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Consider a plane aluminum ik= 120Bmu / h * r - h -^ F) wall at asteady temperature of 650 deg * F In order to increase the heatdissipation rate to surrounding air at 75°F, three aluminum pin-shaped fins per square inch are attached to the wall. Each fin isof 1/8 in diameter and 1 long. Determine the percentageincrease in the heat dissipation rate, if the coefficient of heattransfer is 28 Btu/hr - hbar^ 2 -^ F

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To calculate the percentage increase in the heat dissipation rate, we can use the following formula:

Percentage increase = (New heat dissipation rate - Old heat dissipation rate) / Old heat dissipation rate * 100%

The old heat dissipation rate can be calculated using the following formula:

Q = h * A * (T_w - T_a)

where:

Q is the heat dissipation rate in Btu/hr

h is the coefficient of heat transfer in Btu/hr-ft^2-°F

A is the surface area of the wall in ft^2

T_w is the temperature of the wall in °F

T_a is the temperature of the surrounding air in °F

The new heat dissipation rate can be calculated using the following formula:

Q_f = h * A_f * (T_w - T_a)

where:

Q_f is the heat dissipation rate with fins in Btu/hr

A_f is the surface area of the fins in ft^2

We can calculate the surface area of the fins as follows:

A_f = n * pi * d * L

where:

n is the number of fins per square inch

d is the diameter of the fins in inches

L is the length of the fins in inches

Substituting the given values into the above equations, we get the following results:

Old heat dissipation rate, Q = h * A * (T_w - T_a) = 28 Btu/hr-ft^2-°F * 1 ft^2 * (650°F - 75°F) = 17,150 Btu/hr

New heat dissipation rate, Q_f = h * A_f * (T_w - T_a) = 28 Btu/hr-ft^2-°F * 3 fins/in^2 * pi * 1/8 in * 1 in * (650°F - 75°F) = 46,894 Btu/hr

Percentage increase = (New heat dissipation rate - Old heat dissipation rate) / Old heat dissipation rate * 100% = (46,894 Btu/hr - 17,150 Btu/hr) / 17,150 Btu/hr * 100% = 174.2%

Therefore, the percentage increase in the heat dissipation rate is 174.2%.

User Stefan Hegny
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