The final equilibrium temperature.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 7.13, Problem 92P

a)

To determine

The final equilibrium temperature.

a)

Expert Solution

Answer to Problem 92P

The final equilibrium temperature is 70.2 °C.

Explanation of Solution

Write the expression for the energy balance equation for closed system.

Ein−Eout=ΔEsystem (I)

Here, energy transfer into the control volume is Ein, energy transfer out of control volume is Eout and change in internal energy of system is ΔEsystem.

Write the expression to calculate the mass of the air.

mair=P1vRT1 (II)

Here, mass of the air is mair, initial pressure is P1, specific volume is v, gas constant is R and initial temperature is T1.

Conclusion:

Substitute 0 for Ein, 0 for Eout and ΔUair+ΔUwater for ΔEsystem in Equation (I).

0−0=ΔUwater+ΔUair[mcw(T2−T1)]water+[mcv(T2−T1)]air=0mwatercw(T2−(T1)water)+maircv(T2−(T1)air)=0 (III)

Here, mass of the air is mair, mass of the water is mwater, specific heat at constant pressure is cp and specific heat at constant volume is cv , change in internal energy in air is ΔUair, change in internal energy in water is ΔUwater, initial temperature of water is (T1)water, initial temperature of air is (T1)air, final temperature is T2.

From the Table A-2, “Ideal-gas specific heats of various common gases”, obtain the properties for air.

R=0.2870 kJ/kg⋅Kcp=1.005 kJ/kg⋅Kcv=0.718 kJ/kg⋅K

From the Table A-3, “Properties of common liquids, solids, and foods”, the specific heat of water cw at room temperature is 4.18 kJ/kg⋅K

Substitute 101.3 kPa for P1, 90 m3 for v, 0.2870 kJ/kg⋅K for R and 12 °C for T1 in Equation (II).

mair=(101.3 kPa)(90 m3)(0.2870 kJ/kg⋅K)12 °C=(101.3 kPa)(90 m3)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(12+273)K=111.46 kg

Substitute 45 kg for mwater, 4.18 kJ/kg⋅K for cw, 95 °C for (T1)water, 111.46 kg for mair, 0.718 kJ/kg⋅K for cv and 12 °C for (T1)air in Equation (III).

{[(45 kg)(4.18 kJ/kg⋅K)(T2−95°C)]+[(111.46 kg)(0.718 kJ/kg⋅K)(T2−12°C)]}=0{[(188.1 kJ/K)(T2−95°C)]+[(80.02 kJ/K)(T2−12°C)]}=0T2=70.2 °C

Thus, the final equilibrium temperature is 70.2 °C.

b)

To determine

The amount of heat transfer to the air.

b)

Expert Solution

Answer to Problem 92P

The amount of heat transfer to the air is 4660 kJ.

Explanation of Solution

Write the expression to calculate the heat transfer (Q) to the air.

Q=maircv(T2−(T1)air) (IV)

Conclusion:

Substitute 111.46 kg for mair, 0.718 kJ/kg⋅K for cv and 12 °C for (T1)air in Equation (IV).

Q=(111.46 kg)(0.718 kJ/kg⋅K)(70.2 °C−12 °C)=(111.46 kg)(0.718 kJ/kg⋅°K)[(70.2+273)K−(12 +273)K]=4660 kJ

Thus, the amount of heat transfer to the air is 4660 kJ.

c)

To determine

The entropy generation.

c)

Expert Solution

Answer to Problem 92P

The entropy generation is 1.77 kJ/K.

Explanation of Solution

Write the expression for the entropy balance equation of the system.

Sin−Sout+Sgen=ΔSsystem (V)

Here, rate of net entropy in is Sin, rate of net entropy out is Sout, rate of entropy generation is Sgen and change of entropy of the system is ΔSsystem

Write the expression to calculate the final pressure (P2).

mair=P2vRT2 (VI)

Here, final temparature is T2.

Conclusion:

Substitute 111.46 kg for mair, 90 m3 for v, 0.2870 kJ/kg⋅K for R and 70.2°C for T2 in Equation (VI).

111.46 kg=P2(90 m3)(0.2870 kJ/kg⋅K)70.2°CP2=(111.46 kg)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(70.2+273) K(90 m3)

P2=121.98 kPa=122 kPa

Substitute 0 for Sin, 0 for Sout and ΔSair+ΔSwater for ΔSsystem in Equation (V).

0−0+Sgen=ΔSair+ΔSwaterSgen=mair(cpln(T2(T1)air)−Rln(P2P1))+(mwatercwln(T2(T1)water)) (VII)

Substitute 111.46 kg for mair, 1.005 kJ/kg⋅K for cp, 12 °C for (T1)air, 70.2 °C for T2, 45 kg for mwater, 4.18 kJ/kg⋅K for cw, 95 °C for (T1)water , 0.2870 kJ/kg⋅K for R , 101.3 kPa for P1 and 122 kPa for P2 in Equation (VII).

