Selecting hydraulic motors and a pump for hydrostatic transmission

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So I'm trying to make a GoKart that utilizes a hydraulic drive for the front two wheels and a mechanical driveshaft for the back two wheels but my main questions are selecting the motors and pumps.

I just want to clarify a few things since I'm not sure if I did my math right.

The max RPM for each of the front two wheels is 600RPM, and the max theoretical Horsepower is 2.5 for each wheel. But due to inefficiencies, I'm expecting maybe 300 RPM and 1.5 HP for each wheel.

From this, I can calculate the torque developed by each motor to be 315 in-lbs. However, I'm looking for a small, circular-shaped motor and it seems Orbital motors seem to fit the bill despite their 80% efficiency.

I have found a potential candidate for a motor that supplies more than enough max cont. torque and max cont. RPM at 80% efficiency here (specifically I've chosen the 2.8 cu in/rev motor).

Assuming I do go with this motor, I've been trying to find an axial piston pump (based on my wants that this needs to be a variable displacement pump) that can handle an input shaft speed of 2030RPM (that's the shaft speed of my intermediate shaft in my gearbox, which I intend to use two gears to connect my pump shaft and intermediate shaft), a total system flow rate of 10GPM, and system pressure of 589PSI.

So my two questions are:

A) Does the math make sense/motor selection make sense?
B) What kind of axial piston pumps might meet what I want? I can't find any axial piston pumps that meet my specs without it being overkill and extremely heavy.
 

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  • #2
Baluncore
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I can't find any axial piston pumps that meet my specs without it being overkill and extremely heavy.
I'm surprised you would use power hydraulics in a GoKart. The system efficiency, is the hydraulic efficiency of the motor multiplied by the efficiency of the pump. That is not a happy result.

Anyhow, the required pump will be determined by the hydraulic circuit. That will in turn be decided by the tyres, surface, and wheel slip expected.

If your motors are in parallel, the same torque will be generated on each side, so you have a differential. If the motors are in series, then you have the equivalent of a locked differential. On a hard surface you must provide a relief valve or bleed to limit the slip and tyre wear.

For the series motor combination you need half the flow at twice the pressure. For parallel motor combination, you get twice the flow at half the pressure. It is a pity that it is not the other way around.

So what is your wheel slip and proposed hydraulic circuit ?
 
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  • #3
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I'm surprised you would use power hydraulics in a GoKart. The system efficiency, is the hydraulic efficiency of the motor multiplied by the efficiency of the pump. That is not a happy result.

Anyhow, the required pump will be determined by the hydraulic circuit. That will in turn be decided by the tyres, surface, and wheel slip expected.

If your motors are in parallel, the same torque will be generated on each side, so you have a differential. If the motors are in series, then you have the equivalent of a locked differential. On a hard surface you must provide a relief valve or bleed to limit the slip and tyre wear.

For the series motor combination you need half the flow at twice the pressure. For parallel motor combination, you get twice the flow at half the pressure. It is a pity that it is not the other way around.

So what is your wheel slip and proposed hydraulic circuit ?
I'm just doing it for the challenge.

I thought I'd need to spec my motors off of my RPM and Horsepower and select my pump based on the displacement/flow rate from my two motors?

I can't say I know the specifics of hydraulic circuits yet since I just started. I do know the two motors will be in parallel for the reason you stated above.
 
  • #4
Baluncore
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The max RPM for each of the front two wheels is 600RPM, and the max theoretical Horsepower is 2.5 for each wheel. But due to inefficiencies, I'm expecting maybe 300 RPM and 1.5 HP for each wheel.
The next step in the process is to identify the maximum road speed and front wheel diameter. That will enable calculation of the maximum flow rate through each motor. If at high speed, the hydraulics would not be needed then you could put a one way check valve across the motors so they then run at high speed as circulating pumps.
The hydraulic motors will lock if you try to push backwards or reverse. Is that a problem ?
 
  • #5
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The next step in the process is to identify the maximum road speed and front wheel diameter. That will enable calculation of the maximum flow rate through each motor.
I've already calculated the total system flow rate in Gallons/Minute (10GPM) based on the maximum velocity of each motor and motor displacement of each motor.

