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 132RQ

(a)

To determine

The view factor of the surface.

(a)

Expert Solution
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Explanation of Solution

Given:

The diameter (Da) of the surface is 20cm.

The space (L) between the surfaces is 10cm.

The emissivity (εa) is 0.6.

The temperature (Ta) of the disk is 600°C.

The diameter (Db) of the surface is 40cm.

The emissivity (εb) is 0.8.

The temperature (Tb) of the disk is 200°C.

Calculation:

Calculate the constant (A) using the relation.

    A=Da2L=(20cm×1m100cm)2(10cm×1m100cm)=1

Calculate the constant (B) using the relation.

    B=Db2LB=(40cm×1m100cm)2(10cm×1m100cm)B=2

Calculate the constant (C) using the relation.

    C=1+1+A2B2C=1+1+1222C=1.5

Calculate the view factor (Fab) using the relation.

    Fab=0.5(BA)2{C[C24(AB)2]0.5}=0.5(21)2{1.5[(1.5)24(12)2]0.5}=0.764

Thus, the view factor (Fab) is 0.764.

Calculate the view factor (Fba) using the relation.

  AaFab=AbFba(πDa24)Fab=(πDb24)FbaDa2Fab=Db2Fba(0.2m)2(0.764)=(0.4m)2Fba

  Fba=(0.2m)2(0.764)(0.4m)2Fba=0.191

Thus, the view factor (Fba) is 0.191.

(b)

To determine

The net rate of radiation heat transfer.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

Calculation:

Calculate the surface area (A) using the relation.

    Aa=π[(20cm×1m100cm)(20cm×1m100cm)]24=0.031m2Ab=π[(40cm×1m100cm)(40cm×1m100cm)]24=0.125m2

Calculate the net rate of heat transfer (Q˙ab) using the relation

  Q˙=σ(Ta4Tb4)(1εaεa)1Aa+1AaFab+(1εbεb)1Ab=(5.67×108W/m2K4)[(600°C+273)4(200°C+273)4K4][10.6(0.0314m2)(0.6)+1(0.0314m2)(0.764)+10.8(0.125m2)(0.8)]=464W

Thus, the net rate of heat transfer is 464W.

(c)

To determine

The net rate of heat exchange between the disks.

(c)

Expert Solution
Check Mark

Explanation of Solution

Given:

The emissivity (εc) is 0.7.

Calculation:

Calculate the temperature (Tc) using the relation.

    Q˙ac=Q˙cbσ(Ta4Tc4)(1εaεa)1Aa+1AaFac+(1εcεc)1Ac=σ(Tc4Tb4)(1εcεc)1Ac+1AcFcb+(1εbεb)1Ab(Ta4Tc4)(1εaεa)1Aa+1AaFac+0=(Tc4Tb4)0+1AcFcb+(1εbεb)1Ab[(600°C+273)4(Tc)4K4](10.60.6)1(0.031m2)+1(0.031m2)(1)=[[(Tc)4(200°C+273)4K4][1(0.125m2)+10.8(0.8)(10.125m2)]]Tc=605K

Calculate the net rate of heat transfer (Qab) using the relation.

    Q˙bc=Q˙acQ˙ac=σ(Ta4Tc4)(1εaεa)1Aa+1AaFac+(1εcεc)1AcQ˙ac=(5.67×108W/m2K4)[(600°C+273)4K4(605)4K4](10.60.6)1(0.031m2)+1(0.031m2)(1)+0Q˙ac=477W

Thus, the heat transfer rate is 477W.

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Chapter 21 Solutions

Fundamentals of Thermal-Fluid Sciences

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