Sgen=[(111.46 kg)((1.005 kJ/kg⋅K)ln(70.2 °C12 °C)−(0.2870 kJ/kg⋅K)ln(122 kPa101.3 kPa))+(45 kg(4.18 kJ/kg⋅K)ln(70.2 °C95 °C))]=[(111.46 kg)((1.005 kJ/kg⋅K)ln((70.2+273) K(12+273) K)−(0.2870 kJ/kg⋅K)ln(122 kPa101.3 kPa))+(45 kg(4.18 kJ/kg⋅K)ln((70.2+273) K(95+273) K))]=1.488 kJ/kg+(−13.12 kJ/kg)=1.77 kJ/K

Thus, the entropy generation is 1.77 kJ/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

##2# Superheated steam powers a steam turbine for the production of electrical energy. The steam expands in the turbine and at an intermediate expansion pressure (0.1 Mpa) a fraction is extracted for a regeneration process in a surface regenerator. The turbine has an isentropic efficiency of 90% Design the simplified power plant schematic Analyze it on the basis of the attached figure Determine the power generated and the thermal efficiency of the plant ### Dados in the attached images

### To make a conclusion for a report of an experiment on rockets, in which the openrocket software was used for the construction and modeling of two rockets: one one-stage and one two-stage. First rocket (single-stage) reached a maximum vertical speed of 100 m/s and a maximum height of 500 m The second rocket (two-stage) reached a maximum vertical speed of 50 m/s and a maximum height of 250 m To make a simplified conclusion, taking into account the efficiency of the software in the study of rockets

Determine the coefficients of polynomial for the polynomial function of Cam profile based on the boundary conditions shown in the figure. S a 3 4 5 C₁ (+) Ꮎ В s = q + q { + c f * + q € * + q ( +c+c+c 6 Ꮎ +C5 +C β В В 0 cam angle 0 B 7 (

Answer 2

Question

Chapter 7.13, Problem 92P

a)

To determine

The final equilibrium temperature.

a)

Expert Solution

Answer to Problem 92P

The final equilibrium temperature is 70.2 °C.

Explanation of Solution

Write the expression for the energy balance equation for closed system.

Ein−Eout=ΔEsystem (I)

Here, energy transfer into the control volume is Ein, energy transfer out of control volume is Eout and change in internal energy of system is ΔEsystem.

Write the expression to calculate the mass of the air.

mair=P1vRT1 (II)

Here, mass of the air is mair, initial pressure is P1, specific volume is v, gas constant is R and initial temperature is T1.

Conclusion:

Substitute 0 for Ein, 0 for Eout and ΔUair+ΔUwater for ΔEsystem in Equation (I).

0−0=ΔUwater+ΔUair[mcw(T2−T1)]water+[mcv(T2−T1)]air=0mwatercw(T2−(T1)water)+maircv(T2−(T1)air)=0 (III)

Here, mass of the air is mair, mass of the water is mwater, specific heat at constant pressure is cp and specific heat at constant volume is cv , change in internal energy in air is ΔUair, change in internal energy in water is ΔUwater, initial temperature of water is (T1)water, initial temperature of air is (T1)air, final temperature is T2.

From the Table A-2, “Ideal-gas specific heats of various common gases”, obtain the properties for air.

R=0.2870 kJ/kg⋅Kcp=1.005 kJ/kg⋅Kcv=0.718 kJ/kg⋅K

From the Table A-3, “Properties of common liquids, solids, and foods”, the specific heat of water cw at room temperature is 4.18 kJ/kg⋅K

Substitute 101.3 kPa for P1, 90 m3 for v, 0.2870 kJ/kg⋅K for R and 12 °C for T1 in Equation (II).

mair=(101.3 kPa)(90 m3)(0.2870 kJ/kg⋅K)12 °C=(101.3 kPa)(90 m3)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(12+273)K=111.46 kg

Substitute 45 kg for mwater, 4.18 kJ/kg⋅K for cw, 95 °C for (T1)water, 111.46 kg for mair, 0.718 kJ/kg⋅K for cv and 12 °C for (T1)air in Equation (III).

{[(45 kg)(4.18 kJ/kg⋅K)(T2−95°C)]+[(111.46 kg)(0.718 kJ/kg⋅K)(T2−12°C)]}=0{[(188.1 kJ/K)(T2−95°C)]+[(80.02 kJ/K)(T2−12°C)]}=0T2=70.2 °C

Thus, the final equilibrium temperature is 70.2 °C.

b)

To determine

The amount of heat transfer to the air.

b)

Expert Solution

Answer to Problem 92P

The amount of heat transfer to the air is 4660 kJ.

Explanation of Solution

Write the expression to calculate the heat transfer (Q) to the air.

Q=maircv(T2−(T1)air) (IV)

Conclusion:

Substitute 111.46 kg for mair, 0.718 kJ/kg⋅K for cv and 12 °C for (T1)air in Equation (IV).

Q=(111.46 kg)(0.718 kJ/kg⋅K)(70.2 °C−12 °C)=(111.46 kg)(0.718 kJ/kg⋅°K)[(70.2+273)K−(12 +273)K]=4660 kJ

Thus, the amount of heat transfer to the air is 4660 kJ.

c)

To determine

The entropy generation.

c)

Expert Solution

Answer to Problem 92P

The entropy generation is 1.77 kJ/K.

