Question

Determine the conduction heat transfer through an air layer held between two 10 mm × 10 mm parallel aluminum plates. The plates are at temperatures Ts,1 = 305 K and   Ts, 2 = 295 K, respectively, and the air is at atmospheric pressure. Determine the conduction heat rate for plate spacings of L = 0.8 mm, L = 1.2 μm, and L = 20 nm. Assume a thermal accommodation coefficient of αt = 0.92. For air, the molecular weight is 28.97 kg/kmol and the molecular diameter is 0.372 nm. Determine the conduction heat rate for L= 0.8 mm, in W. q1, Determine the conduction heat rate for L= 1.2 μm, in W q2, and Determine the conduction heat rate for L= 20 nm, in W q3.

Answer 1

A) 0.0325 W

B)21.667 W

C)1300 W

Step-by-step explanation:

A)

THERMAL CONDUCTIVITY OF AIR IS K 0.026 w/m-k

Thermal resistance is given as

`Rth = (L)/(KA)`

and
`\Delta T = Q_1 * Rth`

therefore we have

`305 - 295 = q_1 * (0.8* 10^(-3))/(0.026* 10* 10 * 10^(-6))`

`q_1 = 0.0325 w`

B) WHEN
`L = 1.2 \mu m`

therefore we have

`305 - 295 = q_2 * (1.2* 10^(-6))/(0.026* 10* 10 * 10^(-6))`

`q_2 = 21.667 w`

C) When L is 20 nm

therefore we have
`305 - 295 = q_1 * (20* 10^(-9))/(0.026* 10* 10 * 10^(-6))`

`q_1 = 1300 w`