Chapter 12 THERMODYNAMIC PROPERTY RELATIONS · PDF file12-1 Chapter 12 THERMODYNAMIC PROPERTY RELATIONS Partial Derivatives and Associated Relations 12-1C dz ()z x ()z y y dy

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 the other 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

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

12-2

12-8 Air 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 Air is an ideal gas Properties 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,

( ) kPa 276.1=⋅⋅

==/kgm 0.90

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,

( ) kPa 276.1−=⋅⋅

−=−=23

33

2 /kg)m (0.90/kg)m K)(0.009K)(400/kgmkPa (0.287

v

vdTRdP T

(c) When both v and T increases by 1%, the change in P becomes 0=−+=+= )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 gas Properties 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,

( ) kPa 2319./kgm 0.90

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,

( ) kPa 2319./kg)m (0.90

)m K)(0.009 K)(400/kgmkPa (2.076923

33

2−=

⋅⋅=−=

v

vdTRdP T

(c) When both v and T increases by 1%, the change in P becomes 0)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 v holding P constant yields

RPT

P=

∂∂v

T

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 to P/R, which is proportional to P. Therefore, the high pressure lines are 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,

( )( )R

bbRP

T −=−+=

∂∂ 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 determined using 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)/dT and cv(T) = du(T)/dT. Approximating the differentials as differences about 400 K, the cp and cv values are determined to be

( ) ( )( )

KkJ/kg 1.045 ⋅=

−−

=

−−

=

∆

∆≅

=≅=

K390)(410kJ/kg .011,347)/28(11,932

K390410K 390K 410

)()()K 400(K 400K 400

hh

TTh

dTTdhc

TTp

cp h

T(Compare: Table A-2b at 400 K → cp = 1.044 kJ/kg·K)

( ) ( )( )

KkJ/kg 0.748 ⋅=−

−=

−−

=

∆

∆≅

=≅=

K390)(410kJ/kg 08,104)/28.(8,523

K390410K 390K 410

)()()K400(K 400K 400

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 determined using 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)/dT and cv(T) = du(T)/dT. Approximating the differentials as differences about 600 R, the cp and cv values are determined to be

( ) ( )( )

RBtu/lbm 0.249 ⋅=−

−=

−−

=

∆∆

≅

=

≅=

R580)(620Btu/lbm 8.04,028.7)/2(4,307.1

R580620R 580R 620

)()()R 600(R 600R 600

hh

TTh

dTTdhc

TTp

(Compare: Table A-2Eb at 600 R → cp = 0.248 Btu/lbm·R )

( ) ( )( )

RBtu/lbm 0.178 ⋅=−

−=

−−

=

∆∆

≅

=

≅=

R580)(620Btu/lbm 8.02,876.9)/2(3,075.9

R580620R 580R 620

)()()R 600(R 600R 600

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

kPa 4kPa100)96(

K 4K400)(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,

/kgm 0.0598 3=+=

−−⋅⋅=

−=

∂∂

+

∂∂

=

/kg)m (0.04805/kg)m (0.0117

kPa) (98kPa) 4K)( (402

kPa 98K 4

K)/kgmkPa (0.287

33

23

2PdPTR

PdTR

dPP

dTT

dTP

vvv

(b) Using the ideal gas relation at each state,

/kgm 1.2078kPa 96

K) K)(404/kgmkPa (0.287

/kgm 1.1480kPa 100

K) K)(400/kgmkPa (0.287

33

2

22

33

1

11

=⋅⋅

==

=⋅⋅

==

PRT

PRT

v

v

Thus,

/kgm 0.0598 3=−=−=∆ 1480.12078.112 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 v are 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 by P, v, and T, the cyclic relation can be expressed as

1−=

∂∂

∂∂

∂∂

v

vv P

TT

P

PT

where

( )

Ra

PT

RaPT

PR

Ta

PRT

aP

aRTP

aRTP

P

T

−=

∂∂

→−

=

=

∂∂

→+=

−−=

−

−=

∂∂

→−

=

vv

vv

vvvv

v

)(

2

Substituting,

1−=

−

−−=

∂∂

∂∂

∂∂

Ra

PR

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 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,

( ) ( )

K/kgm 101.0095K/kgm 101.005

C60)(100/kgm0.018404)(0.022442

kPa1000)(1400KkJ/kg1.0458)(1.0056

C60100kPa10001400

3434

3?

kPa1200

C60C100?

