Pump Operating Point Simulator — Intersection of Pump and System Curves
Real-time intersection of the pump performance curve and the system curve. Move the sliders for head and resistance, and the flow, head, hydraulic and shaft power update instantly.
Pump Parameters
Shut-off head H0
m
Pump curve slope kp
×10⁻³
System Parameters
Static head Hstatic
m
System resistance ks
×10⁻³
Results
—
Operating flow Qop
—
Operating head Hop
—
Hydraulic power Pwater
—
Shaft power Pshaft (η=70%)
Pump and System Curves
Pump curve H = H0 − kp·Q²System curve H = Hstatic + ks·Q²Operating point
$$P_{\text{water}}=\rho g Q_{\text{op}} H_{\text{op}},\quad P_{\text{shaft}}=P_{\text{water}}/\eta$$
Q in L/s, H in m, kp / ks in m/(L/s)². Constants: ρ = 1000 kg/m³, g = 9.81 m/s², assumed efficiency η = 0.70.
What is the pump operating point?
When a real pump is installed in a real piping system, the actual operating state is fixed by the intersection of two curves: what the pump can deliver (pump performance curve) and what the system requires (system curve). This intersection is the operating point, and it is the foundation of pump selection.
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Professor, what is the yellow dot where the red and blue curves cross? It moves as I drag the sliders.
🎓
Good catch. That dot is the operating point. The red curve is what the pump can supply at each flow rate Q, the blue curve is what the piping demands at the same Q. The flow at which they balance is the actual flow you would measure — about 68 L/s at 53 m for the default values.
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In the schematic on the right there is a vertical arrow labeled Hstatic. Is that just the difference between the two tank water levels?
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Exactly. Hstatic is the geodetic head you have to climb regardless of flow — a fixed energy "tax". Moving slHst raises the upper tank and shifts the entire system curve up, which slides the operating point sideways too.
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When I press "Sweep ks" the dot slides left and right. What does that correspond to in the field?
🎓
Throttling a valve. Closing it makes ks larger and the system curve steeper, so flow drops. Opening it does the opposite. Watch the hydraulic and shaft power values as you sweep — that tells you how much energy you save by throttling versus, say, slowing the pump down.
Physical model and key equations
Both curves are approximated as parabolas. The physical justification is as follows.
Pump performance curve: $H_{\text{pump}}(Q) = H_0 - k_p Q^2$. The head at Q = 0 is the shut-off head H_0, and kp captures how quickly head drops with flow. It depends on impeller geometry and rotational speed.
System curve: $H_{\text{sys}}(Q) = H_{\text{static}} + k_s Q^2$. H_static lumps geodetic and pressure head, while ks·Q² is the sum of pipe friction (Darcy-Weisbach) and minor losses through fittings and valves.
Operating point: setting the two equal yields $Q_{\text{op}} = \sqrt{(H_0 - H_{\text{static}})/(k_p + k_s)}$ and $H_{\text{op}} = H_{\text{static}} + k_s Q_{\text{op}}^2$. Converting Q to m³/s gives hydraulic power $P_{\text{water}} = \rho g Q_{\text{op}} H_{\text{op}}$, and assuming η = 0.70 the shaft power is $P_{\text{shaft}} = P_{\text{water}}/\eta$.
Real-world applications
Building water supply: sizing the lift pump from a basement tank to a roof tank — compute ks from pipe runs and H_static from the lift, then pick a pump whose operating point is near the BEP.
Industrial cooling loops: in a chiller-tower loop, the heat exchanger pressure drop dominates ks. Sweeping the bypass valve gives an immediate read of how flow and shaft power change with control action.
Irrigation systems: for a farm pump, lift dominates so H_static is the main driver. Adjusting hose diameter changes ks and lets you reach the required flow with the same pump.
Fire protection pumps: hydrant pumps must satisfy static lift plus pipe loss plus nozzle pressure. Two pumps in series can be approximated by simply doubling H_0 to estimate the new operating point.
Common misconceptions
The system, not just the pump, sets the flow: "buying a bigger pump" does not always increase flow. If the system curve is unchanged, a higher pump curve only moves the operating point left, hardly raising Q. Reducing ks (larger pipe diameter) is often more effective.
Static head is a non-negotiable energy floor: when H_static is large, even a very small kp barely increases flow. Look at the numerator H_0 − H_static of Q_op² — if it goes negative, the pump cannot deliver any flow at all.
Efficiency is local, not global: this tool uses a fixed η = 0.70 for a quick estimate, but a real efficiency curve has a peak (the BEP). If your operating point sits far from the BEP, the actual shaft power will deviate from this estimate. Cross-check with the manufacturer's curve.
FAQ
Defaults are H0 = 100 m, kp = 0.010, H_static = 30 m, ks = 0.005. So Q_op² = (100 − 30)/(0.010 + 0.005) = 70/0.015 ≈ 4666.7 and Q_op = √4666.7 ≈ 68.3 L/s. The corresponding head is H_op = 30 + 0.005×4666.7 ≈ 53.3 m.
If H_static is set above H_0, the pump cannot move any liquid and the tool reports Q = 0. Increase H_0 or reduce H_static so that (H_0 − H_static) > 0.
The physical units of kp and ks are m/(L/s)². Realistic values are around 10⁻³, so the slider value is multiplied by 10⁻³ before use. A slider value of 10.0 means kp = 0.0100 m/(L/s)².
This tool focuses on a single pump with a fixed efficiency. For parallel/series operation and BEP efficiency curves, see the related "Pump operating point calculator" and "Centrifugal pump curves simulator" on this site.