Training-Lab MCB Tripping After Compressor Start
A training lab is fed from a 230 V single-phase board through a 32 A Type C MCB and a 25 m run of 4 mm² copper cable in conduit at 40 °C ambient. The connected load is 8 desktop PCs at 300 W each, a 1.5 kW split AC, 300 W of LED lighting, a 500 W printer, and a 1.1 kW compressor motor with a power factor of 0.72 and a locked-rotor current of 48 A. The lab runs continuously for 6 hours. The MCB has tripped twice this week, both times a few seconds after the compressor started. A clamp meter showed 24 A steady-state with the compressor off. The electrician proposes replacing the 32 A MCB with a 40 A unit. The lab manager wants the compressor kept on the same circuit because rewiring is costly.
1. Calculate the steady-state current with all loads running and determine whether the 32 A MCB is correctly sized for continuous duty. Show your working.
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With the compressor off, the listed non-compressor load is 8×300 + 1,500 + 300 + 500 = 4,700 W, which at 230 V and unity power factor is 20.4 A. The measured 24 A is higher, so either the actual power factor is below 1 or there are unrecorded loads; take 24 A as the measured base load. The motor input apparent power is 1,100/0.72 = 1,528 VA, and its current is 1,528/230 = 6.64 A at a phase angle of cos⁻¹0.72 = 43.9°. Adding the motor vectorially to 24 A at unity: I_total = √[(24 + 6.64 cos 43.9°)² + (6.64 sin 43.9°)²] = √[(24 + 4.78)² + 4.61²] = 29.2 A. This is 91% of 32 A, well above the 80% continuous-load guide of 25.6 A, so the circuit is overloaded even before the 48 A locked-rotor surge. If the base load is at 0.85 power factor, the total is nearer 30 A.
2. Is replacing the MCB with a 40 A unit safe and sufficient? Support your answer with cable and protection reasoning.
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No. A 40 A breaker does not protect a 4 mm² cable in conduit at 40 °C, whose tabulated capacity is roughly 26-32 A depending on installation method and is lower after derating. The breaker must be selected for the cable's corrected capacity, not merely to stop tripping. The 48 A locked-rotor current for a few seconds may trip a Type C MCB if it is already warm, but upsizing moves the problem to the cable. Correct measures: give the compressor a dedicated motor-rated circuit with overload protection, correct the power factor, and verify the existing cable's installation method and derating. If the compressor must stay, a 32 A breaker is still too large for the cable; a smaller device may be appropriate after the load is separated. The measured 24 A base load also needs investigation because it exceeds the 20.4 A calculated for the listed equipment.
3. Calculate the capacitor kvar needed to correct the motor to 0.95 power factor and comment on whether this solves the tripping.
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Qc = P(tan φ1 − tan φ2). tan φ1 = tan(43.9°) = 0.963; tan φ2 for 0.95 = 0.329. Qc = 1.1 × (0.963 − 0.329) = 0.70 kvar. This reduces the motor current from 6.64 A to about 1,100/(230×0.95) = 5.03 A and the total current by only about 0.4 A, because the 24 A base load dominates. It does not fix the cable/breaker coordination or the starting surge; it only relieves reactive current. The tripping is driven by the base overload and the compressor start, so power-factor correction is useful but not sufficient.