The change in enthalpy ( h 2 − h 1 ) of water vapor using generalized chart relation. The change in entropy ( s 2 − s 1 ) of water vapor using generalized chart relation. The change in enthalpy ( h 2 − h 1 ) of water vapor using table. The change in entropy ( s 2 − s 1 ) of water vapor table.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 12.6, Problem 71P

To determine

The change in enthalpy (h2−h1) of water vapor using generalized chart relation.

The change in entropy (s2−s1) of water vapor using generalized chart relation.

The change in enthalpy (h2−h1) of water vapor using table.

The change in entropy (s2−s1) of water vapor table.

Expert Solution & Answer

Answer to Problem 71P

The change in enthalpy (h2−h1) of water vapor using generalized chart relation is −423.9 kJ/kg.

The change in entropy (s2−s1) of water vapor using generalized chart relation is −0.2329 kJ/kg⋅K.

The change in enthalpy (h2−h1) of water vapor using table is −426.2 kJ/kg.

The change in entropy (s2−s1) of water vapor table is −0.2355 kJ/kg⋅K.

Explanation of Solution

Write the mean change in enthalpy (h¯2−h¯1)ideal of water vapor.

(h¯2−h¯1)ideal=(h¯2)ideal−(h¯1)ideal (I)

Here, enthalpy of water vapor at temperature of 647.1 K is (h¯2)ideal and enthalpy of water vapor at temperature of 647.1 K is (h¯1)ideal.

Write the change in enthalpy ((h2−h1)ideal) of water vapor table.

(h2−h1)ideal=(h¯2−h¯1)idealMH2O (II)

Here, molar mass of water vapor is MH2O.

Write the mean change in entropy (s¯2−s¯1)ideal for water vapor.

(s¯2−s¯1)ideal=s¯∘2−s¯∘1−RlnP2P1 (III)

Here, gas constant is R, initial pressure is P1, final pressure is P2, entropy at final state is s¯∘2,and entropy at initial state is s¯∘1.

Write the change in entropy (s2−s1)ideal for water vapor table.

(s2−s1)ideal=(s¯2−s¯1)idealMH2O (IV)

Write the reduced temperature (TR1) at initial state.

TR1=T1Tcr (V)

Here, critical temperature is Tcr and initial temperature is T1.

Write the reduced pressure (PR1) at initial state.

PR1=P1Pcr (VI)

Here, critical pressure is Pcr and initial pressure is P1.

Write the reduced temperature (TR2) at final state.

TR2=T2Tcr (VII)

Here, critical temperature is Tcr and final temperature is T2.

Write the reduced pressure (PR2) at initial state.

PR2=P2Pcr (VIII)

Here, critical pressure is Pcr and initial pressure is P1.

Write the change in enthalpy (h2−h1) of water vapor using generalized chart relation.

h2−h1=RTcr(Zh1−Zh2)+(h2−h1)ideal (IX)

Here, change in enthalpy of water vapor is (h2−h1)ideal and gas constant is R.

Write the change in enthalpy (s2−s1) of water vapor using generalized chart relation.

(s2−s1)=R(Zs1−Zs2)+(s2−s1)ideal (X)

Here, change in entropy of water vapor is (s2−s1)ideal.

Conclusion:

Convert the unit of initial temperature °C to K.

T1=600°C=(600+273) K=873 K

Convert the unit of initial temperature °C to K.

T2=400°C=(400+273) K=673 K

Refer table A-23, “Ideal gas properties of water vapor”, obtain the enthalpy of water vapor at temperature of 873 K and 673 K as 30754 kJ/kmol and 23082 kJ/kmol.

(h¯2)ideal=23082 kJ/kmol(h¯1)ideal=30754 kJ/kmol

Substitute 23082 kJ/kmol for (h2)ideal and 30754 kJ/kmol for (h1)ideal in Equation (I).

(h¯2−h¯1)ideal=23082 kJ/kmol−30754 kJ/kmol=−7672 kJ/kmol

Refer table A-1, “Molar mass properties table”, obtain the molar mass MH2O of water vapor as 18.015 kg/kmol.

Refer the table A-20,”Water vapor properties of water vapor table”, select the entropy of water vapor at temperature of 873 K and 673 K as 217.141 kJ/kmol⋅K and 227.109 kJ/kmol⋅K.

s¯∘2=217.141 kJ/kmol⋅Ks¯∘1=227.109 kJ/kmol⋅K

Substitute 217.141 kJ/kmol⋅K for s¯∘2, 227.109 kJ/kmol⋅K for s¯∘1, 8.314 kJ/kg⋅K for R, 500 kPa for P2 and 1000 kPa for P1 in Equation (III).

(s¯2−s¯1)ideal=(217.141−227.109) kJ/kmol⋅K−(8.314 kJ/kg⋅K)ln(1000 kPa500 kPa)=−4.2052 kJ/kmol⋅K

Substitute 873 K for T1 and 647.1 K for Tcr in Equation (V).

TR1=873 K647.1 K=1.349

Substitute 1000 kPa for P1 and 22060 kPa for Pcr in Equation (VI).

PR1=1000 kPa22060 kPa=0.0453

Refer the table A-15, “Nelson-Obert generalized compressibility chart”, select the initial state of compressibility factor Zs1 and Zh1 at reduced pressure and temperature of 1.349 and 0.0453 as 0.0157 and 0.0288.

Zs1=0.0157Zh1=0.0288

Substitute 673 K for T2 and 647.1 K for Tcr in Equation (VII).

TR2=673 K647.1 K=1.040

Substitute 500 kPa for P2 and 22060 kPa for Pcr in Equation (VIII).

PR2=500 kPa22060 kPa=0.0227

Refer the table A-15, “Nelson-Obert generalized compressibility chart”, select the initial state of compressibility factor Zs2 and Zh2 at reduced pressure and temperature of 1.040 and 0.0227 as 0.0146 and 0.0223.

Zs2=0.0146Zh2=0.0223

From the gas constant properties table A-1, select the gas constant of water vapor as 0.1889 kJ/kg⋅K.

Substitute 0.0223 for Zh2, 0.0288 for Zh1, −7672 kJ/kg for (h2−h1)ideal, 8.314 kJ/kg⋅K for R and 647.1 K for Tcr in Equation (IX).

h2−h1=(8.314 kJ/kg⋅K)(647.1 K)(0.0223−0.0288)+24.32 kJ/kg=−7637 kJ/kg

Substitute −7637 kJ/kmol for (h¯2−h¯1)ideal and 18.015 kg/kmol for MH2O in Equation (II).

(h2−h1)ideal=−7637 kJ/kmol18.015 kg/kmol=−423.9 kJ/kg

Thus, the change in enthalpy (h2−h1) of water vapor using generalized chart relation is −423.9 kJ/kg.

Substitute −4.2052 kJ/kg⋅K for (s2−s1)ideal, 0.0146 for Zs2, 0.0157 for Zs1 and 8.314 kJ/kg⋅K for R in Equation (X).

s2−s1=(8.314 kJ/kg⋅K)(0.015−0.0146)+4.2052 kJ/kg⋅K=−4.1961 kJ/kg⋅K

Substitute −4.1961 kJ/kg⋅K for (s¯2−s¯1)ideal and 18.015 kg/kmol for MH2O in Equation (IV).

