The average temperature of the room after 30 min.

Question

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Answer 1

Question

Chapter 8, Problem 189RQ

a)

To determine

The average temperature of the room after 30 min.

a)

Expert Solution

Explanation of Solution

Given:

The dimension of room (vroom) is 4 m×4 m×5 m.

The temperature of the room (Troom) is 10°C.

The volume of the radiator is 15 L.

The pressure of the super-heated vapor (P1) is 200 kPa.

The temperature of the superheated vapor (T1) is 200°C.

The pressure of the steam (P2) is 100 kPa.

The time duration (Δt) is 30 min.

The pressure of the room (Proom) is 100 kPa.

Calculation:

Refer the Table A-6, “Superheated water table”, obtain the following properties of steam at temperature of 200 °C and pressure of 200 kPa.

v1=1.0805 m3/kgu1=2654.4 kJ/kgs1=7.5081 kJ/kg⋅K

Refer the Table A-6, “Saturated water-Pressure table”, obtain the following properties of steam at pressure of 100 kPa.

vf=0.001043 m3/kgvg=1.6941 m3/kg

Here, Saturated liquid specific volume is vf and saturated vapor specific volume is vγ.

Calculate the final vapor quality.

x2=v2−vfvg−vfx2=1.0805 m3/kg−0.001043 m3/kg1.6941 m3/kg−0.001043 m3/kg=0.6376

Refer the Table A-6, “Saturated water-Pressure table”, obtain the following properties of steam at pressure of 100 kPa.

uf=417.40 kJ/kgufg=2088.2 kJ/kg

Calculate the final internal energy of the system.

u2=uf+x2ufg

u2=417.40 kJ/kg+(0.6376)(2088.2 kJ/kg)=1748.7 kJ/kg

Refer the Table A-6, “Saturated water-Pressure table”, obtain the following properties of steam at pressure of 100 kPa.

sf=1.3028 kJ/kg⋅Ksfg=6.0562 kJ/kg⋅K

Calculate the final entropy of the system.

s2=sf+x2sfgs2=1.3028 kJ/kg⋅K+(0.6376)(6.0562 kJ/kg⋅K)=5.1642 kJ/kg⋅K

Calculate the mass of the steam.

m=ν1v1m=0.015 m31.0805 m3/kg=0.01388 kg

Write the expression for the energy balance equation for closed system without air in the room.

Ein−Eout=ΔEsystem (I)

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

Substitute Ein=0, Eout=Qout, and ΔEsystem=ΔU in Equation (I).

0−Qout=ΔU−Qout=m(u2−u1)Qout=m(u1−u2)

Qout=(0.01388 kg)(2654.4 kJ/kg−1748.7 kJ/kg)=12.6 kJ

Refer the Table A-1, “the molar mass, gas constant and critical point properties table”, select the gas constant of air as 0.2870 kJ/kg⋅K.

Calculate the mass of the air using the ideal gas equation.

mair=P1ν1RT1mair=(100 kPa)(4m×4m×5m)(0.2870 kJ/kg⋅K)10°C=(100 kPa)(80 m3)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(10+273)K=98.5 kg

Write the expression to calculate the total work done by the fan.

Wfan,in=W˙fan,in(Δt)Wfan,in=(0.120 kJ/s)(30 min)=(0.120 kJ/s)(30 min)(60 sec1 min)=216 kJ

Refer the Table A-2, “the ideal–gas equation specific heats of various common gases table”, select the specific heat at constant pressure for air as 1.005 kJ/kg⋅K.

Substitute Ein=Qin+Wfan,in, Eout=Wb,out and ΔEsystem=ΔU in Equation (I).

Qin+Wfan,in−Wb,out=ΔUQin+Wfan,in=ΔU+Wb,outQin+Wfan,in=ΔHQin+Wfan,in=mcp(T2−T1)12.6 kJ+216 kJ=(98.5 kg)(1.005 kJ/kg⋅K)(T2−10 °C)T2=12.3°C

Thus, the average temperature of the room after 30 min is 12.3°C.

b)

To determine

The entropy change of the steam.

b)

Expert Solution

Explanation of Solution

Write the expression to calculate the change of entropy of the steam.

ΔSsteam=m(s2−s1)ΔSsteam=(0.01388 kg)(5.1642 kJ/kg⋅K−7.5081 kJ/kg⋅K)=−0.0325 kJ/K

Thus, the entropy change of the steam is −0.0325 kJ/K.

c)

To determine

The entropy change of the air.

c)

Expert Solution

Explanation of Solution

Write the expression to calculate the entropy change of air.

ΔSair=mcplnT2T1−mRlnP2P1

Since, P2=P1.

