×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Are you an
Engineering professional?
Join Eng-Tips Forums!
• Talk With Other Members
• Be Notified Of Responses
• Keyword Search
Favorite Forums
• Automated Signatures
• Best Of All, It's Free!

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

#### Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

#### Jobs

(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

You would need much more information than what you are supplying, starting with transformer impedance and primary source contribution, and maybe throwing in the primary and secondary voltages.
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).

Ohm's Law, baby, V=IZ!!!

where:
I = starting current
Z = tx impedance

(OP)
I disagree, I don't think more info is needed.  The question is how does the steady state load on the transformer impact the instantaneous voltage drop due to the motor starting inrush current.  Since the motor start is a transient condition, I would expect the transient reactance of the transformer to be the primary impedance seen by the inrush current.  So, again, what effect does the transformer steady state current have on the voltage drop?

Frank Starr

The main impact is to cause the pre-start voltage to be lower than if there were no load on the transformer.  This will make the total voltage drop greater.

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.

(OP)
Right, that's the way I see it.  The transformer manufacturer rates his transformer to deliver the design voltage at the design kVA capacity.  So, assuming other things equal (never a safe assumption), the worst-case motor starting voltage dip would be measured starting from a rated-voltage condition.

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

1st derivative of Ohms Law, baby, dv = Zdi!!!

where:
dV = voltage drop due to motor starting
dI = motor starting current
Z = tx impedance

Fstarr, actually, your assumption is incorrect. A transformer is not manufactured to deliver the rated voltage at the rated kVA. The only time the rated voltage on the transformer secondary is present(with rated primary voltage applied) is when there is no load. Increasing load also increases the per unit voltage drop across the transformer impedance, reducing the secondary voltage.
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.

Hi Frank,
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.

Consideration should be given to flicker problem when starting this motor.  Refer to IEC 61000-3-5 TR2 Ed. 1.0 b (1994) :  Electromagnetic compatibility (EMC) - Part 3: Limits - Section 5: Limitation of voltage fluctuations and flicker in low-voltage power supply systems for equipment with rated current greater than 16 A

This is a case of throwing a motor load on a unloaded transformer compared to that on a loaded transformer which has already a secondary voltage drop due to regulation. Routine hand calculations can be done to check the voltage drop during starting in both cases. Parameters like transformer impedence are a must to do this.

Whena large motor is started, there are usually three places where the voltage drop/flicker should be calculated:

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.

(OP)
Right Tinfoil, I agree with your last obversation.  The voltage drop (flicker) is the same for the unloaded and loaded condition of the supply transformer.

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

Nice to hear back from you, Frank (even if it was 15 months after you posted the question).

Fstarr. I'll stand by my original statement, regardless of any specialist you name. Just think about it logically. A transformer has an impedance (Z). There is always a voltage drop across an impedance, directly proportional to the current (V=IxZ). If the transformer were designed for rated voltage at FLA, then it would have to have higher-than-rated voltage at no load. The voltage change between no load and full load is described by this equation:

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

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.

#### Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

#### Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Close Box

# Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

• Talk To Other Members
• Notification Of Responses To Questions
• Favorite Forums One Click Access
• Keyword Search Of All Posts, And More...

Register now while it's still free!