heat transfer is negligible. 2 Kinetic and potential energychanges are negligible.Properties The gas constant of O2 is R = 0.2598 kJ/kg.K (Table A-1). Analysis (a) The compressibility factor of oxygen at theinitial state is determined from the generalized chart to be O2
220 K 10 MPa 15.1and80.0
97.108.5
10
42.18.154
220
11
cr
11
cr
11
h
R
R
ZZ
PP
P
TT
T
Then,
kg5.17/kgm0.00457
m0.08
/kgm0.00457kPa10,000
K)K)(220/kgmkPa98(0.8)(0.25
3
3
1
33
1
vV
vv
m
ZRTP
The specific volume of oxygen remains constant during this process, v2 = v1. Thus,
kPa12,19050804.2
4.20.187.0
577.0kPa)(5080/K)K)(154.8/kgmkPa(0.2598
/kgm0.00457/
615.18.154
250
cr22
2
2
2
3
3
crcr
22
cr
22
PPP
PZZ
PRT
TT
T
R
R
hR
R
vv
(b) The energy balance relation for this closed system can be expressed as
)()()(
112212112212in
12in
systemoutin
TZTZRhhmPPhhmWuumUW
EEE
vv
wherekJ/kg25.3332/)64047275()115.1)(8.154)(2598.0(
)()( ideal1221cr12 hhZZRThh hh
Substituting,kJ393K2200.802500.87K)kJ/kg(0.259833.25kg)5.17(inW
12-42
12-70 The heat transfer and entropy changes of CO2 during a process are to be determined assuming idealgas behavior, using generalized charts, and real fluid (EES) data.Analysis The temperature at the final state is
K2984MPa1MPa8K)273100(
1
212 P
PTT
Using data from the ideal gas property table of CO2 (Table A-20),
kJ/kg9.3386kg/kmol44
kJ/kmol149,024)()(
Kkmol/kJ115.9418ln314.8367.222770.333ln)(
kmol/kJ024,149269,,12293,161)(
ideal12ideal12
1
212ideal12
ideal1,ideal2,ideal12
Mhh
hh
PP
Rssss
hhhh
u
The heat transfer is determined from an energy balance noting that there is no work interaction
kJ/kg2893.7373)-984kJ/kg.K)(2(0.1889-kJ/kg9.3386)()()( 12ideal12ideal12ideal TTRhhuuq
”
The entropy change is
kJ/kg.K2.1390kg/kmol44
kJ/kmol115.94)()( ideal12
ideal12ideal Mss
sss
The compressibility factor and the enthalpy and entropy departures of CO2 at the specified states are determined from the generalized charts to be (we used EES)
and
002685.0and1144.0,009.1083.1
39.78
813.92.304
2985
05987.0and1028.0,976.0135.0
39.71
226.12.304
373
222
cr
22
cr
22
111
cr
11
cr
11
sh
R
R
sh
R
R
ZZZ
PP
P
TT
T
ZZZ
PP
P
TT
T
Thus,
kJ/kg.K2.151
kJ/kg2935.9
1390.2)002685.0(05987.01889.0
)()()(
)3732887)(1889.0)(976.0()1028.01144.0)(2.304)(1889.0(9.3386)()()(
ideal1221chart12chart
12112crideal1212chart
ssZZRsss
TTRZZZRThhuuq
ss
hh
Note that the temperature at the final state in this case was determined from
K2888009.1976.0
MPa1MPa8K)273100(
2
1
1
212 Z
ZPP
TT
The solution using EES built-in property data is as follows:
kJ/kg.K2464.0
kJ/kg614.8
/kgm06885.0
MPa1K373
1
1
31
1
1
s
uPT
v
kJ/kg.K85.1
kJ/kg2754
K2879
/kgm06885.0
MPa8
2
2
2
312
2
s
u
TP
vv
Then
kJ/kg.K2.097
kJ/kg2763
)2464.0(85.1)(
)614.8(2754
12EES12EES
12EES
sssss
uuq
12-43
Review Problems
12-71 For 0, it is to be shown that at every point of a single-phase region of an h-s diagram, the slopeof a constant-pressure line is greater than the slope of a constant-temperature line, but less than the slope of a constant-volume line.Analysis It is given that > 0.
Using the Tds relation:dsdPT
dsdhdPdsTdh vv
(1) P = constant: Tsh
P
(2) T = constant:TT s
PTsh
v
But the 4th Maxwell relation:PT
TsP
v
Substituting: 1TTTsh
PT vv
Therefore, the slope of P = constant lines is greater than the slope of T = constant lines.
(3) v = constant: )(asPT
sh
vv
v
From the ds relation: vv
v dTPdT
Tc
ds
Divide by dP holding v constant: )(or bTP
cT
sP
PT
Tc
Ps
vvvv
v
v
Using the properties P, T, v, the cyclic relation can be expressed as
)(11 cPTT
PP
TTP
TPTP vv
vvv
v vv
where we used the definitions of and . Substituting (b) and (c) into (a),
TcTT
sPT
sh
vvv
vv
Here is positive for all phases of all substances. T is the absolute temperature that is also positive, so is cv. Therefore, the second term on the right is always a positive quantity since is given to be positive.Then we conclude that the slope of P = constant lines is less than the slope of v = constant lines.
