Characteristic Curve Analysis for Centrifugal Pumps
Reference ID: MET-08D8 | Process Engineering Reference Sheets Calculation Guide
Introduction & Context
Centrifugal-pump selection requires matching the pump head-capacity (H-Q) curve to the system head curve.
The intersection defines the operating point where the energy supplied by the impeller exactly balances the static lift plus flow-dependent friction losses.
Knowing this point fixes flow rate, head, efficiency, and power—quantities that govern pipe sizing, motor selection, energy audits, and control-valve specification in water, petrochemical, utility, and food-process plants.
Methodology & Formulas
Convert capacity units
\(Q_{\text{L s^{-1}}} = \dfrac{1000}{3600}\,Q_{\text{m$^{3}$ h$^{-1}$}}\)
System head curve
\(H_{\text{req}}(Q) = H_{\text{static}} + K\,Q^{2}\)
with
\[
K = \dfrac{8}{\pi^{2}\,g\,D^{4}}\left(\dfrac{f\,L}{D}+\sum K_{\text{minor}}\right)\times10^{-6}
\]
where the factor \(10^{-6}\) converts \((\text{m$^{3}$ s$^{-1}$})^{2}\) to \((\text{L s$^{-1}$})^{2}\).
Friction factor (Haaland approximation)
\[
f = \left[-1.8\,\log_{10}\!\left(\dfrac{\varepsilon/D}{3.7}+\dfrac{6.9}{Re}\right)\right]^{-2}
\]
Flow regime
Reynolds number
Laminar
\(Re \le 2300\)
Transitional
\(2300 \lt Re \lt 4000\)
Turbulent
\(Re \ge 4000\)
Linear interpolation of discrete pump data
For \(Q_{i}\le Q_{\text{target}}\le Q_{i+1}\)
\[
Y(Q_{\text{target}})=Y_{i}+\dfrac{Y_{i+1}-Y_{i}}{Q_{i+1}-Q_{i}}\,(Q_{\text{target}}-Q_{i})
\]
applies to both head \(H\) and efficiency \(\eta\).
Operating-point intersection
Equate pump segment \(H = h_{1}+\dfrac{h_{2}-h_{1}}{q_{2}-q_{1}}(Q-q_{1})\) to system curve, giving
\[
a\,Q^{2}+b\,Q+c=0,\quad
a=K,\quad
b=-\dfrac{h_{2}-h_{1}}{q_{2}-q_{1}},\quad
c=H_{\text{static}}-h_{1}+\dfrac{h_{2}-h_{1}}{q_{2}-q_{1}}\,q_{1}
\]
Solve the quadratic; accept the root lying inside the interval \([q_{1},q_{2}]\).
A flat or very shallow head-versus-flow slope (small ΔH/ΔQ) means that a tiny change in flow produces only a small change in head. In practice this makes the loop hard to control with a valve because:
The same valve travel produces a large flow change, so gain is high and the controller tends to oscillate.
Small pressure disturbances in the downstream piping translate into large flow swings.
Best practice is to stay on the steeper part of the curve—typically 60–80 % of BEP flow—where ΔH/ΔQ is larger and the process gain is more forgiving.
Look for the vendor’s “minimum continuous stable flow” (MCSF) mark on the curve. If the curve does not show it, use these rules of thumb:
Stay above 25–30 % of the best-efficiency-point (BEP) flow for standard process pumps.
Check NPSHR at the low-flow end; NPSHR often rises steeply, pushing the pump toward cavitation.
Verify power draw: if power increases at low flow (high head), the pump may overheat; ensure a bypass or orifice will keep flow above MCSF.
The published curve is for cold water (1 cP). For viscous liquids apply the Hydraulic Institute viscosity correction:
Derate head, flow, and efficiency using the HI correction charts or software; typically start corrections above 10–20 cP.
After correction, re-plot the curve and re-check that the new BEP and power still meet system requirements.
Remember NPSHR is not corrected; if viscosity > 100 cP, consider a positive-displacement pump instead.
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Worked Example – Matching a Centrifugal Pump to a Cooling-water Circuit
A process engineer must verify that a new centrifugal pump can deliver 25 L s-1 of cooling water through 150 m of 150 mm pipework to an overhead exchanger. The static lift is 12 m and minor losses are estimated as ΣK = 6.5. The pump manufacturer supplies the quadratic head–capacity curve \( H_p = 50 - 0.00435 Q^2 \) (Q in L s-1, H in m). Determine the operating flow rate, the head the pump will develop, and the hydraulic power it must provide at that point.
Knowns
Static head \( H_{static} = 12.0 \) m
Pipe length \( L = 150.0 \) m
Pipe diameter \( D = 0.150 \) m
Pipe roughness \( \varepsilon = 5 \times 10^{-5} \) m
Sum of minor-loss coefficients \( \Sigma K = 6.5 \)
Water density \( \rho = 998 \) kg m-3
Water dynamic viscosity \( \mu = 0.001 \) Pa s
Pump curve coefficients: \( a = 0.00435 \), \( b = 0 \), \( c = 50 \) (quadratic form \( H_p = c - a Q^2 \))
Step-by-step calculation
Express the system-head curve: major + minor losses plus static lift.
\[
H_{sys}(Q) = H_{static} + \left( f \frac{L}{D} + \Sigma K \right) \frac{V^2}{2g}
\]
with \( V = \frac{4Q}{\pi D^2} \) and Q in m3 s-1.
Convert the target 25 L s-1 to 0.025 m3 s-1 and compute velocity:
\[
V = \frac{4 \times 0.025}{\pi \times 0.15^2} = 1.415 \text{ m s}^{-1}
\]
Calculate Reynolds number:
\[
Re = \frac{\rho V D}{\mu} = \frac{998 \times 1.415 \times 0.15}{0.001} = 211\,359
\]
Use the Colebrook equation (or Swamee–Jain) to find friction factor:
\[
f = 0.020 \quad (\text{fully turbulent, } \varepsilon/D = 0.00033)
\]
Compute total head loss coefficient for the system:
\[
K_{tot} = f \frac{L}{D} + \Sigma K = 0.020 \times \frac{150}{0.15} + 6.5 = 26.5
\]
Write the system curve in terms of Q (L s-1):
\[
H_{sys}(Q) = 12 + \frac{8}{\pi^2 g D^4} \left( f \frac{L}{D} + \Sigma K \right) \left( \frac{Q}{1000} \right)^2
\]
Evaluating constants gives:
\[
H_{sys}(Q) = 12 + 0.00435 Q^2 \quad (Q \text{ in L s}^{-1})
\]
Equate pump and system heads to find the operating point:
\[
50 - 0.00435 Q^2 = 12 + 0.00435 Q^2
\Rightarrow 0.0087 Q^2 = 38
\Rightarrow Q_{op} = \sqrt{38 / 0.0087} = 36.8 \text{ L s}^{-1}
\]
Insert \( Q_{op} \) into either curve to obtain operating head:
\[
H_{op} = 12 + 0.00435 \times 36.8^2 = 17.9 \text{ m}
\]