Explanation of Solution

Write the expression for the entropy balance equation of the system.

Sin−Sout+Sgen=ΔSsystem (V)

Here, rate of net entropy in is Sin, rate of net entropy out is Sout, rate of entropy generation is Sgen and change of entropy of the system is ΔSsystem

Write the expression to calculate the final pressure (P2).

mair=P2vRT2 (VI)

Here, final temparature is T2.

Conclusion:

Substitute 111.46 kg for mair, 90 m3 for v, 0.2870 kJ/kg⋅K for R and 70.2°C for T2 in Equation (VI).

111.46 kg=P2(90 m3)(0.2870 kJ/kg⋅K)70.2°CP2=(111.46 kg)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(70.2+273) K(90 m3)

P2=121.98 kPa=122 kPa

Substitute 0 for Sin, 0 for Sout and ΔSair+ΔSwater for ΔSsystem in Equation (V).

0−0+Sgen=ΔSair+ΔSwaterSgen=mair(cpln(T2(T1)air)−Rln(P2P1))+(mwatercwln(T2(T1)water)) (VII)

Substitute 111.46 kg for mair, 1.005 kJ/kg⋅K for cp, 12 °C for (T1)air, 70.2 °C for T2, 45 kg for mwater, 4.18 kJ/kg⋅K for cw, 95 °C for (T1)water , 0.2870 kJ/kg⋅K for R , 101.3 kPa for P1 and 122 kPa for P2 in Equation (VII).

Sgen=[(111.46 kg)((1.005 kJ/kg⋅K)ln(70.2 °C12 °C)−(0.2870 kJ/kg⋅K)ln(122 kPa101.3 kPa))+(45 kg(4.18 kJ/kg⋅K)ln(70.2 °C95 °C))]=[(111.46 kg)((1.005 kJ/kg⋅K)ln((70.2+273) K(12+273) K)−(0.2870 kJ/kg⋅K)ln(122 kPa101.3 kPa))+(45 kg(4.18 kJ/kg⋅K)ln((70.2+273) K(95+273) K))]=1.488 kJ/kg+(−13.12 kJ/kg)=1.77 kJ/K

Thus, the entropy generation is 1.77 kJ/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 7.13, Problem 92P

a)

To determine

The final equilibrium temperature.

a)

Expert Solution

Answer to Problem 92P

The final equilibrium temperature is 70.2 °C.

Explanation of Solution

Write the expression for the energy balance equation for closed system.

Ein−Eout=ΔEsystem (I)

Here, energy transfer into the control volume is Ein, energy transfer out of control volume is Eout and change in internal energy of system is ΔEsystem.

Write the expression to calculate the mass of the air.

mair=P1vRT1 (II)

Here, mass of the air is mair, initial pressure is P1, specific volume is v, gas constant is R and initial temperature is T1.

Conclusion:

Substitute 0 for Ein, 0 for Eout and ΔUair+ΔUwater for ΔEsystem in Equation (I).

0−0=ΔUwater+ΔUair[mcw(T2−T1)]water+[mcv(T2−T1)]air=0mwatercw(T2−(T1)water)+maircv(T2−(T1)air)=0 (III)

Here, mass of the air is mair, mass of the water is mwater, specific heat at constant pressure is cp and specific heat at constant volume is cv , change in internal energy in air is ΔUair, change in internal energy in water is ΔUwater, initial temperature of water is (T1)water, initial temperature of air is (T1)air, final temperature is T2.

From the Table A-2, “Ideal-gas specific heats of various common gases”, obtain the properties for air.

R=0.2870 kJ/kg⋅Kcp=1.005 kJ/kg⋅Kcv=0.718 kJ/kg⋅K

From the Table A-3, “Properties of common liquids, solids, and foods”, the specific heat of water cw at room temperature is 4.18 kJ/kg⋅K

Substitute 101.3 kPa for P1, 90 m3 for v, 0.2870 kJ/kg⋅K for R and 12 °C for T1 in Equation (II).

mair=(101.3 kPa)(90 m3)(0.2870 kJ/kg⋅K)12 °C=(101.3 kPa)(90 m3)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(12+273)K=111.46 kg

Substitute 45 kg for mwater, 4.18 kJ/kg⋅K for cw, 95 °C for (T1)water, 111.46 kg for mair, 0.718 kJ/kg⋅K for cv and 12 °C for (T1)air in Equation (III).

{[(45 kg)(4.18 kJ/kg⋅K)(T2−95°C)]+[(111.46 kg)(0.718 kJ/kg⋅K)(T2−12°C)]}=0{[(188.1 kJ/K)(T2−95°C)]+[(80.02 kJ/K)(T2−12°C)]}=0T2=70.2 °C

Thus, the final equilibrium temperature is 70.2 °C.

b)

To determine

The amount of heat transfer to the air.

b)

Expert Solution

Answer to Problem 92P

The amount of heat transfer to the air is 4660 kJ.

Explanation of Solution

Write the expression to calculate the heat transfer (Q) to the air.

