I have a substation with two identical transformers 110kv/25kv single phase 50 hz usually one acts as spare. Now for some reason i want to charge my substation from secondary side i.e., 25 kv side from one of the transformer. Let us say My voltage at secondary side is 21 KV, corresponding voltage at primary side will be 120 KV (SAY). Now i want to feed this 120 kv to the primary of second transformer duly keeping the different tap at secondary of second transformer so that out put voltage of second transformer is maintained at 25 KV. Is this arrangement possible? .
Connection of two identical transformers
transformer
Related Solutions
Apparent power is \$V_{RMS}\times I_{RMS}\$ as shown on the phasor diagram below: -
To measure the apparent power you multiply the RMS measurements of voltage and current.
If you had a wattmeter you could also measure the true power and then compute reactive power using \$ \sqrt{P_{apparent}^2 - P_{true}^2}\$.
This enables you to compute iron losses in the transformer.
Side question, if secondary is shorted, will voltage across the primary head towards zero?
Don't short the secondary unless you are connected to a supply that is much, much lower than the normal intended operating voltage of the primary or you might get a fire.
Shorting the secondary (in order to determine copper losses for instance) is usually done by controlling the primary voltage with a variable transformer such that the shorted secondary current is at (and not above) full operating level. This usually means that the primary voltage is down at possibly one-twentieth of its normal operating level.
There are several issues here.
Simulation (C) is failing not because of a mistake in your circuit, but due to a limitation in LTSpice itself. The secondary inductance of your transformer windings is too high! Reduce primary and secondary inductances by a factor of at least 10000 (eg. to 100 and 10000 Henries) and then the results should come out correct.
Your formulas are correct but your calculations for circuit (D) are wrong. From secondary to primary the transformers are stepping down 10:1, so N is 0.1 not 10. Therefore (for example) L3 should be 2.7uH, not 27000uH.
You are driving the circuit with a current generator, so input voltage is directly proportional to impedance. Is that what you really wanted? If not then perhaps you should be using a voltage generator with an appropriate series resistance (eg. 50 Ohms).
Related Topic
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- Electrical – 3 phase transformers and equivalent circuits
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- Electronic – Computing short circuit current of transformer
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Best Answer
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.