Determining primary transfomer current with only partial secondary loading

[Pitt123]

Someone the other day asked me how the transformer primary current could be calculated if the transformer was only partially loaded on the secondary with a purely resistive load?

So for instance lets say for example we had a 100KVA 3phase 480Vpri - 208V sec transformer but the secondary load was only 50kW purely resistive. How would we calculate the current on the primary? I know we simply couldn't reflect the 50kW over to the primary and solve for current since we need to account for the reactive current in the transformer. (Lets ignore internal transformer losses for now).

How would we account for the transformer magnetising current for calculating the primary current on the transformer for the given secondary load?

 

[erickench]

The 100KVA full load rating could be at unity power factor thus making it kilowatts. Why do you think that you can't reflect it to the primary? At 50KW the transformer would be operating at 50% load. Thus:

(.5X100,000)/(480X1.73)=60.2 amps

 

[jim dungar]

Someone the other day asked me how the transformer primary current could be calculated if the transformer was only partially loaded on the secondary with a purely resistive load?

So for instance lets say for example we had a 100KVA 3phase 480Vpri - 208V sec transformer but the secondary load was only 50kW purely resistive. How would we calculate the current on the primary? I know we simply couldn't reflect the 50kW over to the primary and solve for current since we need to account for the reactive current in the transformer. (Lets ignore internal transformer losses for now).

How would we account for the transformer magnetising current for calculating the primary current on the transformer for the given secondary load?

For all intents (i.e. not taking a test) you can ignore the magnetizing current of the transformer as it is typically only a few per cent of the full load rating.
So for your example, I would say the primary current would be about 50kW or 60A

 

[Pitt123]

For all intents (i.e. not taking a test) you can ignore the magnetizing current of the transformer as it is typically only a few per cent of the full load rating.
So for your example, I would say the primary current would be about 50kW or 60A

What is the rule of thumb percentage for transformer magnetizing current?

If we were interested and calculating what this current was how would we go about this? Would we have to be given the magnetizing reactance from the transformer manufacturer? I dont think this could be calculated from the %z or X/R ratio?

If we wanted to account for additional current on primary for transformer losses would we simply divide the calculated current as you mentioned above by the efficiency of the transformer?

 

[jim dungar]

You are correct, you are not given data to calculate magnetizing current. Typically this value comes from testing of the actual transformer. Even a fairly poor transformer is roughly 95% efficient so this test is often not paid for.

 

[gar]

111123-2217 EST

Some experimental data on a small transformer.

See photographs P6-P8 on my website http://beta-a2.com/EE-photos.html

Some measurements on this same transformer with a Kill-A-Watt EZ.

[FONT=Courier New]

Secondary rating 175 VA

 RMS     RMS
Volts  Current  Power   Volt-Amp
  79.0  .02      1.4       2.2
  90.5  .03      1.8       3.6
 100.5  .06      2.3       6.5
 110.4  .10      3.0      11.6
 120.3  .16      3.8      19.6
 130.4  .24      5.0      30.9
[/FONT]

.

 

[gar]

111123-2245 EST

This new forum software is a mess. Very hard to use.

Add to previous post.

[FONT=Courier New]

Volts  Current  Power   Volt-Amp
  79.0  .02      1.4       2.2
  90.5  .03      1.8       3.6
 100.5  .06      2.3       6.5
 110.4  .10      3.0      11.6
 120.3  .16      3.8      19.6
 130.4  .24      5.0      30.9
[/FONT]

An assumption below is that primary and secondary losses are about equal. And actual magnetizing plus load current is not determined so I^2*R are not quite correct. The no load loss has little to do with I^2*R, it is only about 0.16^2*7.9 = 0.2 W of 3.8 W.

The primary resistance = 7.9 ohms. At 120 V full load current is 1.45 A plus a little. Primary I^2*R loss about 16.8 W. Double this and add 4 = about 37 W. 175 + 37 = 212. At full load about 175/212 = 0.83 efficiency.

50% load about 0.72 A plus a little. Total I^2*R = about 8.2, add 4, and result = 12 W. 87 + 12 = 99. Efficiency is about 87/99 = 0.88 .

Check my assumptions and math.

Really need to do actual loads and measurements.

.


.

 

[T.M.Haja Sahib]

What is the rule of thumb percentage for transformer magnetizing current?

For transformers of high efficiency,the no-load current may be taken as magnetizing current.
I hope I am not trolling.

 

[T.M.Haja Sahib]

For transformers of high efficiency,the no-load current may be taken as magnetizing current.
I hope I am not trolling.

Further to above,the no-load current i.e magnetizing current X may be taken as 4 to 5% of full load current of primary and let Y be the primary current for 50 kW load.Then the resultant primary current R =Sq.Root of [(square of X)+(square of Y)].

 

[Besoeker]

Someone the other day asked me how the transformer primary current could be calculated if the transformer was only partially loaded on the secondary with a purely resistive load?

So for instance lets say for example we had a 100KVA 3phase 480Vpri - 208V sec transformer but the secondary load was only 50kW purely resistive. How would we calculate the current on the primary? I know we simply couldn't reflect the 50kW over to the primary and solve for current since we need to account for the reactive current in the transformer. (Lets ignore internal transformer losses for now).

How would we account for the transformer magnetising current for calculating the primary current on the transformer for the given secondary load?

This is a picture I posted on another thread.

It's an example of what we make and the reverse application of what you are asking. It's a low voltage high current (7kA) controlled rectifier. It is current controlled in this case but directly measuring the current on the DC side is a bit difficult so we do it with ACCTs on the primary (the little red items on the right) of the transformer and calibrate accordingly. Yes, it isn't entirely accurate in that it shows up as non zero off load because of the magnetising current but, in the greater scheme of things, it isn't particularly significant.

So, I'm with others here - for most practical purposes, you can ignore it.

 

[iwire]

I hope I am not trolling.

Happy Thanksgiving. :thumbsup:

 

[ronaldrc]

Happy Thanksgiving. :thumbsup:

I don't think I'm trolling but I might be.:roll:



I would divid the secondary current by the primary to secondary turns ratio which in this case is 2.3 which equals around
60.4 amps on the primary.

Hope every one had a great Thanksgiving

 

[Hv&Lv]

T.M.Haja Sahib

I hope I am not trolling.

Happy Thanksgiving. :thumbsup:

:lol:

 

[T.M.Haja Sahib]

I don't think I'm trolling but I might be.:roll:



I would divid the secondary current by the primary to secondary turns ratio which in this case is 2.3 which equals around
60.4 amps on the primary.

Hope every one had a great Thanksgiving

Unfortunately,it does not take into account the no-load i.e magnetizing current,Let us see how it compares with post #9 .The full load current is 121A.Taking magnetizing current to be 4% of full load current ,it is then 5A.Substituting in the formula,we get current in primary as 60.61A .

 

[iwire]

Unfortunately,it does not take into account the no-load i.e magnetizing current,

Wow, thanks that was insightful.:lol:

 

[gar]

111127-1315 EST

T.M.Haja Sahib:

Your equation of post #9 is really not a valid equation because the magnetizing current is not a sine wave. At small values like you present and at 50 % load you might just as well just add the magnetizing current or just ignore it. Not a difference in the real world that would be of much useful consequence.

For a photo of magnetizing current see my photo P8 at http://beta-a2.com/EE-photos.html

To realize there is such a thing as magnetizing current is important, and more important is to understand what happens with magnetizing current at turn on of voltage to the transformer primary.

In the small transformer of my photos the RMS magnetizing current is about 11% of full load current. But magnetizing current is very sensitive to excitation voltage and frequency.

There is more to primary current than just magnetizing current and full load output current reflected to the primary. There is additional current to support the I^2*R losses of the transformer windings.

.

 

[T.M.Haja Sahib]

111127-1315 EST

T.M.Haja Sahib:

Your equation of post #9 is really not a valid equation because the magnetizing current is not a sine wave. At small values like you present and at 50 % load you might just as well just add the magnetizing current or just ignore it. Not a difference in the real world that would be of much useful consequence.

For a photo of magnetizing current see my photo P8 at http://beta-a2.com/EE-photos.html

To realize there is such a thing as magnetizing current is important, and more important is to understand what happens with magnetizing current at turn on of voltage to the transformer primary.

In the small transformer of my photos the RMS magnetizing current is about 11% of full load current. But magnetizing current is very sensitive to excitation voltage and frequency.

There is more to primary current than just magnetizing current and full load output current reflected to the primary. There is additional current to support the I^2*R losses of the transformer windings.

.

Thanks for your reply.The OP stated transformer to have no internal losses.The core then has no hysteresis loss and hence no distortion to transformer primary current i.e under conditions stated by the OP,it is sinusoidal.So my equation is valid under conditions stated by the OP.

 

[iwire]

The OP stated transformer to have no internal losses.

And where did the OP say that?

As a matter of fact it seems the OP is trying to calculate those losses.

 

[T.M.Haja Sahib]

And where did the OP say that?

As a matter of fact it seems the OP is trying to calculate those losses.

[h=2]

[/h]

Someone the other day asked me how the transformer primary current could be calculated if the transformer was only partially loaded on the secondary with a purely resistive load?

So for instance lets say for example we had a 100KVA 3phase 480Vpri - 208V sec transformer but the secondary load was only 50kW purely resistive. How would we calculate the current on the primary? I know we simply couldn't reflect the 50kW over to the primary and solve for current since we need to account for the reactive current in the transformer. (Lets ignore internal transformer losses for now).

How would we account for the transformer magnetising current for calculating the primary current on the transformer for the given secondary load?​

See above.​

 

[T.M.Haja Sahib]

As a matter of fact it seems the OP is trying to calculate those losses.

For a resistive load,it is possible (1)to determine exactly by calculation how much is the transformer magnetizing current is if efficiency,transformer primary and secondary currents and voltages are known and (2) the component of primary current corresponding to its internal losses can also be calculated exactly from the equation for efficiency.
For (1)
From efficiency equation,the resistive component of primary current 'X' can be calculated.If the total primary current is 'Y',the resultant of the difference between it and 'X' can be calculated using equation like in post #9.This is the magnetizing current.
For (2)
The difference between input and output gives the power lost in transformer.By using the equation for transformer input power,the primary current for this power lost in the transformer can also be calculated.