Q=maircv(T2−(T1)air) (IV)

Conclusion:

Substitute 111.46 kg for mair, 0.718 kJ/kg⋅K for cv and 12 °C for (T1)air in Equation (IV).

Q=(111.46 kg)(0.718 kJ/kg⋅K)(70.2 °C−12 °C)=(111.46 kg)(0.718 kJ/kg⋅°K)[(70.2+273)K−(12 +273)K]=4660 kJ

Thus, the amount of heat transfer to the air is 4660 kJ.

c)

To determine

The entropy generation.

c)

Expert Solution

Answer to Problem 92P

The entropy generation is 1.77 kJ/K.

Explanation of Solution

Write the expression for the entropy balance equation of the system.

Sin−Sout+Sgen=ΔSsystem (V)

Here, rate of net entropy in is Sin, rate of net entropy out is Sout, rate of entropy generation is Sgen and change of entropy of the system is ΔSsystem

Write the expression to calculate the final pressure (P2).

mair=P2vRT2 (VI)

Here, final temparature is T2.

Conclusion:

Substitute 111.46 kg for mair, 90 m3 for v, 0.2870 kJ/kg⋅K for R and 70.2°C for T2 in Equation (VI).

111.46 kg=P2(90 m3)(0.2870 kJ/kg⋅K)70.2°CP2=(111.46 kg)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(70.2+273) K(90 m3)

P2=121.98 kPa=122 kPa

Substitute 0 for Sin, 0 for Sout and ΔSair+ΔSwater for ΔSsystem in Equation (V).

0−0+Sgen=ΔSair+ΔSwaterSgen=mair(cpln(T2(T1)air)−Rln(P2P1))+(mwatercwln(T2(T1)water)) (VII)

Substitute 111.46 kg for mair, 1.005 kJ/kg⋅K for cp, 12 °C for (T1)air, 70.2 °C for T2, 45 kg for mwater, 4.18 kJ/kg⋅K for cw, 95 °C for (T1)water , 0.2870 kJ/kg⋅K for R , 101.3 kPa for P1 and 122 kPa for P2 in Equation (VII).

Sgen=[(111.46 kg)((1.005 kJ/kg⋅K)ln(70.2 °C12 °C)−(0.2870 kJ/kg⋅K)ln(122 kPa101.3 kPa))+(45 kg(4.18 kJ/kg⋅K)ln(70.2 °C95 °C))]=[(111.46 kg)((1.005 kJ/kg⋅K)ln((70.2+273) K(12+273) K)−(0.2870 kJ/kg⋅K)ln(122 kPa101.3 kPa))+(45 kg(4.18 kJ/kg⋅K)ln((70.2+273) K(95+273) K))]=1.488 kJ/kg+(−13.12 kJ/kg)=1.77 kJ/K

Thus, the entropy generation is 1.77 kJ/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 7.13, Problem 92P

a)

To determine

The final equilibrium temperature.

a)

Expert Solution

Answer to Problem 92P

The final equilibrium temperature is 70.2 °C.

Explanation of Solution

Write the expression for the energy balance equation for closed system.

Ein−Eout=ΔEsystem (I)

Here, energy transfer into the control volume is Ein, energy transfer out of control volume is Eout and change in internal energy of system is ΔEsystem.

Write the expression to calculate the mass of the air.

mair=P1vRT1 (II)

Here, mass of the air is mair, initial pressure is P1, specific volume is v, gas constant is R and initial temperature is T1.

Conclusion:

Substitute 0 for Ein, 0 for Eout and ΔUair+ΔUwater for ΔEsystem in Equation (I).

0−0=ΔUwater+ΔUair[mcw(T2−T1)]water+[mcv(T2−T1)]air=0mwatercw(T2−(T1)water)+maircv(T2−(T1)air)=0 (III)

Here, mass of the air is mair, mass of the water is mwater, specific heat at constant pressure is cp and specific heat at constant volume is cv , change in internal energy in air is ΔUair, change in internal energy in water is ΔUwater, initial temperature of water is (T1)water, initial temperature of air is (T1)air, final temperature is T2.

From the Table A-2, “Ideal-gas specific heats of various common gases”, obtain the properties for air.

R=0.2870 kJ/kg⋅Kcp=1.005 kJ/kg⋅Kcv=0.718 kJ/kg⋅K

From the Table A-3, “Properties of common liquids, solids, and foods”, the specific heat of water cw at room temperature is 4.18 kJ/kg⋅K

Substitute 101.3 kPa for P1, 90 m3 for v, 0.2870 kJ/kg⋅K for R and 12 °C for T1 in Equation (II).

mair=(101.3 kPa)(90 m3)(0.2870 kJ/kg⋅K)12 °C=(101.3 kPa)(90 m3)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(12+273)K=111.46 kg

Substitute 45 kg for mwater, 4.18 kJ/kg⋅K for cw, 95 °C for (T1)water, 111.46 kg for mair, 0.718 kJ/kg⋅K for cv and 12 °C for (T1)air in Equation (III).

