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 113P

(a)

To determine

The temperature on the two sides of the circuit board.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

Thickness of the board is 0.3 cm.

The height of the board is 12 cm.

The length of the board is 18 cm.

Number of logic chips is 80.

Heat dissipated by each chip is 0.04W.

The thermal conductivity of the board is 30W/mK.

Temperature of the medium is 40°C.

The heat transfer coefficient is 40W/m2K.

Thickness of the aluminum plate is 0.2 cm.

The height of the aluminum plate is 12 cm.

The length of the aluminum plate is 18 cm.

The thermal conductivity of the aluminum plate is 237W/mK.

Number of aluminum fins is 864.

Length of the fins is 2 cm.

Diameter of the fin is 0.25 cm.

Thickness of the epoxy adhesive is 0.02 cm.

Thermal conductivity of the epoxy is 1.8W/mK.

Calculation:

The total heat transfer rate by the chips is,

  Q˙=80(0.04W)=3.2 W

The total thermal resistance is,

  Rtotal=Rboard+Rconv=LkA+1hA=0.003 m(30W/mK)(0.12 m×0.18 m)+1(40W/m2K)(0.12 m×0.18 m)=1.1620°C/W

Calculate the temperature on the two sides of the board.

  Q˙=T1T2Rtotal3.2W=T140°C1.1620°C/WT1=43.72°C

Similarly,

  Q˙=T1T2Rboard3.2W=43.72°CT20.003 m(30W/mK)(0.12 m×0.18 m)T2=43.71°C

Thus, the temperature on the two sides of the circuit board is 43.72°C and 43.71°C.

(b)

To determine

The temperature on the two sides of the circuit board after the aluminum plate is attached.

(b)

Expert Solution
Check Mark

Explanation of Solution

Calculation:

Calculate the value of m.

  m=hpkAc=4hkD=4(40W/m2K)(237W/mK)(0.0025 m)=16.43 m1

The efficiency of the fin is,

  ηfin=tanhmLmL=tanh(16.43×0.02 m)(16.43×0.02 m)=0.9655

Calculate the total area with fins,

  Atotal=Afinned+Aunfinned=ηfinnπDL+(0.12 m×0.18 m)864πD24=(0.9655)(864)π(0.0025 m)(0.02 m)+0.0216 m2864π(0.0025 m)24=0.1484 m2

The total thermal resistance is,

  Rtotal=Rboard+Rconv+Repoxy+Raluminum=(LkA)board+1hAtotal+(LkA)epoxy+(LkA)aluminum=0.003 m(30W/mK)(0.12 m×0.18 m)+1(40W/m2K)(0.1484 m2)+0.0002 m(1.8W/mK)(0.12 m×0.18 m)+0.002 m(237W/mK)(0.12 m×0.18 m)=0.1787°C/W

Calculate the temperature on the two sides of the board.

  Q˙=T1T2Rtotal3.2W=T140°C0.1787°C/WT1=40.57°C

Similarly,

  Q˙=T1T2Rboard3.2W=40.57°CT20.003 m(30W/mK)(0.12 m×0.18 m)T2=40.56°C

Thus, the temperature on the two side of the circuit board after the aluminum plate is attached is 40.57°C and 40.56°C.

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Chapter 17 Solutions

Fundamentals of Thermal-Fluid Sciences

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