(s2−s1)ideal=−4.1961 kJ/kg⋅K18.015 kg/kmol=−0.2329 kJ/kg⋅K

Thus, the change in entropy (s2−s1) of water vapor using generalized chart relation is −0.2329 kJ/kg⋅K.

Refer table A-6, “Superheated water table”, select the inlet enthalpy and exit enthalpy at pressure of 1000 kPa and temperature of 600°C as 3698.6 kJ/kg and 3272.4 kJ/kg.

h1=3698.6 kJ/kgh2=3272.4 kJ/kg

h2−h1=3272.4 kJ/kg−3698.6 kJ/kg=−426.2 kJ/kg

Thus, the change in enthalpy (h2−h1) of water vapor using table is −426.2 kJ/kg

Refer table A-6, “Superheated water table”, select the inlet entropy and exit entropy at pressure of 500 kPa and temperature of 400°C as 8.0311 kJ/kg⋅K and 7.7956 kJ/kg⋅K.

s1=8.0311 kJ/kg⋅Ks2=7.7956 kJ/kg⋅K

(s2−s1)=7.7956 kJ/kg⋅K−8.0311 kJ/kg⋅K=−0.2355 kJ/kg⋅K

Thus, the change in entropy (s2−s1) of water vapor table is −0.2355 kJ/kg⋅K.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Solve this problem and show all of the work. Show how the moments are calculated and draw a diagram

Problem: Textbook Problem 10.52 and 10.53. Determine the moment of inertia of the area about the x- axis and the y-axis. 3 in. 3 in. 6 in. 2 in. 4 in. x

Several reactions are carried out in a closed vessel. The following data are taken for the concentration of compounds A, B, and C [grams per liter] as a function of time [minutes], from the start of the reaction. Show the resulting data and trendlines, with equation and value, on the appropriate graph type (rectilinear, semilog, or log–log) to make the data appear linear.

Answer 2

Question

Chapter 12.6, Problem 71P

To determine

The change in enthalpy (h2−h1) of water vapor using generalized chart relation.

The change in entropy (s2−s1) of water vapor using generalized chart relation.

The change in enthalpy (h2−h1) of water vapor using table.

The change in entropy (s2−s1) of water vapor table.

Expert Solution & Answer

Answer to Problem 71P

The change in enthalpy (h2−h1) of water vapor using generalized chart relation is −423.9 kJ/kg.

The change in entropy (s2−s1) of water vapor using generalized chart relation is −0.2329 kJ/kg⋅K.

The change in enthalpy (h2−h1) of water vapor using table is −426.2 kJ/kg.

The change in entropy (s2−s1) of water vapor table is −0.2355 kJ/kg⋅K.

Explanation of Solution

Write the mean change in enthalpy (h¯2−h¯1)ideal of water vapor.

(h¯2−h¯1)ideal=(h¯2)ideal−(h¯1)ideal (I)

Here, enthalpy of water vapor at temperature of 647.1 K is (h¯2)ideal and enthalpy of water vapor at temperature of 647.1 K is (h¯1)ideal.

Write the change in enthalpy ((h2−h1)ideal) of water vapor table.

(h2−h1)ideal=(h¯2−h¯1)idealMH2O (II)

Here, molar mass of water vapor is MH2O.

Write the mean change in entropy (s¯2−s¯1)ideal for water vapor.

(s¯2−s¯1)ideal=s¯∘2−s¯∘1−RlnP2P1 (III)

Here, gas constant is R, initial pressure is P1, final pressure is P2, entropy at final state is s¯∘2,and entropy at initial state is s¯∘1.

Write the change in entropy (s2−s1)ideal for water vapor table.

(s2−s1)ideal=(s¯2−s¯1)idealMH2O (IV)

Write the reduced temperature (TR1) at initial state.

TR1=T1Tcr (V)

Here, critical temperature is Tcr and initial temperature is T1.

Write the reduced pressure (PR1) at initial state.

PR1=P1Pcr (VI)

Here, critical pressure is Pcr and initial pressure is P1.

Write the reduced temperature (TR2) at final state.

TR2=T2Tcr (VII)

Here, critical temperature is Tcr and final temperature is T2.

Write the reduced pressure (PR2) at initial state.

PR2=P2Pcr (VIII)

Here, critical pressure is Pcr and initial pressure is P1.

Write the change in enthalpy (h2−h1) of water vapor using generalized chart relation.

h2−h1=RTcr(Zh1−Zh2)+(h2−h1)ideal (IX)

Here, change in enthalpy of water vapor is (h2−h1)ideal and gas constant is R.

Write the change in enthalpy (s2−s1) of water vapor using generalized chart relation.

(s2−s1)=R(Zs1−Zs2)+(s2−s1)ideal (X)

Here, change in entropy of water vapor is (s2−s1)ideal.

Conclusion:

Convert the unit of initial temperature °C to K.

T1=600°C=(600+273) K=873 K

Convert the unit of initial temperature °C to K.

T2=400°C=(400+273) K=673 K

Refer table A-23, “Ideal gas properties of water vapor”, obtain the enthalpy of water vapor at temperature of 873 K and 673 K as 30754 kJ/kmol and 23082 kJ/kmol.

(h¯2)ideal=23082 kJ/kmol(h¯1)ideal=30754 kJ/kmol

Substitute 23082 kJ/kmol for (h2)ideal and 30754 kJ/kmol for (h1)ideal in Equation (I).

(h¯2−h¯1)ideal=23082 kJ/kmol−30754 kJ/kmol=−7672 kJ/kmol

Refer table A-1, “Molar mass properties table”, obtain the molar mass MH2O of water vapor as 18.015 kg/kmol.

Refer the table A-20,”Water vapor properties of water vapor table”, select the entropy of water vapor at temperature of 873 K and 673 K as 217.141 kJ/kmol⋅K and 227.109 kJ/kmol⋅K.

s¯∘2=217.141 kJ/kmol⋅Ks¯∘1=227.109 kJ/kmol⋅K

Substitute 217.141 kJ/kmol⋅K for s¯∘2, 227.109 kJ/kmol⋅K for s¯∘1, 8.314 kJ/kg⋅K for R, 500 kPa for P2 and 1000 kPa for P1 in Equation (III).

(s¯2−s¯1)ideal=(217.141−227.109) kJ/kmol⋅K−(8.314 kJ/kg⋅K)ln(1000 kPa500 kPa)=−4.2052 kJ/kmol⋅K

Substitute 873 K for T1 and 647.1 K for Tcr in Equation (V).

TR1=873 K647.1 K=1.349

Substitute 1000 kPa for P1 and 22060 kPa for Pcr in Equation (VI).

PR1=1000 kPa22060 kPa=0.0453

Refer the table A-15, “Nelson-Obert generalized compressibility chart”, select the initial state of compressibility factor Zs1 and Zh1 at reduced pressure and temperature of 1.349 and 0.0453 as 0.0157 and 0.0288.

Zs1=0.0157Zh1=0.0288

Substitute 673 K for T2 and 647.1 K for Tcr in Equation (VII).

TR2=673 K647.1 K=1.040

Substitute 500 kPa for P2 and 22060 kPa for Pcr in Equation (VIII).

PR2=500 kPa22060 kPa=0.0227

Refer the table A-15, “Nelson-Obert generalized compressibility chart”, select the initial state of compressibility factor Zs2 and Zh2 at reduced pressure and temperature of 1.040 and 0.0227 as 0.0146 and 0.0223.

Zs2=0.0146Zh2=0.0223

From the gas constant properties table A-1, select the gas constant of water vapor as 0.1889 kJ/kg⋅K.

Substitute 0.0223 for Zh2, 0.0288 for Zh1, −7672 kJ/kg for (h2−h1)ideal, 8.314 kJ/kg⋅K for R and 647.1 K for Tcr in Equation (IX).

h2−h1=(8.314 kJ/kg⋅K)(647.1 K)(0.0223−0.0288)+24.32 kJ/kg=−7637 kJ/kg

Substitute −7637 kJ/kmol for (h¯2−h¯1)ideal and 18.015 kg/kmol for MH2O in Equation (II).

(h2−h1)ideal=−7637 kJ/kmol18.015 kg/kmol=−423.9 kJ/kg

Thus, the change in enthalpy (h2−h1) of water vapor using generalized chart relation is −423.9 kJ/kg.

Substitute −4.2052 kJ/kg⋅K for (s2−s1)ideal, 0.0146 for Zs2, 0.0157 for Zs1 and 8.314 kJ/kg⋅K for R in Equation (X).

s2−s1=(8.314 kJ/kg⋅K)(0.015−0.0146)+4.2052 kJ/kg⋅K=−4.1961 kJ/kg⋅K

Substitute −4.1961 kJ/kg⋅K for (s¯2−s¯1)ideal and 18.015 kg/kmol for MH2O in Equation (IV).

(s2−s1)ideal=−4.1961 kJ/kg⋅K18.015 kg/kmol=−0.2329 kJ/kg⋅K

Thus, the change in entropy (s2−s1) of water vapor using generalized chart relation is −0.2329 kJ/kg⋅K.

Refer table A-6, “Superheated water table”, select the inlet enthalpy and exit enthalpy at pressure of 1000 kPa and temperature of 600°C as 3698.6 kJ/kg and 3272.4 kJ/kg.

h1=3698.6 kJ/kgh2=3272.4 kJ/kg

h2−h1=3272.4 kJ/kg−3698.6 kJ/kg=−426.2 kJ/kg

Thus, the change in enthalpy (h2−h1) of water vapor using table is −426.2 kJ/kg

Refer table A-6, “Superheated water table”, select the inlet entropy and exit entropy at pressure of 500 kPa and temperature of 400°C as 8.0311 kJ/kg⋅K and 7.7956 kJ/kg⋅K.

s1=8.0311 kJ/kg⋅Ks2=7.7956 kJ/kg⋅K

(s2−s1)=7.7956 kJ/kg⋅K−8.0311 kJ/kg⋅K=−0.2355 kJ/kg⋅K

Thus, the change in entropy (s2−s1) of water vapor table is −0.2355 kJ/kg⋅K.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 12.6, Problem 71P

To determine

The change in enthalpy (h2−h1) of water vapor using generalized chart relation.

The change in entropy (s2−s1) of water vapor using generalized chart relation.

The change in enthalpy (h2−h1) of water vapor using table.

The change in entropy (s2−s1) of water vapor table.

Expert Solution & Answer

Answer to Problem 71P

The change in enthalpy (h2−h1) of water vapor using generalized chart relation is −423.9 kJ/kg.

The change in entropy (s2−s1) of water vapor using generalized chart relation is −0.2329 kJ/kg⋅K.

The change in enthalpy (h2−h1) of water vapor using table is −426.2 kJ/kg.

The change in entropy (s2−s1) of water vapor table is −0.2355 kJ/kg⋅K.

Explanation of Solution

Write the mean change in enthalpy (h¯2−h¯1)ideal of water vapor.

(h¯2−h¯1)ideal=(h¯2)ideal−(h¯1)ideal (I)

Here, enthalpy of water vapor at temperature of 647.1 K is (h¯2)ideal and enthalpy of water vapor at temperature of 647.1 K is (h¯1)ideal.

Write the change in enthalpy ((h2−h1)ideal) of water vapor table.

(h2−h1)ideal=(h¯2−h¯1)idealMH2O (II)

Here, molar mass of water vapor is MH2O.

Write the mean change in entropy (s¯2−s¯1)ideal for water vapor.

(s¯2−s¯1)ideal=s¯∘2−s¯∘1−RlnP2P1 (III)

Here, gas constant is R, initial pressure is P1, final pressure is P2, entropy at final state is s¯∘2,and entropy at initial state is s¯∘1.

Write the change in entropy (s2−s1)ideal for water vapor table.

(s2−s1)ideal=(s¯2−s¯1)idealMH2O (IV)

Write the reduced temperature (TR1) at initial state.

TR1=T1Tcr (V)

Here, critical temperature is Tcr and initial temperature is T1.

Write the reduced pressure (PR1) at initial state.

PR1=P1Pcr (VI)

Here, critical pressure is Pcr and initial pressure is P1.

Write the reduced temperature (TR2) at final state.

TR2=T2Tcr (VII)

Here, critical temperature is Tcr and final temperature is T2.

Write the reduced pressure (PR2) at initial state.

PR2=P2Pcr (VIII)

Here, critical pressure is Pcr and initial pressure is P1.

Write the change in enthalpy (h2−h1) of water vapor using generalized chart relation.

h2−h1=RTcr(Zh1−Zh2)+(h2−h1)ideal (IX)

Here, change in enthalpy of water vapor is (h2−h1)ideal and gas constant is R.

Write the change in enthalpy (s2−s1) of water vapor using generalized chart relation.

(s2−s1)=R(Zs1−Zs2)+(s2−s1)ideal (X)

Here, change in entropy of water vapor is (s2−s1)ideal.

Conclusion:

Convert the unit of initial temperature °C to K.

T1=600°C=(600+273) K=873 K

Convert the unit of initial temperature °C to K.

T2=400°C=(400+273) K=673 K

Refer table A-23, “Ideal gas properties of water vapor”, obtain the enthalpy of water vapor at temperature of 873 K and 673 K as 30754 kJ/kmol and 23082 kJ/kmol.

(h¯2)ideal=23082 kJ/kmol(h¯1)ideal=30754 kJ/kmol

Substitute 23082 kJ/kmol for (h2)ideal and 30754 kJ/kmol for (h1)ideal in Equation (I).

(h¯2−h¯1)ideal=23082 kJ/kmol−30754 kJ/kmol=−7672 kJ/kmol

Refer table A-1, “Molar mass properties table”, obtain the molar mass MH2O of water vapor as 18.015 kg/kmol.

Refer the table A-20,”Water vapor properties of water vapor table”, select the entropy of water vapor at temperature of 873 K and 673 K as 217.141 kJ/kmol⋅K and 227.109 kJ/kmol⋅K.

s¯∘2=217.141 kJ/kmol⋅Ks¯∘1=227.109 kJ/kmol⋅K

Substitute 217.141 kJ/kmol⋅K for s¯∘2, 227.109 kJ/kmol⋅K for s¯∘1, 8.314 kJ/kg⋅K for R, 500 kPa for P2 and 1000 kPa for P1 in Equation (III).

(s¯2−s¯1)ideal=(217.141−227.109) kJ/kmol⋅K−(8.314 kJ/kg⋅K)ln(1000 kPa500 kPa)=−4.2052 kJ/kmol⋅K

Substitute 873 K for T1 and 647.1 K for Tcr in Equation (V).

