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A 30 MVA 3-phase, 50Hz, 13.8 kV, star-connected synchronous generator has positive, negative and zero sequence reactances, 15%, 15% and 5% respectively. A reactance (Xn) is connected between the neutral of the generator and ground. A double line to ground fault takes place involving phases ‘b’ and ‘c’, with a fault impedance of j0.1 p.u. The value of Xn (in p.u) that will limit the positive sequence current to 4270 A is _____________.
Correct answer is '1.1'. Can you explain this answer?
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A 30 MVA 3-phase, 50Hz, 13.8 kV, star-connected synchronous generator ...
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A 30 MVA 3-phase, 50Hz, 13.8 kV, star-connected synchronous generator ...
A and C. The fault impedance is 0.2 + j0.6 ohms.
To determine the fault current, we can use the method of symmetrical components.
The positive sequence current (I1) is given by:
I1 = V1 / (Z1 + Xn)
where V1 is the positive sequence voltage = V / √3 = 13.8 kV / √3 = 7.98 kV
Z1 is the positive sequence impedance = 0.15 + j0.15 ohms
Substituting the values:
I1 = 7.98 kV / (0.15 + j0.15 + Xn)
The negative sequence current (I2) is given by:
I2 = V2 / (Z2 + Xn)
where V2 is the negative sequence voltage = V / √3 = 13.8 kV / √3 = 7.98 kV
Z2 is the negative sequence impedance = 0.15 + j0.15 ohms
Substituting the values:
I2 = 7.98 kV / (0.15 + j0.15 + Xn)
The zero sequence current (I0) is given by:
I0 = V0 / (Z0 + Xn)
where V0 is the zero sequence voltage = V / √3 = 13.8 kV / √3 = 7.98 kV
Z0 is the zero sequence impedance = 0.05 + j0 ohms
Substituting the values:
I0 = 7.98 kV / (0.05 + Xn)
Since the fault is a double line to ground fault, the fault current is the sum of the positive and zero sequence currents:
Ifault = I1 + I0
To solve for Xn, we need to equate the positive sequence current with the fault current:
7.98 kV / (0.15 + j0.15 + Xn) = Ifault
Similarly, we can equate the zero sequence current with the fault current:
7.98 kV / (0.05 + Xn) = Ifault
Solving these two equations simultaneously will give us the value of Xn.
To determine the fault current, we can use the method of symmetrical components.
The positive sequence current (I1) is given by:
I1 = V1 / (Z1 + Xn)
where V1 is the positive sequence voltage = V / √3 = 13.8 kV / √3 = 7.98 kV
Z1 is the positive sequence impedance = 0.15 + j0.15 ohms
Substituting the values:
I1 = 7.98 kV / (0.15 + j0.15 + Xn)
The negative sequence current (I2) is given by:
I2 = V2 / (Z2 + Xn)
where V2 is the negative sequence voltage = V / √3 = 13.8 kV / √3 = 7.98 kV
Z2 is the negative sequence impedance = 0.15 + j0.15 ohms
Substituting the values:
I2 = 7.98 kV / (0.15 + j0.15 + Xn)
The zero sequence current (I0) is given by:
I0 = V0 / (Z0 + Xn)
where V0 is the zero sequence voltage = V / √3 = 13.8 kV / √3 = 7.98 kV
Z0 is the zero sequence impedance = 0.05 + j0 ohms
Substituting the values:
I0 = 7.98 kV / (0.05 + Xn)
Since the fault is a double line to ground fault, the fault current is the sum of the positive and zero sequence currents:
Ifault = I1 + I0
To solve for Xn, we need to equate the positive sequence current with the fault current:
7.98 kV / (0.15 + j0.15 + Xn) = Ifault
Similarly, we can equate the zero sequence current with the fault current:
7.98 kV / (0.05 + Xn) = Ifault
Solving these two equations simultaneously will give us the value of Xn.
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A 30 MVA 3-phase, 50Hz, 13.8 kV, star-connected synchronous generator has positive, negative and zero sequence reactances, 15%, 15% and 5% respectively. A reactance (Xn) is connected between the neutral of the generator and ground. A double line to ground fault takes place involving phases ‘b’ and ‘c’, with a fault impedance of j0.1 p.u. The value of Xn (in p.u) that will limit the positive sequence current to 4270 A is _____________.Correct answer is '1.1'. Can you explain this answer?
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