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 %

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.