How Can I Calculate the Temperature of Hydraulic Oil?

"I calculated the total power loss in my system which came out to be around N=17 kW.

Heat dissipation = 1,44 kW (dissipation at (delta)T = 40 deg Celsius and kp (still air) = 0,012 kW/m2K.

As environment temperature I added 30 deg Celsius."

17 / 1.44 = 11.8

11.8 x 40 = 472

472 + 30 = 502

Sounds about right so I would heavily recommend using the cooler for the system being hydraulic oil starts to brake down at temps over 80 C and hydraulic systems in general rapidly loose service life as the operating temp go up.

So this is correct?

Well I won't be building this at this moment. This is some work i do on paper only. I got an comment that this 2 results (power loss and the temperature) don't sound proportional. The temperature seems to be too high for this power loss.

So i went checking and thought this kp could be the error (am realy a rookie at this stuff). Thought the kp value would go larger at larger (delta)T values.

On the other hand, I couldn't find any other way to calculate this. (Have no other kp value, and am not sure how i'd calculate it by myself).

So you say this kp for (delta)T = 40 deg.C can be used here even though i get as result the temperature would go up high?

Thx

You could just install an inline temperature gauge(s). If runs too hot, shut down and install an oil cooler.

You don't say what kind of system this is, but it sounds like something that should have devices to take periodic temperature readings built in.

This system should've been installed on a vehicle to run a high capacity pump. It won't though cause it is a very large and heavy system, that's not doable. This could be tried out for smaller pumps though. It would require smaller components and it wouldn't heat that much. For now it's only on paper, and for sure it'll stay for a while like this.

Just wanted to check if that what I did was right, and if i CAN use this kp at (delta)T = 40 deg.C.

You give a power loss of 17 kW. How big are flow and pressure and which installed power has the main pump?

I think that your reservoir is too small for the pump flow and if you have such a power loss then you are obliged to use an air cooler.

The system has very large flow value therefore the big loss (that is if i calculated the losses and the temperature right, still trying to figure that one out :) )

The oil tank volume is 5/8 value of the max pump flow, I wouldn't go with any bigger. (although putting a bigger tank wouldn't change the result very much, since the system will not be useable cause of its dimensions and weight). This is just theoretical work. I wanted to see if such a system would be of any use.

Here another question. Just wanna confirm i got it right....

I calculated the power loss this way:

Maybe I should say first what's in the line i calculated the losses from. It's very simple.

A pump runs the fluid through a 4/2 valve towards an motor. Behind the motor is a filter, and by exiting the filter the fluid goes back to the oil tank.

There are some more lines in the simple system, but i ignored them since the oil doesn't flow through the lines when everything is ok in the system. Some very small amount of oil flows through the line when the motors revolution must be regulated, and that's more or less it.

I calculated the pressure drop in the lines which came out to be around 0,5 bar.

Then i calculated the losses in the local resisters (enterance from the oil tank to the line, the reductions, T-parts, exit from line to the oil tank)

By adding this two results together i got the total pressure loss in the lines and resisters.

The power loss is calculated by adding together the power losses in the pipeline, the 4/2 valve and the filter.

that is:

N(line)= (maximum pump flow)*(total pressure loss in lines and resisters)/600

N(valve) = (maximum pump flow)*(total pressure loss in the 4/2 valve)/600

N(filter) = (maximum pump flow)*(total pressure loss in the filter)/600

N=N(line)+N(valve)+N(filter)

No cooler loss in it at this moment, cause the calculation is used to see what cooler should be used.

Can someone check this please?

If that's correct, i'll have to assume the temperature i got from this power loss is correct.

You neglected 2 important losses: internal losses in pump and motor, all frictions generate heat and this heat is transported by the oil flow. Your losses can be even higher than what you computed.

These 2 I ignored. The oil heats up when it moves from high pressure to low pressure without doing any work. In both pump and motor does work, so this was ignored.

Have you ever heard about pumps and motors efficiency?

If not take a look at the graphs in correct catalogues. The internal losses represent 5 to 8 % of input power. In some designs even more !

So you neglect losses about 10% of pump input power ? You risk to choose a too small cooler but anyway it is a theoretical analysis so that it does not matter.

You might be right. I'm a rookie at this field.

Tried calculting it. Additionally adding 10% of the total motor/pump power to the equation would make the power loss and temperature values go sky high.... Values i wouldn't be able to cool down.

I have the feeling that you used the maximal flow and the maximal pressure in your computations. In general first the power is a costant so that if flow goes up pressure goes down and as was very well said the power requirements are NOT all the time equal to the maximal available power so that again it should be thought more into details and not too rough. If other can do why not you why such a negative reaction ?

No negative reaction. :) Thx for all the info you guys gave me. It's very useful

I'm calculating it as we speak.

Yeah am using the data for max pressure and flow. I need the worst case scenario result.

In practical use, the pump would run at a high constant speed, the motor itself is variable.

Some of the energy imparted to the fluid will go to producing mechanical work; not all of it goes into heating the oil. Depending on the system design, if the valves are "open center", much of the time the pump will running at much less than full power, and thus putting less energy into the system.

