Advance Power Systems: Power System: Single Line-To-Ground Fault

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WEEK 17: POWER SYSTEMS: SINGLE LINE-TO-GROUND FAULTS

Sections: Conditions | Network | Example | Problem Set

Conditions

Conditions at Fault. The following are the conditions for single line-to-ground fault for power system:

I_B = 0 I_C = 0 V_A = 0

Current. The following are the current relationships during a single line-to-ground fault for power system:

In matrix form:

I_A0 = I_A1 = I_A2 = I_A / 3

Voltage. The following are the voltage relationships during a single line-to-ground fault for power system:

In matrix form:

V_A0 + V_A1 + V_A2 = -I_A1Z₀ + V_f - I_A1Z₁ - I_A1Z₂

But V_A = V_A0 + V_A1 + V_A2 = 0, thus:

V_f = I_A1 [Z₁ + Z₂ + Z₀]

SLTG Fault at Impedance. The following are the conditions during the fault at impedance in a power system:

I_A1 = I_A2 = I_A0 V_f = I_A1 [ Z₁ + Z₂ + Z₀ + 3Z_f]

Sections: Conditions | Network | Example | Problem Set

Network

Sections: Conditions | Network | Example | Problem Set

Example

A group of identical synchronous motors is connected through a transformer to a 4.16 kV bus at a location remote from the genreating plants of the power system. The motors are rated 600V and operate at 89.5 efficiency when carrying full load at unity power factor and rated voltage. The sum of their output ratings is 4476 kW (6000HP). The reactances in per unit of each motor based on its own input kVA rating are X" = 0.20, X₂ = 0.20 and X₀ = 0.04 and each is grounded through a reactance of 0.02 per unit. The motors are connected to the 4.16 kV bus through a transformer bank composed of 3 single-phase units, each of which is rated 2400/600 kV, 2500 kVA. The 600V windings are connected in delta to the motors and the 2400 V windings are connected in wye. THe leakage reactance of each transformer is 10%.

The power system which supplies the 4.16 kV bus is represented by a Thévenin equivalent generator rated 7500 kVA, 4.16 kV with reactances of X"=X₂=0.10 p.u. and X₀ = 0.05 p.u. X_N from neutral to ground is equal to 0.05 p.u.

Each of the identical motors is supplying an equal share of a total load of 3730 kW (5000 HP) and is operating at rated voltage, 85% p.f. lagging and 88% efficiency when a single line to grounf fault occurs at the low-tension side of the transformer bank. Treat the group of motors as a single equivalent motor. Draw the sequence networks showing values of impedances. Deterrmine the subtransient line currents in all parts of the system with the prefault current neglected.

Solution:
Generator as Base: 7500 kVA, 4.16kV
V_f =1.0 p.u. = prefault voltage of Phase A at fault

One-line diagram of the given power system

Computations for the single-phase transformer:
Ö3 x 2400 = 4160 Volts 3 x 2500 = 7500 kVA
therefore: three-phase rating of the transformer:
7500 kVA, 4160 wye / 600 delta

Therefore, base at Motor circuit: 7,500 kVA, 600V

Input rating of single equivalent motor:
6000 x 0.746 ¸ 0.895 = 5000 kVA.

Reactances of equivalent motor in per unit:
X" = 0.20 (7500¸5000) = 0.30 p.u.
X₂ = 0.20 (7500¸5000) = 0.30 p.u.
X₀ = 0.04 (7500¸5000) = 0.06 p.u.

Reactances of reactor per unit:
Motor: 3X_N = 3[0.20 (7500¸5000)] = 0.09 p.u.
Generator: 3X_N = 3(0.05) = 0.15 p.u.

Equivalent Reactance Diagram

Base Current at Motor Circuit= 7,500,000 ¸ [Ö3) x 600 ] = 7,217 amperes.

Actual Motor Current = (746 x 5,000) ¸ [0.88 x (Ö3) x 600 x 0.85 ] = 4,798 amperes.

