Question

A square (12.5 mm × 12.5 mm) silicon chip is insulated on one side and cooled on the opposite side by atmospheric air in parallel flow at [infinity] 20 m/s and [infinity] 24°C. When in use, electrical power dissipation within the chip maintains a uniform heat flux at the cooled surface. Assume the critical Reynolds number is 5x105.
(a) If the chip temperature may not exceed 80°C at any point on its surface, what is the maximum allowable power?
(b) What is the maximum allowable power if the chip is flush mounted in a substrate that provides for an unheated starting length of 20 mm?

Answer 1

Answer:

a) q = 0.927 W

b) q = 0.73 W

Explanation:

Our assumptions are that:

• there is a steady state condition
• No heat loss during insulation
• No radiation
• Heat flux is uniform

$u_{\infty} = 20 m/s$

$T_{\infty} = 24^{0} C$

Chip temperature, $T_{c} = 80^{0} C$

Temperature of the film, $T_{f} = \frac{T_{\infty}+ T_{c} }{2}$

$T_{f} = \frac{24 + 80}{2} \\T_{f} = 52^{0} C$

$T_{f} = 52 + 273 K \\T_{f} = 325 K$

Atmospheric air is used to insulate the chip. i.e the film is an atmospheric air

1 atm of 325 K of air has the following properties:

viscosity, v = 18.4 × 10⁻⁶ m²/s

k = 0.0282 w/mk

Pr = 0.703

Nuselt number, Nu = $\frac{hL }{k}$..............(i)

Nu = $0.453Re^{0.5} Pr^{0.33}$................(2)

The Reynold number, Re, is calculated by the equation

$Re = \frac{u_{\infty L} }{v}$, L = 12.5 mm = 12.5 * 10⁻³ m

Re = $\frac{20 * 12.5 * 10^{-3} }{18.4 * 10^{-6} }$

Re = 13586.96

Equating (1) and (2) and substituting necessary values

$0.453* 13586.96^{0.5} 0.703^{0.33} = \frac{h * 12.5 * 10^{-3} }{0.0282}$ ..........(3)

47.006 = h * 0.4433

h = 47.006/0.4433

h = 106.05

The maximum allowable power is given by the equation, $q = hA(T_{c}-T_{\infty} )$

Area, A = length * Breadth = 0.0125*0.0125

A = 0.000156 m²

$q = 106.05*0.000156(80-24)\\q = 0.927 W$

b) When the lenght, L = 20 mm = 0.02 m

Reynold number, Re = $\frac{20 * 0.02 }{18.4 * 10^{-6} }$

Re = 21739.13

By replacing Re and L, Equation (3) becomes:

$0.453* 21739.13^{0.5} 0.703^{0.33} = \frac{h * 0.02 }{0.0282}\\59.46 = 0.709h\\h = 59.46/0.709\\h = 83.84$

$q = hA(T_{c}-T_{\infty} )\\q = 83.84*0.000156(80-24 )$

$q = 0.73 W$