Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Like this document? Why not share!

- Ch.15 by R G Sanjay Prakash 7973 views
- Ch.14 by R G Sanjay Prakash 3036 views
- Kota super thermal power plant by R G Sanjay Prakash 5270 views
- Course plan intt_pse_psm_jan_2010_t... by R G Sanjay Prakash 571 views
- Chapter 12 thermodynamics solutio... by Muhammad Abdullah... 1432 views
- Ch.16 by R G Sanjay Prakash 2888 views

No Downloads

Total views

5,151

On SlideShare

0

From Embeds

0

Number of Embeds

13

Shares

0

Downloads

265

Comments

0

Likes

4

No embeds

No notes for slide

- 1. 12-1 Chapter 12 THERMODYNAMIC PROPERTY RELATIONSPartial Derivatives and Associated Relations12-1C z dz ∂x ≡ dx ∂y ≡ dy dz = (∂z ) x + (∂z ) y (∂z)y (∂z)x y y + dy y dy x dx x+dx x12-2C For functions that depend on one variable, they are identical. For functions that depend on two ormore 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 asa result of differential changes in all variables.12-3C (a) (∂x)y = dx ; (b) (∂z) y ≤ dz; and (c) dz = (∂z)x + (∂z) y12-4C Yes.12-5C Yes.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 2. 12-212-6 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 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), ⎛ ∂P ⎞ ⎛ ∂P ⎞ R dT RT dv dP = ⎜ ⎟ dT + ⎜ ⎟ dv = − ⎝ ∂T ⎠v ⎝ ∂v ⎠ T v v2(a) The change in T can be expressed as dT ≅ ΔT = 400 × 0.01 = 4.0 K. At v = constant, (0.287 kPa ⋅ m 3 /kg ⋅ K)(4.0 K) (dP )v = R dT = = 1.276 kPa v 0.90 m 3 /kg(b) The change in v can be expressed as dv ≅ Δv = 0.90 × 0.01 = 0.009 m3/kg. At T = constant, RT dv (0.287 kPa ⋅ m 3 /kg ⋅ K)(400K)(0.009 m 3 /kg) (dP )T =− =− = −1.276 kPa v2 (0.90 m 3 /kg) 2(c) When both v and T increases by 1%, the change in P becomes dP = (dP)v + (dP )T = 1.276 + (−1.276) = 0Thus the changes in T and v balance each other.12-7 Helium 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 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 ), ⎛ ∂P ⎞ ⎛ ∂P ⎞ R dT RT dv dP = ⎜ ⎟ dT + ⎜ ⎟ dv = − ⎝ ∂T ⎠v ⎝ ∂v ⎠ T v v2(a) The change in T can be expressed as dT ≅ ΔT = 400 × 0.01 = 4.0 K. At v = constant, (2.0769 kPa ⋅ m 3 /kg ⋅ K)(4.0 K) (dP )v = R dT = = 9.231 kPa v 0.90 m 3 /kg(b) The change in v can be expressed as d v ≅ Δ v = 0.90 × 0.01 = 0.009 m3/kg. At T = constant, RT dv (2.0769 kPa ⋅ m 3 /kg ⋅ K)(400 K)(0.009 m 3 ) (dP )T =− = = −9.231 kPa v2 (0.90 m 3 /kg) 2(c) When both v and T increases by 1%, the change in P becomes dP = (dP) v + (dP ) T = 9.231 + (−9.231) = 0Thus the changes in T and v balance each other.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 3. 12-312-8 It is to be proven for an ideal gas that the P = constant lines on a T- v diagram are straight lines andthat 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 ⎛ ∂T ⎞ P T ⎜ ⎟ = ⎝ ∂v ⎠ P Rwhich remains constant at P = constant. Thus the derivative P = const(∂T/∂v)P, which represents the slope of the P = const. lineson 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 equalto P/R, which is proportional to P. Therefore, the highpressure lines are steeper than low pressure lines on the T-v vdiagram.12-9 A relation is to be derived for the slope of the v = constant lines on a T-P diagram for a gas that obeysthe van der Waals equation of state.Analysis The van der Waals equation of state can be expressed as 1⎛ ⎞ ⎟(v − b ) a T= ⎜P + 2 R⎝ v ⎠Taking the derivative of T with respect to P holding v constant, ⎛ ∂T ⎞ 1 v −b ⎜ ⎟ = (1 + 0)(v − b ) = ⎝ ∂P ⎠v R Rwhich is the slope of the v = constant lines on a T-P diagram.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 4. 12-412-10 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 ⎛ dh(T ) ⎞ ⎛ Δh(T ) ⎞ cp c p (400 K ) = ⎜ ⎟ ≅⎜ ⎟ ⎝ dT ⎠ T = 400 K ⎝ ΔT ⎠ T ≅ 400 K h h(410 K ) − h(390 K ) = (410 − 390)K (11,932 − 11,347)/28.0 kJ/kg = (410 − 390)K = 1.045 kJ/kg ⋅ K(Compare: Table A-2b at 400 K → cp = 1.044 kJ/kg·K) T ⎛ du (T ) ⎞ ⎛ Δu (T ) ⎞ cv (400K ) = ⎜ ⎟ ≅⎜ ⎟ ⎝ dT ⎠ T = 400 K ⎝ ΔT ⎠ T ≅ 400 K u (410 K ) − u (390 K ) = (410 − 390)K (8,523 − 8,104)/28.0 kJ/kg = = 0.748 kJ/kg ⋅ K (410 − 390)K(Compare: Table A-2b at 400 K → cv = 0.747 kJ/kg·K)PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 5. 12-512-11E 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 aredetermined to be ⎛ dh(T ) ⎞ ⎛ Δh(T ) ⎞ c p (600 R ) = ⎜ ⎟ ≅⎜ ⎟ ⎝ dT ⎠T = 600 R ⎝ ΔT ⎠T ≅ 600 R h(620 R ) − h(580 R ) = (620 − 580)R (4,307.1 − 4,028.7)/28.0 Btu/lbm = = 0.249 Btu/lbm ⋅ R (620 − 580)R(Compare: Table A-2Eb at 600 R → cp = 0.248 Btu/lbm·R ) ⎛ du (T ) ⎞ ⎛ Δu (T ) ⎞ cv (600 R ) = ⎜ ⎟ ≅⎜ ⎟ ⎝ dT ⎠T = 600 R ⎝ ΔT ⎠T ≅ 600 R u (620 R ) − u (580 R ) = (620 − 580)R (3,075.9 − 2,876.9)/28.0 Btu/lbm = = 0.178 Btu/lbm ⋅ R (620 − 580) R(Compare: Table A-2Eb at 600 R → cv = 0.178 Btu/lbm·R)PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 6. 