Poll Results
No votes. Be the first one to vote.
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
To find the resistance that must be placed in the neutral such that the fault current for a single-phase-to-ground fault does not exceed the rated line current in per unit (pu), we can start by determining the rated current of the alternator.
1. Calculate the rated current (I_rated) of the alternator:
– The base power (S_base) = 10 MVA = 10,000 kVA
– The base voltage (V_base) = 11 kV = 11,000 V
Using the formula:
[
I_{rated} = frac{S_{base}}{sqrt{3} times V_{base}}
]
[
I_{rated} = frac{10,000}{sqrt{3} times 11} approx 523.61 text{ A}
]
2. Determine the sequence reactances in per unit:
– Positive sequence reactance (X_1) = 0.25 pu
– Negative sequence reactance (X_2) = 0.40 pu
– Zero sequence reactance (X_0) = 0.10 pu
3. For a single-phase-to-ground fault:
The fault current (I_f) can be expressed as:
[
I_f = frac{V_{ph}}{Z_{total}}
]