BIZOMAT MAKEOS · Hydraulics
Hydraulic Power Unit Sizing (Pump, Motor, Reservoir, Cooling)
What pump displacement and motor power does this hydraulic power unit need, how big should the reservoir be, and does it need a cooler?
Sizing of a hydraulic power unit: pump displacement for the flow at the drive speed and volumetric efficiency, hydraulic power, shaft power with volumetric and mechanical efficiencies (as the hydraulic power tool with the overall efficiency), drive torque, reservoir volume from your multiple of pump flow, heat from the share of power lost and the duty, natural heat dissipation from the reservoir surface, and the cooler capacity still needed.
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How this tool works
Inputs, units and where to find them
| Input | Unit | Typical source |
|---|---|---|
| Flow needed | L/min | Flow the circuit needs from the pump. |
| Working pressure | bar | Highest working pressure. |
| Drive speed | rpm | Speed of the electric motor driving the pump. |
| Pump volumetric efficiency | % | Volumetric efficiency of the pump. |
| Pump mechanical efficiency | % | Mechanical-hydraulic efficiency of the pump. |
| Motor rated power | kW | Rated power of the motor considered. |
| Reservoir volume ÷ pump flow per minute | Reservoir volume as a multiple of pump flow per minute (your design rule; often 3–5 for industrial units, less for mobile). | |
| Share of input power turned to heat | % | Share of the input power that ends up as heat (throttling, relief and leakage; your estimate for the circuit). |
| Share of time on load | % | Share of time the unit runs at the working pressure. |
| Reservoir heat transfer coefficient | W/m²·K | Heat transfer coefficient from the reservoir walls to air (about 10–15 still air, more with a fan). |
| Oil temperature above ambient allowed | K | Oil temperature rise above ambient you allow. |
What each input means
Open an input to see what it is, its units, the worked example and the common mistakes, with a small diagram.
What this means: Flow needed
- What it is
- The oil flow the circuit needs the pump to deliver, for example cylinder area × speed for the fastest movement.
- Units and format
- Enter a number in L/min: litres per minute. The form accepts 0.000001 L/min to 100000 L/min. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 40 L/min as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Pump displacement, hydraulic power and shaft power rise in proportion.
What this means: Working pressure
- What it is
- The highest pressure the pump works against in normal operation, measured as gauge pressure at the pump outlet.
- Units and format
- Enter a number in bar: bar (1 bar = 100 kPa). The form accepts 0.000001 bar to 1000 bar. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 160 bar as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Drive torque, hydraulic power and shaft power rise in proportion; displacement does not change.
- Get it right
- bar, not MPa: 10 bar = 1 MPa.
What this means: Drive speed
- What it is
- The speed of the electric motor that turns the pump, from its data plate.
- Units and format
- Enter a number in rpm: revolutions per minute. The form accepts 0.000001 rpm to 10000 rpm. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 1450 rpm as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- A faster motor needs a smaller pump displacement for the same flow.
What this means: Pump volumetric efficiency
- What it is
- The share of the pump’s geometric displacement that actually reaches the outlet after internal leakage, from the pump data sheet at the working pressure.
- Units and format
- Enter a number in %: percent: the number is read as a percentage, so type 8 for 8 %. The form accepts 50% to 100%. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 92% as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Lower volumetric efficiency raises the displacement and shaft power needed.
- Get it right
- Type a percentage (92 for 92 %), not a fraction.
What this means: Pump mechanical efficiency
- What it is
- The pump’s mechanical-hydraulic efficiency: the share of shaft torque converted to pressure after friction losses, from the pump data sheet.
- Units and format
- Enter a number in %: percent: the number is read as a percentage, so type 8 for 8 %. The form accepts 50% to 100%. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 90% as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Lower mechanical efficiency raises the drive torque and shaft power.
What this means: Motor rated power
- What it is
- The rated power of the electric motor being considered, from its data plate or catalogue. The shaft power needed is compared with it.