12-612-12 The state of an ideal gas is altered slightly. The change in the specific volume of the gas is to bedetermined 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 dT ≅ ΔT = (404 − 400)K = 4 K dP ≅ ΔP = (96 − 100)kPa = −4 kPaThe 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, ⎛ ∂v ⎞ ⎛ ∂v ⎞ R dT RT dP dv = ⎜ ⎟ dT + ⎜ ⎟ dP = − ⎝ ∂T ⎠ P ⎝ ∂P ⎠ T P P2 ⎛ 4K (402 K)(−4 kPa) ⎞ = (0.287 kPa ⋅ m 3 /kg ⋅ K)⎜ − ⎟ ⎜ 98 kPa (98 kPa) 2 ⎟ ⎝ ⎠ = (0.0117 m 3 /kg) + (0.04805 m 3 /kg) = 0.0598 m 3 /kg(b) Using the ideal gas relation at each state, RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K)(400 K) v1 = = = 1.1480 m 3 /kg P1 100 kPa RT2 (0.287 kPa ⋅ m 3 /kg ⋅ K)(404 K) v2 = = = 1.2078 m 3 /kg P2 96 kPaThus, Δv = v 2 − v1 = 1.2078 − 1.1480 = 0.0598 m3 /kgThe two results are identical.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 7. 12-712-13 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 asthe 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 ⎛ ∂P ⎞ ⎛ ∂v ⎞ ⎛ ∂T ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = −1 ⎝ ∂v ⎠T ⎝ ∂T ⎠ P ⎝ ∂P ⎠vwhere RT ⎛ ∂P ⎞ − RT P P= ⎯⎯→ ⎜ ⎟ = =− v −a ⎝ ∂v ⎠ T (v − a )2 v −a RT ⎛ ∂v ⎞ R v= +a ⎯ ⎯→ ⎜ ⎟ = P ⎝ ∂T ⎠ P P P (v − a) ⎛ ∂T ⎞ v −a T= ⎯ ⎯→ ⎜ ⎟ = R ⎝ ∂P ⎠ v RSubstituting, ⎛ ∂P ⎞ ⎛ ∂v ⎞ ⎛ ∂T ⎞ ⎛ P ⎞⎛ R ⎞⎛ v − a ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = ⎜− ⎟⎜ ⎟⎜ ⎟ = −1 ⎝ ∂v ⎠T ⎝ ∂T ⎠ P ⎝ ∂P ⎠v ⎝ v − a ⎠⎝ P ⎠⎝ R ⎠which is the desired result.(b) The reciprocity rule for this gas at v = constant can be expressed as ⎛ ∂P ⎞ 1 ⎜ ⎟ = ⎝ ∂T ⎠v (∂T / ∂P)v P (v − a) ⎛ ∂T ⎞ v −a T= ⎯⎯→ ⎜ ⎟ = R ⎝ ∂P ⎠v R RT ⎛ ∂P ⎞ R P= ⎯ ⎯→ ⎜ ⎟ = v −a ⎝ ∂T ⎠v v − aWe observe that the first differential is the inverse of the second one. Thus the proof is complete.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 8. 12-8The Maxwell Relations12-14 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, ⎛ ∂s ⎞ ? ⎛ ∂v ⎞ ⎜ ⎟ =− ⎜ ⎟ ⎝ ∂P ⎠ T ⎝ ∂T ⎠ P ⎛ Δs ⎞ ? ⎛ Δv ⎞ ⎜ ⎟ ≅−⎜ ⎟ ⎝ ΔP ⎠ T =80°C ⎝ ΔT ⎠ P =1200 kPa ⎛ s1400 kPa − s1000 kPa ⎞ ? ⎛ v 100°C − v 60° C ⎞ ⎜ ⎟ ≅ −⎜ ⎟ ⎜ (1400 − 1000 )kPa ⎟ ⎜ (100 − 60 )°C ⎟ ⎝ ⎠ T =80°C ⎝ ⎠ P =1200kPa (1.0056 − 1.0458)kJ/kg ⋅ K ? (0.022442 − 0.018404)m 3 /kg ≅− (1400 − 1000)kPa (100 − 60)°C − 1.005 × 10 − 4 m 3 /kg ⋅ K ≅ −1.0095 × 10 − 4 m 3 /kg ⋅ Ksince kJ ≡ kPa·m³, and K ≡ °C for temperature differences. Thus the last Maxwell relation is satisfied.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 9. 12-912-15 EES Problem 12-14 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_incrementDELTAP = 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] RightSide=-0.000101 [m^3/kg-K] DELTAs=-0.04026 [kJ/kg-K] s[1]=1.046 [kJ/kg-K] DELTAT=40 [C] s[2]=1.006 [kJ/kg-K] DELTAv=0.004038 [m^3/kg] T=80 [C] LeftSide=-0.0001007 [m^3/kg-K] T[1]=60 [C] P=1200 [kPa] T[2]=100 [C] P[1]=1000 [kPa] T_increment=20 [C] P[2]=1400 [kPa] v[1]=0.0184 [m^3/kg] P_increment=200 [kPa] v[2]=0.02244 [m^3/kg]PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 10. 12-1012-16E 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 thedifferential quantities in the last Maxwell relation with the corresponding finite quantities. Using propertyvalues from the tables about the specified state, ⎛ ∂s ⎞ ? ⎛ ∂v ⎞ ⎜ ⎟ =− ⎜ ⎟ ⎝ ∂P ⎠ T ⎝ ∂T ⎠ P ⎛ Δs ⎞ ? ⎛ Δv ⎞ ⎜ ⎟ ≅− ⎜ ⎟ ⎝ ΔP ⎠ T =800°F ⎝ ΔT ⎠ P = 400psia ⎛ s 450 psia − s 350 psia ⎞ ? ⎛ v 900° F − v 700° F ⎞ ⎜ ⎟ ≅ −⎜ ⎟ ⎜ (450 − 350)psia ⎟ ⎜ (900 − 700 )°F ⎟ ⎝ ⎠ T =800°F ⎝ ⎠ P = 400psia (1.6706 − 1.7009)Btu/lbm ⋅ R ? (1.9777 − 1.6507)ft 3 /lbm ≅− (450 − 350)psia (900 − 700)°F − 1.639 × 10 −3 ft 3 /lbm ⋅ R ≅ −1.635 × 10 −3 ft 3 /lbm ⋅ Rsince 1 Btu ≡ 5.4039 psia·ft3, and R ≡ °F for temperature differences. Thus the fourth Maxwell relation issatisfied.12-17 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. RTAnalysis This equation of state can be expressed as v = + b . Then, P ⎛ ∂v ⎞ R ⎜ ⎟ = ⎝ ∂T ⎠ P PFrom the fourth Maxwell relation, ⎛ ∂s ⎞ ⎛ ∂v ⎞ R ⎜ ⎟ = −⎜ ⎟ =− ⎝ ∂P ⎠ T ⎝ ∂T ⎠ P PPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 11. 12-1112-18 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. RT aAnalysis This equation of state can be expressed as P = + 2 . Then, v −b v ⎛ ∂P ⎞ R ⎜ ⎟ = ⎝ ∂T ⎠ v v − bFrom the third Maxwell relation, ⎛ ∂s ⎞ ⎛ ∂P ⎞ R ⎜ ⎟ =⎜ ⎟ = ⎝ ∂v ⎠ T ⎝ ∂T ⎠ v v − b12-19 Using the Maxwell relations and the ideal-gas equation of state, a relation for (∂s/∂v)T for an idealgas is to be obtained. RTAnalysis The ideal gas equation of state can be expressed as P = . Then, v ⎛ ∂P ⎞ R ⎜ ⎟ = ⎝ ∂T ⎠ v vFrom the third Maxwell relation, ⎛ ∂s ⎞ ⎛ ∂P ⎞ R ⎜ ⎟ =⎜ ⎟ = ⎝ ∂v ⎠ T ⎝ ∂T ⎠ v vPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 12. 12-12 ⎛ ∂P ⎞ k ⎛ ∂P ⎞12-20 It is to be proven that ⎜ ⎟ = ⎜ ⎟ ⎝ ∂T ⎠ s k − 1 ⎝ ∂T ⎠ vAnalysis Using the definition of cv , ⎛ ∂s ⎞ ⎛ ∂s ⎞ ⎛ ∂P ⎞ cv = T ⎜ ⎟ = T⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ v ⎝ ∂P ⎠ v ⎝ ∂T ⎠ v ⎛ ∂s ⎞ ⎛ ∂v ⎞Substituting the first Maxwell relation ⎜ ⎟ = −⎜ ⎟ , ⎝ ∂P ⎠ v ⎝ ∂T ⎠ s ⎛ ∂v ⎞ ⎛ ∂P ⎞ cv = −T ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ s ⎝ ∂T ⎠ vUsing the definition of cp, ⎛ ∂s ⎞ ⎛ ∂s ⎞ ⎛ ∂v ⎞ c p = T⎜ ⎟ = T⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ P ⎝ ∂v ⎠ P ⎝ ∂T ⎠ P ⎛ ∂s ⎞ ⎛ ∂P ⎞Substituting the second Maxwell relation ⎜ ⎟ =⎜ ⎟ , ⎝ ∂v ⎠ P ⎝ ∂T ⎠ s ⎛ ∂P ⎞ ⎛ ∂v ⎞ cp = T⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ s ⎝ ∂T ⎠ PFrom Eq. 12-46, 2 ⎛ ∂v ⎞ ⎛ ∂P ⎞ c p − cv = −T ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ P ⎝ ∂v ⎠ TAlso, k cp = k −1 c p − cvThen, ⎛ ∂P ⎞ ⎛ ∂v ⎞ ⎜ ⎟ ⎜ ⎟ k ⎝ ∂T ⎠ s ⎝ ∂T ⎠ P ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎛ ∂v ⎞ =− = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ k −1 2 ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎝ ∂T ⎠ s ⎝ ∂v ⎠ P ⎝ ∂P ⎠ T ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ P ⎝ ∂v ⎠ TSubstituting this into the original equation in the problem statement produces ⎛ ∂P ⎞ ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎜ ⎟ = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ s ⎝ ∂T ⎠ s ⎝ ∂v ⎠ P ⎝ ∂P ⎠ T ⎝ ∂T ⎠ vBut, according to the cyclic relation, the last three terms are equal to −1. Then, ⎛ ∂P ⎞ ⎛ ∂P ⎞ ⎜ ⎟ =⎜ ⎟ ⎝ ∂T ⎠ s ⎝ ∂T ⎠ sPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 13. 