TR1=873 K647.1 K=1.349

Substitute 1000 kPa for P1 and 22060 kPa for Pcr in Equation (VI).

PR1=1000 kPa22060 kPa=0.0453

Refer the table A-15, “Nelson-Obert generalized compressibility chart”, select the initial state of compressibility factor Zs1 and Zh1 at reduced pressure and temperature of 1.349 and 0.0453 as 0.0157 and 0.0288.

Zs1=0.0157Zh1=0.0288

Substitute 673 K for T2 and 647.1 K for Tcr in Equation (VII).

TR2=673 K647.1 K=1.040

Substitute 500 kPa for P2 and 22060 kPa for Pcr in Equation (VIII).

PR2=500 kPa22060 kPa=0.0227

Refer the table A-15, “Nelson-Obert generalized compressibility chart”, select the initial state of compressibility factor Zs2 and Zh2 at reduced pressure and temperature of 1.040 and 0.0227 as 0.0146 and 0.0223.

Zs2=0.0146Zh2=0.0223

From the gas constant properties table A-1, select the gas constant of water vapor as 0.1889 kJ/kg⋅K.

Substitute 0.0223 for Zh2, 0.0288 for Zh1, −7672 kJ/kg for (h2−h1)ideal, 8.314 kJ/kg⋅K for R and 647.1 K for Tcr in Equation (IX).

h2−h1=(8.314 kJ/kg⋅K)(647.1 K)(0.0223−0.0288)+24.32 kJ/kg=−7637 kJ/kg

Substitute −7637 kJ/kmol for (h¯2−h¯1)ideal and 18.015 kg/kmol for MH2O in Equation (II).

(h2−h1)ideal=−7637 kJ/kmol18.015 kg/kmol=−423.9 kJ/kg

Thus, the change in enthalpy (h2−h1) of water vapor using generalized chart relation is −423.9 kJ/kg.

Substitute −4.2052 kJ/kg⋅K for (s2−s1)ideal, 0.0146 for Zs2, 0.0157 for Zs1 and 8.314 kJ/kg⋅K for R in Equation (X).

s2−s1=(8.314 kJ/kg⋅K)(0.015−0.0146)+4.2052 kJ/kg⋅K=−4.1961 kJ/kg⋅K

Substitute −4.1961 kJ/kg⋅K for (s¯2−s¯1)ideal and 18.015 kg/kmol for MH2O in Equation (IV).

(s2−s1)ideal=−4.1961 kJ/kg⋅K18.015 kg/kmol=−0.2329 kJ/kg⋅K

Thus, the change in entropy (s2−s1) of water vapor using generalized chart relation is −0.2329 kJ/kg⋅K.

Refer table A-6, “Superheated water table”, select the inlet enthalpy and exit enthalpy at pressure of 1000 kPa and temperature of 600°C as 3698.6 kJ/kg and 3272.4 kJ/kg.

h1=3698.6 kJ/kgh2=3272.4 kJ/kg

h2−h1=3272.4 kJ/kg−3698.6 kJ/kg=−426.2 kJ/kg

Thus, the change in enthalpy (h2−h1) of water vapor using table is −426.2 kJ/kg

Refer table A-6, “Superheated water table”, select the inlet entropy and exit entropy at pressure of 500 kPa and temperature of 400°C as 8.0311 kJ/kg⋅K and 7.7956 kJ/kg⋅K.

s1=8.0311 kJ/kg⋅Ks2=7.7956 kJ/kg⋅K

(s2−s1)=7.7956 kJ/kg⋅K−8.0311 kJ/kg⋅K=−0.2355 kJ/kg⋅K

Thus, the change in entropy (s2−s1) of water vapor table is −0.2355 kJ/kg⋅K.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 12.6, Problem 71P

To determine

The change in enthalpy (h2−h1) of water vapor using generalized chart relation.

The change in entropy (s2−s1) of water vapor using generalized chart relation.

The change in enthalpy (h2−h1) of water vapor using table.

The change in entropy (s2−s1) of water vapor table.

Expert Solution & Answer

Answer to Problem 71P

The change in enthalpy (h2−h1) of water vapor using generalized chart relation is −423.9 kJ/kg.

The change in entropy (s2−s1) of water vapor using generalized chart relation is −0.2329 kJ/kg⋅K.

The change in enthalpy (h2−h1) of water vapor using table is −426.2 kJ/kg.

The change in entropy (s2−s1) of water vapor table is −0.2355 kJ/kg⋅K.

Explanation of Solution

Write the mean change in enthalpy (h¯2−h¯1)ideal of water vapor.

(h¯2−h¯1)ideal=(h¯2)ideal−(h¯1)ideal (I)

Here, enthalpy of water vapor at temperature of 647.1 K is (h¯2)ideal and enthalpy of water vapor at temperature of 647.1 K is (h¯1)ideal.

Write the change in enthalpy ((h2−h1)ideal) of water vapor table.

(h2−h1)ideal=(h¯2−h¯1)idealMH2O (II)

Here, molar mass of water vapor is MH2O.

Write the mean change in entropy (s¯2−s¯1)ideal for water vapor.

(s¯2−s¯1)ideal=s¯∘2−s¯∘1−RlnP2P1 (III)

Here, gas constant is R, initial pressure is P1, final pressure is P2, entropy at final state is s¯∘2,and entropy at initial state is s¯∘1.

Write the change in entropy (s2−s1)ideal for water vapor table.

(s2−s1)ideal=(s¯2−s¯1)idealMH2O (IV)

Write the reduced temperature (TR1) at initial state.

TR1=T1Tcr (V)

Here, critical temperature is Tcr and initial temperature is T1.

Write the reduced pressure (PR1) at initial state.

PR1=P1Pcr (VI)

Here, critical pressure is Pcr and initial pressure is P1.

Write the reduced temperature (TR2) at final state.

TR2=T2Tcr (VII)

Here, critical temperature is Tcr and final temperature is T2.

Write the reduced pressure (PR2) at initial state.

PR2=P2Pcr (VIII)

Here, critical pressure is Pcr and initial pressure is P1.

Write the change in enthalpy (h2−h1) of water vapor using generalized chart relation.

h2−h1=RTcr(Zh1−Zh2)+(h2−h1)ideal (IX)

Here, change in enthalpy of water vapor is (h2−h1)ideal and gas constant is R.

Write the change in enthalpy (s2−s1) of water vapor using generalized chart relation.

(s2−s1)=R(Zs1−Zs2)+(s2−s1)ideal (X)

Here, change in entropy of water vapor is (s2−s1)ideal.

Conclusion:

Convert the unit of initial temperature °C to K.

T1=600°C=(600+273) K=873 K

Convert the unit of initial temperature °C to K.

T2=400°C=(400+273) K=673 K

Refer table A-23, “Ideal gas properties of water vapor”, obtain the enthalpy of water vapor at temperature of 873 K and 673 K as 30754 kJ/kmol and 23082 kJ/kmol.

(h¯2)ideal=23082 kJ/kmol(h¯1)ideal=30754 kJ/kmol

Substitute 23082 kJ/kmol for (h2)ideal and 30754 kJ/kmol for (h1)ideal in Equation (I).

