How to Size the Motor and Pump for a Hydraulic Power Unit

To size the motor and pump for a hydraulic power unit, first define the flow and pressure required together during each operating stage. Use the required delivered flow, actual shaft speed, and pump volumetric efficiency to estimate displacement. Then calculate the selected pump’s shaft power and torque at the relevant operating points, and compare them with the electric motor’s available output.

A usable selection also needs checks for starting, duty, supply conditions, pump limits, and the way flow is controlled. A single hydraulic power figure does not cover those requirements. This guide separates the power boundaries and follows one illustrative calculation from required flow to candidate review. The example values are assumptions, not recommended efficiencies, pressure settings, or approved component selections.

Contents

1. Define the required operating points at the pump
2. Calculate pump displacement from delivered flow
3. Separate hydraulic power, shaft power, and electrical input
4. Work through a motor and pump sizing example
5. Check the full cycle and the high-pressure cases
6. Review starting torque and the electrical supply
7. Check pump limits and variable-speed operation
8. Close the selection with a reviewable candidate record
9. Frequently asked questions

Define the required operating points at the pump

Begin with the application’s force, torque, speed, and sequence requirements. Translate these into pressure and delivered flow at a stated boundary. If the requirements are specified at an actuator, establish the additional supply demand and pressure losses before sizing the pump.

For each stage, record what runs simultaneously. A circuit may need high flow during a lightly loaded movement and higher pressure during a slower working stage. Multiplying the highest flow anywhere in the cycle by the highest pressure anywhere can create an operating point that never occurs. Conversely, overlooking a real overlap can underestimate the requirement.

Bosch Rexroth’s sizing assistant accepts a time-dependent pressure/flow cycle. That is a useful structure for the input record, regardless of which manufacturer is eventually selected.

Input	Question to resolve
Delivered pump flow	What must reach the circuit, including simultaneous auxiliary demands?
Pump pressure rise	What outlet-minus-inlet pressure is needed at that flow?
Duration and repetition	How long does each stage last, and how often does it recur?
Control state	Is the pump delivering, unloading, reducing displacement, or stopped?
Drive speed	What actual pump shaft speed applies under load?
Fluid and environment	What conditions apply at startup and during warmed operation?

Distinguish internal pump leakage from circuit flow demand; the volumetric-efficiency calculation below already accounts for pump volumetric loss. Use the power unit specification checklist to collect missing application data before treating a provisional input as a design requirement.

Calculate pump displacement from delivered flow

For a positive-displacement pump at a defined operating condition:

Q [L/min] = V_g [cm³/rev] × n [rpm] × η_v / 1000.

Rearranging gives:

V_g [cm³/rev] = 1000Q [L/min]/(n [rpm] × η_v).

Here, η_v is pump volumetric efficiency as a decimal. Displacement describes geometric volume per revolution; delivered flow also depends on speed and volumetric loss. The divisor 1000 converts cubic centimetres to litres.

Use a performance curve or verified data for the candidate’s pressure, speed, displacement setting, and fluid condition. An efficiency selected only to make the arithmetic convenient remains an assumption. The general flow, torque, and power relationships are given in Danfoss’s sizing equations, printed page 18; that reference does not select a product for this example.

For a directly coupled electric drive, use actual loaded shaft speed. A supply frequency or a synchronous-speed label alone does not establish that value. If there is a gearbox or another transmission, use pump speed and account separately for transmission losses.

With a variable-displacement pump, identify both the installed maximum displacement and the displacement commanded at each stage. A larger maximum displacement does not mean that maximum flow is delivered continuously. Review the flow path in how a hydraulic power unit works before assigning a control state.

Separate hydraulic power, shaft power, and electrical input

Define Δp as pump outlet pressure minus inlet pressure using a consistent reference. A useful steady-state engineering expression is:

P_hydraulic [kW] = Δp [bar] × Q_delivered [L/min]/600.

The pump shaft must also supply pump losses:

P_pump_shaft = P_hydraulic/η_p.

Here η_p is overall pump efficiency at the same operating point. With η_m representing the pump’s mechanical efficiency, η_p = η_v × η_m. These pump efficiency definitions must not be confused with electric motor efficiency.

For displacement in cm³/rev and pressure rise in bar:

T_pump [N·m] = Δp × V_g/(20π × η_m).

A cross-check is P_pump_shaft [kW] = T_pump [N·m] × n [rpm] × π/30,000.

If the motor drives only this pump through an idealized lossless coupling, required motor shaft output equals pump shaft input. Add actual transmission losses or other shaft-driven loads when present. Do not add a loss twice if a supplier’s quoted power already includes it.

Electrical input is a further boundary. With electric motor efficiency η_e:

P_electrical = P_motor_shaft/η_e.