ΔSair=mcplnT2T1ΔSair=(98.5 kg)(1.005 kJ/kg⋅K)ln12.3 °C10 °C=(98.5 kg)(1.005 kJ/kg⋅K)ln(12.3+273)K(10+273)K=0.8013 kJ/K

Thus, the entropy change of the air is 0.8013 kJ/K.

d)

To determine

The entropy generation in the turbine.

d)

Expert Solution

Explanation of Solution

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

Sin−Sout+Sgen=ΔSsystem (II)

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

Substitute Sin=0, Sout=0, and ΔSsystem=ΔSair+ΔSsteam in Equation (II).

0−0+Sgen=ΔSair+ΔSsteamSgen=ΔSair+ΔSsteam

Sgen=0.8013 kJ/K+−0.0325 kJ/K=0.7688 kJ/K

Thus, the total entropy generated during the process is 0.7688 kJ/K.

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Answer 2

Question

Chapter 8, Problem 189RQ

a)

To determine

The average temperature of the room after 30 min.

a)

Expert Solution

Explanation of Solution

Given:

The dimension of room (vroom) is 4 m×4 m×5 m.

The temperature of the room (Troom) is 10°C.

The volume of the radiator is 15 L.

The pressure of the super-heated vapor (P1) is 200 kPa.

The temperature of the superheated vapor (T1) is 200°C.

The pressure of the steam (P2) is 100 kPa.

The time duration (Δt) is 30 min.

The pressure of the room (Proom) is 100 kPa.

Calculation:

Refer the Table A-6, “Superheated water table”, obtain the following properties of steam at temperature of 200 °C and pressure of 200 kPa.

v1=1.0805 m3/kgu1=2654.4 kJ/kgs1=7.5081 kJ/kg⋅K

Refer the Table A-6, “Saturated water-Pressure table”, obtain the following properties of steam at pressure of 100 kPa.

vf=0.001043 m3/kgvg=1.6941 m3/kg

Here, Saturated liquid specific volume is vf and saturated vapor specific volume is vγ.

Calculate the final vapor quality.

x2=v2−vfvg−vfx2=1.0805 m3/kg−0.001043 m3/kg1.6941 m3/kg−0.001043 m3/kg=0.6376

Refer the Table A-6, “Saturated water-Pressure table”, obtain the following properties of steam at pressure of 100 kPa.

uf=417.40 kJ/kgufg=2088.2 kJ/kg

Calculate the final internal energy of the system.

u2=uf+x2ufg

u2=417.40 kJ/kg+(0.6376)(2088.2 kJ/kg)=1748.7 kJ/kg

Refer the Table A-6, “Saturated water-Pressure table”, obtain the following properties of steam at pressure of 100 kPa.

sf=1.3028 kJ/kg⋅Ksfg=6.0562 kJ/kg⋅K

Calculate the final entropy of the system.

s2=sf+x2sfgs2=1.3028 kJ/kg⋅K+(0.6376)(6.0562 kJ/kg⋅K)=5.1642 kJ/kg⋅K

Calculate the mass of the steam.

m=ν1v1m=0.015 m31.0805 m3/kg=0.01388 kg

Write the expression for the energy balance equation for closed system without air in the room.

Ein−Eout=ΔEsystem (I)

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

Substitute Ein=0, Eout=Qout, and ΔEsystem=ΔU in Equation (I).

0−Qout=ΔU−Qout=m(u2−u1)Qout=m(u1−u2)

Qout=(0.01388 kg)(2654.4 kJ/kg−1748.7 kJ/kg)=12.6 kJ

Refer the Table A-1, “the molar mass, gas constant and critical point properties table”, select the gas constant of air as 0.2870 kJ/kg⋅K.

Calculate the mass of the air using the ideal gas equation.

mair=P1ν1RT1mair=(100 kPa)(4m×4m×5m)(0.2870 kJ/kg⋅K)10°C=(100 kPa)(80 m3)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(10+273)K=98.5 kg

Write the expression to calculate the total work done by the fan.

Wfan,in=W˙fan,in(Δt)Wfan,in=(0.120 kJ/s)(30 min)=(0.120 kJ/s)(30 min)(60 sec1 min)=216 kJ

Refer the Table A-2, “the ideal–gas equation specific heats of various common gases table”, select the specific heat at constant pressure for air as 1.005 kJ/kg⋅K.

Substitute Ein=Qin+Wfan,in, Eout=Wb,out and ΔEsystem=ΔU in Equation (I).

Qin+Wfan,in−Wb,out=ΔUQin+Wfan,in=ΔU+Wb,outQin+Wfan,in=ΔHQin+Wfan,in=mcp(T2−T1)12.6 kJ+216 kJ=(98.5 kg)(1.005 kJ/kg⋅K)(T2−10 °C)T2=12.3°C

Thus, the average temperature of the room after 30 min is 12.3°C.

b)

To determine

The entropy change of the steam.

b)

Expert Solution

Explanation of Solution

Write the expression to calculate the change of entropy of the steam.