(h¯2−h¯1)ideal=23082 kJ/kmol−30754 kJ/kmol=−7672 kJ/kmol

Refer table A-1, “Molar mass properties table”, obtain the molar mass MH2O of water vapor as 18.015 kg/kmol.

Refer the table A-20,”Water vapor properties of water vapor table”, select the entropy of water vapor at temperature of 873 K and 673 K as 217.141 kJ/kmol⋅K and 227.109 kJ/kmol⋅K.

s¯∘2=217.141 kJ/kmol⋅Ks¯∘1=227.109 kJ/kmol⋅K

Substitute 217.141 kJ/kmol⋅K for s¯∘2, 227.109 kJ/kmol⋅K for s¯∘1, 8.314 kJ/kg⋅K for R, 500 kPa for P2 and 1000 kPa for P1 in Equation (III).

(s¯2−s¯1)ideal=(217.141−227.109) kJ/kmol⋅K−(8.314 kJ/kg⋅K)ln(1000 kPa500 kPa)=−4.2052 kJ/kmol⋅K

Substitute 873 K for T1 and 647.1 K for Tcr in Equation (V).

TR1=873 K647.1 K=1.349

Substitute 1000 kPa for P1 and 22060 kPa for Pcr in Equation (VI).

PR1=1000 kPa22060 kPa=0.0453

Refer the table A-15, “Nelson-Obert generalized compressibility chart”, select the initial state of compressibility factor Zs1 and Zh1 at reduced pressure and temperature of 1.349 and 0.0453 as 0.0157 and 0.0288.

Zs1=0.0157Zh1=0.0288

Substitute 673 K for T2 and 647.1 K for Tcr in Equation (VII).

TR2=673 K647.1 K=1.040

Substitute 500 kPa for P2 and 22060 kPa for Pcr in Equation (VIII).

PR2=500 kPa22060 kPa=0.0227

Refer the table A-15, “Nelson-Obert generalized compressibility chart”, select the initial state of compressibility factor Zs2 and Zh2 at reduced pressure and temperature of 1.040 and 0.0227 as 0.0146 and 0.0223.

Zs2=0.0146Zh2=0.0223

From the gas constant properties table A-1, select the gas constant of water vapor as 0.1889 kJ/kg⋅K.

Substitute 0.0223 for Zh2, 0.0288 for Zh1, −7672 kJ/kg for (h2−h1)ideal, 8.314 kJ/kg⋅K for R and 647.1 K for Tcr in Equation (IX).

h2−h1=(8.314 kJ/kg⋅K)(647.1 K)(0.0223−0.0288)+24.32 kJ/kg=−7637 kJ/kg

Substitute −7637 kJ/kmol for (h¯2−h¯1)ideal and 18.015 kg/kmol for MH2O in Equation (II).

(h2−h1)ideal=−7637 kJ/kmol18.015 kg/kmol=−423.9 kJ/kg

Thus, the change in enthalpy (h2−h1) of water vapor using generalized chart relation is −423.9 kJ/kg.

Substitute −4.2052 kJ/kg⋅K for (s2−s1)ideal, 0.0146 for Zs2, 0.0157 for Zs1 and 8.314 kJ/kg⋅K for R in Equation (X).

s2−s1=(8.314 kJ/kg⋅K)(0.015−0.0146)+4.2052 kJ/kg⋅K=−4.1961 kJ/kg⋅K

Substitute −4.1961 kJ/kg⋅K for (s¯2−s¯1)ideal and 18.015 kg/kmol for MH2O in Equation (IV).

(s2−s1)ideal=−4.1961 kJ/kg⋅K18.015 kg/kmol=−0.2329 kJ/kg⋅K

Thus, the change in entropy (s2−s1) of water vapor using generalized chart relation is −0.2329 kJ/kg⋅K.

Refer table A-6, “Superheated water table”, select the inlet enthalpy and exit enthalpy at pressure of 1000 kPa and temperature of 600°C as 3698.6 kJ/kg and 3272.4 kJ/kg.

h1=3698.6 kJ/kgh2=3272.4 kJ/kg

h2−h1=3272.4 kJ/kg−3698.6 kJ/kg=−426.2 kJ/kg

Thus, the change in enthalpy (h2−h1) of water vapor using table is −426.2 kJ/kg

Refer table A-6, “Superheated water table”, select the inlet entropy and exit entropy at pressure of 500 kPa and temperature of 400°C as 8.0311 kJ/kg⋅K and 7.7956 kJ/kg⋅K.

s1=8.0311 kJ/kg⋅Ks2=7.7956 kJ/kg⋅K

(s2−s1)=7.7956 kJ/kg⋅K−8.0311 kJ/kg⋅K=−0.2355 kJ/kg⋅K

Thus, the change in entropy (s2−s1) of water vapor table is −0.2355 kJ/kg⋅K.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 12.6, Problem 71P

To determine

The change in enthalpy (h2−h1) of water vapor using generalized chart relation.

The change in entropy (s2−s1) of water vapor using generalized chart relation.

The change in enthalpy (h2−h1) of water vapor using table.

The change in entropy (s2−s1) of water vapor table.

Expert Solution & Answer

Answer to Problem 71P

The change in enthalpy (h2−h1) of water vapor using generalized chart relation is −423.9 kJ/kg.

The change in entropy (s2−s1) of water vapor using generalized chart relation is −0.2329 kJ/kg⋅K.

The change in enthalpy (h2−h1) of water vapor using table is −426.2 kJ/kg.

The change in entropy (s2−s1) of water vapor table is −0.2355 kJ/kg⋅K.

Explanation of Solution

Write the mean change in enthalpy (h¯2−h¯1)ideal of water vapor.

(h¯2−h¯1)ideal=(h¯2)ideal−(h¯1)ideal (I)

Here, enthalpy of water vapor at temperature of 647.1 K is (h¯2)ideal and enthalpy of water vapor at temperature of 647.1 K is (h¯1)ideal.

Write the change in enthalpy ((h2−h1)ideal) of water vapor table.

(h2−h1)ideal=(h¯2−h¯1)idealMH2O (II)

Here, molar mass of water vapor is MH2O.

Write the mean change in entropy (s¯2−s¯1)ideal for water vapor.

(s¯2−s¯1)ideal=s¯∘2−s¯∘1−RlnP2P1 (III)

Here, gas constant is R, initial pressure is P1, final pressure is P2, entropy at final state is s¯∘2,and entropy at initial state is s¯∘1.

Write the change in entropy (s2−s1)ideal for water vapor table.

(s2−s1)ideal=(s¯2−s¯1)idealMH2O (IV)

Write the reduced temperature (TR1) at initial state.

TR1=T1Tcr (V)

Here, critical temperature is Tcr and initial temperature is T1.

Write the reduced pressure (PR1) at initial state.

PR1=P1Pcr (VI)

Here, critical pressure is Pcr and initial pressure is P1.

Write the reduced temperature (TR2) at final state.

TR2=T2Tcr (VII)

Here, critical temperature is Tcr and final temperature is T2.

Write the reduced pressure (PR2) at initial state.

PR2=P2Pcr (VIII)

Here, critical pressure is Pcr and initial pressure is P1.

Write the change in enthalpy (h2−h1) of water vapor using generalized chart relation.

h2−h1=RTcr(Zh1−Zh2)+(h2−h1)ideal (IX)

Here, change in enthalpy of water vapor is (h2−h1)ideal and gas constant is R.

