Fundamentals Of Thermal-fluid Sciences In Si Units
Fundamentals Of Thermal-fluid Sciences In Si Units
5th Edition
ISBN: 9789814720953
Author: Yunus Cengel, Robert Turner, John Cimbala
Publisher: McGraw-Hill Education
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Chapter 20, Problem 22P

(a)

To determine

The plate midpoint temperature for a highly polished surface.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The length of thin vertical plate (l) is 0.5m.

The heat flux (Q) is 1000W/m2.

The temperature of air (T) is 5°C.

The film temperature (Tf) is 30°C.

The emissivity is (ε) is 0.02.

Calculation:

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

Obtain the following properties of air corresponding to the temperature of 30°C as follows:

k=0.02588W/mKv=1.608×105m2/sPr=0.7282

Calculate the midpoint temperature (TL/2) using the relation.

    TL/2=2TfT=2(30°C)5°C=55°C

Calculate Characteristic length (Lc) using the relation.

    Lc=l=0.5m

Calculate the Rayleigh number (Ra) using the relation.

  Ra=gβ(TL/2T)Lc3v2Pr=g(1Tf)(TL/2T)Lc3v2Pr=(9.81m/s2)(1(30°C+273)K)((55°C+273)K(5°C+273)K)(0.5m)3(1.608×105m2/s)2(0.7282)=5.699×108

Calculate the Nusselt number (Nu) for vertical plate using the relation.

    Nu=[0.825+0.387Ra1/6[1+(0.492Pr)9/16]8/27]2=[0.825+0.387(5.699×108)1/6[1+(0.4920.7282)9/16]8/27]2=103.7

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

    h=kLcNu=0.02588W/mK(0.5m)(103.7)=5.368W/m2K

Calculate the plate midpoint temperature (TL/2) using the relation.

  Q=h(TL/2T)+εσ(TL/24Tsurr4)1000W/m2=[(5.368W/m2K)((TL/2+273)K(5°C+273)K)+(5.67×108W/m2K4)(0.02)((TL/2+273)4K4(5°C+273)4K4)]Tf=(420.2K273)°C=147.2°C

Thus, the plate midpoint temperature for a highly polished surface is 147.2°C.

(b)

To determine

The plate midpoint temperature for a black oxidized surface.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

The length of thin vertical plate (l) is 0.5m.

The heat flux (Q) is 1000W/m2.

The temperature of air (T) is 5°C.

The film temperature (Tf) is 30°C.

The emissivity is (ε) is 0.78.

Calculation:

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

Obtain the following properties of air corresponding to the temperature of 30°C as follows:

k=0.02588W/mKv=1.608×105m2/sPr=0.7282

Calculate the midpoint temperature (TL/2) using the relation.

    TL/2=2TfT=2(30°C)5°C=55°C

Calculate Characteristic length (Lc) using the relation.

    Lc=l=0.5m

Calculate the Rayleigh number (Ra) using the relation.

  Ra=gβ(TL/2T)Lc3v2Pr=g(1Tf)(TL/2T)Lc3v2Pr=(9.81m/s2)(1(30°C+273)K)((55°C+273)K(5°C+273)K)(0.5m)3(1.608×105m2/s)2(0.7282)=5.699×108

Calculate the Nusselt number (Nu) for vertical plate using the relation.

    Nu=[0.825+0.387Ra1/6[1+(0.492Pr)9/16]8/27]2=[0.825+0.387(5.699×108)1/6[1+(0.4920.7282)9/16]8/27]2=103.7

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

    h=kLcNu=0.02588W/mK(0.5m)(103.7)=5.368W/m2K

Calculate the plate midpoint temperature (TL/2) using the relation.

  Q=h(TL/2T)+εσ(TL/24Tsurr4)1000W/m2=[(5.368W/m2°K)((TL/2+273)K(5°C+273)K)+(5.67×108W/m2K4)(0.78)((TL/2+273)4K4(5°C+273)4K4)]Tf=(361.5K273)°C=88.5°C

Thus, the plate midpoint temperature for a highly polished surface is 88.5°C.