12-44
12-72 Using the cyclic relation and the first Maxwell relation, the other three Maxwell relations are to beobtained.Analysis (1) Using the properties P, s, v, the cyclic relation can be expressed as
1sP P
ssP v
vv
Substituting the first Maxwell relation,vv s
PT
s,
PsPssPs sPTs
PT
PsT v
vv
vv11
(2) Using the properties T, v, s, the cyclic relation can be expressed as
1v
vv T
ss
T
Ts
Substituting the first Maxwell relation,vv s
PT
s,
vvvv vvv
TPs
sTP
Ts
ssP
TTT11
(3) Using the properties P, T, v, the cyclic relation can be expressed as
1TP P
TTP v
vv
Substituting the third Maxwell relation,vv T
Ps
T,
PTPTTPT TPsT
Ps
PTs v
vv
vv11
12-73 It is to be shown that the slope of a constant-pressure line on an h-s diagram is constant in thesaturation region and increases with temperature in the superheated region. Analysis For P = constant, dP = 0 and the given relation reduces to dh = Tds, which can also be expressed as
Tsh
Ph
P = const.
s
Thus the slope of the P = constant lines on an h-s diagram is equal tothe temperature.(a) In the saturation region, T = constant for P = constant lines, and theslope remains constant.(b) In the superheat region, the slope increases with increasing temperature since the slope is equal temperature.
12-45
12-74 The relations for u, h, and s of a gas that obeys the equation of state (P+a/v2)v = RT for an isothermal process are to be derived.Analysis (a) For an isothermal process dT = 0 and the general relation for u reduces to
2
1
2
1
2
112
v
v v
v
v vv vv dP
TPTdP
TPTdTcuuu
T
T
For this gas the equation of state can be expressed as
vvv v
RTPaRTP
2
Thus,
Substituting,
212
22
112
1 vvv
v
vvvv
v
v
v
adau
aaRTRTPTPT
(b) The enthalpy change h is related to u through the relation
where
vv
vv
aRTP
PPuh 1122
Thus,
Substituting,
21
21121122
112
11
vv
vvvvvv
ah
aaRTaRTPP
(c) For an isothermal process dT = 0 and the general relation for s reduces to
2
1
2
1
2
112
v
v v
v
v v
v vv dTPd
TPdT
Tc
sssT
T
Substituting ( P/ T)v = R/v,2
1 1
2lnv
v v
vv
vRdRs
12-46
12-75 It is to be shown that
vv
vTP
TTc
s and
Psp TT
PTc v
Analysis Using the definition of cv ,
vvvv T
PPsT
TsTc
Substituting the first Maxwell relationsTP
s v
v
,
vv
vTP
TTc
s
Using the definition of cp,
PPPp T
sTTsTc v
v
Substituting the second Maxwell relation sP T
Psv
,
Psp TT
PTc v
12-76 The Cp of nitrogen at 300 kPa and 400 K is to be estimated using the relation given and itsdefinition, and the results are to be compared to the value listed in Table A-2b.Analysis (a) We treat nitrogen as an ideal gas with R = 0.297 kJ/kg·K and k = 1.397. Note that PT-k/(k-1) = C = constant for the isentropic processes of ideal gases. The cp relation is given as
)1(111)1/()1/(1)1/()1/(
kTkPTPT
kkCT
kk
TPCTP
PR
TPRT
TTPTc
kkkkkk
s
kk
P
Psp
vv
v
Substituting,
KkJ/kg1.0451397.1
K)kJ/kg71.397(0.291)1( k
kRPR
kTkPTc p
(b) The cp is defined as cp = PT
h . Replacing the differentials by differences,
KkJ/kg1.045K390410
kJ/kg/28.011,34711,932K390410
K390410
kPa300
hKhThc
Pp
(Compare: Table A-2b at 400 K cp = 1.044 kJ/kg·K)
12-47
12-77 The temperature change of steam and the average Joule-Thompson coefficient during a throttlingprocess are to be estimated.Analysis The enthalpy of steam at 4.5 MPa and T = 300 C is h = 2944.2 kJ/kg. Now consider a throttlingprocess from this state to 2.5 MPa. The temperature of the steam at the end of this throttling process is
C72.273kJ/kg2.2944
MPa5.22T
hP
Thus the temperature drop during this throttling process is C26.2830072.27312 TTT
The average Joule-Thomson coefficient for this process is determined from
C/MPa13.14MPa4.52.5
C003273.72
kJ/kg7.3204hh PT
PT
