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Hydraulic power equals pressure difference times volume flow, or density times gravity, flow and head.
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Formula
P in W; Δp in Pa; Q in m3/s; ρ in kg/m3; g in m/s2; H in m
| Symbol | Meaning | Unit |
|---|---|---|
| Power transferred to the fluid | W | |
| Pressure difference between inlet and outlet | Pa | |
| Volume flow rate | m3/s | |
| Fluid density | kg/m3 | |
| Head corresponding to Δp/(ρg) | m |
P_h=\Delta p\,Q=\rho g Q HPressure is energy per volume, so pressure difference multiplied by volume per time gives power. The head form uses H=Δp/(ρg) and an incompressible fluid with representative density. Electrical or shaft input exceeds hydraulic output when pump efficiency is below one. With elevation or velocity changes, use the full energy balance rather than pressure difference alone.
Pressure is energy per volume, so pressure difference multiplied by volume per time gives power. The head form uses H=Δp/(ρg) and an incompressible fluid with representative density. Electrical or shaft input exceeds hydraulic output when pump efficiency is below one. With elevation or velocity changes, use the full energy balance rather than pressure difference alone.
Hydraulic power equals pressure difference times volume flow, or density times gravity, flow and head.
Base equation for the target quantity
Preserve signs, boundary conditions and units when rearranging.
Work through the numerical example for “Hydraulic power of a flow machine” and check its unit and magnitude.
Water at Q=0.010 m3/s and H=15 m needs Ph=1000·9.81·0.010·15≈1.47 kW of hydraulic power.
Substituting a related quantity that shares the same symbol.
Use variable definitions and check: P in W; Δp in Pa; Q in m3/s; ρ in kg/m3; g in m/s2; H in m.
Formula cluster
Hydraulic power equals pressure difference times volume flow, or density times gravity, flow and head.
Water at Q=0.010 m3/s and H=15 m needs Ph=1000·9.81·0.010·15≈1.47 kW of hydraulic power.
Calculate fluid power transferred by pumps and flow machines.
Here is how to work through a typical Hydraulic power of a flow machine (Ph = Δp·Q = ρgQH) task step by step:
Task
Work through the numerical example for “Hydraulic power of a flow machine” and check its unit and magnitude.
Solution path
Water at Q=0.010 m3/s and H=15 m needs Ph=1000·9.81·0.010·15≈1.47 kW of hydraulic power.
Pressure is force per area and therefore energy per volume. Multiplying pressure difference by volume delivered per second gives energy per second, or watts. The unit check is Pa·m3/s=(N/m2)·m3/s=N·m/s=W. Hydraulic power is delivered to the fluid, not necessarily electrical input or shaft power. The head form assumes representative constant density and H=Δp/(ρg). Pump efficiency, elevation change and velocity-head change may require a fuller system energy calculation. Finally check units, assumptions, measurement conditions and reference quantities against the problem before using the result.
Only an ideal lossless pump would have equal input and fluid power. In practice, shaft power is greater and depends on efficiency: Ph=ηPinput when η is defined accordingly. Motor losses can further separate electrical input from pump-shaft power. Hydraulic power is delivered to the fluid, not necessarily electrical input or shaft power. The head form assumes representative constant density and H=Δp/(ρg). Pump efficiency, elevation change and velocity-head change may require a fuller system energy calculation. Finally check units, assumptions, measurement conditions and reference quantities against the problem before using the result.
The form ΔpQ captures pressure work between states but not automatically changes in elevation and velocity head or heat and shaft transfers. For pumps in a real system, use a control-volume energy balance and account consistently for datum, speeds, losses and efficiency. Hydraulic power is delivered to the fluid, not necessarily electrical input or shaft power. The head form assumes representative constant density and H=Δp/(ρg). Pump efficiency, elevation change and velocity-head change may require a fuller system energy calculation. Finally check units, assumptions, measurement conditions and reference quantities against the problem before using the result.
Write the quantity and unit from the variable list next to every value: P in W; Δp in Pa; Q in m3/s; ρ in kg/m3; g in m/s2; H in m. Substitute only after converting to a consistent unit system, and state sign or reference direction explicitly. The example Water at Q=0.010 m3/s and H=15 m needs Ph=1000·9.81·0.010·15≈1.47 kW of hydraulic power. is a calculation check: result, unit and order of magnitude must fit the described situation. Round after intermediate calculations so small contributions are not lost prematurely. Hydraulic power is delivered to the fluid, not necessarily electrical input or shaft power. The head form assumes representative constant density and H=Δp/(ρg). Pump efficiency, elevation change and velocity-head change may require a fuller system energy calculation. Finally check units, assumptions, measurement conditions and reference quantities against the problem before using the result.
The background states this starting condition: Pressure is energy per volume, so pressure difference multiplied by volume per time gives power. The head form. That limits when the result is a valid model prediction. Compare the problem’s boundary conditions with this assumption before substituting values; real components, material properties or loads can introduce additional effects. The cited reference supports the relation used here but does not remove the need to check whether the specific case fits the model. Hydraulic power is delivered to the fluid, not necessarily electrical input or shaft power. The head form assumes representative constant density and H=Δp/(ρg). Pump efficiency, elevation change and velocity-head change may require a fuller system energy calculation. Finally check units, assumptions, measurement conditions and reference quantities against the problem before using the result.
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Which formula describes Hydraulic power of a flow machine?
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How do you rearrange Hydraulic power of a flow machine for Base equation for the target quantity?
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Which common mistake happens with Hydraulic power of a flow machine?
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