12-1312-21 It is to be shown how T, v, u, a, and g could be evaluated from the thermodynamic function h = h(s,P).Analysis Forming the differential of the given expression for h produces ⎛ ∂h ⎞ ⎛ ∂h ⎞ dh = ⎜ ⎟ ds + ⎜ ⎟ dP ⎝ ∂s ⎠ P ⎝ ∂P ⎠ sSolving the dh Gibbs equation gives dh = Tds + vdPComparing the coefficient of these two expressions ⎛ ∂h ⎞ T =⎜ ⎟ ⎝ ∂s ⎠ P ⎛ ∂h ⎞ v =⎜ ⎟ ⎝ ∂P ⎠ sboth of which can be evaluated for a given P and s. From the definition of the enthalpy, ⎛ ∂h ⎞ u = h − Pv = h-P⎜ ⎟ ⎝ ∂P ⎠ sSimilarly, the definition of the Helmholtz function, ⎛ ∂h ⎞ ⎛ ∂h ⎞ a = u − Ts = h − P⎜ ⎟ − s⎜ ⎟ ⎝ ∂P ⎠ s ⎝ ∂s ⎠ Pwhile the definition of the Gibbs function gives ⎛ ∂h ⎞ q = h − Ts = h − s⎜ ⎟ ⎝ ∂s ⎠ PAll of these can be evaluated for a given P and s and the fundamental h(s,P) equation.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 14. 12-14The Clapeyron Equation12-22C 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 assumed that vfg ≅ vg ≅ RT/P, and hfg ≅ constant for small temperature intervals.12-24 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, ⎛ dP ⎞ h fg = Tv fg ⎜ ⎟ ⎝ dT ⎠ sat ⎛ ΔP ⎞ ≅ T (v g − v f ) @300 kPa ⎜ ⎟ ⎝ ΔT ⎠ sat, 300 kPa ⎛ (325 − 275)kPa ⎞ = Tsat @300 kPa (v g − v f ) @300 kPa ⎜ ⎟ ⎜ Tsat @325 kPa − Tsat @275 kPa ⎟ ⎝ ⎠ 3 ⎛ 50 kPa ⎞ = (133.52 + 273.15 K)(0.60582 − 0.001073 m /kg)⎜ ⎜ (136.27 − 130.58)°C ⎟ ⎟ ⎝ ⎠ = 2159.9 kJ/kgThe tabulated value of hfg at 300 kPa is 2163.5 kJ/kg.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 15. 12-1512-25 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, ⎛ dP ⎞ h fg = Tv fg ⎜ ⎟ ⎝ dT ⎠ sat ⎛ ΔP ⎞ ≅ T (v g − v f ) @120°C ⎜ ⎟ ⎝ ΔT ⎠ sat,120°C ⎛ Psat @125°C − Psat @115°C ⎞ = T (v g − v f ) @120°C ⎜ ⎜ ⎟ ⎟ ⎝ 125°C − 115°C ⎠ ⎛ (232.23 − 169.18)kPa ⎞ = (120 + 273.15 K)(0.89133 − 0.001060 m 3 /kg)⎜ ⎜ ⎟ ⎟ ⎝ 10 K ⎠ = 2206.8 kJ/kgAlso, h fg 2206.8 kJ/kg s fg = = = 5.6131 kJ/kg ⋅ K T (120 + 273.15) KThe tabulated values at 120°C are hfg = 2202.1 kJ/kg and sfg = 5.6013 kJ/kg·K.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 16. 12-1612-26E [Also solved by EES on enclosed CD] The hfg of refrigerant-134a at a specified temperature is to becalculated using the Clapeyron equation and Clapeyron-Clausius equation and to be compared to thetabulated data.Analysis (a) From the Clapeyron equation, ⎛ dP ⎞ h fg = Tv fg ⎜ ⎟ ⎝ dT ⎠ sat ⎛ ΔP ⎞ ≅ T (v g − v f ) @ 50°F ⎜ ⎟ ⎝ ΔT ⎠ sat, 50° F ⎛ Psat @ 60 ° F − Psat @ 40 °F ⎞ = T (v g − v f ) @ 50°F ⎜ ⎜ ⎟ ⎟ ⎝ 60°F − 40°F ⎠ ⎛ (72.152 − 49.776) psia ⎞ = (50 + 459.67 R)(0.79136 − 0.01270 ft 3 /lbm)⎜ ⎜ ⎟ ⎟ ⎝ 20 R ⎠ = 444.0 psia ⋅ ft 3 /lbm = 82.16 Btu/lbm (0.2% error)since 1 Btu = 5.4039 psia·ft3.(b) From the Clapeyron-Clausius equation, ⎛P ⎞ h fg ⎛ 1 1 ⎞ ln⎜ 2 ⎜P ⎟ ≅ ⎟ ⎜ − ⎟ ⎜T T ⎟ ⎝ 1 ⎠ sat R ⎝ 1 2 ⎠ sat ⎛ 72.152 psia ⎞ h fg ⎛ 1 1 ⎞ ⎜ 49.776 psia ⎟ ≅ 0.01946 Btu/lbm ⋅ R ⎜ 40 + 459.67 R − 60 + 459.67 R ⎟ ln⎜ ⎟ ⎝ ⎠ ⎝ ⎠ h fg = 93.80 Btu/lbm (14.4% error)The tabulated value of hfg at 50°F is 82.00 Btu/lbm.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 17. 12-1712-27 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 ΔP h fg ,Clapeyron = Tv fg ΔTAt 100ºC, for an increment of 5ºC, we obtain T1 = T − Tincrement = 100 − 5 = 95°C T2 = T + Tincrement = 100 + 5 = 105°C P1 = Psat @ 95°C = 84.61 kPa P2 = Psat @ 105°C = 120.90 kPa ΔT = T2 − T1 = 105 − 95 = 10°C ΔP = P2 − P1 = 120.90 − 84.61 = 36.29 kPa v f @ 100°C = 0.001043 m 3 /kg v g @ 100°C = 1.6720 m 3 /kg v fg = v g − v f = 1.6720 − 0.001043 = 1.6710 m 3 /kgSubstituting, ΔP 36.29 kPa h fg ,Clapeyron = Tv fg = (100 + 273.15 K)(1.6710 m 3 /kg) = 2262.8 kJ/kg ΔT 10 KThe enthalpy of vaporization from steam table is h fg @ 100°C = 2256.4 m 3 /kgThe percent error in using Clapeyron equation is 2262.8 − 2256.4 PercentError = × 100 = 0.28% 2256.4We repeat the analysis over the temperature range 10 to 200ºC using EES. Below, the copy of EESsolution 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]"PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 18. 12-18"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 hfg,Clapeyron PercentError T [kJ/kg] [kJ/kg] [%] [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 2600 2500 hfg calculated by Clapeyron equation 2400 hfg [kJ/kg] 2300 hfg calculated by EES 2200 2100 2000 1900 1800 1700 0 50 100 150 200 250 T [C]PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 19. 