EEE 470/591: Electric Power Devices

Homework #4
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Problem 1 (25 Points)

In Figure 1, V = 277 volts, L = 2 mH, R = 0.4 Ω, and system frequency = 60 Hz. Determine (a) the RMS
symmetrical fault current. (b) The RMS asymmetrical fault current at the instant the switch closes,
assuming maximum dc offset. (c) The RMS asymmetrical fault current five cycles after the switch closes,
assuming maximum dc offset. (d) The dc offset as a function of time if the switch closes when the
instantaneous source voltage is 300 volts.

Figure 1. Circuit for Problem 1

Solution to Problem 1
(a)
Absolute value of complex impedance:
𝑍 = √𝑅2 + (𝜔𝐿)2 = √0.42 + (2𝜋 × 60 × 2 × 10−3 )2 = 0.8535 Ω

Apply Equation (7.1.9), the RMS symmetrical (ac) fault current is:
𝑉 𝑉 277
𝐼𝑎𝑐 = = = = 324.54 𝐴
𝑍 √𝑅2 +(𝜔𝐿)2 0.8535

(b)
For the above circuit, time constant T is:
𝐿 2×10−3
𝑇=𝑅= = 0.005 𝑠𝑒𝑐
0.4

When the switch closes, the time instant t = 0 sec.

Apply Equation (7.1.10), the RMS asymmetrical fault current at t = 0 sec, assuming maximum dc offset:
𝐼𝑟𝑚𝑠 (𝑡 = 0) = 𝐼𝑎𝑐 √1 + 2𝑒 −2𝑡/𝑇 = 324.54 × √1 + 2𝑒 −2×0/0.005 = 324.54√3 = 562.1198 𝐴
(c)
5 𝑐𝑦𝑐𝑙𝑒𝑠 5 𝑐𝑦𝑐𝑙𝑒𝑠
For five cycles after the switch closes, the time instant 𝑡 = = = 0.0833 𝑠𝑒𝑐.
𝑓 60 𝐻𝑧

Apply Equation (7.1.10), the RMS asymmetrical fault current at t = 0.0833 sec, assuming maximum dc
offset is:
𝐼𝑟𝑚𝑠 (𝑡 = 0.0833) = 𝐼𝑎𝑐 √1 + 2𝑒 −2𝑡/𝑇 = 324.54 × √1 + 2𝑒 −2×0.0833/0.005 = 324.54 𝐴

(d)
𝜔𝐿 2𝜋×60×2×10−3
Apply Equation (7.1.6) and obtain angle 𝜃: 𝜃 = tan−1 = tan−1 = 62.0533°
𝑅 0.4

The switch opens (at time t = 0 sec) when the instantaneous source voltage is 300 V ➔
Instantaneous source voltage 𝑒(𝑡 = 0) = √2𝑉𝑠𝑖𝑛(𝜔𝑡 + 𝛼 ) = √2 × 277 × sin(2𝜋 × 60 × 0 + 𝛼 ) =
300
300 ➔ 𝛼 = sin−1 ( ) = 49.98°
√2×277

Apply Equation (7.1.4), dc offset as a function of time under this situation is:
𝑡 𝑡
√2𝑉 √2×277
𝑖𝑑𝑐 (𝑡) = − sin(𝛼 − 𝜃) 𝑒 −𝑇 = − sin(49.98° − 62.0533°) 𝑒 −0.005 = 95.98𝑒 −𝑡/0.005
𝑍 0.8535

Problem 2 (25 Points)

A 1000-MVA, 20-kV, 60-Hz, three-phase generator is connected through a 1000-MVA, 20-kV, Delta-Y
transformer to a 345-kV circuit breaker and a 345-kV transmission line. The transformer rated voltage on
the high side (Y side) is 345 kV. The generator reactances are 𝑋𝑑′′ = 0.17 𝑝𝑢, 𝑋𝑑′ = 0.30 𝑝𝑢, 𝑋𝑑 = 1.5 𝑝𝑢,
and its time constants are 𝑇𝑑′′ = 0.05 𝑠, 𝑇𝑑′ = 1.0 𝑠, 𝑇𝐴 = 1.10 𝑠. The transformer series reactance is 0.1
pu; transformer losses and exciting current are neglected. A three-phase short-circuit occurs on the line
side of the circuit breaker when the generator is operated at rated terminal voltage and at no-load. The
breaker interrupts the fault three cycles after fault inception. Determine (a) the subtransient current through
the breaker in per unit and in kA RMS. (b) The RMS asymmetrical fault current the breaker interrupts,
assuming maximum dc offset. Neglect the effect of the transformer on the time constants.

