BIZOMAT MAKEOS · Hydraulics
Hydraulic Power and Pump Input
What fluid power and shaft input does this duty require?
Calculate hydraulic power from differential pressure and delivered flow, with an entered overall pump efficiency.
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How this tool works
Inputs, units and where to find them
| Input | Unit | Typical source |
|---|---|---|
| Pressure rise | bar | Pump outlet minus inlet pressure under the specified duty. |
| Delivered flow | L/min | Actual delivered flow at the stated pressure. |
| Overall pump efficiency | % | Combined volumetric and mechanical efficiency at this duty, from supplier data. |
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: Pressure rise
- What it is
- The increase in pressure the pump creates: outlet pressure minus inlet pressure at the specified duty.
- Units and format
- Enter a number in bar: bar (1 bar = 100 kPa). The form accepts 0 bar to 10000 bar. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 150 bar as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Hydraulic power is directly proportional to pressure rise.
- Get it right
- If the inlet is near atmospheric, outlet gauge pressure is close; use the actual working pressure, not the relief setting.
What this means: Delivered flow
- What it is
- The oil flow the pump actually delivers at that pressure.
- Units and format
- Enter a number in L/min: litres per minute. The form accepts 0 L/min to 1000000 L/min. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 20 L/min as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Hydraulic power is directly proportional to flow.
- Get it right
- Delivered flow is lower than theoretical displacement times speed because of internal leakage.
What this means: Overall pump efficiency
- What it is
- The overall pump efficiency at this duty: the share of shaft input power that becomes hydraulic power, combining volumetric and mechanical losses.
- Units and format
- Enter a number in %: percent: the number is read as a percentage, so type 8 for 8 %. The form accepts 0.001% to 100%. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 85% as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Shaft input power is inversely proportional to efficiency.
- Get it right
- Type a percentage (85 for 85 %). Use supplier data at this pressure and speed.
Method and formulas
- Fluid power (kW) = pressure rise (bar) × flow (L/min) / 600.
- Shaft input = fluid power / (efficiency / 100).
- Pump loss = shaft input − fluid power. Example: 150 bar and 20 L/min gives 5 kW fluid power, 5.882353 kW shaft input at 85%, and 0.882353 kW pump loss.
Limitations
- Efficiency is specific to the pump and operating point; the default is illustrative.
- Loss here excludes downstream throttling and motor electrical losses. It is not total system cooling demand.
- Example values are illustrations. Replace them with values for your actual task.
Typical uses
- pump power
- hydraulic kW
- pump loss
Who it is for
- Designers
- Maintenance engineers
What you get
- Hydraulic power
- Pump shaft input
- Estimated pump loss
Interpreting the result
- Size the motor and cooling system for the full duty cycle and additional losses.
References
- Machinery’s Handbook, 29th edition (Industrial Press, 2012), printed pp. 186, 365–368; CD pp. 3416–3420. Topic reference; no source tables or recommended settings are reproduced.
- Parker: Basic Hydraulic Formulas — https://www.parker.com/content/dam/Parker-com/Literature/Accumulator---Cooler-Division---Americas/Catalogs---Bulletins/GAD-HY10-1630-5-2012/ACD_CoolerCatalog_HY10_1700.pdf
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