Furnace Batch Heating and Quench-Pump Motor Protection
A heat-treatment bay has an electrically heated tempering furnace rated 6 kW at 415 V three-phase, and a 5.5 kW three-phase induction motor driving the quench-oil pump. The motor nameplate gives 88% efficiency and 0.85 power factor. The furnace is used to heat a 10 kg batch of steel parts from 30 °C to 850 °C; the specific heat of steel is taken as 0.50 kJ/(kg·K) and furnace efficiency is 60%. Electricity costs ₹8.50 per kWh. The pump motor is protected by a 10 A miniature circuit breaker on an existing 2.5 mm² circuit. During start-up the breaker trips about once a week; the electrician measures a starting current of 60 A for roughly 4 s and a running current near 10 A. The maintenance head wants to replace the 10 A breaker with a 32 A cartridge fuse and a 4 mm² cable.
1. Calculate the useful heat energy, the electrical energy input at 60% efficiency, the cost, and the heating time at the 6 kW rating.
Read the worked answer
ΔT = 850 − 30 = 820 K. Q = mcΔT = 10 × 0.50 × 820 = 4,100 kJ = 4.100 MJ. Convert: 4,100/3,600 = 1.139 kWh useful. Input = 1.139/0.60 = 1.898 kWh. Cost = 1.898 × ₹8.50 = ₹16.13 per batch. Time at 6 kW input = 1.898/6 = 0.316 h = 19.0 min. Judgement: this is the theoretical minimum with the furnace at full rating and no standing losses. A real furnace has wall losses and a heat-up period, so expect 25–35 min and a higher bill. The 60% efficiency already includes some loss, but the 6 kW rating may not be delivered continuously at 415 V. Meter the actual consumption for one batch before quoting a cost per batch to the customer.
2. Calculate the motor's input power and line current at the nameplate values. Explain why the 10 A breaker trips and whether the measured 60 A starting current is abnormal.
Read the worked answer
Input power = output/efficiency = 5.5/0.88 = 6.25 kW. Three-phase current: I = P/(√3 × V × cos φ) = 6250/(1.732 × 415 × 0.85) = 6250/610.9 = 10.23 A. The normal running current is already 10.2 A, so a 10 A breaker sits at or below the load current; it will trip thermally even without a start. The measured running current near 10 A confirms this. Starting current 60 A = 60/10.23 = 5.9 × full-load current, which is normal for a direct-on-line induction motor (typically 6–8 × FLC). It is not a fault. Judgement: the breaker is both undersized and probably the wrong characteristic. Replace it with a 16 A motor-rated device, but the motor still needs a separate overload relay set to about 10.2 A (or the next standard setting); a breaker alone does not give correct overload protection.
3. Evaluate the proposal to fit a 32 A cartridge fuse and 4 mm² cable. State the correct protection arrangement and give the reason.
Read the worked answer
A 32 A fuse will carry 10.2 A indefinitely and will not open on a sustained 15–20 A overload - exactly the condition that burns out a motor winding. It protects the cable, not the motor. The 4 mm² cable is electrically adequate for 10.2 A, but upsizing the cable does not justify upsizing the fuse. Correct arrangement: a 16 A motor-rated MCB (type D, or an MCCB with adjustable magnetic setting) for short-circuit protection, plus a contactor with a thermal overload relay set at the motor full-load current, about 10.2 A (set 10–11 A). The existing 2.5 mm² cable is adequate if its installed current rating exceeds the protective device rating and voltage drop is acceptable; check derating for conduit and ambient temperature. Judgement: the intermittent trip is a sizing/characteristic problem, not a reason to fit a bigger fuse. A 32 A fuse would let the motor run hot and fail, and the fuse would then be blamed. Also check that the 4 s start is not lengthened by a mechanical fault - a stiff bearing or low supply voltage can both raise starting current and start time.