Solution to Problem 2
(a)
3𝜙 𝐺𝑒𝑛
Select three-phase power base 𝑆𝐵 = 1000 𝑀𝑉𝐴, line-to-line voltage base for the generator 𝑉𝐵−𝐿𝐿 =
20𝑘𝑉.

Under the above base selection, the generator is operated at rated terminal voltage ➔ the per unit RMS
generator terminal voltage 𝐸𝑔 = 1.0 𝑝𝑢.

′′ 𝐸𝑔 1.0
Subtransient current in pu: 𝐼𝑝𝑢 = ′′ = = 3.704 𝑝𝑢
𝑋𝑑 +𝑋𝑥𝑓𝑚𝑟 0.17+0.1
Line-to-line voltage base for the transmission line section (or high-side of the transformer):
𝐿𝑖𝑛𝑒 𝑋𝑓𝑚𝑟 𝐿𝑖𝑛𝑒
𝑉𝐵−𝐿𝐿 𝑉𝑅𝑎𝑡𝑒𝑑−𝐻𝑖𝑔ℎ 𝑉𝐵−𝐿𝐿 345 𝐿𝑖𝑛𝑒
𝐺𝑒𝑛 = 𝑋𝑓𝑚𝑟 ➔ = ➔ 𝑉𝐵−𝐿𝐿 = 345 𝑘𝑉
𝑉𝐵−𝐿𝐿 𝑉𝑅𝑎𝑡𝑒𝑑−𝐿𝑜𝑤 20 20

Line current base for the transmission line section:

3𝜙
𝑆𝐵 1000
𝐼𝐵𝐿𝑖𝑛𝑒 = 𝐿𝑖𝑛𝑒 = = 1.673 𝑘𝐴
√3×𝑉𝐵−𝐿𝐿 √3×345

Subtransient current in kA: 𝐼 ′′ = 𝐼𝑝𝑢

′′
× 𝐼𝐵𝐿𝑖𝑛𝑒 = 3.704 × 1.673 = 6.198 𝑘𝐴

(b)
The breaker interrupts the fault three cycles after fault inception ➔ When the breaker interrupts, time
3 𝑐𝑦𝑐𝑙𝑒𝑠 3 𝑐𝑦𝑐𝑙𝑒𝑠
instant t = = = 0.05 𝑠𝑒𝑐
𝑓 60 𝐻𝑧

RMS ac (symmetrical) fault current when breaker interrupts (at t = 0.05 sec):
1 1 ′′ 1 1 ′ 1
𝐼𝑎𝑐 (𝑡 = 0.05) = 𝐸𝑔 [(𝑋 ′′ +𝑋 − 𝑋 ′ +𝑋 ) 𝑒 −𝑡/𝑇𝑑 + (𝑋 ′ +𝑋 −𝑋 ) 𝑒 −𝑡/𝑇𝑑 + 𝑋 ]=
𝑑 𝑥𝑓𝑚𝑟 𝑑 𝑥𝑓𝑚𝑟 𝑑 𝑥𝑓𝑚𝑟 𝑑 +𝑋𝑥𝑓𝑚𝑟 𝑑 +𝑋𝑥𝑓𝑚𝑟
1 1 1 1 1
1.0 × [(0.17+0.1 − 0.3+0.1) 𝑒 −0.05/0.05 + (0.3+0.1 − 1.5+0.1) 𝑒 −0.05/1.0 + 1.5+0.1] = 2.851 pu

Apply Equation (7.2.5), the maximum dc offset:

√2𝐸𝑔 √2×1.0
𝑖𝑑𝑐 (𝑡 = 0.05) = 𝑋 ′′ +𝑋 𝑒 −𝑡/𝑇𝐴 = 0.17+0.1 𝑒 −0.05/1.10 = 5.0051 pu
𝑑 𝑥𝑓𝑚𝑟

The RMS asymmetrical fault current in pu:

𝐼𝑟𝑚𝑠 (𝑡 = 0.05) = √𝐼𝑎𝑐 (𝑡)2 + 𝑖𝑑𝑐 (𝑡)2 = √𝐼𝑎𝑐 (𝑡 = 0.05)2 + 𝑖𝑑𝑐 (𝑡 = 0.05)2 = √2.8512 + 5.00512 =
5.7601 𝑝𝑢

The RMS asymmetrical fault current in kA = 5.7601 × 𝐼𝐵𝐿𝑖𝑛𝑒 = 5.7601 × 1.673 = 9.6367 𝑘𝐴

Problem 3 (25 Points)

For Problem 2, determine (a) the instantaneous symmetrical fault current in kA in phase A of the generator
as a function of time, assuming maximum dc offset occurs in this generator phase. (b) The maximum dc
offset current in kA as a function of time that can occur in any one generator phase.

Solution to Problem 3
(a)
The maximum dc offset in any phase is obtained when 𝛼 = 0 (Equatin 7.2.5).
Apply Equation (7.2.1), the per-unit instantaneous symmetrical (ac) fault current at maximum dc offset
(𝛼 = 0) in phase A of the generator, as a function of time t:
𝑝𝑢 1 1 ′′ 1 1 ′ 1
𝑖𝑎𝑐 (𝑡) = √2𝐸𝑔 [( − ) 𝑒 −𝑡/𝑇𝑑 + ( − ) 𝑒 −𝑡/𝑇𝑑 + ] sin (𝜔𝑡 +
𝑋𝑑′′ +𝑋𝑥𝑓𝑚𝑟 𝑋𝑑′ +𝑋𝑥𝑓𝑚𝑟 𝑋𝑑′ +𝑋𝑥𝑓𝑚𝑟 𝑋𝑑 +𝑋𝑥𝑓𝑚𝑟 𝑋𝑑 +𝑋𝑥𝑓𝑚𝑟
𝜋 1 1 1 1 1
𝛼 − 2 ) = √2 × 1.0 × [(0.17+0.1 − 0.3+0.1) 𝑒 −𝑡/0.05 + (0.3+0.1 − 1.5+0.1) 𝑒 −𝑡/1.0 + 1.5+0.1] sin (2𝜋 × 60 ×
𝑡 𝑡
𝜋 𝜋
𝑡 + 0 − 2 ) = √2[1.204𝑒 −0.05 + 1.875𝑒 −1.0 + 0.625] sin (377𝑡 − 2 ) pu

Current base for the generator section:

3𝜙
𝑆𝐵 1000
𝐼𝐵𝐺𝑒𝑛 = 𝐺𝑒𝑛 = = 28.87 𝑘𝐴
√3×𝑉𝐵−𝐿𝐿 √3×20

The instantaneous symmetrical (ac) fault current at maximum dc offset in phase A of the generator, in
kA:
𝑡 𝑡
𝑝𝑢 𝜋
𝑖𝑎𝑐 (𝑡) = 𝑖𝑎𝑐 (𝑡) × 𝐼𝐵𝐺𝑒𝑛 = 28.87 × √2 [1.204𝑒 −0.05 + 1.875𝑒 −1.0 + 0.625] sin (377𝑡 − 2 ) =
𝑡 𝑡
𝜋
40.83 [1.204𝑒 −0.05 + 1.875𝑒 −1.0 + 0.625] sin (377𝑡 − 2 ) kA