ΔSsteam=m(s2−s1)ΔSsteam=(0.01388 kg)(5.1642 kJ/kg⋅K−7.5081 kJ/kg⋅K)=−0.0325 kJ/K

Thus, the entropy change of the steam is −0.0325 kJ/K.

c)

To determine

The entropy change of the air.

c)

Expert Solution

Explanation of Solution

Write the expression to calculate the entropy change of air.

ΔSair=mcplnT2T1−mRlnP2P1

Since, P2=P1.

ΔSair=mcplnT2T1ΔSair=(98.5 kg)(1.005 kJ/kg⋅K)ln12.3 °C10 °C=(98.5 kg)(1.005 kJ/kg⋅K)ln(12.3+273)K(10+273)K=0.8013 kJ/K

Thus, the entropy change of the air is 0.8013 kJ/K.

d)

To determine

The entropy generation in the turbine.

d)

Expert Solution

Explanation of Solution

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

Sin−Sout+Sgen=ΔSsystem (II)

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

Substitute Sin=0, Sout=0, and ΔSsystem=ΔSair+ΔSsteam in Equation (II).

0−0+Sgen=ΔSair+ΔSsteamSgen=ΔSair+ΔSsteam

Sgen=0.8013 kJ/K+−0.0325 kJ/K=0.7688 kJ/K

Thus, the total entropy generated during the process is 0.7688 kJ/K.

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Answer 3

Question

Chapter 8, Problem 189RQ

a)

To determine

The average temperature of the room after 30 min.

a)

Expert Solution

Explanation of Solution

Given:

The dimension of room (vroom) is 4 m×4 m×5 m.

The temperature of the room (Troom) is 10°C.

The volume of the radiator is 15 L.

The pressure of the super-heated vapor (P1) is 200 kPa.

The temperature of the superheated vapor (T1) is 200°C.

The pressure of the steam (P2) is 100 kPa.

The time duration (Δt) is 30 min.

The pressure of the room (Proom) is 100 kPa.

Calculation:

Refer the Table A-6, “Superheated water table”, obtain the following properties of steam at temperature of 200 °C and pressure of 200 kPa.

v1=1.0805 m3/kgu1=2654.4 kJ/kgs1=7.5081 kJ/kg⋅K

Refer the Table A-6, “Saturated water-Pressure table”, obtain the following properties of steam at pressure of 100 kPa.

vf=0.001043 m3/kgvg=1.6941 m3/kg

Here, Saturated liquid specific volume is vf and saturated vapor specific volume is vγ.

Calculate the final vapor quality.

x2=v2−vfvg−vfx2=1.0805 m3/kg−0.001043 m3/kg1.6941 m3/kg−0.001043 m3/kg=0.6376

Refer the Table A-6, “Saturated water-Pressure table”, obtain the following properties of steam at pressure of 100 kPa.

uf=417.40 kJ/kgufg=2088.2 kJ/kg

Calculate the final internal energy of the system.

u2=uf+x2ufg

u2=417.40 kJ/kg+(0.6376)(2088.2 kJ/kg)=1748.7 kJ/kg

Refer the Table A-6, “Saturated water-Pressure table”, obtain the following properties of steam at pressure of 100 kPa.

sf=1.3028 kJ/kg⋅Ksfg=6.0562 kJ/kg⋅K

Calculate the final entropy of the system.

s2=sf+x2sfgs2=1.3028 kJ/kg⋅K+(0.6376)(6.0562 kJ/kg⋅K)=5.1642 kJ/kg⋅K

Calculate the mass of the steam.

m=ν1v1m=0.015 m31.0805 m3/kg=0.01388 kg

Write the expression for the energy balance equation for closed system without air in the room.

Ein−Eout=ΔEsystem (I)

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

Substitute Ein=0, Eout=Qout, and ΔEsystem=ΔU in Equation (I).

0−Qout=ΔU−Qout=m(u2−u1)Qout=m(u1−u2)

Qout=(0.01388 kg)(2654.4 kJ/kg−1748.7 kJ/kg)=12.6 kJ

Refer the Table A-1, “the molar mass, gas constant and critical point properties table”, select the gas constant of air as 0.2870 kJ/kg⋅K.

Calculate the mass of the air using the ideal gas equation.

mair=P1ν1RT1mair=(100 kPa)(4m×4m×5m)(0.2870 kJ/kg⋅K)10°C=(100 kPa)(80 m3)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(10+273)K=98.5 kg

Write the expression to calculate the total work done by the fan.

Wfan,in=W˙fan,in(Δt)Wfan,in=(0.120 kJ/s)(30 min)=(0.120 kJ/s)(30 min)(60 sec1 min)=216 kJ

Refer the Table A-2, “the ideal–gas equation specific heats of various common gases table”, select the specific heat at constant pressure for air as 1.005 kJ/kg⋅K.

