Pressure Compensated Pump - Heat Generated During Stand-by

[HydraulicsGuy]

Think about a pressure compensated variable displacement pump. The pump discharge has reached the compensator setting, the pump has fully compensated, and there is no flow going out to the system, only case drain flow. The pump has de-stroked and is in stand-by. Since there is no useful work being done, there is only case drain flow, all of the power input to the pump is going into heat generation.

Example:

Pump de-stroked in stand-by
Pump pressure compensator setting = 2500 psi
Tank pressure = 0 psi
Pump case drain flow = 1 GPM (assumed)
Hydraulic HP = (2500 psi – 0 psi) x 1 GPM / 1714 = 1.46 HP <-- This is NOT the heat generated
Pump input HP = Motor output HP = Hydraulic HP / pump efficiency % = 1.46 HP / ___% = ____ HP <-- This is the heat generated

In order to calculate heat generated in stand-by, I need either pump efficiency or pump input power. The pumps we use almost exclusively here at my company don't publish this data. I asked their tech support, and they couldn't help me. They don't even publish case drain flows in their catalog, but they were able to give those to me. How do you calculate heat generated in stand-by, in lieu of manufacturer data? I have found a pump manufacturer, Duplomatic, that publishes this data. Running the numbers on their data using my calculation steps above, I back-calculated a typical stand-by efficiency of about 20% for their pumps. Does 20% seem reasonable to use for other manufacturer's pumps?

Thank you.

 

[HydraulicsGuy]

Ted, that's a good idea. But by the time we test, all the components including oil cooler have been bought and installed. They wouldn't let me measure V and A to the motor with the rest of the system installed and almost ready to be shipped out to the customer, and then run back and do the calculations and buy the oil cooler. But that would at least be a data point that I could use for future applications. I was just hoping for a way to calculate it up front using case drain flow rate, cut-off pressure, and pump efficiency at cut-off. But again, that is a good idea, I like it, and I will try to have V and A measured the next time we do final testing on a unit before shipping it out.

 

[HydraulicsGuy]

LittleInch, customer is not having overheating issues anymore, because every new unit we do for them has the kidney loop pump sending reservoir fluid through an oil cooler and back into the reservoir. This has presumably solved any overheating problem like the problem they had on a past system. They may very well be leaving the motor & pump on for a very long time in the pressure compensated state. They are holding a certain target pressure in a cylinder against a workpiece, and if that pressure drops off, a pressure-reducing valve opens and the pump goes on stroke. So they need that motor on and that pump standing by in the pressure compensated state, ready to deliver high pressure at a second's notice.

So let me ask you, how would you calculate heat generated, for ultimately sizing an oil cooler, for an application like I have here? You have to assume the motor & pump stay on for long periods of time, with the pump in the pressure compensated state. All you have for cooling is the kidney loop pump circulating fluid from reservoir, through cooler, back to reservoir. How would you size the oil cooler?

 

[LittleInch]

Well I think we've gone through that above.

Either take the figure from a similar vendor like your duplomatic vendor or use the last known power figure from your current pumps. Or measure it even as a test as it's going out the door for next time.

For a cooler it shouldn't matter that much to get it precise.

By kidney loop do you mean a lobe pump?



Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[HydraulicsGuy]

Don't have a power figure from current pumps. Don't have any existing data points. All I have is Duplomatic and their generosity in providing comprehensive data. We never use Duplomatic pumps. I just stumbled across their data while searching this topic one day.

So when the next unit comes up, what do you suggest I do? Just use the Duplomatic data for a similar size pump?

"Kidney loop" is the function it performs: offline circulation of fluid for filtration, cooling, or both. In our case, we use gear pumps.

 

[hydtools]

Can you share with us the brand and model number of the pump you use?

Ted

 

[TugboatEng]

It seems that if you sized your reservoir correctly for the rated horsepower of the pump you shouldn't have any capacity issues when operating in compensation mode.

 

[hydtools]

It is a poor design choice to depend on reservoir size for temperature control.

Ted

 

[TugboatEng]

The size of the reservoir is normally determined by the heat dissipation requirements. External coolers allow for a smaller reservoir.

 

[hydtools]

Large reservoirs add a lot of thermal inertia to hydraulic systems. Quiet oil in a large reservoir transfers heat slowly.

Ted

 

[LittleInch]

In the absence of specific data use the best data you can find and add a bit.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[HydraulicsGuy]

Can you share with us the brand and model number of the pump you use?

It's Fluidyne (A)A10VSO Series 31. Mostly 18, 28, 45, or 71 cc. Almost always with only the "DR" pressure compensator control. Very similar to the Rexroth pumps of the same model.

 

[HydraulicsGuy]

It seems that if you sized your reservoir correctly for the rated horsepower of the pump you shouldn't have any capacity issues when operating in compensation mode.

Sizing a reservoir for all of the heat dissipation, even in the compensation mode data I've seen from Duplomatic pumps, would require a very large volume, many multiples of what we normally use in these systems. That would go over like a lead zeppelin here at my company. In the first couple oil cooler sizing calculations I did when I got here, I included reservoir cooling, per the formulas in the IFPS study manual and the lightning reference book. I used a reservoir size somewhere in the range of 2.5 - 5 times the pump max flowrate, which is the range where most of my company's past reservoir sizes fall. It turned out to be such a small percentage of the heat generated that I just ignored it in future calculations.

