An array of eight aluminum alloy long fins, each 3 mm wide, 0.4 mm thick, and 40 mm long, is used to cool a transistor. When the base is at 340 K and the ambient air is at 300 K, how much power do they dissipate if the combined convection and radiation heat transient coefficient is estimated to be 8 W/m2K? The alloy has a conductivity of 175 W/mKand the heat transfer from the tip is negligible.

0.08704 W

Explanation:

converting the mm to m (1000mm = 1m)

cross-sectional area of the fins, Ac = (0.003) (0.0004) = 0.0000012‬m^2

The wetted perimeter of the cross-section, P = 2 (0.003 + 0.0004) = 0.0068‬m

Thickness of solid in direction of heat flow, B^2 = (heat transient coefficient, h) (The wetted perimeter of the cross-section, P) ÷ (Thermal conductivity, k) (cross-sectional area of the fins, Ac)

B^2 = (8 W/m2K)(0.0068‬m) ÷ (175 W/mK)(0.0000012‬m^2)

=259.0476m^-2

B= square root of the result

B = 16.09m^-1

we now look for:

The Coordinate, x = B, multiplied by Length, L

x = (16.09m^-1) (0.04m) = 0.6436‬

 finding the side area of a fin =  P multiplied by Length, L

= 0.0068‬m X 0.04m = 0.000272m^2

Neglecting inefficiency, assuming the fins are all 100% efficient, the power they would dissipate =

h, Heat-transfer coefficient (PL) (temperature of at the base - temperature at the ambient air)

= (8) (0.000272m^2)(340 K- 300k)

= 0.08704 W

unlike the threaded fasteners which are used to clamp the machine members, power screws are used to transmit power. common applications of power screws are: screw jacks: to lift weight. lead screw of lathe: for axial movement of tool and its precise positioning.

answer: report them

explanation: