Short-Cycling Duplex Lift Station
Your duplex submersible lift station has a 2.4 m inside-diameter wet well. The lead pump starts at 1.50 m and stops at 1.25 m; the lag pump starts at 1.85 m; high-level alarm is 2.10 m. Influent averages 22 L/s. The pump curve shows 34 L/s at 11.0 m total head, and static head is 8.8 m. The SCADA trend shows 24 starts per hour. The manufacturer allows a maximum of 10 starts per hour. The check valve was replaced last month. The force main is 150 mm and must carry at least 0.6 m/s to keep solids moving. The operations supervisor suggests simply raising the high-level alarm to stop nuisance alarms.
1. Calculate cycle time and starts per hour from the given data and compare them with the SCADA trend and the manufacturer's limit.
Read the worked answer
Wet-well area = π(1.2)^2 = 4.52 m². Working volume = 4.52 × 0.25 m = 1.13 m³ = 1130 L.
Fill time = 1130/22 = 51 s. Drawdown rate = 34 − 22 = 12 L/s; drawdown time = 1130/12 = 94 s.
Cycle = 51 + 94 = 145 s = 2.4 min; starts/hr = 60/2.4 = 25.
SCADA shows 24, consistent with variable inflow. Both exceed the manufacturer's 10 starts/hr. Because static head (8.8 m) is below the curve head (11.0 m), the pump may run out beyond 34 L/s, shortening drawdown and raising starts further. Check actual pump flow and run current.
2. Determine the minimum operating differential (lead start to off) needed to meet 10 starts per hour at 22 L/s, and explain whether simply raising the high-level alarm is acceptable.
Read the worked answer
For 10 starts/hr, cycle = 360 s. Let V be working volume: V/22 + V/12 = 360 → V(0.0455 + 0.0833) = 360 → V = 2.79 m³.
Depth = 2.79/4.52 = 0.62 m.
So the lead start-to-off differential must be about 0.62 m, not 0.25 m. Raising the high-level alarm only delays the alarm; it does not change cycling and leaves the pumps exposed to excessive starts. Set lead start at 1.50 m and off at about 0.88 m, provided the pump minimum submergence and wet-well volume allow. Verify lag and alarm settings and check for check-valve leakage or float drift.
3. A VFD is proposed to reduce pump flow to 26 L/s. Evaluate the effect on starts per hour and on force-main velocity, using the 150 mm force main.
Read the worked answer
At 26 L/s, drawdown = 26 − 22 = 4 L/s. With V = 1.13 m³: fill = 51 s, drawdown = 283 s, cycle = 334 s, starts = 10.8/hr - still slightly above 10.
Force-main velocity (150 mm, A = 0.0177 m²): v = 0.026/0.0177 = 1.47 m/s, above the 0.6 m/s minimum, so solids transport is acceptable.
A VFD can help, but check the pump curve: reducing speed also reduces head. If the reduced-speed shutoff head falls below 8.8 m static head, the pump will not deliver 26 L/s. Combine a modest differential increase to 0.30 m with the VFD: V = 1.36 m³, fill 62 s, drawdown 339 s, cycle 401 s ≈ 9 starts/hr. Do not use a VFD to solve a storage problem if it cannot overcome static head.