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#1

# Voltage Frequency Calculations

I am having an issue finding any documentation on calculations for 50hz systems. Here is my example:

If i have to design a system around 30kw heaters supplied with 480 volts from a 3phase service.

Purely resistive loads typically have a PF of 1. So 30,000\480 x 1.732 = 36 amps

What would the current draw be for 50hz 480v 3phase? What is the formula to solve for this?

I read something somewhere, which i cant for the life of me find, about a volt frequency ratio; any help would be appreciated.

09-15-2013 03:00 PM
Top #2
Frequency is not a factor in this calculation. P = U * I * SQRT(3)
09-15-2013 05:03 PM
Top #3
You may be thinking of the motor formulas for running a 50 hz motor at 60 hz or any other voltage than what the motor is wound for.

Are the heaters actually three phase or are they single phase heaters and just the service is three phase?
09-15-2013 07:36 PM
Top #4
This is my concern. In america we assume with our calculations 60hz. If a system is designed and spec'd at 60hz what happens when you install said system into a facility who's power supply is rated at 50hz, for example almost the rest of the world...

Does this affect only motors or both motors and purely resistive loads?

So I found this equation X(l) = 2pi x F x L where f is the frequency and l is henries. 2 is the constant for a single phase system i confirmed this using a known value for X(l)

So now lets move to a 3phase system. X(l) = 1.732 x pi x F x L

Lets use my question above as the example. 30kw \ 480 x 1.732 = 36A

480 x 1.732 \ 36 = 23

So lets plug in 23 for X(l) so 23 = 1.732 pi x F x L

23 = 5.43848 x 60 x L

23 = 326.3088 x L

L = 23 \ 326.3088

L = .070485381

Now we have the all of the variables solved for lets check if this makes sense.

Lets cut the frequency in half. X(l) = 5.43848 x 30 x .070485381

X(l) = 11.5 so lets this in.

480 x 1.732 \ 11.5 = 72A

We have doubled the current by reducing the frequency by half, this makes sense to me.

So now lets plug in 50hz

X(l) = 5.43848 x 50 x .070485381

X(l) = 19.17

480 x 1.732 \ 19.17 = 43 amps.

This again makes sense to me

However my concern is if this affects a system designed for 60hz put in service at 50hz, again it will be for a purely resistive load (heaters)

It is to my understanding it will work but it will draw more current, and this is the math ive used to solve for it.
09-15-2013 10:26 PM
Top #5
Inductuve resistance has nothing to do with simple resistance, you are going in completely the wrong direction. As the frequency is important for inductive and capacitive resistance, it is totally of no influence for resistive loads,
09-16-2013 12:38 AM
Top #6
As a collegue also pointed out, acording to IEC-60038, for 50Hz systems 480V is not a standard value, 50Hz systems only use 230/400V, 400/690V or 1000V as standard values.

http://cmapspublic2.ihmc.us/rid=1L6G6428L-XZ32Y6-1H9G/iec60038.pdf
09-16-2013 03:22 AM
Top #7
Ok so a change in frequency has no influence on a purely resistive load. that was my main concern. Thanks for the input!

I know 480 isnt a standard, I typically see 380, 575, and 600V for our foreign customers. I was just using a random voltage for the calculation.

Thanks again to all who responded.
09-16-2013 06:02 AM
Top #8
As Mr. David said , purely resistive load has no concern with frequency. Out of your question and just to support ideas: