Question

A square, thin electronic chip is mounted to a substrate that is installed in a computer box whose interior walls are maintained at Tr as shown. The chip is heated by an electrical circuit with power input Pelec and cooled by convection and radiation to the surrounding air and walls. The maximum allowable chip surface temperature is 85 °C. Tsur 25°C Substrate Air rad Chip, Pelec T 85°C, E 0.60 L-15 mm T=25°c h-4.2(T,-Ta)4 or qconv h=250 W/m2-K a) Write a detailed energy balance equation for the steady-state chip that can be used to calculate the power input Pae to the chip. Expand each term in the energy balance using the variables defined in the diagram. h 4.2(T-To)4 , b) If the chip is cooled by natural convection where find the power input Patec to the chip. c) If a cooling fan within the computer box is now used so that the forced convection with h= 250 W/m2K cools the chip, find the power input Pate to the chip

Answer 1

Since we are given that the heat transfer only takes place by modes of convection and radiation, the total heat emitted from the chip will be partially carried away by the air and partially emitted by the surface as radiation, the walls of the interior will have no part to play in these heat transfer interactions, as there is no conduction heat transfer, and also anyways the walls are already at a temperature same as that of the ambient air.

Considering this background, the energy balance equation would be,

Energy coming into the chip = energy going out of the chip in the form of convection + energy going out of the chip in the form of radiation

i.e. P = hA∆T + eSA(Ts)^4

where,

∆T = Ts - Tair , Ts - surface temperature of chip and Tair - ambient temperature

h - convective heat transfer coefficient, A - surface area of the chip exposed to air

e - emissivity of the chip surface, S - Stephan Boltzmann constant (5.67x10^-8)

Since, emissivity is not given, we assume it to be 1 . Also we assume that only one surface of the square chip is exposed to the air.

A = (0.15 x 0.15) x 10^-6 = 0.225 X 10^-6 m^2

h = 4.2(Ts - Tair)^(1/4) + 250, if both natural and forced cooling are taking place. If only anyone is taking place then we can simply ignore the other in the formula for h.

a)So, the energy balance equation will be ( assuming cooling is taking place by natural as well as forced convection)

P= [4.2(Ts - Tair)^(1/4) + 250] x (0.225 x 10^-6) x (Ts - Tair) + (1)(5.67 x 10^-8)(A)(Ts)^4

b) Considering only natural convection,

P = [4.2(Ts - Tair)^(1/4)] x (0.225 x 10^-6) x (Ts - Tair) + (5.67 X 10^-8)(A)(Ts)^4

Putting, Ts = 85 + 273 = 358 K and Tair = 25 + 273 = 298 K we get,

P = 0.1578 x 10^-3 + 0.2096 x 10^-3 = 0.3674 x 10^-3 Watts

c) Considering both natural as well as forced convection

P = (250 x 0.225 x 10^-6) + 0.3674 x 10^-3 = 0.4237 x 10^-3 Watts