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 20, Problem 34P
To determine

The equilibrium temperature of the absorber plate.

Expert Solution & Answer
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Explanation of Solution

Given:

The length of the plate (l) is 1.2m.

The width of plate  is (w) is 0.8m.

The temperature of air (T) is 25°C.

The temperature of sky (Tsky) is 10°C.

The surface temperature (Ts) is 115°C.

The solar radiation (q) is 600W/m2.

The emissivity (ε) is 0.98.

The absorptivity (α) is 0.98.

Calculation:

Calculate the film temperature (Tf) using the relation.

  Tf=Ts+T2=115°C+25°C2=70°C

Refer Table A-22 “Properties of air at 1 atm pressure”.

Obtain the following properties of air corresponding to the temperature of 70°C as follows:

k=0.02881W/mKv=1.995×105m2/sPr=0.7177

Calculate Characteristic length (Lc) using the relation.

    Lc=Ap=lw2(l+w)=(1.2m)(0.8m)2[(1.2m)+(0.8m)]=0.24m

Calculate the Rayleigh number (Ra) using the relation.

  Ra=gβ(TsT)Lc2v2Pr=g(1Tf)(TsT)Lc3v2Pr=(9.81m/s2)(1(70°C+273)K)((115°C+273)K(25°C+273)K)(0.24m)3(1.995×105m2/s)2(0.7177)=6.414×107

Calculate the Nusselt number (Nu) using the relation.

    Nu=0.15Ra1/3=0.15(6.414×107)1/3=60.04

Calculate the heat transfer coefficient (h) using the relation.

    h=kLcNu=0.02881W/mK(0.24m)(60.04)=7.208W/m2K

Calculate the rate of heat loss (Q) using the relation.

    Q=αqA=(0.87)(600W/m2)(0.8m)(1.2m)=501.1W

Calculate the surface temperature (Ts) using the relation.

    Q=hA(TsT)+εAσ[((Ts+273)K)4((Tsky+273)K)4]501.1W=[7.208W/m2K((0.8m)(1.2m))(Ts(25°C+273)K)+(0.09)((0.8m)(1.2m))(5.67×108W/m2K4)[((Ts+273)K)4((10+273)K)4]]Ts=(362.7K273)°C=89.7°C

This is not nearby assumed temperature of 115°C so repeat the calculation using Ts=95°C.

 Calculate the film temperature (Tf) using the relation.

  Tf=Ts+T2=95°C+25°C2=60°C

Refer Table A-22 “Properties of air at 1 atm pressure”.

Obtain the following properties of air corresponding to the temperature of 70°C as follows:

k=0.02808W/mKv=1.896×105m2/sPr=0.7202

Calculate Characteristic length (Lc) using the relation.

    Lc=Ap=lw2(l+w)=(1.2m)(0.8m)2[(1.2m)+(0.8m)]=0.24m

Calculate the Rayleigh number (Ra) using the relation.

  Ra=gβ(TsT)Lc2v2Pr=g(1Tf)(TsT)Lc3v2Pr=(9.81m/s2)(1(60°C+273)K)((95°C+273)K(25°C+273)K)(0.24m)3(1.995×105m2/s)2(0.7177)=5.711×107

Calculate the Nusselt number (Nu) using the relation.

    Nu=0.15Ra1/3=0.15(5.711×107)1/3=57.77

Calculate the heat transfer coefficient (h) using the relation.

    h=kLcNu=0.02808W/mK(0.24m)(57.77)=6.759W/m2K

Calculate the rate of heat loss (Q) using the relation.

    Q=αqA=(0.87)(600W/m2)(0.8m)(1.2m)=501.1W

Calculate the surface temperature (Ts) using the relation.

    Q=hA(TsT)+εAσ[((Ts+273)K)4((Tsky+273)K)4]501.1W=[6.759W/m2K((0.8m)(1.2m))(Ts(25°C+273)K)+(0.09)((0.8m)(1.2m))(5.67×108W/m2K4)[((Ts+273)K)4((10°C+273)K)4]]Ts=(366.5K273)°C=93.5°C

Repeat the whole calculation with α=0.28 and ε=0.07.

Calculate the rate of heat loss (Q) using the relation.

    Q=αqA=(0.28)(600W/m2)(0.8m)(1.2m)=161.3W

Calculate the surface temperature (Ts) using the relation.

    Q=hA(TsT)+εAσ[((Ts+273)K)4((Tsky+273)K)4]161.3W=[6.629W/m2K((0.8m)(1.2m))(Ts(25°C+273)K)+(0.07)((0.8m)(1.2m))(5.67×108W/m2K4)×[((Ts+273)K)4((10+273)K)4]]Ts=(320.8K273)°C=47.8°C

Thus, the equilibrium temperature of the absorber plate is 47.8°C.

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Chapter 20 Solutions

Fundamentals of Thermal-Fluid Sciences

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