A drive converter and other electrical loads introduce their own losses or demands. Comparing pump shaft demand with a motor’s rated mechanical output does not require dividing that demand by electric motor efficiency; the additional division estimates electrical input.

Likewise, when delivered flow and overall pump efficiency are used, do not divide by volumetric efficiency again. That would count the volumetric loss twice. The hydraulic system energy and flow article provides the underlying energy picture.

Work through a motor and pump sizing example

Assume a circuit requires 30 L/min delivered at the pump outlet during one working stage. Its required pump pressure rise is 180 bar. For this example only, assume a directly coupled drive with negligible coupling loss, actual loaded speed of 1450 rpm, pump volumetric efficiency of 0.90, and overall pump efficiency of 0.85.

These efficiencies are arithmetic inputs, not predictions for a particular pump. There are no additional shaft-driven auxiliaries in the example.

Example input	Assumed value
Required delivered flow	30 L/min
Pump pressure rise	180 bar
Loaded pump speed	1450 rpm
Volumetric efficiency η_v	0.90
Overall pump efficiency η_p	0.85

The initial displacement calculation is:

V_required = 1000 × 30/(1450 × 0.90) = approximately 22.989 cm³/rev.

Now consider a hypothetical 25 cm³/rev fixed-displacement candidate. Under the same assumptions, it delivers:

Q_candidate = 25 × 1450 × 0.90/1000 = 32.625 L/min.

This exceeds the specified 30 L/min. The circuit must have an acceptable way to accommodate or control that difference. If the whole candidate delivery is pumped through the stated pressure rise, calculate power using 32.625 L/min:

P_hydraulic = 180 × 32.625/600 = 9.7875 kW.

P_pump_shaft = 9.7875/0.85 = approximately 11.515 kW.

Using only the original 30 L/min would give approximately 10.588 kW shaft demand and would miss the increased demand from this candidate under the stated full-flow condition.

The consistent mechanical-efficiency assumption is η_m = 0.85/0.90 = approximately 0.94444. Therefore:

T_pump = 180 × 25/(20π × 0.94444) = approximately 75.833 N·m.

The torque-speed calculation returns approximately 11.515 kW, confirming the same shaft-power result.

A hypothetical motor limited to 11 kW continuous shaft output would not meet this continuous operating case. A candidate rated for 15 kW shaft output at the required speed and conditions exceeds this calculated demand, but still needs the remaining reviews. Neither comparison approves startup, overload duration, installation, or thermal performance.

Finally, assume electric motor efficiency of 0.92 at this load. The estimated motor electrical input is 11.515/0.92, approximately 12.516 kW, before any converter or separate electrical loads. That electrical-input estimate is distinct from the required 11.515 kW shaft output. See the complete power unit selection process for the rest of the assembly.

Check the full cycle and the high-pressure cases

Repeat the calculation for every required state using the candidate’s actual delivery and pressure. Include legitimate high-pressure, transition, standby, and auxiliary-load cases with their durations. The purpose is to establish the load the selected arrangement can actually impose.

For an additional illustrative check, suppose the same candidate must deliver its full assumed flow at a 210 bar pressure rise. Holding the example efficiencies constant only for comparison:

P_pump_shaft = 210 × 32.625/(600 × 0.85) = approximately 13.434 kW.

This is a separate case, not an instruction to set a relief valve to 210 bar. Whether such a case belongs in a real design depends on the circuit, controls, pressure-limiting behavior, and expected operation. Real efficiencies also need updating with the operating point.

A fixed-displacement pump sending full flow across a high pressure drop still loads the motor even when the actuator is stationary. A suitably controlled unloading or variable-displacement arrangement changes the operating point, but it does not justify assuming zero standby loss. Obtain data for the actual control configuration.

Do not use average hydraulic power as automatic permission to reduce motor size. Peak torque, stage duration, starts, cooling, and thermal history matter. ABB’s motor guide, section 4.7, distinguishes continuous, short-time, and intermittent duties. Match the declared duty to the actual cycle.

Any surplus flow dissipated across restrictions also belongs in the heat assessment. The hydraulic system overheating guide explains why adequate installed power alone does not establish acceptable oil temperature.

Review starting torque and the electrical supply

The running-power calculation does not describe the entire acceleration from rest. Record whether pressure is present when starting, whether an unloading function is part of the design, the driven inertia, the required acceleration, and the expected number of starts.

Compare the pump load and acceleration demand with the motor-and-starter combination’s available torque over the starting process. An unloading arrangement can change that demand, but its operation and failure behavior must be established by the circuit design. Do not assume that selecting a larger nameplate kW solves every starting problem.

