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Determine the fault current in the system following a double line to ground short circuit fault at the terminal of a star connected synchronous generator operating initially on an open circuit voltage of 1.0 pu. The positive, negative, and zero sequences reactance of the generator are, respectively j0.35, j0.25, and j0.20, and the star point is isolated from the ground.
- a)-2.165 pu
- b)-2.887 pu
- c)-3.75 pu
- d)-5 pu
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
Determine the fault current in the system following a double line to g...
Concept:
Here star point is isolated so the flow of fault current is not possible in neutral or ground ( also zero sequence current is zero because zero-sequence current flow through the ground and here the ground is not available) so current is absent in neutral but current available in phase so this fault is considered as line to line (L-L) fault.
Here star point is isolated so the flow of fault current is not possible in neutral or ground ( also zero sequence current is zero because zero-sequence current flow through the ground and here the ground is not available) so current is absent in neutral but current available in phase so this fault is considered as line to line (L-L) fault.
Positive sequence current (i1) for L-L fault is given by;
i1 = voc/(x1 + x2)
Where;
voc = open circuit voltage, x1 = Positive sequence reactance, x2 = Negative sequence reactance
Fault current (if) for L-L fault is given by:
if = √3 × i1
i1 = voc/(x1 + x2)
Where;
voc = open circuit voltage, x1 = Positive sequence reactance, x2 = Negative sequence reactance
Fault current (if) for L-L fault is given by:
if = √3 × i1
Calculation:
Given that:
voc = 1.0 pu, x1 = j0.35, x2 = j0.25 pu, xo = 0.20pu, xn = ∞ pu (since Neutral is isolated from ground)
Here, voc = Open circuit voltage, x1 = Positive sequence reactance, x2 = Negative sequence reactance, xo = Zero sequence reactance, xn = Neutral reactance
Here L-L (line to line) fault happened so we will first calculate positive sequence current (i1) then calculate fault current (if) by help of positive sequence current.
Calculation of Positive sequence current (i1);
i1 = 1/(j0.35+j0.25)
i1 = -j1.6666 pu
So fault current (if) ;
if = √3 × i1
If = -j2.887 pu
Given that:
voc = 1.0 pu, x1 = j0.35, x2 = j0.25 pu, xo = 0.20pu, xn = ∞ pu (since Neutral is isolated from ground)
Here, voc = Open circuit voltage, x1 = Positive sequence reactance, x2 = Negative sequence reactance, xo = Zero sequence reactance, xn = Neutral reactance
Here L-L (line to line) fault happened so we will first calculate positive sequence current (i1) then calculate fault current (if) by help of positive sequence current.
Calculation of Positive sequence current (i1);
i1 = 1/(j0.35+j0.25)
i1 = -j1.6666 pu
So fault current (if) ;
if = √3 × i1
If = -j2.887 pu
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Determine the fault current in the system following a double line to g...
To determine the fault current in the system following a double line to ground short circuit fault at the terminal of a star-connected synchronous generator, we need to calculate the positive sequence fault current (I1), negative sequence fault current (I2), and zero sequence fault current (I0) separately.
Positive Sequence Fault Current (I1):
The positive sequence fault current is calculated using the formula:
I1 = Vf / (Z1 + Z0)
Where,
Vf = Fault voltage (1.0 pu)
Z1 = Positive sequence reactance (j0.35 pu)
Z0 = Zero sequence reactance (j0.20 pu)
Substituting the given values, we get:
I1 = 1.0 / (j0.35 + j0.20)
To simplify the calculation, we can convert the reactances into admittances:
Y1 = 1 / Z1 = 1 / (j0.35) = -j2.857
Y0 = 1 / Z0 = 1 / (j0.20) = -j5.0
Now, we can rewrite the formula for I1 as:
I1 = Vf * (Y1 + Y0)
Substituting the values, we get:
I1 = 1.0 * (-j2.857 - j5.0)
= -j2.857 - j5.0
= -j7.857
Negative Sequence Fault Current (I2):
The negative sequence fault current is calculated using the formula:
I2 = Vf / (Z2 + Z0)
Where,
Z2 = Negative sequence reactance (j0.25 pu)
Substituting the given values, we get:
I2 = 1.0 / (j0.25 + j0.20)
= 1.0 / (j0.45)
= -j2.222
Zero Sequence Fault Current (I0):
Since the star point is isolated from the ground, there is no zero sequence path. Therefore, the zero sequence fault current is zero (0).
Overall Fault Current:
The overall fault current is the vector sum of the positive, negative, and zero sequence fault currents. Since the zero sequence fault current is zero (0), the overall fault current is equal to the positive sequence fault current:
Overall Fault Current = I1 = -j7.857
Therefore, the fault current in the system following a double line to ground short circuit fault at the terminal of the star-connected synchronous generator is -j7.857 pu.
Positive Sequence Fault Current (I1):
The positive sequence fault current is calculated using the formula:
I1 = Vf / (Z1 + Z0)
Where,
Vf = Fault voltage (1.0 pu)
Z1 = Positive sequence reactance (j0.35 pu)
Z0 = Zero sequence reactance (j0.20 pu)
Substituting the given values, we get:
I1 = 1.0 / (j0.35 + j0.20)
To simplify the calculation, we can convert the reactances into admittances:
Y1 = 1 / Z1 = 1 / (j0.35) = -j2.857
Y0 = 1 / Z0 = 1 / (j0.20) = -j5.0
Now, we can rewrite the formula for I1 as:
I1 = Vf * (Y1 + Y0)
Substituting the values, we get:
I1 = 1.0 * (-j2.857 - j5.0)
= -j2.857 - j5.0
= -j7.857
Negative Sequence Fault Current (I2):
The negative sequence fault current is calculated using the formula:
I2 = Vf / (Z2 + Z0)
Where,
Z2 = Negative sequence reactance (j0.25 pu)
Substituting the given values, we get:
I2 = 1.0 / (j0.25 + j0.20)
= 1.0 / (j0.45)
= -j2.222
Zero Sequence Fault Current (I0):
Since the star point is isolated from the ground, there is no zero sequence path. Therefore, the zero sequence fault current is zero (0).
Overall Fault Current:
The overall fault current is the vector sum of the positive, negative, and zero sequence fault currents. Since the zero sequence fault current is zero (0), the overall fault current is equal to the positive sequence fault current:
Overall Fault Current = I1 = -j7.857
Therefore, the fault current in the system following a double line to ground short circuit fault at the terminal of the star-connected synchronous generator is -j7.857 pu.
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Determine the fault current in the system following a double line to ground short circuit fault at the terminal of a star connected synchronous generator operating initially on an open circuit voltage of 1.0 pu. The positive, negative, and zero sequences reactance of the generator are, respectively j0.35, j0.25, and j0.20, and the star point is isolated from the ground.a)-2.165 pub)-2.887 puc)-3.75 pud)-5 puCorrect answer is option 'B'. Can you explain this answer?
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