--- title: "Pumps in series and parallel on a system" subject: "Fluid Machines" chapter: 4 key: "semester_5/fluid_machines:pumps-series-parallel-numerical" source: https://www.bctengineer.com/q/semester_5/fluid_machines/pumps-series-parallel-numerical site: BCT Engineering --- # Pumps in series and parallel on a system Fluid Machines · Chapter 4: Pump - Asked in: year not given, 8 marks - Video: [Combining Pumps in Series and Parallel](https://www.youtube.com/watch?v=A611jYuKwaU) — MECH 241 - Fluid Mechanics I - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/fluid_machines/pumps-series-parallel-numerical - Whole chapter: https://www.bctengineer.com/md/semester_5/fluid_machines/chapter-4 ## Question Two identical centrifugal pumps each have the head-discharge characteristic H = 40 - 2500 Q² (H in m, Q in m³/s). They deliver water to a system whose head requirement is H = 12 + 1500 Q². Find the operating point (discharge and head) when (a) one pump runs alone, (b) two pumps run in series, and (c) two pumps run in parallel. Comment on which arrangement gives more discharge, and state when each arrangement is preferred. ## Short answer Equate pump and system curves. 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. Parallel gives more discharge for this flat system curve. Series is for high head, parallel for large discharge or variable demand. ## Answer **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 | Arrangement | Q (m³/s) | H (m) | Change in Q over single pump | |---|---|---|---| | Single pump | 0.0837 | 22.50 | - | | Series | 0.1023 | 27.69 | +22.2 % | | Parallel | 0.1148 | 31.76 | +37.2 % | 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the operating point of a pump in a system found? A. Where the pump curve cuts the power curve B. Where the pump head curve equals the system head curve C. At zero discharge D. At the maximum of the pump curve Answer: B. The intersection of H-Q pump curve and system curve gives the operating Q and H. (Key point: Operating point method) 2. How is the combined curve obtained for two identical pumps in series? A. Heads add at the same discharge B. Discharges add at the same head C. Both heads and discharges add D. Heads are averaged Answer: A. In series the same Q flows through both so H = 2(40 - 2500 Q^2); this gives Q = 0.1023 m^3/s. (Key point: Series combination) 3. For two identical pumps in parallel, each delivers Q/2 at the common head. What is the combined curve? A. H = 80 - 5000 Q^2 B. H = 40 - 625 Q^2 C. H = 40 - 1250 Q^2 D. H = 20 - 2500 Q^2 Answer: B. Substituting Q/2 gives 40 - 2500 (Q/2)^2 = 40 - 625 Q^2, leading to Q = 0.1148 m^3/s. (Key point: Parallel combination) 4. When is a series arrangement preferred? A. When a large discharge is needed at the same head B. When the system needs a high head, such as a large static lift C. When standby capacity is needed D. When demand varies widely Answer: B. Series pumps share the head (as in multistage pumps), while parallel pumps suit large discharge and varying demand. 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