Write the change in enthalpy (s2−s1) of water vapor using generalized chart relation.

(s2−s1)=R(Zs1−Zs2)+(s2−s1)ideal (X)

Here, change in entropy of water vapor is (s2−s1)ideal.

Conclusion:

Convert the unit of initial temperature °C to K.

T1=600°C=(600+273) K=873 K

Convert the unit of initial temperature °C to K.

T2=400°C=(400+273) K=673 K

Refer table A-23, “Ideal gas properties of water vapor”, obtain the enthalpy of water vapor at temperature of 873 K and 673 K as 30754 kJ/kmol and 23082 kJ/kmol.

(h¯2)ideal=23082 kJ/kmol(h¯1)ideal=30754 kJ/kmol

Substitute 23082 kJ/kmol for (h2)ideal and 30754 kJ/kmol for (h1)ideal in Equation (I).

(h¯2−h¯1)ideal=23082 kJ/kmol−30754 kJ/kmol=−7672 kJ/kmol

Refer table A-1, “Molar mass properties table”, obtain the molar mass MH2O of water vapor as 18.015 kg/kmol.

Refer the table A-20,”Water vapor properties of water vapor table”, select the entropy of water vapor at temperature of 873 K and 673 K as 217.141 kJ/kmol⋅K and 227.109 kJ/kmol⋅K.

s¯∘2=217.141 kJ/kmol⋅Ks¯∘1=227.109 kJ/kmol⋅K

Substitute 217.141 kJ/kmol⋅K for s¯∘2, 227.109 kJ/kmol⋅K for s¯∘1, 8.314 kJ/kg⋅K for R, 500 kPa for P2 and 1000 kPa for P1 in Equation (III).

(s¯2−s¯1)ideal=(217.141−227.109) kJ/kmol⋅K−(8.314 kJ/kg⋅K)ln(1000 kPa500 kPa)=−4.2052 kJ/kmol⋅K

Substitute 873 K for T1 and 647.1 K for Tcr in Equation (V).

TR1=873 K647.1 K=1.349

Substitute 1000 kPa for P1 and 22060 kPa for Pcr in Equation (VI).

PR1=1000 kPa22060 kPa=0.0453

Refer the table A-15, “Nelson-Obert generalized compressibility chart”, select the initial state of compressibility factor Zs1 and Zh1 at reduced pressure and temperature of 1.349 and 0.0453 as 0.0157 and 0.0288.

Zs1=0.0157Zh1=0.0288

Substitute 673 K for T2 and 647.1 K for Tcr in Equation (VII).

TR2=673 K647.1 K=1.040

Substitute 500 kPa for P2 and 22060 kPa for Pcr in Equation (VIII).

PR2=500 kPa22060 kPa=0.0227

Refer the table A-15, “Nelson-Obert generalized compressibility chart”, select the initial state of compressibility factor Zs2 and Zh2 at reduced pressure and temperature of 1.040 and 0.0227 as 0.0146 and 0.0223.

Zs2=0.0146Zh2=0.0223

From the gas constant properties table A-1, select the gas constant of water vapor as 0.1889 kJ/kg⋅K.

Substitute 0.0223 for Zh2, 0.0288 for Zh1, −7672 kJ/kg for (h2−h1)ideal, 8.314 kJ/kg⋅K for R and 647.1 K for Tcr in Equation (IX).

h2−h1=(8.314 kJ/kg⋅K)(647.1 K)(0.0223−0.0288)+24.32 kJ/kg=−7637 kJ/kg

Substitute −7637 kJ/kmol for (h¯2−h¯1)ideal and 18.015 kg/kmol for MH2O in Equation (II).

(h2−h1)ideal=−7637 kJ/kmol18.015 kg/kmol=−423.9 kJ/kg

Thus, the change in enthalpy (h2−h1) of water vapor using generalized chart relation is −423.9 kJ/kg.

Substitute −4.2052 kJ/kg⋅K for (s2−s1)ideal, 0.0146 for Zs2, 0.0157 for Zs1 and 8.314 kJ/kg⋅K for R in Equation (X).

s2−s1=(8.314 kJ/kg⋅K)(0.015−0.0146)+4.2052 kJ/kg⋅K=−4.1961 kJ/kg⋅K

Substitute −4.1961 kJ/kg⋅K for (s¯2−s¯1)ideal and 18.015 kg/kmol for MH2O in Equation (IV).

(s2−s1)ideal=−4.1961 kJ/kg⋅K18.015 kg/kmol=−0.2329 kJ/kg⋅K

Thus, the change in entropy (s2−s1) of water vapor using generalized chart relation is −0.2329 kJ/kg⋅K.

Refer table A-6, “Superheated water table”, select the inlet enthalpy and exit enthalpy at pressure of 1000 kPa and temperature of 600°C as 3698.6 kJ/kg and 3272.4 kJ/kg.

h1=3698.6 kJ/kgh2=3272.4 kJ/kg

h2−h1=3272.4 kJ/kg−3698.6 kJ/kg=−426.2 kJ/kg

Thus, the change in enthalpy (h2−h1) of water vapor using table is −426.2 kJ/kg

Refer table A-6, “Superheated water table”, select the inlet entropy and exit entropy at pressure of 500 kPa and temperature of 400°C as 8.0311 kJ/kg⋅K and 7.7956 kJ/kg⋅K.

s1=8.0311 kJ/kg⋅Ks2=7.7956 kJ/kg⋅K

(s2−s1)=7.7956 kJ/kg⋅K−8.0311 kJ/kg⋅K=−0.2355 kJ/kg⋅K

Thus, the change in entropy (s2−s1) of water vapor table is −0.2355 kJ/kg⋅K.

Answer 6

Chapter 12.6, Problem 71P

To determine

The change in enthalpy (h2−h1) of water vapor using generalized chart relation.

The change in entropy (s2−s1) of water vapor using generalized chart relation.

The change in enthalpy (h2−h1) of water vapor using table.

The change in entropy (s2−s1) of water vapor table.

Expert Solution & Answer

Answer to Problem 71P

The change in enthalpy (h2−h1) of water vapor using generalized chart relation is −423.9 kJ/kg.

The change in entropy (s2−s1) of water vapor using generalized chart relation is −0.2329 kJ/kg⋅K.

The change in enthalpy (h2−h1) of water vapor using table is −426.2 kJ/kg.

The change in entropy (s2−s1) of water vapor table is −0.2355 kJ/kg⋅K.

Explanation of Solution

Write the mean change in enthalpy (h¯2−h¯1)ideal of water vapor.

(h¯2−h¯1)ideal=(h¯2)ideal−(h¯1)ideal (I)

Here, enthalpy of water vapor at temperature of 647.1 K is (h¯2)ideal and enthalpy of water vapor at temperature of 647.1 K is (h¯1)ideal.

Write the change in enthalpy ((h2−h1)ideal) of water vapor table.

(h2−h1)ideal=(h¯2−h¯1)idealMH2O (II)

Here, molar mass of water vapor is MH2O.

Write the mean change in entropy (s¯2−s¯1)ideal for water vapor.

(s¯2−s¯1)ideal=s¯∘2−s¯∘1−RlnP2P1 (III)

Here, gas constant is R, initial pressure is P1, final pressure is P2, entropy at final state is s¯∘2,and entropy at initial state is s¯∘1.

