Learn: Simple system curve
A simple pump-system curve adds a static head to flow-dependent losses. The static part remains even when flow stops; the quadratic part grows as liquid moves faster through the system.
What the formula is saying
Square discharge Q, multiply by the fitted system coefficient K, and add static head Hs. K groups the chosen friction and local-loss behavior into one study coefficient.
Read the symbols in plain language
- Hstatic
- Static head
Static head. Static and flow-dependent heads must use the same sign convention and units.
mMetres measure length; 1 m = 1000 mm.
- K
- System coefficient
Combined quadratic system-loss coefficient in H = Hs + KQ². Its unit is s²/m⁵ here, unlike a dimensionless minor-loss K.
s²/m⁵Use s²/m⁵ as the base unit shown here. Q is m³/s; therefore K must be s²/m⁵ so that KQ² is m. Hs and the total required head are also m. This K is not the dimensionless K used for a single fitting.
- Q
- Flow rate
Flow rate. Quadratic system losses depend on discharge squared rather than discharge alone.
m³/sUse m³/s as the base unit shown here. Q is m³/s; therefore K must be s²/m⁵ so that KQ² is m. Hs and the total required head are also m. This K is not the dimensionless K used for a single fitting.
- H
- Result to find
Simple system curve. Static and flow-dependent heads must use the same sign convention and units.
m
Sort out the units first
Q is m³/s; therefore K must be s²/m⁵ so that KQ² is m. Hs and the total required head are also m. This K is not the dimensionless K used for a single fitting.
Assumptions before calculating
Assume a fixed system configuration and an approximately constant quadratic-loss coefficient over the flow range of interest. Positive Q denotes the chosen flow direction; Hs can be signed when the receiving level is lower.
Let’s solve one together
Read the given values, follow each operation, then check what the result means.
Read the supplied values as one complete study case. Find H and explain the result in the stated output unit.
- Hstatic · Static head
- 15 m
- K · System coefficient
- 5000 s²/m⁵
- Q · Flow rate
- 0.05 m³/s
Square the selected discharge
Quadratic system losses depend on discharge squared rather than discharge alone.
(0.05)^2 = 0.0025 m⁶/s²Calculate the flow-dependent loss
The dimensional coefficient converts squared discharge into metres of head.
(5000) × (0.0025) = 12.5 mAdd the static component
Static and flow-dependent heads must use the same sign convention and units.
(15) + (12.5) = 27.5 m
Does this worked answer make sense?
At Q = 0, the answer is Hs. Doubling Q multiplies only the loss part KQ² by four, not necessarily the total head.
A second worked example — different values
A second case uses different data. Predict which way the answer will change, then calculate it without reusing the first answer.
- Hstatic · Static head
- 10 m
- K · System coefficient
- 4000 s²/m⁵
- Q · Flow rate
- 0.04 m³/s
Square the selected discharge
Quadratic system losses depend on discharge squared rather than discharge alone.
(0.04)^2 = 0.0016 m⁶/s²Calculate the flow-dependent loss
The dimensional coefficient converts squared discharge into metres of head.
(4000) × (0.0016) = 6.4 mAdd the static component
Static and flow-dependent heads must use the same sign convention and units.
(10) + (6.4) = 16.4 m
Now try your own values
Change a value or its unit. The same method will show your calculation, step by step.
Results update only when you calculate. The lesson example above stays unchanged.
Your turn — check your understanding
Solve this separate case yourself. Use only the values below; the two worked examples use different data. Give the requested result in the selected unit.
- Hstatic · Static head
- 12 m
- K · System coefficient
- 6000 s²/m⁵
- Q · Flow rate
- 0.03 m³/s
Find: Learn: Simple system curve
A hint, not the answer
Square discharge Q, multiply by the fitted system coefficient K, and add static head Hs. K groups the chosen friction and local-loss behavior into one study coefficient.
Q is m³/s; therefore K must be s²/m⁵ so that KQ² is m. Hs and the total required head are also m. This K is not the dimensionless K used for a single fitting.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Square the selected discharge
Quadratic system losses depend on discharge squared rather than discharge alone.
(0.03)^2 = 0.0009 m⁶/s²Calculate the flow-dependent loss
The dimensional coefficient converts squared discharge into metres of head.
(6000) × (0.0009) = 5.4 mAdd the static component
Static and flow-dependent heads must use the same sign convention and units.
(12) + (5.4) = 17.4 m
Avoid the common trap
Do not enter Q in L/s with a coefficient fitted for m³/s. Do not square Hs or add this coefficient directly to a dimensionless local-loss coefficient.
When this method applies — and when it does not
This is a simplified system curve, not a pump performance curve. Changes in valves, pipe routing, friction-factor regime or reservoir levels change the model. A negative calculated head indicates the chosen system may supply rather than require head at that flow.
For study and understanding, not approval of a real structure, site operation or design. Apply the correct standard, National Annex and professional review to actual engineering work.
One idea understood. Keep going.
This optional checkmark is saved only in this browser. It is your own progress note, not a certificate.
Sources & further reading
References open in a new tab and explain the underlying principles. The teaching text and examples here are SimpleFlick’s own; the source organisations have not endorsed this calculator.
Reading focus: Simple system curve. Water-measurement principles; for discharge devices, read the orifice/weir chapters and the installation and head-measurement conditions, not only the coefficient formula.
- U.S. Bureau of Reclamation — Water Measurement Manual, 3rd edition (1997; revised reprint 2001)
- Dawei Han, University of Bristol — Concise Hydraulics (2008, Ventus Publishing)
Lesson updated: · Both examples and the separate practice case are checked against an independent high-precision numerical implementation. This verifies arithmetic for the stated model, not engineering certification.