Sizing a Distribution Transformer for a Growing Workshop Feeder
A diploma trainee is asked to recommend a distribution transformer for a small industrial feeder. Connected load is 85 kW of three-phase induction motors (total), plus 12 kW of lighting and 9 kW of small single-phase loads. Measured demand factor from a similar feeder is 0.65; overall power factor is 0.78 lagging. The utility requires the transformer to be loaded to no more than 80 percent of rating at peak. Standard sizes available are 63, 100, 160 and 250 kVA. The trainee proposes a 100 kVA unit, arguing that 100 kVA exceeds the 85 kW motor total alone. The supervisor notes voltage drop at the far end of the 180 m feeder reaches 6 percent during motor starting and asks for a review of both transformer rating and the voltage-drop issue.
1. Evaluate the trainee's 100 kVA proposal and determine the minimum standard transformer rating that satisfies the 80 percent loading rule.
Read the worked answer
Step 1 - Peak demand in kW. Total connected load = 85 + 12 + 9 = 106 kW. Apply demand factor: 106 × 0.65 = 68.9 kW.
Step 2 - Convert to apparent power at actual power factor. S = P / pf = 68.9 / 0.78 = 88.3 kVA.
Step 3 - Apply the 80 percent loading limit. Required nameplate = 88.3 / 0.80 = 110.4 kVA.
Step 4 - Select standard size. 63 kVA → 140% loaded (fails). 100 kVA → 88.3/100 = 88% loaded (exceeds 80%, fails). 160 kVA → 55% loaded (passes). Minimum compliant size = 160 kVA.
The trainee's reasoning is flawed because kW of motors is not kVA, demand factor was ignored, and the 80 percent rule was not applied. A 100 kVA unit would run at 88 percent at peak with no margin for growth or motor starting inrush.
2. Explain why the 6 percent voltage drop during motor starting is a separate problem from transformer sizing, and propose two corrective measures with justification.
Read the worked answer
Transformer rating affects steady-state voltage regulation only weakly; a 6 percent dip during starting is dominated by the starting current flowing through the feeder impedance (cable resistance and reactance over 180 m) plus transformer impedance. Upsizing the transformer alone would reduce the dip only partially, because most of the impedance is in the long feeder.
Corrective measures:
- Increase conductor cross-section on the 180 m run. Voltage drop is proportional to conductor resistance, so a larger cable directly reduces the dip at starting current, where R dominates.
- Reduce starting current using a star-delta or soft starter on the largest motor. Starting current typically falls to about one-third with star-delta, cutting the dip roughly in proportion.
A third option - a capacitor bank for power factor correction - improves steady-state regulation and frees transformer capacity but does little for the starting dip. The answer should distinguish steady-state regulation from transient starting dip.