Transformer change in impedance related to change in voltage? (Engineering Pro Guides Full Exam Question 32)

[akyip]

Hey guys,

On Engineering Pro Guides' Full Exam, there is one question regarding transformer change in impedance that I don't quite understand.

Question 32 of this exam gives a 3-phase transformer rated at 480 V/120 V with an actual impedance of 10<30° ohms. If the transformer is run at a voltage of 400 V, then what will be the change in impedance?

The solution states that the impedance is reduced by the square of the voltage change:

Z new = 10<30° * (400 V/480 V)^2 = 6.94<30° ohms

The change in impedance is 10<30° - 6.94<30° = 3.06<30° ohms.

What I don't quite follow is where that formula comes from...

Z new = Z old * (Vnew / Vold)^2

Z new / Z old = (Vnew / Vold)^2

The best explanation I can think of is this...

S = V^2 / Z --> (V new^2 / Z new) = (V old^2 / Z old) --> (V new^2 / V old^2) = (V new / V old)^2 = (Z new / Z old)

The problem I think i see with that is if the transformer primary voltage is decreased, I would expect the apparent power S to decrease instead of the transformer impedance Z...

Can anyone help provide an explanation for this? Thanks for any input!

 

[DLD PE]

I would say for problems such as this, assume power to remain the same,  (Pin=Pout), so for this problem, S will remain constant.  The impedance will be reduced by the square of the voltage change.

 

[Dothracki PE]

I think the power of a 3-phase transformer is constant no matter what voltage is applied, like what Duran said, so your explanation makes sense. Only in problems that take a single phase transformer and convert to an auto-transformer does any change in power occur. That solution looks very similar to the base change equation below on p.35 of the handbook, but since power is constant, than that second value can be neglected.

 

[Zach Stone P.E.]

I think the power of a 3-phase transformer is constant no matter what voltage is applied, like what Duran said, so your explanation makes sense. Only in problems that take a single phase transformer and convert to an auto-transformer does any change in power occur. That solution looks very similar to the base change equation below on p.35 of the handbook, but since power is constant, than that second value can be neglected.

View attachment 19179

If anyone is looking for an explanation of where this formula comes from, including the "short cut" version, here is a nice explainer video:







If you'd like to dig deeper, here is the article the first video is from, there is a second example video with an actual practice problem that is worked out with numbers: https://www.electricalpereview.com/base-changing-percent-impedance-and-per-unit-impedance/

 

[akyip]

Hm, so the transformer's apparent power VA should remain the same regardless of voltage... So I guess my derivation using S = V^2 / Z and (Vold^2 / Zold) = (Vnew^2 / Znew) does make some sense here... thank you for the input, all!!!

 

[jd5191]

What I don't quite follow is where that formula comes from...

Z new = Z old * (Vnew / Vold)^2

The actual formula is Znew = Zold * (Vold / Vnew)^2, so is the answer they're giving you incorrect?

If in S = V^2/Z, S stays the same and V decreases, than Z must increase ... and in the solution Z decreases.

 

[jd5191]

If in S = V^2/Z, S stays the same and V decreases, than Z must increase ... and in the solution Z decreases.

Actually, Z must also decrease ... ooops.

 

[justin-hawaii]

@akyip  I don't check engineerboards as often.  You can email me for quicker responses.  Anybody else who has purchased an Engineering Pro Guides product, please email me and I can get back to you quicker.  Thanks again for the support!

 

[xnuke]

The solution manual and all of the responses above are wrong: a transformer's impedance in ohms does not change when the applied voltage changes. It is a function of core and winding construction only. The short-circuit test is used on a transformer to determine this value, and nominal voltage does not play a part in this test.

The video includes an equation that proves the impedance in ohms doesn't change, though it isn't mentioned. The equation of p.u._new = p.u._old*Base_old/Base_new actually comes from the fact that Z_ohms = p.u._new*Base_new = p.u._old*Base_old. If Z_ohms changed, this equality wouldn't apply, and the derived change-of-base formulas wouldn't work correctly.

Changing of per unit or percent impedance does occur when the applied voltage or power changes as shown using the change-of-base equation above and in the video, but not the impedance in ohms.

 

[akyip]

The solution manual and all of the responses above are wrong: a transformer's impedance in ohms does not change when the applied voltage changes. It is a function of core and winding construction only. The short-circuit test is used on a transformer to determine this value, and nominal voltage does not play a part in this test.

The video includes an equation that proves the impedance in ohms doesn't change, though it isn't mentioned. The equation of p.u._new = p.u._old*Base_old/Base_new actually comes from the fact that Z_ohms = p.u._new*Base_new = p.u._old*Base_old. If Z_ohms changed, this equality wouldn't apply, and the derived change-of-base formulas wouldn't work correctly.

Changing of per unit or percent impedance does occur when the applied voltage or power changes as shown using the change-of-base equation above and in the video, but not the impedance in ohms.

Honestly, that was my initial thought too. My initial thought was that a transformer's impedance should not change (much) since it is mainly a function of the windings that constitute the transformer. It's why I started to question this exam solution and see what the experts better than me have to say about this.

My initial thoughts were if voltage applied to a transformer changes, then apparent power should change instead of the transformer impedance... I stated all this in my first post...

I am not much of an expert on this as others on this board, which is why I raised this question in the first place.