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 131RQ

(a)

To determine

The surface temperature on the two sides of the circuit board.

(a)

Expert Solution
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Explanation of Solution

Given:

Thermal conductivity (k) is 12 W/mK.

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

Thickness of the circuit board (t) is 0.2 cm.

Height of the circuit board (H) is 10 cm.

Length of the circuit board (L) is 15 cm.

Calculation:

Determine the thermal resistance of the board.

  Rboard =LkA=0.002 m(12 W/m°C)(0.1 m)(0.15 m)=0.011 °C/W

Determine the thermal resistance of the convection.

  Rconv=1hA=1(45 W/m°C)(0.1 m)(0.15 m)=1.481 °C/W

Determine the total thermal resistance.

  Rtotal =Rboard +Rconv =0.011 °C/W+1.481 °C/W=1.492 °C/W

Determine the surface temperature on one side.

  Q˙=T1TRtotal T1=T+Q˙Rtotal =37°C+(15 W)(1.492 °C/W)=59.4°C

Thus, the surface temperature on one side of the circuit board is 59.4°C_.

Determine the surface temperature on the other side.

  Q˙=T1T2Rboard T2=T1Q˙Rbourd =59.4°C(15 W)(0.011 °C/W)=59.2°C

Thus, the surface temperature on the other side of the circuit board is 59.2°C_.

(b)

To determine

The new temperature on the two sides of the circuit board.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

Thermal conductivity of the aluminum plate (k) is 237 W/mK.

Thickness of the aluminum plate (t) is 0.1 cm.

Height of the aluminum plate (H) is 10 cm.

Length of the aluminum plate (L) is 15 cm.

Thermal conductivity of epoxy adhesive (k) is 1.8 W/mK.

Calculation:

Determine the value of m.

  m=hpkAch(2w)k(tv)=2hkt=2(45 W/m2°C)(237 W/m°C)(0.002 m)

  m=13.78m1

Determine the efficiency of fin.

  ηfin =tanhmLmL=tanh(13.78 m1×0.02 m)13.78 m1×0.02 m=0.975

Determine the finned surface area.

  Afinned =(20)2w(L+t2)=20×2(0.15 m)(0.02 m+0.002 m2)=0.126 m2

Determine the unfinned surface area.

  Aunfinned =(0.1 m)(0.15 m)20(0.002 m)(0.15 m)=0.0090 m2

Determine the expression of finned heat transfer.

  Q˙finned =ηfin, max=ηfin hAfin (Tbase T)

Determine the expression of unfinned heat transfer.

  Q˙unfinned =hAunfinned (Tbase T)

Determine the expression of total heat transfer.

  Q˙total =Q˙unfinned +Q˙finned =h(Tbase T)(ηfin Afin +Aunfinned )

Determine the base temperature of the finned surface.

  Tbase=T+Q˙totalh(ηfinAfin+Aunfinmed)=37°C+15 W(45 W/m2°C)[(0.975)(0.126 m2)+(0.0090 m2)]=39.5°C

Determine the thermal resistance of aluminum surface.

  Raluminum =LkA=0.001 m(237 W/m°C)(0.1 m)(0.15 m)=0.00028 °C/W

Determine the thermal resistance of epoxy surface.

  Repoxy =LkA=0.0003 m(1.8 W/m°C)(0.1 m)(0.15 m)=0.01111 °C/W

Determine the new temperature on one side of the circuit board.

  Q˙=T1TbaseRaluminum +Repoxy +Rboard 15 W=(T139.5°C)0.00028°C/W+0.01111°C/W+0.011°C/WT1=39.8°C

Thus, the new temperature on one side of the circuit board is 39.8°C_.

Determine the new temperature on the other side of the circuit board.

  Q˙=T1T2Rboard T2=T1Q˙Rboard T2=39.8°C(15 W)(0.011°C/W)T2=39.6°C

Thus, the new temperature on the other side of the circuit board is 39.6°C_.

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Chapter 17 Solutions

Fundamentals of Thermal-Fluid Sciences

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