Worked Example: Voltage Dip During Motor Starting for a 55 kW DOL Motor

When a large induction motor starts direct-on-line (DOL), it draws a starting current typically 6–8 times its full-load current. This surge current causes a voltage dip at the motor terminals and on the supply bus, potentially affecting other connected loads. If the voltage dip is too severe, the motor may fail to accelerate, contactors may drop out, and sensitive equipment may malfunction.

This worked example calculates the percentage voltage dip at the motor terminals during DOL starting of a 55 kW three-phase induction motor and verifies it is within the commonly accepted 15% limit per IEC 60034-12 and AS/NZS 3000 Clause 2.5.3.

The motor full-load current for a three-phase motor is:

IFL = P / (√3 × V × PF × η)  — (Eq. 1)

IFL = 55,000 / (√3 × 400 × 0.87 × 0.93)
IFL = 55,000 / (1.732 × 400 × 0.87 × 0.93)
IFL = 55,000 / (1.732 × 400 × 0.8091)
IFL = 55,000 / 560.6
IFL = 98.1 A

With a starting current ratio of 7.0:

Istart = (Is/In) × IFL  — (Eq. 2)

Istart = 7.0 × 98.1
Istart = 686.7 A

This is the current drawn at the instant of starting (locked rotor condition) at rated voltage. The actual starting current will be slightly less because the voltage at the motor terminals will be reduced due to the voltage dip itself — but the calculation conservatively uses the full rated-voltage starting current.

The source impedance as seen at the switchboard can be derived from the prospective short-circuit level:

Zsource = V2 / Sfault  — (Eq. 3)

Where:
  V = 400 V (line-to-line)
  Sfault = 25 MVA = 25,000,000 VA

Zsource = 400² / 25,000,000
Zsource = 160,000 / 25,000,000
Zsource = 0.00640 Ω
Zsource = 6.40 mΩ

For typical utility supplies, the X/R ratio is approximately 5:1 at this fault level. Therefore:

Rsource = Zsource / √(1 + (X/R)²)
Rsource = 6.40 / √(1 + 25)
Rsource = 6.40 / 5.099
Rsource = 1.26 mΩ

Xsource = Rsource × (X/R)
Xsource = 1.26 × 5
Xsource = 6.28 mΩ

Check: Z = √(1.26² + 6.28²) = √(1.588 + 39.44) = √41.03 = 6.41 mΩ ✓

The cable impedance per phase for the 50 m route:

Rcable = r × L = 0.524 × 50 = 26.20 mΩ  — (Eq. 4)

Xcable = x × L = 0.077 × 50 = 3.85 mΩ

Zcable = √(Rcable² + Xcable²)
Zcable = √(26.20² + 3.85²)
Zcable = √(686.44 + 14.82)
Zcable = √701.26
Zcable = 26.48 mΩ

The total impedance from the source to the motor terminals (per phase):

Rtotal = Rsource + Rcable = 1.26 + 26.20 = 27.46 mΩ
Xtotal = Xsource + Xcable = 6.28 + 3.85 = 10.13 mΩ

Ztotal = √(Rtotal² + Xtotal²)  — (Eq. 5)
Ztotal = √(27.46² + 10.13²)
Ztotal = √(754.05 + 102.62)
Ztotal = √856.67
Ztotal = 29.27 mΩ

The voltage drop across the source and cable impedances during motor starting is calculated using the starting current and accounting for the starting power factor:

Phase voltage: Vph = 400 / √3 = 230.9 V

Voltage drop per phase (using complex impedance):
  ΔVph = Istart × (Rtotal × cosφstart + Xtotal × sinφstart)  — (Eq. 6)

Where:
  cosφstart = 0.30
  sinφstart = √(1 − 0.30²) = √(1 − 0.09) = √0.91 = 0.954

  ΔVph = 686.7 × (27.46 × 10&supmin;³ × 0.30 + 10.13 × 10&supmin;³ × 0.954)
  ΔVph = 686.7 × (8.238 × 10&supmin;³ + 9.664 × 10&supmin;³)
  ΔVph = 686.7 × 17.902 × 10&supmin;³
  ΔVph = 12.29 V (per phase)

Convert to line-to-line voltage dip and percentage:

ΔVLL = √3 × ΔVph = 1.732 × 12.29 = 21.29 V

Alternatively, express as a percentage of supply voltage:
  ΔV% = (ΔVph / Vph) × 100  — (Eq. 7)
  ΔV% = (12.29 / 230.9) × 100
  ΔV% = 5.32%

Motor terminal voltage during starting:
  Vmotor = 400 − 21.29 = 378.7 V
  Vmotor% = (378.7 / 400) × 100 = 94.7%

The commonly accepted voltage dip limits for motor starting are:

Calculated voltage dip: ΔV% = 5.32%

  5.32% < 15%   → DOL starting acceptable (general industrial)
  5.32% < 10%   → DOL starting acceptable (sensitive equipment)

Motor terminal voltage: 94.7% of rated
  94.7% > 85%   → Motor will develop sufficient starting torque

Conclusion: DOL starting of the 55 kW motor is acceptable. The voltage dip of 5.32% is well within the 15% general limit and even within the 10% limit for installations with sensitive equipment. The cable impedance (26.48 mΩ) dominates over the source impedance (6.40 mΩ), which is typical for motor circuits fed from strong supply networks. For weaker supplies or longer cable runs, a soft starter or star-delta starter may be necessary.

• IEC 60034-12 — Starting performance of single-speed three-phase cage induction motors
• AS/NZS 3000:2018, Clause 2.5.3 — Requirements for starting of motors
• IEC 60228 — Conductors of insulated cables (resistance data)
• IEC 61000-4-11 — Voltage dips and short interruptions immunity tests
• IEEE Std 141 (Red Book) — Electric power distribution for industrial plants, Chapter 9

Use the ECalPro Motor Current Calculator to evaluate voltage dip for your motor installations. Enter the motor rating, cable details, and source fault level — the calculator determines starting current, voltage dip percentage, and recommends the appropriate starting method (DOL, star-delta, soft starter, or VFD).