Silicon carbide nanowires of diameter D = 15 nm can be grown onto a solid siliconcarbide surface by carefully depositing droplets of catalyst liquid onto a flat silicon carbide substrate. Siliconcarbide nanowires grow upward from the deposited drops, and if the drops are deposited in a pattern, an array ofnanowire fins can be grown, forming a silicon carbide nano-heat sink. Consider finned and unfinned electronicspackages in which an extremely small, 10 μm × 10 μm electronics device is sandwiched between two d = 100-nm-thick silicon carbide sheets. In both cases, the coolant is a dielectric liquid at 20°C. A heat transfer coefficient of h= 1 × 10^5 W/m^2 · K exists on the top and bottom of the unfinned package and on all surfaces of the exposed siliconcarbide fins, which are each L = 300 nm long. Each nano-heat sink includes a 200 × 200 array of nanofins. Required:
Determine the maximum allowable heat rate that can be generated by the electronic device so that its temperatureis maintained at Tt < 85°C for the unfinned and finned packages.

For unfined package = 1.30 × 10^-3 W.

Dor fined package = 8.64 × 10^-3 W.

Explanation:

STEP ONE: The first thing to do is to determine the cross sectional area and input the value into the convectional resistance formula/equation in order to determine the heat rate for the unfined package.

Thus, Area = (dimension of the electronic device) ^2. = (10 × 10^-6)^2 m.

Convectional Resistance = (100 × 10^-9)÷ (10 × 10^-6)^2 × 490 = 2.04 K/W.

Heat transfer coefficient = 1/ (10 × 10^-7)^2 × 10^5. = 1 × 10^5 K/W.

Thus, the heat rate for the unfined package = 2{(85 - 20) ÷ (10^5 + 2.04)} = 1.30 × 10^-3 W.

STEP TWO: Determine the surface area of each fin and the prime area.

Surface area of each fin = {300 + 15/4} × 10^-9 × 10^-9 × π × 15 = 1.43 × 10^-14 m^2.

Prime area = (10^-6 × 10)^2 - (200 × 200) × π × [ ( 15 × 10^-9)^2 ÷ 4] = 9.29 × 10^-11 m^2.

STEP THREE: Determine the efficiency

G ={ [300 + 15/4) × 10^9 } × [ 4 × 10^5÷ 490 × 15 × 10^-9] = 7.09 × 10^-2.

Efficiency= tanh ( 7.09 × 10^-2) ÷ 7.09 × 10^-2 = 0.998.

Thus, the coefficient = 1 - (1.43 × 10^-14 × (200 × 200) ÷ (6.65 × 10^-10) × (1 - 0.998) = 0.99.

STEP FOUR: Determine the heat rate for the finned package.

Thermal resistance = 1/ (10^5 × 0.999 × 10^-10 × 6.65 = 1.50 × 10^4 K/W.

The heat rate for the finned package = 2 { 85 - 20} ÷ { 1.50 × 10^4 + 2.04} = 8.64 × 10^-3 W

answer:

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