12-1912-28 A substance is heated in a piston-cylinder device until it turns from saturated liquid to saturatedvapor at a constant pressure and temperature. The boiling temperature of this substance at a differentpressure is to be estimated.Analysis From the Clapeyron equation, ⎛ dP ⎞ h fg (5 kPa ⋅ m 3 )/(0.002 kg) ⎜ ⎟ = = = 14.16 kPa/K Weight ⎝ dT ⎠ sat Tv fg (353 K)(1× 10 −3 m 3 )/(0.002 kg)Using the finite difference approximation, ⎛ dP ⎞ ⎛ P − P1 ⎞ 200 kPa ⎜ ⎟ ≈⎜ 2⎜ ⎟ ⎟ 80°C Q ⎝ dT ⎠ sat ⎝ T2 − T1 ⎠ sat 2 gramsSolving for T2, Sat. liquid P2 − P1 (180 − 200)kPa T2 = T1 + = 353 K + = 351.6 K dP / dT 14.16 kPa/K12-29 A substance is heated in a piston-cylinder device until it turns from saturated liquid to saturatedvapor at a constant pressure and temperature. The saturation pressure of this substance at a differenttemperature is to be estimated.Analysis From the Clapeyron equation, Weight 3 ⎛ dP ⎞ h fg (5 kPa ⋅ m )/(0.002 kg) ⎜ ⎟ = = = 14.16 kPa/K ⎝ dT ⎠ sat Tv fg (353 K)(1× 10 −3 m 3 )/(0.002 kg) 200 kPaUsing the finite difference approximation, 80°C Q 2 grams ⎛ dP ⎞ ⎛ P − P1 ⎞ Sat. liquid ⎜ ⎟ ≈⎜ 2 ⎟ ⎝ dT ⎠ sat ⎜ T2 − T1 ⎟ sat ⎝ ⎠Solving for P2, dP P2 = P1 + (T2 − T1 ) = 200 kPa + (14.16 kPa/K )(373 − 353)K = 483.2 kPa dTPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 20. 12-2012-30 A substance is heated in a piston-cylinder device until it turns from saturated liquid to saturatedvapor at a constant pressure and temperature. The sfg of this substance at a different temperature is to beestimated.Analysis From the Clapeyron equation, Weight ⎛ dP ⎞ h fg s fg ⎜ ⎟ = = ⎝ dT ⎠ sat Tv fg v fg 200 kPaSolving for sfg, 80°C Q h fg (5 kJ)/(0.002 kg) 2 grams s fg = = = 7.082 kJ/kg ⋅ K Sat. liquid T 353 KAlternatively, ⎛ dP ⎞ 1× 10 -3 m 3 s fg = ⎜ ⎟ v fg = (14.16 kPa/K ) = 7.08 kPa ⋅ m 3 /kg ⋅ K = 7.08 kJ/kg ⋅ K ⎝ dT ⎠ sat 0.002 kg ⎛ ∂ (h fg / T ) ⎞12-31 It is to be shown that c p , g − c p , f = T ⎜ ⎟ + v fg ⎛ ∂P ⎞ . ⎜ ⎟ ⎜ ∂T ⎟ ⎝ ∂T ⎠ sat ⎝ ⎠PAnalysis The definition of specific heat and Clapeyron equation are ⎛ ∂h ⎞ cp = ⎜ ⎟ ⎝ ∂T ⎠ P ⎛ dP ⎞ h fg ⎜ ⎟ = ⎝ dT ⎠ sat Tv fgAccording to the definition of the enthalpy of vaporization, h fg hg hf = − T T TDifferentiating this expression gives ⎛ ∂h fg / T ⎞ ⎛ ∂h / T ⎞ ⎛ ∂h / T ⎞ ⎜ ⎟ =⎜ g ⎟ −⎜ f ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ P ⎝ ∂T ⎠ P ⎝ ∂T ⎠P 1 ⎛ ∂h g ⎞ ⎛ ∂h ⎞ ⎟ − g −1⎜ f h h = ⎜ ⎟ + f T ⎜ ∂T ⎟ 2 T ⎜ ∂T ⎟ 2 ⎝⎠P T ⎝ ⎠P T c p, g c p, f h g − h f = − − T T T2Using Clasius-Clapeyron to replace the last term of this expression and solving for the specific heatdifference gives ⎛ ∂ (h fg / T ) ⎞ c p, g − c p, f = T ⎜ ⎟ + v fg ⎛ ∂P ⎞ ⎜ ⎟ ⎜ ∂T ⎟ ⎝ ∂T ⎠ sat ⎝ ⎠PPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 21. 12-2112-32E A table of properties for methyl chloride is given. The saturation pressure is to be estimated at twodifferent temperatures.Analysis The Clapeyron equation is ⎛ dP ⎞ h fg ⎜ ⎟ = ⎝ dT ⎠ sat Tv fgUsing the finite difference approximation, ⎛ dP ⎞ ⎛ P2 − P1 ⎞ h fg ⎜ ⎟ ≈⎜ ⎟ ⎜ T − T ⎟ = Tv ⎝ dT ⎠ sat ⎝ 2 1 ⎠ sat fgSolving this for the second pressure gives for T2 = 110°F h fg P2 = P1 + (T2 − T1 ) Tv fg 154.85 Btu/lbm ⎛ 5.404 psia ⋅ ft 3 ⎞ = 116.7 psia + ⎜ ⎟(110 − 100)R (560 R)(0.86332 ft 3 /lbm) ⎜ ⎝ 1 Btu ⎟ ⎠ = 134.0 psiaWhen T2 = 90°F h fg P2 = P1 + (T2 − T1 ) Tv fg 154.85 Btu/lbm ⎛ 5.404 psia ⋅ ft 3 ⎞ = 116.7 psia + ⎜ ⎟(90 − 100)R (560 R)(0.86332 ft 3 /lbm) ⎜ ⎝ 1 Btu ⎟ ⎠ = 99.4 psiaPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 22. 12-2212-33 Saturation properties for R-134a at a specified temperature are given. The saturation pressure is to beestimated at two different temperatures.Analysis From the Clapeyron equation, ⎛ dP ⎞ h fg 225.86 kPa ⋅ m 3 /kg ⎜ ⎟ = = = 2.692 kPa/K ⎝ dT ⎠ sat Tv fg (233 K)(0.36010 m 3 /kg)Using the finite difference approximation, ⎛ dP ⎞ ⎛ P − P1 ⎞ ⎜ ⎟ ≈⎜ 2⎜ ⎟ ⎟ ⎝ dT ⎠ sat ⎝ T2 − T1 ⎠ satSolving for P2 at −50°C dP P2 = P1 + (T2 − T1 ) = 51.25 kPa + (2.692 kPa/K )(223 − 233)K = 24.33 kPa dTSolving for P2 at −30°C dP P2 = P1 + (T2 − T1 ) = 51.25 kPa + (2.692 kPa/K )(243 − 233)K = 78.17 kPa dTPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 23. 12-23General Relations for du, dh, ds, cv, and cp12-34C Yes, through the relation ⎛ ∂c p ⎞ ⎛ 2 ⎞ ⎜ ⎟ = −T ⎜ ∂ v ⎟ ⎜ ∂P ⎟ ⎜ ∂T 2 ⎟ ⎝ ⎠T ⎝ ⎠PPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 24. 12-2412-35 General expressions for Δu, Δh, and Δs for a gas whose equation of state is P(v-a) = RT for anisothermal process are to be derived.Analysis (a) A relation for Δu is obtained from the general relation T2 v2 ⎛ ⎛ ∂P ⎞ ⎞ Δu = u 2 − u1 = ∫ T1 cv dT + ∫v 1 ⎜T ⎜ ⎟ ⎜ ⎝ ∂T ⎟ − P ⎟dv ⎝ ⎠v ⎠The equation of state for the specified gas can be expressed as RT ⎛ ∂P ⎞ R P= ⎯⎯→ ⎜ ⎟ = v −a ⎝ ∂T ⎠v v − aThus, ⎛ ∂P ⎞ RT T⎜ ⎟ −P = −P = P−P = 0 ⎝ ∂T ⎠v v −a T2Substituting, Δu = ∫T1 cv dT(b) A relation for Δh is obtained from the general relation ⎛ ∂v ⎞ ⎞ ⎜v − T ⎛ T2 P2 Δh = h2 − h1 = ∫ T1 c P dT + ∫ P1 ⎜ ⎝ ⎜ ⎟ ⎟dP ⎝ ∂T ⎠ P ⎟ ⎠The equation of state for the specified gas can be expressed as RT ⎛ ∂v ⎞ R v= +a ⎯ ⎯→ ⎜ ⎟ = P ⎝ ∂T ⎠ P PThus, ⎛ ∂v ⎞ R v −T⎜ ⎟ = v − T = v − (v − a ) = a ⎝ ∂T ⎠ P PSubstituting, c p dT + a(P2 − P1 ) T2 P2 T2 Δh = ∫T1 c p dT + ∫ P1 a dP = ∫T1(c) A relation for Δs is obtained from the general relation T2 cp P2 ⎛ ∂v ⎞ Δs = s 2 − s1 = ∫ T1 T dT − ∫P1 ⎜ ⎟ dP ⎝ ∂T ⎠ PSubstituting (∂v/∂T)P = R/T, T2 cp P2 ⎛R⎞ T2 cp P2 Δs = ∫ T1 T dT − ∫P1 ⎜ ⎟ dP = ⎝ P ⎠P ∫T1 T dT − R ln P1For an isothermal process dT = 0 and these relations reduce to P2 Δu = 0, Δh = a(P2 − P1 ), and Δs = − Rln P1PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 25. 12-2512-36 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 ⎛ ⎛ ∂P ⎞ ⎞ du = cv dT + ⎜ T ⎜ ⎟ ⎜ ⎝ ∂T ⎟ − P ⎟dv ⎝ ⎠v ⎠Differentiating each term in this equation with respect to P at T = constant yields ⎛ ∂u ⎞ ⎛ ⎛ ∂P ⎞ ⎞⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎛ ∂v ⎞ ⎛ ∂v ⎞ ⎜ ⎟ = 0 + ⎜T ⎜ ⎟ ⎜ ⎝ ∂T ⎠ − P ⎟⎜ ∂P ⎟ = T ⎜ ∂T ⎟ ⎜ ∂P ⎟ − P⎜ ∂P ⎟ ⎟ ⎝ ∂P ⎠ T ⎝ v ⎠⎝ ⎠T ⎝ ⎠v ⎝ ⎠T ⎝ ⎠TUsing the properties P, T, v, the cyclic relation can be expressed as ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎛ ∂v ⎞ ⎛ ∂v ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = −1 ⎯ ⎯→ ⎜ ⎟ ⎜ ⎟ = −⎜ ⎟ ⎝ ∂T ⎠v ⎝ ∂v ⎠ P ⎝ ∂P ⎠ T ⎝ ∂T ⎠ v ⎝ ∂P ⎠ T ⎝ ∂T ⎠ PSubstituting, we get ⎛ ∂u ⎞ ⎛ ∂v ⎞ ⎛ ∂v ⎞ ⎜ ⎟ = −T ⎜ ⎟ − P⎜ ⎟ ⎝ ∂P ⎠ T ⎝ ∂T ⎠ P ⎝ ∂P ⎠ TThe general relation for dh is ⎛ ⎛ ∂v ⎞ ⎞ dh = c p dT + ⎜v − T ⎜ ⎜ ⎟ ⎟dP ⎝ ⎝ ∂T ⎠ P ⎟ ⎠Differentiating each term in this equation with respect to v at T = constant yields ⎛ ∂h ⎞ ⎛ ⎛ ∂v ⎞ ⎞⎛ ∂P ⎞ ⎛ ∂P ⎞ ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎜ ⎟ = 0 + ⎜v − T ⎜ ⎜ ⎟ ⎟⎜⎟⎝ ∂v ⎟ = v ⎜ ∂v ⎟ − T ⎜ ∂T ⎟ ⎜ ∂v ⎟ ⎝ ∂v ⎠ T ⎝ ⎝ ∂T ⎠ P ⎠ ⎠T ⎝ ⎠T ⎝ ⎠P ⎝ ⎠TUsing the properties v, T, P, the cyclic relation can be expressed as ⎛ ∂v ⎞ ⎛ ∂T ⎞ ⎛ ∂P ⎞ ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = −1 ⎯ ⎯→ ⎜ ⎟ ⎜ ⎟ = −⎜ ⎟ ⎝ ∂T ⎠ P ⎝ ∂P ⎠ v ⎝ ∂v ⎠ T ⎝ ∂T ⎠ P ⎝ ∂v ⎠ T ⎝ ∂P ⎠ vSubstituting, we get ⎛ ∂h ⎞ ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎜ ⎟ =v⎜ ⎟ +T⎜ ⎟ ⎝ ∂v ⎠ T ⎝ ∂v ⎠ T ⎝ ∂P ⎠ vPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 26. 12-2612-37E 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 2 ⎛ ∂v ⎞ ⎛ ∂P ⎞ c p − cv = −T ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ P ⎝ ∂v ⎠ TApproximating differentials by differences about the specified state, 2 ⎛ Δv ⎞ ⎛ ΔP ⎞ c p − cv ≅ −T ⎜ ⎟ ⎜ ⎟ ⎝ ΔT ⎠ P =1000 psia ⎝ Δv ⎠ T =150°F 2 ⎛ v 175° F − v 125° F ⎞ ⎛ (1500 − 500)psia ⎞ = −(150 + 459.67 R )⎜ ⎟ ⎜ ⎟ ⎜ (175 − 125)°F ⎟ ⎜ ⎟ ⎝ ⎠ P =1000psia ⎝ v 1500 psia − v 500 psia ⎠ T =150° F 2 ⎛ (0.016427 − 0.016177)ft 3 /lbm ⎞ ⎛ 1000 psia ⎞ = −(609.67 R)⎜ ⎟ ⎜ ⎟ ⎜ 50 R ⎟ ⎜ (0.016267 − 0.016317)ft 3 /lbm ⎟ ⎝ ⎠ ⎝ ⎠ = 0.3081 psia ⋅ ft 3 /lbm ⋅ R = 0.0570 Btu/lbm ⋅ R (1 Btu = 5.4039 psia ⋅ ft 3 )12-38 The volume expansivity β and the isothermal compressibility α of refrigerant-134a at 200 kPa and30°C are to be estimated.Analysis The volume expansivity and isothermal compressibility are expressed as 1 ⎛ ∂v ⎞ 1 ⎛ ∂v ⎞ β= ⎜ ⎟ and α = − ⎜ ⎟ v ⎝ ∂T ⎠ P v ⎝ ∂P ⎠ TApproximating differentials by differences about the specified state, 1 ⎛ Δv ⎞ 1 ⎛ v 40°C − v 20°C ⎞ β≅ ⎜ ⎟ = ⎜ ⎟ v ⎝ ΔT ⎠ P = 200kPa v ⎜ (40 − 20)°C ⎝ ⎟ ⎠ P = 200 kPa 1 ⎛ (0.12322 − 0.11418) m 3 /kg ⎞ = ⎜ ⎟ 0.11874 m 3 /kg ⎜ ⎝ 20 K ⎟ ⎠ = 0.00381 K −1and 1 ⎛ Δv ⎞ 1 ⎛ v 240 kPa − v 180 kPa ⎞ α ≅− ⎜ ⎟ =− ⎜ ⎟ v ⎝ ΔP ⎠ T =30°C v ⎜ (240 − 180)kPa ⎝ ⎟ ⎠ T =30°C 1 ⎛ (0.09812 − 0.13248)m 3 /kg ⎞ =− ⎜ ⎟ 0.11874 m /kg ⎜ 3 ⎝ 60 kPa ⎟ ⎠ = 0.00482 kPa −1PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 27. 12-27 ⎛ ∂P ⎞ ⎛ ∂v ⎞12-39 It is to be shown that c p − cv = T ⎜ ⎟ ⎜ ⎟ . ⎝ ∂T ⎠ v ⎝ ∂T ⎠ PAnalysis We begin by taking the entropy to be a function of specific volume and temperature. Thedifferential of the entropy is then ⎛ ∂s ⎞ ⎛ ∂s ⎞ ds = ⎜ ⎟ dT + ⎜ ⎟ dv ⎝ ∂T ⎠ v ⎝ ∂v ⎠ T ⎛ ∂s ⎞ cSubstituting ⎜ ⎟ = v from Eq. 12-28 and the third Maxwell equation changes this to ⎝ ∂T ⎠ v T cv ⎛ ∂P ⎞ ds = dT + ⎜ ⎟ dv T ⎝ ∂T ⎠vTaking the entropy to be a function of pressure and temperature, ⎛ ∂s ⎞ ⎛ ∂s ⎞ ds = ⎜ ⎟ dT + ⎜ ⎟ dP ⎝ ∂T ⎠ P ⎝ ∂P ⎠ T ⎛ ∂s ⎞ cpCombining this result with ⎜ ⎟ = from Eq. 12-34 and the fourth Maxwell equation produces ⎝ ∂T ⎠ P T cp ⎛ ∂v ⎞ ds = dT − ⎜ ⎟ dP T ⎝ ∂T ⎠ PEquating the two previous ds expressions and solving the result for the specific heat difference, ⎛ ∂v ⎞ ⎛ ∂P ⎞ (c p − cv )dT = T ⎜ ⎟ dP + ⎜ ⎟ dv ⎝ ∂T ⎠ P ⎝ ∂T ⎠ vTaking the pressure to be a function of temperature and volume, ⎛ ∂P ⎞ ⎛ ∂P ⎞ dP = ⎜ ⎟ dT + ⎜ ⎟ dv ⎝ ∂T ⎠ v ⎝ ∂v ⎠ TWhen this is substituted into the previous expression, the result is ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎡⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎛ ∂P ⎞ ⎤ (c p − cv )dT = T ⎜ ⎟ ⎜ ⎟ dT + T ⎢⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎥ dv ⎝ ∂T ⎠ P ⎝ ∂T ⎠ v ⎣⎝ ∂T ⎠ P ⎝ ∂v ⎠ T ⎝ ∂T ⎠ v ⎦According to the cyclic relation, the term in the bracket is zero. Then, canceling the common dT term, ⎛ ∂P ⎞ ⎛ ∂v ⎞ c p − cv = T ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ v ⎝ ∂T ⎠ PPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 28. 12-2812-40 It is to be proven that the definition for temperature T = (∂u / ∂s ) v reduces the net entropy change oftwo constant-volume systems filled with simple compressible substances to zero as the two systemsapproach thermal equilibrium.Analysis The two constant-volume systems form an isolated system shown hereFor the isolated system dS tot = dS A + dS B ≥ 0Assume S = S (u , v ) VA =const. Isolated TA systemThen, boundary ⎛ ∂s ⎞ ⎛ ∂s ⎞ ds = ⎜ ⎟ du + ⎜ ⎟ dv VB =const. ⎝ ∂u ⎠ v ⎝ ∂v ⎠ u TBSince v = const. and dv = 0 , ⎛ ∂s ⎞ ds = ⎜ ⎟ du ⎝ ∂u ⎠ vand from the definition of temperature from the problem statement, du du = (∂u / ∂s ) v TThen, du A du B dS tot = m A + mB TA TBThe first law applied to the isolated system yields E in − E out = dU 0 = dU ⎯ ⎯→ m A du A + m B du B = 0 ⎯ ⎯→ m B du B = −m A du ANow, the entropy change may be expressed as ⎛ 1 1 ⎞ ⎛ T − TA ⎞ dS tot = m A du A ⎜ ⎜T − T ⎟ = m A du A ⎜ B ⎟ ⎜ T T ⎟ ⎟ ⎝ A B ⎠ ⎝ A B ⎠As the two systems approach thermal equilibrium, lim dS tot = 0 T A → TBPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 29. 12-2912-41 The internal energy change of air between two specified states is to be compared for two equationsof states.Assumptions Constant specific heats for air can be used.Properties For air at the average temperature (20+300)/2=160°C=433 K, cv = 0.731 kJ/kg⋅K (Table A-2b).Analysis Solving the equation of state for P gives RT P= v −aThen, ⎛ ∂P ⎞ R ⎜ ⎟ = ⎝ ∂T ⎠v v − aUsing equation 12-29, ⎡ ⎛ ∂P ⎞ ⎤ du = cv dT + ⎢T ⎜ ⎟ − P ⎥ dv ⎣ ⎝ ∂T ⎠v ⎦Substituting, ⎛ RT RT ⎞ du = cv dT + ⎜ − ⎟dv ⎝ v −a v −a⎠ = cv dTIntegrating this result between the two states with constant specific heats gives u 2 − u1 = cv (T2 − T1 ) = (0.731 kJ/kg ⋅ K)(300 − 20)K = 205 kJ/kgThe ideal gas model for the air gives du = cv dTwhich gives the same answer.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 30. 12-3012-42 The enthalpy change of air between two specified states is to be compared for two equations ofstates.Assumptions Constant specific heats for air can be used.Properties For air at the average temperature (20+300)/2=160°C=433 K, cp = 1.018 kJ/kg⋅K (Table A-2b).Analysis Solving the equation of state for v gives RT v= +a PThen, ⎛ ∂v ⎞ R ⎜ ⎟ = ⎝ ∂T ⎠ P PUsing equation 12-35, ⎡ ⎛ ∂v ⎞ ⎤ dh = c p dT + ⎢v − T ⎜ ⎟ ⎥ dP ⎣ ⎝ ∂T ⎠ P ⎦Substituting, ⎛ RT RT ⎞ dh = c p dT + ⎜ +a− ⎟dP ⎝ P P ⎠ = c p dT + adPIntegrating this result between the two states with constant specific heats gives h2 − h1 = c p (T2 − T1 ) + a ( P2 − P1 ) = (1.018 kJ/kg ⋅ K)(300 − 20)K + (0.10 m 3 /kg)(600 − 100)kPa = 335.0 kJ/kgFor an ideal gas, dh = c p dTwhich when integrated gives h2 − h1 = c p (T2 − T1 ) = (1.018 kJ/kg ⋅ K)(300 − 20)K = 285.0 kJ/kgPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 31. 12-3112-43 The entropy change of air between two specified states is to be compared for two equations of states.Assumptions Constant specific heats for air can be used.Properties For air at the average temperature (20+300)/2=160°C=433 K, cp = 1.018 kJ/kg⋅K (Table A-2b)and R = 0.287 kJ/kg⋅K (Table A-1).Analysis Solving the equation of state for v gives RT v= +a PThen, ⎛ ∂v ⎞ R ⎜ ⎟ = ⎝ ∂T ⎠ P PThe entropy differential is dT ⎛ ∂v ⎞ ds = c p −⎜ ⎟ dP T ⎝ ∂T ⎠ P dT dP = cp −R T Pwhich is the same as that of an ideal gas. Integrating this result between the two states with constantspecific heats gives T2 P s 2 − s1 = c p ln − R ln 2 T1 P1 573 K 600 kPa = (1.018 kJ/kg ⋅ K)ln − (0.287 kJ/kg ⋅ K)ln 293 K 100 kPa = 0.1686 kJ/kg ⋅ KPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 32. 12-3212-44 The internal energy change of helium between two specified states is to be compared for twoequations of states.Properties For helium, cv = 3.1156 kJ/kg⋅K (Table A-2a).Analysis Solving the equation of state for P gives RT P= v −aThen, ⎛ ∂P ⎞ R ⎜ ⎟ = ⎝ ∂T ⎠v v − aUsing equation 12-29, ⎡ ⎛ ∂P ⎞ ⎤ du = cv dT + ⎢T ⎜ ⎟ − P ⎥ dv ⎣ ⎝ ∂T ⎠v ⎦Substituting, ⎛ RT RT ⎞ du = cv dT + ⎜ − ⎟dv ⎝v − a v − a ⎠ = cv dTIntegrating this result between the two states gives u 2 − u1 = cv (T2 − T1 ) = (3.1156 kJ/kg ⋅ K)(300 − 20)K = 872.4 kJ/kgThe ideal gas model for the helium gives du = cv dTwhich gives the same answer.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 33. 12-3312-45 The enthalpy change of helium between two specified states is to be compared for two equations ofstates.Properties For helium, cp = 5.1926 kJ/kg⋅K (Table A-2a).Analysis Solving the equation of state for v gives RT v= +a PThen, ⎛ ∂v ⎞ R ⎜ ⎟ = ⎝ ∂T ⎠ P PUsing equation 12-35, ⎡ ⎛ ∂v ⎞ ⎤ dh = c p dT + ⎢v − T ⎜ ⎟ ⎥ dP ⎣ ⎝ ∂T ⎠ P ⎦Substituting, ⎛ RT RT ⎞ dh = c p dT + ⎜ +a− ⎟dP ⎝ P P ⎠ = c p dT + adPIntegrating this result between the two states gives h2 − h1 = c p (T2 − T1 ) + a ( P2 − P1 ) = (5.1926 kJ/kg ⋅ K)(300 − 20)K + (0.10 m 3 /kg)(600 − 100)kPa = 1504 kJ/kgFor an ideal gas, dh = c p dTwhich when integrated gives h2 − h1 = c p (T2 − T1 ) = (5.1926 kJ/kg ⋅ K)(300 − 20)K = 1454 kJ/kgPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 34. 