Transformer impedance and different loads 3

[Morteza2]

Hi all,

Can someone please help me understand nameplate transformer impedance values better, and how it can be used for varying loads.

I realize that the nameplate transformer impedance (let's call it Z%) is by definition the voltage (in pu) that causes full load current circulation in the secondary, if the secondary is short circuited.

I also understand that this Z% can then be taken as the transformer voltage drop in pu (let's call it Vdrop%), while operating at rated current, or in other words at rated load.
(My understanding is that this is because this is the voltage that is basically being dropped over the transformer impedances due to iron/copper/core losses while there is rated current circulating as set up in the experiment, as this was the voltage needed to overcome these impedances while there's no load in the circuit)

My question is, how can we know what the Z%, or in other words the Vdrop% is, if we are operating at a current different than the rated current?

Thank you in advance

 

[davidbeach]

You're confusing regulation and impedance. Voltage drop (regulation) will be much better (less drop) for normal load currents than you'd get by assuming Z% voltage drop. For normal load the resistive component of the transformer impedance provides voltage drop based on the component of the current that is in phase with the voltage and the reactive component of the transformer impedance produces voltage drop based on the component of the current in quadrature to the voltage. For normal load you have the a small resistance interacting with much of the current and a much larger reactance interacting with a smaller amount of the current.

 

[Marmite]

Both the ABB transformer handbook and the J&P transformer book are excellent references which give a full discussion of regulation and in particular how it is significantly influenced by the power factor of the load.
Regards
Marmite

 

[Morteza2]

Thank you for the comments;

I will reword my question as follows:

The only impedance information available on most transformer nameplates is a Z%.

Let's say the transformer nameplate has the following information:

kVA = 1000kVA
Vpri = 4160V
Vsec = 480V
f = 60Hz
3phase
Z% = 10%

If we know the value of our operating current, how can we calculate the voltage drop over the transformer?

i.e. what would be the voltage drop at:

a. 100% of FLA
b. 50% of FLA
c. 25% of FLA

Thank you

 

[davidbeach]

If you don't have a test report you'll have to use an assumed X/R for the transformer. Then you need to calculate your %R and %X. With that, the voltage drop across the transformer is the same calculation as the voltage drop across any other impedance.

 

[Morteza2]

Thank you davidbeach, so from nameplate alone, we won't know the full details needed to do the calculations.

I'm still confused with the '%' unit for the impedance, and it is my understanding that is based on rated current.

Assuming an X/R ratio of 1 (for simplicity, let's assume Z% = X%), how would you calculate voltage drop at different loads (i.e. 100% FLA, 50% FLA, etc.)

 

[Morteza2]

correction to my previous post: let's assume R% = X%*

 

[Marmite]

See attached from J&P Transformer book 12th Edition.
Regards
Marmite

 

[prc]

In Power Transformers %R is very small, compared to %X and hence for all practical purposes, voltage regulation(the drop in secondary terminal voltage due to loading) is %X sin phi( even %Z sin phi ok) where cos phi is the power fcator of the load.

 

[jkristinn]

You are correct that the % value of the Z equals the voltage (as % of rated voltage) induced on the primary side when the secondary is shorted (and any other windings open-circuited) and the rated current flows into the primary terminals. But as soon as you have some other load than the short-circuit, the power factor of the load influences the voltage drop and you have to include the X/R ratio of the transformer's Z.

The reason why the z impedance is expressed as % is because it is defined (in IEC 60076-1) as z[%] = 100 (Z/Zref) where Z is the ohm value of the impedance, Zref = U^2/Sr and U is the rated voltage and Sr is the rated power of the unit (this is the Per-Unit system representation, see here if you are not familiar with this representation.

 

[Morteza2]

Thank you all.

Based on the J&P document Marmite attached, I can see that there is a load value dependence aside from power factor, as the simplified model in the document is expressed as:

percentage regulation = a[Vr cos(phi) + Vx sin(pi)]

Where a is the load factor (expressed as a multiple/fraction of FLA)

So prc, I think the further-simplified equation you cited would be:

voltage regulation = a [%Z sin (phi)]

Where as explained above, a is the loading factor.

Would you agree?

 

[prc]

Yes,Sure. In another way %z varies directly as the load.

 

[Morteza2]

Thank you

 

[wroggent]

I have a feeling that a lot of people get this confused. I too thought that the %Z would cause a voltage drop equal to the %Z at full load. I made an excel sheet showing various loading situations on a transformer to help myself understand this. It should be attached to this post; take a look if you'd like.

 

[Morteza2]

this is great thanks