Unit protection (differential relays) is used to protect transformers. Please see attachment.
There are two cases in this question. Case 1 is when circuit breaker A is open and B is closed, Case 2 is when A is closed and B is open.
The question asks for the fault levels to a base of 10 MVA at the 11kV and 3.3kV busbars for a fault at T1 transformer. Current settings and time multiplier settings for the relays are also to be determined if the operating time of the directional overcurrent relays on the 3.3kV side of the transformers is 0.5s. 0.4s discrimination interval is to be used.
My attempt:
Let's consider case 1 first.
At 11kV busbar, after reducing the circuit to a single impedance I've got X per unit = 0.15pu and fault level = 66.67 MVA.
At 3.3kV busbar, X pu = 0.2pu, fault level = 50 MVA.
Is that correct so far?
To calculate current settings I believe we need the transformer full load current which I would say: \(\frac{5000}{sqrt{3}*3.3}\) provided there is no generator. And the full load current on the primary side would be the same but multiplied by 11 in the denominator instead of 3.3.
How do the generators affect full load currents through the transformers? How do I calculate them?
Once I've got the full load current I would calculate current settings as follows:
Current setting for inverse time overcurrent relay on the 11kV side: \(110%*\frac{full load current on primary side}{250}\)
Current setting for directional relay on the 3.3kV side: \(110%*\frac{full load current}{1000}\)
Is this correct?
Now, the time multiplier settings:
I think we need the fault current in each transformer, which I don't know how to determine. However, knowing that I would continue:
For 3.3kV side:
Multiple of current setting = fault current in each transformer / 250.
Then time for full travel = 3 / log multiple of current setting
Time multiplier setting = operating time 0.5s / time for full travel.
The same drift for 11kV side but taking into account discrimination interval.
Does this all make any sense?
There are two cases in this question. Case 1 is when circuit breaker A is open and B is closed, Case 2 is when A is closed and B is open.
The question asks for the fault levels to a base of 10 MVA at the 11kV and 3.3kV busbars for a fault at T1 transformer. Current settings and time multiplier settings for the relays are also to be determined if the operating time of the directional overcurrent relays on the 3.3kV side of the transformers is 0.5s. 0.4s discrimination interval is to be used.
My attempt:
Let's consider case 1 first.
At 11kV busbar, after reducing the circuit to a single impedance I've got X per unit = 0.15pu and fault level = 66.67 MVA.
At 3.3kV busbar, X pu = 0.2pu, fault level = 50 MVA.
Is that correct so far?
To calculate current settings I believe we need the transformer full load current which I would say: \(\frac{5000}{sqrt{3}*3.3}\) provided there is no generator. And the full load current on the primary side would be the same but multiplied by 11 in the denominator instead of 3.3.
How do the generators affect full load currents through the transformers? How do I calculate them?
Once I've got the full load current I would calculate current settings as follows:
Current setting for inverse time overcurrent relay on the 11kV side: \(110%*\frac{full load current on primary side}{250}\)
Current setting for directional relay on the 3.3kV side: \(110%*\frac{full load current}{1000}\)
Is this correct?
Now, the time multiplier settings:
I think we need the fault current in each transformer, which I don't know how to determine. However, knowing that I would continue:
For 3.3kV side:
Multiple of current setting = fault current in each transformer / 250.
Then time for full travel = 3 / log multiple of current setting
Time multiplier setting = operating time 0.5s / time for full travel.
The same drift for 11kV side but taking into account discrimination interval.
Does this all make any sense?
