Per unit impedance 1

[CAS81]

How do you find the respective complex powers provided by two transformers that are supplying a load in parallel when you have the per unit impedances of the transformers w.r.t the total power of the load, the load and the power factor of the load?

Regards,

Chris

 

[waross]

I am not sure what you mean by "complex powers" but I won't let that stop me from commenting. The real power will be divided in the inverse ratio of the resistances of the transformers.
The reactive power will be divided in the inverse ratio of the reactances of the transformers.
Now comes the D'uh;
To use per unit impedance you must also know the base KVA ratings.
But, the impedance won't help you anyway, at least not directly. The impedance is only a factor when the transformer is short circuited. It is used to select switchgear with the proper interupting ratings.
You need to know the resistances and reactances of the transformers.
If you know the base KVAs (not given) and the regulation, you can derive the missing units.
If you know the base KVAs and the X/R ratios you can derive the missing units.
Or, if this is a lower level class and you haven't yet studied the "Rest of the story" you may use this method.
Use the power factor to determine the actual current of the load.
Use the base KVAs and the per unit impedances to calculate the impedance of each transformer.
Consider the load current to be supplied in inverse proportion to the impedances.
This will give a correct answer if the X/R ratios are the same, and an acceptable wrong answer if the X/Rs are different but you have not yet studied X/R ratios.

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

 

[magoo2]

Start with some basics.

Assume you understand per unit enough to put them on the same MVA base.

What do you know about the circuit?
- If they're in parallel, then the voltage drop will be the same.

So I1 * Z1 = I2 * Z2

and I1 + I2 = Itotal

Itotal should be known, so you have enough to solve for I1 and I2.

 

[waross]

Doesn't the voltage drop depend on the regulation?
Isn't the pu impedance often several times greater than the regulation??
Doesn't the power factor of non unity loads complicate this just a little more???

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

 

[davidbeach]

Bill, nope, the per-unit impedance of the transformer is fully specified with the impedance, but that has to include the angle of the impedance (X/R, etc.). "Regulation" is how the very reactive transformer impedance interacts with the very resistive load currents. Because load generally produces very little voltage drop across the transformer reactance there appears to be a difference between "impedance" and "regulation", but there isn't. On the other hand, if you only consider the magnitude of the impedance and ignore the angle, then, yes, you will find that "regulation" is something very different from "impedance" but that's because half the information of impedance is being thrown away. Always, incomplete information will lead to incomplete results.

 

[waross]

Thanks David. I think that you have made my point.
The OP only mentioned P.U. impedance. The smaller distribution transformers that I am most familiar with only show the %imp voltage with no X/R information.
The regulation describes the voltage drop under normal loading.
P.U. impedance relates to the current under steady state short circuit conditions.
With a resistive load at normal loading, the transformer resistance is the predominant factor in voltage drop and load sharing.
Under short circuit conditions the current is determined by the transformer impedance which is dominated by the transformer inductive reactance.
If the transformer X/R ratios are not equal, things become complicated.
If a economy model transformer is paralleled with a high efficiency transformer the impedances may be equal while the regulation may not be equal.
Calculating the load sharing based on P.U. impedances often yeilds acceptable results, but it is well to be aware that this approximation may have significant errors under some conditions. With equal P.U. impedances but unequal X/R ratios the transformers may share the load properly but the sum of the transformer currents may be more than the load current.

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