KEVOS® 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. |
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.
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