Pump Operating Point on System Curve
Find the operating point where the pump curve intersects the system curve using linear interpolation. Includes pump and system curve data tables.
| Flow (gpm) | Head (ft) |
|---|---|
| 0 | 110 |
| 400 | 100 |
| 800 | 80 |
| 1,000 | 68 |
| 1,200 | 52 |
| Flow (gpm) | Head (ft) |
|---|---|
| 0 | 40 |
| 400 | 52 |
| 800 | 72 |
| 1,000 | 82 |
| 1,200 | 95 |
Answer Choices
A) Approximately 875 gpm at approximately 76 ft TDH Correct
Correct: The operating point occurs where pump head equals system head. Between 800 and 1,000 gpm, the pump drops from 80 to 68 ft while the system rises from 72 to 82 ft; solving by linear interpolation gives Q ≈ 873 gpm and H ≈ 75.6 ft (rounds to about 875 gpm at 76 ft). Sanity check: at 800 gpm the pump head (80 ft) is higher than the system head (72 ft), and at 1,000 gpm the pump head (68 ft) is lower than the system head (82 ft), so the intersection must fall between them.
B) 800 gpm at 80 ft TDH
Wrong: This is a pump-curve table point, not the intersection. At 800 gpm the system curve shows about 72 ft, not 80 ft, so the pump would still have excess head and the flow would increase above 800 gpm until the curves meet.
C) 900 gpm at 74 ft TDH
Wrong: 900 gpm is close to the day’s influent when converting 1.3 MGD to gpm, but the operating point is determined by the pump and system curves, not today’s influent. Also, at 900 gpm the pump head is about 74 ft while the system head is about 77 ft, so they do not match.
D) 1,000 gpm at 82 ft TDH
Wrong: This is a system-curve table point, not the intersection. At 1,000 gpm the pump curve indicates about 68 ft, so the pump cannot meet the 82 ft required by the system at that flow; the actual operating flow must be lower.
Step-by-Step Solution
- Given tables, check bracket for intersection: at Q = 800 gpm, H_pump = 80 ft and H_system = 72 ft (pump > system); at Q = 1,000 gpm, H_pump = 68 ft and H_system = 82 ft (pump < system) ⇒ intersection is between 800–1,000 gpm.
- Pump curve slope between 800–1,000 gpm: (68 ft − 80 ft) ÷ (1,000 gpm − 800 gpm) = (−12 ft) ÷ (200 gpm) = −0.06 ft/gpm.
- System curve slope between 800–1,000 gpm: (82 ft − 72 ft) ÷ (1,000 gpm − 800 gpm) = (10 ft) ÷ (200 gpm) = 0.05 ft/gpm.
- Let Q = 800 gpm + x gpm. Then H_pump = 80 ft + (−0.06 ft/gpm)(x gpm) and H_system = 72 ft + (0.05 ft/gpm)(x gpm).
- Set heads equal: 80 ft − 0.06x ft = 72 ft + 0.05x ft.
- Solve for x: 80 ft − 72 ft = 0.11x ft ⇒ 8 ft = 0.11x ft ⇒ x = 72.73 gpm.
- Flow at operating point: Q = 800 gpm + 72.73 gpm = 872.73 gpm ≈ 875 gpm.
- Head at operating point: H = 72 ft + (0.05 ft/gpm)(72.73 gpm) = 75.64 ft ≈ 76 ft TDH.
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