Pumps in series and parallel on a system

Method: the operating point is the intersection of the pump curve and the system curve $H_{sys}=12+1500Q^2$.

(a) Single pump

$$40-2500Q^2=12+1500Q^2\ \Rightarrow\ 4000Q^2=28\ \Rightarrow\ Q=\sqrt{\frac{28}{4000}}=0.0837\ \text{m}^3/\text{s}$$ $$H=12+1500\times0.0837^2=22.50\ \text{m}$$

(b) Two pumps in series

For series operation the same discharge flows through both pumps and the heads add:

$$H_{ser}=2(40-2500Q^2)=80-5000Q^2$$ $$80-5000Q^2=12+1500Q^2\ \Rightarrow\ 6500Q^2=68\ \Rightarrow\ Q=0.1023\ \text{m}^3/\text{s}$$ $$H=12+1500\times0.1023^2=27.69\ \text{m}\ \ (13.85\ \text{m per pump})$$

(c) Two pumps in parallel

For parallel operation the heads are equal and the discharges add. Each pump delivers $Q/2$:

$$H_{par}=40-2500\left(\frac{Q}{2}\right)^2=40-625Q^2$$ $$40-625Q^2=12+1500Q^2\ \Rightarrow\ 2125Q^2=28\ \Rightarrow\ Q=0.1148\ \text{m}^3/\text{s}$$ $$H=12+1500\times0.1148^2=31.76\ \text{m}\ \ (\text{each pump delivers }0.0574\ \text{m}^3/\text{s})$$

Comparison

For this system the parallel arrangement gives the larger discharge, because the system curve is flat (small friction constant 1500 compared with the pump curve constant 2500). Note that two pumps never give double the discharge, since the system head rises with the flow.

When to use:

• Series: when the system needs a high head (steep system curve, large static lift); the pumps share the head, as in multistage pumps.
• Parallel: when a large discharge at nearly the same head is needed or the demand changes widely (one pump can be switched off); it also gives standby capacity.

Answer: Single: $Q=0.0837$ m³/s, $H=22.50$ m; series: $Q=0.1023$ m³/s, $H=27.69$ m; parallel: $Q=0.1148$ m³/s, $H=31.76$ m.