{[(45 kg)(4.18 kJ/kg⋅K)(T2−95°C)]+[(111.46 kg)(0.718 kJ/kg⋅K)(T2−12°C)]}=0{[(188.1 kJ/K)(T2−95°C)]+[(80.02 kJ/K)(T2−12°C)]}=0T2=70.2 °C

Thus, the final equilibrium temperature is 70.2 °C.

b)

To determine

The amount of heat transfer to the air.

b)

Expert Solution

Answer to Problem 92P

The amount of heat transfer to the air is 4660 kJ.

Explanation of Solution

Write the expression to calculate the heat transfer (Q) to the air.

Q=maircv(T2−(T1)air) (IV)

Conclusion:

Substitute 111.46 kg for mair, 0.718 kJ/kg⋅K for cv and 12 °C for (T1)air in Equation (IV).

Q=(111.46 kg)(0.718 kJ/kg⋅K)(70.2 °C−12 °C)=(111.46 kg)(0.718 kJ/kg⋅°K)[(70.2+273)K−(12 +273)K]=4660 kJ

Thus, the amount of heat transfer to the air is 4660 kJ.

c)

To determine

The entropy generation.

c)

Expert Solution

Answer to Problem 92P

The entropy generation is 1.77 kJ/K.

Explanation of Solution

Write the expression for the entropy balance equation of the system.

Sin−Sout+Sgen=ΔSsystem (V)

Here, rate of net entropy in is Sin, rate of net entropy out is Sout, rate of entropy generation is Sgen and change of entropy of the system is ΔSsystem

Write the expression to calculate the final pressure (P2).

mair=P2vRT2 (VI)

Here, final temparature is T2.

Conclusion:

Substitute 111.46 kg for mair, 90 m3 for v, 0.2870 kJ/kg⋅K for R and 70.2°C for T2 in Equation (VI).

111.46 kg=P2(90 m3)(0.2870 kJ/kg⋅K)70.2°CP2=(111.46 kg)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(70.2+273) K(90 m3)

P2=121.98 kPa=122 kPa

Substitute 0 for Sin, 0 for Sout and ΔSair+ΔSwater for ΔSsystem in Equation (V).

0−0+Sgen=ΔSair+ΔSwaterSgen=mair(cpln(T2(T1)air)−Rln(P2P1))+(mwatercwln(T2(T1)water)) (VII)

Substitute 111.46 kg for mair, 1.005 kJ/kg⋅K for cp, 12 °C for (T1)air, 70.2 °C for T2, 45 kg for mwater, 4.18 kJ/kg⋅K for cw, 95 °C for (T1)water , 0.2870 kJ/kg⋅K for R , 101.3 kPa for P1 and 122 kPa for P2 in Equation (VII).

Sgen=[(111.46 kg)((1.005 kJ/kg⋅K)ln(70.2 °C12 °C)−(0.2870 kJ/kg⋅K)ln(122 kPa101.3 kPa))+(45 kg(4.18 kJ/kg⋅K)ln(70.2 °C95 °C))]=[(111.46 kg)((1.005 kJ/kg⋅K)ln((70.2+273) K(12+273) K)−(0.2870 kJ/kg⋅K)ln(122 kPa101.3 kPa))+(45 kg(4.18 kJ/kg⋅K)ln((70.2+273) K(95+273) K))]=1.488 kJ/kg+(−13.12 kJ/kg)=1.77 kJ/K

Thus, the entropy generation is 1.77 kJ/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 7.13, Problem 92P

a)

To determine

The final equilibrium temperature.

a)

Expert Solution

Answer to Problem 92P

The final equilibrium temperature is 70.2 °C.

Explanation of Solution

Write the expression for the energy balance equation for closed system.

Ein−Eout=ΔEsystem (I)

Here, energy transfer into the control volume is Ein, energy transfer out of control volume is Eout and change in internal energy of system is ΔEsystem.

Write the expression to calculate the mass of the air.

mair=P1vRT1 (II)

Here, mass of the air is mair, initial pressure is P1, specific volume is v, gas constant is R and initial temperature is T1.

Conclusion:

Substitute 0 for Ein, 0 for Eout and ΔUair+ΔUwater for ΔEsystem in Equation (I).

0−0=ΔUwater+ΔUair[mcw(T2−T1)]water+[mcv(T2−T1)]air=0mwatercw(T2−(T1)water)+maircv(T2−(T1)air)=0 (III)

Here, mass of the air is mair, mass of the water is mwater, specific heat at constant pressure is cp and specific heat at constant volume is cv , change in internal energy in air is ΔUair, change in internal energy in water is ΔUwater, initial temperature of water is (T1)water, initial temperature of air is (T1)air, final temperature is T2.

From the Table A-2, “Ideal-gas specific heats of various common gases”, obtain the properties for air.

R=0.2870 kJ/kg⋅Kcp=1.005 kJ/kg⋅Kcv=0.718 kJ/kg⋅K

From the Table A-3, “Properties of common liquids, solids, and foods”, the specific heat of water cw at room temperature is 4.18 kJ/kg⋅K

Substitute 101.3 kPa for P1, 90 m3 for v, 0.2870 kJ/kg⋅K for R and 12 °C for T1 in Equation (II).

mair=(101.3 kPa)(90 m3)(0.2870 kJ/kg⋅K)12 °C=(101.3 kPa)(90 m3)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(12+273)K=111.46 kg

Substitute 45 kg for mwater, 4.18 kJ/kg⋅K for cw, 95 °C for (T1)water, 111.46 kg for mair, 0.718 kJ/kg⋅K for cv and 12 °C for (T1)air in Equation (III).

