Centrifugal Pump and System Operating Point: 04-BS-7, December 2019, Q5

Question 5: Centrifugal Pump and System Operating Point (5 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Pump curve constants A / B	0.000060 / 1.6
System curve constant D	10
Pump speed, N	900 rev/min
Static head, C	20 m

Find. The pump curve $H(Q)$, the system curve $H(Q)$, and their operating (intersection) point.

Approach. Substitute $N=900$ into the pump curve, set it equal to the system curve (both give $H$ as a function of the same $Q$), and solve the resulting quadratic in $Q^2$ for the operating point.

Pump characteristic at N = 900 rev/min. $$H_{pump}=AN^2-BQ^2=(0.000060)(900)^2-1.6Q^2=\boxed{48.6-1.6Q^2}$$
System characteristic. $$H_{sys}=C+DQ^2=\boxed{20+10Q^2}$$
Operating point (intersection). $$48.6-1.6Q^2=20+10Q^2\ \Rightarrow\ 28.6=11.6Q^2\ \Rightarrow\ Q^2=2.4655$$ $$\boxed{Q=1.570\ \text{m}^3\text{/s}}\qquad H=20+10(2.4655)=\boxed{44.66\ \text{m}}$$

Pump head falls with Q; system head rises with Q. Their intersection at Q=1.57 m³/s, H=44.66 m is the operating point.

Quantity	Value
Pump curve	$H=48.6-1.6Q^2$
System curve	$H=20+10Q^2$
Operating point	Q = 1.570 m³/s, H = 44.66 m

Concept & Theory

Topic. Pump/system curve matching to find the operating point.

Key relations. Centrifugal pump head falls with the square of flow at fixed speed; a system with static lift plus friction losses rises with the square of flow; the operating point is where the two curves intersect.

Why this works. A centrifugal pump cannot independently set both its flow and head — the piping system "pushes back" harder as flow increases (frictional losses scale with $Q^2$), so the ONLY flow at which the pump's delivered head matches what the system demands is the unique intersection of the two curves; any other point is not a steady operating state.

Common pitfall. Solving $H_{pump}=H_{sys}$ for $Q$ directly without first confirming both are expressed as functions of the SAME variable $Q$ (not $Q^2$ vs $Q$) — here both curves are naturally quadratic in $Q$, so solving in $Q^2$ first, then taking the square root, avoids sign errors.

Source. White, Fluid Mechanics, Ch. 11 (Turbomachinery, pump/system matching).

Question 5 of 13: Centrifugal Pump and System Operating Point

Question 5: Centrifugal Pump and System Operating Point (5 marks)