A long, circular aluminum rod is attached at one end to a heated wall and transfers heat by convection to a cold fluid. If the diameter of the rod is doubled, by how much would the rate of heat removal change

Complete question is;

A long, circular aluminum rod is attached at one end to a heated wall and transfers heat by convection to a cold fluid. If the diameter of the rod is doubled, by how much would the rate of heat removal change?

182.84 %

Explanation:

Formula for rate of heat transfer of an infinite log fin is given as;

q_f1 = (π/2) × (hk)^(½)) × D^(3/2)) × (T_b - T_∞)

Where D is diameter.

Now, if the diameter of the rod is doubled, it means Diameter is now 2D.

Thus;

q_f2 = (π/2) × (hk)^(½)) × (2D^(3/2)) × (T_b - T_∞)

To find how much the rate of heat removal will change, we will calculate as follows;

((q_f2/q_f1) - 1) × 100

Plugging in the relevant expressions, we have;

([[(π/2) × (hk)^(½)) × ((2D)^(3/2)) × (T_b - T_∞)]/[(π/2) × (hk)^(½)) × (D^(3/2)) × (T_b - T_∞)]] - 1) × 100

Upon simplifying, we have;

(((2D)^(3/2))/(D^(3/2)) - 1) × 100

((2^(3/2)) - 1) × 100

This gives;

182.84 %

work: w = fd (given)

work done by chet   =50*1 = 50 j

work done by mika =40*2 = 80 j

work done by sara =30*1.5 = 45 j

work done by bill =60*0.5 = 30 j

students in order from the greatest amount of work done to the least is mika> chet> sara> bill

hence option (b) is correct.