12-78 The initial state and the final temperature of argon contained in a rigid tank are given. The mass of the argon in the tank, the final pressure, and the heat transfer are to be determined using the generalizedcharts.Analysis (a) The compressibility factor of argon at the initial state is determined from the generalized chart to be
Q
Ar-100 C1 MPa
18.0and95.0206.0
86.41
146.10.151
173
1
1
1
1
cr
1
cr
1
h
R
R
ZZ
PP
P
TT
T
Then,
kg135./kgm0.0342
m1.2
/kgm0342.0kPa1000
K)K)(173/kgmkPa081(0.95)(0.2
3
3
33
vV
vv
m
PZRTZRTP
(b) The specific volume of argon remains constant during this process, v2 = v 1. Thus,
kPa1531)4860)(315.0(
099.0
315.0
29.5kPa)K)(4860K)(151/kgmkPa(0.2081
/kgm0.0342/
808.10.151
273
cr2
2
3
3
crcr
2
cr
2
2
2
2
2
2
PPP
ZZ
P
PRT
TT
T
R
h
R
R
R
vv
(c) The energy balance relation for this closed system can be expressed as
)()()(
112212112212in
12in
systemoutin
TZTZRhhmPPhhmQuumUQ
EEE
vv
wherekJ/kg69.5710005203.0018.01512081.0ideal1212 21
hhZZRThh hhcr
Thus,kJ1251K)(0.95)(173)(0.99)(273K)kJ/kg(0.208157.69kg35.1inQ
12-48
12-79 Argon enters a turbine at a specified state and leaves at another specified state. Power output of theturbine and exergy destruction during this process are to be determined using the generalized charts.Properties The gas constant and critical properties of Argon are R = 0.2081 kJ/kg.K, Tcr = 151 K, and Pcr= 4.86 MPa (Table A-1).Analysis (a) The enthalpy and entropy departures of argon at the specified states are determined from thegeneralized charts to be
0044.1
86.47
97.3151600
11
1
1
cr
1
cr
1
sh
R
R
ZandZ
PP
P
TT
T
T0 = 25 CW·
P1 = 7 MPaT1 = 600 K V1 = 100 m/s
Ar = 5 kg/sm·
60 kW
Thus argon behaves as an ideal gas at turbine inlet. Also,
02.0and04.0206.0
86.41
85.1151280
22
2
2
cr
2
cr
2
sh
R
R
ZZ
PP
P
TT
T
P2 = 1 MPaT2 = 280 K V2 = 150 m/s Thus,
kJ/kg8.1676002805203.004.001512081.0ideal1212 21
hhZZRThh hhcr
The power output of the turbine is to be determined from the energy balance equation,
out
21
22
12out
outout2
222
11
outinsystemoutin
2)(
)2/()2/(
(steady)0
QVV
hhmW
WQVhmVhm
EEEEE
Substituting,
kW747.8kJ/s60/sm1000
kJ/kg12
m/s)(100m/s)(1508.167)kg/s5(
22
22
outW
(b) Under steady conditions, the rate form of the entropy balance for the turbine simplifies to
( ),
S S S S
ms ms QT
S S m s s QTb
in out gen system
out
outgen gen
out
0
1 2 2 20
0
0
The exergy destroyed during a process can be determined from an exergy balance or directly from its definition ,X Tdestroyed gen0S
0
out220gen0destroyed )(
TQ
ssmTSTX
where ideal1212 21ssZZRss ss
and KkJ/kg0084.071ln2081.0
600280ln5203.0lnln
1
2
1
2ideal12 P
PR
TT
css p
Thus, KkJ/kg0042.00084.0)02.0(0)2081.0(ideal1212 21ssZZRss ss
Substituting, kW366.K298
kW60KkJ/kg0.0042kg/s5K298destroyedX
12-49
12-80 EES Problem 12-79 is reconsidered. The problem is to be solved assuming steam is the workingfluid by using the generalized chart method and EES data for steam. The power output and the exergydestruction rate for these two calculation methods against the turbine exit pressure are to be plotted.Analysis The problem is solved using EES, and the results are tabulated and plotted below.
" Input Data "T[1]=600 [K] P[1]=7000 [kPa] Vel[1]=100 [m/s] T[2]=455 [K] P[2]=1000 [kPa] Vel[2]=150 [m/s] Q_dot_out=60 [kW] T_o=25+273 "[K]"m_dot=5 [kg/s] Name$='Steam_iapws' T_critical=647.3 [K] P_critical=22090 [kPa] Fluid$='H2O'
R_u=8.314M=molarmass(Fluid$)R=R_u/M
"****** IDEAL GAS SOLUTION ******""State 1"h_ideal[1]=enthalpy(Fluid$,T=T[1]) "Enthalpy of ideal gas"s_ideal[1]=entropy(Fluid$, T=T[1], P=P[1]) "Entropy of ideal gas""State 2"h_ideal[2]=enthalpy(Fluid$,T=T[2]) "Enthalpy of ideal gas"s_ideal[2]=entropy(Fluid$, T=T[2], P=P[2]) "Entropy of ideal gas"
