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I am working on an indirect ammonia chiller that is used to cool the ice in an arena. The system consists of 3 reciprocating compressors, an evaporative condenser with a vfd on the fan, a float throttling valve and a plate heat exchanger acting as the evaporator.

With one compressor operating, the condensing pressure of the ammonia ranges from 140 psi to 180 psi depending on the speed of the fan. When the condensing pressure is higher, the compressor is doing more work. When the fan speeds up, the head pressure lowers which corresponds to a drop in compressor work.

I know that lowering the condensing pressure reduces compressor work and am familiar with the pressure-enthalpy chart that is used to describe the refrigeration cycle. My question is, how does increasing the fan speed cause a drop in compressor work? The fan kicks on and rejects more heat from the gas, but what is the process or mechanism that occurs as you follow the refrigeration cycle that results in the compressor lowering the condensing pressure?

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  • $\begingroup$ You're asking about the fan on the cooling side (non-ammonia side) of the condenser increasing in speed and that causing a drop in the work done by the compressors pumping the ammonia in the chiller, correct? $\endgroup$
    – Trevor Archibald
    Apr 4, 2015 at 2:57
  • $\begingroup$ The fan is on the ammonia side. The ammonia is the refrigerant and the plate heat exchanger acts as the evaporator which transfers heat from brine (which is circulated under the ice) to the ammonia on the chiller side. The fan is cooling the ammonia that travels through the coils in the cooilng tower (it is an evaporative condensor baltimore aircoil cooling tower). Does that clear things? $\endgroup$
    – Prevost
    Apr 4, 2015 at 19:06
  • $\begingroup$ Way too much hand waving. I can't figure out what is connected to what and where eactly this "fan" is within the system. Also, why you need a vacuum flourescent dispaly on the fan is beyond me. $\endgroup$
    – Olin Lathrop
    Apr 5, 2015 at 13:02
  • $\begingroup$ @OlinLathrop FYI, he means "variable frequency drive" by vfd. $\endgroup$
    – regdoug
    Apr 13, 2015 at 2:27
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    $\begingroup$ @reg: Perhaps so, but the point is that it's irresponsible to use abbreviations without defining them unless they are very broadly used across engineering disciplines. As I showed, "VFD" (also note that such TLAs should be written in caps), can have at least one other meaning. There is no excuse for not spending the extra second to type out the full name once. That and the wild hand waving make this a very poor question. $\endgroup$
    – Olin Lathrop
    Apr 13, 2015 at 13:44

2 Answers 2

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I think @regdoug is on the right track, but this is how I think of it: The work done by the compressor is equal to the difference in enthalpy of the ammonia between the compressor inlet and outlet; $W_c=H_{out}-H_{in}$ On the following pressure-enthalpy diagram, points 2 and 3 are the states at the compressor inlet and outlet, respectively:

enter image description here

Now, if the fan speed in the condenser is increased, it will draw more heat out of the ammonia, which will cause the temperature and pressure of the ammonia in the condenser to drop. Consequently, the enthalpy of the ammonia will also drop, as indicated by the dashed line in the pressure-enthalpy diagram (points 3' and 4'). Lower outlet enthalpy of the ammonia results in less work done by the compressor, according to the previous equation.

By the way, the other points in the diagram are as follows:

1->2 Evaporator

2->3 Compressor

3->4 Condenser

4->1 Throttle

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EDIT: I still think that I'm on the right track, but I have realized some errors that I made (specifically I have drawn my diagram with the wrong area shaded for work, I labeled the diagram h instead of T, and I skimmed over the specific volume ratio of the compressor) so until I get this straightened out, consider the answer as tentative at best.

As far as I understand, it works as follows:

  1. By increasing the fan speed, the heat rejected by the evaporator is increased
  2. The increased heat rejected can only be accomplished by "widening" the points on the evaporation line
  3. The increased quality at the output of the evaporator causes the compressor to compress to a lower pressure.

That may not be right, but it seems to make sense on the diagram I drew.

Vapor-compression refrigeration cycle

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    $\begingroup$ The heat rejected by the condenser is equal to the heat absorbed in the evaporator as well as the heat added by the compressor doing work on the refrigerant. The lower condensing pressure is then achieved as the system attempts to balance itself out (you said the increased heat rejection is accomplished by widening the points). But I don't think it's the quality at the output. The system lowers the condenser pressure to the pressure that can handle the increased amount of heat rejection while satisfying the refrigeration cycle (because the system has to work). Does that make sense? $\endgroup$
    – Prevost
    May 30, 2015 at 2:56

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