Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
8th Edition
ISBN: 9780073398174
Author: Yunus A. Cengel Dr., Michael A. Boles
Publisher: McGraw-Hill Education
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Chapter 6.11, Problem 138RP
To determine

The time required to cool the truck from 25°C to 5°C.

Expert Solution & Answer
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Answer to Problem 138RP

The time required to cool the truck from 25°C to 5°C is 3.95min_.

Explanation of Solution

Determine the mass of the air in the truck.

mair=ρair×νtruck=ρair×(l×b×h)truck (I)

Here, the density of air is ρair, the volume of the truck is νtruck, the length of the truck is l, the breath of the truck is b, and the height of the truck is h.

Determine the amount of heat removed from the truck.

Qcooling,air=(maircpΔT)=maircp(T1T2) (II)

Here, the specific heat at the average temperature is cp, the initial temperature of the truck is T1, and the final temperature of the truck is T2.

Determine the average temperature in the truck during precooling.

Tavg=T1+T22 (III)

Determine the average rate of heat gain by transmission.

Q˙transmission,avg=UAΔT=UA(T1Tavg) (IV)

Here, the rate of transmission heat gain occurs is UA.

Determine the time required to cool the truck.

Q˙refrig×Δt=Qcooling,air+Q˙transmission×ΔtΔt=Qcooling,airQ˙refrigQ˙transmission (V)

Conclusion:

From the Table A-2, “Ideal-gas specific heats of various common gases” to obtain value of density of air and the specific heat at the average temperature of 15°C as 1.2kg/m3 and 1.0kJ/kg°C.

Substitute 1.2kg/m3 for ρair, 12 m for l, 2.3 m for b, and 3.5 m for h in Equation (I).

mair=(1.2kg/m3)×(12m×2.3m×3.5m)truck=(1.2kg/m3)×(96.6m3)=115.92kg116kg

Substitute 116 kg for mair, 1.0kJ/kg°C for cp, 25°C for T1, and 5°C for T2 in Equation (II).

Qcooling,air=(116kg)×(1.0kJ/kg°C)×(255)°C=(116kg)×(1.0kJ/kg°C)×(20°C)=2,320kJ

Substitute 25°C for T1, and 5°C for T2 in Equation (III).

Tavg=(25+5)°C2=15°C

Substitute 120W/°C for UA, 25°C for T1, and 15°C for Tavg in Equation (IV).

Q˙transmission,avg=(120W/°C)×(25°C15°C)=(120W/°C)×(10°C)=1200W=1200W×(103kJ/s1W)

                  =1.2kJ/s

Substitute 2320 kJ for Qcooling,air, 11kJ/s for Q˙refrig, 1.2kJ/s for Q˙transmission in Equation (V).

Δt=2320kJ(111.2)kJ/s=2320kJ9.8kJ/s=236.73s=236.73s×(1min60sec)

    =3.9455min3.95min

Thus, the time required to cool the truck from 25°C to 5°C is 3.95min_.

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Chapter 6 Solutions

Thermodynamics: An Engineering Approach

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