THERMODYNAMICS (LL)-W/ACCESS >CUSTOM<
THERMODYNAMICS (LL)-W/ACCESS >CUSTOM<
9th Edition
ISBN: 9781266657610
Author: CENGEL
Publisher: MCG CUSTOM
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Chapter 2.8, Problem 100P
To determine

The plotting of rate of heat transfer against the convection heat transfer coefficient for the surface emissivities of 0.1, 0.5, 0.8, and 1. Also, discuss the results.

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Answer to Problem 100P

The plotting of rate of heat transfer against the convection heat transfer coefficient for the surface emissivities of 0.1, 0.5, 0.8, and 1 are shown in Figure (1) and results are discussed as below.

Explanation of Solution

Calculate the rate of heat transfer by convection.

Q˙conv=hAΔT=h(πD2)(TsTo) (I)

Here, change in the temperature is ΔT, surface temperature is TS, final temperature is To, the diameter of a spherical ball is D, the convection heat transfer coefficient is h, and area of a spherical ball is A.

Calculate the rate of heat transfer by radiation.

Q˙rad=εσA(Ts4To4)=εσ(πD2)(Ts4To4) (II)

Here, surface temperature is TS, exit temperature is To, emissivity of the ball is ε, and Stefan-Bolzmann constant is σ.

Calculate the total rate of heat transfer from the ball.

Q˙total=Q˙conv+Q˙rad (III)

Conclusion:

Let us solve for ε=0.8.

Substitute 15W/m2°C for h, 9 cm for D, 110°C for TS, and 20°C for To in Equation (I).

Q˙conv=15W/m2°C(π×(9cm)2)(110°C20°C)=15W/m2(π×81cm2×(0.01mcm)2)(90)=34.35W

Substitute 5.67×108W/m2K4 for σ, 9 cm for D, 110°C for TS, and 20°C for To, and 0.8 for ε in Equation (II).

Q˙rad=0.8(5.67×108W/m2K4)(π×(9cm)2)((110°C)4(20°C)4)=[0.8×5.67×108Wm2K4×π×(9cm×0.01m1cm)2[{(110+273)K}4{(20+273)K}4]]=16.33W

Substitute 34.35 W for Q˙conv and 16.33 W for Q˙rad in Equation (III).

Q˙total=34.35W+16.33W=50.7W

Follow the above process to calculate the rate of heat transfer against the convection heat transfer coefficient for the surface emissivities of 0.1 and 0.5 using spreadsheet including equations (I), (II), and (III) as in table (1).

h,(W/m2K4)Q˙convQ˙rad@ε=0.1Q˙totalQ˙rad@ε=0.5Q˙total
511.45112.0412713.492410.206421.6575
7.517.17672.0412719.217910.206427.383
1022.90222.0412724.943510.206433.1086
12.528.62782.0412730.66910.206438.8341
1534.35332.0412736.394610.206444.5597
17.540.07892.0412742.120110.206450.2852
2045.80442.0412747.845710.206456.0108
22.551.532.0412753.571210.206461.7363
2557.25552.0412759.296810.206467.4619
27.562.98112.0412765.022410.206473.1874
3068.70662.0412770.747910.206478.913

Continue table (1) for ε=0.8and1.0 as in Table (2).

Q˙rad@ε=0.8Q˙totalQ˙rad@ε=1Q˙total
16.330227.781320.412731.8638
16.330233.506820.412737.5894
16.330239.232420.412743.3149
16.330244.957920.412749.0405
16.330250.683520.412754.766
16.330256.409120.412760.4916
16.330262.134620.412766.2172
16.330267.860220.412771.9427
16.330273.585720.412777.6683
16.330279.311320.412783.3938
16.330285.036820.412789.1194

Show the plotting of rate of heat transfer against the convection heat transfer coefficient for the surface emissivities of 0.1, 0.5, 0.8, and 1.0 using Table (1) and (2) as in Figure (1).

THERMODYNAMICS (LL)-W/ACCESS >CUSTOM<, Chapter 2.8, Problem 100P

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