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 129RQ
To determine

The heat transfer rate from a single fin and the increase in the rate of heat transfer.

Expert Solution & Answer
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Explanation of Solution

Given:

Thermal conductivity of the fins (k) is 230 W/mK.

Heat transfer coefficient (h) is 45 W/m2K.

Length of the fin (L) is 25 mm.

Diameter of the fin (D) is 4 mm.

Calculation:

Determine the value of mL.

  mL=4hkDL=4(45 W/m2°C)(230 W/m°C)(0.004 m)(0.025 m)=0.3497

Determine the area of fin.

  Afin =πD496L2{[16(LD)2+1]3/21}=π(0.004 m)496(0.025 m)2{[16(0.025 m0.004 m)2+1]3/21}=2.099×104m2

Determine the efficiency of fin.

  ηfin =32mLI1(43mL)I0(43mL)ηfin =32(0.3497)I1(43(0.3497))I0(43(0.3497))ηfin =0.9738

Determine the heat transfer rate from a single fin.

  Q˙fin=ηfinhAfin(TbT)=(0.9738)(45 W/m2°C)(2.099×104m2)(200°C25°C)=1.61 W

Thus, the heat transfer rate from a single fin is 1.61 W_.

Determine the heat transfer rate for 100 fins.

  Q˙fin total =(100)(1.61 W)=161 W

Determine the surface area of the unfinned portion.

  Aunfin =1 m2100(π4D2)=1100(π4(0.004 m)2)=0.9987 m2

Determine the heat transfer from the unfinned portion.

  Q˙unfin =hAunfin (TbT)=(0.9987 m2)(45 W/m2°C)(200°C25°C)=7865W

Determine the total heat transfer from the surface.

  Q˙total =Q˙fin total +Q˙unfin =161 W+7865 W=8026 W

Determine the heat transfer rate if there was no fin at the surface.

  Q˙nofin =hAunfin (TbT)=(1 m2)(45 W/m2°C)(200°C25°C)=7875 W

Determine the increase in heat transfer rate.

  Q˙increase =Q˙total Q˙nofin =8026 W7875 W=151 W

Thus, the increase in the rate of heat transfer is 151 W_.

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Chapter 17 Solutions

Fundamentals of Thermal-Fluid Sciences

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