Some of the principal hydraulic manufacturers should have some information on calculating oil coolers for various system conditions.

You sure those temperatures shouldn't be in absolute units (ie. K)?

@Tornado: The pump would work at it's max capacity. The motor is the one that should directly regulate its own speed.

@Guest: It wouldn't make any difference if I put the environment temperature in K there, cause when i again subtract the 273,15 later, i would get the same value as now where I used the deg.C.

Not subtract, meant add....

Sorry. Early in the morning... :)

Hi,

Whilst it is possible to calculate the heat energy wasted it is mostly done by hydraulic engineers by experience. I assume this is a mobile system i.e. on a vehicle. and that the duty is intermittant.

A crude but effective first concept is what is the installed power hp or kw. What are the pumps/ actuators. Piston pumps, gear pumps, cylinders and/or motors. If say it is a gear pump to a gear motor then efficiency is likely to be 85% for the pump (at best) and 75% for the motor. therefore 0.75 x 0.85 = 64% efficent i.e. 36% of the installed power will go to heat. So if you are putting in 100kw's then you have to get rid of 36 in heat energy. Your tank is too small even by mobile standards, you should have a capacity equal to the pump output in one minute. This is not for heat dissipation by to allow for de-aeration etc. Your tank will never get rid of more than a few kw's. Industrial tanks of say 600 litres capacity will get rid of some 6kw's on a continous basis. The next question you have to ask is how long does the system operate and how long is it at rest. If it runs for two minutes once a day you can get away with a lot but if it runs for 30 minutes buy a cooler.

Regards

Oliver Dunthorne

Hydraulic engineer

Both pump and motor are axial piston types. The problem is the manufacturer Bosch-Rexroth didn't put any info of the efficiency. The standard values used for their calculations seem to be 0,8-0,85 for both motor and pump.

so i'd have to add the following to the losses equation:

(total power of pump) x 0,15 + (total power of motor) x 0,15

Not liking the idea...

Hi, Bosch Rexroth would normally give such detail, I am suprised you cannot find it. But it is likely to be 85% plus for both pump and motor so say 0.85 x 0.85 = 72% so you have 28% (of power input)going into heat. So the next question is how long do you run and how frequently.

Oliver Dunthorne

Obviously there is some kind of mistake in measuring units or in your assumptions for calculation. You NEVER will get such temperatute in a hyd. circuit unless you deliberatell give heat to it. At 500 ° the whole trhing will be fried!!!. think in hoses, seals, oil viscosity, etc.

You should review deeply what you did.

Good luck!!!

I would install a full line bypass and two diverting valves on either side of the bypass valve in your piping. If you realize later that you need a cooler (or a large one), you can install to the diverting valves and close the bypass.

If you are preparing a proposal, you can add that you think a cooler might be needed, but you will leave it as an option that can be purchased later. To easily accomodate the installation of a cooler, you will provide the full line bypass valve / diverter assembly.

This gives you and the customer a chance to determine the real heat load to the system - taking into account any variations due to environmental conditions

Both arses are covered.

Regards,

Paul

Hi,

Following my earlier comments. I should have said open or closed loop circuit.

Regards

Oliver Dunthorne

Thx to all! I added the pump and motor power losses to the equation, the rest seems to be in order :)

I got another question though.

About calculating the temperature using this formula I added in my first post:

T= (N/(kp*As) + Te

Where T is the temperature, N the power loss, kp the Thermal conductivity of the chosen oil tank, As the surface area of the tank, and Te the environment temperature.

Since I didn't have any idea what the temperature would be at the time i calculated this, I used the kp that was added in the data sheet of the chosen oil tank that was for delta T = 40 deg.C.

What I want to do next is add the Thermal conductivity of the pipes to the equation.

If i'm not misstaken the equation should look like this then :

T= (N/((kptank*Astank) + (kppipe*Aspipe)) + Te

SO my question is. Is there some table i could pull the kp for the pipes out?

Tried to calculate it using equations from thermodynamics, but the weird thing is, when you calculate the radiation from the pipe to the environment, you don't add any numerical data in the equation that invoves the material of the pipe (oil resistant rubber). I'd guess that would be an important component in calculating thermal conductivity of the pipe? So...where's tha catch? :/

the losses from a hydraulic piston pump to a hydraulic motor approach 25-30%. In industrial installations we install we "NORMALLY" base the cooler on the full input horsepower. i.e. 100 hp electric equivalent 100 hp cooling based on flow! This looks like overkill but when you loose 10-50% efficiency in plugged lines in the cooler (lack of maintenance) the system may still function OK. Cooling loss from pipe work is negligible and almost null thru hoses. Tank cooling is also only about 1-5% depending on size. If the system does not have idle dwell time sufficient to cool down the oil then a cooler is the only way.

You mentioned some small line that you did not consider significant. If these are leakage lines from the piston pump and the piston motor they are very significant as the are deliberate orifices in the units used for lubrication and internal case cooling. As they are leakage lines they usually have a high temperature associated with them. i.e. HEAT

The power to weight ratio for this type of device is still extremely good (maybe the best), but the efficiency is still a bit offensive.