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 27P
To determine

The thickness of the insulation that needed to be used.

The time taken for the insulation to pay for itself from the energy it saves.

Expert Solution & Answer
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Explanation of Solution

Given:

The height of the side surface of the cubic industrial surface (h) is 3m.

The outer surface temperature of the section (Ts) is 110°C.

The temperature of the furnace room including the surface (T) is 30°C.

The effectiveness of the outer surface of furnace (ε) is 0.7.

The thermal conductivity of the glass wool insulation (kins) is 0.038W/mK.

The effectiveness of reflective sheet (εr) is 0.2.

The reduction in heat loss (R) is 90%.

The efficiency of the furnace (η) is 78%.

The price of natural gas (P) is $1.10/therms.

The continuous operation time of furnace (Δt) is 1year.

The cost of the installation of insulation (Cinsu) is $550.

Calculation:

Calculate the film temperature (Tf) using the relation.

    Tf=Ts+T2=110°C+30°C2=70°C

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

Obtain the following values of different properties corresponding film temperature 70°C to the as follows:

k=0.02881W/mKPr=0.7177v=1.995×105m2/s

Calculate the volume expansion coefficient (β) using the relation.

    β=1T=1(70°C+273)K=1343K=0.002915K1

Calculate the characteristics length (Lc) is using the relation.

    Lc=h=3m

Calculate the Rayleigh number (Ra) using the relation.

    Ra=gβ(TsT)(Lc)3v2Pr=(9.81m/s2)(0.002915K1)(110°C30°C)(3m)3(1.995×105m2/s)2(0.7177)=(0.02859m/s2K)[(110°C+273)K(30°C+273)K](27m3)(1.995×105m2/s)2(0.7177)=1.114×1011

Calculate the Nusselt number (Nu) using the relation.

    Nu=[0.825+0.389(Ra)1/6[1+(0.492/Pr)9/16]8/27]2=[0.825+0.389(1.114×1011)1/6[1+(0.492/0.7177)9/16]8/27]2=[0.825+26.9831.19]2=552.24

Calculate the surface area of heat transfer (A) using the relation.

    A=4(L)2=4(3m)2=4(9m2)=36m2

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

    h=kLcNu=0.02881W/mK3m(552.24)=5.303W/m2K

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

    Q˙=hA(TsT)+εAσ(Ts4Tsurr4)=[(5.303W/m2K)(36m2)((110°C+273)K(25°C+273)K)+(0.7)(36m2)(5.67×108W/m2K4)[((110°C+273)K)4((25°C+273)K)4]]=(190.908W/K)(85K)+(1.428×106W/K4)(1.363×1010K4)=35690.8W

Calculate the rate of heat transfer after insulation (Q˙saved) using the relation.

    Q˙saved=RQ˙=(90%)(35690.8W)=(0.9)(35690.8W)=32121.72W

Calculate the rate of heat loss after insulation (Q˙loss) using the relation.

    Q˙loss=(1R)Q˙=(190%)(35690.8W)=(0.1)(35690.8W)=3569.08W

Calculate the energy saved by installing the insulation (Esaved) using the relation.

  Esaved=Q˙saved(Δt)=(32121.72W×103kJ/s1W)(1year×365day1year)=(32.12172kJ/s)(365day×86400s1days)=1.013×109kJ

Calculate the total money saved (Msaved) using the relation.

    Msaved=(Esaved)(P)=(1.013×109kJ×1therms105500kJ)($1.10/therms)=$10562.08

Means the money saved due to insulation in a year is $10562.08.

Calculate the time taken by the insulation to pay for the cost (tins) using the relation.

    tins=CinsMsaved=$550$10562.081/year×1year365days=19days

Thus, the time taken for the insulation to pay for itself from the energy it saves is 19days.

Insulation will reduce the temperature of the surface which reduce the Rayleigh number, Nusselt number, hence for the convection coefficient the temperature of the surface is consider as 50°C.

Calculate the film temperature (Tf) using the relation.

    Tf=Ts+T2=50°C+30°C2=80°C2=40°C

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

Obtain the following values of different properties corresponding film temperature 40°C to the as follows:

k=0.02662W/mKPr=0.7255v=1.702×105m2/s

Calculate the volume expansion coefficient (β) using the relation.

    β=1T=1(40°C+273)K=1313K=0.003195K1

Calculate the Rayleigh number (Ra) using the relation.

    Ra=gβ(TsT)(Lc)3v2Pr=(9.81m/s2)(0.003195K1)(50°C30°C)(3m)3(1.702×105m2/s)2(0.7255)=(0.03134m/s2K)[(50°C+273)K(30°C+273)K](27m3)(1.702×105m2/s)2(0.7255)=4.238×1010

Calculate the Nusselt number (Nu) using the relation.

    Nu=[0.825+0.389(Ra)1/6[1+(0.492/Pr)9/16]8/27]2=[0.825+0.389(4.238×1010)1/6[1+(0.492/0.7255)9/16]8/27]2=[0.825+22.9691.19]2=405.08

Calculate the surface area of heat transfer (A) using the relation.

    A=4(L)(L+2t)=4(3m)(3m+2t)=12m(3m+2t)

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

    h=kLcNu=0.02662W/mK3m(405.08)=3.594W/m2K

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

    Q˙=hA(TsT)+εAσ(Ts4Tsurr4)35690.8W=[(3.594W/m2K)A(Ts(30°C+273)K)+(0.7)A(5.67×108W/m2K4)[Ts4((30°C+273)K)4]]35690.8W=[(3.594W/m2K)A(Ts(30°C+273)K)+(3.969×108W/m2K)A[Ts4((30°C+273)K)4]]

The heat loss from the insulation by radiation and convection is equal to heat conduction through insulation.

Calculate the rate of heat transfer through insulation (Q˙cond) using the relation.

    Q˙cond=kinsA(Ts(furnace)Ts)t35690.8W=(0.038W/mK)A((110°C+273)KTst)

By use of trial and error method the surface temperature of insulation obtained is 41.2°C and thickness of the insulation is 2.84cm.

Thus, the time taken for the insulation to pay for itself from the energy it saves is 19days and thickness of the insulation is 2.84cm.

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Chapter 20 Solutions

Fundamentals of Thermal-Fluid Sciences

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