Substitute Ein=Qin+Wfan,in, Eout=Wb,out and ΔEsystem=ΔU in Equation (I).

Qin+Wfan,in−Wb,out=ΔUQin+Wfan,in=ΔU+Wb,outQin+Wfan,in=ΔHQin+Wfan,in=mcp(T2−T1)12.6 kJ+216 kJ=(98.5 kg)(1.005 kJ/kg⋅K)(T2−10 °C)T2=12.3°C

Thus, the average temperature of the room after 30 min is 12.3°C.

b)

To determine

The entropy change of the steam.

b)

Expert Solution

Explanation of Solution

Write the expression to calculate the change of entropy of the steam.

ΔSsteam=m(s2−s1)ΔSsteam=(0.01388 kg)(5.1642 kJ/kg⋅K−7.5081 kJ/kg⋅K)=−0.0325 kJ/K

Thus, the entropy change of the steam is −0.0325 kJ/K.

c)

To determine

The entropy change of the air.

c)

Expert Solution

Explanation of Solution

Write the expression to calculate the entropy change of air.

ΔSair=mcplnT2T1−mRlnP2P1

Since, P2=P1.

ΔSair=mcplnT2T1ΔSair=(98.5 kg)(1.005 kJ/kg⋅K)ln12.3 °C10 °C=(98.5 kg)(1.005 kJ/kg⋅K)ln(12.3+273)K(10+273)K=0.8013 kJ/K

Thus, the entropy change of the air is 0.8013 kJ/K.

d)

To determine

The entropy generation in the turbine.

d)

Expert Solution

Explanation of Solution

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

Sin−Sout+Sgen=ΔSsystem (II)

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

Substitute Sin=0, Sout=0, and ΔSsystem=ΔSair+ΔSsteam in Equation (II).

0−0+Sgen=ΔSair+ΔSsteamSgen=ΔSair+ΔSsteam

Sgen=0.8013 kJ/K+−0.0325 kJ/K=0.7688 kJ/K

Thus, the total entropy generated during the process is 0.7688 kJ/K.

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Answer 4

Question

Chapter 8, Problem 189RQ

a)

To determine

The average temperature of the room after 30 min.

a)

Expert Solution

Explanation of Solution

Given:

The dimension of room (vroom) is 4 m×4 m×5 m.

The temperature of the room (Troom) is 10°C.

The volume of the radiator is 15 L.

The pressure of the super-heated vapor (P1) is 200 kPa.

The temperature of the superheated vapor (T1) is 200°C.

The pressure of the steam (P2) is 100 kPa.

The time duration (Δt) is 30 min.

The pressure of the room (Proom) is 100 kPa.

Calculation:

Refer the Table A-6, “Superheated water table”, obtain the following properties of steam at temperature of 200 °C and pressure of 200 kPa.

v1=1.0805 m3/kgu1=2654.4 kJ/kgs1=7.5081 kJ/kg⋅K

Refer the Table A-6, “Saturated water-Pressure table”, obtain the following properties of steam at pressure of 100 kPa.

vf=0.001043 m3/kgvg=1.6941 m3/kg

Here, Saturated liquid specific volume is vf and saturated vapor specific volume is vγ.

Calculate the final vapor quality.

x2=v2−vfvg−vfx2=1.0805 m3/kg−0.001043 m3/kg1.6941 m3/kg−0.001043 m3/kg=0.6376

Refer the Table A-6, “Saturated water-Pressure table”, obtain the following properties of steam at pressure of 100 kPa.

uf=417.40 kJ/kgufg=2088.2 kJ/kg

Calculate the final internal energy of the system.

u2=uf+x2ufg

u2=417.40 kJ/kg+(0.6376)(2088.2 kJ/kg)=1748.7 kJ/kg

Refer the Table A-6, “Saturated water-Pressure table”, obtain the following properties of steam at pressure of 100 kPa.

sf=1.3028 kJ/kg⋅Ksfg=6.0562 kJ/kg⋅K

Calculate the final entropy of the system.

s2=sf+x2sfgs2=1.3028 kJ/kg⋅K+(0.6376)(6.0562 kJ/kg⋅K)=5.1642 kJ/kg⋅K

Calculate the mass of the steam.

m=ν1v1m=0.015 m31.0805 m3/kg=0.01388 kg

Write the expression for the energy balance equation for closed system without air in the room.

Ein−Eout=ΔEsystem (I)

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

Substitute Ein=0, Eout=Qout, and ΔEsystem=ΔU in Equation (I).

0−Qout=ΔU−Qout=m(u2−u1)Qout=m(u1−u2)

Qout=(0.01388 kg)(2654.4 kJ/kg−1748.7 kJ/kg)=12.6 kJ

Refer the Table A-1, “the molar mass, gas constant and critical point properties table”, select the gas constant of air as 0.2870 kJ/kg⋅K.

