Dissipation of heat from hydraulic systems

[subberzz]

Hey everyone,
I want to know what considerations must go into selecting a air-cooled radiator for a open loop hydraulic system. My understanding is the amount of energy put into the system is determined by your prime mover and in this case its a 30KW e-motor. If the pump is a variable displacement piston pump with a max operating displacement of 66.5LPM based on max swash plate angle displacement and motor RPM, what is the general rule when considering removing heat from the system? How much of the input energy should be considered as heat?

Regards,
s

 

[BrianPetersen]

All of the input mechanical energy should be considered the heat load.

The mechanical efficiency of hydraulic systems in general is pretty poor, and all of the input energy has to go somewhere...heat, eventually.

 

[hydtools]

For our mobile systems we use 50% of our rated hydraulic power to be rejected at ambient temperature of 100F. Our design hydraulic temperature was 140F. E.g., 50% of 8 gpm @ 2000 psi. We never had an overheated system.
We had also a thermostatic bypass valve to bypass the heat exchanger at colder temperatures.

Ted

 

[subberzz]

@brianpetersen, so a cooling circuit should be able to remove the total energy input in the form of heat? It was my understanding that if a system is operating at below 90% efficiency it should be taken out of service. For example if 20KW of heat was dissipated from a 100KW system then less than 80% is turned into useful work.

Is there a general design rule when choosing cooling solutions? 50%, 75%, 100% of input energy?

 

[LittleInch]

I don't know what the nech efficiency of a swash plate pump is but my guess would be 70 to 80%.

Worst case is full power so to be on the safe side 40%of max power would seem to be a good guess without getting more data from the suppliers. Also how long does it ooerate at max power? Continuous for long periods, use 40%. Short bursts, 30%. IMHO.

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

 

[BrianPetersen]

Few things to be mindful of.

The "rated" power is usually not the "actual" power. Hopefully, the "actual" power will be less, because you're not always going to be demanding full flow, and hopefully the pump isn't sized right to the limit and the motor that's driving it isn't sized right to the limit. So it's quite conceivable that the actual power (energy) being put into the hydraulic pump could be 50% give or take of "rated" on some sort of long-term average.

As for how much efficiency you can get ...

If you're lifting something with a cylinder, if you have any sort of flow control, that's a pressure drop (i.e. an energy loss). If it lifts the load up, and it has to come down again, it's practically always dissipated through a flow control / counterbalance valve / pressure regulator / "pressure drop" of some sort. I've yet to see a hydraulic system of this sort that is capable of "regenerative braking". They always throw it away as heat (due to flow through a pressure drop through an orifice / flow control / counterbalance valve / etc).

There can be some applications where the efficiency isn't terrible, e.g. your hydraulic pump is operating a hydraulic motor that's mechanically driving something else, and there are no flow controls causing a pressure drop anywhere in the system - just the pump, and some hoses, and a hydraulic motor - effectively, a hydraulic transmission.

 

[hydtools]

The prime mover puts no heat into the hydraulic system. So it is not the basis of hydraulic system heat generation. Start with the delivered hydraulic power. The losses in the hydraulic system generate system heating. What is your system temperature limit you want to have at what ambient temperature(s)? How inefficient is your hydraulic system and for what time interval, work cycle.

Ted

 

[73lafuite]

Hello,
Any hydraulic power (flow x pressure) that is not transformed into mechanical power (force x speed) is transformed into oil heating.
A pressure drop of 100bar in a component without producing mechanical energy (flow limiter or pressure reducer or pressure limiter etc.) produces a temperature rise of 5.5°C.
For example a wood splitter with a horizontal cylinder, a simple gear pump and an open center distributor will not need a cooler. The small pressure drops in the pipes, the friction of the seals of the cylinder and the internal friction and leaks of the pump cause low power losses compared to the power of the motor driving the pump. The heat exchanges by conduction of the system (mainly tank and cylinder) are sufficient to control the temperature. But if the cylinder stalls at 200bar because it cannot split the wood: the pressure limiter will spit the oil towards the tank at 11°C more than at the inlet.
A system's cooling calculation is so dependent on your application that there is no fixed percentage of cooling power to install.

 

[subberzz]

Hey all,
Thankyou for your insight and responses.

