## Motor Start Voltage Drop re. Transformer Loading

## Motor Start Voltage Drop re. Transformer Loading

(OP)

Given: A 40 HP elevator motor fed from a 300 kVA transformer. Assume across-the-line starter and neglect line impedances, etc. The question has come up re. how does the motor starting voltage drop change from the unloaded transformer condition to the loaded transformer condition? Assume no load in the first case and, say, a 200 kVA steady state load for the second case.

Thanks, Frank

Thanks, Frank

## RE: Motor Start Voltage Drop re. Transformer Loading

The 40HP motor at start would be approximately 180-190kVA inrush, at least.

You could definitely assume a larger voltage drop if there was 200kVA of added load on the secondary of the transformer when you attempted to start the motor(unless that added load was a capacitor, in which case it should be described as 200kVAR).

## RE: Motor Start Voltage Drop re. Transformer Loading

where:

I = starting current

Z = tx impedance

## RE: Motor Start Voltage Drop re. Transformer Loading

Frank Starr

## RE: Motor Start Voltage Drop re. Transformer Loading

But it isn't quite that simple because it depends on the nature of the other loads. Motor loads tend to look like constant kVA loads, so a voltage dip during starting will cause current to the other motors to increase. Resistive loads will see a current decrease.

Motor starting calculations require iteration for an exact solution, like any power flow problem.

## RE: Motor Start Voltage Drop re. Transformer Loading

In the case at hand, the other loads, assumed at 200 kVA, are primarily lighting and miscellaneous loads with some small motors. (Its a 5-story condominium). Assume the other loads are primarily resistive. So it would seem that the steady state loading has little effect other than the initial voltage condition.

Frank

## RE: Motor Start Voltage Drop re. Transformer Loading

where:

dV = voltage drop due to motor starting

dI = motor starting current

Z = tx impedance

## RE: Motor Start Voltage Drop re. Transformer Loading

The larger the load, the lower the voltage, and this includes any added load during your motor start.

As I stated originally, there will definitely be a larger voltage drop if there was 200kVA of added load on the secondary of the transformer when you attempted to start the motor.

Following your assumption, how much added load would you believe would be needed before the transformer is overloaded?

## RE: Motor Start Voltage Drop re. Transformer Loading

To find out the difference in voltage drop at no-load v/s full loaded transformer can be found by performing transit voltage drop calculations. To perform calculations it requires, the fault level at this transformer & impedance of transformer.

## RE: Motor Start Voltage Drop re. Transformer Loading

## RE: Motor Start Voltage Drop re. Transformer Loading

## RE: Motor Start Voltage Drop re. Transformer Loading

1) At the motor

2) At the main service panel

3) At the primary terminals of the utility transformer

Number 3 is of particular concern to a utility as this flicker affects other nearly customers.

Number 2 is a concern because this flicker (always more than the flicker at point 3) affects all other loads in the plant.

Number 1 is a concern because this flicker (always greater than at point 2) affects the voltage available to start the motor. Low voltages increase the currents and start times on a motor, which generates more heat.

For proper calcualtrions, you need to know the motor start parameters, and the impedances of all of the other components (wire(s) from motor to main panel, impedance of utility transformer, impedance of utility grid on primary of xmfr).

In practice, for flicker calculations, the preloaded condition of the system is often neglected as long as the steady-state load with the motor running does not exceed the rating of the plant system or transformer (in extreme cases, the core of the transformer can begin to show saturation effects).

I agree that the overall voltage will be lower in the loaded case (which will impact the actual voltage available to start the motor), but the magnitude of the transient voltage drop (i.e. the 'flicker') will be about the same as in the unloaded case.

## RE: Motor Start Voltage Drop re. Transformer Loading

For the rest of you guys who said more information is needed to answer the question, you need to step back and see the big picture. This is a before and after calculation where only one variable is changed. The question does not ask for actual values, it just asks how changing one variable (transformer load current) affects another variable (dV/dt for starting a given motor). Sure, you can run a bunch of calcs and prove that there is a 2% difference between the two conditions because of some secondary effect or other, but nobody cares about that in the real world. Don't make the problem harder than it is.

And DanDel, a transformer is manufactured to provide the rated voltage at the rated kVA. I suggest you check with your local Sauare D engineering specialist. I did.

The application that led to this question is a mid-size condo that is experiencing (alledgedly) lamp flicker whenever the elevator starts. The elevator, which is a hydraulic unit, is fed from the main service transformer. Because I was brought in as an expert witness after everybody (including the serving utility) got lawyered up, I haven't been allowed access onto the property to physically observe and measure the actual installation. So I'm left to deal with theoretical 'what-ifs' at this point, hence the question.

Ironically, there is a condo a block away, with a smaller service transformer and a larger elevator motor that doesn't have lamp flicker. Ironic, because I am the design engineer of record on that condo!

Frank Starr, P.E.

Seattle

## RE: Motor Start Voltage Drop re. Transformer Loading

## RE: Motor Start Voltage Drop re. Transformer Loading

% Regulation = 100 x (Vnl - Vfl)/Vfl,

where Vnl = no-load voltage and Vfl = full-load voltage.

Transformers are manufactured with no-load turns ratios, meaning that the 480V to 208/120V tap on a 480delta/208wye transformer is almost exactly(within 0.5%) 480/120 = 4.000.

Hopefully, you are familiar with the Open-Circuit and Short-Circuit transformer tests. These two tests are used to find the no-load and full-load losses. If there were no losses in the transformer, any application of voltage to the primary winding during the Short-circuit test would theoretically allow infinite current on the shorted secondary. The Short-circuit test is also used to measure the %Z of the transformer, and the losses at full load are usually in the 2-5% range.

It's obvious that the answer to your original question involves a larger voltage drop for the loaded transformer than the unloaded transformer; I believe everyone in this thread agrees with that (basic electric theory). You should also agree that to get the actual values, you will need the actual impedances, loads, turns ratio, and voltage applied.