More VD through Transformers 1

[bjenks]

Using IEEE Grey book for Volt Drop I get the following:
(sorry about the notications)
VD = I[RCosq + XSinq]
PF = Cosq
Sinq = Sin(Cos-1q)

If I am given the z% of a transformer and I wanted to figure the voltage drop by using the IEEE formula I would first come up with:
z=sqr(R*R +X*X)
R=Zcos(q)
X=Zsin(q)

If I substitute R above with the IEEE formula I get
VD = IZCosq*cosq + IZsinq*sinq
VD = IZ[cosq*cosq +sinq*sinq]
VD = IZ[1]
VD = IZ

So a 30kVA transformer would at 480V 3 phase have 36A
If Z=.053
VD = 36*.053 =1.91 is that correct? Is that a percentage or actual VD? Is that per phase so I multiply it times 3 or is it L-N and mulitply by 1.7321 or is that it?

Am I getting confused between Z of the trig triangle of X/R and Z%. Since the PF changes based on the circuit impedance, but Z is fixed based on the manufacture spec. Maybe they are the same since it is the only item being considered in the VD and in most designs we have a load in addition to the transformer. Maybe I did everything right above, but the flaw is that I have to add the load R and X then do VD and that is why I am confused.

It seems I could do the Voltage drop on knowing just the Z% and the load, but I feel like I am missing something...

I think I have gotten too use to SKM PowerTools...

 

[dpc]

You need to know the actual power factor - angle between current and voltage of the circuit. The impedance of the transformer is only part of the circuit.

Voltage drop across the transformer will be at one angle and the other voltage drops in the circuit (conductor, load) will all be at different angles. You have to use the phasor sum of the voltage drops to determine the total voltage drop in the circuit.



"An 'expert' is someone who has made every possible mistake in a very narrow field of study." -- Edward Teller

 

[waross]

You have to know either the R and X of the transformer, or work back from the X:R ratio.
I don't see real field values for these in your equations. Did I miss something?
Under short circuit conditions the Z of the transformer is assumed to be the only impedance in the circuit. Under full load conditions, the complete circuit contains the transformer inductive reactance, the transformer resistance, the load resistance, the load inductive reactance and the load capacitive reactance. (include the secondary conductors in the load on the transformer terminals).
To calculate the transformer voltage drop at full load you need the load power factor and the transformer X:R ratio.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[bjenks]

OK, I got the part now where I have to add the impendance downstream of the transformer. Also if you short circuit the transformer you have in effect disregaurded anything downstream. maybe I can be more direct based on my original thought I mentioned below:

VD = IZCosq*cosq + IZsinq*sinq
VD = IZ[cosq*cosq +sinq*sinq]
VD = IZ[1]
VD = IZ

Is this Z mentioned above the %Z of the transfomer or is the Z above the square root of the squares of the X and R. where X and R come from the X/R of the transformer which is then used multiplied by the respective Cosq or Sinq?

If a transformer was called %Visc (% voltage on input to give rated current on output at short circuit) I wouldn't be asking this question. However maybe I want to make sure %Z really only has to do with a factor in calculating short circuit.

Is X/R of the transformer alsways equal to the Tan(q) and that is how you get the VD?

 

[waross]

You cannot assume to directly sum impedances. Impedances may only be summed directly when the phase angle of both is is the same.
Draw a right triangle. Label the base R, the altitude X, and the hypotenuse Z. Now apply all the trig you want and apply Pythagoras' theorem to your hearts content.
The R, X and Z relationships in a transformer may be described by a right triangle
R is the actual resistance of the transformer. (At operating temperature)
x is the inductive reactance of the transformer.
Z is the impedance. % Impedance is defined as the percentage of rated voltage required to force full load current through a transformer with the secondary short circuited. (At operating temperature.)
To calculate the voltage drop at any load, you must consult the manufacturer or the transformer specs to determine either the X:R ratio or the actual X and R. You must also Know either the phase angle or the resistance and net reacance of the load. Then, depending on what informetion you have, use vector addition of the impedances or sum the resistances and sum the reactances and use Pythagoras to calculate the total inductance.


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[bjenks]

I didn't do a good job of my question as I found my answer:
%Z=IZ(100/V)

I is full load current of the transformer
Z is the impedance in ohms of the transformer
V is the voltage rating of the transformer
All above are referenced to either the primary or secondary side of the transformer.

Further %VD through the transformer is (Iload/Irated)/Z%

which shows %Z does not equal Z which I think you all said indirectly.

I am still missing how to substitute the above information into the IEEE formula. To get R and X do I just take the Z and mulitply it times cos and sin of the phase angle? If I do then I plug it into the IEEE formula? or do I have to take the arc Tan of the phase angle and use the X/R ratio and then plug into IEEE formula? or is this two methods to get to the same results?

Sorry for the confusion on my questions. I am sure I am over complicated this.

 

[waross]

%Z is an abbreviation for "Percent Impedance Voltage", or the percentage of rated voltage to force full load current through a shorted transformer.
When you are first getting used to it it is easy to confuse the abbreviation for percent impedance voltage with the impedance in ohms.
The %Z is given on the transformer name plate and is used to select switch gear of adequate fault current ratings.
When you consider the current through a shorted transformer, it is composed of an active current and reactive current, through the resistance and inductive reactance respectively, of the transformer. X generally predominates the impedance of a distribution transformer. Now add a resistive load. You have drastically increased the resistance of the circuit. Now the total resistance of the circuit (load and transformer resistances) is much greater than the inductive reactance of the transformer. As a result, the voltage drop of a transformer at full load is less than would be indicated by calculations based on %Z without considering the power factor of the load. %Z describes the current in a short circuited transformer. Regulation describes the voltage drop of a transformer under full load at stated power factor!
Let's look at an example:
10KVA transformer
100V
100A
5%Z
X:R ratio=6
Z=0.2 Ohms
X=0.197 Ohm
R=0.033 Ohm
You can see that the inductive voltage drop is 6 time greater than the resistive voltage drop. The voltage drop through a shorted transformer is highly reactive or the reactive component predominates.
Now lets add a resistive load.
R_l=1 Ohm.
V=100V
A=100A
The total impedance is now determined by the total resistance of 1 Ohm plus 0.033 Ohms and 0.197 Ohms inductive reactance.
Our total ohms is now root(1.033+0.197)= 1.034 Ohms.
The inductive reactance acts at right angles to the resistance and when the total resistance is more than 5 times the inductive reactance, resistance predominates.
Our actual open circuit voltage of the transformer would be of the order of 103.4 volts, dropping to 100 volts under load. The regulation would be about 3.4% rather than the 5% indicated by the %Z value.
Change the power factor of the load and this all changes.
I am sure that some of you will notice that I have simplified the values somewhat. Please consider that the errors that my short cuts have introduced are insignificant compared to the effects being illustrated.
Most of the errors stem from ignoring the small difference between the rated voltage of the transformer and the open circuit voltage of the transformer until it was convenient to use them.
%Z, Short circuit current.
Regulation, Voltage drop.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[jghrist]

If I am given the z% of a transformer and I wanted to figure the voltage drop by using the IEEE formula I would first come up with:
z=sqr(R*R +X*X)
R=Zcos(q)
X=Zsin(q)

Your problem is that the q in the voltage drop equation is the load current angle. The q in your equation for R and X is the impedance angle. These are two different things which messes things up in your substitution.

 

[bjenks]

Yes, I was missing the relationship between Z and Z% which I now will replace %Z=IZ(100/V) to Z=(V*V/VA)*(Z%/100). Once I have Z I am able to get R and X from the sin and cos of arctan(X/R) times Z. The sin and cos of the load power factor is multiplied against the calculated R and X which is then multiplied times the load current for the VD. I have since put a hypothical example together in SKM and was able to hand calculate a matching answer.

Thanks to everyone for helping make sense out of things. Like I said earlier, I got too use to plugging data into SKM.