Question

A 20 mm 3 20 mm silicon chip is mounted such thatthe edges are flush in a substrate. The substrate provides anunheated starting length of 20 mm that acts as turbulator. Airflow at 25°C (1 atm) with a velocity of 25 m/s is used to coolthe upper surface of the chip. If the maximum surface temperature of the chip cannot exceed 75°C, determine the maximumallowable power dissipation on the chip surface.

Answer 1

q= 1.77 W

Step-by-step explanation:

Given data, Dimensions = 20mm x 20mm, Unheated starting length (ε) = 20 mm, Air flow temperature = 25 C, Velocity of flow = 25m/s, Surface Temperature = 75 C

To find maximum allowed power dissipated use the formula Q = hA(ΔT), where Q = Maximum allowed power dissipated from surface , h = heat transfer coefficiant, A = Area of Surface, ΔT = Temperature difference between surface and surrounding

we need to find h, which is given by (Nu x K)/x, where Nu is the Nusselt's Number, k is the thermal conductivity at film temperature and x is the length of the substrate.

For Nu use the Churchill and Ozoe relation used for parallel flow over the flat plate , Nu = (Nux (ε=0))/[1-(ε/x)^3/4]^1/3

Nux (ε=0) = 0.453 x Re^0.5 x Pr^1/3, where Re is the Reynolds number calculated by [Rex = Vx/v, where V is the velocity and v is the kinematic velocity, x being the length of the substrate] and Pr being the Prandtl Number

The constants, namel Pr, k and v are temperature dependent, so we need to find the film temperature

Tfilm = (Tsubstrate + Tmax)/2 = (25 + 75)/2 = 50 C

At 50 C, Pr = 0.7228, k = 0.02735w/m.k , v = 1.798 x 10^-5 m2/s

First find Rex and keep using the value in the subsequent formulas until we reach Q

Rex = Vx/v = 25 x (0.02 + 0.02)/ 1.798 x 10^-5 = 55617.35 < 5 x 10^5, thus flow is laminar, (x = L + ε)

Nux = (Nux (ε=0))/[1-(ε/x)^3/4]^1/3 = 0.453 x Re^0.5 x Pr^1/3/[1-(ε/x)^3/4]^1/3

= 0.453 x 55617.35 ^0.5 x 0.7228^1/3/ [1 - (0.02/0.04)^3/4]^1/3 = 129.54

h = (Nu x K)/x = (129.54 x 0.02735)/0.04 = 88.75W/m2.k

Q = 88.75 x (0.02)^2 x (75-25) = 1.77 W