Question

In a high-performance computing system 100 CPU chips, each dissipating 25 W, are attached to one surface of a 200 mm by 200 mm copper heat spreader. The copper plate is cooled on its opposite surface by water, which is tripped at the leading edge of the copper so that the boundary layer is turbulent throughout. The plate may be assumed to be isothermal due to the high thermal conductivity of the copper. The water velocity and temperature are 2 m/s and 17˚C. What is the temperature of the copper plate?

Answer 1

Given Information:

Velocity of water flow = 2 m/s

Temperature of water = 17° C

Heat dissipation = 2500 W

Area of copper plate = 0.04 m²

Required Information:

Temperature of copper plate  = ?

Answer:

$T_{p} = 27$° $C$

Explanation:

Each chip dissipates 25 W so 100 chips will dissipate 25*100 = 2500 W

Area of copper plate = 0.2*0.2 = 0.04 m²

According to the convection rate equation

$T_{p}= T_{w} + \frac{q}{hA}$

Where Tp is the temperature of copper plate, Tw is the temperature of water, q is the the heat dissipation of chips, A is the area of copper plate and h is the convection coefficient

The convection coefficient is given by turbulent flow correlation

$h = Nu_{L}(k/L) =0.037Re_{L}^{4/5}P_{r}^{1/3}(k/L)$

Where Nu is Nusselt number, Re is Reynolds number, Pr = 5.2 is Prandtl number and k = 0.620 W/m.K

$Re_{L}= uL/v$

Where u = 2 m/s and L = 0.2 m and  v = 0.96x10⁻⁶m² /s

$Re_{L}= 2*0.2/0.96x10^{-6}$

$Re_{L}= 416666.66$

$h = 0.037(416666.66)^{4/5}(5.2)^{1/3}(0.620/0.2)$

$h = 6223.89$ $W/m^{2}K$

$T_{p}= 17 + \frac{2500}{(6223.89)0.04}$

$T_{p} = 27$° $C$

Therefore, the temperature of the copper plate is 27° C