@LittleInch, I haven't actually seen a manufacturer put the mech-efficiency in the spec sheets. Maybe i'm not looking hard enough or its just system dependent. The system is near continuous use with short overhauls maybe two a year, it consists of a variable displacement piston pump that drives a hagglunds motor. The system operating pressure is on average 125 bar with periodic fluctuations(+- 25bar) based on the controller and sensor inputs. Unfortunately I am unable to put a flow meter in-line so have to determine a line of best fit so to speak on system flow.

@BrianPetersen , As I understand it the input power from the motor is a good place to start when determining the overall maximum amount of KW power in the system. When I say maximum I mean the system cannot provide more KW than inputted.
Since the pump is variable and isn't using the total rotational input unless max pressure I am toying with determining the KW used by the system using KW=((l/min x dP)/60000 with dP in kPa. I figure since the system pressure is approximately half of max capacity 30 LPM is a good place to start, which gives 6.3KW of input energy in service. So if the system had a efficiency of 80%, 1.26KW of energy would be lost to heat. Please advise if i'm misunderstanding anything, i'm still on a journey of learning.

@hydtools , thanks for your input bud, i feel you put me in the right direction. System runs in ambient ranging from 10C - 50C, runs continuously at 125 bar. I'm not exactly sure how to determine the inefficiencies ? volumetric efficiency? Please point me in the right direction as i am willing to read and understand.

@73lafuite, thanks for your input. As above for particulars of the system, only thing that I have to add is a Danfoss control block is used to change the signal pressure for the differential spring setting on the pump controller so the standby can be changed on demand. The block controls flow to the motor and is the only orifice between the pump and motor.

 

[hydtools]

@subberzz here is a link to an older but useful Womack series. The heat exchanger example is shell and tube, but the process would be the same for any type of heat exchanger. The units are imperial.

https://www.womackmachine.com/engin...heets/cooling-an-overheated-hydraulic-system/

Ted

 

[73lafuite]

A diagram, photos or Danfoss pump/motor/block references would allow you to calculate this.
At Hagglunds the Vicking engines have a large internal leak and good mechanical performance. Marathon engines have low leakage and good mechanical performance. Compact motors have little leakage but a slightly lower mechanical efficiency.
It is therefore assumed that your engine must never retain a load, so that it never works as a pump.
From what I understand your Load sensing pump delivers 10/15 bar more than the engine pressure. In this case the power lost in the Danfoss block is 10bar*30l/mn/600=0.5kW. The power lost at the pump is around 30l/min x125bar/600 x15% = 0.9kW. The power lost in the Hagglunds engine depends a lot on the model. A Vicking engine of 38000cm3/t with 30l/mn at the inlet has an efficiency which tends towards 0....
Now you are talking to us about putting on a cooling. Why today? Could it be because you have wear and internal leaks that are becoming more important?

 

[LittleInch]

I think you need to really work out the maximum shaft power of the pump - so max flow (66 LPM) x your max DP the pump is running at. If this is 125 bar, then I calculate with a pump efficiency of 70% you have a shaft power of about 19kW. I don't recognise where your 60,000 denominator comes from, but would appear to assume some sort of pump efficiency?

Then take min 20% of that as heat to be rejected - 3.8kW min

If your liquid tank is big and the length of time it operates flat out is small then you can reduce that, but risk that if you go flat out for more than say 10 mins, your oil temperature will start to climb.

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

 

[IceStationZebra]

Another thing to consider is how long will the system be active? You don't mention if this has to be running indefinitely but you will see some systems with little to no cooling. In some cases they are meant to run intermittently and there is sufficient thermal mass that they can run long enough to get the job done. And in some cases there is sufficient radiant cooling, but if you do the calcs the effect is usually fairly small.

As far as sizing goes I will echo what everyone above said. You have to determine the energy balance: energy in - net useful work = cooling load. Be cognizant of operations that convert hydraulic energy into potential energy, then from potential to hydraulic to heat. A good example would be a forklift that raises a load. But in a short period of time also lowers a load. (I've had to convince other engineers on this)

ISZ

 

[BrianPetersen]

Yes...you may have a circuit that is 80% efficient at raising the weight of that load...and then it's lowered through a flow control and counterbalance valve. 0% net efficiency, it's just that the heat load is shifted in time and space.

Or it's 80% efficient at max load at fluid pressure 100 bar...but then when it's only carrying half load, the regulated pressure is still the same but the diff is made up in a flow control valve.