The average temperature of the room after 30 min.

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Answer 1

Question

Chapter 7.13, Problem 206RP

a)

To determine

The average temperature of the room after 30 min.

a)

Expert Solution

Answer to Problem 206RP

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

Explanation of Solution

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.

Write the expression to calculate the final vapor quality.

x2=v2−vfvg−vf (II)

Here, final vapor quality is x2, final molar volume is v2, saturated liquid molar volume is vf and saturated vapor molar volume is vg.

Write the expression to calculate the final internal energy of the system.

u2=uf+x2ufg (III)

Here, final internal energy of the system is u2, saturated liquid internal energy is uf, final vapor quality is x2, and evaporation internal energy is ufg.

Write the expression to calculate the final entropy of the system.

s2=sf+x2sfg (IV)

Here, final entropy of the system is s2, saturated liquid entropy is sf, final vapor quality is x2, and evaporation entropy is sfg.

Write the expression to calculate the mass of the steam.

m=ν1v1 (V)

Here, mass of the steam is m, volume of the steam is ν1 and initial specific volume of steam is v1.

Write the expression to calculate the ideal gas equation, to find the mass of the air.

mair=P1ν1RT1 (VI)

Here, mass of the air is mair, initial pressure is P1, initial volume is ν1, gas constant is R and initial temperature is T1.

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

Wfan,in=W˙fan,in(Δt) (VII)

Here, rate of work done by fan is W˙fan,in and time is Δt.

Conclusion:

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

0−Qout=ΔU−Qout=m(u2−u1)Qout=m(u1−u2) (VIII)

Here, change in internal energy of system is ΔU, heat transfer out is Qout, mass of the steam is m, initial internal energy is u1 and final internal energy is u2.

From 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

From 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.

Substitute 1.0805 m3/kg for v2, 0.001043 m3/kg for vf and 1.6941 m3/kg for vg in Equation (II).

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

From 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

Here, Saturated liquid internal energy is uf and Evaporation internal energy is ufg.

Substitute 417.40 kJ/kg for uf, 0.6376 for x2 and 2088.2 kJ/kg for ufg in Equation (III).

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

From 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

Here, Saturated liquid entropy is sf and evaporation entropy is sfg.

Substitute 1.3028 kJ/kg⋅K for sf, 0.6376 for x2 and 6.0562 kJ/kg⋅K for sfg in Equation (IV).

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

Substitute 0.015 m3 for ν1 and 1.0805 m3/kg for v1 in Equation (V).

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

Substitute 0.01388 kg for m, 2654.4 kJ/kg for u1 and 1748.7 kJ/kg for u2 in Equation (VIII).

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

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

Qin+Wfan,in−Wb,out=ΔUQin+Wfan,in=ΔU+Wb,outQin+Wfan,in=ΔHQin+Wfan,in=mcp(T2−T1) (IX)

Here, heat transfer in is Qin, work done by the fan is Wfan,in, mass of the air is m, specific heat at constant pressure is cp, initial temperature is T1 and final temperature is T2.

From 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.

Substitute 100 kPa for P1, (4m×4m×5m) for ν1, 0.2870 kJ/kg⋅K for R and 10 °C for T1 in Equation (VI).

mair=(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

Substitute 0.120 kJ/s for W˙fan,in and 30 min for Δt in Equation (VII).

Wfan,in=(0.120 kJ/s)(30 min)=(0.120 kJ/s)(30 min)(60 sec1 min)=216 kJ

From 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 12.6 kJ for Qin, 216 kJ for Wfan,in, 98.5 kg for m, 1.005 kJ/kg⋅K for cp and 10 °C for T1 in Equation (IX).

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

Answer to Problem 206RP

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

Explanation of Solution

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

ΔSsteam=m(s2−s1) (X)

Here, mass of the steam is m, final entropy is s2 and initial entropy is s1.

Conclusion:

Substitute 0.01388 kg for m, 5.1642 kJ/kg⋅K for s2 and 7.5081 kJ/kg⋅K for s1 in Equation (X).

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

Answer to Problem 206RP

The entropy change of the air is 0.8013 kJ/K.

Explanation of Solution

Write the expression to calculate the entropy change of air.

ΔSair=mcplnT2T1−mRlnP2P1

Since, P2=P1.

ΔSair=mcplnT2T1 (XI)

Conclusion:

Substitute 98.5 kg for m, 1.005 kJ/kg⋅K for cp, 12.3 °C for T2 and 10 °C for T1 in Equation (XI).

ΔSair=(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

Answer to Problem 206RP

The entropy generation in the turbine is 4.01 kW/K.

Explanation of Solution

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

Sin−Sout+Sgen=ΔSsystem (XII)

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

Conclusion:

Substitute 0 for Sin, 0 for Sout and ΔSair+ΔSsteam for ΔSsystem in Equation (XII).

0−0+Sgen=ΔSair+ΔSsteamSgen=ΔSair+ΔSsteam (XIII)

Here, entropy change of air is ΔSair and entropy change of steam is ΔSstean.

Substitute 0.8013 kJ/K for ΔSair and −0.0325 kJ/K for ΔSsteam in Equation (XIII).

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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