C80

kPa 1000kPa 1400

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 at the 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_increment P[1]=P-P_increment T[2]=T+T_increment T[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 Gibbs function 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/lbmft 101.635R/lbmft 101.639

F00)7(900/lbmft.6507)1(1.9777

psia0)35(450RBtu/lbm1.7009)(1.6706

F007900psia035450

3333

3?

psia400

F700F900?

F800

psia 350psia 450

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 is satisfied. 12-19 Using the Maxwell relations, a relation for (∂s/∂P)T for a gas whose equation of state is P(v-b) = RT is to be obtained.

Analysis This equation of state can be expressed as bP

RT+=v . 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,

b

RTP

−=

∂∂

vv

From the third Maxwell relation,

b

R−

=

∂∂

=

∂∂

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

R=

∂∂

=

∂∂

TPs

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 specified temperature is to be estimated and to be compared to the tabulated data. Analysis From the Clapeyron equation,

kJ/kg 163.00=

−−+=

°−°

−−=

∆∆

−≅

=

°°°

°°

K 4kPa.68)963(1072.8

/kg)m 0.00087202K)(0.01995 273.15(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 function of temperature over the temperature range -20 to 80°C by using the Clapeyron equation and the refrigerant 134-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_increment T[1]=T-T_increment P[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 diagram and 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 -20 205.96 206.56 0.2895 -10 198.60 199.05 0.2283 0 190.73 191.07 0.1776 10 182.27 182.52 0.1394 20 173.08 173.28 0.1154 30 163.00 163.18 0.1057 40 151.79 151.96 0.1081 50 139.10 139.26 0.1166 60 124.32 124.47 0.1195 70 106.35 106.45 0.09821 80


12-12

12-27 Using the Clapeyron equation, the enthalpy of vaporization of steam at a specified pressure is to be estimated and to be compared to the tabulated data. Analysis From the Clapeyron equation,

kJ/kg 2161.1=

°−

−+=

−−

−=

∆∆

−≅

=

C130.58)(136.27kPa 50

/kg)m 0.001073K)(0.60582 273.15(133.52

kPa75)2(325)(

)(

3

kPa @275satkPa 253@satkPa 003@kPa 300@sat

kPa 300 sat,kPa 300@

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 equation and to be compared to the tabulated data. Analysis From the Clapeyron equation,

kgkJ 82206 /.K 10

kPa.18)169(232.23/kg)m 0.001060K)(0.89133 273.15(120

C115C125)(

)(

3

C115@satC125@satC120@

C120sat,C120@

sat

=

−−+=

°−°

−−=

∆∆

−≅

=

°°°

°°

PPT

TPT

dTdPTh

fg

fg

fgfg

vv

vv

v

Also,

Kkg/kJ 6131.5 ⋅=+

==K273.15)(120

kJ/kg 2206.8T

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 the tabulated data. Analysis (a) From the Clapeyron equation,

error) (0.2%/lbmftpsia 444.0

R 20psia 49.776)(72.152

/lbm)ft 0.01270R)(0.79136 459.67(50

F40F60)(

)(

3

3

F 40 @satF 60 @satF50 @

F50 sat,F50 @

sat

Btu/lbm 82.16=⋅=

−−+=

°−°

−−=

∆∆

−≅

=

°°°

°°

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%

R 67.459601

R 67.459401

RBtu/lbm 0.01946psia 776.49psia 72.152

ln

11lnsat21sat1

2

Btu/lbm 93.80=

+−

+⋅≅

−≅

fg

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 equation and 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

/kgm 6710.1001043.06720.1

/kgm 6720.1

/kgm 001043.0kPa 29.3661.8490.120∆

C1095105∆kPa 90.120

kPa 61.84C1055100

C955100

3

3C100 @

3C100 @

12

12

C105 @sat2

C95 @sat1

increment2

increment1

=−=−=

=

=

=−=−=°=−=−=

====

°=+=+=°=−=−=

°

°

°

°

fgfg

g

f

PPPTTT

PPPP

TTTTTT

vvv

v

v

Substituting,

kJ/kg 2262.8=+=∆∆

=K 10kPa 36.29/kg)m K)(1.6710 15.273100( 3

Clapeyron, TPTh fgfg v

The enthalpy of vaporization from steam table is

/kgm 2256.4 3=°C100 @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 diagram and 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]

PercentError

[%]

T [C]