(b)
Apply Equation (7.2.5), maximum dc offset in any generator phase (in per unit), as a function of time t:
𝑝𝑢 √2𝐸𝑔 √2×1.0
𝑖𝑑𝑐 (𝑡) = 𝑋 ′′ +𝑋 𝑒 −𝑡/𝑇𝐴 = 0.17+0.1 𝑒 −𝑡/1.10 = 5.2382𝑒 −𝑡/1.10 pu
𝑑 𝑥𝑓𝑚𝑟

Maximum dc offset in any one generator phase (in kA):

𝑝𝑢
𝑖𝑑𝑐 (𝑡) = 𝑖𝑑𝑐 (𝑡) × 𝐼𝐵𝐺𝑒𝑛 = 28.87 × 5.2382𝑒 −𝑡/1.10 = 151.2𝑒 −𝑡/1.10 𝑘𝐴

Problem 4 (25 Points)

Equipment ratings for the four-bus system shown in Figure 2 are as follows:

Generator G1: 500 MVA, 13.8 kV, 𝑋′′ = 0.20 pu.

Generator G2: 750 MVA, 18 kV, 𝑋′′ = 0.18 pu.
Generator G3: 1000 MVA, 20 kV, 𝑋′′ = 0.17 pu.
Three-phase Delta-Y transformer T1: 500 MVA, 13.8(Δ)/500(Y) kV, X = 0.12 pu.
Three-phase Delta-Y transformer T2: 750 MVA, 18(Δ)/500(Y) kV, X = 0.10 pu.
Three-phase Delta-Y transformer T3: 1000 MVA, 20(Δ)/500(Y) kV, X = 0.10 pu.
Each 500kV line: 𝑋1 = 50 Ω.
A three phase short circuit occurs at bus 1, where the prefault voltage is 525 kV. Prefault load current is
neglected. Draw the positive-sequence reactance diagram in per unit on a 1000-MVA, 20-kV base in the
zone of generator G3. Determine (a) the subtransient fault current in per unit and in kA RMS. (b)
Contributions to the fault current from generator G1 and from line 1-2 in per unit. For both (a) and (b),
perfault currents are neglected.

Figure 2. Oneline diagram for Problem 4.

Solution to Problem 4 (Method 1 – Superposition)

3𝜙 𝐺3
Select three-phase power base 𝑆𝐵 = 1000 𝑀𝑉𝐴, line-to-line-voltage base for generator G3 𝑉𝐵−𝐿𝐿 =
20𝑘𝑉.

𝑇𝑟𝑎𝑛𝑠
Line-to-line voltage base of transmission system 𝑉𝐵−𝐿𝐿 :
20 𝑉 𝐺3
𝐵−𝐿𝐿 20 𝑇𝑟𝑎𝑛𝑠
= 𝑉 𝑇𝑟𝑎𝑛𝑠 = 𝑉 𝑇𝑟𝑎𝑛𝑠 => 𝑉𝐵−𝐿𝐿 = 500 𝑘𝑉
500 𝐵−𝐿𝐿 𝐵−𝐿𝐿

Line-to-line voltage base of G1:

20 𝑉 𝐺1
𝐵−𝐿𝐿 13.8 𝐺1
= 𝑉 𝑇𝑟𝑎𝑛𝑠 = => 𝑉𝐵−𝐿𝐿 = 13.8 𝑘𝑉
500 𝐵−𝐿𝐿 500

Impedance base for transmission system 𝑍𝐵𝑇𝑟𝑎𝑛𝑠 :

𝑇𝑟𝑎𝑛𝑠 2
𝑉𝐵−𝐿𝐿 5002
𝑍𝐵𝑇𝑟𝑎𝑛𝑠 = 3𝜙 = 1000 = 250 Ω
𝑆𝐵

Per unit reactance of each transmission line:

𝑝𝑢 1 𝑋 50
𝑋1 = 𝑍 𝑇𝑟𝑎𝑛𝑠 = 250 = 0.2 𝑝𝑢
𝐵

Per unit reactance of G1 in system base:

