Question

An aluminum heat sink (k = 240 W/m-K) used to cool an array of electronic chips consists of a square channel of inner width w = 30 mm, through which liquid flow may be assumed to maintain a uniform surface temperature of T1 = 25 °C. The outer width and length of the channel are W = 50 mm and L = 200 mm, respectively.

If N = 100 chips attached to the outer surfaces of the heat sink maintain an approximately uniform surface temperature of T2 = 60 °C and all of the heat dissipated by the chips is assumed to be transferred to the coolant, what is the heat dissipation per chip? If the contact resistance between each chip and the heat sink is R12 = 0.2 K/W, what is the chip temperature?

Answer 1

The answers to the question are;

The heat dissipation per chip is 84 W

The chip temperature is 76.8 ° C

Step-by-step explanation:

T1 = 25 °

T2 = 60°

ka = 240 W/(m×k)

The formula for heat transferred by conduction is given by

Q = (ka×A×(T1-T2))/La

Area A is given by

A = Length×Width = 50 mm × 200 mm = 10000 mm^2 = 0.01 m^2

La = (Outer width - Inner width)/2 = (50 mm - 30 mm)/2

= 10 mm = 0.01 mm

Therefore Q = (240×0.01×(60-25))/0.01 = 8400 W

Heat dissipated per chip = Q/N = 8400 W/100

= 84 W

Contact resistance is given by

Rcontact = (T3-T2)/(Q/A)

Where Q/A is the heat dissipated per each chip

Therefore

0.2 K/W = (T3-60)K/84

Or 84×0.2 = 16.8 = T3 - 60

T3 = 60 + 16.8 = 76.8 ° C

The chip temperature = 76.8 ° C

Answer 2

Explanation: see attachment below