Can someone provide the formula for calculating the output, in amps, of a three phase transformer? The primary is 240 delta and the secondary is 480 delta. The nameplate indicates the transformer is rated at 15kW. Any help will be greatly appreciated.

 

Isn't it
W/V=I
No matter if its Primary or secondary
32 amps for Primary
31 amps for secondary

 

I = .577*KVA/V

 

OK, here's the deal: if you tell me for what power factor are those 15kW, I'll tell you for what current is the transformer rated. Normally the transformers' power is rated in kVA, but if you say yours gives it in kW, then it should also say the power factor for the load.

The theory is like this:
Power = (square root of 3) x the voltage (480V) x the current x the power factor of the load if you want the result to be in kW

 

I=KVA/E*1.73

The current is equal to the KVA (15 KW) of the transformer divided by the line voltage (480 VAC) times the square root of 3 (1.73). This will give you the current for each phase.

[email protected]

 

Transformers are rated in KVA but the formula you are looking for is:

AMPS = (KVAx1000)/VOLTSx1.732

That is 36 amps per phase @ 240 volts and 18 amps per phase @480 volts. This is for a balanced 3 phase load. If powering a single phase load you will be limited to 5 KVA. A three phase transformer is 3 single phase transformers, in your case, connected in a DELTA connection.

I hope this helps

You can find electrical formulas at:
http://www.electrician.com

 

First of all ratings of a transformer are normally given in kVA. I'm assuming it to be as kVA.

Formula is

Power (kVA) = Sqrt 3 * V (line) * I (line)

For you:

15 kVA = 1.732 * 0.240 * I (line)
I (line - pr) = 36 Amps

15 kVA = 1.732 * 0.480 * I (line)
I (line - sec) = 18 Amps

Note for delta system phase current = line current / sqrt 3

Diwaker Basnet

 

((KW*1000)/(VOLTS*1.732))*EFFICIENCY (USUALY .95 AT THE SECONDARY)
IE: PRIMARY = ((15*1000)/(240*1.732))=36 AMPS

SECONDARY = (((15*1000)/(480*1.732))*.95)=17 AMPS

 

VA = Amps * Volts * sqrt(3), so Amps = VA / Volts / sqrt (3). In your case, A = 15,000 / 480 / 1.732 = 18.0 amps. Are you sure the output is 480 and the input is 240? Transformers are more commonly used for reduction of voltage than increase. A transformer designed for reduction can be wired backwards and used in a "boost" mode, but you should derate the KVA capacity about 10-15%, which in your case would give an output of about 16 amps at 480V.

 

No, it is not.

It's P/(1.73*U)=I (simplifying things of course).
So primary is ca. 36A and secondary is ca. 18A.

Regards,

Jacek Dobrowolski, Ms. Sc. E. E.
Software Engineer

 

how to solve the eqvation if apmere is 600 and volts is 430 and cable distance is 200 ft now we find the thikness of the cable in mm what is the formula to solve this eqvation

and what is the formula to find the three phase motor efficiency of the motor

 

This is the last post I got via the A-list for electrical formulas relateed to this thread, look it up in the respective electircal code book - according to NEC the wire size for 600 amps at 480vac can be solved by running two 500kcmil cables for each phase (500kcmil copper is rated 380 amps), plus for the distance your going that would be the best solution to run two conductors for each phase - you may have to consider 750kcmil cable as well, but check with your electrical code which will provide you with the cable size in mm or awg or kcmil.

And if you working with this type of voltage and current - you had better become very familair with the electrical code of what ever country your in, otherwise hire a professional electrican

Matt Hyatt
technical consultants
[email protected]

 

The correct cable size should be determined based on expected, or allowable, voltage drop for a given load. Assuming the load power-factor is known, then, the correct formula for voltage-drop of a specific wire-size is:

Delta U = sqrt(3) x I x L x [R cos(phi) + X sin(phi)], where:

U = voltage drop in Volts.
I = current in Amperes.
R = resistance of wire selected in ohm/km.
X = reactance of wire selected in ohms/km.
L = length of cable in meters.
phi = load power-factor angle in degrees.

Useful wire-size conversion factors are:

mmq = 0.5067 x kcmil.
kcmil = 1.9735 x mmq.

Regards,
Phil Corso, PE
Boca Raton, FL
[[email protected]] ([email protected]) {[email protected]}

 

Sir,
I have some clarification on your answer

Length of the cable is in metres,But the R and X value of the cable is per ohms/ Km. I think the cable length is to be converted into Km.
To select the cables percentage voltage drop to be calculated for steady state and transient state.
For steady state the % voltage drop should be less than 3%. For transient state the %voltage drop should be less than 10%.
cos phi=0.8 for steady state ,0.3 for transient state.

We are calculating as per this formula only.If there is any other answer please let me know.

P.S.Pandi

 

Dear

Please give me the formula if I know the current (load) and percent voltage drop acceptedand and the distance between the supply and load. I needonly calculate the size of cable

 

CM = (2 x K x L x I) / VD
then refer to Table 8 - Conductor Properties of the NEC to determine your wire size. in the 2002 NEC, it is on page 70-625.

 

It is unlikely that a transformer is rated in KW. They are usually rated in KVA However... 15KW 240 volt 45amps 480 volt 23amps

 

Take a look at the following URL:
http://www.tenntech.com/formula_index.htm
Make it a Favourite!

This covers AC Transformers for both single and 3Phase. pf values etc.

 

<p>For Anonymous:

<p>The following formula will yield the minimum wire size to satisfy a voltage-drop requirement. However, it will not provide the size based on construction, installation, and operating conditions:
<pre>
S = Kp x Kc x L x A / Ed, where,

S = cable area (units shown below).
L = cable circuit length (units shown below).
A = line current, Amperes.
Ed = acceptable voltage drop, Volts.
Kp = circuit constant: 2 for 1-ph; sqrt(3) for 3-ph.
Kc = material conductivity constant.

If conductor is copper, then,

Country Kc L Size
USA 96E-3 feet kcmil
Metric 17E-3 meters mmq
UK 26E-3 yards sq in
</pre>
<p>If additional detail is required, such as determining line current given apparent power (kVA), input power (kWI), output power (kWo), or horsepower (Hp), let me know.

<p>Regards,<br>
Phil Corso, PE<br>
Boca Raton, FL<br>
[[email protected]] ([email protected]) {[email protected]}

 

check this out
http://www.elec-toolbox.com/Formulas/Transformer/xfmrform.htm