2477.20 2508.09 1.247 10 2429.82 2451.09 0.8756 30 2381.95 2396.69 0.6188 50 2333.04 2343.47 0.4469 70 2282.51 2290.07 0.3311 90 2229.68 2235.25 0.25 110 2173.73 2177.86 0.1903 130 2113.77 2116.84 0.1454 150 2014.17 2016.15 0.09829 180 1899.67 1900.98 0.06915 210 1765.50 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 equation and 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 in Table A-4). Also, the enthalpy of sublimation of water at -30ºC is determined from Table A-8 to be 2838.4 kJ/kg. Substituting,

kPa 0.03799=

+−−=

−=

CCsub,

CCsub,

211

CCsub,

K 15.273301

K 16.2731

kJ/kg.K 0.4615kJ/kg 4.2838

kPa 6117.0ln

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 an incompressible substance it also depends on pressure. Analysis The change in enthalpy is expressed as

dPT

TdTcdhP

P

∂∂

−+=v

v

For an ideal gas v = RT/P. Then,

Thus, dTcdh

PRT

TT

p

P

=

=−=

−=

∂∂

− 0vvvv

v

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)/( and 2

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 and pressure.


12-18

12-34 General expressions for ∆u, ∆h, and ∆s for a gas that obeys the van der Waals equation of state 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

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

vv

vv

ab

RTP −−

=

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

Rs−−

=−

=∆ ∫1

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,

0=−=−−

=−

∂∂

−=

∂∂

→−

=

PPPa

RTP

TP

T

aR

TP

aRT

P

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 ln and , ,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 P

PPT

PTP

PTPT

Pu

∂∂

−

∂∂

∂∂

=

∂∂

−

∂∂

+=

∂∂ vvv

vv

0

Using the properties P, T, v, the cyclic relation can be expressed as

PTTP TPT

PP

TTP

∂∂

−=

∂∂

∂∂

→−=

∂∂

∂∂

∂∂ vvv

v vv

1

Substituting, we get

TPT P

PT

TPu

∂∂

−

∂∂

−=

∂∂ vv

The general relation for dh is

dPT

TdTcdhP

p

∂∂

−+=v

v

Differentiating each term in this equation with respect to v at T = constant yields

TPTTPT

PT

TPPT

Th

∂∂

∂∂

−

∂∂

=

∂∂

∂∂

−+=

∂∂

vv

vv

vv

vv

0

Using the properties v, T, P, the cyclic relation can be expressed as

vv v

vv

v

∂∂

−=

∂∂

∂∂

→−=

∂∂

∂∂

∂∂

PTP

TP

PT

T TPTP1

Substituting, we get

vv

vv

∂∂

+

∂∂

=

∂∂

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

TP

pP

TTcc

∂∂

∂∂

−=−v

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 =−

+ v

v 2 . Then,

( ) ( )

( )b

TRba

aPR

baR

TbaPR

TP

−+

−=

++−

−=

∂∂

→−

+=

vv

vv

vvv

vv

3

232

2

1211

Inverting, ( )b

TRbaT 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, 0=− vcc 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

TP

pP

TTcc

∂∂

∂∂

−=−v

vv

2

Approximating differentials by differences about the specified state,

Kkg/kJ 0.3214 ⋅=⋅⋅=

−

−−=

−−

°−

−+−=

∆∆

∆∆

−≅−

°==

°°

°==

K/kgmkPa 3114.0

/kgm0.0010244)(0.0010199kPa 10,000

K 40/kgm0.0010105)(0.0010361

)K 15.353(

MPa)10(20C)60100(

)K 15.27380(

3

3

23

C80MPa10MPa20

2

MPa 15

C60C100

C80

2

MPa 15

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

TP

pP

TTcc

∂∂

∂∂

−=−v

vv

2


)ftpsia 5.4039Btu (1R/lbmftpsia 3081.0

/lbmft0.016317)(0.016267psia 1000

R 50/lbmft0.016177)(0.016427

R) 609.67(

psia500)(1500F)125175(

)R 67.459150(

33

3

23

F150psia500psia1500

2

psia1000

F125F175

F150

2

psia1000

⋅=⋅=⋅⋅=

−

−−=

−−

°−

−+−=

∆∆

∆∆

−≅−

°==

°°

°==

RBtu/lbm 0.0570

TP

TPp

PT

Tcc

vv

vv

vv

v


12-23

12-40 Relations for the volume expansivity β and the isothermal compressibility α 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 PT

∂∂

−=α

∂∂

=βv

vv

v1and1

(a) For an ideal gas v = RT/P. Thus,

P11

T11

2=

−−=α→−=−=

∂∂

==β→=

∂∂

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

−=

−−−=α→

−−=−=

∂∂

+==β→=

∂∂

1

1

2

12-41 The volume expansivity β and the isothermal compressibility α of refrigerant-134a at 200 kPa and 30°C are to be estimated. Analysis The volume expansivity and isothermal compressibility are expressed as