Calculate the mass of the air using the ideal gas equation.

mair=P1ν1RT1mair=(100 kPa)(4m×4m×5m)(0.2870 kJ/kg⋅K)10°C=(100 kPa)(80 m3)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(10+273)K=98.5 kg

Write the expression to calculate the total work done by the fan.

Wfan,in=W˙fan,in(Δt)Wfan,in=(0.120 kJ/s)(30 min)=(0.120 kJ/s)(30 min)(60 sec1 min)=216 kJ

Refer the Table A-2, “the ideal–gas equation specific heats of various common gases table”, select the specific heat at constant pressure for air as 1.005 kJ/kg⋅K.

Substitute Ein=Qin+Wfan,in, Eout=Wb,out and ΔEsystem=ΔU in Equation (I).

Qin+Wfan,in−Wb,out=ΔUQin+Wfan,in=ΔU+Wb,outQin+Wfan,in=ΔHQin+Wfan,in=mcp(T2−T1)12.6 kJ+216 kJ=(98.5 kg)(1.005 kJ/kg⋅K)(T2−10 °C)T2=12.3°C

Thus, the average temperature of the room after 30 min is 12.3°C.

b)

To determine

The entropy change of the steam.

b)

Expert Solution

Explanation of Solution

Write the expression to calculate the change of entropy of the steam.

ΔSsteam=m(s2−s1)ΔSsteam=(0.01388 kg)(5.1642 kJ/kg⋅K−7.5081 kJ/kg⋅K)=−0.0325 kJ/K

Thus, the entropy change of the steam is −0.0325 kJ/K.

c)

To determine

The entropy change of the air.

c)

Expert Solution

Explanation of Solution

Write the expression to calculate the entropy change of air.

ΔSair=mcplnT2T1−mRlnP2P1

Since, P2=P1.

ΔSair=mcplnT2T1ΔSair=(98.5 kg)(1.005 kJ/kg⋅K)ln12.3 °C10 °C=(98.5 kg)(1.005 kJ/kg⋅K)ln(12.3+273)K(10+273)K=0.8013 kJ/K

Thus, the entropy change of the air is 0.8013 kJ/K.

d)

To determine

The entropy generation in the turbine.

d)

Expert Solution

Explanation of Solution

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

Sin−Sout+Sgen=ΔSsystem (II)

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

Substitute Sin=0, Sout=0, and ΔSsystem=ΔSair+ΔSsteam in Equation (II).

0−0+Sgen=ΔSair+ΔSsteamSgen=ΔSair+ΔSsteam

Sgen=0.8013 kJ/K+−0.0325 kJ/K=0.7688 kJ/K

Thus, the total entropy generated during the process is 0.7688 kJ/K.

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Answer 5

Chapter 8, Problem 189RQ

a)

To determine

The average temperature of the room after 30 min.

a)

Expert Solution

Explanation of Solution

Given:

The dimension of room (vroom) is 4 m×4 m×5 m.

The temperature of the room (Troom) is 10°C.

The volume of the radiator is 15 L.

The pressure of the super-heated vapor (P1) is 200 kPa.

The temperature of the superheated vapor (T1) is 200°C.

The pressure of the steam (P2) is 100 kPa.

The time duration (Δt) is 30 min.

The pressure of the room (Proom) is 100 kPa.

Calculation:

Refer the Table A-6, “Superheated water table”, obtain the following properties of steam at temperature of 200 °C and pressure of 200 kPa.

v1=1.0805 m3/kgu1=2654.4 kJ/kgs1=7.5081 kJ/kg⋅K

Refer the Table A-6, “Saturated water-Pressure table”, obtain the following properties of steam at pressure of 100 kPa.

vf=0.001043 m3/kgvg=1.6941 m3/kg

Here, Saturated liquid specific volume is vf and saturated vapor specific volume is vγ.

Calculate the final vapor quality.

x2=v2−vfvg−vfx2=1.0805 m3/kg−0.001043 m3/kg1.6941 m3/kg−0.001043 m3/kg=0.6376

Refer the Table A-6, “Saturated water-Pressure table”, obtain the following properties of steam at pressure of 100 kPa.

uf=417.40 kJ/kgufg=2088.2 kJ/kg

Calculate the final internal energy of the system.

u2=uf+x2ufg

u2=417.40 kJ/kg+(0.6376)(2088.2 kJ/kg)=1748.7 kJ/kg

Refer the Table A-6, “Saturated water-Pressure table”, obtain the following properties of steam at pressure of 100 kPa.

sf=1.3028 kJ/kg⋅Ksfg=6.0562 kJ/kg⋅K

Calculate the final entropy of the system.

s2=sf+x2sfgs2=1.3028 kJ/kg⋅K+(0.6376)(6.0562 kJ/kg⋅K)=5.1642 kJ/kg⋅K

Calculate the mass of the steam.

m=ν1v1m=0.015 m31.0805 m3/kg=0.01388 kg

Write the expression for the energy balance equation for closed system without air in the room.

