HW8, problem 43

The efficiency is defined differently in this problem than in the case for a heat engine. The reason for this difference is that what we are putting in is a work (the work required to refrigerate) rather than a heat (the boiler in the Carnot cycle, for example). Here, the efficiency is closer to the "isentropic efficiency" we discussed in class for the turbine. For cases in which we expect the nonideal work to be less than the ideal work (as in a turbine), we have:

eta = (actual work required) / (work required in the ideal reversible case)

On the other hand, for cases in which we expect the nonideal work to be greater (as in refrigeration), we have:

eta = (work required in the ideal reversible case) / (actual work required)

Regardless, eta is always defined so that it is less than one.

In general, efficiencies tend to measure works relative to an ideal case. In the special case of a heat engine (like a Carnot engine), the efficiency we discussed is relative to a sort-of "super ideal" case in which all of the boiler heat is converted to work.

So the general way in which one treats problems like this is almost always the same:
(1) Solve the system as if it were ideal (reversible processes) -- in this case, it is a straightforward application of the open energy balance and the entropy generation equation.
(2) Use the efficiency to relate the actual work to the ideal work.

MSS