The hydraulic motors will lock if you try to push backwards or reverse. Is that a problem ?
It won't since I have no intention of going backwards for this version.
 
  • #6
Baluncore
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I've already calculated the total system flow rate in Gallons/Minute (10GPM) based on the maximum velocity of each motor and motor displacement of each motor.
Your gallons may be different to my gallons = 4.546 litre. I need the pressure specified in bar, or kPa. I need flow rate in litres per second. Then the numbers come out in kW without any horse power conversions.
You have specified a motor that appears to be 3 times bigger than is needed. By not running full pressure you are carting around all that extra mass. More flow is less efficient when you have hydraulic lines, so you need to use bigger hoses and more expensive fittings. Increase the maximum pressure, reduce the flow.

If you set a variable flow pump to regulate output pressure, you will have constant torque on the front wheels, no matter what speed within the design range. That must be a big pump because your motors are overly high displacement. A pressure regulating pump eliminates the need for a pressure regulating bypass valve, and so increases the efficiency.

Will you have independent front wheel bearings, or were you planning to use the motor as the axle?
Which port version of the motor were you planning to use. The side port positions make driving steering wheels difficult. If you use the end connections you will have a lower maximum pressure rating and so a higher displacement, heavier system.

The max RPM for each of the front two wheels is 600RPM, and the max theoretical Horsepower is 2.5 for each wheel. But due to inefficiencies, I'm expecting maybe 300 RPM and 1.5 HP for each wheel.
The inefficiencies are the extra energy you must put in to get 600 RPM out. Is the maximum wheel speed you required 600 RPM or 300 RPM? How much power can you take from the engine to drive the pump?
 
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  • #7
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Your gallons may be different to my gallons = 4.546 litre. I need the pressure specified in bar, or kPa. I need flow rate in litres per second. Then the numbers come out in kW without any horse power conversions.
Ok I'll include SI units then. 37.8541liter/min for the new motor, 58.9502 Bar (system pressure)

You have specified a motor that appears to be 3 times bigger than is needed. By not running full pressure you are carting around all that extra mass. More flow is less efficient when you have hydraulic lines, so you need to use bigger hoses and more expensive fittings. Increase the maximum pressure, reduce the flow.
So then I'll need to select a motor with a max continuous RPM/Torque values near my theoretical RPM/Torque values then? I was kinda going overboard with the range of torque/rpm. There's a few motors I've found that match that, like this that looks promising?

If you set a variable flow pump to regulate output pressure, you will have constant torque on the front wheels, no matter what speed within the design range. That must be a big pump because your motors are overly high displacement. A pressure regulating pump eliminates the need for a pressure regulating bypass valve, and so increases the efficiency.
Okay, so I'm a bit confused now. If I go with a variable displacement vane pump, it seems like I can go with a variable displacement, pressure-compensated vane pump, or just a variable displacement vane pump right? But I originally thought a variable displacement pump would adjust it's flow rate based off downstream pressure and now it seems like that's the job for a variable displacement pressure compensated pump?


Will you have independent front wheel bearings, or were you planning to use the motor as the axle?
Which port version of the motor were you planning to use. The side port positions make driving steering wheels difficult. If you use the end connections you will have a lower maximum pressure rating and so a higher displacement, heavier system.
I plan to attach the motors to my uprights, which is then connected to my hubs (hubs will probably be designed so it fits the keyway of my motor.

I'm not sure if I have the freedom to select how my ports are oriented for certain motors. The motor I attached to this post has end connections.
 
  • #8
Baluncore
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I'm not sure if I have the freedom to select how my ports are oriented for certain motors. The motor I attached to this post has end connections.
Then you must specify a motor with the port connection orientation needed to fit the application.

Ok I'll include SI units then. 37.8541liter/min for the new motor, 58.9502 Bar (system pressure)
Pressure * flow rate = power; In SI units; Pa * m3/sec = kPa * litre/sec = watts.
Here is an example; First convert to SI units;
58.9502 Bar = 58.9502 bar * 100 kPa/bar = 5895.02 kPa
37.8541 litre/min = 37.8541 / 60 = 0.631 litre/sec
5895.02 * 0.631 = 3719.75 watt = 3.719 kW = 5.0 HP.
But you have not clearly specified if that is for one motor, or for two, or for the pump?