Ein−Eout=ΔEsystem (I)

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

Substitute Ein=0, Eout=Qout, and ΔEsystem=ΔU in Equation (I).

0−Qout=ΔU−Qout=m(u2−u1)Qout=m(u1−u2)

Qout=(0.01388 kg)(2654.4 kJ/kg−1748.7 kJ/kg)=12.6 kJ

Refer the Table A-1, “the molar mass, gas constant and critical point properties table”, select the gas constant of air as 0.2870 kJ/kg⋅K.

Calculate the mass of the air using the ideal gas equation.

mair=P1ν1RT1mair=(100 kPa)(4m×4m×5m)(0.2870 kJ/kg⋅K)10°C=(100 kPa)(80 m3)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(10+273)K=98.5 kg

Write the expression to calculate the total work done by the fan.

Wfan,in=W˙fan,in(Δt)Wfan,in=(0.120 kJ/s)(30 min)=(0.120 kJ/s)(30 min)(60 sec1 min)=216 kJ

Refer the Table A-2, “the ideal–gas equation specific heats of various common gases table”, select the specific heat at constant pressure for air as 1.005 kJ/kg⋅K.

Substitute Ein=Qin+Wfan,in, Eout=Wb,out and ΔEsystem=ΔU in Equation (I).

Qin+Wfan,in−Wb,out=ΔUQin+Wfan,in=ΔU+Wb,outQin+Wfan,in=ΔHQin+Wfan,in=mcp(T2−T1)12.6 kJ+216 kJ=(98.5 kg)(1.005 kJ/kg⋅K)(T2−10 °C)T2=12.3°C

Thus, the average temperature of the room after 30 min is 12.3°C.

b)

To determine

The entropy change of the steam.

b)

Expert Solution

Explanation of Solution

Write the expression to calculate the change of entropy of the steam.

ΔSsteam=m(s2−s1)ΔSsteam=(0.01388 kg)(5.1642 kJ/kg⋅K−7.5081 kJ/kg⋅K)=−0.0325 kJ/K

Thus, the entropy change of the steam is −0.0325 kJ/K.

c)

To determine

The entropy change of the air.

c)

Expert Solution

Explanation of Solution

Write the expression to calculate the entropy change of air.

ΔSair=mcplnT2T1−mRlnP2P1

Since, P2=P1.

ΔSair=mcplnT2T1ΔSair=(98.5 kg)(1.005 kJ/kg⋅K)ln12.3 °C10 °C=(98.5 kg)(1.005 kJ/kg⋅K)ln(12.3+273)K(10+273)K=0.8013 kJ/K

Thus, the entropy change of the air is 0.8013 kJ/K.

d)

To determine

The entropy generation in the turbine.

d)

Expert Solution

Explanation of Solution

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

Sin−Sout+Sgen=ΔSsystem (II)

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

Substitute Sin=0, Sout=0, and ΔSsystem=ΔSair+ΔSsteam in Equation (II).

0−0+Sgen=ΔSair+ΔSsteamSgen=ΔSair+ΔSsteam

Sgen=0.8013 kJ/K+−0.0325 kJ/K=0.7688 kJ/K

Thus, the total entropy generated during the process is 0.7688 kJ/K.

Answer 6

Chapter 8, Problem 189RQ

a)

To determine

The average temperature of the room after 30 min.

a)

Expert Solution

Explanation of Solution

Given:

The dimension of room (vroom) is 4 m×4 m×5 m.

The temperature of the room (Troom) is 10°C.

The volume of the radiator is 15 L.

The pressure of the super-heated vapor (P1) is 200 kPa.

The temperature of the superheated vapor (T1) is 200°C.

The pressure of the steam (P2) is 100 kPa.

The time duration (Δt) is 30 min.

The pressure of the room (Proom) is 100 kPa.

Calculation:

Refer the Table A-6, “Superheated water table”, obtain the following properties of steam at temperature of 200 °C and pressure of 200 kPa.

v1=1.0805 m3/kgu1=2654.4 kJ/kgs1=7.5081 kJ/kg⋅K

Refer the Table A-6, “Saturated water-Pressure table”, obtain the following properties of steam at pressure of 100 kPa.

vf=0.001043 m3/kgvg=1.6941 m3/kg

Here, Saturated liquid specific volume is vf and saturated vapor specific volume is vγ.