"Conservation of Energy, Steady-flow: ""E_dot_in=E_dot_out"
m_dot*(h_ideal[1]+Vel[1]^2/2*convert(m^2/s^2,kJ/kg))=m_dot*(h_ideal[2]+Vel[2]^2/2*convert(m^2/s^2,kJ/kg))+Q_dot_out+W_dot_out_ideal
"Second Law analysis:""S_dot_in-S_dot_out+S_dot_gen = 0"m_dot*s_ideal[1] - m_dot*s_ideal[2] - Q_dot_out/T_o + S_dot_gen_ideal = 0
"Exergy Destroyed:"X_dot_destroyed_ideal = T_o*S_dot_gen_ideal
"***** COMPRESSABILITY CHART SOLUTION ******""State 1"Tr[1]=T[1]/T_criticalPr[1]=P[1]/P_criticalZ[1]=COMPRESS(Tr[1], Pr[1]) DELTAh[1]=ENTHDEP(Tr[1], Pr[1])*R*T_critical "Enthalpy departure"h_chart[1]=h_ideal[1]-DELTAh[1] "Enthalpy of real gas using charts"DELTAs[1]=ENTRDEP(Tr[1], Pr[1])*R "Entropy departure"s_chart[1]=s_ideal[1]-DELTAs[1] "Entropy of real gas using charts""State 2"
12-50
Tr[2]=T[2]/T_criticalPr[2]=P[2]/P_criticalZ[2]=COMPRESS(Tr[2], Pr[2]) DELTAh[2]=ENTHDEP(Tr[2], Pr[2])*R*T_critical "Enthalpy departure"DELTAs[2]=ENTRDEP(Tr[2], Pr[2])*R "Entropy departure"h_chart[2]=h_ideal[2]-DELTAh[2] "Enthalpy of real gas using charts"s_chart[2]=s_ideal[2]-DELTAs[2] "Entropy of real gas using charts"
"Conservation of Energy, Steady-flow: ""E_dot_in=E_dot_out"
m_dot*(h_chart[1]+Vel[1]^2/2*convert(m^2/s^2,kJ/kg))=m_dot*(h_chart[2]+Vel[2]^2/2*convert(m^2/s^2,kJ/kg))+Q_dot_out+W_dot_out_chart
"Second Law analysis:""S_dot_in-S_dot_out+S_dot_gen = 0"m_dot*s_chart[1] - m_dot*s_chart[2] - Q_dot_out/T_o + S_dot_gen_chart = 0
"Exergy Destroyed:"
X_dot_destroyed_chart = T_o*S_dot_gen_chart"[kW]"
"***** SOLUTION USING EES BUILT-IN PROPERTY DATA *****""At state 1"h_ees[1]=enthalpy(Name$,T=T[1],P=P[1])s_ees[1]=entropy(Name$,T=T[1],P=P[1])"At state 2"h_ees[2]=enthalpy(Name$,T=T[2],P=P[2])s_ees[2]=entropy(Name$,T=T[2],P=P[2])
"Conservation of Energy, Steady-flow: ""E_dot_in=E_dot_out"
m_dot*(h_ees[1]+Vel[1]^2/2*convert(m^2/s^2,kJ/kg))=m_dot*(h_ees[2]+Vel[2]^2/2*convert(m^2/s^2,kJ/kg))+Q_dot_out+W_dot_out_ees
"Second Law analysis:""S_dot_in-S_dot_out+S_dot_gen = 0"m_dot*s_ees[1] - m_dot*s_ees[2] - Q_dot_out/T_o + S_dot_gen_ees= 0
"Exergy Destroyed:"X_dot_destroyed_ees = T_o*S_dot_gen_ees
12-51
P2
[kPa]T2
[K]Woutchart
[kW]Woutees
[kW]Woutideal
[kW]Xdestroyedchart
[kW]Xdestroyedees
[kW]Xdestroyedideal
[kW]100 455 713.3 420.6 1336 2383 2519 2171200 455 725.2 448.1 1336 1901 2029 1694300 455 737.3 476.5 1336 1617 1736 1416400 455 749.5 505.8 1336 1415 1523 1218500 455 761.7 536.1 1336 1256 1354 1064600 455 774.1 567.5 1336 1126 1212 939700 455 786.5 600 1336 1014 1090 833800 455 799.1 633.9 1336 917.3 980.1 741.2900 455 811.8 669.3 1336 831 880.6 660.2
1000 455 824.5 706.6 1336 753.1 788.4 587.7
100 200 300 400 500 600 700 800 900 1000400
800
1200
1600
P[2] [kPa]
Wout;ees
[kW] Solution Method
EESEESChartChartIdeal gasIdeal gas
100 200 300 400 500 600 700 800 900 1000400
800
1200
1600
2000
2400
2800
P[2] [kPa]
Xdestroyed
[kW]
Solution MethodEESEESChartsChartsIdeal GasIdeal Gas
12-52
12-81E Argon gas enters a turbine at a specified state and leaves at another specified state. The poweroutput of the turbine and the exergy destruction associated with the process are to be determined using thegeneralized charts. Properties The gas constant and critical properties of argon are R = 0.04971 Btu/lbm.R, Tcr = 272 R, and Pcr = 705 psia (Table A-1E).Analysis (a) The enthalpy and entropy departures of argon at the specified states are determined from thegeneralized charts to be
0and0418.1
7051000
68.3272
1000
11
1
1
cr
1
cr
1
sh
R
R
ZZ
PP
P
TT
T
W·
P1 = 1000 psia T1 = 1000 R V1 = 300 ft/s
m·Ar
= 12 lbm/s
80 Btu/s
Thus argon behaves as an ideal gas at turbine inlet. Also,
02.0and04.0213.0
705150
P
838.1272500
22
2
2
cr
2
cr
2
sh
R
R
ZZP
P
TT
T
P2 = 150 psia T2 = 500 R V2 = 450 ft/s
Thus ,
Btu/lbm2.6310005001253.004.0027204971.0ideal12cr12 21
hhZZRThh hh
The power output of the turbine is to be determined from the energy balance equation,
out
21
22
12outoutout2
222
11
outinsystemoutin
2)()2/()2/(
(steady)0
QVV
hhmWWQVhmVhm
EEEEE
hp922Btu/s651.4
Btu/s80/sft25,037
Btu/lbm12
ft/s)(300ft/s)(4502.63)lbm/s21(
22
22
outW
(b) Under steady conditions, the rate form of the entropy balance for the turbine simplifies to
0
out22gengen
out,
out21
0systemgenoutin
)(0
0
TQ
ssmSSTQ
smsm
SSSS
b
The exergy destroyed during a process can be determined from an exergy balance or directly from its definition ,X Tdestroyed gen0S
0
out220gen0destroyed )(
TQ
ssmTSTX
where ideal1212 21ssZZRss ss
and RBtu/lbm00745.01000150ln04971.0
1000500ln1253.0lnln
1
2
1
2ideal12 P
PR
TT
css p
Thus RBtu/lbm00646.000745.0)02.0(0)04971.0(ideal1212 21ssZZRss ss
Substituting, sBtu5121 /.R535
Btu/s80RBtu/lbm0.00646lbm/s12R535destroyedX
12-53
12-82 An adiabatic storage tank that is initially evacuated is connected to a supply line that carries nitrogen. A valve is opened, and nitrogen flows into the tank. The final temperature in the tank is to bedetermined by treating nitrogen as an ideal gas and using the generalized charts, and the results are to be compared to the given actual value. Assumptions 1 Uniform flow conditions exist. 2 Kinetic and potential energies are negligible.Analysis We take the tank as the system, which is a control volume since mass crosses the boundary.Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: m m m m m m miin out system out initial (since )2 0
Energy balance: E E E m h m ui iin out system 0 2 2
N2
V1 = 0.2 m3
Initiallyevacuated
10 MPa225 K
Combining the two balances: u2 = hi
(a) From the ideal gas property table of nitrogen, at 225 K we read
u h hi2 6 537@225 K kJ / kmol,
The temperature that corresponds to this u value is 2
(7.4% error)T2 314.8 K
(b) Using the generalized enthalpy departure chart, hi is determined to be
9.095.2
39.310
78.12.126
225ideal,
,
cr,
cr,
cru
iiih
iiR
iiR
TRhh
Z
PP
P
TT
T (Fig. A-29)
Thus,
and
kJ/kmol593,5
kJ/kmol593,52.126314.89.0537,69.0
2
crideal,
i
uii
hu
TRhh
Try T2 = 280 K. Then at PR2 = 2.95 and TR2 = 2.22 we read Z2 = 0.98 and cr2,2 /)( TRhh uideal = 0.55
Thus,
kJ/kmol283,5280314.898.0564,7
kJ/kmol564,72.126314.855.0141,855.0
222
crideal2,2
TZRhu
TRhh
u
u
Try T2 = 300 K. Then at PR2 = 2.95 and TR2 = 2.38 we read Z2 = 1.0 and cr2,2 /)( TRhh uideal = 0.50
Thus,
kJ/kmol704,5300314.80.1198,8
kJ/kmol198,82.126314.850.0723,850.0
222
crideal2,2
TZRhu
TRhh
u
u
By linear interpolation, (0.6% error)K294.72T
12-54
12-83 It is to be shown that dPdTdvv . Also, a relation is to be obtained for the ratio of specific
volumes v 2/ v 1 as a homogeneous system undergoes a process from state 1 to state 2. Analysis We take v = v (P, T). Its total differential is
dPP
dTT
dTP
vvv
Dividing by v,
dPP
dTT
d
TP
vv
vvv
v 11
Using the definitions of and ,
dPdTdvv
Taking and to be constants, integration from 1 to 2 yields
12121
2ln PPTTv
v
which is the desired relation.
12-84 It is to be shown that dPdTdvv . Also, a relation is to be obtained for the ratio of specific
volumes v 2/ v 1 as a homogeneous system undergoes an isobaric process from state 1 to state 2. Analysis We take v = v (P, T). Its total differential is
dPP
dTT
dTP
vvv
which, for a constant pressure process, reduces to
dTT
dP
vv
Dividing by v,
dTT
d
P
vvv
v 1
Using the definition of ,
dTdvv
Taking to be a constant, integration from 1 to 2 yields
or
121
2
121
2
exp
ln
TT
TT
v
v
v
v
which is the desired relation.
12-55
12-85 The volume expansivity of water is given. The change in volume of water when it is heated atconstant pressure is to be determined.Properties The volume expansivity of water is given to be 0.207 10-6 K-1 at 20 C.Analysis We take v = v (P, T). Its total differential is
dPP
dTT
dTP
vvv
which, for a constant pressure process, reduces to
dTT
dP
vv
Dividing by v and using the definition of ,
dTdTT
d
P
vvv
v 1
Taking to be a constant, integration from 1 to 2 yields
or
121
2
121
2
exp
ln
TT
TT
v
v
v
v
Substituting the given values and noting that for a fixed mass V2/V1 = v2/v1,
3
1631212
m00000414.1
C1030K10207.0expm1exp TTVV
Therefore,3cm4.143
12 m00000414.0100000414.1= VVV
12-56
12-86 The volume expansivity of copper is given at two temperatures. The percent change in the volume of copper when it is heated at atmospheric pressure is to be determined.Properties The volume expansivity of copper is given to be 49.2 10-6 K-1 at 300 K, and be 54.2 10-6 K-1 at 500 K Analysis We take v = v (P, T). Its total differential is
dPP
dTT
dTP
vvv
which, for a constant pressure process, reduces to
dTT
dP
vv
Dividing by v and using the definition of ,
dTdTT
d
P
vvv
v 1
Taking to be a constant, integration from 1 to 2 yields
or
121
2
121
2
exp
ln
TT
TT
v
v
v
v
The average value of is 1666
21ave K107.512/102.54102.492/
Substituting the given values,
0104.1K300500K107.51expexp 1612
1
2 TTv
v
Therefore, the volume of copper block will increase by 1.04 percent.
12-57
12-87 It is to be shown that the position of the Joule-Thompson coefficient inversion curve on the T-Pplane is given by ( Z/ T)P = 0. Analysis The inversion curve is the locus of the points at which the Joule-Thompson coefficient is zero,
01v
v
Pp TT
c
which can also be written as
0P
ZRTT
TP
v (a)
since it is given that
PZRT
v (b)
Taking the derivative of (b) with respect to T holding P constant gives
ZT
TPR
TPZRT
T PPP
Z/v
Substituting in (a),
0
0
0
P
P
P
TZ
ZZTZT
PZRTZ
TZT
PTR
which is the desired relation.
12-58
12-88 It is to be shown that for an isentropic expansion or compression process Pv k = constant. It is also tobe shown that the isentropic expansion exponent k reduces to the specific heat ratio cp/cv for an ideal gas. Analysis We note that ds = 0 for an isentropic process. Taking s = s(P, v), the total differential ds can be expressed as
0vvv
dsdPPsds
P (a)
We now substitute the Maxwell relations below into (a)
to get
0
and
vv
vv
v
dTPdP
T
TPs
TPs
ss
sPs
Rearranging,
00 vv
vv
dPdPdTPTdP
sss
Dividing by P, 01v
vdP
PPdP
s (b)
We now define isentropic expansion exponent k as
s
PP
kv
v
Substituting in (b), 0vvdk
PdP
Taking k to be a constant and integrating,
Thus,constant
constantlnconstantlnln
k
k
P
PkP
v
vv
To show that k = cp/cv for an ideal gas, we write the cyclic relations for the following two groups of variables:
)(11,,
)(11,,
dPT
sP
Tc
PT
sP
TsPTs
cTsT
cTsT
sTs
sT
p
sTP
sTsT vv
vv
v v
v
where we used the relations P
p TsTc
TsTc and
vv
Setting Eqs. (c) and (d) equal to each other,
sTsT
p TsT
cPT
sP
Tc
vvv
or,
sTsTsTsT
p PP
TTP
sPsT
sTP
Ps
cc
vv
vv
vv
v
butPP
PRTP TT
vv /
Substituting, kPPc
c
s
p
vv
v
which is the desired relation.
12-59
12-89 EES The work done by the refrigerant 134a as it undergoes an isothermal process in a closed systemis to be determined using the tabular (EES) data and the generalized charts.
Analysis The solution using EES built-in property data is as follows:
kJ/kg.K2035.1kJ/kg35.280
MPa1.0C60
kJ/kg.K4828.0kJ/kg65.135
MPa3C60
2
2
2
2
1
1
1
1
su
PT
su
PT
kJ/kg95.40)65.13535.280(1.240)(
kJ/kg11.240)kJ/kg.K7207.0)(K15.27360(
kJ/kg.K0.72074828.02035.1
12EESEES
EES1EES
12EES
uuqw
sTq
sss
For the generalized chart solution we first determine the following factors using EES as
02281.0and03091.0,988.002464.0
059.41.0
8903.02.37415.333
383.4and475.4,1292.07391.0
059.43
8903.02.37415.333
2222
2
22
1111
1
11
sh
crR
crR
sh
crR
crR
ZZZ
PPP
TTT
ZZZ
PPP
TTT
Then,
kJ/kg.K3572.0kJ/kg.K)08148.0)(383.4(kJ/kg43.136K)74.2kJ/kg.K)(308148.0)(475.4(
11
cr11
RZsRTZh
s
h
kJ/kg.K001858.0kJ/kg.K)08148.0)(02281.0(kJ/kg94.0K)74.2kJ/kg.K)(308148.0)(03091.0(
22
cr22
RZsRTZh
s
h
KkgkJ2771.03
0.1kJ/kg.K)ln08148.0(ln1
2ideal /
PP
Rs
KkJ/kg0.6324)3572.0001858.0(2771.0)( 12idealchart ssss
kJ/kg70.210)kJ/kg.K6324.0)(K15.27360(chart1chart sTq
kJ/kg98.5317.11270.210
kJ/kg17.112)333)(08148.0)(1292.0()333)(08148.0)(988.0()43.13694.0(0)()(
chartchartchart
112212idealchart
uqw
RTZRTZhhhu
The copy of the EES solution of this problem is given next.
