Also will be heat sinking the three resistors as well as the LM338T

Sorry about that omission!

What matters is the number of watts the regulator will be dissipating. Likely the maximum wattage will be during the 2.5 ampere position but without knowing your source voltage and load it will be impossible to tell how many watts and therefore how much of an air heat sink you'll require.

Its simple, the cooler the elctronics, the longer they last, therefore put the largest heatsink you can on it. No need for calculating anything....

To check its working OK, use the Mark I index finger, if its getting warm, add extra cooling! Check at the highest current for at least twice as long as you will ever use it....or an hour if you don't know how long....

At 32V DC input, heat dissipation for your current regulation configuration (because that's what your schematic is about) will be too high, and internal IC regulator heat protection will emerge very quickly, with heat sink or not. If your max current will be 2.5A, either use lower input voltage, or use a ((32-1.3)/2.5)Ohm that's about 12 Ohm high power resistor in series with input voltage. The regulation, voltages etc will be identical, but most of the heat will be dissipated on the resistor, not the IC. S.M.

The worst case scenario is 2.5A continuously from 32V into a short circuit = 80W. The dissipation across the 0.5 ohm resistor is 0.625W, so, apart from using a 2W resistor, we can neglect that. If you are certain about the 8 second-3 second timing then the dissipation drops to 58 watts. The thermal resistance to open air of the LM338T is 50 degrees/watt, so yes, you will need a heatsink. If you don't use a heatsink, the internal protective circuitry will operate to limit the current.
The calculation of the heatsink depends on your choice of maximum temperature. The LM338 will operate up to 150 degrees. It would be nice to have a lower working temperature, but you are rather close to the limits. 150 is about 125 above ambient, so the heatsink rating needs to be 125/80 = 1.56 degrees/watt. That is unachievable with the LM338T, where the thermal resistance to case is already 4 degrees/watt. The LM338K in the TO-3 package has a thermal resistance of only 1, but a heatsink of 0.5 can only be achieved with fan cooling. It looks slightly better with the 58 watt dissipation, but not much.
The obvious question is "Why do you want to run it from 32V?" Would 12V not do?

"The obvious question is "Why do you want to run it from 32V?" Would 12V not do?"

That leads to an important question: all assumptions base on a 0-Ohm-Load (short). If the load has R>0 then, depending on the load resistance, perhaps most of the power will be dissipated in the load plus the resistors instead of the LM338T.

If the thermal resistance of the LM338T will be too high to dissipate the whole heat another possibility would be to take a smaller regulator and add n power transistors which supply the current while having a better thermal resistance in sum because the single thermal resistors will be divided by n. This solution will do with only passive cooling without a fan.

Apart from that: the power in the 0.5 Ohm resistor will be 3.125 W instead of 0.6W.

regards Uwe

If the load has R>0 then, depending on the load resistance, perhaps most of the power will be dissipated in the load plus the resistors instead of the LM338T

Exactly! That's why I suggested experimentation back at post #3
Del

I think you will find that the resistance of a plating bath is measured in fractions of an ohm, which is why I suggested designing for a short circuit.

No. This is a plating bath, not a plasma chamber. Getting ions to move in a fluid requires power. More power will be dissipated in the bath than moving electrons in the conductors attached to the bath. If we knew the geometry of the bath, the bath temperature and the ion concentrations then we maybe able to calculate the anticipated electrical load.

The circuit is also a current limiter circuit, not a constant current design.

The figure for resistivity I have found for a chrome plating bath is about 3 ohm-cm, see

http://www.finishing.com/104/97.shtml

For a 10cm^2 plating area and a 2 cm anode-cathode gap that works out as a bath resistance of 0.6 ohms. This is in agreement with my previous suggestion that the load is a fraction of an ohm. I doubt the resistivity of a gold or silver plating bath is an order of magnitude different, but I will bow to an engineer's opinion.

So it sounds to me if I lower the voltage I should still be able to get the current I desire. It's what's most important in plating anyway. I'll just add a voltage divider before the input or find a different supply.