You must clearly specify exactly what you mean by a number. Only then might rational analysis and design be possible. Your proposed system takes energy from the engine and transfers it to the front wheels.
1. You must specify the available power from the engine to the hydraulic pump.
2. You must specify the pump minimum and maximum RPM.
3. You must specify the maximum motor RPM required.
Carve those three values into stone.
Do not continue to obfuscate and hide those numbers and units.

Your proposed alternative motor has a capacity of; 1.93 cu” = 1.93 * 2.543 = 31.6 cc/rev
That is very similar to the old motor capacity of 36 cc/rev. I see no advantage.
If you keep grabbing motor specs without constraint, you will build a heavy and expensive system, it will be inefficient, and may stall the engine.

Once you specify the power available from the engine to the pump, and the maximum motor RPM required for the front wheel drive, everything else will follow. It will be possible to maximise the motor torque available and minimise the weight and cost. From that will come the motor and pump selection parameters.
 
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But you have not clearly specified if that is for one motor, or for two, or for the pump?
2.5 HP for each motor.

1. You must specify the available power from the engine to the hydraulic pump.
5 HP is available for the pump

2. You must specify the pump minimum and maximum RPM.
Max pump RPM 2030. Minimum is 600RPM.

3. You must specify the maximum motor RPM required.
600 RPM theoretical, not accounting for losses.

If you keep grabbing motor specs without constraint, you will build a heavy and expensive system, it will be inefficient, and may stall the engine.
I've been basing my motor specs off of the 5 HP the pump will get from the engine and maximum 600 RPM of each wheel, by finding whatever motor provides max continuous torque of about 262 in lbs, and max continuous RPM of 600 RPM. Apparently, that's wrong since it sounds like you're alluding that there's a different, more correct method which I'm not aware of.
 
  • #10
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I may be wrong, but it seems to me that the synchrony of torque and rpm’s between front and rear axles may be a problem.
The rear axle receives power directly from the engine via roller chain #1 or belt: reduction #1.
The front axles receive power from the same engine via roller chain #2, pump and hydraulic motors: shouldn’t reduction #2 equal reduction #1?
If so, how could that synchrony be achieved?
 
  • #11
Baluncore
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I may be wrong, but it seems to me that the synchrony of torque and rpm’s between front and rear axles may be a problem.
The rear wheel drive is through a step ratio gearbox. So ratio depends on gear selected.
But the pump runs from lay-shaft in gearbox, so is proportional to gear independent engine RPM.

The variable geometry pump generates sufficient volume to maintain a set pressure. The pressure to the front motors is therefore controlled. With a set pressure applied to the two motors in parallel, the fixed displacement motors will generate a torque proportional to pressure, independent of road speed.

The front wheel torque can be set by adjusting the pump output pressure. The compensation for the different gear ratios is therefore automatic.
 
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  • #12
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Your proposed alternative motor has a capacity of; 1.93 cu” = 1.93 * 2.543 = 31.6 cc/rev
That is very similar to the old motor capacity of 36 cc/rev. I see no advantage.
So on what basis makes you say this motor is too powerful for the specs I listed?
 
  • #13
Baluncore
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That motor could be too powerful, if it was fully utilised. It has too much flow, and so would need to operate at too low a pressure. But focussing on that motor is designing backwards, basing specification decisions on what you find in the marketplace, rather than looking for and finding what is rational and optimum. Scale errors at the start of a design can become very expensive by the end.

The cost of hydraulics is low for “agricultural pressures” below about 135 bar, (2000 psi). The design strategy should be to use the full available pressure range, and so to minimise the maximum flow rate required. That will minimise the cost and weight of the system.

I'll try to do more calculations, but I must wake up and finish work here first.
 
  • #14
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The rear wheel drive is through a step ratio gearbox. So ratio depends on gear selected.
But the pump runs from lay-shaft in gearbox, so is proportional to gear independent engine RPM.

The variable geometry pump generates sufficient volume to maintain a set pressure. The pressure to the front motors is therefore controlled. With a set pressure applied to the two motors in parallel, the fixed displacement motors will generate a torque proportional to pressure, independent of road speed.

