Hydraulic Motor Power Confusion

[BEMPE16524]

Dear All,

I have a confusion here. I have a working system like this,

Electrical motor:
RPM: 1500 @50Hz
Full load torque: 1651
Power: 250kW

Hyd pump:
Model: Eaton PVWS 360M (variable displacement)
flow: 360cc
Power consumption: 315kW @ 350Bar
Driving torque: 2005 @ 350Bar

above pump will drive a hydraulic motor below:

Hydraulic motor:
Model: Rextroth A6V (variable displacement)
displacement: 355cc
Max Torque: 1978 Nm
Max Speed: 2950RPM

So, our operating pressure is 220 Bar.
lpm @ 220 Bar of hydraulic pump = cc/rev x RPM / 1000 = 540lpm
@540lpm, the hydraulic motor will deliver 1344Nm and 2525RPM

When I calculate the hydraulic motor power base on this calculation they gave in their catalogue,
Power = 2 x 3.142 x T x RPM / 60,000 = 646HP = 482kW.

This Hydraulic motor is driving a submersible 3 stages water pump with following spec:
Head: 171m
speed: 2920 rpm
(see attachment for detail)

My question is, can that be possible? Hydraulic motor power is 482kW but the electrical motor is only 250kW? Sorry if I sound stupid, but that is a working system from a manufacturer which I can't trace them anymore. Or did I make false assumption some where?

TIA.

R.Efendy

 

[3DDave]

It is not possible. The electric motor input is easy to find/confirm by measuring voltage and amperage. Not all 250kW ends up at the shaft so the pump is starting with less.

Note that the peak torque the pump can handle is more than the motor can supply; similarly the power input allowable to the pump is far more than the motor can supply.

The main assumption error is that the components have only their peak operating condition power and torque. The pump cannot take in more than the motor can deliver.

Per

https://hydraulicsonline.com/wp-content/uploads/2019/12/Bosch-Rexroth-A6VM-Series-63.pdf

page 8

Vg = 355 cc/rev

dP = 220 bar

n = 1651 RPM
= 1651

Qv =vg*n/1000
= 586.105 lpm

T = vg*dp*1/20/pi
= 1243.00010554770257235498 N-m

Power = 2 * pi*T*n/60000
= 214.90516666666666666667 kW

So, maybe the flow rate can be higher because the torque required is less than the motor can produce at maximum, but it's clear the power input is less than the motor peak output. I did not include efficiency factors.

Something seems really bad if operating the motor below the full torque RPM.

 

[TugboatEng]

It's a common mistake to assume maximum flow and pressure simultaneously on a centrifugal pump. You posted the pump curves. What do they tell you about your operating point.

 

[LittleInch]

I suspect what the manufacturer did was start at the end user (the submersible pump) and work backwards.

So I suggest you do the same. Find out the actual flow and head required and then translate that back to a speed and power from your pump graph.

So yes, the electric motor is the weak link in the system if you wanted to increase flow or head. Quite why they used such oversized hydraulic units isn't clear but that's what it looks like from here. [tt][/tt]

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

 

[73lafuite]

Good morning,
The error is that the motor speed is 360/355*1500RPM=1521RPM

 

[BEMPE16524]

Thank you all, I got what you all mean.

R.Efendy

 

[BEMPE16524]

found my mistake, I used wrong units and water pump and hydraulic motor info were wrong.
please allow me to keep some info here for my future reference.

Electrical motor:
RPM: 1500 @50Hz, 1800 @60Hz
Full load torque Nm: 1651 @50Hz, 1362 @60Hz
Power: 250kW

Main Pump
Model: Eaton PVWS 360M (variable displacement)
swept volume (displacement): 360cc/rev
Power consumption: 315kW @ 350Bar
Driving torque: 2005 @ 350Bar

Assuming operating displacement = 0.8Vg max = 0.8 x 360 = 288 cc/rev
@220 Bar, driving torque: Bar x cc / (62.8 x mech hydraulic efficiency) = 220 x 288 / (62.8 x 0.9) = 1121 Nm
Flow @ 1500 RPM: cc x RPM / 1000 = 288 x 1500 /1000 = 432 lpm
Flow @ 1800 RPM: cc x RPM / 1000 = 288 x 1800 /1000 = 518 lpm

If I assume full kW deliver from the electric motor (250kW) @ 220Bar, and from this link using 0.8 total efficiency,

https://www.hk-hydraulik.com/en/hydraulics-calculator

I should get the pump flow as 545 lpm

Hydraulic motor:
Model: Rextroth A6V (variable displacement)
displacement: 250cc
Max Torque: 1391 Nm
Max Speed: 3600RPM

Vg = 0.8Vgmax = 0.8 x 250 = 200cc
P = 220 Bar
Volumetric efficiency = 0.94
mech hydraulic efficiency = 0.9

If Flow = 432 lpm (from the main pump)

Hyd motor Speed RPM = flow x 1000 x Volumetric efficiency / Vg = 432 x 1000 x 0.94 / 200 = 2030 RPM
Torque = Vg x P x mech hydraulic efficiency / (20 x 3.142) = 630 Nm
Power = 2 x 3.142 x T x RPM / 60,000 = 6.284 x 630 x 2030 / 60,000 = 134kW

If Flow = 518 lpm (from the main pump)

Hyd motor Speed RPM = flow x 1000 x Volumetric efficiency / Vg = 518 x 1000 x 0.94 / 200 = 2435 RPM
Torque = Vg x P x mech hydraulic efficiency / (20 x 3.142) = 630 Nm
Power = 2 x 3.142 x T x RPM / 60,000 = 6.284 x 630 x 2435 / 60,000 = 161kW

If Flow = 545 lpm (from the main pump)

Hyd motor Speed RPM = flow x 1000 x Volumetric efficiency / Vg = 545 x 1000 x 0.94 / 200 = 2561 RPM
Torque = Vg x P x mech hydraulic efficiency / (20 x 3.142) = 630 Nm
Power = 2 x 3.142 x T x RPM / 60,000 = 6.284 x 630 x 2561 / 60,000 = 169kW

Water Pump Spec:

R.Efendy