2 3𝜙
′′ ′′
𝐺1−𝑜𝑙𝑑
𝑉𝐵−𝐿𝐿 𝑆𝐵 13.8 2 1000
𝑋𝐺1−𝑝𝑢 = 𝑋𝐺1 ×( 𝐺1 ) ×( 3𝜙−𝐺1−𝑜𝑙𝑑 ) = 0.20 × (13.8) × ( 500 ) = 0.40 𝑝𝑢
𝑉𝐵−𝐿𝐿 𝑆𝐵

Per unit reactance of T1 in system base:

𝑇1−𝐻𝑖𝑔ℎ−𝑜𝑙𝑑 2 3𝜙
′′ ′′ 𝑉𝐵−𝐿𝐿 𝑆𝐵 500 2 1000
𝑋𝑇1−𝑝𝑢 = 𝑋𝑇1 ×( 𝑇𝑟𝑎𝑛𝑠 ) ×( 3𝜙−𝑇1−𝑜𝑙𝑑 ) = 0.12 × (500) × ( 500 ) = 0.24 𝑝𝑢
𝑉𝐵−𝐿𝐿 𝑆𝐵
Per unit reactance of G2 in system base:
2 3𝜙
′′ ′′
𝐺2−𝑜𝑙𝑑
𝑉𝐵−𝐿𝐿 𝑆𝐵 18 2 1000
𝑋𝐺2−𝑝𝑢 = 𝑋𝐺2 ×( 𝐺2 ) ×( 3𝜙−𝐺2−𝑜𝑙𝑑 ) = 0.18 × ( ) × ( ) = 0.24 𝑝𝑢
𝑉𝐵−𝐿𝐿 𝑆𝐵 18 750
Per unit reactance of T2 in system base:
𝑇2−𝐻𝑖𝑔ℎ−𝑜𝑙𝑑 2 3𝜙
′′ ′′ 𝑉𝐵−𝐿𝐿 𝑆𝐵 500 2 1000
𝑋𝑇2−𝑝𝑢 = 𝑋𝑇2 ×( 𝑇𝑟𝑎𝑛𝑠 ) ×( 3𝜙−𝑇2−𝑜𝑙𝑑 ) = 0.10 × (500) × ( 750 ) = 0.1333 𝑝𝑢
𝑉𝐵−𝐿𝐿 𝑆𝐵

Per unit reactance of G3 in system base:

2 3𝜙
′′ ′′
𝐺3−𝑜𝑙𝑑
𝑉𝐵−𝐿𝐿 𝑆𝐵 20 2 1000
𝑋𝐺3−𝑝𝑢 = 𝑋𝐺3 ×( 𝐺3 ) ×( 3𝜙−𝐺3−𝑜𝑙𝑑 ) = 0.17 × (20) × (1000 ) = 0.17 𝑝𝑢
𝑉𝐵−𝐿𝐿 𝑆𝐵

Per unit reactance of T3 in system base:

𝑇3−𝐻𝑖𝑔ℎ−𝑜𝑙𝑑 2 3𝜙
′′ ′′ 𝑉𝐵−𝐿𝐿 𝑆𝐵 500 2 1000
𝑋𝑇3−𝑝𝑢 = 𝑋𝑇3 ×( 𝑇𝑟𝑎𝑛𝑠 ) ×( 3𝜙−𝑇3−𝑜𝑙𝑑 ) = 0.10 × (500) × (1000 ) = 0.10 𝑝𝑢
𝑉𝐵−𝐿𝐿 𝑆𝐵

The per unit reactance diagram is shown below (fault is applied at bus 1):

Apply Superposition method, the above circuit = Circuit 1 + Circuit 2 (when fault is applied at bus 1).

Circuit 1 (post-fault circuit with no pre-fault voltage sources):

Circuit 2 (pre-fault circuit):

In Circuit 1, pre-fault voltage at fault location (bus 1):

𝑝𝑢 𝑉𝐹 525∠0°
𝑉𝐹 = 525∠0° 𝑘𝑉 ➔ 𝑉𝐹 = 𝑇𝑟𝑎𝑛𝑠 = = 1.05∠0° 𝑝𝑢
𝑉𝐵−𝐿𝐿 500
In Circuit 1:
′′ 𝑉𝐹 1.05∠0°
𝐼𝐹−𝐺1 = = = −𝑗1.64 pu
𝑗0.40+𝑗0.24 𝑗0.40+𝑗0.40