Prefault current drawn by motor = (4,798 ¸ 7,217) /-cos¹ 85 = 0.665/-31.8°
Prefault current drawn by motor = 0.565 -j0.35 p.u.

If prefault current is neglected, E"g and E"m are equal to 1.0/0°, the Thé equivalent of the positive-sequence network is:

I_A1 = V_f ¸ [ Z₁ + Z₂ + Z₀ ] = 1 + j0 ¸ [ j0.12 + j0.12 + j0.15 ] = -j2.564 p.u.

I_A1 = I_A2 = I_A0 = -j2.564 p.u.

Current in the fault: I_f = 3I_A0 = 3(-j2.564 p.u.) = -j7.692 p.u.

Current I_A1
From Transformer to Point P: (-j2.564 x j0.30) ¸ j0.50 = -j1.538 p.u.
From Motor to Point P: (-j2.564 x j0.20) ¸ j0.50 = -j1.026 p.u.

Current I_A2
From Transformer to Point P: (-j2.564 x j0.30) ¸ j0.50 = -j1.538 p.u.
From Motor to Point P: (-j2.564 x j0.20) ¸ j0.50 = -j1.026 p.u.

Currents in the lines at the fault
From Transformer to Point P:

From Motor to Point P:

But at Transformer:
I_a1 = jI_A1I_a2 = -jI_A2
thus:
I_A1 = -jI_a1 = -j(-j1.538) = -1.538 p.u.
I_A2 = jI_a2 = j(-j1.538_ = 1.538 p.u.
I_A0 = 0 p.u.

I_A = I_A1 + I_A2 = 0 p.u.
I_B = a²I_A1 + aI_A2 = (-1.538/240°) + (-1.538/120°)
I_B = (0.769 +j1.332) + (-0.769 +j1.332) = j2.664 p.u.
I_C = aI_A1 + a²I_A2 = (-1.538/120°) + (-1.538/240°)
I_C = (0.769 -j1.332) + (-0.769 -j1.332) = -j2.664 p.u.

Base Current at High Tension Side = 7,500 kVA ¸ [Ö3) x 4.16 kV] = 1,041 amperes.

Current in the Fault: I_f = 3I_A0 = 7,692 X 7,217 = 55,500 Amperes.

Current in line between Transformer and the Fault:
I_a = 3.076 X 7,217 = 22,200 Amperes.
I_b = 1.538 X 7,217 = 11,100 Amperes.
I_c = 1.538 X 7,217 = 11,100 Amperes.

Current between Motor and the Fault:
I_A = 4.616 X 7,217 = 33,300 Amperes.
I_B = 1.538 X 7,217 = 11,100 Amperes.
I_C = 1.538 X 7,217 = 11,100 Amperes.

Current between 4.16 kV Bus and Transformer:
I_A = 0 Ampere
I_B = 2.664 X 1,041 = 2,773 Amperes.
I_C = 2.664 X 1,041 = 2,773 Amperes.

Sections: Conditions | Network | Example | Problem Set

Problems

Solve the following problems
1. A 60-hz turbogenerator is rated 500MVA, 22kV. It is Y-connected and solidly grounded and is operating at rated voltage at no load. It is disconnected from the rest of the system. Its reactances are X" = X₂ = 0.15 and X₀ = 0.05 per unit. Find the ratio of the subtransient line current for a single line-to-ground fault to the subtransient line current for a symmetrical three-phase fault.

2. Determine the ohms of inductive reactance to be inserted in the neutral connection of the Generator in Problem no. 1 to limit the subtransient line current for a single line-to-ground fault to that for a three-phase fault.

3. How many ohms of resistance in the neutral connection of the generator in Problem No. 1 would limit the subtransient line current for a single line-to-ground fault to that for a three-phase fault?

4. A generator rated 100 MVA, 20 kV has X" = X₂ = 20% and X₀ = 5 %. Its neutral is grounded through a reactor of 0.32 W. The generator is operating at rated voltage without load and is disconnected from the system when a single line-to-gound fault occurs at its terminals. Find the subtransient current in the faulted phase.