Parallel pumps: solve total flow at the system operating point

ReasonSpan™. P E Architectural Engineering. Parallel-pump operating points.

Two identical pumps in parallel do not automatically produce twice the original operating flow. The governing method is to combine flow at a common head, then find the intersection with the system curve. In this example, we will calculate both operating points and keep total flow separate from each pump's flow.

Each pump's head is fifty-five minus zero point zero zero four times its own flow squared. The system head is fifteen plus zero point zero zero one five times total system flow squared. Head is in meters and flow is in liters per second, with coefficients defined for those units. The identical pumps run at the same fixed speed and share total flow equally. These simplified instructional curves are not manufacturer selection data.

With one pump, its flow is the total flow. Set the pump head equal to the system head. Moving the terms gives forty equal to zero point zero zero five five times flow squared. The positive square root gives eighty-five point two eight liters per second. Substituting into the system curve gives an operating head of twenty-five point nine one meters. Static head alone is not the operating head.

With two identical pumps, each carries half the total flow at the same head. Substitute total flow divided by two into the individual pump equation. The parallel curve becomes fifty-five minus zero point zero zero one times total flow squared. We have added flow at a common head. We have not doubled head; that would confuse parallel operation with head addition in series.

Now intersect the parallel curve with the unchanged system curve.

Forty equals zero point zero zero two five times total flow squared. Total flow becomes one hundred and twenty-six point four nine liters per second, at thirty-nine meters of head. Each pump carries approximately sixty-three point two five liters per second. The total has increased about forty-eight point three two percent, while each pump's flow is lower than the original one-pump flow.

Change the assumption. If system head stayed constant at fifteen meters, with no friction-dependent increase, one pump would supply one hundred liters per second under this model. Two identical pumps would each supply that same flow at the same head, for two hundred liters per second total. So the original result depends on its rising system curve. Do not turn this example into a rule that parallel flow can never double.

For an actual selection, verify manufacturer curves and their published range, controls, minimum flow, efficiency, motor loading and available versus required suction head. When one pump stops, the remaining pump moves to its one-pump intersection. Open the complete ReasonSpan lesson to inspect the exact curves and test your reasoning, then try the separate free Architectural questions. The links are in the description.

Yes, in this idealized constant-head case. One pump supplies q = √((55 − 15)/0.004) = 100 L/s at 15 m. Two identical pumps each supply that same 100 L/s at the same head, giving 200 L/s total. The original non-doubling result depended on its rising system curve; it is not a universal prohibition on doubling.