{[(45 kg)(4.18 kJ/kg⋅K)(T2−95°C)]+[(111.46 kg)(0.718 kJ/kg⋅K)(T2−12°C)]}=0{[(188.1 kJ/K)(T2−95°C)]+[(80.02 kJ/K)(T2−12°C)]}=0T2=70.2 °C

Thus, the final equilibrium temperature is 70.2 °C.

b)

To determine

The amount of heat transfer to the air.

b)

Expert Solution

Answer to Problem 92P

The amount of heat transfer to the air is 4660 kJ.

Explanation of Solution

Write the expression to calculate the heat transfer (Q) to the air.

Q=maircv(T2−(T1)air) (IV)

Conclusion:

Substitute 111.46 kg for mair, 0.718 kJ/kg⋅K for cv and 12 °C for (T1)air in Equation (IV).

Q=(111.46 kg)(0.718 kJ/kg⋅K)(70.2 °C−12 °C)=(111.46 kg)(0.718 kJ/kg⋅°K)[(70.2+273)K−(12 +273)K]=4660 kJ

Thus, the amount of heat transfer to the air is 4660 kJ.

c)

To determine

The entropy generation.

c)

Expert Solution

Answer to Problem 92P

The entropy generation is 1.77 kJ/K.

Explanation of Solution

Write the expression for the entropy balance equation of the system.

Sin−Sout+Sgen=ΔSsystem (V)

Here, rate of net entropy in is Sin, rate of net entropy out is Sout, rate of entropy generation is Sgen and change of entropy of the system is ΔSsystem

Write the expression to calculate the final pressure (P2).

mair=P2vRT2 (VI)

Here, final temparature is T2.

Conclusion:

Substitute 111.46 kg for mair, 90 m3 for v, 0.2870 kJ/kg⋅K for R and 70.2°C for T2 in Equation (VI).

111.46 kg=P2(90 m3)(0.2870 kJ/kg⋅K)70.2°CP2=(111.46 kg)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(70.2+273) K(90 m3)

P2=121.98 kPa=122 kPa

Substitute 0 for Sin, 0 for Sout and ΔSair+ΔSwater for ΔSsystem in Equation (V).

0−0+Sgen=ΔSair+ΔSwaterSgen=mair(cpln(T2(T1)air)−Rln(P2P1))+(mwatercwln(T2(T1)water)) (VII)

Substitute 111.46 kg for mair, 1.005 kJ/kg⋅K for cp, 12 °C for (T1)air, 70.2 °C for T2, 45 kg for mwater, 4.18 kJ/kg⋅K for cw, 95 °C for (T1)water , 0.2870 kJ/kg⋅K for R , 101.3 kPa for P1 and 122 kPa for P2 in Equation (VII).

Sgen=[(111.46 kg)((1.005 kJ/kg⋅K)ln(70.2 °C12 °C)−(0.2870 kJ/kg⋅K)ln(122 kPa101.3 kPa))+(45 kg(4.18 kJ/kg⋅K)ln(70.2 °C95 °C))]=[(111.46 kg)((1.005 kJ/kg⋅K)ln((70.2+273) K(12+273) K)−(0.2870 kJ/kg⋅K)ln(122 kPa101.3 kPa))+(45 kg(4.18 kJ/kg⋅K)ln((70.2+273) K(95+273) K))]=1.488 kJ/kg+(−13.12 kJ/kg)=1.77 kJ/K

Thus, the entropy generation is 1.77 kJ/K.

Answer 6

Chapter 7.13, Problem 92P

a)

To determine

The final equilibrium temperature.

a)

Expert Solution

Answer to Problem 92P

The final equilibrium temperature is 70.2 °C.

Explanation of Solution

Write the expression for the energy balance equation for closed system.

Ein−Eout=ΔEsystem (I)

Here, energy transfer into the control volume is Ein, energy transfer out of control volume is Eout and change in internal energy of system is ΔEsystem.

Write the expression to calculate the mass of the air.

mair=P1vRT1 (II)

Here, mass of the air is mair, initial pressure is P1, specific volume is v, gas constant is R and initial temperature is T1.

Conclusion:

Substitute 0 for Ein, 0 for Eout and ΔUair+ΔUwater for ΔEsystem in Equation (I).

0−0=ΔUwater+ΔUair[mcw(T2−T1)]water+[mcv(T2−T1)]air=0mwatercw(T2−(T1)water)+maircv(T2−(T1)air)=0 (III)

Here, mass of the air is mair, mass of the water is mwater, specific heat at constant pressure is cp and specific heat at constant volume is cv , change in internal energy in air is ΔUair, change in internal energy in water is ΔUwater, initial temperature of water is (T1)water, initial temperature of air is (T1)air, final temperature is T2.

