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 130RQ
To determine

The increase in heat transfer from the tube per meter of its length.

Expert Solution & Answer
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Explanation of Solution

Given:

Thermal conductivity of the fins (k) is 180 W/mK.

Heat transfer coefficient (h) is 60 W/m2K.

Space between the fins (s) is 3 mm.

Outer diameter of the tube (D1) is 3 cm.

Outer diameter of the fin (D2) is 6 cm.

Number of fins (n) is 200.

Calculation:

Determine the surface area in case of no fins.

  Ano fin =πD1L=π(0.03 m)(1 m)=0.0942 m2

Determine the heat transfer in case of no fins.

  Q˙no fin =hAno fin (TbT)=(60 W/m2°C)(0.0942 m2)(120°C25°C)=537 W

Determine the length.

  L=(D2D1)2=(0.06 m0.03 m)2=0.015 m

Determine the following factor.

  r2+(t2)r1=0.03 m+(0.002 m2)0.015 m=2.07

Determine the following factor.

  Lc3/2(hkAp)1/2=(L+t2)hkt=(0.015 m+0.002 m2)60 W/m2°C(180 W/m°C)(0.002 m)=0.207 m

Obtain the efficiency of these circular fins from the efficiency curve of Figure 17-44 as 0.96.

Determine the area of fin.

  Afin=2π(r22r12)+2πr2t=2π((0.03 m)2(0.015 m)2)+2π(0.03 m)(0.002 m)=0.004624 m2

Determine the heat transfer from a single fin.

  Q˙fin =ηfin Q˙fin, max =ηfin hAfin (TbT)=0.96(60 W/m2°C)(0.004624 m2)(120°C25°C)=25.3 W

Determine the surface area of a single unfinned portion of the tube.

  Aunfin =πD1s=π(0.03 m)(0.003 m)=0.000283 m2

Determine the heat transfer from a single unfinned portion of the tube.

  Q˙unfin =hAunfin (TbT)=(60 W/m2°C)(0.000283 m2)(120°C25°C)=1.6 W

Determine the total heat transfer from the finned tube.

  Q˙total fin =n(Q˙fin +Q˙unfin )=200(25.3 W+1.6 W)=5380 W

Determine the increase in heat transfer from the tube.

  Q˙increase =Q˙total fin Q˙no fin =5380 W537 W=4843 W

Thus, the increase in heat transfer from the tube is 4843 W_.

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Chapter 17 Solutions

Fundamentals of Thermal-Fluid Sciences

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