The front wheel torque can be set by adjusting the pump output pressure. The compensation for the different gear ratios is therefore automatic.
I know that you are busy now, Baluncore, but I would like you to explain this a little further whenever you have the time.
Doesn't the rpm's of the lay-shaft in gearbox depend on the rpm's of the engine?
(Could you @AielloJ clarify this point?)

The way I see it, perhaps incorrectly:
Torque supplied to each front tire will be limited by the normal force on the contact patch: too much and tires spin.
The rotational speed of that same tire will depend on the fixed ratio of displacement of pump and motor.

If that pump is directly linked (via fixed gearbox reduction) to the rpm's of the engine, the displaced flow will vary with it.
Wouldn't that fact induce a mismatch of tangential velocities among rear (modulated by gearbox) and front tires?

I appreciate your previous explanation.
 
  • #15
Baluncore
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Here are my initial calculations, ignoring hydraulic efficiency.

5 HP is available for the pump drive. That is 3.7285 kw.
If the pump output pressure is set at 135 Bar, (2000 psi), = 13500. kPa.
Then the flow volume needs to be 3.7285 kW / 13500 kPa = 276.2 ml/sec.
With the pump at 2000 RPM = 33.3 rev/sec;
At lower pump RPM, the road speed will be lower so motor flow will be less.
The pump displacement needs to be 276.2 / 33.3 = 8.29 ml/rev.

The motors will share that 276.2 ml/sec, as 138 ml/sec each.
The max motor speed is 600 RPM = 10 rev/sec;
So each motor will have a displacement of about 138 / 10 = 13.8 ml/rev.

Now you can start to look for motors with a displacement of 13.8 ml/rev. (2.5 HP).
You can look for a pump with a displacement of 8.3 ml/rev. (5 HP).
The pump needs an adjustable, output pressure, self regulating.

The inefficiency of power hydraulics is due to two effects.
1. Pressure loss due to flow through passages, valves, lines, and connectors.
2. Fluid leakage due to clearances in components and seals.
Higher viscosity fluid reduces leakage losses, but increases pressure loss due to flow.
The optimum fluid viscosity is always a compromise.

As the hot fluid carries away the wasted energy it must be cooled to maintain fluid characteristics and quality. The tank return line, and the fluid reservoir, are thermal radiators. If the area available cannot radiate sufficient heat, then a fluid radiator must be provided. Plan to radiate 25% of the input energy, unless you can show it will be less. For a 5 HP system expect to radiate 1.25 HP = 940 watt.

The fluid reservoir will need a mesh sieve and a spill proof air vent. There will need to be a low-pressure high-flow fluid filter in the pump input line. Maybe you also need provision for a vacuum gauge at the pump input, so you can monitor the filter as it becomes blocked.
 
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  • #16
Baluncore
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The rotational speed of that same tire will depend on the fixed ratio of displacement of pump and motor.
The engine turns the gearbox input shaft, which is geared with constant ratio to the lay-shaft and a pressure compensated hydraulic pump. So the volumetric pump ratio is NOT fixed.

The solution comes from the adjustable displacement pump. The displacement is controlled by the outlet pressure so as to maintain the set pressure required.
https://ifp1.com/pressure-compensated-pump-controls/
The pressure compensated pump must be adjusted to the pressure where there is little wheel slip.

Imagine the inverted electrical equivalent.
DC motors at front wheels, with a constant current, have fixed torque. Speed is not important.
Hydraulic motors at front wheels, have constant pressure, so have fixed torque at any road speed.

At maximum RPM, in top gear, pump must provide sufficient volume for motors at top road speed.
At half RPM, in top gear, pump must provide sufficient volume for motors at half road speed.
But those are both in highest gear, and they are proportional, (on a straight line).
For low road speeds, in lower gears, the pump will adjust displacement to regulate pressure.

Shunting the flow around the motors, while stopping flow from the pump, will stop the forward torque, allow free-wheel, and reduce hydraulic power consumption.
 
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  • #17
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The engine turns the gearbox input shaft, which is geared with constant ratio to the lay-shaft and a pressure compensated hydraulic pump. So the volumetric pump ratio is NOT fixed.