 

[HydraulicsGuy]

In the absence of specific data use the best data you can find and add a bit.

Fair enough. The best data I have at this time is the Duplomatic data I keep referencing. As Ted (hydtools) suggested, I'll try to measure volts and amps to the motor on future HPUs with a pressure compensated pump that we commission test before shipping out.

 

[hydtools]

When I was involved in the design of portable power units, we used 50% of system power as the design heat load. The reservoir was sized to 1x the flow rate and roto-molded plastic.

I have never seen an over-cooled hydraulic system.

Ted

 

[HydraulicsGuy]

Spoke to the lead tech here and he says he can measure motor volts and amps, and he will involve me in that when the next HPU with pressure compensated pump is being tested.

 

[HydraulicsGuy]

For those interested:

Yesterday during a test, we measured amps in the pressure-compensated condition on an HPU we had just finished building.

Data and calculations:
[ul]
[li]Pump is Fluidyne, 45cc[/li]
[li]Pressure compensator setting 2800 psi[/li]
[li]Motor is 3-phase, 50 HP, Premium Efficiency[/li]
[li]V = 480 Volts (known from the generator we use)[/li]
[li]I = 25 Amps (measured with amp meter)[/li]
[li]Power Factor = 60% (per Figure 4-5 in PREMIUM EFFICIENCY MOTOR SELECTION AND APPLICATION GUIDE, US Dept of Energy document)[/li]
[li]Motor Efficiency = 91% at 15 HP, which is 30% full load (per Figure 4-4 in same document)[/li]
[li]*Note that no published data exists for the above 2 numbers for the particular motor we used on the HPU, so i had to find references for them*[/li]
[li]Motor output HP calculated = 15.2 HP[/li]
[/ul]

So the motor is outputting 15.2 HP. This is also the pump input power. Since none of this power is used for useful work, this all goes into hydraulic system heat. Therefore the hydraulic system must be capable of rejecting 15.2 HP when the pump is in the pressure-compensated condition for long periods of time.

For the Duplomatic pumps that I've mentioned in this thread, I looked at their similar sized pump. Its "input power at full cut-off" for the cut-off pressure we were at in the test is about 2.5 kW, or 3.4 HP. So apparently it is much more efficient in the pressure-compensated state than the Fluidyne we use.

 

[HydraulicsGuy]

I'm actually not sure about the above 15.2 HP now.

I just found that the performance curves for a similar Bosch Rexroth pump, their (A)A10VS, show about 3-5 HP at zero flow (pressure-compensated condition) for their 45 cc pump. (Hard to determine exact HP value from their pitiful graphs; they should be ashamed of publishing graphs like that). Anyway, this 3-5 HP is on par with what Duplomatic shows for their 46 cc pump (about 3.4 HP).

The Fluidyne is much cheaper than the Rexroth, so perhaps its efficiency is much worse than the Rexroth at pressure compensation, and therefore requires much more motor power. In which case my calculation might be correct.

Or maybe the motor efficiency and/or power factor I used are too high, resulting in a falsely high calculated motor power. In which case my calculation is wrong.

Anybody have any comments?

 

[hydtools]

Add an air to oil cooler rated for 3 to 4hp heat rejection; take into consideration ambient temperatures. Get the FluiDyne pump with a through shaft, add a small gear pump to circulate oil in a kidney cooling loop. Test it.

In your calculations, have you considered the pump's pilot flow noted in the literature? That may represent the bulk of your wasted hp.

Ted

 

[HydraulicsGuy]

We already have a kidney cooling loop:

customer is not having overheating issues anymore, because every new unit we do for them has the kidney loop pump sending reservoir fluid through an oil cooler and back into the reservoir. This has presumably solved any overheating problem like the problem they had on a past system.

Just want to be able to calculate motor power (= heat load) accurately in the design phase from now on. 3-5 HP motor power is what Duplomatic and Rexroth publish for their ~45cc pumps in the pressure-compensated state. The motor power I calculated for the Fluidyne was much higher, but there were some assumptions. I calculated the motor side of things, nothing on the pump side. The pump side of things seems to be too difficult to quantify. Too difficult to calculate a motor power from the pump's parameters in the pressure-compensated state. Even a Fluidyne technical rep couldn't tell me what motor power it takes to run their pumps in the pressure-compensated state. I think it's better to use the motor side, as you suggested earlier:

Go about this another way. Do you know the efficiency of the driving motor? I assume it is electric. Measure the input electric power when the pump is in stand-by mode. Calculated the electric motor output shaft power which is equal to the pump shaft input power. That would be your worst case wasted power.

Although the motor side also comes with its uncertainties, like efficiency and power factor. But at least I found representative data on those 2 factors in some pretty authoritative literature, the Dept of Energy document I mentioned.

 

[skream47]

I think you're trying to calculate the impossible as there is so little work #(fluid flow) being done by the pump that efficiency would be in single figures.

IMO Pumps like to pump fluids, not sit spinning going nowhere.