Calculation
P = Δp · Q
High pressure and high flow increase fluid power; pump efficiency is applied afterwards.
P = Δp·Q
Hydraulic Power Calculator: Pressure and Flow
High pressure and high flow increase fluid power; pump efficiency is applied afterwards.
MINTSI
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Result
Select a target and calculate.
Understand the inputs
- Fluid power P — The ideal hydraulic or pneumatic useful power contained in the fluid – still before pump or motor efficiency. Computed as a result from Δp·Q.
- Pressure difference Δp — The pressure difference between inlet and outlet of the component considered (e.g. pump, cylinder, valve). Source: two pressure measurements, or from a pump curve.
- Volumetric flow Q — The volumetric flow through the component. Source: a flow meter, or the pump curve at the operating point considered.
Example
10 bar · 6 l/min = 100 W.
Assumptions and limits
Steady incompressible estimate; efficiency and compressibility are excluded.
Understand Hydraulic Power Calculator: Pressure and Flow
This calculator finds the missing quantity among fluid power P, pressure difference Δp and volumetric flow Q from the other two. Using its direct analogy to electrical power, it shows why pressure in fluid power plays the same role as voltage in electrical engineering.
What does this quantity describe?
The ideal hydraulic or pneumatic useful power of a fluid stream is the product of pressure difference and volumetric flow: P = Δp·Q. It describes power transferred in the fluid before pump or motor efficiency is applied.
Fluid power follows the same mathematical structure as electrical power P = U·I: pressure takes the role of voltage, volumetric flow the role of electric current. This analogy is more than a mnemonic — for laminar flow through a restriction, Δp = R_hydraulic·Q holds approximately, the same linear relationship as Ohm's law U = R·I for an electrical resistor.
Formula and variables
P = Δp · Q
P = Δp · QΔp = P / QQ = P / Δp
| Symbol / input | Meaning |
|---|---|
| Fluid power P | The ideal hydraulic or pneumatic useful power contained in the fluid – still before pump or motor efficiency. Computed as a result from Δp·Q. |
| Pressure difference Δp | The pressure difference between inlet and outlet of the component considered (e.g. pump, cylinder, valve). Source: two pressure measurements, or from a pump curve. |
| Volumetric flow Q | The volumetric flow through the component. Source: a flow meter, or the pump curve at the operating point considered. |
Choose the inputs correctly
Enter two of the three quantities. Fluid power requires pressure difference and flow, pressure difference requires power and flow, flow requires power and pressure difference. Δp is the actually usable pressure difference between inlet and outlet.
How to use the calculator
Select the target quantity and enter the two known values with units. To go from ideal fluid power to a pump's required drive power, divide the result by pump efficiency.
Worked example
Given Δp = 10 bar and Q = 6 l/min, substitution gives P = 10 bar · 6 l/min = 100 W (after unit conversion).
Understand the result and units
A fluid power of 100 W is the ideal power transferred in the fluid. The electrical drive power a pump actually draws sits above this, since pump efficiencies typically run 60-90%.
Pressure difference in bar or Pa, volumetric flow in l/min or m³/s, power in watts (W). 1 W = 1 Pa·m³/s; using bar and l/min requires a unit conversion before calculating.
How far does the fluid-power/electrical-power analogy extend?
The correspondence is structurally very close: volume corresponds to electric charge, volumetric flow to current, pressure to voltage, and fluid power P = Δp·Q to electrical power P = U·I. For laminar flow through a restriction or capillary, an approximately linear Δp-Q relationship also holds, analogous to Ohm's law U = R·I. Under turbulent flow, though, this relationship becomes nonlinear (Δp grows roughly with Q²), so the analogy to a linear electrical resistor no longer applies directly there.
Typical applications
The formula is used to size hydraulic pumps, compressors, pneumatic cylinders, and to roughly estimate drive power for fluid power systems.
Assumptions, limits and common mistakes
The formula gives ideal fluid power without pump or motor losses, and assumes incompressible, steady flow. For gases with noticeable compressibility, or unsteady events (pressure surges, start-up transients), actual power draw deviates from this ideal value.
Common mistake: Do not equate ideal fluid power with the actually required electrical drive power — that additionally requires dividing by pump or motor efficiency. Also, take care converting between bar and pascal and between l/min and m³/s, since both conversion factors enter the result.
Frequently asked questions
Why is fluid power often compared to electrical power?
Because both follow the same mathematical structure: pressure corresponds to voltage, volumetric flow to electric current, and their product gives power in each case.
How do I get from ideal fluid power to required pump drive power?
Divide ideal fluid power by pump efficiency, typically around 60-90% depending on design and operating point.
Does the Ohm's-law analogy still hold under turbulent flow?
Only to a limited extent. Under turbulent flow, pressure loss grows roughly with the square of volumetric flow, not linearly as with an electrical resistor.