12-60
"Input data"T_critical=T_CRIT(R134a) "[K]"P_critical=P_CRIT(R134a) "[kpa]"T[1]=60+273.15"[K]"T[2]=T[1]"[K]"P[1]=3000"[kPa]"P[2]=100"[kPa]"R_u=8.314"[kJ/kmol-K]"M=molarmass(R134a)R=R_u/M"[kJ/kg-K]"
"***** SOLUTION USING EES BUILT-IN PROPERTY DATA *****"
"For the isothermal process, the heat transfer is T*(s[2] - s[1]):"DELTAs_EES=(entropy(R134a,T=T[2],P=P[2])-entropy(R134a,T=T[1],P=P[1]))q_EES=T[1]*DELTAs_EES
s_2=entropy(R134a,T=T[2],P=P[2])s_1=entropy(R134a,T=T[1],P=P[1])
"Conservation of energy for the closed system:"DELTAu_EES=intEnergy(R134a,T=T[2],p=P[2])-intEnergy(R134a,T=T[1],P=P[1])q_EES-w_EES=DELTAu_EESu_1=intEnergy(R134a,T=T[1],P=P[1])u_2=intEnergy(R134a,T=T[2],p=P[2])
"***** COMPRESSABILITY CHART SOLUTION ******""State 1"Tr[1]=T[1]/T_criticalpr[1]=p[1]/p_criticalZ[1]=COMPRESS(Tr[1], Pr[1]) DELTAh[1]=ENTHDEP(Tr[1], Pr[1])*R*T_critical"Enthalpy departure"Z_h1=ENTHDEP(Tr[1], Pr[1]) DELTAs[1]=ENTRDEP(Tr[1], Pr[1])*R "Entropy departure"Z_s1=ENTRDEP(Tr[1], Pr[1])
"State 2"Tr[2]=T[2]/T_criticalPr[2]=P[2]/P_criticalZ[2]=COMPRESS(Tr[2], Pr[2]) DELTAh[2]=ENTHDEP(Tr[2], Pr[2])*R*T_critical"Enthalpy departure"Z_h2=ENTHDEP(Tr[2], Pr[2]) DELTAs[2]=ENTRDEP(Tr[2], Pr[2])*R "Entropy departure"Z_s2=ENTRDEP(Tr[2], Pr[2])
"Entropy Change"DELTAs_ideal= -R*ln(P[2]/P[1]) DELTAs_chart=DELTAs_ideal-(DELTAs[2]-DELTAs[1])
"For the isothermal process, the heat transfer is T*(s[2] - s[1]):"q_chart=T[1]*DELTAs_chart
"Conservation of energy for the closed system:"DELTAh_ideal=0DELTAu_chart=DELTAh_ideal-(DELTAh[2]-DELTAh[1])-(Z[2]*R*T[2]-Z[1]*R*T[1])q_chart-w_chart=DELTAu_chart
12-61
SOLUTION
DELTAh[1]=136.43DELTAh[2]=0.94DELTAh_ideal=0DELTAs[1]=0.3572DELTAs[2]=0.001858DELTAs_chart=0.6324 [kJ/kg-K] DELTAs_EES=0.7207 [kJ/kg-K] DELTAs_ideal=0.2771 [kJ/kg-K] DELTAu_chart=112.17DELTAu_EES=144.7M=102 [kg/kmol]P[1]=3000 [kPa] P[2]=100 [kPa] pr[1]=0.7391Pr[2]=0.02464P_critical=4059 [kpa] q_chart=210.70 [kJ/kg] q_EES=240.11 [kJ/kg] R=0.08148 [kJ/kg-K
R_u=8.314 [kJ/kmol-K]s_1=0.4828 [kJ/kg-K] s_2=1.2035 [kJ/kg-K] T[1]=333.2 [K] T[2]=333.2 [K] Tr[1]=0.8903Tr[2]=0.8903T_critical=374.2 [K] u_1=135.65 [kJ/kg] u_2=280.35 [kJ/kg] w_chart=98.53 [kJ/kg] w_EES=95.42 [kJ/kg]Z[1]=0.1292Z[2]=0.988Z_h1=4.475Z_h2=0.03091Z_s1=4.383Z_s2=0.02281
12-62
12-90 The heat transfer, work, and entropy changes of methane during a process in a piston-cylinderdevice are to be determined assuming ideal gas behavior, using generalized charts, and real fluid (EES) data.Analysis The ideal gas solution: (Properties are obtained from EES)
State 1:
/kgm04834.0kPa4000
K15.273100)kJ/kg.K5182.0(
kJ/kg4685)15.273100)(5182.0()4492(kJ/kg.K22.10MPa4C,100
kJ/kg4492C100
3
1
11
111
111
11
PT
R
RThusPT
hT
v
State 2:
/kgm08073.0kPa4000
K15.273350)kJ/kg.K5182.0(
kJ/kg4093)15.273350)(5182.0()3770(kJ/kg.K68.11MPa4C,350
kJ/kg3770C350
3
2
22
222
222
22
PT
R
RThusPT
hT
v
kJ/kg129.56/kg0.04834)m-73kPa)(0.0804000()( 312ideal vvPw
kJ/kg721.70)4685()4093(56.129)( 12idealideal uuwq
kJ/kg1.4622.1068.1112ideal sss
For the generalized chart solution we first determine the following factors using EES as
2555.0and4318.0,9023.05413.0
39.74
227.12.304
373
111
cr
11
cr
11
sh
R
R
ZZZ
PP
P
TT
T
06446.0and1435.0,995.05413.0
39.74
048.22.304
623
222
cr
22
cr
22
sh
R
R
ZZZ
PP
P
TT
T
State 1:
kJ/kg4734)15.373)(5182.0)(9023.0()4560(kJ/kg456007.68)4492(