From the Table A-2, “Ideal-gas specific heats of various common gases”, obtain the properties for air.

R=0.2870 kJ/kg⋅Kcp=1.005 kJ/kg⋅Kcv=0.718 kJ/kg⋅K

From the Table A-3, “Properties of common liquids, solids, and foods”, the specific heat of water cw at room temperature is 4.18 kJ/kg⋅K

Substitute 101.3 kPa for P1, 90 m3 for v, 0.2870 kJ/kg⋅K for R and 12 °C for T1 in Equation (II).

mair=(101.3 kPa)(90 m3)(0.2870 kJ/kg⋅K)12 °C=(101.3 kPa)(90 m3)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(12+273)K=111.46 kg

Substitute 45 kg for mwater, 4.18 kJ/kg⋅K for cw, 95 °C for (T1)water, 111.46 kg for mair, 0.718 kJ/kg⋅K for cv and 12 °C for (T1)air in Equation (III).

{[(45 kg)(4.18 kJ/kg⋅K)(T2−95°C)]+[(111.46 kg)(0.718 kJ/kg⋅K)(T2−12°C)]}=0{[(188.1 kJ/K)(T2−95°C)]+[(80.02 kJ/K)(T2−12°C)]}=0T2=70.2 °C

Thus, the final equilibrium temperature is 70.2 °C.

Answer 7

Chapter 7.13, Problem 92P

a)

To determine

The final equilibrium temperature.

Answer 8

a)

Expert Solution

Answer to Problem 92P

The final equilibrium temperature is 70.2 °C.

Answer 9

a)

Expert Solution

Answer 10

a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write the expression for the energy balance equation for closed system.

Ein−Eout=ΔEsystem (I)

Here, energy transfer into the control volume is Ein, energy transfer out of control volume is Eout and change in internal energy of system is ΔEsystem.

Write the expression to calculate the mass of the air.

mair=P1vRT1 (II)

Here, mass of the air is mair, initial pressure is P1, specific volume is v, gas constant is R and initial temperature is T1.

Conclusion:

Substitute 0 for Ein, 0 for Eout and ΔUair+ΔUwater for ΔEsystem in Equation (I).

0−0=ΔUwater+ΔUair[mcw(T2−T1)]water+[mcv(T2−T1)]air=0mwatercw(T2−(T1)water)+maircv(T2−(T1)air)=0 (III)

Here, mass of the air is mair, mass of the water is mwater, specific heat at constant pressure is cp and specific heat at constant volume is cv , change in internal energy in air is ΔUair, change in internal energy in water is ΔUwater, initial temperature of water is (T1)water, initial temperature of air is (T1)air, final temperature is T2.

From the Table A-2, “Ideal-gas specific heats of various common gases”, obtain the properties for air.

R=0.2870 kJ/kg⋅Kcp=1.005 kJ/kg⋅Kcv=0.718 kJ/kg⋅K

From the Table A-3, “Properties of common liquids, solids, and foods”, the specific heat of water cw at room temperature is 4.18 kJ/kg⋅K

Substitute 101.3 kPa for P1, 90 m3 for v, 0.2870 kJ/kg⋅K for R and 12 °C for T1 in Equation (II).

mair=(101.3 kPa)(90 m3)(0.2870 kJ/kg⋅K)12 °C=(101.3 kPa)(90 m3)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(12+273)K=111.46 kg

Substitute 45 kg for mwater, 4.18 kJ/kg⋅K for cw, 95 °C for (T1)water, 111.46 kg for mair, 0.718 kJ/kg⋅K for cv and 12 °C for (T1)air in Equation (III).

{[(45 kg)(4.18 kJ/kg⋅K)(T2−95°C)]+[(111.46 kg)(0.718 kJ/kg⋅K)(T2−12°C)]}=0{[(188.1 kJ/K)(T2−95°C)]+[(80.02 kJ/K)(T2−12°C)]}=0T2=70.2 °C

Thus, the final equilibrium temperature is 70.2 °C.

Answer 13

b)

To determine

The amount of heat transfer to the air.

b)

Expert Solution

Answer to Problem 92P

The amount of heat transfer to the air is 4660 kJ.

Explanation of Solution

Write the expression to calculate the heat transfer (Q) to the air.

Q=maircv(T2−(T1)air) (IV)

Conclusion:

Substitute 111.46 kg for mair, 0.718 kJ/kg⋅K for cv and 12 °C for (T1)air in Equation (IV).

Q=(111.46 kg)(0.718 kJ/kg⋅K)(70.2 °C−12 °C)=(111.46 kg)(0.718 kJ/kg⋅°K)[(70.2+273)K−(12 +273)K]=4660 kJ

Thus, the amount of heat transfer to the air is 4660 kJ.

Answer 14

b)

To determine

The amount of heat transfer to the air.

Answer 15

b)

Expert Solution

Answer to Problem 92P

The amount of heat transfer to the air is 4660 kJ.

Answer 16

b)

Expert Solution

Answer 17

b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Write the expression to calculate the heat transfer (Q) to the air.

