Fundamentals of Thermal-Fluid Sciences
Fundamentals of Thermal-Fluid Sciences
5th Edition
ISBN: 9780078027680
Author: Yunus A. Cengel Dr., Robert H. Turner, John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 8, Problem 97P

a)

To determine

The final equilibrium temperature.

a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The mass of the water (m) is 45kg.

The initial temperature of the water (T1) is 95°C.

The volume of the room (v) is 90 m3.

The initial temperature of the room (T1) is 12°C.

Calculation;

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

  R=0.2870kJ/kgKcp=1.005kJ/kgKcv=0.718kJ/kgK

Calculate the mass of the air.

  mair=P1vRT1mair=(101.3kPa)(90m3)(0.2870kJ/kgK)12 °C=(101.3kPa)(90m3)(0.2870kJ/kgK)(1kPam31kJ)(12+273)K=111.46kg

Refer the Table A-3, “Properties of common liquids, solids, and foods”,

The specific heat of water cw at room temperature is 4.18kJ/kgK

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

EinEout=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.

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

  00=ΔUwater+ΔUair[mcw(T2T1)]water+[mcv(T2T1)]air=0mwatercw(T2(T1)water)+maircv(T2(T1)air)=0

  {[(45kg)(4.18kJ/kgK)(T295°C)]+[(111.46kg)(0.718kJ/kgK)(T212°C)]}=0{[(188.1kJ/K)(T295°C)]+[(80.02kJ/K)(T212°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
Check Mark

Explanation of Solution

Calculate the heat transfer (Q) to the air.

  Q=maircv(T2(T1)air)Q=(111.46kg)(0.718kJ/kgK)(70.2°C12 °C)=(111.46kg)(0.718kJ/kg°K)[(70.2+273)K(12 +273)K]=4660kJ

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

c)

To determine

The entropy generation.

c)

Expert Solution
Check Mark

Explanation of Solution

Calculate the final pressure (P2).

  mair=P2vRT2

  111.46kg=P2(90m3)(0.2870kJ/kgK)70.2°CP2=(111.46kg)(0.2870kJ/kgK)(1kPam31kJ)(70.2+273)K(90m3)P2=121.98kPa=122kPa

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

  SinSout+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+ΔSwater in Equation (II).

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

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

Thus, the entropy generation is 1.77kJ/K.

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Chapter 8 Solutions

Fundamentals of Thermal-Fluid Sciences

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