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 21, Problem 137RQ
To determine

The rate of heat transfer by natural convection.

Expert Solution & Answer
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Explanation of Solution

Given:

The temperature of top surface is (T1) is 700K.

The temperature of base surface is (T2) is 950K.

The emissivity of base surface ε2 is 0.90.

The temperature of side surface is (T3) is 450K.

Calculation:

The below figure represents the cubical surface.

Fundamentals of Thermal-Fluid Sciences, Chapter 21, Problem 137RQ

Figure-(1)

Consider the top surface is surface 1, the base surface is surface 2 and side surface is surface 3.

The base surface is flat; therefore the value of view factor will be zero.

    F11=0

Calculate the view factor from the base or top of side surface.

    F11+F12+F13=10+0.20+F13=1F13=0.80

The value of other view factor will be as follows.

    F23=F13=0.80F21=F12=0.20

Use the energy balance for surface 3.

    σT34=J3(5.67×108W/m2K4)(450K)4=J3J3=2325.05W/m2

Use the energy balance for surface 1.

    σT14=J1+1ε1ε1[F12(J1J2)+F13(J1J3)](5.67×108W/m2K4)(700K)4=J1+1ε1ε1[0.20(J1J2)+0.80(J12325.05W/m2)]        (I)

Use the energy balance for surface 2.

    σT24=J2+1ε2ε2[F21(J2J1)+F23(J2J3)](5.67×108W/m2K4)(950K)4=J2+10.900.90[0.20(J2J1)+0.80(J22325.05W/m2)]        (II)

Calculate the heat flux at surface 2.

    Q˙2=A2[F21(J2J1)+F23(J2J3)]=(9m2)[0.20(J2J1)+0.80(J2J3)]        (III)

Solve equation (I), (II) and (III).

ε1=0.44J1=11736W/m2J2=41985W/m2J3=2325W/m2

Calculate the rate of heat transfer between the bottom and top surface.

    Q˙21=A2F21(J2J1)=(9m2)(0.20)(41985W/m211736W/m2)=54.4kW

Calculate the rate of heat transfer between the bottom and side surface.

    Q˙23=A2F23(J2J1)=(9m2)(0.80)(41985W/m22325W/m2)=285.6kW

Thus, the emmisivity at the top surface is 0.44, the heat transfer between the bottom and top surface is 54.4kW and the transfer between the bottom and side surface is 285.6kW.

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Chapter 21 Solutions

Fundamentals of Thermal-Fluid Sciences

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