EBK THERMODYNAMICS: AN ENGINEERING APPR
EBK THERMODYNAMICS: AN ENGINEERING APPR
8th Edition
ISBN: 8220100257056
Author: CENGEL
Publisher: YUZU
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Chapter 3.8, Problem 131RP
To determine

The average temperature of the air in the balloon for the atmospheric temperature of 15°C and 30°C.

Expert Solution & Answer
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Answer to Problem 131RP

The average temperature of the air in the balloon for the atmospheric temperature of 15°C is 306.5K.

The average temperature of the air in the balloon for the atmospheric temperature of 30°C is found to be 323.6K.

Explanation of Solution

Write the expression to obtain the buoyancy force acting on the hot-air balloon (FB).

FB=ρcool airgVballoon (I)

Here, density of the cool air is ρcool air, acceleration due to gravity is g, and volume of the hot-air balloon is Vballoon.

Write the expression to obtain the volume of the hot-air balloon (Vballoon).

Vballoon=4πr33 (II)

Here, radius of the hot-air balloon is r.

Write the expression to obtain the density of the cool air (ρcool air).

ρcool air=PRT (III)

Here, atmospheric pressure is P, and atmospheric air temperature is T.

Write the expression to obtain the vertical force balance acting on the hot-air balloon.

FB=Whot air+Wcage+Wpeople=mhot airg+mcageg+mpeopleg=(mhot air+mcage+mpeople)g (IV)

Here, weight of the hot air is Whot air, weight of the cage is Wcage, weight of the people is Wpeople, mass of the hot air is mhot air, mass of the cage is mcage, and mass of the people is mpeople.

Write the expression to obtain the average temperature of the air in the balloon (T) .

T=PVmhot airR (V)

Conclusion:

Consider that the air behaves as an ideal gas.

Refer Table A-1, “Molar mass, gas constant, and critical-point properties”, obtain the gas constant of air (R) as 0.287kPam3/kgK.

Substitute 10m for r in Equation (II).

Vballoon=4π(10m)33=4,189m3

Substitute 90kPa for P, 0.287kPam3/kgK for R, and 15°C for T in Equation (III).

ρcool air=90kPa(0.287kPam3/kgK)(15°C)=90kPa(0.287kPam3/kgK)(273+15K)=90kPa(0.287kPam3/kgK)(288K)=1.089kg/m3

Substitute 1.089kg/m3 for ρcool air, 9.8m/s2 for g, and 4,189m3 for Vballoon in Equation (I).

FB=(1.089kg/m3)(9.8m/s2)(4,189m3)=44,700kgm/s2=44,700kgm/s2(1N1kgm/s2)=44,700N

Substitute 44,700N for FB, 80kg for mcage, 195kg for mpeople, and 9.8m/s2 for g in Equation (IV).

44,700N=(mhot air+80kg+195kg)(9.8m/s2)44,700N=(mhot air+275kg)(9.8m/s2)(1N1kgm/s2)44,700N=(mhot air(9.8m/s2)(1N1kgm/s2))+275kg(9.8m/s2)(1N1kgm/s2)44,700N=(mhot air(9.8m/s2)(1N1kgm/s2))+2,695N

mhot air=42,005N9.8m/s2(1kgm/s21N)=4,287kg

Substitute 90kPa for P, 4,189m3 for Vballoon, 4,287kg for mhot air, and 0.287kPam3/kgK for R in Equation (V).

T=(90kPa)(4,189m3)(4,287kg)(0.287kPam3/kgK)=306.5K

Thus, the average temperature of the air in the balloon for the atmospheric temperature of 15°C is 306.5K.

Similarly the average temperature of the air in the balloon for the atmospheric temperature of 30°C is calculated in the same manner.

Thus, the average temperature of the air in the balloon for the atmospheric temperature of 30°C is found to be 323.6K.

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Chapter 3 Solutions

EBK THERMODYNAMICS: AN ENGINEERING APPR

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