TP PT

∂∂

−=α

∂∂

=βv

vv

v1and1


1K 003810 −

=

°°

=

=

−=

°−

−=

∆∆

≅β

.K 20

/kgm0.11418)(0.12322/kgm 0.11874

1

C)2040(11

3

3

kPa200

C20C40

kPa200 PPTvv

vv

v

and

1kPa 004820 −

°=°=

=

−−=

−

−−=

∆∆

−≅α

.kPa 60

/kgm0.13248)(0.09812/kgm 0.11874

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=

∂∂

→+=v

v

Substituting into the Joule-Thomson coefficient relation,

( ) 0111<−=+−−=

−−=

∂∂

−−=µpppPp c

aacP

RTcT

Tc

vvvv

v

Therefore, this gas cannot be cooled by throttling since µ 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(

22

11 2

vv

vv

vv

vRTac

aaRT

RTcT

Tc ppPp −

−=

−−−=

∂∂

−−=µ

The temperature at µ = 0 is the inversion temperature,

00)2(

2=→=

−−=µ v

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/kg 3.2994=

∆∆

≅

∂∂

=µhh P

TPT

Considering a throttling process from 3.5 MPa to 2.5 MPa at h = 2994.3 kJ/kg, the Joule-Thomson coefficient is determined to be

C/MPa 12.3 °=−

°−=

−

−=µ

=MPa )5.2(3.5

C)2943.306(MPa )5.2(3.5

kJ/kg 3.2994

MPa 2.5MPa 3.5

h

TT

(b) The enthalpy of steam at 6 MPa and 500°C is h = 3423.1 kJ/kg. Approximating differentials by differences about the specified state, the Joule-Thomson coefficient is expressed as

kJ/kg 1.3423=

∆∆

≅

∂∂

=µhh P

TPT


C/MPa 4.9 °=−

°−=

−

−=

=MPa 5.0)(7.0

C)1.4958.504(MPa 5.0)(7.0

kJ/kg 1.3423

MPa 5.0MPa 7.0

h

TTµ

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/lbm 564.10−=

∆∆

≅

∂∂

=µhh P

TPT

Considering a throttling process from 210 psia to 190 psia at h = -10.564 Btu/lbm, the Joule-Thomson coefficient is determined to be

R/psia 0.0297=−−

=

−

−=µ

−=psia )210(190

R 500.296)(499.703psia 210)(190

Btu/lbm 564.10

psia 210psia 190

h

TT

(b) The enthalpy of nitrogen at 2000 psia and 400 R is, from EES, h = -55.321 Btu/lbm. Approximating differentials by differences about the specified state, the Joule-Thomson coefficient is expressed as

Btu/lbm 321.55−=

∆∆

≅

∂∂

=µhh P

TPT

Considering a throttling process from 2010 psia to 1990 psia at h = -55.321 Btu/lbm, the Joule-Thomson coefficient is determined to be

R/psia 0.0213=−

=

−

−=µ

−=psia 2010)(1990

R 400.213)-(399.786psia 2010)(1990

Btu/lbm 321.55

psia 2001psia 1999

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/lbm

P [psia] µ [R/psia] 100 0.003675 275 0.003277 450 0.002899 625 0.00254 800 0.002198 975 0.001871

1150 0.001558 1325 0.001258 1500 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]

µ[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. Approximating differentials by differences about the specified state, the Joule-Thomson coefficient is expressed as

kJ/kg 53.288=

∆∆

≅

∂∂

=µhh P

TPT


C/MPa 18.1 °=−

°−=

−

−=µ

=MPa0.6)(0.8

C.19)48(51.81MPa0.6)(0.8

kJ/kg 53.288

MPa 0.6MPa 0.8

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 temperature of the steam at the end of this throttling process will be

C52.297kJ/kg 6.3051

MPa 8.02 °=

==

ThP

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/kmol 8.5083

2N === Mh kJ/kg 181.48

N2 175 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/kmol 28.013kJ/kmol 3405.0

kJ/kmol 0.34052.126314.86.18.5083crideal

kJ/kg 121.6===

=−=−=

Mhh

TRZhh uh


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Chapter 12 THERMODYNAMIC PROPERTY RELATIONS · PDF file12-1 Chapter 12 THERMODYNAMIC PROPERTY RELATIONS Partial Derivatives and Associated Relations 12-1C dz ()z x ()z y y dy