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

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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.

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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.

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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.

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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.

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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.