LOGOS Engineering Workspace

LOGOS Learning · 10

Minimum flow recirculation in pump networks: the controlled bypass

Minimum flow recirculation keeps a centrifugal pump above its minimum flow when the consumers close: Q_pump = Q_A + Q_B + Q_R, and the controller opens the recirculation valve only if Q_pump would drop below Q_min. While the branches consume enough, the valve stays shut and wastes no energy.

Close the valves on branches A and B. When pump flow would drop below the minimum, the controller opens the recirculation valve just enough to hold the pump at minimum flow.

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Quick test

Three questions about this lesson. Got one wrong? The explanation shows right away.

1. Advantage of controlled recirculation over a fixed plate:

2. Why can the pump sit slightly above the minimum-flow setpoint?

3. When the branches draw plenty, the controlled recirculation valve stays…

0 of 3 answered

Why this happens

Arrows compare with the same A and B positions without recirculation.

Close the branches until flow drops below minimum

Continuity at the node

Qpump = QA + QB + QR
59.3 = 33.9 + 25.4 + 0.0 m³/h

The pump sees three paths: the two consumer branches and the recirculation. Pump flow is the sum of the three.

Controller rule

xR = 0 if Qpump ≥ Qmín; else Qpump = Qmín
59.3 → 59.3 m³/h · SP ≥ 40 m³/h · MV 0.0 %

FIC-01 measures pump flow and opens the recirculation valve only when it would fall below the minimum. The pump stays protected without wasting energy when the branches consume plenty, unlike the fixed plate of the minimum-flow lesson.

Energy at the node

Hnode = ztop,R + hf,R + hvalve,R
H_node = 23.43 m · z_topo,R = 4 m

The recirculation climbs to a 4 m crest and drops back into the source tank. Whatever node head is left after elevation and friction is burned in the recirculation valve: that is how it "adjusts the curve" the pump sees.

Rangeability

Kv,mín = Kv,100 / R
Kv_mín ≈ Kv100 / R = 2.8

An equal-percentage valve with R = 50 cannot control below about 2 % of its maximum Kv. When only a little flow is missing, it opens at its minimum stable point and the pump sits slightly above the setpoint.

Teaching pump with an illustrative curve.

Formulas in plain text

Continuity at the node
Q_pump = Q_A + Q_B + Q_R
Q_pump = pump flow (m3/h) · Q_A, Q_B = consumer branch flows (m3/h) · Q_R = recirculation flow back to the tank (m3/h)
Controller rule
x_R = 0 if Q_pump >= Q_min ; else Q_pump = Q_min
x_R = recirculation valve opening (%) · Q_min = pump minimum flow setpoint (m3/h)
Energy at the node
H_node = z_top,R + hf,R + h_valve,R
H_node = head at the node (m) · z_top,R = crest elevation of the recirculation line (m) · hf,R = friction loss in the recirculation line (m) · h_valve,R = head burned in the recirculation valve (m)
Valve rangeability
Kv_min = Kv_100 / R
Kv_min = smallest controllable Kv (m3/h at 1 bar) · Kv_100 = Kv fully open (m3/h at 1 bar) · R = rangeability (-)

Frequently asked questions

Why does a centrifugal pump need minimum flow recirculation?

At very low flow the pump heats the liquid it is churning and suffers internal recirculation, vibration and seal wear. A recirculation line returns part of the flow to the suction tank so the pump never runs below the minimum flow set by its manufacturer.

Controlled recirculation valve or fixed orifice?

A fixed orifice recirculates all the time and wastes energy even at full demand. A controlled valve, driven by a flow loop or an automatic recirculation valve, opens only when the pump flow would fall below the minimum.

Why does the pump sit slightly above the setpoint at small openings?

A control valve cannot regulate below its minimum controllable Kv, about Kv_100/R. With an equal-percentage valve of R = 50 that is about 2 % of the maximum Kv, so when only a little flow is missing the valve opens at that point and slightly overshoots.

The link opens the lesson exactly as it is now.