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 17, Problem 111P

(a)

To determine

The average surface temperature of the pipe disregarding the flanges.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

Length of the pipe is 3 m.

Thickness of the pipe is 0.4 cm.

Thermal conductivity of the pipe is 52W/mK.

Outer diameter of the pipes is 10 cm.

Thickness of the flanges is 1 cm.

Outer diameter of the flange is 20 cm.

Temperature of steam is 200°C.

The heat transfer coefficient of the inside is 180W/m2K.

Temperature of ambient air is 12°C.

The heat transfer coefficient of the outside is 25W/m2K.

Calculation:

The total thermal resistance is,

  Rtotal=Ri+Rcond+Ro=1hiAi+ln(r2/r1)2πkL+1hoAo=1(180W/m2K)(π×0.092 m×6 m)+ln(5/4.6)2π(52W/mK)(6 m)+1(25W/m2K)(π×0.1 m×6 m)=0.02447°C/W

The rate of heat transfer is,

  Q˙=T1T2Rtotal=(20012)°C0.02447°C/W=7684 W

Calculate the surface temperature of the pipe.

  Q˙=T2T2Ro7684 W=(T212)°C1(25W/m2K)(π×0.1 m×6 m)T2=175°C

Thus, the average surface temperature of the pipe disregarding the flanges is 175°C.

(b)

To determine

The efficiency of the fin and the rate of heat transfer from the flanges.

(b)

Expert Solution
Check Mark

Explanation of Solution

Calculation:

Calculate the value of r2cr1.

  r2cr1=r2+t2r1=0.1+0.0220.05=2.2

Calculate the value of ε.

  ε=Lc3/2hkAp=(L+t2)3/2hkt=(0.05 m+0.022 m)3/225W/m2K(52W/mK)(0.02 m)=0.29

Obtain the efficiency of the fin from Figure 17 – 44 corresponding to the above values as ηfin=0.93

Thus, the efficiency of the fin is 0.93.

Calculate the heat transfer rate from the flanges.

  Q˙finned=ηfinQ˙max=ηfinhAfin(TbT)=ηfinh[2π(r22r12)+2πr2t](TbT)=(0.93)(25W/m2K)2π[(0.1m)2(0.05m)2+(0.1 m)(0.02 m)](17512)K=226W

Thus, the heat transfer rate from the flanges is 226W.

(c)

To determine

The length of the pipe equivalent to the flanges for heat transfer purpose.

(c)

Expert Solution
Check Mark

Explanation of Solution

Calculation:

The heat transfer rate per unit length of the pipe is,

  7684 W6 m=1271 W/m

The equivalent length of the pipe is,

  Leqv=226 W1281W/m=0.176 m=17.6 cm

Thus, the length of the pipe equivalent to the flanges for heat transfer purpose is 17.6 cm.

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Chapter 17 Solutions

Fundamentals of Thermal-Fluid Sciences

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