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.

 

[hydtools]

Why do you say the hydraulic hp is not the energy loss = heat generated?

Ted

 

[HydraulicsGuy]

Because there is always a pump inefficiency that makes the input power higher than the hydraulic output power. Because no useful work is being done, that means all of the input power becomes heat. In other words, all the power being put into the hydraulic system by the electric motor is being wasted because it's just being used to generate high pressure fluid and send it back to tank, with no useful output work being done (eg, no cylinders delivering mechanical output power).

 

[hydtools]

So what you are now saying is that total input energy in standby mode is wasted, generating heat. Therefore you do not need to know electric motor efficiency when the pump is in standby mode.

I thought your original question was only about hydraulic system heat generation. Very little electric motor heat makes its way into the hydraulic system.

Ted

 

[HydraulicsGuy]

Right, I'm saying input power to the hydraulic system, meaning motor output power, meaning pump input power (all the same thing) is all wasted and becomes heat generated. Pump inefficiency is heat generated. Case drain flow flowing from high pressure to low pressure is heat generated. There is no useful work being performed. All input power becomes heat.

Simple math example: Let's say input power to the hydraulic system is 10 HP (input power to hydr system = motor output shaft power = pump input shaft power). Let's say pump efficiency is 50%. So 5 HP goes to heat generated by the pump. The other 5 HP goes to pressurizing the case drain flow, which then just flows from high pressure (pump case) to zero pressure (tank), which creates 5 HP of heat in the process. Total heat generated = 10 HP = input power to the hydraulic system = motor output shaft power = pump input shaft power. I'm trying to calculate heat generated, knowing compensator setting, case drain flow, but not knowing pump efficiency.

 

[1gibson]

Pump efficiency is 0% when there is no flow. The pump is not doing any useful work. It is a little conservative to say it all goes into heat, because some energy creates vibration, noise, and there may be a small amount of flow. But for what you need to know, efficiency can be considered 0%.

What is the miles per gallon of a car that just sits in park with the engine running? 0 mpg.

 

[hydtools]

Piston pump efficiency is generally in the range of 90% to 96%. FWIW

Ted

 

[HydraulicsGuy]

Ted, but what is the efficiency when the pump is in stand-by? That's what I'm trying to get at, so I can calculate input power, which I believe equals heat generated.

Pump overall efficiency can vary widely for a pump, and it depends on discharge pressure. Every time we do an HPU, I look at the pump curves for the pump we're going to use, and I do the calculations based on the curves. I can tell you that pump overall efficiency for the pumps we deal with can vary between 20-25% at very low discharge pressure, up to about 85% at "sweet spot" discharge pressure (for lack of a better term). 90% is very generous. 96% must be a very expensive well-made pump, because I have not seen anywhere close to that overall efficiency in any of the pumps curves I've dealt with. Unless you're talking about volumetric efficiency only? In which case, yes, 90%-96%, even up to 99%. But overall efficiency varies widely, and it depends on discharge pressure.

Gibson, pump efficiency cannot be 0% in stand-by, because then input power would be infinite. Pump input power = pump output power / pump overall efficiency.

Ted, I think you and I don't agree on whether input power = heat generated for a pump in stand-by. I believe it does. I believe that all input power goes to heat generated. My simple example above demonstrates why I think so. Am I correct to say that you believe input power =/ heat generated? If that's a fair characterization, how would you calculate heat generated for a pump in stand-by? As I mentioned before, there is a manufacturer called Duplomatic that publishes what they call "input power at full cut-off" (meaning pump has compensated, de-stroked to no system flow). If you Google "Duplomatic VPPL pump", the first or second search return is a PDF. In that PDF, go down to the pump curves and you can see "input power at full cut-off". I say this is the heat generated. Seeing these curves, if you were using or analyzing this pump, how would you calculate heat generated for this pump "at full cut-off", meaning "in stand-by"?

 

[HydraulicsGuy]

I may not be using the right terminology. By "stand-by", I mean the pump discharge has reached the compensator setting, and the pump has de-stroked. Maybe the better definition of "stand-by" means pump at low discharge pressure, and de-stroked. But I'm talking about high discharge pressure and de-stroked. Sorry for any confusion.

 

[hydtools]

Total efficiency on the efficiency curve is 90% at 210 bar. So hydraulic heat would be generated by .9 x input power.

Depending on the pump size, the charts show cut-off flow rates from 1.0 to about 1.5 l/min at 210 bar. So what power loss do you have for QxP for your size pump?

Ted

 

[HydraulicsGuy]

Ok, you say 0.9 x input power is hydraulic heat. I agree so far. The other 0.1 x input power is.....also heat. That's the pump inefficiency, the other 10%. It manifests as hydraulic heat. If you have it, see IFPS hydraulic specialist study manual, page 3-56, review question 3.9.2.1. The pump inefficiency contributes to the heat generated. So all 100% of input power turns into heat.

So if you look at the Duplomatic "input power at full cut-off" curves, the input power is much higher than the QxP, where Q = drain flow rate and P = cut-off pressure. If you calculate QxP / "input power at full cut-off", you get an efficiency of like 53% for their -016 pump, and it goes down to like 23% for their -046 pump. Assuming this is a valid way of looking at it??

The 90% efficiency at 210 bar only applies to the pump operating at full displacement, not at full cut-off. If you try to take the pump efficiency as 90% at full cut-off, they give you the input power at full cut-off, so you can back-calculate the QxP. QxP = input power at full cut-off x efficiency. What would this calculated QxP mean? It can't be "drain flow x cut-off pressure", because "drain flow x cut-off pressure" is way too low.

To answer your question, I don't know what power loss I have at full cut-off for the pumps we use. That's what I'm trying to figure out. As I said before, the pumps we use almost exclusively here at my company don't publish any data of what's going on at full cut-off, and couldn't help me with my request.

 

[hydtools]

I don't understand the charted increase in input power as the pump moves to shut-off. Just spinning the pump at little or no output flow should be a low power condition, even at 210 bar.

Ted

 

[HydraulicsGuy]

I know! That's why i'm thinking the pump's efficiency must be very low in this condition. If it's taking 3.5 HP (2.6 kW) input at 1800 RPM to spin the -046 pump at cut-off, the pump must be very inefficient and there must be a lot of heat generated in this condition! That's why I mentioned the 20% efficiency number in my original post, which I had calculated for their -046 pump using Q = their case drain flow, P = 2500 psi cut-off pressure, and their graphed input power. Since I imagine their pumps aren't significantly different in performance than other pumps, I'm hoping to extrapolate their numbers at cut-off to other pump brands, to get the heat generated when the pump is sitting there pressure compensated for 45 minutes or an hour at a time while a cylinder is holding force against a workpiece.

(I think the input power isn't increasing as the pump moves to cut-off. I think what those curves are saying is: for the range of different cut-off pressure settings that you might set this pump at, from ~15 bar to 210 bar, here is your input power at any of those pressure settings.)

 

[HydraulicsGuy]

If nobody has any other comments or insights, I'm next going to post my question in the Fluid Power sub-forum here. Thank you for discussing this with me. It's nice to discuss and get the thoughts of others in hydraulics engineering.

 

[LittleInch]

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.

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

I also think your quoted duty is a little odd - 45 minutes sitting at 210 bar with no fluid flow? Why don't you use an accumulator to hold pressure and turn the pump off to stop the hydraulic fluid boiling?

But sure the hydraulics forum probably has some more information on this issue.

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

 

[HydraulicsGuy]

Duplomatic, a pump manufacturer that we don't use, publishes the input power to the pump with the pump in the pressure compensated condition. This input power to the pump is equal to the heat being generated, which I have to dissipate through the oil cooler. I think efficiency when the pump has compensated to full cut-off is in the double digits, but from 50% down to possibly the high teens. At least that's what I calculate from the Duplomatic data, using their stated input power, Q = their stated case drain flow rate, and P = cut-off pressure. I'm trying to use the Duplomatic back-calculated efficiency, the case drain flow rate for the pumps we use, the pressure cut-off valve for the application, and use these pieces of information to calculate the input power = heat generated to size and select an oil cooler.

My company's chief engineer and owner has designed a hydraulic circuit for a recurring customer for several of that customer's units. When they want more of the units, I have to use the previous designs as a go-by, and not stray too far from them or the boss will ask me what the heck I'm doing. We as staff engineers don't always have the luxury of designing fancy systems or knowing exactly what the customer's duty cycle is. Our customers are always looking for the up-front cheapest simplest solution. And almost to a man, they can't articulate or write out how they are operating the system, or maybe the salesman just doesn't ask. I don't like to be pushy or step on toes, so I just go along with the information provided and work within those confines. I don't know when the customer is turning off the pump. Maybe he leaves it running the whole work day. I don't know. He probably doesn't even know, because his workers might each do it differently. I have to plan for the worst case, which is that he leaves the pump on, and the pump has compensated, while he is holding cylinder force against the workpiece and waiting for the plastic to cure.

 

[hydtools]

Case drain flow delivers heat to the reservoir. Energy other than hydraulic loss heat goes into heating the pump. Until the pump temp exceeds the case drain flow temp, no pump component heat conducts to the case drain flow fluid. Unless the pump is submerged in the reservoir, the only way for pump component heat to enter the system is via fluid flow of the case drain and that will not happen until pump component temp exceeds fluid temp.
Why do you want to know how much heat is generated in cut-off mode? Is system temp rising to an unacceptable level? Are you wanting to change the heat rejection capacity of the system?

Ted

 

[HydraulicsGuy]

I don't know the full history because I'm fairly new to the company, but apparently a past system (either customer's original or one my company provided) had gotten too hot according to the customer. On any new system we do for them, we are using a small dedicated kidney loop pump (through-drive off the main system pump) to take fluid from the reservoir, pump it through the oil cooler and filter, and right back into the reservoir. Since doing all new systems this way, the customer has not reported any overheating issues.

When I get handed one of their new systems to design and spec all the components (using an older system hydraulic schematic and BOM as a template), I want to go ahead and calculate the heat I need to reject, and size and select the appropriate oil cooler. Let's just say the full engineering record on the older projects isn't always possible to come by, and leave it at that. So I need to do my diligence on the oil cooler sizing and spec'ing.

What you explained makes sense. However, if you have it, please look at the IFPS hydraulic specialist study manual, page 3-56, review question 3.9.2.1. I can try to upload the page here if you want.
According to what's on that page, the "pump component heat" as you say contributes to the heat generated, regardless of temperatures. They don't take into account temperature differences. They just say, calculate all the power losses from pressure drops and inefficiencies, add them up, and that's your heat generated that you need to reject.

 

[hydtools]

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.

Ted

 

[LittleInch]

The thing is all these figures come from actual physical tests of a pump.

It's just that one of them has bothered to measure the power input at effectively zero forward flow, but the others haven't.

If they leave it running at full pressure but no flow I think you would have had complaints about overheating but I understand your issue.

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