A Submersible Pump That Trips Its Overload Every Twenty Minutes
A 3.7 kW, 415 V three-phase submersible pump irrigates a plot from a borewell. The nameplate gives 7.6 A, power factor 0.83 and efficiency 0.82. After about 20 minutes of running, the thermal overload relay in the starter trips. The starter panel stands in direct sun and reaches 48 degrees C inside. Measured while running: line voltages at the panel 412 V, 410 V, 411 V; at the motor terminals 396 V between phases; line currents 9.4 A, 9.5 A and 9.2 A. The overload relay is a Class 10 type set at 7.5 A. The cable is 4 sq mm copper, 70 m long. Winding resistance is balanced to within 2 percent, and insulation resistance to earth is 40 Mohm after the motor was lifted and dried. The farmer says the water has been sandy since the monsoon and suggests setting the relay to 10 A to finish the season.
1. Check whether the overload relay setting is correct by calculating full-load current from the nameplate. Then decide whether the relay or the motor is at fault.
Read the worked answer
Full-load current: I = P / (root 3 x V x pf x efficiency) = 3700 / (1.732 x 415 x 0.83 x 0.82) Denominator: 1.732 x 415 = 718.8; x 0.83 = 596.6; x 0.82 = 489.2 I = 3700 / 489.2 = 7.56 A.
So the nameplate current is 7.56 A and the relay is set at 7.5 A, which is essentially 100 percent of full load. That is correct practice: the relay is set at nameplate full-load current, never above it.
The measured current is 9.4 A, which is 9.4 / 7.56 = 124 percent of full load. A Class 10 relay at 1.24 times its setting will trip in the order of 10 to 30 minutes, which matches the observed 20 minutes exactly.
Conclusion: the relay is correctly set and is doing its job. The fault is not in the protection. The motor is genuinely drawing about 24 percent more than rated current, and the insulation resistance of 40 Mohm with balanced winding resistance says the motor itself is electrically healthy. The problem is in the load the motor is being asked to drive, plus a small supply-voltage contribution.
2. Break the 24 percent excess current into its likely contributions and explain why setting the relay to 10 A is unsafe.
Read the worked answer
Split the excess into three parts.
Undervoltage. The motor sees 396 V against a 415 V rating, i.e. 95.4 percent. For the same shaft power the current rises roughly in inverse proportion, about 1 / 0.954 = 4.8 percent, which is 0.36 A. So 396 V explains only about 5 of the 24 percent.
Real shaft load. The remaining roughly 19 percent must be mechanical or hydraulic: sandy water increasing friction in the bearings and impeller clearances, worn wear rings letting the pump run beyond its design flow (the power curve rises with flow), a delivery valve left fully open, or a partly choked foot valve and strainer causing cavitation and vibration.
Unbalance is negligible here: average current 9.37 A, maximum deviation 0.13 A, about 1.4 percent, below the 2 percent figure at which motors must be derated.
Why 10 A is unacceptable: 10 A is 132 percent of full-load current. Winding temperature rise climbs roughly with the square of current, so 1.32 squared is about 1.74 times the design rise. Insulation life halves for every 10 degrees C of extra temperature, so the motor may fail within a season, and the winding will be burnt beyond rewinding. It also removes the only protection the pump has, since a 16 or 20 A fuse or MCB will never see 10 A. If the pump then seizes or runs dry, the fire and shock risk passes to the cable and the starter. It also voids any warranty.
3. List the measurements and checks you would carry out next, using the cable and voltage data given, and say what each one would prove.
Read the worked answer
Cable drop first, to see whether the supply is really the villain. R per conductor = 0.0175 x 70 / 4 = 0.306 ohm. Three-phase drop = 1.732 x 9.4 x 0.306 = 5.0 V. But the measured panel-to-motor drop is 412 - 396 = 16 V. So about 11 V is being lost somewhere that is not the cable: loose terminal screws, pitted contactor tips, a corroded joint, a fuse-carrier contact or undersized internal panel wiring. Check the drop across the contactor (should be well under 1 V per phase) and feel or thermally scan for hot terminals after 15 minutes of running. This one check may recover most of the undervoltage.
Flow and head: measure the actual discharge with a bucket or weir and compare with the pump curve at that head. If flow is well above the design point, the pump is running out on its curve and the overload is hydraulic; correct the valve setting or replace worn rings. Sandy water calls for a strainer check and a sand separator.
Panel temperature: 48 degrees C in the sun derates a thermal relay and shortens contactor life; a sunshade and ventilation are cheap fixes.
Insulation and winding resistance: already acceptable (40 Mohm, 2 percent balance), so record them as the baseline and re-test after any rewinding.
Only after these checks should the relay setting be touched, and then only to the nameplate 7.6 A.