Pump Flow Calculator

If shaft input power, pump efficiency, fluid density, gravity, and pump head are known, first calculate hydraulic power as efficiency times shaft input power. Then calculate flow from Q = P_hyd/(ρgH). The result represents a theoretical duty relationship at the entered conditions, not a unique prediction of actual operating flow without a pump and system curve.

For an incompressible liquid using head, Q = ηP_shaft/(ρgH), where η is pump efficiency, P_shaft is pump shaft input power, ρ is density, g is gravitational acceleration, and H is pump head. If hydraulic power is known directly, the formula becomes Q = P_hyd/(ρgH).

Convert mechanical horsepower to watts, determine the pump shaft power basis, multiply shaft power by pump efficiency to obtain hydraulic power, then divide by ρgH. If the horsepower is actually motor electrical input, motor efficiency must be applied first. The calculator uses 745.699872 W per mechanical horsepower.

Convert kW to watts, then use the correct power basis. For shaft power, hydraulic power is ηP_shaft. Divide that hydraulic power by ρgH to estimate theoretical pump flow. If the entered kW is motor electrical input, use motor efficiency to obtain shaft power before applying pump efficiency.

Head and power alone are not enough unless efficiency and fluid density are also known or appropriately assumed. For shaft power, the simplified relationship is Q = ηP_shaft/(ρgH). Actual flow must still be checked against the specific pump curve and system curve because pump power does not uniquely determine an operating point.

No. Horsepower provides an available power scale, but actual flow also depends on pump head, efficiency, pump design, speed, impeller diameter, system resistance, fluid properties, and the operating point. Two pumps with the same horsepower can produce very different flow and head combinations because their performance curves are different.

Yes. In the theoretical energy relationship, higher head requires more hydraulic power for the same flow. For fixed shaft power and efficiency, increasing head reduces the theoretical flow obtained from Q = ηP/(ρgH). In a real system, the actual flow is set by the pump curve and system curve intersection.

Yes. At fixed shaft input power and head, higher pump efficiency converts more shaft power into hydraulic power, increasing the theoretical flow available from the energy relationship. Pump efficiency itself is not constant across the full operating range, so use the manufacturer curve or field measurements for the actual operating point.

Yes as a theoretical duty calculation when head, efficiency, and fluid density are also known. Convert horsepower to watts, apply pump efficiency to obtain hydraulic power, calculate flow in m³/s, and then convert to US GPM. Power alone cannot uniquely determine actual GPM for a real pump.

Use the pump's shaft input power, pump efficiency, head, and fluid density to calculate Q = ηP/(ρgH). Convert the resulting m³/s to US gallons per minute using the exact US-gallon conversion. Be careful not to confuse US GPM with Imperial GPM, because the gallon sizes are different.

Hydraulic power is the useful rate of mechanical energy transferred to the fluid. For an incompressible liquid using head, P_hyd = ρgQH. If pressure rise is known directly, P_hyd = ΔP × Q. Hydraulic power is lower than shaft input power whenever pump efficiency is below 100%.

Pump flow calculations relate power, head, efficiency, and hydraulic duty. Pipe-flow calculations focus on how a known or assumed flow behaves in the piping, including velocity, Reynolds number, friction factor, and pressure drop. In an actual system, pump performance and pipe/system resistance interact to determine the operating flow.

You can calculate flow if hydraulic power is known directly, because Q = P_hyd/(ρgH). If only shaft or motor input power is known, pump efficiency is needed to determine how much of that input becomes hydraulic power. Without efficiency or direct hydraulic power, the energy balance is incomplete.

Yes when the pump hydraulic pressure rise and hydraulic power are known, because P_hyd = ΔP × Q, so Q = P_hyd/ΔP. If only input power is known, efficiency is still required to determine hydraulic power. Pressure rise must represent the relevant hydraulic difference across the pump.

For US gallons, convert using the exact US-gallon volume of 0.003785411784 m³ and the time conversion from minutes to hours. This calculator performs the conversion internally. Do not use an Imperial-gallon conversion for US GPM, because an Imperial gallon is larger and produces a different result.

Convert litres per minute to cubic metres per second or directly divide by the litre equivalent of a US gallon. The calculator keeps US GPM and Imperial GPM as separate units to avoid ambiguity. Confirm which gallon definition is required by the equipment documentation or project standard.

For an incompressible fluid, pressure rise and head are related by ΔP = ρgH. Head represents energy per unit weight, while pressure is force per unit area. Because density appears in the conversion, the same pressure difference corresponds to different head values for fluids with different densities.

