Hydraulic Motor Overheating

[lightbulb22uk]

Dear All,

I am currently experiencing overheating of a hydraulic motor casing in a closed loop hydraulic system. The motor reaches around 100 degC in approximately 15 minutes when working at designed maximum speed and pressure (1600RPM and 300 BAR). The increase in temperature shows no sign of plateauing when it reaches 100 degC.

The system does not feature motor casing flushing (only a case drain), this is suggested by the motor OEM when motor speeds of 1800RPM are exceeded. Firstly would the requirement for case drain flushing be dependent upon motor power and efficiency and not speed? Secondly I would like to understand how theoretically I can calculate if motor flushing should be required and at what flow rate. Presumably calculations would be based on heat generation through motor efficiencies and heat lost through convection?

Any help would be much appreciated as I am currently racking my brain trying to remember 1st/2nd fluid mechanics!

Many thanks.

 

[hydroman247]

Motors will get hotter is they are converting more power as a percentage of the power input will be lost as heat. More power in, more heat out. You should not run most components with normal hydraulic fluid over 70-80 degrees C as it will lose its lubricating properties and will the damage it causes is exponential.

What other components are in this system?
What is the oil temperature in relation to the case of the motor?

 

[lightbulb22uk]

Thanks for the response Hydroman,

In addition to the motor are: 2 pumps and a purge valve in the working circuit, a cooler supplied by the purge valve and the pump case drains and a reservoir and header tank. The case drain from the motor returns direct to the reservoir tank.

The working oil temperature has not exceeded 40 degC. The motor overheats before the working oil temperature reaches a temperature at which the cooler will have an effect. Oil temperatures in the case drain line are in the region of 80 - 90 degC once the motor casing reaches 100 degC. Case drain temperatures also show no signs of plateauing.

 

[RobertHasty]

Hello lightbulb22uk,

The system does not feature motor casing flushing (only a case drain), this is suggested by the motor OEM when motor speeds of 1800RPM are exceeded.

What is the motor manufacturer? Is there datasheet available?

 

[lightbulb22uk]

Robert,

Thanks for the reply. The motor manufacturer is Parker and it is a F12-250 frame size motor. See below link for product literature, see page 67 for information detailing a recomendation for case flushing above 1800 RPM.

http://www.parker.com/parker/jsp/documentdisplay.jsp?mgmtid=dd7173c509c63110VgnVCM10000048021dacRCRD

Many thanks,

Pat

 

[RobertHasty]

Excellent catalogue.
Please tell me where do you measure temperature? Datasheet states following: The temperature should be measured at the utilized drain port.
If so, please check also temp. reccomendations for drain circuit (FKM 115 °C, NBR 90 °C).

 

[Oldhydroman]

Hi Pat

There's nothing wrong with your motor selection - that is a fine motor with a good pedigree.

The datasheet invites you to ask Parker for more specific efficiency information for your particular frame size - please do this and then post the graphs for this forum. Somebody here will be able to show you how to interpret the graphs and estimate whether or not you need to flush the motor case. Also let us know what oil you are using because this may have a bearing on the matter.

Actually, you already know that you do need to flush the motor case because of the temperatures you are experiencing. The datasheet suggests you don't need flushing but 1600 rpm is very like 1800 rpm and your working pressure is quite high. It's not so black and white as 1799 rpm means flushing is NOT required and 1801 rpm means flushing IS required. It may be the case that the particular build up of manufacturing tolerances has meant you received a motor which is a bit tight (internal friction is a little high - resulting in more power loss but also resulting in a smallish case drain flow - so there isn't enough case flow to take away the waste power - and that is why you are getting such a high case temperature).

According to the datasheet you are allowed to go up to 90 deg C for a motor with NBR seals (115 deg C for a motor with Viton seals) but, personally, I would prefer not to run at these temperatures for the following reasons:

Oil temperatures above 60 deg C can result in scalding/scarring if any escaping fluid contacts the skin (below this temperature getting soaked with the oil is just irritating, above this temperature it constitutes an injury).

The oxidation rate of the oil is very temperature dependant, doubling for every 10 deg C increase in temperature. Having very hot oil shortens its service life.

The very hot oil in the case will have a low viscosity and this will rapidly decrease the service life of the motor bearings. In a bent axis motor the shaft bearings work very hard.

DOL

 

[LittleInch]

I think you've taken the brochure wording a little bit too literally. The wording states that continuous operation "may require case flushing" and the limits are those which flushing is "usually required". Given you're at 1600 rpm versus 1800rpm for the "usual" limit, then it would appear that the vendor has covered himself.

The amount of heat generated will be approximately the power input times (1-efficiency). There are equations in the first part of the book to work this put which needs flow as well as pressure. Given that there are no cooling vanes, whilst a lot of the heat will escape via the hydraulic fluid, enough will reomain to heat up the casing quite quickly as you've discovered.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[lightbulb22uk]

All,
Thanks again for your responses.

Robert,
Temperature is currently being measured on the physical motor case and at the drain port, both using an infrared temperature gun.

Oldhydroman,
I have received efficiency graphs from Parker for the 250 size motor, I have however been unable to upload them on to this thread. Reading the graphs I have estimated a volumetric flow efficiency of 97%, a mechanical efficiency of 97% and total efficiency of 94%. This does however seem a little high? From this and the pumps theoretical power output (174kW) I have estimated that 11kW of heat will be generated. What I am now trying to calculate is how much heat will be removed through discharge of fluid from the motor and how much will be left in the motor casing through convective heat transfer. Or perhaps I am going about this in the wrong way?

We are using OM33 hydraulic fluid.

LittleInch,
I fully appreciate that the Parker manual uses terms such as "may" and "usually" and that it is very much application specific. And It is clear that a flushing system is required. I would however like to apply some theory behind the application to ensure that my seniors are confident that this application is being investigated and justified thoroughly. What steps would you advise to follow once efficiency has been calculated to calculate heat dissipated by the fluid and heat remaining in the casing?

Many thanks,

Pat

 

[LittleInch]

Looks like you're nearly there. Measure temp in versus temp out of the hydraulic fluid x flow x thermal capacity and you have a good guess as to heat being carried away by the fluid. What's left is basicaly heating the case up, so even if it's 1kW, that 1kW into a relatively small thing with not much heat loss capacity other than convection - which won't be much. You could get really scientific and apply some sort of water cooling blanket to calcualte the heat being generated in the casing and then show that without additional cooling the motor would simply continue to heat up.

If you had a 175 kW electrical motor and no fan on it then it would also get red hot. Electrical motor efficicieny at that size is about the same. Would you think that you could install an electrical motor that big and not cool it? Ok it doesn't have fluid taking some of the heat away, but even so...

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[Oldhydroman]

Here’s the quick and dirty calculation of drain oil temperature (there’s too many unknowns in the actual performance of your particular motor and too little precision in your efficiency data to warrant doing a more sophisticated calculation).

Your motor is 250 cc/rev and running at 1600 rpm so your theoretical input flow would be 400 L/min. But your volumetric efficiency is 97% so your actual input flow will be 400/0.97 = 412 L/min and we can assume that the extra 12 L/min you needed becomes your case drain flow. (You could quite easily measure your case drain flow using a measuring flask and a stopwatch.)

This case drain oil came from your 300 bar supply so will be warm because it will have reduced its pressure to 0 bar by passing through the clearances in the motor. Assume a temperature rise of 6.8 deg C per 100 bar which means your case drain oil “source” is now 20.4 deg hotter than the supply.

The theoretical shaft output power from your remaining 400 L/min at 300 bar should be 200 kW but your mechanical efficiency is 97% so you lose 3% of 200 kW, i.e., 6 kW. This heat has to be carried away by your 12 L/min case drain flow.

The specific heat capacity of a typical mineral oil is 1.67 J/g/K and the typical density is 0.88 g/cc. Your flow of 12 L/min is 200 cc/sec so that’s 176 g/sec. Your heat input is 6 kW, i.e., 6000 J/sec so the energy input to the oil is 34 J/g. The temperature rise because of this energy input will be 34/1.67 = 20.4 deg C (yes, it’s the same as the temperature rise from the pressure drop but that’s because both of your efficiencies are numerically equal).

Your case drain oil will be approximately 41 deg C hotter than your input flow. If your cooler doesn’t come on until the bulk oil temperature reaches 60 deg C then that’s you breaking the 100 deg C mark.

As I stated earlier, this is a horribly simplistic calculation because no account is taken of: the compressibility of the oil, heat taken away from the motor by the outlet [return] flow, conduction through whatever the motor is bolted to, convection from the motor case etc. All I wanted to do was show you how the numbers stack up.

If you were to apply some motor case flushing flow this would be taken from the low pressure side of the circuit (say 20 bar) which would be from the output side of the motor. So let’s assume a bulk oil temperature of 60 deg C and the temperature of the motor supply oil would be about the same. Call the motor outlet temp 62 deg C so that will give you a flushing flow inlet temp of about 64 deg C. A flushing flow of, say, 20 L/min at this 64 deg C would add to your case drain “source” flow of 12 L/min at 80.4 deg C to give you 32 L/min at ~70 deg C. This combined flow now has to absorb the 6 kW wasted mechanical power; the resulting temperature rise will be ~7.6 deg C and your case drain oil would be coming out of the motor at just under 80 deg C.

If you added a separate hot oil shuttle valve to your motor circuit with an adjustable throttle valve on the shuttle outlet you could vary the amount of flushing flow until you brought the case drain exit oil temperature back into line. Do check, however, that your boost pump has enough capacity to do this – if you take too much flow off the motor outlet circuit then the boost pressure may drop too low for the pump’s comfort. Alternately, you could redirect some of the flow from your existing purge valve if it is close enough to the motor. Remember that increasing the case flushing flow will decrease the flow through your cooler so its cooling performance will be reduced. Is it possible that you could redirect the motor case drain flow through the cooler as well (as long as the motor case and pump case can take the cooler back pressure)? If you could do this the cooler performance would increase because of the greater flow and the greater average temperature difference.

DOL

 

[lightbulb22uk]

LittleInch, Oldhydroman, thanks for the responses.

I have had the following presented to me by a colleague, however I am not entirely bought into this as a solution just yet:

Motor Power = 174kW
Motor Flow Rate = 408 l/min
Heat generated through volumetric inefficiency = approximately 4% of 174kW = 6.96kW (4% based on 96% volumetric efficiency)

Actual heat rejected = Q = m Cp delt T
where:
m = (0.04*408) l/min *.9 kg/l / 60 = 0.24 kg/sec
Cp = 1.7 kj/kg K
delt T = 100 - 40 = 60 K
Therefore Q = 0.24 * 1.7 * 60 = 24.48 kW

Since actual heat transfer (24.48 kW) is significantly greater than the calculated heat generated through volumetric inefficiencies (6.96 kW), this would suggest that there is a problem within the hydraulic motor and the motor should be replaced.

What do you guys think?

Many thanks,

Pat

 

[LittleInch]

Your colleague seems to be taking everything as conservative. You say the Vol eff is 97%, so flow is less. Also density will be lower at higher temp, say 0.8. This gives 0.16kg/sec. Can't you measure actual case drain flow then weigh it??

You say drain oil temp is 80 to 90, not 100. Let's be good and say 80C. so Q = .16 x 40 x 1.7 = 10.88kw.

As hydroman says, vol losses AND mechanical losses are 6 kw each, this totals 12 kW. Colleague appears to have forgotten about the Mech eff losses...

A few more tests would seem to be in order to me.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[lightbulb22uk]

Oldhydroman,

Reference the following point: "This case drain oil came from your 300 bar supply so will be warm because it will have reduced its pressure to 0 bar by passing through the clearances in the motor. Assume a temperature rise of 6.8 deg C per 100 bar which means your case drain oil “source” is now 20.4 deg hotter than the supply."

Can you confirm how you estimated the 6.8 deg C per 100 bar. I have been trying to understand the relationship between reducing pressure and a temperature increase. Or how can I estimate the heat generated through a pressure reduction, do i need to know delta T to calculate this?

If anyone else could help on the point above, that would be appreciated.

Thanks,

Pat

 

[Oldhydroman]

Suppose you had a flow of 60 L/min at 100 bar, the [theoretical] power you would need to pump this would be: 60 x 100 / 600 = 10 kW. If you then let this source of pressurised oil de-pressurise by passing over a jet (or some other clearance passage within a hydraulic component) then no “real” work will have been done and the original power input of 10 kW will all be converted to heat.

60 L/min = 1 Litre/sec = 0.88 kg/sec (assuming that the fluid density is 880 kg/m³). Your input power of 10 kW is 10 kJ/sec so you will be putting 10 kJ/sec into 0.88 kg/sec, i.e., 10 kJ into 0.88 kg. This equates to 11.36 kJ/kg.

The specific heat capacity of the oil is typically 1.67 kJ/kg/K so the temperature rise will be 11.36/1.67 = 6.8 K for a 100 bar pressure drop. If the pressure drop were higher the energy input would be higher and the temperature rise would be higher. Similary a lower pressure drop creates a lower temperature rise. To be more specific the temperature rise is 0.068 K/bar.

I’ve used real numbers in this example, but you should be able to see that the flow rate is actually of no consequence. If the flow had been 120 L/min then the input power would have been 20 kW … but this would have been dissipated into 1.76 kg/sec so the specific energy input would still have been 11.36 kJ/kg.

Feel free to fiddle with the numbers to take account of fluid compressibility, changes of density with temperature and pressure, changes of specific heat capacity with temperature and pressure etc. but I don’t believe it’s worth the effort.

If your motor is new then you can think of it behaving just like a rebuilt engine that hasn't fully run in yet. It will be "tight" so there will be a poor mechanical efficiency and a good volumetric efficiency. The overall efficiency may still be good [this is a measure of the motor's abilty to convert the hydraulic input power (pressure x flow) into mechanical output power (torque x shaft speed)]. The [poorish] mechanical efficiency means that some of the mechanical power is lost inside the motor - but the [high] volumetric efficiency means that there isn't much leakage flow available to take the wasted power away - hence your high case drain temperature.

You really need to measure your actual case drain flow before you can get any further with your analysis of the system.

DOL