THERMODYNAMICS: ENG APPROACH LOOSELEAF
THERMODYNAMICS: ENG APPROACH LOOSELEAF
9th Edition
ISBN: 9781266084584
Author: CENGEL
Publisher: MCG
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Chapter 8.8, Problem 49P

(a)

To determine

The amount of heat lost to the surroundings.

(a)

Expert Solution
Check Mark

Answer to Problem 49P

The amount of heat lost to the surroundings is 2563kJ.

Explanation of Solution

Express the final volume of the air.

νa2=νa1νw (I)

Here, initial volume of air is νa1 and volume of water is νw.

Express the mass of air in the large tank.

ma=P1νa1RTa1 (II)

Here, initial pressure is P1, universal gas constant of air is R and initial temperature of air is Ta1.

Express the pressure of air at the final state.

Pa2=maRTa2νa2 (III)

Here, final temperature of air is Ta2.

Express the mass of water.

mw=ρwνw (IV)

Here, density of water is ρw.

Express the amount of heat lost to the surroundings.

Qout=[mwcw(T2Tw1)+macva(T2Ta1)] (V)

Here, specific heat of water is cw, final temperature is T2, initial temperature of water is Tw1, specific heat constant of air is cva and initial temperature of air is Ta1.

Conclusion:

Refer Table A-2, “ideal gas specific heats of various gases”, and write the properties of air at room temperature.

R=0.287kPam3/kgKcp=1.005kJ/kgKcva=0.718kJ/kgK

Refer Table A-3 (a), “properties of common liquids, solids and foods”, and write the properties of water at room temperature.

ρw=997kg/m3cw=4.18kJ/kgK

Substitute 0.04m3 for νa1 and 15L for νw in Equation (I).

νa2=0.04m315L=0.04m315L[m31000L]=0.04m30.015m3=0.025m3

Substitute 100kPa for P1, 0.04m3 for νa1, 0.287kPam3/kgK for R and 22°C for Ta1 in Equation (II).

ma=(100kPa)(0.04m3)(0.287kPam3/kgK)(22°C)=(100kPa)(0.04m3)(0.287kPam3/kgK)(22+273)K=(100kPa)(0.04m3)(0.287kPam3/kgK)(295K)=0.04724kg

Substitute 0.04724kg for ma, 0.287kPam3/kgK for R, 44°C for Ta2 and 0.025m3 for νa2 in Equation (III).

Pa2=(0.04724kg)(0.287kPam3/kgK)(44°C)0.025m3=(0.04724kg)(0.287kPam3/kgK)(44+273)K0.025m3=(0.04724kg)(0.287kPam3/kgK)(317K)0.025m3=171.9kPa

Substitute 997kg/m3 for ρw and 15L for νw in Equation (IV).

mw=(997kg/m3)(15L)=(997kg/m3)[15L×m31000L]=(997kg/m3)(0.015m3)=14.96kg

Substitute 14.96kg for mw, 4.18kJ/kgK for cw, 44°C for T2, 85°C for Tw1, 0.04724kg for ma, 0.718kJ/kgK for cva and 22°C for Ta1 in Equation (V).

Qout=[(14.96kg)(4.18kJ/kgK){(44°C85°C)K}+(0.04724kg)(0.718kJ/kgK){(44°C22°C)K}]=[(14.96kg)(4.18kJ/kgK)(41K)+(0.04724kg)(0.718kJ/kgK)(22K)]=2563kJ

Hence, the amount of heat lost to the surroundings is 2563kJ.

(b)

To determine

The exergy destruction during the process.

(b)

Expert Solution
Check Mark

Answer to Problem 49P

The exergy destruction during the process is 318kJ.

Explanation of Solution

Express entropy change in water.

ΔSw=mwcwlnTw1T2 (VI)

Express entropy change in air.

ΔSa=ma[cplnTa1T2RlnPa1P2] (VII)

Here, initial pressure of air is Pa1.

Express net internal energy of water.

ΔUw=mwcw(T1wT2) (VIII)

Express net internal energy of air.

ΔUa=macva(T1aT2) (IX)

Express the exergy destruction during the process.

Xdest=ΔUwT0ΔSw+ΔUaT0ΔSa (X)

Here, surrounding temperature is T0.

Conclusion:

Substitute 14.96kg for mw, 4.18kJ/kgK for cw, 44°C for T2 and 85°C for Tw1 min Equation (VI).

ΔSw=(14.96kg)(4.18kJ/kgK)ln85°C44°C=(14.96kg)(4.18kJ/kgK)[ln358K317K]=7.6059kJ/K

Substitute 0.04724kg for ma, 1.005kJ/kgK for cp, 44°C for T2, 22°C for Ta1, 0.287kPam3/kgK for R, 100kPa for Pa1 and 171.9kPa for P2 in Equation (VII).

ΔSa=(0.04724kg)[(1.005kJ/kgK)ln22°C44°C(0.287kPam3/kgK)ln100kPa171.9kPa]=(0.04724kg)[(1.005kJ/kgK)ln295K317K(0.287kPam3/kgK)(0.5417)]=0.003931kJ/K

Substitute 14.96kg for mw, 4.18kJ/kgK for cw, 44°C for T2 and 85°C for Tw1 min Equation (VIII).

ΔUw=(14.96kg)(4.18kJ/kgK)[(85°C44°C)K]=2564kJ

Substitute 0.04724kg for ma, 0.718kJ/kgK for cva and 22°C for Ta1, and 44°C for T2 in Equation (IX).

ΔUa=(0.04724kg)(0.718kJ/kgK)[(22°C44°C)K]=0.7462kJ

Substitute 2564kJ for ΔUw, 22°C for T0, 7.6059kJ/K for ΔSw, 0.7462kJ for ΔUa and 0.003931kJ/K for ΔSa in Equation (X).

Xdest=[(2564kJ)(22°C)(7.6059kJ/K)+(0.7462kJ)(22°C)(0.003931kJ/K)]=[(2564kJ)(22+273)K(7.6059kJ/K)+(0.7462kJ)(22+273)K(0.003931kJ/K)]=[(2564kJ)(295K)(7.6059kJ/K)+(0.7462kJ)(295K)(0.003931kJ/K)]=318kJ

Hence, the exergy destruction during the process is 318kJ.

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Chapter 8 Solutions

THERMODYNAMICS: ENG APPROACH LOOSELEAF

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