5 HP Irrigation Pump 370 m From the Transformer
A 5 HP (3.73 kW), 415 V three-phase submersible pump for a farmer's field is 370 m from the 100 kVA transformer. The existing connection is 2.5 mm² aluminium three-core cable, direct buried, with two joints. At the transformer the voltage is 425 V; at the pump terminal box while running it is 352 V and the current is 9.8 A per phase. The pump discharges about half of its rated flow and the motor body is hot after 40 minutes. The nameplate gives full-load current 7.5 A and power factor 0.80. The farmer has been told to 'change the overload relay setting and add a capacitor'. You are asked to write the recommendation for a vocational project report.
1. Confirm by calculation whether the cable explains the measurements. Show the voltage drop and explain the 9.8 A reading.
Read the worked answer
Resistance of the run: aluminium 2.5 mm² = 28.3/2.5 = 11.3 Ω/km at 20 °C; a buried, loaded cable runs near 70 °C, so apply ≈1.2: 13.6 Ω/km. For 0.37 km: R = 13.6 × 0.37 = 5.03 Ω per conductor.
Running drop: ΔV ≈ √3 × I × R × cos φ = 1.732 × 9.8 × 5.03 × 0.80 = 68 V. The measured drop is 425 − 352 = 73 V; the extra 5 V is consistent with the two joints, which add resistance.
Conclusion: the cable, not the motor, explains the measurement. The drop is 73/425 = 17%, more than three times the usual 5% limit.
Why 9.8 A: at 352 V the motor cannot produce its rated torque without drawing more current. A submersible pump is a roughly constant-torque load, so current rises as voltage falls - 9.8 A is 131% of FLC. That excess current, not a faulty relay, is what heats the motor body and what the overload relay is (correctly) seeing. The reduced terminal voltage also explains the roughly half-rated flow: the motor cannot reach its rated speed and the pump runs off its design point.
2. Select a replacement conductor and check both running and starting voltage. Explain why 6 mm² is not sufficient.
Read the worked answer
Candidate 10 mm² aluminium: 28.3/10 = 2.83 Ω/km × 1.2 = 3.4 Ω/km; over 0.37 km, R = 1.26 Ω.
Running: with the correct voltage the current returns to about 7.5 A, so ΔV = 1.732 × 7.5 × 1.26 × 0.80 = 13 V, i.e. 3.1% of 415 V - comfortably inside the 5% limit.
Starting (DOL, ≈6 × FLC = 45 A, starting pf ≈0.35): ΔV = 1.732 × 45 × 1.26 × 0.35 = 34 V, terminal voltage ≈ 425 − 34 = 391 V - the contactor will hold in and the pump will accelerate.
Why not 6 mm²: 28.3/6 = 4.72 Ω/km × 1.2 = 5.66 Ω/km, R = 2.09 Ω over the run. Running drop = 1.732 × 7.5 × 2.09 × 0.80 = 21.7 V = 5.2%, already over the limit with no margin for the joints or for a future larger pump; starting drop = 1.732 × 45 × 2.09 × 0.35 = 57 V, leaving about 368 V at start, which is marginal for a submersible set. Paying twice for cable is worse than paying once.
Recommendation: lay 10 mm² aluminium (or 6 mm² copper if the budget allows a smaller section), replace the two buried joints with a single sealed junction box using crimped lugs and proper insulation, and record the insulation resistance and earth continuity before energizing. If a larger pump is planned within a few years, lay 16 mm² now - the trenching costs more than the extra metal.
3. Evaluate the farmer's two suggestions - raising the overload relay setting and adding a capacitor - and state what else must be checked.
Read the worked answer
Raising the overload setting: wrong in principle. The current is high because the voltage is low, not because the relay is set too low. The relay is doing its job by sensing 131% of FLC. Raising the setting would let the winding run at that current until it fails, converting a cable problem into a burnt motor and a rewind bill. The setting stays at the nameplate FLC of 7.5 A (or the starter manufacturer's figure).
Adding a capacitor: power-factor correction does reduce line current - bringing pf from 0.80 to about 0.95 cuts current by roughly 15%, which would trim the drop by perhaps 8–10 V. That is 10 V out of a 73 V problem. It is a partial, symptomatic measure that leaves the cable grossly undersized, the joints hot and the starting voltage unacceptable. Capacitors on a rural pump also need switching and protection, or they cause over-voltage at light load.
Further checks before energizing:
- Insulation resistance of the motor and cable (a cable that has been hot and buried may already be degraded).
- Earth continuity at the pump and the earth electrode resistance.
- The two joints: feel for heat after 30 minutes of running, or scan; replace regardless.
- Actual head and flow against the pump curve after the voltage is corrected - the half-flow symptom may be partly the system, not only the supply.
- The starter's contactor and relay condition after months of operating at 131% current.