75 kVA transformer load with no load

Status
Not open for further replies.
M

MCBC

Member
Hi all,

I have a Square D 75 kVA Cat # C75T79H transformer in my facility.

Specs are as follows;

HV: 600, HVA 75, WT 500
Class AA
LV 208Y/120, LVA 208

With all of the panel breakers shut off I get no load on the LV side.
But I get load on the HV side.

Readings with no load from left to right are as follows;

X3 = 0.0A
H2 = 2.6A
X0 = 0.0A
X2 = 0.0A
H1 = 2.4A
H3 = 4.4A
X1 = 0.0A

Is this normal for the 600 V side to draw current when there is no load on the 120 V side?

Thanks in advance for the reply.

MCBC
 
What you are reading is magnitizing current. As long as the transformer primary is connected to a source it will draw this current. It is the current drawn by the transformer to establish the magnetic feild. In your case it is about 1% of the transformer full load current.
 
Last edited:
Thanks for the reply Guy's.

From what I understand here, the transformer will always increase my electricity bill.

How can I balance the primary?

MCBC
 
MCBC said:
Thanks for the reply Guy's.

From what I understand here, the transformer will always increase my electricity bill.

How can I balance the primary?

MCBC
Click to expand...

Yes to the best of my knowledge a X-fmr that has no load on it will use a little amount of power but I'm not sure how much it would use or how to calculate it.
 
ha ha the jokes on me.

I already balance to load on the secondary.

Thanks for all the info guy's.

MCBC
 
081117-1930 EST

MCBC:

Call the manufacturer and ask for a curve of the noload power dissipation vs voltage for your particular model.

Otherwise you need a wattmeter with clamp-on current sensors. I suspect you are not an electrician and lack experience with highpower medium voltage and thus the best idea is to get the information from the manufacturer.

Your power losses are not worse than about (2.6+2.4+4.4)*600/1.732 = 3.25 KW. The not worse is because this is the input VA and much of the current is inductive.

If the transformer is inside it helps heat the building.

.
 
Gar,

You are right in all cases, I worked in Automation, mainly LV systems.
About the cals you did, does this mean I'm using 3500 W/h no mater what?
Yes it does help heat the building.

Thanks for the info.

MCBC
 
081118-0829 EST

MCBC:

It means you are using someting less than 3.25 KW all the time a power source is connected to the transformer primary.

Virtually no I^2*R losses because the current is low. All the losses are eddy current and hysteresis.

I will guess at maybe 1 to 2 KW of power loss.

.
 
The KVA that the transformer is using is what the clamp reading indicate.

But the power factor of transformer magnetizing current is quite low, so the KW that the transformer is using is significantly less than the KVA that the transformer is using.

-Jon
 
081118-0911 EST

iwire:

The input current to an inductor can be broken into two components -- the inphase (resistive component) and the quadrature (90 deg out of phase reactive component).

The resistive component is the power loss component.

There is a good discussion on core loss in "Alternating-Current Machinery", by Bailey and Gault, McGraw-Hill, 1951. In chapter 1, Transformers, see pages 16 thru 22.

An example 10 KVA transformer is discussed. It has a measured core loss of 61 W or 0.6%. I could not find find what was the noload VA, but it will be much larger because of the magnetizing current (the quadrature current).

Depending upon the application of a transformer the design may favor a much higher core loss in order to reduce copper loss at high loads. If little iron is used, then noload core losses are reduced and full load I^2*R losses are high. If lots of iron is used, then core losses are high and full load resistive losses are low and full load efficiency will be good.

.
 
winnie said:
The KVA that the transformer is using is what the clamp reading indicate.

But the power factor of transformer magnetizing current is quite low, so the KW that the transformer is using is significantly less than the KVA that the transformer is using.

-Jon
Click to expand...

I am still lost.

If clamp meter reads it so will the KWH meter Yes/ No?
 
gar said:
081118-0911 EST

iwire:

The input current to an inductor can be broken into two components -- the inphase (resistive component) and the quadrature (90 deg out of phase reactive component).

The resistive component is the power loss component.

There is a good discussion on core loss in "Alternating-Current Machinery", by Bailey and Gault, McGraw-Hill, 1951. In chapter 1, Transformers, see pages 16 thru 22.

An example 10 KVA transformer is discussed. It has a measured core loss of 61 W or 0.6%. I could not find find what was the noload VA, but it will be much larger because of the magnetizing current (the quadrature current).

Depending upon the application of a transformer the design may favor a much higher core loss in order to reduce copper loss at high loads. If little iron is used, then noload core losses are reduced and full load I^2*R losses are high. If lots of iron is used, then core losses are high and full load resistive losses are low and full load efficiency will be good.

.
Click to expand...

Gar, that is well beyond me.

In simple terms, if the clamp meter reads the current won't the KWH meter read it as well?
 
081118-1020 EST

iwire:

If I could get a perfect inductor, one with no losses, then the current measurement could be some value like 10 A, and the power measurement would be 0 because there are no losses, and there would be no heat dissipated in the inductor.

With real world capacitors we can get much closer to an ideal reactive component than with real world inductors.

Get a wattmeter, an ammeter, voltmeter, and a suitable transformer for the range of the wattmeter. Run the experiment.

Here are my results from a small, about 170 VA, transformer. 120 V 60 Hz input, no secondary load, I = 0.213 A, calculated input volt-amperes = 120*0.213 = 25.6 VA, and the wattmeter reading was 3 W. Thus, PF = 0.12 .

.
 
From a chart in Cutler-Hammer's Engineering Guide, a 75 kVA dry type transformer has no load losses of 400W. These are standard trasnformers, not necessarily energy efficient.

You are only paying for the 400 watts loss 24/7. A more expensive transformer might save about 150 watts when unloaded.

Full load losses varied from 1950 watts to 3000 watts, depending on the transformer temperature rating and efficiency.
 
gar said:
081118-1020 EST

iwire:

If I could get a perfect inductor, one with no losses, then the current measurement could be some value like 10 A, and the power measurement would be 0 because there are no losses, and there would be no heat dissipated in the inductor.

With real world capacitors we can get much closer to an ideal reactive component than with real world inductors.

Get a wattmeter, an ammeter, voltmeter, and a suitable transformer for the range of the wattmeter. Run the experiment.

Here are my results from a small, about 170 VA, transformer. 120 V 60 Hz input, no secondary load, I = 0.213 A, calculated input volt-amperes = 120*0.213 = 25.6 VA, and the wattmeter reading was 3 W. Thus, PF = 0.12 .

.
Click to expand...

No help at all, trying to get a simple answer from you is like dealing with the IRS. :grin:

I asked a what I though to be a simple question.

If he can read the primary current with his amp clamp won't the KWH meter read that as well?
 
rcwilson said:
From a chart in Cutler-Hammer's Engineering Guide, a 75 kVA dry type transformer has no load losses of 400W. These are standard trasnformers, not necessarily energy efficient.

You are only paying for the 400 watts loss 24/7. A more expensive transformer might save about 150 watts when unloaded.

Full load losses varied from 1950 watts to 3000 watts, depending on the transformer temperature rating and efficiency.
Click to expand...

Bob, maybe you can explain it to me.

The OP is reading H1 = 2.4A, H2 = 2.6A, H3 = 4.4A at 480 volts, that sounds to me to be a lot more power then 400 watts
 
Status
Not open for further replies.
Top