3 replies to this topic

[Zorgon]

We have a pump that circulates a very cold heat transfer media through the evaporator of a refrigeration machine and out to the users. The pump was mis-specified and has a much higher differential head than is required. The result is that we have to throttle the discharge valve to prevent the pump from running out on its curve and shutting the motor down on high current. We can't replace the impeller to match the process requirements so we will eventually replace the pump. The refrigeration capacity is tight and this "wasted" energy started me thinking about how much heat is being added to the system by the pump.

 

I understand that the inefficient power of the pump is directly converted to heat (1 minus fractional pump efficiency times the calculated hydraulic horsepower). I also understand that the frictional pressure drop in the piping and equipment downstream of the pump will be converted to heat. But now we are converting the pressure drop across the discharge valve directly into heat by the difference between the calculated hydraulic HP based on the head and flow upstream of the throttled discharge valve and the head and flow downstream of the valve. This is correct, right? My question is how much more heat is the pump adding to the system? I have read that the entire hydraulic HP of the pump will be eventually converted into heat but I don't understand that. It seems that the head required to pump the heat transfer media from the elevation of the pump discharge to the elevation of the users would be useful work and would not be converted into heat.

 

The differential head for any pump needs to be based on the system pressure drop plus the elevation of the highest user. But what if the highest elevation user requires only a small fraction of the total flow? Will all the energy imparted to the larger "low elevation flow" get immediately converted to heat somewhere downstream of the pump? Can someone please help me to understand?

[breizh]

Hi,

You should understand the concept reading the document attached .

hope this is helping you

Breizh

[Saml]

Just move a step back and look at the system as a whole. System is in steady state. So whatever is entering it is leaving it.

 

- Your are adding heat from the users and power to your pump.

- You are taking out heat at your refrigeration machine evaporator

 

All the energy you put in the pump ends up as a load to the refrigeration loop. So you are right. you have to use the most efficient pump you can use.

 

Your second question: you have a single user with a much higher height or pressure drop.  If pumps were free you would have one pump matched to every single user to deliver just the right amount of refrigerant without the need to throttle. In reality you compromise by throttling some users.

 

Now if a single user needs a high head and low flow and you find that you are throttling the majority of your flow, you may consider using a main circulation pump and a dedicated small flow high head for that single service. But it is a a technical and economic analysis that has to be done professionally. We cannot do it on a forum.

 

And one more thing.

 

 

I understand that the inefficient power of the pump is directly converted to heat (1 minus fractional pump efficiency times the calculated hydraulic horsepower).

 

What you said  is true when you consider just pump inlet and outlet. But since you are in a closed loop in steady state, all the energy is eventually converted to heat as the pressure is dropping back to the initial state throughout the system. An inefficient pump simply adds more energy.

[Zorgon]

Thank you very much breizh for the reference and thank you very much Saml for the detailed explanation. I think I understand the big picture that all the energy added by the pump is eventually converted into heat. But my skull is pretty thick, so I need to ask at what point does this conversion occur? I consider that the pump does useful work when it pumps a liquid to a certain elevation. I clearly understand that the temperature at the pump discharge will be higher than the temperature at the suction by the “inefficient power” adsorbed by the pump.

 

Assume that the discharge nozzle of the pump is at a reference elevation of zero meters and the discharge line runs vertically to a user at a high elevation. Further assume that the line is perfectly insulated, and I have installed ultra-accurate temperature measuring devices every meter of vertical elevation. I can understand that the measured temperature at each meter would be ever so slightly higher than the previous meter as the pressure drop is converted to heat. But is the work required to raise the liquid meter by meter also converted to heat meter by meter? Would the temperature at the 10-meter mark equal the temperature at the zero-meter mark plus 10 meters of pressure drop plus 10 meters of elevation?

 

I’m sorry that I am being so dense, but the light hasn’t come on fully for me yet. I would greatly appreciate any more insight that you could provide.