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 103RQ
To determine

The temperature of the aluminum tube when equilibrium is established.

Expert Solution & Answer
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Explanation of Solution

Given:

The inner diameter of tank (di) is 5cm.

The outer diameter of tank (do) is 7cm.

The solar radiation absorbed (Q) is 20W.

The temperature of air is (T) is 30°C.

The surface temperature is (Ts) is 33°C.

Calculation:

Calculate the film temperature (Tf) using the relation.

  Tf=Ts+T2=30°C+33°C2=31.5°C

Calculate Characteristic length (Lc) using the relation.

    Lc=do=7cm×1m100cm=0.07m

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

Obtain the following properties of air corresponding to the temperature of 31.5°C as follows:

  k=0.02599W/mKv=1.622×105m2/sPr=0.7278

Calculate the Rayleigh number (Ra) using the relation.

  Ra=gβ(TsT)Lc3v2Pr=g(1Tf)(TsT)d3v2Pr=[(9.81m/s2)(1(31.5°C+273)K)((33°C+273)K(30°C+273)K)(7cm×1m100cm)3](1.622×105m2/s)2(0.7278)=91700

Calculate the Nusselt number (Nu) using the relation.

    Nu=[0.6+0.387Ra1/6[1+(0.559Pr)9/16]8/27]2=[0.6+0.387(91700)1/6[1+(0.5590.7278)9/16]8/27]2=7.626

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

    h=kdoNu=0.02599W/mK(7cm×1m100cm)(7.626)=2.832W/m2K

Calculate rate of heat transfer by convection (Qc) using the relation.

  Q˙c=hA(TsT)=(h)(πdoL)(TsT)=(2.832W/m2K)(π(7cm×1m100cm)(1m))[(Ts°C+273)K(30°C+273)K]=(2.832W/m2K)(0.2199m2)(Ts30)K

Calculate rate of heat transfer by radiation (Qr) using the relation.

    Q˙r=εAsσ(Ts4Tsurr4)=ε(πdoL)σ(Ts4Tsurr4)=[(1)(π(7cm×1m100cm)(1m))(5.67×108W/m2K4)×[((Ts+273)K)4((20+273)K)4]]

Calculate total rate of heat transfer by radiation (Q) using the relation.

    Q˙=Q˙c+Q˙rQ˙=[(2.832W/m2K)(0.2199m2)(Tg(30°C+273)K)+{(1)(π(7cm×1m100cm)(1m))×(5.67×108W/m2K)×((Tg+273K)4(20+273K)4)}]20W=[(2.832W/m2K)(0.2199m2)(Ts(30°C+273)K)+{(1)(π(7cm×1m100cm)(1m))×(5.67×108W/m2K4)×((Ts+273K)4(20+273K)4)}]Ts=(306.34K273)°CTs=33.34°C

Assume the aluminum tube temperature to be 45°C.

Calculate the film temperature (Tf) using the relation.

  Tf=Ts+T2=45°C+33.34°C2=39.17°C

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

Obtain the following properties of air corresponding to the temperature of 39.17°C as follows:

  k=0.02656W/mKv=1.694×105m2/sPr=0.7257

Calculate Characteristic length (Lc) using the relation.

    Lc=d=d0di2=(75)cm2=1cm

Calculate the Rayleigh number (Ra) using the relation.

  Ra=gβ(TsT)Lc3v2Pr=g(1Tf)(TsT)d3v2Pr=[(9.81m/s2)(1(39.17°C+273)K)((45+273)°K(33.34+273)°K)(1cm×1m100cm)3](1.694×105m2/s)2(0.7257)=926.5

Calculate geometric factor for concentric cylinders (Fcyl) using the relation.

    Fcyl=[ln(dodi)]4Lc3(di3/5+do3/5)5=[ln(7cm×1m100cm5cm×1m100cm)]4(1cm×1m100cm)3((5cm×1m100cm)3/5+(9cm×1m100cm)3/5)5=0.08085

Calculate effective thermal conductivity (keff) using the relation.

    keff=0.386k(Pr0.861+Pr)1/4(FcylRa)1/4=0.368(0.02706W/mK)(0.72380.861+0.7238)1/4((0.08085)(926.5))1/4=0.02480W/mK

Calculate the heat loss from the collector per meter length of the tube (Q) using the relation.

    Q˙c=2πkeffln(dodi)(TiTo)=2πkeffln(dodi)(TiTo)=2π(0.02480W/mK)[(Ttube°C+273)K(33.34°C+273)K]ln(7cm×1m100cm5cm×1m100cm)=2π(0.02480W/mK)(Ttube33.34K)ln(0.07m0.05m)

Calculate rate of heat transfer by radiation (Qr) using the relation.

    Q˙r=εAsσ(Ts4Tsurr4)=ε(πdiL)σ(Ts4Tsurr4)=[(1)(π(5cm×1m100cm)(1m))(5.67×108W/m2K4)×[((Ttube°C+273)K)4((33.34°C+273)K)4]]

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

    Q˙=Q˙c+Q˙r20W=[2π(0.02480W/m°C)(Ttube33.34K)ln(0.07m0.05m)+(1)(0.1517m2)(5.67×108W/m2K4)((Ttube+273)4K4+(33.34°C+273)4K4)]Ttube=(319.3K273)°CTtube=46.3°C

Thus, the temperature of the aluminum tube when equilibrium is established is 46.3°C.

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Chapter 20 Solutions

Fundamentals of Thermal-Fluid Sciences

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