Calculate the final vapor quality.

x2=v2−vfvg−vfx2=1.0805 m3/kg−0.001043 m3/kg1.6941 m3/kg−0.001043 m3/kg=0.6376

Refer the Table A-6, “Saturated water-Pressure table”, obtain the following properties of steam at pressure of 100 kPa.

uf=417.40 kJ/kgufg=2088.2 kJ/kg

Calculate the final internal energy of the system.

u2=uf+x2ufg

u2=417.40 kJ/kg+(0.6376)(2088.2 kJ/kg)=1748.7 kJ/kg

Refer the Table A-6, “Saturated water-Pressure table”, obtain the following properties of steam at pressure of 100 kPa.

sf=1.3028 kJ/kg⋅Ksfg=6.0562 kJ/kg⋅K

Calculate the final entropy of the system.

s2=sf+x2sfgs2=1.3028 kJ/kg⋅K+(0.6376)(6.0562 kJ/kg⋅K)=5.1642 kJ/kg⋅K

Calculate the mass of the steam.

m=ν1v1m=0.015 m31.0805 m3/kg=0.01388 kg

Write the expression for the energy balance equation for closed system without air in the room.

Ein−Eout=ΔEsystem (I)

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

Substitute Ein=0, Eout=Qout, and ΔEsystem=ΔU in Equation (I).

0−Qout=ΔU−Qout=m(u2−u1)Qout=m(u1−u2)

Qout=(0.01388 kg)(2654.4 kJ/kg−1748.7 kJ/kg)=12.6 kJ

Refer the Table A-1, “the molar mass, gas constant and critical point properties table”, select the gas constant of air as 0.2870 kJ/kg⋅K.

Calculate the mass of the air using the ideal gas equation.

mair=P1ν1RT1mair=(100 kPa)(4m×4m×5m)(0.2870 kJ/kg⋅K)10°C=(100 kPa)(80 m3)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(10+273)K=98.5 kg

Write the expression to calculate the total work done by the fan.

Wfan,in=W˙fan,in(Δt)Wfan,in=(0.120 kJ/s)(30 min)=(0.120 kJ/s)(30 min)(60 sec1 min)=216 kJ

Refer the Table A-2, “the ideal–gas equation specific heats of various common gases table”, select the specific heat at constant pressure for air as 1.005 kJ/kg⋅K.

Substitute Ein=Qin+Wfan,in, Eout=Wb,out and ΔEsystem=ΔU in Equation (I).

Qin+Wfan,in−Wb,out=ΔUQin+Wfan,in=ΔU+Wb,outQin+Wfan,in=ΔHQin+Wfan,in=mcp(T2−T1)12.6 kJ+216 kJ=(98.5 kg)(1.005 kJ/kg⋅K)(T2−10 °C)T2=12.3°C

Thus, the average temperature of the room after 30 min is 12.3°C.

Answer 7

Chapter 8, Problem 189RQ

a)

To determine

The average temperature of the room after 30 min.

Answer 8

a)

Expert Solution

Explanation of Solution

Given:

The dimension of room (vroom) is 4 m×4 m×5 m.

The temperature of the room (Troom) is 10°C.

The volume of the radiator is 15 L.

The pressure of the super-heated vapor (P1) is 200 kPa.

The temperature of the superheated vapor (T1) is 200°C.

The pressure of the steam (P2) is 100 kPa.

The time duration (Δt) is 30 min.

The pressure of the room (Proom) is 100 kPa.

Calculation:

Refer the Table A-6, “Superheated water table”, obtain the following properties of steam at temperature of 200 °C and pressure of 200 kPa.

v1=1.0805 m3/kgu1=2654.4 kJ/kgs1=7.5081 kJ/kg⋅K

Refer the Table A-6, “Saturated water-Pressure table”, obtain the following properties of steam at pressure of 100 kPa.

vf=0.001043 m3/kgvg=1.6941 m3/kg

Here, Saturated liquid specific volume is vf and saturated vapor specific volume is vγ.

Calculate the final vapor quality.

x2=v2−vfvg−vfx2=1.0805 m3/kg−0.001043 m3/kg1.6941 m3/kg−0.001043 m3/kg=0.6376

Refer the Table A-6, “Saturated water-Pressure table”, obtain the following properties of steam at pressure of 100 kPa.

uf=417.40 kJ/kgufg=2088.2 kJ/kg

Calculate the final internal energy of the system.

u2=uf+x2ufg

u2=417.40 kJ/kg+(0.6376)(2088.2 kJ/kg)=1748.7 kJ/kg

Refer the Table A-6, “Saturated water-Pressure table”, obtain the following properties of steam at pressure of 100 kPa.

sf=1.3028 kJ/kg⋅Ksfg=6.0562 kJ/kg⋅K

Calculate the final entropy of the system.

s2=sf+x2sfgs2=1.3028 kJ/kg⋅K+(0.6376)(6.0562 kJ/kg⋅K)=5.1642 kJ/kg⋅K

Calculate the mass of the steam.

m=ν1v1m=0.015 m31.0805 m3/kg=0.01388 kg

Write the expression for the energy balance equation for closed system without air in the room.

