Motor starting & voltage sag
The key idea
Switch an induction motor straight onto the supply. It then pulls about six times its rated current until it reaches full speed. That starting current flows through the grid impedance like any other current. The bus voltage therefore falls, and every other load on the bus sees the fall.
The idea
A motor at standstill does not yet behave as a motor. It behaves as a short-circuited transformer. The stator is energized, the rotor does not turn, and only the winding impedance limits the current. The machine draws its starting current, which is typically six times its rated current. The power factor is approximately 0.2, so almost all of the current is reactive.
The network must supply that current. The network's Thévenin impedance lies between the motor terminals and the rest of the power system. During the acceleration the motor behaves as a fixed impedance. The two impedances divide the supply voltage. A stiff network holds the bus voltage near nominal. A weak network lets the bus voltage fall much further.
That fall is the voltage sag, and it affects more than the motor. All the loads on a bus share one voltage. The lighting, the contactors, the drives and the neighbor's process all see the lower voltage. They see it for the full duration of the start, which is seconds, not cycles. This is the voltage drop effect, but the load connects all at once instead of slowly.
The obvious answer is to start the motor at less than full voltage. Reduced-voltage starting does exactly that. Star–delta connects the windings in star for the start. Each winding then receives the phase voltage instead of the line voltage, and the machine draws one third of the direct-on-line current. A soft starter does the same electronically. It holds the current at a set limit and increases the voltage slowly.
But reduced voltage has a cost, and that cost is the whole engineering decision. Torque depends on the square of the voltage at the motor. If you cut the starting current to one third, you also cut the starting torque to one third. A fan gives very little resistance at standstill, so it still accelerates. A loaded conveyor or a crusher does not accelerate. It stays at zero speed and draws starting current until the protection trips.
Try it
Start with the defaults. A 2 MW motor on a 100 MVA bus, direct on line, gives a 12% dip that recovers in a few seconds. Then reduce the fault level toward 20 MVA. The trace no longer recovers. Try star–delta on that weak bus and see whether it helps.
during the start: 0.876 pu
- Starting current
- 6× rated · 14.1 MVA
- Bus voltage during start
- 0.876 pu · dip 12.4% · below 0.90 pu
- Torque margin
- torque 1.23 pu vs load 0.30 pu
The model treats the motor as a fixed impedance while it accelerates. The bus voltage is therefore a pure-reactance divider: v = S_fault / (S_fault + k · S_motor). Here S_motor = MW / 0.85, and k is the starting multiple: 6 direct on line, 2 in star–delta and 3 on the soft starter. Torque follows v² times the method factor, against a 0.30 pu fan load. The widget draws the run-up as a fixed 6 s. A real study integrates the torque curves of the motor and the load to find the true duration.
Why it matters
- The dip is a connection condition, not a nuisance. Network operators limit how far a single start may pull a shared bus. The limit is commonly a few percent on the public supply. If you exceed it, you must fit a reduced-voltage starter, a dedicated transformer or a larger connection.
- The rest of the bus carries the effect of the start. Contactor coils drop out at approximately 0.7 pu, drives trip on undervoltage, and discharge lighting goes out. A sag that is deep enough to disconnect other loads turns one motor start into a plant-wide stoppage.
- Reduced voltage gives a shallower sag and takes away torque. Torque follows the square of the voltage, so the trade is never one for one in your favor. Check the torque that the driven machine needs at zero speed before you choose the method.
- A starting method cannot repair a weak grid. If the bus voltage collapses, the motor will not start direct on line. Star–delta usually does not help either. The sag is shallower, but the available torque falls as fast. The real solutions are a stiffer supply, a smaller motor, or a starter that adds torque instead of removing it.
The math, if you want itOptional — the page reads completely without it
The widget models the motor during the start as a fixed impedance. That impedance draws k times the rated apparent power of the motor. The widget models the network as a pure reactance that comes from its fault level. The bus voltage is then a divider:
bus voltage during the start
v = SfaultSfault + k · Smotor
Here Smotor = Pmotor / 0.85, and the 0.85 combines power factor and efficiency into one factor. The multiple k is 6 direct on line and 2 in star–delta. It is 3 for a soft starter that limits the current to half the direct-on-line value.
The motor then develops this accelerating torque, in per unit of the rated torque:
torque at reduced voltage
T = 1.6 · v² · m
Here 1.6 pu is the locked-rotor torque at full voltage. The factor m is the method factor. It is 1 direct on line, 1/3 in star–delta, and (3/6)² = 0.25 for the soft starter. The torque of a soft starter falls with the square of its current limit. The load needs 0.30 pu at all speeds, which is a fan-type demand. The motor stalls when T falls to that value.
The star–delta factors come from the same 3× relationship that the wye and delta page derives. In star, each winding receives VLL/√3. The winding therefore draws 1/√3 of the current at 1/√3 of the voltage. The result is one third of the power and one third of the line current. Torque follows the square of the voltage, so the result is also one third of the torque.
This model makes two approximations. First, it treats the motor as a constant impedance for the whole start, but the impedance and the power factor change continuously with speed. Second, it draws the run-up as a fixed 6 seconds. The real duration comes from the integral of the difference between the motor torque and the load torque over the speed range. A real motor-starting study does that integration with both torque curves and the inertia of the machine. The study tells you whether a marginal start takes 8 seconds or never finishes.
See it in Phasor
In Phasor, a motor start is a power flow with the motor at its starting impedance. Add the machine and solve the network. The voltage results show the sag at every bus, not only at the bus of the motor. That model also includes the transformer between the motor and the source, and the other loads that already draw current. It shows you how far across the network the sag reaches.