Fundamentals Of Thermal-fluid Sciences In Si Units
Fundamentals Of Thermal-fluid Sciences In Si Units
5th Edition
ISBN: 9789814720953
Author: Yunus Cengel, Robert Turner, John Cimbala
Publisher: McGraw-Hill Education
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Chapter 21, Problem 96P
To determine

The net rate of radiation heat transfer between two surfaces and the horizontal surface and the surroundings.

Expert Solution & Answer
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Explanation of Solution

Given:

The length of edge (w) is 1.6m.

The width of horizontal surface (L1) is 0.8m.

The width of vertical surface (L2) is 1.2m.

The emissivity of horizontal surface (ε1) is 0.75.

The emissivity of vertical surface is (ε2) is 1.

The emissivity of surrounding (ε3) is 0.85.

The temperature of horizontal surface (T1) is 400K.

The temperature of vertical surface (T2) is 550K.

The temperature of surrounding (T3) is 290K.

Calculation:

Calculate the ratio of length to width of horizontal surface (r1) using the relation.

    r1=L1w=0.8m1.6m=0.5

Calculate the ratio of length to width of vertical surface (r2) using the relation.

    r2=L2w=1.2m1.6m=0.75

Refer Table  21-35 “View factor between coaxial parallel disks.”.

Obtain the following values of view factors of different surfaces corresponding to the ratio r1=0.5 and r2=0.75 as follows:

  F12=0.27

Calculate the surface area of horizontal surface (A1) using the relation.

    A1=L1×w=(0.8m)×(1.6m)=1.28m2

Calculate the surface area of vertical surface (A1) using the relation.

    A2=L2×w=(1.2m)×(1.6m)=1.92m2

Calculate the surface area of third connecting surface (A3) using the relation.

    A3=2L1×L22+(L12+L22)(w)=2(0.8m)×(1.2m)2+((0.8m2)2+(1.2m)2)(1.6m)=0.96m2+2.307m2=3.26m2

Calculate the view factor from surface 2 to 1 (F12) using the relation.

    A2F21=A1F12(1.92m2)F21=(1.28m2)(0.27)F12=0.18

Calculate the view factor from surface 1 to 3 (F13) using the relation.

    F11+F12+F13=10+0.27+F13=1F13=10.27F13=0.73

Calculate the view factor from surface 2 to 3 (F23) using the relation.

    F21+F22+F23=10.18+0+F23=1F23=10.18=0.82

Calculate the view factor from surface 3 to 1 (F31) using the relation.

    A1F12=A3F31(1.28m2)(0.73)=(3.26m2)F31F31=0.9344m23.26m2F13=0.29

Calculate the view factor from surface 3 to 2 (F32) using the relation.

    A2F23=A3F32(1.92m2)(0.82)=(3.26m2)F32F32=1.574m23.26m2F13=0.48

Calculate the radiosity of vertical surface (J2) using the relation.

    J2=σT24=(5.67×108W/m2K4)(550K)4=(5.67×108W/m2K4)(9.150×1010K4)=5188W/m2

Calculate the radiosity of horizontal surface (J1) using the relation.

    σT14=J1+1ε1ε1[F12(J1J2)+F13(J1J3)](5.67×108W/m2K4)(400K)4=J1+10.750.75[0.27(J1J2)+0.73(J1J3)]1451.52W/m2=1.33J10.33[0.27(5188W/m2)+0.73J3]J1=1439W/m2+0.181J3

Calculate the radiosity for third surface (J3) using the relation.

  σT34=J3+1ε3ε3[F31(J3J1)+F32(J3J2)](5.67×108W/m2K4)(290K)4=1.22J30.176[0.77J1+0.48(5188W/m2)]401.03W/m2=1.22J30.135(1439W/m2+0.181J3)0.084J2J3=811.5W/m2

Calculate the radiosity for horizontal surface (J1) using the relation.

    J3=1439W/m2+0.181(811.5W/m2)=1587W/m2

Calculate the net rate of heat transfer between two surfaces (Q˙12) using the relation.

    Q˙12=A1F12(J1J2)=(1.28m2)(0.27)(1587W/m25188W/m2)=(0.3456m2)(3601W/m2)=1245W

Calculate the net rate of heat transfer between horizontal surface and surrounding (Q˙13) using the relation.

    Q˙13=A1F13(J1J3)=(1.28m2)(0.73)(1587W/m2811.5W/m2)=725W

Thus, the net rate of radiation heat transfer between two surfaces is 1245W and the net rate of radiation heat transfer between the horizontal surface and the surroundings is 725W.

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Chapter 21 Solutions

Fundamentals Of Thermal-fluid Sciences In Si Units

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