- Units and format
- Enter a number in kW: kilowatts. The form accepts 0.000001 kW to 10000 kW. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 15 kW as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
What this means: Reservoir volume ÷ pump flow per minute
- What it is
- The reservoir volume expressed as so many minutes of pump flow: reservoir litres divided by pump litres per minute.
- Units and format
- Enter a plain number with no unit. The form accepts 0.5 to 10. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 3 as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- A larger multiple gives a bigger tank with more surface to shed heat.
- Get it right
- Enter the multiple, not the volume in litres; the tool works out the volume.
What this means: Share of input power turned to heat
- What it is
- Your estimate of the share of the pump input power that ends up as heat in the oil, from throttling, relief flow and leakage in this circuit.
- Units and format
- Enter a number in %: percent: the number is read as a percentage, so type 8 for 8 %. The form accepts 0% to 100%. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 25% as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Heat load rises in proportion.
- Get it right
- Circuits that throttle or blow off over a relief valve turn most of their power into heat.
What this means: Share of time on load
- What it is
- The share of operating time the unit runs at the working pressure. The tool counts heat only for this share.
- Units and format
- Enter a number in %: percent: the number is read as a percentage, so type 8 for 8 %. The form accepts 0% to 100%. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 50% as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Heat load rises in proportion.
- Get it right
- Type a percentage (50 for 50 %), not a fraction.
What this means: Reservoir heat transfer coefficient
- What it is
- The rate at which the reservoir walls pass heat to the surrounding air, per square metre per degree, from your data or reference values for still or moving air.
- Units and format
- Enter a number in W/m²·K: watts per square metre per kelvin. The form accepts 0.000001 W/m²·K to 100 W/m²·K. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 12 W/m²·K as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Natural dissipation rises in proportion, lowering the cooler capacity needed.
What this means: Oil temperature above ambient allowed
- What it is
- How far above the surrounding air temperature you allow the oil in the reservoir to run.
- Units and format
- Enter a number in K: kelvin (a difference of 1 K equals 1 °C). The form accepts 0.000001 K to 80 K. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 30 K as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- A larger allowed rise lets the reservoir shed more heat in proportion.
- Get it right
- This is a rise above ambient, not the oil temperature itself.
Method and formulas
- Displacement V_g = Q × 1000 ÷ (n η_v) cm³/rev; drive torque = V_g p ÷ (20π η_m) N·m.
- Hydraulic power = p Q ÷ 600 kW; shaft power = hydraulic ÷ (η_v η_m) (as the hydraulic power tool).
- Reservoir = multiple × Q; surface ≈ 6.5 V^(2/3) m² (V in m³, a near-cubic tank); dissipation = k A ΔT.
- Heat = shaft power × loss share × duty; cooler = heat − dissipation (if positive).
- Example: 40 L/min at 160 bar needs a 29.985 cm³/rev pump and 12.882 kW; about 1.041 kW of cooling is still needed.
Limitations
- Heat share is an estimate; throttling circuits turn most of the power into heat.
- Reservoir surface uses a near-cubic tank; long, flat tanks dissipate more.
- Example values are illustrations. Replace them with values for your actual task.
Typical uses
- Size a power pack for a press or machine
- Check whether a reservoir needs a cooler
- Choose a motor for a pump at a pressure
Who it is for
- Hydraulic engineers
- Machine designers
- Maintenance engineers
What you get
- Pump displacement and drive torque
- Hydraulic and shaft power against the motor
- Reservoir, heat load and cooler capacity
Interpreting the result
- A pressure-compensated or load-sensing pump can cut the heat share sharply compared with a fixed pump on relief.
References
- Machinery’s Handbook, 29th edition (Industrial Press, 2012), printed pp. 3421–3425, 371–378. Topic reference; no source tables or recommended settings are reproduced.
Learn the Concepts Behind This Tool
These BIZOMAT Library readings explain the method this tool uses, with worked examples.
Method explained here · BIZOMAT original book
Leakage through clearances
What it covers: Calculate laminar leakage through annular clearances, see why it follows the cube of the gap and falls with viscosity, track a pump's efficiency through the year and size it for hot oil.
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