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?
Answer:
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 %
