Voltage Drop, Breaker Tripping, and Motor Current
A 1.5 hp, 230 V, single-phase, 3-wire submersible pump is set 480 ft deep in a well. The motor nameplate current is 9.8 A at 230 V. The branch circuit uses 12 AWG copper conductors run 500 ft from the pressure switch to the wellhead, plus 20 ft of 12 AWG submersible cable down the well. With the pump running, voltage at the pressure switch is 242 V and voltage at the motor is 218 V; current is 11.2 A. The 20 A breaker trips after about 3 minutes. Insulation resistance and winding resistance are normal. The control box capacitor tests good.
1. Calculate the voltage drop and determine whether the conductor size is acceptable.
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Total one-way length ≈ 500 + 20 = 520 ft, so the round-trip length is about 1,040 ft. 12 AWG copper has about 1.93 ohms per 1,000 ft at 20°C. R ≈ 1.93 × 1,040/1,000 ≈ 2.01 ohms. At 11.2 A, drop ≈ 2.01 × 11.2 ≈ 22.5 V, which agrees with the measured 242 − 218 = 24 V. Percent drop ≈ 24/242 × 100 ≈ 9.9%. That greatly exceeds the common 3% branch-circuit recommendation and is too much for a submersible motor. The motor draws higher current at reduced voltage, which explains the 11.2 A versus the 9.8 A nameplate value.
2. The homeowner suggests replacing the 20 A breaker with a 30 A breaker. How should you respond?
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Do not solve this by increasing the breaker. The breaker is responding to a real overcurrent condition caused largely by excessive voltage drop. Upsizing the breaker without correcting the conductor size can allow the motor to run hotter and fail, and it may exceed the branch-circuit protection limits for the motor. First correct the voltage drop. Then verify the motor overload protection in the control box, the capacitor, the start winding, and the pump load. A 20 A breaker may hold a 9.8 A motor under normal voltage; if a larger breaker is needed, size it per the motor branch-circuit rules, not by guesswork.
3. What conductor size would you recommend, and what else should be checked before restarting?
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To stay near 3% drop at 242 V, allowable drop is about 7.3 V. Required resistance is 7.3/11.2 ≈ 0.65 ohm over 1,040 ft, or about 0.63 ohm/1,000 ft. 8 AWG is about 0.76 ohm/1,000 ft and gives about 8.5 V drop, roughly 3.5%. 6 AWG is about 0.49 ohm/1,000 ft and gives about 5.5 V drop, roughly 2.3%. I would use 6 AWG to provide margin, or 8 AWG if the total drop including the service is acceptable. Before restart, check all splices, the well seal, the control box, the pressure switch, and the actual running current after the voltage is corrected. Record voltage and current at the motor.