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Chapter 20 Solutions

Fundamentals Of Thermal-fluid Sciences In Si Units

Ch. 20 - A 10 cm × 10 cm plate has a constant surface...Ch. 20 - Prob. 12PCh. 20 - Prob. 13PCh. 20 - Prob. 14PCh. 20 - Prob. 15PCh. 20 - A 0.2-m-long and 25-mm-thick vertical plate (k =...Ch. 20 - A 0.2-m-long and 25-mm-thick vertical plate (k =...Ch. 20 - A 0.5-m-long thin vertical plate is subjected to...Ch. 20 - Prob. 21PCh. 20 - Prob. 22PCh. 20 - Prob. 24PCh. 20 - Consider a 2-ft × 2-ft thin square plate in a room...Ch. 20 - Prob. 27PCh. 20 - A 50-cm × 50-cm circuit board that contains 121...Ch. 20 - Prob. 29PCh. 20 - Prob. 30PCh. 20 - Prob. 32PCh. 20 - Consider a thin 16-cm-long and 20-cm-wide...Ch. 20 - Prob. 34PCh. 20 - Prob. 35PCh. 20 - Prob. 36PCh. 20 - Prob. 37PCh. 20 - Flue gases from an incinerator are released to...Ch. 20 - In a plant that manufactures canned aerosol...Ch. 20 - Reconsider Prob. 20–39. In order to reduce the...Ch. 20 - Prob. 41PCh. 20 - Prob. 42PCh. 20 - Prob. 43PCh. 20 - Prob. 44PCh. 20 - A 10-m-long section of a 6-cm-diameter horizontal...Ch. 20 - Prob. 46PCh. 20 - Prob. 47PCh. 20 - Prob. 48PCh. 20 - Prob. 49PCh. 20 - Prob. 50PCh. 20 - Prob. 52PCh. 20 - Prob. 53PCh. 20 - Prob. 54PCh. 20 - Prob. 55PCh. 20 - Prob. 57PCh. 20 - Prob. 58PCh. 20 - Prob. 59PCh. 20 - Prob. 60PCh. 20 - Prob. 61PCh. 20 - Prob. 62PCh. 20 - Prob. 63PCh. 20 - Prob. 64PCh. 20 - Prob. 65PCh. 20 - Prob. 66PCh. 20 - Prob. 67PCh. 20 - Prob. 68PCh. 20 - Prob. 69PCh. 20 - Prob. 70PCh. 20 - Prob. 71PCh. 20 - Prob. 72PCh. 20 - Prob. 73PCh. 20 - Prob. 75PCh. 20 - Prob. 77PCh. 20 - Prob. 78PCh. 20 - Prob. 79PCh. 20 - Prob. 81PCh. 20 - An electric resistance space heater is designed...Ch. 20 - Prob. 83RQCh. 20 - A plate (0.5 m × 0.5 m) is inclined at an angle of...Ch. 20 - A group of 25 power transistors, dissipating 1.5 W...Ch. 20 - Prob. 86RQCh. 20 - Prob. 87RQCh. 20 - Consider a flat-plate solar collector placed...Ch. 20 - Prob. 89RQCh. 20 - Prob. 90RQCh. 20 - Prob. 91RQCh. 20 - Prob. 92RQCh. 20 - Prob. 93RQCh. 20 - Prob. 94RQCh. 20 - Prob. 95RQCh. 20 - Prob. 96RQCh. 20 - Prob. 97RQCh. 20 - Prob. 98RQCh. 20 - Prob. 99RQCh. 20 - Prob. 100RQCh. 20 - Prob. 101RQCh. 20 - A solar collector consists of a horizontal copper...Ch. 20 - Prob. 103RQCh. 20 - Prob. 104RQ
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