′′ 𝐹 𝑉 1.05∠0° 1.05∠0°
𝐼𝐹−𝐿𝑖𝑛𝑒12 = 𝑗0.2+[(𝑗0.20+𝑗0.1333+𝑗0.24)//(𝑗0.20+𝑗0.10+𝑗0.17)] = 𝑗0.20+[𝑗0.5733//0.47] = 𝑗0.20+𝑗0.2583 =
−𝑗2.29 𝑝𝑢

(a)
Subtransient fault current in per unit RMS, neglecting pre-fault current (𝐼𝐹′′ in Circuit 1 in per unit):
𝐼𝐹′′ = 𝐼𝐹−𝐺1
′′ ′′
+ 𝐼𝐹−𝐿𝑖𝑛𝑒12 = −𝑗1.64 − 𝑗2.29 = −𝑗3.93 𝑝𝑢

Current base in transmission system:

3𝜙
𝑇𝑟𝑎𝑛𝑠 𝑆𝐵 1000
𝐼𝐵𝑎𝑠𝑒 = 𝑇𝑟𝑎𝑛𝑠 = = 1.1547 𝑘𝐴
√3 𝑉𝐵−𝐿𝐿 √3×500

Subtransient fault current in kA RMS, neglecting pre-fault current (𝐼𝐹′′ in Circuit 1 in kA):
′′
𝐼𝐹−𝑘𝐴 = 𝐼𝐹′′ × 𝐼𝐵𝑎𝑠𝑒
𝑇𝑟𝑎𝑛𝑠
= (−𝑗3.93) × 1.1547 = −𝑗4.54 𝑘𝐴

(b)
Contributions to the fault current from generator G1 (in per unit), neglecting per-fault currents:
′′
𝐼𝐹−𝐺1 = −𝑗1.64 pu

Contributions to the fault current from Line 1-2 (in per unit), neglecting per-fault currents:
′′
𝐼𝐹−𝐿𝑖𝑛𝑒12 = −𝑗2.29 pu

Solution to Problem 4 (Method 2 – Zbus Method – This is an alternative method for this problem)
3𝜙 𝐺3
Select three-phase power base 𝑆𝐵 = 1000 𝑀𝑉𝐴, line-to-line-voltage base for generator G3 𝑉𝐵−𝐿𝐿 =
20𝑘𝑉.

𝑇𝑟𝑎𝑛𝑠
Line-to-line voltage base of transmission system 𝑉𝐵−𝐿𝐿 :
20 𝑉 𝐺3
𝐵−𝐿𝐿 20 𝑇𝑟𝑎𝑛𝑠
= 𝑉 𝑇𝑟𝑎𝑛𝑠 = 𝑉 𝑇𝑟𝑎𝑛𝑠 => 𝑉𝐵−𝐿𝐿 = 500 𝑘𝑉
500 𝐵−𝐿𝐿 𝐵−𝐿𝐿

Line-to-line voltage base of G1:

20 𝑉 𝐺1
𝐵−𝐿𝐿 13.8 𝐺1
= 𝑉 𝑇𝑟𝑎𝑛𝑠 = => 𝑉𝐵−𝐿𝐿 = 13.8 𝑘𝑉
500 𝐵−𝐿𝐿 500

Impedance base for transmission system 𝑍𝐵𝑇𝑟𝑎𝑛𝑠 :

 𝑇𝑟𝑎𝑛𝑠 2
𝑉𝐵−𝐿𝐿 5002
𝑍𝐵𝑇𝑟𝑎𝑛𝑠 = 3𝜙 = 1000 = 250 Ω
𝑆𝐵

Per unit reactance of each transmission line:

𝑝𝑢 𝑋
1 50
𝑋1 = 𝑍 𝑇𝑟𝑎𝑛𝑠 = 250 = 0.2 𝑝𝑢
𝐵

Per unit reactance of G1 in system base:

