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 19, Problem 55P
To determine

The amount of heat lost from the steam and the amount of money this facility will save a year as a result of insulating the steam pipes.

Expert Solution & Answer
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Explanation of Solution

Given:

The diameter of the section (D) is 10cm.

The length of the section (L) is 12m.

The temperature of wind (T) is 5°C.

The temperature of surface (Ts) is 75°C.

The emissivity  (ε) is 0.8.

The time of heat transfer (ΔT)  is 10h/day.

The average temperature of the surfaces (Ta) is 0°C.

The efficiency of steam generator (η) is 0.8.

The Unit cost of energy is (U) is $1.05/therm.

The velocity of air is (V) is 10km/hr.

Calculation:

Calculate the film temperature (Tf) using the relation.

  Tf=Ts+T2=75°C+5°C2=40°C

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

Obtain the following properties of air corresponding to the temperature of 40°C as follows:

k=0.02662W/mKPr=0.7255v=1.702×105m2/s

Calculate the Reynolds number (Re) using the relation.

  Re=VDv=(10km/hr×1000m1km×1hr3600s)(10cm×1m100cm)1.702×105m2/s=1.632×104

Calculate the Nusselt number (Nu) using the relation.

    Nu=0.3+0.62Re0.5Pr1/3[1+(0.4/Pr)2/3]1/4[1+(Re282000)5/8]4/5=0.3+0.62(1.632×104)0.5(0.7255)1/3[1+(0.4/0.7255)2/3]1/4[1+(1.632×104282000)5/8]4/5=71.19

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

    h=kD(Nu)=0.02662W/mK(10cm×1m100cm)(71.19)=18.95W/m2K

Calculate the heat loss by convection (qc) using the relation.

    qc=hA(TsT)=hπDL(TsT)=18.95W/m2K×π(10cm×1m100cm)(12m)((75°C+273)K(5°C+273)K)=5000.78W

Calculate the heat loss by radiation (qr) using the relation.

    qc=εAσ(Ts4Ta4)=επDLσ(Ts4Ta4)=(0.8)π(10cm×1m100cm)(12m)(5.67×108W/m2K4)((75°C+273)4K4(0°C+273)4K4)=1558.11W

Calculate the total heat loss (Q) using the relation.

    Q=qc+qr=5001.78W+1558.11W=6559.89W

Calculate the amount of heat loss from the steam during a 10-hour work day using the relation.

    Qd=Q×ΔT=6559.89W(1J/s1W)(10h/day×3600s/hr)=2.361×108J/day

Calculate the total amount of heat (QT)  loss from the steam per year using the relation.

    QT=Qd(no.ofdays)=2.361×108J/day(365days/yr)=8.618×1010J/yr=8.618×1010J/yr(1kJ1000J)=8.618×107kJ/yr

Calculate the amount of gas (G) used using the relation.

    G=QTη=8.618×107kJ/yr0.80(1therm105500kJ)=1021.09therm/yr

Calculate the money saved (M) using the relation.

    M=(0.9)(G)(U)=0.9(1021therm/yr)($1.05/therm)=$964.93

Thus, The amount of heat lost from the steam is 2.361×108J/day and the amount of money this facility will save a year as a result of insulating the steam pipes is $964.93.

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Chapter 19 Solutions

Fundamentals of Thermal-Fluid Sciences

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