Q=maircv(T2−(T1)air) (IV)

Conclusion:

Substitute 111.46 kg for mair, 0.718 kJ/kg⋅K for cv and 12 °C for (T1)air in Equation (IV).

Q=(111.46 kg)(0.718 kJ/kg⋅K)(70.2 °C−12 °C)=(111.46 kg)(0.718 kJ/kg⋅°K)[(70.2+273)K−(12 +273)K]=4660 kJ

Thus, the amount of heat transfer to the air is 4660 kJ.

Answer 20

c)

To determine

The entropy generation.

c)

Expert Solution

Answer to Problem 92P

The entropy generation is 1.77 kJ/K.

Explanation of Solution

Write the expression for the entropy balance equation of the system.

Sin−Sout+Sgen=ΔSsystem (V)

Here, rate of net entropy in is Sin, rate of net entropy out is Sout, rate of entropy generation is Sgen and change of entropy of the system is ΔSsystem

Write the expression to calculate the final pressure (P2).

mair=P2vRT2 (VI)

Here, final temparature is T2.

Conclusion:

Substitute 111.46 kg for mair, 90 m3 for v, 0.2870 kJ/kg⋅K for R and 70.2°C for T2 in Equation (VI).

111.46 kg=P2(90 m3)(0.2870 kJ/kg⋅K)70.2°CP2=(111.46 kg)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(70.2+273) K(90 m3)

P2=121.98 kPa=122 kPa

Substitute 0 for Sin, 0 for Sout and ΔSair+ΔSwater for ΔSsystem in Equation (V).

0−0+Sgen=ΔSair+ΔSwaterSgen=mair(cpln(T2(T1)air)−Rln(P2P1))+(mwatercwln(T2(T1)water)) (VII)

Substitute 111.46 kg for mair, 1.005 kJ/kg⋅K for cp, 12 °C for (T1)air, 70.2 °C for T2, 45 kg for mwater, 4.18 kJ/kg⋅K for cw, 95 °C for (T1)water , 0.2870 kJ/kg⋅K for R , 101.3 kPa for P1 and 122 kPa for P2 in Equation (VII).

Sgen=[(111.46 kg)((1.005 kJ/kg⋅K)ln(70.2 °C12 °C)−(0.2870 kJ/kg⋅K)ln(122 kPa101.3 kPa))+(45 kg(4.18 kJ/kg⋅K)ln(70.2 °C95 °C))]=[(111.46 kg)((1.005 kJ/kg⋅K)ln((70.2+273) K(12+273) K)−(0.2870 kJ/kg⋅K)ln(122 kPa101.3 kPa))+(45 kg(4.18 kJ/kg⋅K)ln((70.2+273) K(95+273) K))]=1.488 kJ/kg+(−13.12 kJ/kg)=1.77 kJ/K

Thus, the entropy generation is 1.77 kJ/K.

Answer 21

c)

To determine

The entropy generation.

Answer 22

c)

Expert Solution

Answer to Problem 92P

The entropy generation is 1.77 kJ/K.

Answer 23

c)

Expert Solution

Answer 24

c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Write the expression for the entropy balance equation of the system.

Sin−Sout+Sgen=ΔSsystem (V)

Here, rate of net entropy in is Sin, rate of net entropy out is Sout, rate of entropy generation is Sgen and change of entropy of the system is ΔSsystem

Write the expression to calculate the final pressure (P2).

mair=P2vRT2 (VI)

Here, final temparature is T2.

Conclusion:

Substitute 111.46 kg for mair, 90 m3 for v, 0.2870 kJ/kg⋅K for R and 70.2°C for T2 in Equation (VI).

111.46 kg=P2(90 m3)(0.2870 kJ/kg⋅K)70.2°CP2=(111.46 kg)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(70.2+273) K(90 m3)

P2=121.98 kPa=122 kPa

Substitute 0 for Sin, 0 for Sout and ΔSair+ΔSwater for ΔSsystem in Equation (V).

0−0+Sgen=ΔSair+ΔSwaterSgen=mair(cpln(T2(T1)air)−Rln(P2P1))+(mwatercwln(T2(T1)water)) (VII)

Substitute 111.46 kg for mair, 1.005 kJ/kg⋅K for cp, 12 °C for (T1)air, 70.2 °C for T2, 45 kg for mwater, 4.18 kJ/kg⋅K for cw, 95 °C for (T1)water , 0.2870 kJ/kg⋅K for R , 101.3 kPa for P1 and 122 kPa for P2 in Equation (VII).

Sgen=[(111.46 kg)((1.005 kJ/kg⋅K)ln(70.2 °C12 °C)−(0.2870 kJ/kg⋅K)ln(122 kPa101.3 kPa))+(45 kg(4.18 kJ/kg⋅K)ln(70.2 °C95 °C))]=[(111.46 kg)((1.005 kJ/kg⋅K)ln((70.2+273) K(12+273) K)−(0.2870 kJ/kg⋅K)ln(122 kPa101.3 kPa))+(45 kg(4.18 kJ/kg⋅K)ln((70.2+273) K(95+273) K))]=1.488 kJ/kg+(−13.12 kJ/kg)=1.77 kJ/K