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 94P
To determine

The rates of radiation heat transfer between the cylinders and between the hot cylinder and the surroundings.

Expert Solution & Answer
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Explanation of Solution

Given:

The diameter of cylinders (D) is 20cm.

The distance between cylinders (s) is 30cm.

The temperature of cylinder 1 (T1) is 425K.

The temperature of cylinder 2 (T2) is 275K.

The temperature of surrounding back body (T3) is 300K.

The length of the cylinder (L) is 1m.

The emissivity (ε) is 1.

Calculation:

Calculate the length 1 (L1) using the relation.

  L12=s2+L2L1=s2+L2=(30cm×102m1cm)2+(1m)2=1.04m

Calculate the length 2 (L2) using the relation.

    L12=s2+L2L1=s2+L2=(30cm×102m1cm)2+(1m)2=1.04m

Calculate the surface of string (P) using the relation.

    P=π2D=π2(20cm×102m1cm)=0.314m

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

    F12=(Crossed strings)(Uncrossed strings)2P=(L1+L2)(s+s)2P=(1.04m+1.04m)2(30cm×102m1cm)2(0.314m)=2.35

Calculate the view factor between hot cylinder and surrounding (F13) using the relation.

    F13=1F12=12.35=1.35

Calculate the area of the cylinder (A) using the relation.

    A=π2DL=π2(20cm×102m1cm)(1m)=π20.2m2=0.314m2

Calculate the rate of heat transfer between cylinders (Q˙12) using the relation.

    Q˙12=AF12σ(T14T24)=(0.314m2)(2.355)(5.67×108W/m2K4)[(425K)4(275K)4]=(0.74m2)(5.67×108W/m2K4)(2.7×1010K)=1132.86W

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

    A1=πDL=π(20cm×102m1cm)(1m)=π(20cm2×102m21cm2)=0.628m2

Calculate the rate of heat transfer between cylinder surface and surroundings.

  Q˙13=A1F13σ(T14T34)=(0.628m2)(1.35)(5.67×108W/m2K4)[(425K)4(275K)4]=(4.8×108W/K4)(2.452×1010K4)=1178.68W

Thus, the radiation heat transfer between the cylinders is 1132.86W and between the hot cylinder and the surroundings is 1178.68W.

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Chapter 21 Solutions

Fundamentals of Thermal-Fluid Sciences

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