Maximum power from a transformer

[tim9000]

TL;DR Summary

Above nameplate kVA, theoretical implications.

Hi,
I was having a discussion with a colleague today now I'm second guessing myself.
My understanding of nameplate kVA is that it is based on 'ambient temperature, temperature rise, heat transfer, thermal mass, thermal time constant, materials used, and expected transformer life'.

However, when I do fault current calculations I use the nominal power and impedance of the transformer. This makes sense to me, as the net flux in the core won't ever change much from O/C to fault. This makes sense as there will be a voltage drop on the leakage impedance and resistance of the primary side. And as the secondary side is a short, most of the current will flow through the secondary side, as the magnetising branch is relatively a larger impedance then the secondary windings, as per the model of an ideal transformer. So in my mind it is the windings which limit the power of the transformer under high load. I don't see core saturation being a factor.
In the case of a fault I always imagined the current rising into the kA while the voltage on the secondary side drops right down, I always imagined that the power was still around nameplate power.

But I know that some transformers if you put a fan on to force cool the fins, you can extract above passive kVA rating.

So I've had concepts of Voltage regulation and Thevinin maximum transfer theorum running around my head. So my question is, does the power of the transformer rise under a fault or stay around nominal? Is the secondary voltage going to drop right down under a bolted fault? Maybe the voltage drop on the impedance of the windings isn't as much of a factor as I expected.
If the power output can just rise based on increasing load (assuming ammpacity of secondary conductors is not limiting) if you were cooling a transformer, what limits the amount of power it will output?

Thanks!

 

[alan123hk]

TL;DR Summary: Above nameplate kVA, theoretical implications.

In the case of a fault I always imagined the current rising into the kA while the voltage on the secondary side drops right down, I always imagined that the power was still around nameplate power.

I think when the transformer fails, the output power may decrease or increase. In the simplest case. If the primary winding is partially short-circuited the output voltage rises, if the output winding is partially short-circuited the output voltage drops, so the output power at constant load will rise or fall accordingly.

TL;DR Summary: Above nameplate kVA, theoretical implications.

If the power output can just rise based on increasing load (assuming ammpacity of secondary conductors is not limiting) if you were cooling a transformer, what limits the amount of power it will output?

The power supply system of the power company has overload protection, and the primary winding resistance and secondary winding resistance of the transformer will limit the output power.

 

[DaveE]

The power delivered to the load will increase as the load current increases until the load impedance roughly matches the source impedance (roughly because these are complex impedances and probably won't actually match), then it will decrease to zero. The source impedance will be primarily composed of the leakage inductance and the winding resistance.

This is a good simplistic model, but you'll also want to add the resistance part too:

https://www.physicsforums.com/threa...-coupling-high-low-power.1001193/post-6478419

 

[tim9000]

I think when the transformer fails, the output power may decrease or increase. In the simplest case. If the primary winding is partially short-circuited the output voltage rises, if the output winding is partially short-circuited the output voltage drops, so the output power at constant load will rise or fall accordingly.

The power supply system of the power company has overload protection, and the primary winding resistance and secondary winding resistance of the transformer will limit the output power.

Thanks for the reply but the OP is wrt a fault external to the transformer, with the assumption that the supply is an infinite bus. Cheers

 

[tim9000]

The power delivered to the load will increase as the load current increases until the load impedance roughly matches the source impedance (roughly because these are complex impedances and probably won't actually match), then it will decrease to zero. The source impedance will be primarily composed of the leakage inductance and the winding resistance.

This is a good simplistic model, but you'll also want to add the resistance part too:

https://www.physicsforums.com/threa...-coupling-high-low-power.1001193/post-6478419

Thanks for the reply. I appreciate the derivation of your matrix in your post #62, however I failed to derive any intuitive understanding in a fault. Because it implies to me that as i1 and i2 increase, so will V2. Anyway putting this aside, in your post #42 you wrote a nice summary explanation of the core: "The magnetizing current has nothing to do with the secondary current or power delivered (again, ideal model), but there is some reactive "power" associated with it. The value of the core is to redirect the flux in the windings to increase coupling and increases the shunt inductance. It doesn't really change that flux, nor does that flux change the core (again, ideal case)." Agreed.

