ENGINEERING EXPLAINED / SERIES
Two pumps.
Not twice
the head.
Heads add at the same flow. Your system decides which flow you actually get.
Find the crossing ↓REFLUX FOR YOUR WORK
From a model
to a real process.
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01 / ADD THE HEADS. THEN FIND THE FLOW.
The crossing
is the answer.
The orange curve is two identical pumps in series. Its height doubles at each fixed flow. The operating point moves to where that curve meets the green system curve.
At the old flow, the combined head is 51 m. At the new crossing, it is only 34.23 m.
Why the lift slider does not change 1.34× in this model
For this deliberately simple quadratic example, Hpump = 50 − 0.005Q² and Hsystem = Hstatic + kQ². With n identical stages, Q² = (50n − Hstatic)/(0.005n + k). Static lift changes flow and total head, but the ratio H₂/H₁ = 2(0.005 + k)/(0.010 + k) cancels the static term while both arrangements have forward-flow operating points. Real vendor curves need not behave this way.
The baseline k = 25.5/70² gives Q₁ = 70, H₁ = 25.5 m; Q₂ = 81.10, H₂ = 34.23 m. Increase friction to make the head gain approach 2×. “About 1.3×” is this example, not a universal rule. Flow uses arbitrary consistent units. No efficiency, motor, NPSH or allowable operating-range constraints are included.
02 / WHAT THE SECOND CASING FEELS
It starts
under pressure.
Liquid enters the impeller eye, moves outward through rotating blades, and collects in the volute. The shaft supplies energy; the casing helps convert velocity into pressure.
Pump 2 takes pump 1’s discharge as its inlet. Its casing and seal must tolerate that inlet pressure plus the pressure it develops. The checks include the relevant maximum and transient conditions.
pout,2 ≈ pin,1 + ρg(H₁ + H₂)
Equal elevations and pipe velocities; connecting-pipe losses neglected. Head is energy per unit weight. For water, 10 m is about 0.98 bar.
ILLUSTRATIVE WATER CASE
Baseline duty: each stage adds 17.11 m. These are gauge pressures with a 1 bar(g) starting pressure, not the tutorial case below.
03 / THE SIZING CASE
Three smaller pumps
lost to one bigger pump.
One tutorial compared three different equipment selections. It does not prove that series is inherently worse. The selected series train simply fell short of the required discharge pressure.
bar discharge · required ≥40 bar
Case provenance and the separate loss-of-stage issue
CHE480 Tutorial 3 — Pump Arrangements Summary, provided course material: 25 wt% aqueous DEA, 40 °C and 20 bar inlet. Reported discharge pressures were 39.2954, 43.7435 and 46.3613 bar. These are discharge pressures, not differential heads. The single, parallel and series cases used different pump models, speeds or impellers; the comparison is not a controlled arrangement-only experiment.
The running series train already missed the specification. Losing a stage is a separate operability risk, not the explanation for its 39.3 bar result. A stopped stage can add resistance, and the remaining duty depends on the system, bypasses, check valves and operating sequence. The bars above show only the reported running cases; no outage pressure is invented.
04 / DRAW THE SYSTEM TOO
A discharge pressure
is only one point.
A basic pump block can report a discharge pressure. To size the train, you also need the pump curves, system curve and operating limits. Some simulation tools can model the full hydraulic network; a single block result does not establish that those checks have been done.
Sources: KSB series operation · KSB operating point. Pump artwork is an original generated teaching illustration; motion is slowed for visibility.