Ein−Eout=ΔEsystem (I)

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

Substitute Ein=0, Eout=Qout, and ΔEsystem=ΔU in Equation (I).

0−Qout=ΔU−Qout=m(u2−u1)Qout=m(u1−u2)

Qout=(0.01388 kg)(2654.4 kJ/kg−1748.7 kJ/kg)=12.6 kJ

Refer the Table A-1, “the molar mass, gas constant and critical point properties table”, select the gas constant of air as 0.2870 kJ/kg⋅K.

Calculate the mass of the air using the ideal gas equation.

mair=P1ν1RT1mair=(100 kPa)(4m×4m×5m)(0.2870 kJ/kg⋅K)10°C=(100 kPa)(80 m3)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(10+273)K=98.5 kg

Write the expression to calculate the total work done by the fan.

Wfan,in=W˙fan,in(Δt)Wfan,in=(0.120 kJ/s)(30 min)=(0.120 kJ/s)(30 min)(60 sec1 min)=216 kJ

Refer the Table A-2, “the ideal–gas equation specific heats of various common gases table”, select the specific heat at constant pressure for air as 1.005 kJ/kg⋅K.

Substitute Ein=Qin+Wfan,in, Eout=Wb,out and ΔEsystem=ΔU in Equation (I).

Qin+Wfan,in−Wb,out=ΔUQin+Wfan,in=ΔU+Wb,outQin+Wfan,in=ΔHQin+Wfan,in=mcp(T2−T1)12.6 kJ+216 kJ=(98.5 kg)(1.005 kJ/kg⋅K)(T2−10 °C)T2=12.3°C

Thus, the average temperature of the room after 30 min is 12.3°C.

Answer 9

a)

Expert Solution

Answer 10

a)

Expert Solution

Answer 11

Expert Solution

Answer 12

b)

To determine

The entropy change of the steam.

b)

Expert Solution

Explanation of Solution

Write the expression to calculate the change of entropy of the steam.

ΔSsteam=m(s2−s1)ΔSsteam=(0.01388 kg)(5.1642 kJ/kg⋅K−7.5081 kJ/kg⋅K)=−0.0325 kJ/K

Thus, the entropy change of the steam is −0.0325 kJ/K.

Answer 13

b)

To determine

The entropy change of the steam.

Answer 14

b)

Expert Solution

Explanation of Solution

Write the expression to calculate the change of entropy of the steam.

ΔSsteam=m(s2−s1)ΔSsteam=(0.01388 kg)(5.1642 kJ/kg⋅K−7.5081 kJ/kg⋅K)=−0.0325 kJ/K

Thus, the entropy change of the steam is −0.0325 kJ/K.

Answer 15

b)

Expert Solution

Answer 16

b)

Expert Solution

Answer 17

Expert Solution

Answer 18

c)

To determine

The entropy change of the air.

c)

Expert Solution

Explanation of Solution

Write the expression to calculate the entropy change of air.

ΔSair=mcplnT2T1−mRlnP2P1

Since, P2=P1.

ΔSair=mcplnT2T1ΔSair=(98.5 kg)(1.005 kJ/kg⋅K)ln12.3 °C10 °C=(98.5 kg)(1.005 kJ/kg⋅K)ln(12.3+273)K(10+273)K=0.8013 kJ/K

Thus, the entropy change of the air is 0.8013 kJ/K.

Answer 19

c)

To determine

The entropy change of the air.

Answer 20

c)

Expert Solution

Explanation of Solution

Write the expression to calculate the entropy change of air.

ΔSair=mcplnT2T1−mRlnP2P1

Since, P2=P1.

ΔSair=mcplnT2T1ΔSair=(98.5 kg)(1.005 kJ/kg⋅K)ln12.3 °C10 °C=(98.5 kg)(1.005 kJ/kg⋅K)ln(12.3+273)K(10+273)K=0.8013 kJ/K

Thus, the entropy change of the air is 0.8013 kJ/K.

Answer 21

c)

Expert Solution

Answer 22

c)

Expert Solution

Answer 23

Expert Solution

Answer 24

d)

To determine

The entropy generation in the turbine.

d)

Expert Solution

Explanation of Solution

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

Sin−Sout+Sgen=ΔSsystem (II)

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

Substitute Sin=0, Sout=0, and ΔSsystem=ΔSair+ΔSsteam in Equation (II).

0−0+Sgen=ΔSair+ΔSsteamSgen=ΔSair+ΔSsteam

Sgen=0.8013 kJ/K+−0.0325 kJ/K=0.7688 kJ/K

Thus, the total entropy generated during the process is 0.7688 kJ/K.

Answer 25

d)

To determine

The entropy generation in the turbine.

Answer 26

d)

Expert Solution

Explanation of Solution

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

Sin−Sout+Sgen=ΔSsystem (II)

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