There is not a simple answer to this question (as always). It strongly depends on the specific cooling cycle you are using and more importantly, most coolers will be quite at the end of their load line (i.e. cooling power versus temperature). If you intend to use a regenerative cooler such as pulse-tube or GM you could use two data points to extrapolate to a somewhat higher temperature. You would like to have the zero-load temperature and the cooling power at 4.5 K.
Then simply take a straight line and you have a fair impression of what the cooler will do at a higher temperature. If, however, you intend to use Joule-Thomson, the story is different and much more complicated. The temperature is determined by the pressure at the low-pressure side (boiling point) and the cooling power primarily by the mass-flow rate and also by the pressure difference or better usually the high pressure.
You could follow a general approach and that is relating the cooler to the Carnot efficiency and assuming that the cooler operates at a fixed percentage of Carnot. Then is you know that percentage at 4.5 K you can use that (assuming that the input power remains the same) to evaluate the cooling power at a higher temperature.
However, please do realize that these are all approximations.
I’d like to refer to a review paper we published some years ago: Brake, H.J.M. ter & Wiegerinck, G.F.M. (2002). Low-power cryocooler survey. Cryogenics, 42, 705-718
2 Comments
Goran Perinic
August 23, 20121g/s liquefaction @ 4.5 K is equivalent to 101 W refrigeration power @
4.5 K. (This assumes return of the evaporated gas at room temperature.)
Marcel ter Brake
August 23, 2012There is not a simple answer to this question (as always). It strongly depends on the specific cooling cycle you are using and more importantly, most coolers will be quite at the end of their load line (i.e. cooling power versus temperature). If you intend to use a regenerative cooler such as pulse-tube or GM you could use two data points to extrapolate to a somewhat higher temperature. You would like to have the zero-load temperature and the cooling power at 4.5 K.
Then simply take a straight line and you have a fair impression of what the cooler will do at a higher temperature. If, however, you intend to use Joule-Thomson, the story is different and much more complicated. The temperature is determined by the pressure at the low-pressure side (boiling point) and the cooling power primarily by the mass-flow rate and also by the pressure difference or better usually the high pressure.
You could follow a general approach and that is relating the cooler to the Carnot efficiency and assuming that the cooler operates at a fixed percentage of Carnot. Then is you know that percentage at 4.5 K you can use that (assuming that the input power remains the same) to evaluate the cooling power at a higher temperature.
However, please do realize that these are all approximations.
I’d like to refer to a review paper we published some years ago: Brake, H.J.M. ter & Wiegerinck, G.F.M. (2002). Low-power cryocooler survey. Cryogenics, 42, 705-718