ABB’s starting discussion, section 4.6, treats load torque, inertia, and motor torque as relevant to starting time and identifies thermal effects from repeated starts. Use the selected equipment’s data for the actual assessment.

Provide the electrical designer with supply voltage, frequency, phase arrangement, allowable starting demand, control method, environment, and the proposed motor duty. Current capacity, protective devices, cables, and drive selection require their own checks; a hydraulic kW calculation is not an electrical installation design.

For a battery-fed compact unit, also establish the operating voltage under load, required on-time, and available supply capability. Do not transfer an AC motor example directly into a DC installation. Keep these unresolved inputs visible in the power unit specification checklist.

Check pump limits and variable-speed operation

A displacement and power match does not confirm that the pump can operate throughout the proposed range. Check the selected product’s speed limits, permitted pressures, inlet conditions, fluid viscosity, temperature, case-drain requirements where applicable, and shaft loading.

Danfoss’s Series 45 performance information states the conditions associated with its curves and presents inlet-pressure/speed operating ranges. Use that principle when reviewing candidate data; do not transfer one pump family’s limits to another construction.

Candidate check	Evidence to request
Delivered flow and shaft demand	Curves at the required pressure, speed, and fluid conditions
Inlet and case conditions	Applicable pressure limits and installation requirements
Speed range	Permitted continuous, transient, and startup conditions
Shaft and coupling	Torque capacity, alignment, mounting, and applicable loads
Control response	Behavior during demand changes, unloading, and stopping
Motor availability	Output and torque capability for the actual duty and environment

Variable speed offers another way to adjust pump delivery, but low hydraulic flow does not necessarily mean low required torque. At fixed displacement and pressure rise, the torque equation remains approximately independent of speed apart from changing losses. Power falls with speed in that simplified comparison.

Do not apply a centrifugal-pump cubic power rule to this positive-displacement case. Also, do not infer that the pump can operate arbitrarily slowly. Both lubrication and controlled performance require the manufacturer’s operating limits.

The electric motor has its own speed-dependent constraints. ABB’s low-speed guidance, section 10.11, explains that a shaft-driven fan provides less cooling at low speed. Check the motor/drive combination’s loadability and cooling provisions across the required range.

Then revisit matching the pump, valve, and cylinder: a change in flow can change valve losses, actuator speed, and return-flow demand, requiring another sizing iteration.

Close the selection with a reviewable candidate record

The final record should connect application requirements to a specific proposed pump and motor configuration. A spreadsheet containing only displacement and kW leaves the most consequential assumptions hidden.

Record section	Required content
Application cycle	Simultaneous pressures, flows, durations, and control states
Candidate pump	Product, displacement, rotation, speed, and curve revisions
Calculated results	Delivered flow, hydraulic output, shaft input, and torque by case
Candidate motor	Rated shaft output, speed, duty, supply, and relevant capability data
Interfaces	Coupling, shaft, mounting, ports, installation, and auxiliary loads
Open issues	Provisional efficiency, starting, thermal, control, and supply questions
Verification	Agreed acceptance conditions, measurements, reviewer, and revision

Keep provisional values clearly marked. If a supplier changes displacement, speed, control type, or the basis of quoted power, repeat the affected calculations. A similar catalog description does not demonstrate equivalent behavior over the cycle.

The acceptance plan should identify the conditions under which delivered flow, pressure, speed, and power will be verified, plus the applicable limits for thermal and starting behavior. If electrical input is measured, account for the conversion boundary before comparing it with a shaft-power prediction. Agreement at one operating point does not close all other cases.

Use the complete power unit selection process to coordinate the remaining assembly requirements. Selection is complete only when the responsible engineering review has resolved the necessary conditions and evidence.

Frequently asked questions

Does a higher-kW motor automatically produce more pump flow?

No. At unchanged pump displacement, speed, and operating efficiency, the calculated delivery is unchanged. A larger motor may change available drive capacity; it does not by itself specify more flow.

Can I select the pump only from the required pressure?

Pressure capability is one requirement. You also need delivered flow, speed, power, duty, fluid compatibility, and the required control behavior.

Which efficiency belongs in the motor shaft-power calculation?

Use overall pump efficiency when starting from delivered hydraulic power, then account for separate transmission or shaft loads. Electric motor efficiency is needed when estimating electrical input.

Can a VFD make a smaller motor adequate?

It can change the operating profile when the application permits, but it does not remove the required torque or power at a specified working point. Verify the complete pump, motor, and drive envelope.

Does failure to build pressure prove the motor is too small?

No. The cause needs diagnosis; see power unit not building pressure. For selection, retain the duty table, candidate calculations, manufacturer data, and unresolved items so the supplier and engineering reviewer can evaluate the proposed combination.