2 3𝜙
′′ ′′
𝐺1−𝑜𝑙𝑑
𝑉𝐵−𝐿𝐿 𝑆𝐵 13.8 2 1000
𝑋𝐺1−𝑝𝑢 = 𝑋𝐺1 ×( 𝐺1 ) ×( 3𝜙−𝐺1−𝑜𝑙𝑑 ) = 0.20 × (13.8) × ( 500 ) = 0.40 𝑝𝑢
𝑉𝐵−𝐿𝐿 𝑆𝐵

Per unit reactance of T1 in system base:

𝑇1−𝐻𝑖𝑔ℎ−𝑜𝑙𝑑 2 3𝜙
′′ ′′ 𝑉𝐵−𝐿𝐿 𝑆𝐵 500 2 1000
𝑋𝑇1−𝑝𝑢 = 𝑋𝑇1 ×( 𝑇𝑟𝑎𝑛𝑠
𝑉𝐵−𝐿𝐿
) ×( 3𝜙−𝑇1−𝑜𝑙𝑑 ) = 0.12 × (500) × ( 500 ) = 0.24 𝑝𝑢
𝑆𝐵

Per unit reactance of G2 in system base:

2 3𝜙
′′ ′′
𝐺2−𝑜𝑙𝑑
𝑉𝐵−𝐿𝐿 𝑆𝐵 18 2 1000
𝑋𝐺2−𝑝𝑢 = 𝑋𝐺2 ×( 𝐺2 ) ×( 3𝜙−𝐺2−𝑜𝑙𝑑 ) = 0.18 × (18) × ( 750 ) = 0.24 𝑝𝑢
𝑉𝐵−𝐿𝐿 𝑆𝐵

Per unit reactance of T2 in system base:

𝑇2−𝐻𝑖𝑔ℎ−𝑜𝑙𝑑 2 3𝜙
′′ ′′ 𝑉𝐵−𝐿𝐿 𝑆𝐵 500 2 1000
𝑋𝑇2−𝑝𝑢 = 𝑋𝑇2 ×( 𝑇𝑟𝑎𝑛𝑠 ) ×( 3𝜙−𝑇2−𝑜𝑙𝑑 ) = 0.10 × (500) × ( 750 ) = 0.1333 𝑝𝑢
𝑉𝐵−𝐿𝐿 𝑆𝐵

Per unit reactance of G3 in system base:

2 3𝜙
′′ ′′
𝐺3−𝑜𝑙𝑑
𝑉𝐵−𝐿𝐿 𝑆𝐵 20 2 1000
𝑋𝐺3−𝑝𝑢 = 𝑋𝐺3 ×( 𝐺3 ) ×( 3𝜙−𝐺3−𝑜𝑙𝑑 ) = 0.17 × (20) × (1000 ) = 0.17 𝑝𝑢
𝑉𝐵−𝐿𝐿 𝑆𝐵

Per unit reactance of T3 in system base:

𝑇3−𝐻𝑖𝑔ℎ−𝑜𝑙𝑑 2 3𝜙
′′ ′′ 𝑉𝐵−𝐿𝐿 𝑆𝐵 500 2 1000
𝑋𝑇3−𝑝𝑢 = 𝑋𝑇3 ×( 𝑇𝑟𝑎𝑛𝑠 ) ×( 3𝜙−𝑇3−𝑜𝑙𝑑 ) = 0.10 × (500) × (1000 ) = 0.10 𝑝𝑢
𝑉𝐵−𝐿𝐿 𝑆𝐵

Pre-fault voltage at fault location (bus 1):