kJ/kg07.68K)04.2kJ/kg.K)(35182.0)(4318.0(
1111
1ideal1,1
cr11
RTZhuhhh
RTZh h
/kgm04362.04000
15.373)5182.0)(9023.0( 3
1
111 P
TRZv
kJ/kg.K09.101324.022.10kJ/kg.K1324.0kJ/kg.K)5182.0)(2555.0(
11,ideal1
11
sssRZs s
State 2:
kJ/kg4114)15.623)(5182.0)(995.0()3793(kJ/kg379362.22)3770(
kJ/kg62.22K)04.2kJ/kg.K)(35182.0)(1435.0(
2222
2ideal2,2
cr22
RTZhuhhh
RTZh h
12-63
/kgm08033.04000
15.623)5182.0)(995.0( 3
2
222 P
TRZv
kJ/kg.K65.1103341.068.11kJ/kg.K03341.0kJ/kg.K)5182.0)(06446.0(
22,ideal2
22
sssRZs s
Then,
kJ/kg146.84/kg0.04362)m-33kPa)(0.0804000()( 312chart vvPw
kJ/kg766.84)4734()4114(84.146)( 12chartchart uuwq
kJ/kg1.5609.1065.1112chart sss
The solution using EES built-in property data is as follows:
kJ/kg.K439.1kJ/kg82.39
/kgm04717.0
MPa4C100
1
1
31
1
1
su
PT
v
kJ/kg.K06329.0kJ/kg52.564
/kgm08141.0
MPa4C350
2
2
32
2
2
su
PT
v
kJ/kg136.96/kg0.04717)m-41kPa)(0.0814000()( 312EES vvPw
kJ/kg741.31)82.39(52.56497.136)( 12EESEES uuwq
kJ/kg1.50)439.1(06329.012EES sss
12-64
Fundamentals of Engineering (FE) Exam Problems
12-91 A substance whose Joule-Thomson coefficient is negative is throttled to a lower pressure. Duringthis process, (select the correct statement)(a) the temperature of the substance will increase.(b) the temperature of the substance will decrease.(c) the entropy of the substance will remain constant.(d) the entropy of the substance will decrease.(e) the enthalpy of the substance will decrease.
Answer (a) the temperature of the substance will increase.
12-92 Consider the liquid-vapor saturation curve of a pure substance on the P-T diagram. The magnitudeof the slope of the tangent line to this curve at a temperature T (in Kelvin) is(a) proportional to the enthalpy of vaporization hfg at that temperature,(b) proportional to the temperature T,(c) proportional to the square of the temperature T,(d) proportional to the volume change vfg at that temperature,(e) inversely proportional to the entropy change sfg at that temperature,
Answer (a) proportional to the enthalpy of vaporization hfg at that temperature,
12-93 Based on the generalized charts, the error involved in the enthalpy of CO2 at 350 K and 8 MPa if it isassumed to be an ideal gas is (a) 0 (b) 20% (c) 35% (d) 26% (e) 65%
Answer (d) 26%
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).
T=350 "K"P=8000 "kPa"Pcr=P_CRIT(CarbonDioxide)Tcr=T_CRIT(CarbonDioxide)Tr=T/TcrPr=P/PcrZ=COMPRESS(Tr, Pr) hR=ENTHDEP(Tr, Pr)
12-65
12-94 Based on data from the refrigerant-134a tables, the Joule-Thompson coefficient of refrigerant-134aat 0.8 MPa and 100 C is approximately(a) 0 (b) -5 C/MPa (c) 11 C/MPa (d) 8 C/MPa (e) 26 C/MPa
Answer (c) 11 C/MPa
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).
T1=100 "C"P1=800 "kPa"h1=ENTHALPY(R134a,T=T1,P=P1)Tlow=TEMPERATURE(R134a,h=h1,P=P1+100)Thigh=TEMPERATURE(R134a,h=h1,P=P1-100)JT=(Tlow-Thigh)/200
12-95 For a gas whose equation of state is P(v - b) = RT, the specific heat difference cp – cv is equal to(a) R (b) R – b (c) R + b (d) 0 (e) R(1 + v/b)
Answer (a) R
Solution The general relation for the specific heat difference cp - cv is
TPp
PT
Tccv
vv
2
For the given gas, P(v - b) = RT. Then,
bP
bRTP
bRTP
PR
Tb
PRT
T
P
vvvv
vv
2)(
Substituting,
RbP
TRb
PPRTcc p )(
22
vvv
12-96 ··· 12-98 Design and Essay Problems