How to Calculate the Voltage Drop

Voltage drop Calculation:-

1. Open internet - any browser

2. Search - Google-voltage drop calculation

3. Read the websites in the results.

4. Follow the procedure mentioned in the sites.

5. May be You can do it without any help in future after trying this way.

Voltage Drop= Length (metres) x Current (Amps) x VC (millivolt per ampere metre)/1000

Are you asking about voltage drop due to the length and size of the feeders powering this beast of a motor or the impact on the transformer from the starting of the motor?

North, I agree, I think he's talking about the drop due to the transformer, a question we can't answer without some more data about the xfrmr,,,, also, why is the PF of the motor specified? It's synchronous, so one vary the PF considerably.

iam asking about the impact on the transformer.

THERE ARE SOFTWARES AVAILABLE IN THE MARKET NAMED E-TAP,CYME,EDSA ETC,USE ANY OF THIS ALSO YOU HAVE TO GIVE THE CABLE SIZE AND LENGTH USED AND ALSO THE STARTING DATA EG STARTING CURRENT AND PF .

A JUDICIOUSLY PROVIDED INPUTS WILL PROVIDE YOU THE CORRECT ANSWER.

OTHERRWISE THE FORMULA OF VOLTAGE DROP CALCULATION IS

RT3XI(COSphi-jSINphi)(RCOSphi+jSINphi)

Oi! Keep the NOISE down, Mate. Some of us are trying to get some kip! <Wheeze, splutter>

How do you do

Please clarify voltage level.

voltage level is 33/11 kv

Well which is it? 33kv or 11kv, is the motor wired star or delta, how is the transformer configured? What is the primary side voltage, seondary voltage, size and length of feeder wire to the motor?

132/11 kv

see #9

Give operting voltage of motor,

Size and length of cable

Swastik Electrical engineering co.

Nirbhaiy Modha

As I understand it, you want the voltage drop at the transformer (i.e., at the transformer, ignoring voltage drop in the cables). (Although I studied electrical engineering, it was focused on electronics and computers, so the following may be wrong, but I did a quick google (on ["percent impedance" "voltage drop"]) and found the following:

[PDF]
Document Reference (AN060019H2) Understanding Percent Impedance ...
File Format: PDF/Adobe Acrobat - Quick View
Percent Impedance or Percent IZ (%IZ) is the voltage drop due to impedance, at rated current, expressed in a percent of the rated voltage. Discussion: ...

www.baldormotion.com/.../H2%20Application%20Note%20Percent%20Impedance.pdf - Similar

This sounds like a homework problem?

Maybe not.

It seems you first have to know the rated current of the 50 MVA transformer, which, if this is a real life problem, should be on the transformer. If not, you can calculate the rated voltage of the motor by using the rated current and power of the motor:

P = IE, so E = P / I, equals 12.5 MW / 720 amps, equals 17,361 volts--oops, that size motor is almost certainly 3 phase, so also divide that by 1.732--equals ~10,000 volts, so I'm going to assume the motor is rated at 10 kv (is that a standard motor voltage rating?).

Based on that, the rated current of the transformer would be:

50 MVA--oh, wait , it's even easier--this is four times the rating of the motor, so the full current rating on the transformer is 4 x 720 amps = 2880 amps. (You could also calculate this using P = IE (= P / (1.732 x E) (again).)

The 12.64% impedance means that at full load, the (voltage) drop at the transformer is 12.64% x 10,000 volts = 1264 volts. (Hmm, that seems high (to me), but that is irrelevant to this question--the question is the voltage drop at the transformer when starting the (i.e., one) motor.)

To really answer the question, you need to decide whether one motor is running when the other is started or not--hmm, I'm remembering though, that if one motor is running, some of the starting current for the 2nd motor is contributed by the running motor, resulting in less than the full 1720 amps being drawn from the transformer--

I can't remember offhand--was there some rule of thumb to guesstimate what came from the other motor instead of the transformer--I think there is, but I won't remember it (I thin it comes into particular use when you are calculating short circuit currents in the system--you have to account for the contribution from the other motor (and other inductive loads) when you check the interrupt rating for circuit breakers or fuses in the system...

Anyway, let me make it simpler (for myself): without the other motor running, the load on the transformer when starting one motor will be 1720 amps, which is 1720 / 2880 equals .597 or 59.7% of the full load on the transformer, thus the voltage drop will be 59.7% of the transformer's voltage drop at full load, so .597 x 1264 volts equals 755 volts. (This is starting one motor while the other is not running.)

I suppose for a worst case calculation of the voltage drop when starting one motor while the other is running you could ignore the starting current coming from the running motor, and assume that other motor is running at full load (672 amps)--then the load on the transformer in that condition is 1720 + 672 = 2392 amps.

Then, the voltage drop in that condition would be 2392 / 2880 (=.83) x 1264 volts = 1050 volts. (This is starting one motor when the other is running at full load, and ignoring the starting current contribution from the running motor.)

Hmm, I wonder if I'm anywhere close to right ;-)

Do you want to calculate voltage drop on the primary or secondary side of the transformer?

As 50 MW transformer supplies 12 MW I would not think about trifles like that at all. Also synchronous motor need to be rev. up to synchronous speed during starting up. What is the method to start it up? In general the voltage drop would be within voltage tolerance for the system.

It seems that your synchronous motor is configured as asynchronous for starting up and then it is re-switched to synchronous configuration. Is it true?

If it starts as asynchronous you might expect voltage drop 50% as this is for max power transfer. When it reached 85% of the speed its re-switched for synchronous operation and then its voltage drop would be negligible, but if you run 20 or more motors like that, you should be sure that motors do not start up at the same time so you have some sequence of starting.

You did not mention the method of starting or starting current.Why?.