Chiller Capacity and System Heat Generation 1

[CENSMR]

I am trying to use a chiller to cool a hydraulic system, and am trying to verify if I have enough cooling capacity in my chiller. The stated maximum cooling capacity is 10.5KW.

My system starts with a 50HP motor (~37KW), and my first question is how much of that in worst case scenario can become heat? As for my system, the motor drives a pump, which runs a hydraulic motor, which runs a generator. This is a test setup and I realize highly inefficient, but are these inefficiencies translating to heat in my hydraulic oil? If I load my generator I can get say 10KW - does that mean that the remaining 27 KW would be heat? What about when there is no load on the generator - is it all heat? It is likely safe to assume that the pump swashplate is fully stroked for this application.

My initial instinct and testing is telling me that the 10.5KW chiller is not adequate, but I do not want to oversize the chiller. Any help for a worst case cooling scenario?

SMR

 

[RossABQ]

All power going into the room emerges as heat in your case.

 

[hollandhvac]

Why would do you want to use a chiller for this purpose?

I would use a heat exchanger ( radiator like in a car ) and a fan. Is a lot cheaper than using a chiller.

 

[MintJulep]

How much power does the pump absorb?

Just because you have a 50 HP motor doesn't mean that it producing 50 HP.

Beyond that, what are you trying to cool, or what are you trying to accomplish by cooling?

 

[CENSMR]

Thanks for your replies. As for the requested clarifications, I am using a heat exchanger but it is not air cooled. The exchanger is part of the hydraulic system I have, but requires chilled water input, and thus why I have the chiller... to cool this water.

As for the electric motor and pump efficiencies, the motor is ~94% and the pump ~85%. The energy in the pressurized fluid is approximately 29kW. So the 37kW (50HP) is already ~29kW by the time it is in the form of pressurized fluid.

Does this help any? Am I looking at a worst case of needing to remove 29kW of heat?

Steve

 

[RossABQ]

Not quite. You need to remove 29kW/.85/.94 = 36 kW, if I understand correctly what you mean by "The energy in the pressurized fluid is approximately 29kW". I assume that you are saying the hydraulic power needed to circulate the cooling water to the hydraulic skid is 29kW? If that number already includes the pump and motor efficiencies, then you are correct. To get total chiller load then you have to add in the cooling needed for the hydraulic system, and any heat lost to the room from the piping and heat exchanger.

 

[MintJulep]

Ok, so you are apparently trying to remove heat from the hydraulic fluid.

When? Before it enters the pump, or after it leaves the pump?

What is the goal? What temperature decrease do you need to achieve, or what is the maximum allowed temperature at the exit of the cooler?

Ross is correct that all of the energy input will eventually leave the hydraulic system as heat - but you don't have to remove all of it with the chiller. Lots of it will leave all by itself.

You will lose heat in the piping. You will lose heat in the devices. You will lose heat to doing work (which will eventually become heat in whatever room the work is done).

Your cooler needs to be sized to match fluid flow x fluid heat capacity x required temperature drop.

 

[CENSMR]

I am trying to remove the heat from the hydarulic fluid after it leaves the pump... in fact, after it is done doing whatever work I am asking of it (sometimes none), before returning to tank.

MintJulep - I agree/understand that I have to remove less heat from the fluid when doing work , as this heat leaves the system as power from my hydraulic drive generator to an external load bank. However when there is no load on my alternator this pressurized fluid generates heat thru a pressure drop on a compensated control. How much of the loss due to inefficiencies (8kW from the 94% & 85%) end up as heat in the hydraulic fluid (that I have to remove), and how much ends going into the environment which I do not need to remove from the fluid?

The 29kW is a number I can arrive at two ways which gives some confidence:

1. 37kW (50HP) *.94(motor)*.85(pump) = ~29kW

2. The result of system gives me ~22GPM @ 3000 PSI. 3000*22/1714 = 38.5HP 38.5/1.341 = ~29kW

This makes sense that all of my inital power, minus inefficiencies to pressurize the fluid, ends up as energy in the pressurized fluid. If no work is done valves/fittings/motor heats up as well as the hydraulic oil... less when doing work as mentioned. I need to remove it from the hydraulic oil, but not the heat that hot valves/fittings give off to the environment. As mentioned, some of the 8kW efficiency loss is given to the environment which I do not need to remove.

Does this help? As for temperature ranges, I am trying to keep it 110-130F. When I run it now I have my chiller on and go right past 130F and need to shut down prior to reaching 150.