12-3412-46 The entropy change of helium between two specified states is to be compared for two equations ofstates.Properties For helium, cp = 5.1926 kJ/kg⋅K and R = 2.0769 kJ/kg⋅K (Table A-2a).Analysis Solving the equation of state for v gives RT v= +a PThen, ⎛ ∂v ⎞ R ⎜ ⎟ = ⎝ ∂T ⎠ P PThe entropy differential is dT ⎛ ∂v ⎞ ds = c p −⎜ ⎟ dP T ⎝ ∂T ⎠ P dT dP = cp −R T Pwhich is the same as that of an ideal gas. Integrating this result between the two states gives T2 P s 2 − s1 = c p ln − R ln 2 T1 P1 573 K 600 kPa = (5.1926 kJ/kg ⋅ K)ln − (2.0769 kJ/kg ⋅ K)ln 293 K 100 kPa = −0.2386 kJ/kg ⋅ K12-47 An expression for the volume expansivity of a substance whose equation of state is P (v − a) = RTis to be derived.Analysis Solving the equation of state for v gives RT v= +a PThe specific volume derivative is then ⎛ ∂v ⎞ R ⎜ ⎟ = ⎝ ∂T ⎠ P PThe definition for volume expansivity is 1 ⎛ ∂v ⎞ β= ⎜ ⎟ v ⎝ ∂T ⎠ PCombining these two equations gives R β= RT + aPPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 35. 12-35 v ⎛ T T T ⎞12-48 The Helmholtz function of a substance has the form a = − RT ln − cT0 ⎜1 − ⎟ ⎜ T + T ln T ⎟ . It is to v0 ⎝ 0 0 0 ⎠be shown how to obtain P, h, s, cv, and cp from this expression.Analysis Taking the Helmholtz function to be a function of temperature and specific volume yields ⎛ ∂a ⎞ ⎛ ∂a ⎞ da = ⎜ ⎟ dT + ⎜ ⎟ dv ⎝ ∂T ⎠ v ⎝ ∂v ⎠ Twhile the applicable Helmholtz equation is da = − Pdv − sdTEquating the coefficients of the two results produces ⎛ ∂a ⎞ P = −⎜ ⎟ ⎝ ∂v ⎠ T ⎛ ∂a ⎞ s = −⎜ ⎟ ⎝ ∂T ⎠ vTaking the indicated partial derivatives of the Helmholtz function given in the problem statement reducesthese expressions to RT P= v v T s = R ln + c ln v0 T0The definition of the enthalpy (h = u + Pv) and Helmholtz function (a = u−Ts) may be combined to give h = u + Pv = a + Ts + Pv ⎛ ∂a ⎞ ⎛ ∂a ⎞ = a −T⎜ ⎟ −v ⎜ ⎟ ⎝ ∂T ⎠ v ⎝ ∂v ⎠ T v ⎛ T T T ⎞ v T = − RT ln − cT0 ⎜1 − ⎟ ⎜ T + T ln T ⎟ + RT ln v − cT ln T + RT v0 ⎝ 0 0 0 ⎠ 0 0 = cT0 + cT + RT ⎛ ∂s ⎞ cAccording to ⎜ ⎟ = v given in the text (Eq. 12-28), ⎝ ∂T ⎠ v T ⎛ ∂s ⎞ c cv = T ⎜ ⎟ =T =c ⎝ ∂T ⎠ v TThe preceding expression for the temperature indicates that the equation of state for the substance is thesame as that of an ideal gas. Then, c p = R + cv = R + cPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 36. 12-3612-49 An expression for the volume expansivity of a substance whose equation of state RT ais P = − is to be derived. v − b v (v + b)T 1 / 2Analysis The definition for volume expansivity is 1 ⎛ ∂v ⎞ β= ⎜ ⎟ v ⎝ ∂T ⎠ PAccording to the cyclic relation, ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = −1 ⎝ ∂T ⎠ P ⎝ ∂v ⎠ T ⎝ ∂P ⎠ vwhich on rearrangement becomes ⎛ ∂P ⎞ ⎜ ⎟ ⎛ ∂v ⎞ ⎝ ∂T ⎠ v ⎜ ⎟ =− ⎝ ∂T ⎠ P ⎛ ∂P ⎞ ⎜ ⎟ ⎝ ∂v ⎠ TProceeding to perform the differentiations gives ⎛ ∂P ⎞ R a ⎜ ⎟ = + ⎝ ∂T ⎠ v v − b 2v (v + b)T 3 / 2and ⎛ ∂P ⎞ RT a ⎡ 1 1 ⎤ ⎜ ⎟ =− + 1/ 2 ⎢ 2 − 2 ⎥ ⎝ ∂v ⎠ T (v − b) 2 bT ⎢v ⎣ (v + b) ⎥⎦ RT a 2v + b =− + (v − b) 2 T 1 / 2 v 2 (v + b) 2Substituting these results into the definition of the volume expansivity produces R a + 1 v − b 2v (v + b)T 3 / 2 β=− v − RT a 2v + b 2 + 1/ 2 2 (v − b) T v (v + b) 2PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 37. 12-3712-50 An expression for the specific heat difference of a substance whose equation of state RT ais P = − is to be derived. v − b v (v + b)T 1 / 2Analysis The specific heat difference is expressed by 2 ⎛ ∂v ⎞ ⎛ ∂P ⎞ c p − cv = −T ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ P ⎝ ∂v ⎠ TAccording to the cyclic relation, ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = −1 ⎝ ∂T ⎠ P ⎝ ∂v ⎠ T ⎝ ∂P ⎠ vwhich on rearrangement becomes −1 ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎛ ∂P ⎞ ⎜ ⎟ = −⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ P ⎝ ∂T ⎠ v ⎝ ∂v ⎠ TSubstituting this result into the expression for the specific heat difference gives 2 −1 ⎛ ∂P ⎞ ⎛ ∂P ⎞ c p − cv = −T ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ v ⎝ ∂v ⎠ TThe appropriate partial derivatives of the equation of state are ⎛ ∂P ⎞ R a/2 ⎜ ⎟ = + ⎝ ∂T ⎠v v − b v (v + b)T 3 / 2 ⎛ ∂P ⎞ RT a ⎡ 1 1 ⎤ ⎜ ⎟ =− + 1/ 2 ⎢ 2 − 2⎥ ⎝ ∂v ⎠ T (v − b) 2 bT ⎢v ⎣ (v + b) ⎥ ⎦ RT a 2v + b =− 2 + 1/ 2 (v − b) T v (v + b) 2 2The difference in the specific heats is then 2 ⎡ R a/2 ⎤ −T ⎢ + 3/ 2 ⎥ ⎢ v − b v (v + b)T ⎣ ⎥ ⎦ c p − cv = RT a 2v + b − + (v − b) 2 T 1 / 2 v 2 (v + b) 2PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 38. 12-38 RT a12-51 An expression for the volume expansivity of a substance whose equation of state is P = − v − b v 2Tis to be derived.Analysis The definition for volume expansivity is 1 ⎛ ∂v ⎞ β= ⎜ ⎟ v ⎝ ∂T ⎠ PAccording to the cyclic relation, ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = −1 ⎝ ∂T ⎠ P ⎝ ∂v ⎠ T ⎝ ∂P ⎠ vwhich on rearrangement becomes ⎛ ∂P ⎞ ⎜ ⎟ ⎛ ∂v ⎞ ⎝ ∂T ⎠ v ⎜ ⎟ =− ⎝ ∂T ⎠ P ⎛ ∂P ⎞ ⎜ ⎟ ⎝ ∂v ⎠ TProceeding to perform the differentiations gives ⎛ ∂P ⎞ R a ⎜ ⎟ = + ⎝ ∂T ⎠ v v − b v 2 T 2and ⎛ ∂P ⎞ RT 2a ⎜ ⎟ =− + 3 ⎝ ∂v ⎠ T (v − b) 2 v TSubstituting these results into the definition of the volume expansivity produces R a + 1 v − b v 2T 2 β=− v − RT 2a 2 + 3 (v − b) v TPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 39. 