Challenge: Will This Motor Start Without Crashing the Network?

The Scenario

A food processing plant powered by a single 500 kVA transformer (11kV/415V, 5% impedance). Existing load: 300 kVA at 0.85 pf.

The plant manager wants to add a 90 kW refrigeration compressor with DOL starting:

• Rated current: 157A at 415V
• Starting current: 6× FLC = 942A
• Starting kVA: √3 × 415 × 942 = 677 kVA
• Starting duration: 5 seconds
• Motor power factor at start: 0.25 lagging

Adjacent equipment on the same bus:

• 45 kW VFD-driven conveyor — trips at <85% voltage (undervoltage protection)
• PLC control system — requires >90% voltage for reliable operation

The Challenge

1. Calculate the voltage dip during motor starting
2. Determine if the VFD and PLC will survive
3. Propose a solution if they won't

The Solution

Voltage Dip Calculation

The voltage dip during DOL motor starting:

ΔV% = (S_start / S_sc) × 100%

Where:

• S_start = starting kVA of the motor = 677 kVA
• S_sc = short-circuit power of the supply = S_tx / Z% = 500 / 0.05 = 10,000 kVA

ΔV% = (677 / 10,000) × 100% = 6.8%

Voltage during starting: 415 × (1 − 0.068) = 386.8V (93.2% of nominal)

Impact Assessment

Both pass — barely. The PLC has only 3.2% margin. But we haven't accounted for cable impedance.

Including Cable Impedance

The motor is 80m from the distribution board via 70mm² cable. The cable impedance adds to the source impedance:

Z_cable = 80m × (0.000319 + j0.000309) = 0.0255 + j0.0247 Ω

Total impedance at motor terminals during start: Z_total = Z_tx + Z_cable = (0.0086 + j0) + (0.0255 + j0.0247) Ω

The voltage dip at the motor is now larger (more impedance), and the bus voltage depends on the cable impedance ratio.

Using the impedance divider:

ΔV_bus% ≈ (Z_motor_start / (Z_source + Z_motor_start)) × Z_source/Z_total × 100%

Simplified: the additional cable impedance increases the total system impedance, but most of the voltage drop occurs in the transformer. The bus voltage dip remains approximately 6.8%, but the motor terminal voltage drops further.

However, there's another factor: the motor starting power factor of 0.25 means the starting kVA is mostly reactive. The transformer impedance is mostly reactive (X >> R). Reactive current through reactive impedance causes maximum voltage drop.

Adjusting for the reactive nature: ΔV% = (S_start × sinφ_start) / S_sc = (677 × 0.968) / 10,000 = 6.55% (reactive component)

This is slightly lower than the simple calculation because some of the starting current is resistive (real power component), which causes less voltage drop through the reactive transformer impedance.

Revised Assessment with Safety Factor

With commissioning measurement uncertainty (±2%) and supply voltage variation (±6%):

Worst case: 94% supply × (1 − 0.068) = 87.6%

Now the VFD passes (87.6% > 85%) but the PLC fails (87.6% < 90%).

Solution: Soft Starter

Replace DOL starting with a soft starter set to 3× FLC current limit:

Starting current: 3 × 157A = 471A Starting kVA: √3 × 415 × 471 = 339 kVA

ΔV% = 339 / 10,000 = 3.4%

Worst-case bus voltage: 94% × (1 − 0.034) = 90.8% — PLC passes with margin.

Alternative: Star-delta starting gives approximately 33% starting current (1/3 of DOL), similar to the soft starter result.

Model the starting: Calculate motor starting voltage dip with the Motor Calculator and verify transformer adequacy with the Transformer Calculator.

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Frequently Asked Questions

What is the maximum allowable voltage drop?

AS/NZS 3000 Clause 3.3.4 allows 5% for lighting and heating, IEC 60364-5-52 allows 4% for lighting (3% recommended), BS 7671 allows 3% for lighting and 5% for other uses. Always check local regulations.

Does conductor temperature affect voltage drop?

Yes significantly. Copper resistance increases with temperature per IEC 60228 (α = 0.00393/°C). A cable at 10°C has ~20% lower resistance than at 90°C, affecting voltage drop calculations for cold-start conditions.

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