Motor-starting voltage dip calculator

Bus voltage while a motor starts across the line, from the source, the transformer and the motor's locked-rotor kVA.

Result

Bus voltage, percent of nominal: 92.92%
Voltage dip: 7.08%

Bus voltage on a 0 to 100% scale. Ticks mark 85% (above the bar) and 90% (below it) of nominal.

Z_s (source): 0.006pu
Z_t (transformer): 0.0575pu
Z_m (motor, locked): 0.8333pu

Per unit on the transformer kVA, with the bus at 1.0 pu before the start and no other load.

What it calculates

The bus voltage at the moment an induction motor is started across the line, and how far it falls below nominal. At standstill the motor draws locked-rotor current, so it looks like a small fixed impedance connected to the bus. That impedance, the transformer and the utility source form a voltage divider. All three are put on the transformer kVA as the base:

Z_s = \dfrac{S_\text{base}}{S_\text{sc}}

The motor's share of the divider is the bus voltage, and the dip is what is left over:

V_\text{bus} = \dfrac{Z_m}{Z_s + Z_t + Z_m}

Locked-rotor kVA from current

Enter the motor's locked-rotor kVA directly, or open the helper and enter the locked-rotor current and the line-to-line voltage it was measured or quoted at:

S_\text{LR} = \dfrac{\sqrt{3}\,V_\text{LL}\,I_\text{LR}}{1000}

Worked example

A 1,500 kVA transformer with 5.75% impedance feeds a motor whose locked-rotor draw is 1,800 kVA. With an infinite source (no utility impedance):

\begin{aligned} Z_m &= \dfrac{1{,}500}{1{,}800} = 0.8333 \\[4pt] V_\text{bus} &= \dfrac{0.8333}{0.0575 + 0.8333} \\[4pt] &= \dfrac{0.8333}{0.8908} = 0.9355 \\[4pt] \text{dip} &= (1 - 0.9355) \times 100\% = 6.45\% \end{aligned}

Now add a utility source with 250 MVA of available short-circuit power. The calculator opens on this case. The source adds 0.006 pu in series, so the bus sags further:

\begin{aligned} Z_s &= \dfrac{1{,}500}{250 \times 1{,}000} = 0.006 \\[4pt] V_\text{bus} &= \dfrac{0.8333}{0.006 + 0.0575 + 0.8333} \\[4pt] &= \dfrac{0.8333}{0.8968} = 0.9292 \\[4pt] \text{dip} &= (1 - 0.9292) \times 100\% = 7.08\% \end{aligned}

Compared with the shortcut

A common shortcut multiplies the transformer %Z by the ratio of locked-rotor kVA to transformer kVA. For the infinite-source case above, that gives 5.75% × 1,800 / 1,500 = 6.9%, while the divider gives 6.45%. The shortcut leaves the motor's own impedance out of the denominator, as if the motor still drew its full locked-rotor current at nominal voltage. In fact the motor's current falls as the bus voltage falls, so the shortcut overstates the dip. It also has no place for the utility source impedance.

Assumptions

The impedances are added as magnitudes, as if they all had the same angle. In practice the source and transformer are mostly reactive and the motor at locked rotor has a lower power factor, so the true dip is somewhat smaller than this adds up to.
There is no other load on the bus before the start, and the bus is at 1.0 pu before it.
The motor is a constant impedance at its locked-rotor value for the whole calculation. It does not model the motor speeding up, so it gives the dip at the first instant of the start, which is the worst case for an across-the-line start.
The transformer is the only series impedance between the source and the motor, and its taps are at nominal. Cable impedance between the transformer and the motor is not included. To add it, see the voltage drop calculator.
Locked-rotor kVA is taken from the manufacturer's data. This tool does not estimate it from a code letter.

Questions

How large a voltage dip is acceptable when a motor starts?

It depends on what else is on the bus. A common design target is a dip of no more than about 10 to 15% at the motor and less at buses that feed lighting or sensitive controls. Check the requirements of the equipment and the utility you are connected to.

Where do I find the locked-rotor kVA?

From the motor's nameplate code letter and horsepower, or from the manufacturer's locked-rotor current. This page's helper converts locked-rotor amps at the motor's rated voltage into kVA.

Why is this result slightly different from the %Z × LR kVA / kVA shortcut?

The shortcut assumes the transformer impedance is small next to the motor's. The impedance divider here keeps the motor's own impedance in the denominator, so it gives a slightly smaller dip, and it can add the utility source impedance.

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PhasorPrep is an independent study resource. It is not affiliated with, endorsed by, or sponsored by NCEES. Questions are original and written for practice; they are not actual exam questions. Calculators are for learning, not for engineering design.

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