Transformer Capacity and Motor-Starting Dip for a Plant Expansion
A food-processing plant in 10th of Ramadan City is fed at 11 kV through one 1000 kVA, 11/0.4 kV, Dyn11, ONAN transformer with Z = 6 %. Metering shows a present demand of 620 kW at 0.86 power factor. The owner wants to add a 355 kW, 400 V, four-pole induction motor driving a new refrigeration compressor (efficiency 0.95, power factor 0.89, locked-rotor current 6.5 × FLC) together with 120 kW of new lighting and small-power load at 0.90 power factor. The plant manager argues that no new transformer is needed because 'the unit is only at about 72 % load', and he does not want the capital tied up before the season starts. The same 0.4 kV board also feeds PLCs and a 200 kVA group of variable-speed drives, and the operations manager states flatly that this equipment 'cannot ride through voltage dips'. The utility confirms the 11 kV source fault level is 250 MVA.
1. Using the demand figures given, determine whether the existing transformer can carry the expansion. Show the kVA arithmetic and state your recommendation to the client.
Read the worked answer
Arithmetic (0.4 kV side, apparent power):
| Load | kW | pf | kVA |
|---|---|---|---|
| Existing | 620 | 0.86 | 721 |
| Motor input 355/0.95 = 373.7 | 373.7 | 0.89 | 420 |
| New miscellaneous | 120 | 0.90 | 133 |
| Total | 1114 | - | ≈ 1274 |
Vector cross-check: ΣP = 1114 kW; ΣQ = 721×0.510 + 420×0.456 + 133×0.436 = 368 + 191 + 58 = 617 kvar; S = √(1114² + 617²) ≈ 1273 kVA. The two methods agree, so the coincidence of power factors is not hiding anything.
Verdict: demand becomes about 127 % of nameplate, i.e. 1.27 pu. That is outside any sustained loading permitted by IEC 60076-7 for an ONAN unit, and it leaves no margin for the 45 °C summer ambient, for harmonic losses from the 200 kVA of drives, or for the loss of one cooling path. The manager's 72 % figure is arithmetically correct for the present load (721/1000) but says nothing about the future one, and it is a kVA figure while he is quoting it as if it were kW headroom.
Recommendation: refuse the connection as proposed. Either install a second transformer with a split busbar (preferred: keeps the drives and the compressor on separate sections), or reduce the new load before re-applying. Note that the answer does not depend on exact diversity assumptions: even at an optimistic 0.9 demand factor the figure is 1147 kVA, still over 100 %.
2. Estimate the voltage dip at the 0.4 kV busbar if the compressor is started direct-on-line with the plant at its present load, and state whether the PLC and VSD loads are at risk. Give at least two remedies with their trade-offs.
Read the worked answer
Source strength at the 0.4 kV bus: transformer 1000/0.06 = 16.7 MVA in series with the utility 250 MVA → 1/(1/16.7 + 1/250) = 15.6 MVA.
Motor starting apparent power: rated input S = 373.7/0.89 = 420 kVA → FLC = 420 000/(√3 × 400) = 606 A. At 6.5 × FLC = 3941 A, S_start = √3 × 400 × 3941 ≈ 2.73 MVA.
Dip from the motor alone: ΔV ≈ S_start/(S_sc + S_start) = 2.73/(15.6 + 2.73) = 14.9 %. Adding the 721 kVA already flowing through the same impedance pushes the estimate to roughly 18 %, i.e. about 0.82 pu at the busbar and lower at the motor terminals.
Consequence: the motor itself is borderline - NEMA MG-1 expects a standard motor to accelerate at 80 % terminal voltage - but the sensitive loads are not. Most drives trip when the DC link falls below about 85–90 % for a few cycles, and PLC power supplies will drop out. A 4–5 s compressor start would therefore stop the line: a production failure, not just a motor problem.
Remedies and trade-offs:
- VFD on the compressor: start current falls to ~1.0–1.5 pu, dip becomes a few percent; costs more, adds harmonics (mitigate with a line reactor or active front end) and a VFD is itself a sensitive load unless on a separate section.
- Soft starter: current ~2.5–3 pu → dip ~6–7 %; torque falls with the square of voltage (0.65² ≈ 42 %), so verify the compressor's breakaway torque with the unloaded-start valve open.
- Dedicated transformer or parallel operation: fixes the dip and the overload together but is the capital option the client is resisting; parallel operation also raises the LV fault level, which must be re-checked against the switchgear rating.
3. The contractor proposes fitting forced cooling (ONAF) to the existing transformer to gain 25 % capacity instead of buying a new unit. Assess this proposal technically.
Read the worked answer
ONAF on the same core and windings typically raises the rating to about 1250 kVA (+25 %), so 1274 kVA/1250 kVA = 102 % - still overloaded, and that is before any ambient correction. At the site's summer ambient the IEC 60076-7 derating would remove another several percent, so the true margin is negative.
The proposal also fails on two points the capacity figure does not address:
- Impedance is unchanged. Cooling raises the thermal rating only; the ohmic impedance stays at 6 %, so the 24 kA LV fault level and, critically, the motor-starting dip of ~15–18 % are exactly as calculated before. The sensitive PLC and VSD loads gain nothing.
- Reliability. The extra capacity exists only while the fans and their control circuit run. On fan failure or a control supply loss the unit reverts to 1000 kVA while the plant is drawing 1274 kVA, and the winding temperature rises with no alarm path unless winding temperature devices are wired to trip or alarm. For a process plant that is an unacceptable single point of failure.
Position to give the client: ONAF is a legitimate way to recover margin on a unit that is marginally overloaded; it is not a way to add 27 % new load. Either reduce the compressor to a smaller frame with a VFD, or install the second transformer. If the client insists on ONAF as an interim measure, require in writing: winding temperature protection connected, fan-failure alarm, a load-restriction scheme interlocked to the fans, and a re-run of the load flow after the season.