𝑝𝑢 𝑉 525∠0°
𝑉𝐹 = 525∠0° 𝑘𝑉 ➔ 𝑉𝐹 𝐹
= 𝑉 𝑇𝑟𝑎𝑛𝑠 = = 1.05∠0° 𝑝𝑢
𝐵−𝐿𝐿 500

The per unit reactance diagram is shown below (fault is applied at bus 1):
Obtain Ybus for the above 4-bus system:
1 1 1 1 1
𝑌11 = (𝑗0.4+𝑗0.24 + 𝑗0.2) = -j6.5625 pu, 𝑌22 = (𝑗0.2 + 𝑗0.2 + 𝑗0.2) = −𝑗15 𝑝𝑢
1 1 1 1
𝑌33 = (𝑗0.2 + 𝑗0.1+𝑗0.17) = −𝑗8.7037 𝑝𝑢, 𝑌44 = (𝑗0.2 + 𝑗0.1333+𝑗0.24) = −𝑗7.6788 𝑝𝑢
1 1
𝑌12 = 𝑌21 = − 𝑗0.2 = 𝑗5 pu, 𝑌13 = 𝑌31 = 0, 𝑌14 = 𝑌41 = 0, 𝑌23 = 𝑌32 = − 𝑗0.2 = 𝑗5 𝑝𝑢,
1
𝑌24 = 𝑌42 = − 𝑗0.2 = 𝑗5 pu, 𝑌34 = 𝑌43 = 0.
𝑌11 𝑌12 𝑌13 𝑌14 −j6.5625 𝑗5 0 0
𝑌21 𝑌22 𝑌23 𝑌24 𝑗5 −𝑗15 𝑗5 𝑗5
𝑌𝑏𝑢𝑠 =[ ]=[ ]
𝑌31 𝑌32 𝑌33 𝑌34 0 𝑗5 −𝑗8.7037 0
𝑌41 𝑌42 𝑌43 𝑌44 0 𝑗5 0 −𝑗7.6788

Obtain Zbus for the above system:

𝑗0.2670 𝑗0.1505 𝑗0.0865 𝑗0.0980
−1 𝑗0.1505 𝑗0.1975 𝑗0.1135 𝑗0.1286
𝑍𝑏𝑢𝑠 = 𝑌𝑏𝑢𝑠 =[ ]
𝑗0.0865 𝑗0.1135 𝑗0.1801 𝑗0.0739
𝑗0.0980 𝑗0.1286 𝑗0.0739 𝑗0.2140
(a)
When fault is applied at bus 1, subtransient fault current in per unit (𝐼𝐹′′ ):
𝑉𝐹 1.05∠0°
𝐼𝐹′′ = = = −𝑗3.93 pu
𝑍11 𝑗0.2670

Current base in transmission system:

3𝜙
𝑇𝑟𝑎𝑛𝑠 𝑆𝐵 1000
𝐼𝐵𝑎𝑠𝑒 = 𝑇𝑟𝑎𝑛𝑠 = = 1.1547 𝑘𝐴
√3 𝑉𝐵−𝐿𝐿 √3×500

Subtransient fault current in kA (𝐼𝐹′′ in Circuit 1 in kA):

′′
𝐼𝐹−𝑘𝐴 = 𝐼𝐹′′ × 𝐼𝐵𝑎𝑠𝑒
𝑇𝑟𝑎𝑛𝑠
= (−𝑗3.93) × 1.1547 = −𝑗4.54 𝑘𝐴

(b)
Post-fault voltage at bus 1 (neglecting pre-fault currents):
𝑍
𝐸1 = (1 − 𝑍11 ) 𝑉𝐹 = (1 − 1) × 1.05∠0° = 0 pu
11

Post-fault voltage at bus 2 (neglecting pre-fault currents):

𝑍 𝑗0.1505
𝐸2 = (1 − 𝑍21 ) 𝑉𝐹 = (1 − 𝑗0.2670) × 1.05∠0° = 0.4581∠0° pu
11

Contributions to the fault current from Line 1-2 (in per unit), neglecting per-fault currents:
′′ 𝐸2 −𝐸1 0.4581∠0°−0
𝐼𝐹−𝐿𝑖𝑛𝑒12 = = = −𝑗2.29 𝑝𝑢
𝑧12 𝑗0.2

Contributions to the fault current from generator G1 (in per unit), neglecting per-fault currents:
′′
𝐼𝐹−𝐺1 = 𝐼𝐹′′ − 𝐼𝐹−𝐿𝑖𝑛𝑒12
′′
= −𝑗3.93 − (−𝑗2.29) = - j1.64 pu