Write the change in entropy (s2−s1)ideal for water vapor table.

(s2−s1)ideal=(s¯2−s¯1)idealMH2O (IV)

Write the reduced temperature (TR1) at initial state.

TR1=T1Tcr (V)

Here, critical temperature is Tcr and initial temperature is T1.

Write the reduced pressure (PR1) at initial state.

PR1=P1Pcr (VI)

Here, critical pressure is Pcr and initial pressure is P1.

Write the reduced temperature (TR2) at final state.

TR2=T2Tcr (VII)

Here, critical temperature is Tcr and final temperature is T2.

Write the reduced pressure (PR2) at initial state.

PR2=P2Pcr (VIII)

Here, critical pressure is Pcr and initial pressure is P1.

Write the change in enthalpy (h2−h1) of water vapor using generalized chart relation.

h2−h1=RTcr(Zh1−Zh2)+(h2−h1)ideal (IX)

Here, change in enthalpy of water vapor is (h2−h1)ideal and gas constant is R.

Write the change in enthalpy (s2−s1) of water vapor using generalized chart relation.

(s2−s1)=R(Zs1−Zs2)+(s2−s1)ideal (X)

Here, change in entropy of water vapor is (s2−s1)ideal.

Conclusion:

Convert the unit of initial temperature °C to K.

T1=600°C=(600+273) K=873 K

Convert the unit of initial temperature °C to K.

T2=400°C=(400+273) K=673 K

Refer table A-23, “Ideal gas properties of water vapor”, obtain the enthalpy of water vapor at temperature of 873 K and 673 K as 30754 kJ/kmol and 23082 kJ/kmol.

(h¯2)ideal=23082 kJ/kmol(h¯1)ideal=30754 kJ/kmol

Substitute 23082 kJ/kmol for (h2)ideal and 30754 kJ/kmol for (h1)ideal in Equation (I).

(h¯2−h¯1)ideal=23082 kJ/kmol−30754 kJ/kmol=−7672 kJ/kmol

Refer table A-1, “Molar mass properties table”, obtain the molar mass MH2O of water vapor as 18.015 kg/kmol.

Refer the table A-20,”Water vapor properties of water vapor table”, select the entropy of water vapor at temperature of 873 K and 673 K as 217.141 kJ/kmol⋅K and 227.109 kJ/kmol⋅K.

s¯∘2=217.141 kJ/kmol⋅Ks¯∘1=227.109 kJ/kmol⋅K

Substitute 217.141 kJ/kmol⋅K for s¯∘2, 227.109 kJ/kmol⋅K for s¯∘1, 8.314 kJ/kg⋅K for R, 500 kPa for P2 and 1000 kPa for P1 in Equation (III).

(s¯2−s¯1)ideal=(217.141−227.109) kJ/kmol⋅K−(8.314 kJ/kg⋅K)ln(1000 kPa500 kPa)=−4.2052 kJ/kmol⋅K

Substitute 873 K for T1 and 647.1 K for Tcr in Equation (V).

TR1=873 K647.1 K=1.349

Substitute 1000 kPa for P1 and 22060 kPa for Pcr in Equation (VI).

PR1=1000 kPa22060 kPa=0.0453

Refer the table A-15, “Nelson-Obert generalized compressibility chart”, select the initial state of compressibility factor Zs1 and Zh1 at reduced pressure and temperature of 1.349 and 0.0453 as 0.0157 and 0.0288.

Zs1=0.0157Zh1=0.0288

Substitute 673 K for T2 and 647.1 K for Tcr in Equation (VII).

TR2=673 K647.1 K=1.040

Substitute 500 kPa for P2 and 22060 kPa for Pcr in Equation (VIII).

PR2=500 kPa22060 kPa=0.0227

Refer the table A-15, “Nelson-Obert generalized compressibility chart”, select the initial state of compressibility factor Zs2 and Zh2 at reduced pressure and temperature of 1.040 and 0.0227 as 0.0146 and 0.0223.

Zs2=0.0146Zh2=0.0223

From the gas constant properties table A-1, select the gas constant of water vapor as 0.1889 kJ/kg⋅K.

Substitute 0.0223 for Zh2, 0.0288 for Zh1, −7672 kJ/kg for (h2−h1)ideal, 8.314 kJ/kg⋅K for R and 647.1 K for Tcr in Equation (IX).

h2−h1=(8.314 kJ/kg⋅K)(647.1 K)(0.0223−0.0288)+24.32 kJ/kg=−7637 kJ/kg

Substitute −7637 kJ/kmol for (h¯2−h¯1)ideal and 18.015 kg/kmol for MH2O in Equation (II).

(h2−h1)ideal=−7637 kJ/kmol18.015 kg/kmol=−423.9 kJ/kg

Thus, the change in enthalpy (h2−h1) of water vapor using generalized chart relation is −423.9 kJ/kg.

Substitute −4.2052 kJ/kg⋅K for (s2−s1)ideal, 0.0146 for Zs2, 0.0157 for Zs1 and 8.314 kJ/kg⋅K for R in Equation (X).

s2−s1=(8.314 kJ/kg⋅K)(0.015−0.0146)+4.2052 kJ/kg⋅K=−4.1961 kJ/kg⋅K

Substitute −4.1961 kJ/kg⋅K for (s¯2−s¯1)ideal and 18.015 kg/kmol for MH2O in Equation (IV).

(s2−s1)ideal=−4.1961 kJ/kg⋅K18.015 kg/kmol=−0.2329 kJ/kg⋅K

Thus, the change in entropy (s2−s1) of water vapor using generalized chart relation is −0.2329 kJ/kg⋅K.

Refer table A-6, “Superheated water table”, select the inlet enthalpy and exit enthalpy at pressure of 1000 kPa and temperature of 600°C as 3698.6 kJ/kg and 3272.4 kJ/kg.

h1=3698.6 kJ/kgh2=3272.4 kJ/kg

h2−h1=3272.4 kJ/kg−3698.6 kJ/kg=−426.2 kJ/kg

Thus, the change in enthalpy (h2−h1) of water vapor using table is −426.2 kJ/kg

Refer table A-6, “Superheated water table”, select the inlet entropy and exit entropy at pressure of 500 kPa and temperature of 400°C as 8.0311 kJ/kg⋅K and 7.7956 kJ/kg⋅K.

s1=8.0311 kJ/kg⋅Ks2=7.7956 kJ/kg⋅K

(s2−s1)=7.7956 kJ/kg⋅K−8.0311 kJ/kg⋅K=−0.2355 kJ/kg⋅K

Thus, the change in entropy (s2−s1) of water vapor table is −0.2355 kJ/kg⋅K.

Answer 7

Chapter 12.6, Problem 71P

To determine

The change in enthalpy (h2−h1) of water vapor using generalized chart relation.

The change in entropy (s2−s1) of water vapor using generalized chart relation.

The change in enthalpy (h2−h1) of water vapor using table.

The change in entropy (s2−s1) of water vapor table.

Answer 8

Expert Solution & Answer

Answer to Problem 71P

The change in enthalpy (h2−h1) of water vapor using generalized chart relation is −423.9 kJ/kg.

The change in entropy (s2−s1) of water vapor using generalized chart relation is −0.2329 kJ/kg⋅K.

