My calculation went like this...
Refering to the schematic of the Bangood PSU, which Music sent to me in a conversation a few days ago:
We are concerned with the dissipation in potentiometer P1.
The voltage across P1 is the putput voltage of op-amp U1.
R5 = R6 = 10k
D8 is a 5.1v zener.
To calculate the output voltage (Vout) of the op-amp:
First find an experssion for the voltage at pin2 of the op-amp V2 = Vout - 5.1
And find an expression for the voltage at pin3 of the op-amp V3 = Vout x R6/(R5+R6)
If the op-amp is working correctly V2 = V3 (or very near enough, I am not going to argue about it).
So, that gives:
Vout - 5.1 = Vout x R6/(R5 + R6)
using basic algebra, this can be kicked around to give Vout = 10.2 volts.
The power dissipated in P1 can be calculated using W = V^2 /R.
The voltage across P2 will be less than that across P1, and can be calculated by considering the effects of R17 and R18.
And that is how I did it.
Nigels quick and dirty method also has merit, using that method he found that the dissipation in the pot was reasonably low and could be tolerated by most 10 turn panel mounted potentiometers, end of problem.
If quick and dirty gave a high power disipation which would entail buying an expensive pot, then it is time to go sophisticated and get the true value of dissipation.
JimB