The solution comes from the adjustable displacement pump. The displacement is controlled by the outlet pressure so as to maintain the set pressure required.
https://ifp1.com/pressure-compensated-pump-controls/
The pressure compensated pump must be adjusted to the pressure where there is little wheel slip.
It is very clear now.
Thank you for your time.
 
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  • #18
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Here are my initial calculations, ignoring hydraulic efficiency.

5 HP is available for the pump drive. That is 3.7285 kw.
If the pump output pressure is set at 135 Bar, (2000 psi), = 13500. kPa.
Then the flow volume needs to be 3.7285 kW / 13500 kPa = 276.2 ml/sec.
With the pump at 2000 RPM = 33.3 rev/sec;
At lower pump RPM, the road speed will be lower so motor flow will be less.
The pump displacement needs to be 276.2 / 33.3 = 8.29 ml/rev.

The motors will share that 276.2 ml/sec, as 138 ml/sec each.
The max motor speed is 600 RPM = 10 rev/sec;
So each motor will have a displacement of about 138 / 10 = 13.8 ml/rev.

Now you can start to look for motors with a displacement of 13.8 ml/rev. (2.5 HP).
You can look for a pump with a displacement of 8.3 ml/rev. (5 HP).
The pump needs an adjustable, output pressure, self regulating.
Oh wow okay this is helpful, thanks. The math makes sense when I know the HP available for the pump and the pressure, but where did you come up with the output pressure? Is there a standard output pressure for certain pumps doing performing certain tasks?

and just to be clear, your pump output pressure would be the max. operating pressure?
 
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  • #19
Baluncore
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Is there a standard output pressure for certain pumps doing performing certain tasks?
and just to be clear, your pump output pressure would be the max. operating pressure?
"Agricultural" pressures are maximum of about 135 bar = 2000 psi.
If you soak yourself in hydraulic oil you soon learn those values used on old tractors. The parts for those pressures are reliable, and available at low cost.

Those are maximum working pressures.

By specifying input power, the efficiency calculation is largely eliminated. Notice that inefficiency = pressure loss, is accumulated as it flows from the pump shaft to the motors and reduces the final pressure and so torque. Hopefully there will not be too much fluid bleed that would reduce wheel RPM.
 
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  • #20
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The motors will share that 276.2 ml/sec, as 138 ml/sec each.
From my understanding, the displacement for the pump tells you how much volume of fluid is moved per unit of time through the internal cavities of the pump.

Since flow volume is the amount of fluid "spat out" from pump per revolution, that was why you halved the flow volume, rather than displacement for each motor right?

Now you can start to look for motors with a displacement of 13.8 ml/rev. (2.5 HP).
You can look for a pump with a displacement of 8.3 ml/rev. (5 HP).
The math works out right for this, but I'm having trouble wrapping my head around the fact that each motor requires a higher displacement value than the pump conceptually. The way I see this is that with two motors, I need a displacement of 27.6 ml/rev but I'll be looking for a pump of 8.3 ml/rev?
 
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  • #21
Baluncore
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From my understanding, the displacement for the pump tells you how much volume of fluid is moved per unit of time through the internal cavities of the pump.
The displacement is the volume swept per revolution.
The flow rate is the displacement multiplied by the revolutions per second.
The pump rotates faster (2000 RPM) than the motors (600 RPM), which explains the displacement ratio.
 
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Off the top of your head, what sites do you recommend looking at for pumps? Sites like Eaton or Parker have pumps that appear to have pretty high max pressure than what's needed.
 
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  • #23
Baluncore
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I would use Google to find suppliers, then look at what they stock.
Distributors like, for example; https://www.whitehouseproductsltd.com
Then I would get data from the manufacturers site.

Start design with the motors, select a pump, and repeat, juggle the choices until you have a solution. For example, you might slightly slow the gear drive to the pump, and increase the pump displacement, so as to maintain the motor RPM and flow rate.

Look for Chinese clone products, then find the original design from the West.

Edited:
Look for a "pressure compensated" pump such as 10YCY14, for example;
https://www.aliexpress.com/item/4000671061004.html
https://www.amazon.com/dp/B07HFBYX1J/?tag=pfamazon01-20
https://www.hudsunindustry.com/product/variable-displacement-axial-piston-pump-ycy/
 
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