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 90RQ

(a)

To determine

The heat transfer from the outer surface due to convection.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The power (Q˙) of the system is 180W.

The dimensions of the box are 6in×6in.

The ambient temperature (Ta) is 85°F.

The outlet temperature (To) is 100°F.

The flow rate (ε) of the box is 22cfm.

Calculation:

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

Obtain the following properties of air corresponding to the temperature of 100°F.

k=0.015Btu/hftRv=0.1809×105ft2/sPr=0.726ρ=0.072lbm/ft3cp=0.2404Btu/lbmR

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

    β=1Tm=[1(100°F+460)R]=0.0017R1

Calculate the mass flow rate (m˙) using the relation.

    m˙=ρV˙=(0.072lbm/ft3)(22ft3/min)=1.602lbm/min

Calculate the heat transfer (Q˙forced) using the relation.

    Q˙forced=m˙cp(ToutTin)=(1.602lbm/min(60min1h))(0.2404Btu/lbmR)((100°F+460)R(85°F+460)R)=346.6Btu/h

Calculate the heat transfer (Q˙conv) using the relation.

    Q˙conv=Q˙totalQ˙forced=(180×3.412)Btu/h(346.6Btu/h)=268Btu/h

Thus, the heat transfer through convection is 268Btu/h.

(b)

To determine

The average temperature of the duct.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

Calculation:

Calculate the characteristic length (Lc) using the relation.

    Lc=Asp=((4ft)×(6in×1ft12in))2((4ft)+(6in×1ft12in))=0.22ft

Assume surface temperature to be 120°F

Calculate the Rayleigh number (Ra) using the relation.

  Ra= ( gβ(TsTa)Lc3v2Pr)=[(32.2ft/s2)(0.0017R1)((120°F+460)R(80°F+460)R)(0.22ft)3(0.18×103ft2/s)2(0.726)]=5.599×105

Calculate the nusselt number (Nu) using the relation.

    Nu=0.54RaL1/4=0.54(5.559×105)1/4=14.77

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

    h=kLcNu=0.015Btu/hftR0.22ft(14.77)=1.016Btu/hft2R

Calculate the surface area of top (Atop) using the relation.

    Atop=((4ft)×(6in×1ft12in))=2ft2

Calculate the Nusselt number (Nu) using the relation.

    Nu=0.27RaL1/4=0.27(5.559×105)1/4=7.386

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

    h=kLcNu=0.015Btu/hftR0.22ft(7.386)=0.5082Btu/hft2R

Calculate for vertical side surface:

Calculate the Rayleigh number (Ra) using the relation.

  Ra= ( gβ(TsTa)Lc3v2Pr)=[(32.2ft/s2)(0.0017R1)((120°F+460)R(80°F+460)R)((6in×1ft12in))3(0.18×103ft2/s)2(0.726)]=6.379×106

Calculate the Nusselt number (Nu) using the relation.

    Nu=[0.825+0.387Ra1/6[1+(0.492Pr)9/16]8/27]2=[0.825+0.387(6.379×106)1/6[1+(0.4920.726)9/16]8/27]2=27.57

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

    h=kLcNu=0.015Btu/hftR(6in×1ft12in)(27.57)=0.843Btu/hft2R

Calculate the surface Area (Aside) using the relation.

    Aside=2((4ft)×(6in×1ft12in))=4ft2

Calculate the surface temperature (Ts) using the relation

    Q=Q˙top+Q˙bottom+Q˙side=[(hA)top+((hA)bottom+(hA)side)][TsTa]268Btu/h=[(1.016Btu/hft2R×2ft2)+(0.508Btu/hft2R×4ft2)+(0.843Btu/hft2R×4ft2)][Ts(80°F+460)R]Ts=(581.7R460)°F=121.7°F

Thus the surface temperature is 121.7°F.

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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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