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 94RQ
To determine

The rate of heat loss from all surfaces of the tank by natural convection and radiation.

Expert Solution & Answer
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Explanation of Solution

Given:

The diameter of the cylinder (d) is 40cm.

The height of the cylinder (l) is 110cm.

The temperature of the surrounding (To) is 20°C.

The temperature of the tank surface (Ts) is 44°C.

The emissivity of the tank (ε) is 0.4.

Calculation:

Calculate the bulk mean temperature (Tm) using the relation.

    Tm=To+Ts2=(20°C)+(44°C)2=32°C

Refer table A-22 “properties of air at 1atm pressure”.

Obtain the following properties of air corresponding to the temperature of 32°C.

k=0.02603W/mKv=1.627×105m2/sPr=0.7276

Calculate the temperature coefficient (β) using the relation

    β=1Tm=1(32°C+273)K=0.003279K1

Calculate the Rayleigh number (Ra) using the relation

    Ra=gβ(TsTo)D3υ2Pr=(9.81m/s2)(0.003279K1)((44°C+273)K(20°C+273)K)(1.1m)3(1.627×105m2/s)2(0.7276)=2.334×109

Calculate the Nusslet number (Nu) using the relation.

    Nu={0.825+0.387Ra16[1+(0.492Pr)916]827}2={0.825+0.387(2.334×109)16[1+(0.4920.7276)916]827}2=170.2

The heat transfer coefficient (h) can be calculated using the relation

    h=kLNu=0.02603W/mK1.1m(170.2)=4.027W/m2K

The surface area (As) can be calculated using the relation

    As=πDL=π(40cm×102m1cm)(110cm×102m1cm)=1.382m2

The convective heat transfer from the curved surface is (Qside) by using the relation

    Qside=hAs(TsTo)=(4.027W/m2K)(1.382m2)[(44°C+273)K(20°C+273)K]=133.6W

The characteristic length (Lc) can be calculated using the relation.

    Lc=d4=0.4m4=0.1m

The Rayleigh number (Ra) can be calculated using the relation

     Ra=gβ(TsTo)D3υ2Pr=(9.81m/s2)(0.003279K1)((44°C+273)K(20°C+273)K)(0.1m)3(1.627×105m2/s)2(0.7276)=2.123×106

The Nusslet number (Nu) can be calculated using the relation

    Nu=0.54Ra1/4=0.54(2.123×106)1/4=20.61

The convective heat transfer coefficient (h) can be calculated using the relation

    h=kLcNu=0.02603W/mK0.1m(20.61)=5.635W/m2K

The surface area of the top surface (Atop) can be calculated as using the relation

    Atop=π4d2=π4×(0.4m)2=0.1257m2

The convective heat transfer from the top surface (Qtop) can be calculated as using the relation

    Qtop=hAtop(TsTo)=(5.635W/m2K)(0.1257m2)[(44°C+273)K(20°C+273)K]=16.2W

The Nusslet number for the bottom surface (Nu) can be calculated by using the relation

    Nu=0.27Ra1/4=0.27(2.123×106)14=10.31

The heat transfer coefficient (h) can be calculated as using the relation

    h=kLcNu=0.02603W/mK0.1m(10.31)=2.683W/m2K

The heat loss from the bottom (Qbottom) can be calculated as using the relation

    Qbottom=hAbottom(TsTo)=(2.683W/m2K)(0.1257m2)[(44°C+273)K(20°C+273)K]=8.1W

The total heat loss by convection (Qconv) can be calculated as using the relation,

    Qconv=Qside+Qtop+Qbottom=133.6W+16.2W+8.1W=157.9W

The heat transfer by radiation (Qrad) can be calculated as using the relation

    Qrad=ε(As+Atop+Abottom)(Ts4To4)=[0.4×(1.382m2+0.1257m2+0.1257m2)[{(44°C+273)K}4{(20°C+273)K}4]]=101.1W

Thus, the rate of heat transfer by convection from all surfaces is 157.9W and by means of radiation is 101.1W.

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Chapter 20 Solutions

Fundamentals of Thermal-Fluid Sciences

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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