Using distribution transformers vs direct connection to primary switch gear

There is no physical reason you can't back-energise your 110 kV (HV) system from 25 kV (LV). However, watch out for the following issues:

• Protection settings do not usually expect reverse power flow, or back-energisation. You may find that the LV protection trips on reverse power or directional overcurrent, or the LV overcurrent protection trips on magnetising inrush current.

• Any power transmitted from 120 kV -> 25 kV -> 120kV will suffer two lots of voltage drop. You can compensate for this using the transformer's tap-changers.

If you do not understand the above, this indicates that you should not be working on HV equipment, unless closely supervised by someone who does understand the above.

More so than usual for my other posts, this is not professional engineering advice. In case of doubt, consult a professional engineer who is familiar with your installation to provide detailed advice.

Do not mess with HV equipment unless you have been trained, deemed competent, and authorised by the owner/operator of the equipment. Do not perform HV switching unless your switching sheet has been checked and authorised by a person who is familiar with the equipment, and trained, deemed competent, and authorised by the person taking charge of electrical works.

You will kill someone if you do not know what you are doing.

Are you really saying that you have a 220 V supply that intermittently gives 440 V? How does anything electrical survive in your location?

You can't series connect two primaries on different cores like that unless you parallel the secondaries. Let's think about why:

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. (a) Due to different secondary currents the primary impedances will not match. (b) and (c) will work because the secondary currents will be the same, therefore the primary impedances and currents will be the same.

• A transformer's primary current will depend on it's secondary current (which will vary with the load). \$ \frac {N_P}{N_S} = \frac {I_S}{I_P} \$ where N is the number of turns, P is primary, S is secondary and I is current.
• With no secondary current a transformer's impedance (AC resistance, if you like) is quite high - otherwise a high current would flow.
• On the transformer with secondary current the primary impedance would be lower.
• It should now be clear that you no longer have half the supply voltage on each primary. The one that is loaded will have a much lower voltage.

Connecting the secondaries in series or parallel ensures that both secondaries pass the same current and the primary impedances will match.

This only gives you enough theory to understand why your proposed solution is poor. You need to fix your incoming power supply.

Impedance

I don't quite understand how impedance of transformer will have to change?

Given that \$ Z_P = N^2 Z_S \$ where \$ N \$ is the turns ration we can see that if a seconday is open circuit then Z is infinity on both sides of the transformer.

I think the connection shown in Fig. 1a will induced voltage tending to infinity across secondary, similar to case in current transformers.

It would if you could force current through the primary. You can't however. You only have your mains voltage and an infinite impedance. The primary of XFMR2 will appear to be an open-circuit to an AC supply. No current will flow. Full mains voltage will be across XFMR1 primary and XFMR1 secondary will have a voltage \$ \frac {1}{N} \$ times that.

Remember that we're dealing with ideal transformers for this discussion. Real ones will have some losses and leakage so the impedance will not be infinite.