Use the density corresponding to the actual water temperature and composition. The calculator's 998.2 kg/m³ default is only an approximate reference near 20°C. Hot water, brine, glycol mixtures, dissolved solids, or other process conditions can have materially different density and should use project-specific property data.

Temperature can affect water density and viscosity, and it can also affect pump/system behavior. The simplified hydraulic power equation uses density directly. For more accurate work, use fluid properties at the operating temperature and compare the duty with the manufacturer's pump curve under the relevant conditions.

Yes, it is especially useful for preliminary centrifugal-pump duty calculations involving flow, head, hydraulic power, and efficiency. However, centrifugal-pump flow varies strongly with the pump and system curves. Final selection or troubleshooting should use the manufacturer's performance data and the actual system resistance curve.

The basic energy relationship between hydraulic power, pressure/head, and flow still provides useful context, but positive-displacement pump flow is often governed primarily by displacement and speed, with slip and efficiency effects. Use pump-specific performance data and the appropriate positive-displacement model for final design or capacity prediction.

The calculation assumes a specified head, efficiency, power, and density. Real flow may differ because the actual head is different, the pump operates at another point on its curve, efficiency changes, system resistance changes, valves move, speed or impeller diameter differs, wear is present, or measurements contain uncertainty.

The operating point is the flow and head at which the pump curve and system curve intersect for the current configuration. It represents the balance between what the pump can produce and what the system requires. Changing speed, valve position, pipe resistance, static head, or impeller diameter can move that point.

A pump curve shows the relationship between flow and head for a specific pump, speed, and impeller configuration, often with efficiency, power, and NPSHR information. It is the correct reference for determining how a real pump behaves across its operating range rather than assuming one fixed flow.

A system curve represents the head or pressure the piping system requires at different flow rates. It can include static head, pipe friction, fittings, valves, equipment losses, and control conditions. The intersection of the system curve and pump curve establishes the operating point for a typical centrifugal-pump system.

The Best Efficiency Point is the region of a pump's performance curve where its efficiency is highest for the specified speed and configuration. It does not mean every pump should always operate at exactly one flow, but it is an important reference when evaluating performance, wear, vibration, and energy use.

Not necessarily. At fixed head and efficiency, the theoretical energy relationship shows more shaft power can support more flow. In a real system, however, flow depends on the pump curve, system curve, speed, impeller diameter, and control conditions. Additional power may instead appear as increased head, losses, or operation at another point.

Yes. Pumps with the same motor or shaft-power rating can have very different hydraulic designs, speed, impeller diameter, efficiency, and head-flow curves. One pump may be designed for high flow at low head, while another produces lower flow at much higher head.

Yes. Pipe friction increases the system head required at a given flow and therefore shifts the system curve. In a centrifugal-pump system, the new system curve intersects the pump curve at a different operating point. Higher resistance typically reduces actual operating flow unless the pump speed or control strategy changes.

Yes. Throttling valves add system resistance and increase the required system head at a given flow, shifting the operating point. Control valves, bypasses, check valves, and other components can also change the effective system curve. The exact effect depends on the pump and system characteristics.

Yes. For many rotodynamic pumps, changing speed shifts the pump performance curve and therefore changes flow, head, and power. Variable-speed drives are commonly used to control pump duty. Exact performance should be verified with manufacturer data and appropriate affinity-law assumptions for the pump and operating range.

Yes. Changing impeller diameter alters the pump's head-flow and power characteristics. Trimmed impeller curves supplied by the manufacturer should be used for accurate prediction, because simple proportional assumptions may not capture efficiency changes, hydraulic limits, or all effects of trimming.

Only after accounting for motor efficiency. Pump efficiency uses mechanical shaft input power as the denominator. If the available measurement is motor electrical input, calculate shaft power as electrical input multiplied by motor efficiency, then apply pump efficiency to determine hydraulic power.

Motor efficiency describes electrical-to-mechanical conversion from motor input to shaft output. Pump efficiency describes mechanical-to-hydraulic conversion from shaft input to hydraulic power delivered to the fluid. Combining them gives an overall wire-to-water efficiency for the represented motor and pump stages.

No. The calculator gives a theoretical energy-balance flow for the entered head, power, efficiency, and density. Actual operating flow depends on the pump performance curve, the system resistance curve, valve positions, speed, impeller configuration, fluid conditions, and the actual operating point. Use manufacturer curves and field measurements for final engineering decisions.