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 133RQ

(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 40cm.

The emissivity (εa) is 0.8.

The temperature (Ta) of the disk is 800°C.

The diameter (Db) of the surface is 60cm.

The emissivity (εb) is 0.4.

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

Calculation:

Calculate the constant (A) using the relation.

    A=DL=(20cm×1m100cm)(40cm×1m100cm)=0.5

Calculate the constant (B) using the relation.

    B=DLB=(60cm×1m100cm)(40cm×1m100cm)B=1.5

Calculate the view factor (F) using the relation.

    Fab=12A{[(B+A)2+4]0.5[(BA)2+4]0.5}=12A{[(1.5+0.5)2+4]0.5[(1.50.5)2+4]0.5}=0.592

Thus, the view factor is 0.592.

(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=(20cm×1m100cm)(20cm×1m100cm)=0.04m2Ab=(60cm×1m100cm)(60cm×1m100cm)=0.36m2

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

  Q˙=σ(Ta4Tb4)(1εaεa)1Aa+1AaFab+(1εbεb)1Ab=(5.67×108W/m2K4)[(800°C+273)4(200°C+273)4K4][10.8(0.04m2)(0.8)+1(0.04m2)(0.592)+10.4(0.36m2)(0.4)]=1374W

Thus, the net rate of heat transfer is 1374W.

(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 0.1.

The diameter (D) is 2.0m

The length (L) is 20cm.

Calculation:

Calculate the constant (A) using the relation.

    A=DL=(20cm×1m100cm)(20cm×1m100cm)=1

Calculate the constant (C) using the relation.

    C=DL=(2m)(20cm×1m100cm)=10

Calculate the view factors (F) using the relation.

    Fab=12A{[(C+A)2+4]0.5[(CA)2+4]0.5}=12×1{[(10+1)2+4]0.5[(101)2+4]0.5}=0.98

Calculate the constant (B) using the relation.

    B=DLB=(60cm×1m100cm)(20cm×1m100cm)B=3

Calculate the view factors (F) using the relation.

    Fbc=12A{[(C+B)2+4]0.5[(CB)2+4]0.5}=12×1{[(10+3)2+4]0.5[(103)2+4]0.5}=0.59

Calculate the view factor (F) using the relation.

    AbFbc=AcFcb(0.36)(0.979)=(4)FcbFcb=0.08

Calculate the temperature (Tc) using the relation.

    Q˙ac=Q˙cbσ(Ta4Tc4)(1εaεa)1Aa+1AaFac+(1εcεc)1Ac=σ(Tc4Tb4)(1εcεc)1Ac+1AcFcb+(1εbεb)1Ab[(800°C+273)4(Tc)4][(10.80.8)1(0.04m2)+1(0.04m2)(0.981)+(10.10.1)14m2]=[[(Tc)4(200°C+273)4K][1(4m2)10.1(0.1)+(1(4m2)(0.0881))+(10.40.4)10.36m2]]Tc=((754K273))°CTc=481°C

Thus, the temperature is 481°C.

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Chapter 21 Solutions

Fundamentals of Thermal-Fluid Sciences

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