Adding Loads to a 25 kVA Workshop Transformer
The electrical board of a small printing unit is fed by a dedicated 415 V, 50 Hz, 25 kVA three-phase distribution transformer whose iron loss is 400 W and full-load copper loss is 700 W. The connected load is a 7.5 kW three-phase motor (efficiency 88%, power factor 0.82 at full load) plus single-phase lighting and socket circuits taken phase-to-neutral at 240 V: 3 kW on the red phase, 2 kW on the yellow phase and 1 kW on the blue phase, all effectively resistive. The neutral link at the distribution board was recently found discoloured and slightly loose, and the electrician simply tightened it and moved on. The owner now wants to add a second 5 kW motor (same efficiency and power factor) on the same board and asks whether the transformer can take it. The utility meter records kVA demand, not kW, and the connection is four-wire with the neutral solidly earthed at the transformer.
1. Calculate the neutral current with the existing lighting load and explain why the discoloured neutral link is a serious matter.
Read the worked answer
Phase voltage = 415/√3 = 240 V; all lighting loads are resistive, so: IR = 3000/240 = 12.5 A, IY = 2000/240 = 8.33 A, IB = 1000/240 = 4.17 A, displaced 120° apart. The balanced motor contributes no neutral current. IN = √(IR² + IY² + IB² − IRIY − IYIB − IBIR) = √(156.3 + 69.4 + 17.4 − 104.2 − 34.7 − 52.1) = √52.1 ≈ 7.2 A. So the neutral carries 7.2 A continuously - it is not a zero-current wire, and a discoloured link means it was carrying that current through a poor joint, which raises local resistance and heat and can end in an open circuit. An open neutral is worse: the star point floats, the phase with the highest resistance (the 1 kW load, 57.6 Ω) receives the highest voltage and its lamps and sockets are destroyed, while the 3 kW phase voltage collapses. The neutral must be sized as a current-carrying conductor, terminated properly, and never fused or switched on its own; the link should be replaced, not just tightened.
2. Calculate the transformer loading before and after the second motor is added, and give a reasoned recommendation.
Read the worked answer
Existing motor input = 7.5/0.88 = 8.52 kW at pf 0.82 (φ = 34.9°), so Q = 8.52 × tan 34.9° = 5.95 kvar. Lighting P = 6 kW, Q = 0. Total P = 14.52 kW, Q = 5.95 kvar → S = √(14.52² + 5.95²) = 15.7 kVA, i.e. 63% of 25 kVA. New motor: input = 5/0.88 = 5.68 kW, Q = 3.97 kvar → P = 20.2 kW, Q = 9.92 kvar → S = √(20.2² + 9.92²) = 22.5 kVA = 90% of rating. Judgement: the transformer can carry the steady load - 90% is acceptable for a distribution transformer - but leaves almost no margin for motor starting. A DOL start of the 5 kW motor (about 6 × 10 A = 60 A) will dip the voltage and disturb the printing machines and lighting. Options: start the new motor with star-delta or a soft starter, correct the power factor to release kVA, and check the tap position. If further growth is expected, plan a 40 or 63 kVA unit rather than overloading this one.
3. At what load does this transformer reach maximum efficiency, and what is that efficiency at the existing 0.82 power factor?
Read the worked answer
Maximum efficiency occurs when variable (copper) loss equals constant (iron) loss: x² × 700 = 400 → x = √(400/700) = 0.756, i.e. 75.6% of 25 kVA = 18.9 kVA. At that load: copper loss = 0.756² × 700 = 400 W, iron loss = 400 W, total 800 W. Output at pf 0.82 = 18.9 × 0.82 = 15.5 kW, so η = 15.5/(15.5 + 0.8) = 0.951, about 95.1%. The transformer is therefore already close to its best-efficiency band at present (15.7 kVA, just below optimum), and will run beyond it once the second motor is added (22.5 kVA), where copper loss rises as the square of the load and efficiency falls. Note that the kVA at which maximum efficiency occurs is fixed by the loss ratio, but the efficiency value itself depends on the load power factor, because useful output is kVA × pf.