MVA method and calculating 3-phase fault apparent power and 3-phase fault current when X/R ratio is involved: PPI Exam 2 Question 152

[akyip]

Hey guys,

PPI Exam 2 Question 152 is the only question I've seen so far in all the practice exams and questions I have that asks for solving the 3-phase fault apparent power and fault current with the X/R ratio involved. So I'm not quite understanding how/why the solution uses the X/R ratio the way it does.

I know that the problem statement states that the X/R ratio is the same between the generator and the transformer. But why specifically is the generator's p.u. impedance equal to the transformer's p.u impedance times the ratio of transformer rated MVA / generator rated MVA?

Also, why in the solution is S base the transformer's rated MVA instead of the generator's rated MVA? Is it because the transformer has a lower rated MVA than the generator? If the generator had a lower MVA rating, would the S base be the generator rated MVA then? Or is the transformer used as S base because it links the primary side and the secondary side together?

I really hope that someone is able to answer these questions about the X/R ratio being used to calculate 3-phase fault power (VA) and current. This is the only question I've seen so far where X/R ratio is involved in calculating 3-phase fault power and current, and I'm not quite understanding how the X/R ratio is used here... Thanks for any input on this!

 

[CeeYem]

It does not really matter which MVA rating is considered as the base. If you are choosing Gen rating as a base, then you will need to convert the impedance to the new base when transferring to the 460V side. Since Primary voltage is not given in the question, the conversion seems impossible, and it is always a good idea to use Transformer rating as the base for calculation.

 

[Zach Stone P.E.]

Hey guys,

PPI Exam 2 Question 152 is the only question I've seen so far in all the practice exams and questions I have that asks for solving the 3-phase fault apparent power and fault current with the X/R ratio involved. So I'm not quite understanding how/why the solution uses the X/R ratio the way it does.

I know that the problem statement states that the X/R ratio is the same between the generator and the transformer.

Many fault current problems include the assumption of equal X/R ratios, it helps to simplify the calculations. Generators and transformers are mostly inductive devices but they do have resistance (and I²R copper losses). The typical %Z rating of a transformer is the magnitude of the percent impedance. Likewise anytime you see a percent reactance rating %R, that is the per-unit reactance of the machine expressed in base values equal to the machine's ratings.

To explain equal X/R ratios, think of two complex impedances in series circuit. To find the total equivalent impedance, you would have to use complex math and add the resistance and reactance of each one:

If: Z1 = R1 + jX1, and Z2 = R2 + jX2​

Then: Z1 + Z2 = (R1 + R2) + j(X1 + X2).​

And: |Z1 + Z2| = sqrt[ (R1 + R2)^2 + (X1 + X2)^2 ]​

However, if the X/R ratios of both complex impedances are equal, then it means that both impedances have equal impedance angles since:

TanΘ = X/R​

Θ = Tan-1(X/R)​

When the impedance angles are equal, you can add the magnitudes directly without worrying about complex math. For example:

If Z1 = |Z1|<Θ, and Z2 = |Z2|<Θ​

Then: Z1 + Z2 = (|Z1|+|Z2|)<Θ.​

And: |Z1 + Z2| = |Z1|+|Z2|​

Saying that the generator, transformer, and system all have equal X/R ratios means you can use the MVA method, or the per-unit method without having to worry about complex numbers, just use magnitudes and ignore any angles.

In this problem and in most worse case three-phase fault current problems intended to be solved with the MVA method, or the per-unit method, complex impedances of each device are not given. Just the percent impedance (or the percent reactance).

Why specifically is the generator's p.u. impedance equal to the transformer's p.u impedance times the ratio of transformer rated MVA / generator rated MVA?

I'm scratching my head a little here at their solution and I'm wondering if there is a mistake here.

The ratio of the transformer rated MVA to the generator rated MVA is simply just the impedance per unit base changing formula with equal old voltage base and new voltage base values. The voltage bases cancel since they are equal.

If you're not totally familiar with this, here is a short video-based article that sums it up pretty quickly: Electrical PE Review - Base Changing Percent Impedance and Per Unit Impedance

The part that has me scratching my head is that they are treating the generator as if it has a per unit impedance of 1, or a percent impedance of 100% because they are using the ratio directly, without multiplying the generator's old per unit impedance to base change it to the transformer base values.

In other words:

Zpu-gen = %Zg*(S_xfmr/S_gen)​

Zpu-gen = %Zg*(2 MVA/20 MVA)​

Zpu-gen = %Zg*0.1 pu​

They are using (2 MVA/20 MVA) = 0.1 pu as the per-unit impedance of the generator, but this value needs to be multiplied by the generator's percent impedance first.

Nothing in the problem alludes to this, even considering equal X/R ratios.

In fact, you can solve this problem using the MVA method, with a Z = 100% or Z = 1 pu value for the 20 MVA generator and get the same answer of 16.7 kA.

A generator with a Z = 1 pu expressed in its own base values is..... never seen that before so I'm wondering if this is just a mistake.

Also, why in the solution is S base the transformer's rated MVA instead of the generator's rated MVA? Is it because the transformer has a lower rated MVA than the generator? If the generator had a lower MVA rating, would the S base be the generator rated MVA then? Or is the transformer used as S base because it links the primary side and the secondary side together?

The ratio of MVA values with the transformer's 2MVA on top of the generator's 20MVA is just from base changing the generator's per unit impedance in base values of the generator's ratings, to the base values of the transformer's ratings.

See the link in the above paragraph for a video of a worked out example that shows where this relationship comes from.

They've selected the transformer's base values without outright saying that they have.

Since they are solving for short circuit current in amps (and not per unit), then really it does not matter what base values you initially use. Once you base changing out of the per-unit system and back into actual values, it will be the same answer regardless of the new system base values used to get from actual units to the per-unit system.

I really hope that someone is able to answer these questions about the X/R ratio being used to calculate 3-phase fault power (VA) and current. This is the only question I've seen so far where X/R ratio is involved in calculating 3-phase fault power and current, and I'm not quite understanding how the X/R ratio is used here... Thanks for any input on this!

Just about anytime the MVA method or per unit method is used for worse case three-phase faults, it's under the assumption of equal X/R ratios. If you check the NCEES practice exam you'll see similar wording.

Hope this helps!