Coming to:

The source impedance will be primarily composed of the leakage inductance and the winding resistance.

I agree, so to me this says that when the fault value is equal to the impedance of the winding resistance and leakage reactance (I assume this is the lumped impedance equivalent sum of primary and secondary windings), the power output will be the maximum deliverable. The voltage will be higher below this value, and the current will be higher below this impedance match, but at the impedance match the power will peak. Please correct me if this is wrong.

The significance of the transformer impedance is that it is the impedance when you perform a short circuit test at rated secondary current. So, talking about current - how close is nameplate rated current to the current which would flow in the secondary when the secondary impedance matches the transformer impedance?
I assume you would design the transformer such that when it is operating at nominal power, the impedance of the transformer is slightly lower than the load impedance, but not too much. That way if the load impedance drops a bit, there is still some room for the transformer to output more power, as the internal and external impedances are closer to matching.

Since under a fault condition the impedance of the fault is going to be lower than the internal impedance of the transformer, this implies that the power output of the transformer is lower than peak power output.Thanks

 

[DaveE]

how close is nameplate rated current to the current which would flow in the secondary when the secondary impedance matches the transformer impedance?
I assume you would design the transformer such that when it is operating at nominal power, the impedance of the transformer is slightly lower than the load impedance, but not too much. That way if the load impedance drops a bit, there is still some room for the transformer to output more power, as the internal and external impedances are closer to matching.

This all depends on the design of the transformer. I don't believe there are universally useful rules to relate the nameplate ratings, which are really about reliability and safety, to voltage regulation, output impedance, and such. There are lots of different ways to make a transformer.

 

[Babadag]

In order to calculate the power wasted in a transformer short-circuit case you need the following data:
S=rated kVA, uk%-short-circuit voltage, kxr= the X on R ratio[X/R].
The transformer impedance [short-circuit impedance] Z=VLL^2/S*uk%/100
where Z=R+Xi and R=Z/sqrt(1+kxr^2)
Then Isc=VLL/Z for single phase system and VLL/sqrt(3)/Z for three-phase system.

The short-circuit involving ground fault, phase-to-ground and phase-to-ground-to phase, it is more complicate, and it could be more elevated in some cases.
In short-circuit case, due to a huge voltage drop in transformer, the part of voltage destined to magnetic circuit will be close to zero, then we can neglect the magnetic losses.
Then Psc will be R*Isc^2 [ or 3*R*Isc^2]
There are standards for maximum permissible fault duration before fault clearance and how many times, per transformer life. the transformer may overcome a short-circuit.
See:
IEC 60076-5 Ability to withstand short-circuit.
IEEE Std C57.12.90/2006 12.1 Short-circuit tests

 

[Babadag]

The apparent power [kVA or MVA] on the name plate it is not the power “consumed” in the transformer, but only the transferred power to the load. The power consumed in the transformer it is the transformer losses-electrical and magnetic.
Let's take an example.
ABB 33/11 kV 10 MVA transformer presents 8.1 kW magnetic losses and 65 kW load losses. That means the temperature rise of 65 oC is produced when the transformer expels 65 kW "consumed" in copper and iron.
The transformer uk%=7.82% and kxr=15.The rated current Irat=10/sqrt(3)/33=0.175 kA
Then the Isc=0.175*100/7.82=2.238 kA
Z=33^2/10*7.82/100=8.516 ohm
R=8.516/sqrt(1+15^2)=0.566 ohm
If the rated temperature of copper in transformer it is around 90oC, in short-circuit case could be 200oC
The resistance rise factor will be (234.5+200)/(234.5+90)=1.339
Then the power “consumed” in this transformer will be 3*0.566*1.339*2.238^2=11.39 MW. That means 11.39/0.065=175 times.

 

[Babadag]

I forgot to say that 65 kW is evacuated in the air, while
11.39 MW still remain in the transformer because the exhaust begins after 5-10 seconds when the outside temperature of the transformer is high enough.