The change in enthalpy (h2−h1) of water vapor using table is −426.2 kJ/kg.

The change in entropy (s2−s1) of water vapor table is −0.2355 kJ/kg⋅K.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Write the mean change in enthalpy (h¯2−h¯1)ideal of water vapor.

(h¯2−h¯1)ideal=(h¯2)ideal−(h¯1)ideal (I)

Here, enthalpy of water vapor at temperature of 647.1 K is (h¯2)ideal and enthalpy of water vapor at temperature of 647.1 K is (h¯1)ideal.

Write the change in enthalpy ((h2−h1)ideal) of water vapor table.

(h2−h1)ideal=(h¯2−h¯1)idealMH2O (II)

Here, molar mass of water vapor is MH2O.

Write the mean change in entropy (s¯2−s¯1)ideal for water vapor.

(s¯2−s¯1)ideal=s¯∘2−s¯∘1−RlnP2P1 (III)

Here, gas constant is R, initial pressure is P1, final pressure is P2, entropy at final state is s¯∘2,and entropy at initial state is s¯∘1.

Write the change in entropy (s2−s1)ideal for water vapor table.

(s2−s1)ideal=(s¯2−s¯1)idealMH2O (IV)

Write the reduced temperature (TR1) at initial state.

TR1=T1Tcr (V)

Here, critical temperature is Tcr and initial temperature is T1.

Write the reduced pressure (PR1) at initial state.

PR1=P1Pcr (VI)

Here, critical pressure is Pcr and initial pressure is P1.

Write the reduced temperature (TR2) at final state.

TR2=T2Tcr (VII)

Here, critical temperature is Tcr and final temperature is T2.

Write the reduced pressure (PR2) at initial state.

PR2=P2Pcr (VIII)

Here, critical pressure is Pcr and initial pressure is P1.

Write the change in enthalpy (h2−h1) of water vapor using generalized chart relation.

h2−h1=RTcr(Zh1−Zh2)+(h2−h1)ideal (IX)

Here, change in enthalpy of water vapor is (h2−h1)ideal and gas constant is R.

Write the change in enthalpy (s2−s1) of water vapor using generalized chart relation.

(s2−s1)=R(Zs1−Zs2)+(s2−s1)ideal (X)

Here, change in entropy of water vapor is (s2−s1)ideal.

Conclusion:

Convert the unit of initial temperature °C to K.

T1=600°C=(600+273) K=873 K

Convert the unit of initial temperature °C to K.

T2=400°C=(400+273) K=673 K

Refer table A-23, “Ideal gas properties of water vapor”, obtain the enthalpy of water vapor at temperature of 873 K and 673 K as 30754 kJ/kmol and 23082 kJ/kmol.

(h¯2)ideal=23082 kJ/kmol(h¯1)ideal=30754 kJ/kmol

Substitute 23082 kJ/kmol for (h2)ideal and 30754 kJ/kmol for (h1)ideal in Equation (I).

(h¯2−h¯1)ideal=23082 kJ/kmol−30754 kJ/kmol=−7672 kJ/kmol

Refer table A-1, “Molar mass properties table”, obtain the molar mass MH2O of water vapor as 18.015 kg/kmol.

Refer the table A-20,”Water vapor properties of water vapor table”, select the entropy of water vapor at temperature of 873 K and 673 K as 217.141 kJ/kmol⋅K and 227.109 kJ/kmol⋅K.

s¯∘2=217.141 kJ/kmol⋅Ks¯∘1=227.109 kJ/kmol⋅K

Substitute 217.141 kJ/kmol⋅K for s¯∘2, 227.109 kJ/kmol⋅K for s¯∘1, 8.314 kJ/kg⋅K for R, 500 kPa for P2 and 1000 kPa for P1 in Equation (III).

(s¯2−s¯1)ideal=(217.141−227.109) kJ/kmol⋅K−(8.314 kJ/kg⋅K)ln(1000 kPa500 kPa)=−4.2052 kJ/kmol⋅K

Substitute 873 K for T1 and 647.1 K for Tcr in Equation (V).

TR1=873 K647.1 K=1.349

Substitute 1000 kPa for P1 and 22060 kPa for Pcr in Equation (VI).

PR1=1000 kPa22060 kPa=0.0453

Refer the table A-15, “Nelson-Obert generalized compressibility chart”, select the initial state of compressibility factor Zs1 and Zh1 at reduced pressure and temperature of 1.349 and 0.0453 as 0.0157 and 0.0288.

Zs1=0.0157Zh1=0.0288

Substitute 673 K for T2 and 647.1 K for Tcr in Equation (VII).

TR2=673 K647.1 K=1.040

Substitute 500 kPa for P2 and 22060 kPa for Pcr in Equation (VIII).

PR2=500 kPa22060 kPa=0.0227

Refer the table A-15, “Nelson-Obert generalized compressibility chart”, select the initial state of compressibility factor Zs2 and Zh2 at reduced pressure and temperature of 1.040 and 0.0227 as 0.0146 and 0.0223.

Zs2=0.0146Zh2=0.0223

From the gas constant properties table A-1, select the gas constant of water vapor as 0.1889 kJ/kg⋅K.

Substitute 0.0223 for Zh2, 0.0288 for Zh1, −7672 kJ/kg for (h2−h1)ideal, 8.314 kJ/kg⋅K for R and 647.1 K for Tcr in Equation (IX).

h2−h1=(8.314 kJ/kg⋅K)(647.1 K)(0.0223−0.0288)+24.32 kJ/kg=−7637 kJ/kg

Substitute −7637 kJ/kmol for (h¯2−h¯1)ideal and 18.015 kg/kmol for MH2O in Equation (II).

(h2−h1)ideal=−7637 kJ/kmol18.015 kg/kmol=−423.9 kJ/kg

Thus, the change in enthalpy (h2−h1) of water vapor using generalized chart relation is −423.9 kJ/kg.

Substitute −4.2052 kJ/kg⋅K for (s2−s1)ideal, 0.0146 for Zs2, 0.0157 for Zs1 and 8.314 kJ/kg⋅K for R in Equation (X).

s2−s1=(8.314 kJ/kg⋅K)(0.015−0.0146)+4.2052 kJ/kg⋅K=−4.1961 kJ/kg⋅K

Substitute −4.1961 kJ/kg⋅K for (s¯2−s¯1)ideal and 18.015 kg/kmol for MH2O in Equation (IV).

(s2−s1)ideal=−4.1961 kJ/kg⋅K18.015 kg/kmol=−0.2329 kJ/kg⋅K

Thus, the change in entropy (s2−s1) of water vapor using generalized chart relation is −0.2329 kJ/kg⋅K.

Refer table A-6, “Superheated water table”, select the inlet enthalpy and exit enthalpy at pressure of 1000 kPa and temperature of 600°C as 3698.6 kJ/kg and 3272.4 kJ/kg.

h1=3698.6 kJ/kgh2=3272.4 kJ/kg

h2−h1=3272.4 kJ/kg−3698.6 kJ/kg=−426.2 kJ/kg

Thus, the change in enthalpy (h2−h1) of water vapor using table is −426.2 kJ/kg

Refer table A-6, “Superheated water table”, select the inlet entropy and exit entropy at pressure of 500 kPa and temperature of 400°C as 8.0311 kJ/kg⋅K and 7.7956 kJ/kg⋅K.