12-3912-52 An expression for the isothermal compressibility of a substance whose equation of state RT ais P = − is to be derived. v − b v 2TAnalysis The definition for the isothermal compressibility is 1 ⎛ ∂v ⎞ α=− ⎜ ⎟ v ⎝ ∂P ⎠ TNow, ⎛ ∂P ⎞ − RT 2a ⎜ ⎟ = + 3 ⎝ ∂v ⎠ T (v − b) 2 v TUsing the partial derivative properties, 1 1 α=− =− ⎛ ∂P ⎞ − RTv 2a v⎜ ⎟ 2 + ⎝ ∂v ⎠ T (v − b) v 2T ⎛ ∂P ⎞12-53 It is to be shown that β = α ⎜ ⎟ . ⎝ ∂T ⎠ vAnalysis The definition for the volume expansivity is 1 ⎛ ∂v ⎞ β= ⎜ ⎟ v ⎝ ∂T ⎠ PThe definition for the isothermal compressibility is 1 ⎛ ∂v ⎞ α=− ⎜ ⎟ v ⎝ ∂P ⎠ TAccording to the cyclic relation, ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = −1 ⎝ ∂T ⎠ P ⎝ ∂v ⎠ T ⎝ ∂P ⎠ vwhich on rearrangement becomes ⎛ ∂v ⎞ ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎜ ⎟ = −⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ P ⎝ ∂P ⎠ T ⎝ ∂T ⎠ vWhen this is substituted into the definition of the volume expansivity, the result is 1 ⎛ ∂v ⎞ ⎛ ∂P ⎞ β=− ⎜ ⎟ ⎜ ⎟ v ⎝ ∂P ⎠ T ⎝ ∂T ⎠v ⎛ ∂P ⎞ = − α⎜ ⎟ ⎝ ∂T ⎠ vPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 40. 12-40 cp vα12-54 It is to be demonstrated that k = =− . cv (∂v / ∂P )sAnalysis The relations for entropy differential are dT ⎛ ∂P ⎞ ds = cv +⎜ ⎟ dv T ⎝ ∂T ⎠ v dT ⎛ ∂v ⎞ ds = c p −⎜ ⎟ dP T ⎝ ∂T ⎠ PFor fixed s, these basic equations reduce to dT ⎛ ∂P ⎞ cv = −⎜ ⎟ dv T ⎝ ∂T ⎠ v dT ⎛ ∂v ⎞ cp =⎜ ⎟ dP T ⎝ ∂T ⎠ PAlso, when s is fixed, ∂v ⎛ ∂v ⎞ =⎜ ⎟ ∂P ⎝ ∂P ⎠ sForming the specific heat ratio from these expressions gives ⎛ ∂v ⎞ ⎛ ∂T ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ P ⎝ ∂P ⎠ v k=− ⎛ ∂v ⎞ ⎜ ⎟ ⎝ ∂P ⎠ sThe cyclic relation is ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = −1 ⎝ ∂T ⎠ P ⎝ ∂v ⎠ T ⎝ ∂P ⎠ vSolving this for the numerator of the specific heat ratio expression and substituting the result into thisnumerator produces ⎛ ∂v ⎞ ⎜ ⎟ ⎝ ∂P ⎠ T vα k= =− ⎛ ∂v ⎞ ⎛ ∂v ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ∂P ⎠ s ⎝ ∂P ⎠ sPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 41. 12-41 RT a12-55 An expression for the specific heat difference of a substance whose equation of state is P = − v − b v 2Tis to be derived.Analysis The specific heat difference is expressed by 2 ⎛ ∂v ⎞ ⎛ ∂P ⎞ c p − cv = −T ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ P ⎝ ∂v ⎠ TAccording to the cyclic relation, ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = −1 ⎝ ∂T ⎠ P ⎝ ∂v ⎠ T ⎝ ∂P ⎠ vwhich on rearrangement becomes −1 ⎛ ∂v ⎞ ⎛ ∂P ⎞ ⎛ ∂P ⎞ ⎜ ⎟ = −⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ P ⎝ ∂T ⎠ v ⎝ ∂v ⎠ TSubstituting this result into the expression for the specific heat difference gives 2 −1 ⎛ ∂P ⎞ ⎛ ∂P ⎞ c p − cv = −T ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ v ⎝ ∂v ⎠ TThe appropriate partial derivatives of the equation of state are ⎛ ∂P ⎞ R a ⎜ ⎟ = + 2 2 ⎝ ∂T ⎠ v v − b v T ⎛ ∂P ⎞ RT 2a ⎜ ⎟ =− + 3 ⎝ ∂v ⎠ T (v − b) 2 v TThe difference in the specific heats is then 2 ⎡ R a ⎤ ⎢v − b + 2 2 ⎥ ⎣ v T ⎦ c p − cv = −T RT 2a − 2 + 3 (v − b) v TPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 42. 12-42The Joule-Thomson Coefficient12-56C It represents the variation of temperature with pressure during a throttling process.12-57C The line that passes through the peak points of the constant enthalpy lines on a T-P diagram iscalled the inversion line. The maximum inversion temperature is the highest temperature a fluid can becooled by throttling.12-58C No. The temperature may even increase as a result of throttling.12-59C Yes.12-60C No. Helium is an ideal gas and h = h(T) for ideal gases. Therefore, the temperature of an idealgas remains constant during a throttling (h = constant) process.12-61E [Also solved by EES on enclosed CD] The Joule-Thompson coefficient of nitrogen at two states isto be estimated.Analysis (a) The enthalpy of nitrogen at 200 psia and 500 R is, from EES, h = -10.564 Btu/lbm. Note thatin 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 isexpressed as ⎛ ∂T ⎞ ⎛ ΔT ⎞ μ=⎜ ⎟ ≅⎜ ⎟ ⎝ ∂P ⎠ h ⎝ ΔP ⎠ h = −10.564 Btu/lbmConsidering a throttling process from 210 psia to 190 psia at h = -10.564 Btu/lbm, the Joule-Thomsoncoefficient is determined to be ⎛ T190 psia − T210 psia ⎞ (499.703 − 500.296) R μ=⎜ ⎟ = = 0.0297 R/psia ⎜ ⎟ (190 − 210) psia ⎝ (190 − 210) psia ⎠ h = −10.564 Btu/lbm(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 ⎛ ∂T ⎞ ⎛ ΔT ⎞ μ=⎜ ⎟ ≅⎜ ⎟ ⎝ ∂P ⎠ h ⎝ ΔP ⎠ h = −55.321 Btu/lbmConsidering a throttling process from 2010 psia to 1990 psia at h = -55.321 Btu/lbm, the Joule-Thomsoncoefficient is determined to be ⎛ T1999 psia − T2001 psia ⎞ (399.786 - 400.213) R ⎜ μ=⎜ ⎟ = = 0.0213 R/psia ⎟ (1990 − 2010) psia ⎝ (1990 − 2010) psia ⎠ h = −55.321 Btu/lbmPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 43. 12-4312-62E EES Problem 12-61E is reconsidered. The Joule-Thompson coefficient for nitrogen over thepressure 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 + dPP[2] = P - dPT[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 0.004 0.003 0.002 h = 100 Btu/lbm 0.001 μ [R/psia] 0 -0.001 -0.002 h = 175 Btu/lbm -0.003 -0.004 h = 225 Btu/lbm -0.005 -0.006 0 200 400 600 800 1000 1200 1400 1600 P [psia]PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.

No public clipboards found for this slide

×
### Save the most important slides with Clipping

Clipping is a handy way to collect and organize the most important slides from a presentation. You can keep your great finds in clipboards organized around topics.

Be the first to comment