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3) A 200MW generator is connected to a 138kV system with two transmission lines as shown below. With the strongest line out of service the 138kV system thevenin impedance is j0.20 per unit on a 100MVA base. Following closing into a close-in three-phase fault on the strongest line, determine the critical clearing time for this fault condition. (20 pts.) To be conservative assume the system equivalent voltage is at 1.0 / 0 deg. Vpu. Assume the power flow is 200MW and 0Mvar at the equivalent system voltage. GSU: 20kV to 138kV, Impedance = 9% @ 200MVA Generator: 20KV, 200MW, X'd = 20% @ 200MVA, H=4.0 @ 200MVA

          3) A 200MW generator is connected to a 138kV system with two transmission lines
as shown below. With the strongest line out of service the 138kV system
thevenin impedance is j0.20 per unit on a 100MVA base. Following closing into
a close-in three-phase fault on the strongest line, determine the critical clearing
time for this fault condition. (20 pts.)
To be conservative assume the system equivalent voltage is at 1.0 / 0 deg. Vpu.
Assume the power flow is 200MW and 0Mvar at the equivalent system voltage.
GSU: 20kV to 138kV, Impedance = 9% @ 200MVA
Generator:
20KV, 200MW, X'd = 20% @ 200MVA, H=4.0 @ 200MVA

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3) A 200MW generator is connected to a 138kV system with two transmission lines
as shown below. With the strongest line out of service the 138kV system
thevenin impedance is j0.20 per unit on a 100MVA base. Following closing into
a close-in three-phase fault on the strongest line, determine the critical clearing
time for this fault condition. (20 pts.)
To be conservative assume the system equivalent voltage is at 1.0 / 0 deg. Vpu.
Assume the power flow is 200MW and 0Mvar at the equivalent system voltage.
GSU: 20kV to 138kV, Impedance = 9% @ 200MVA
Generator:
20KV, 200MW, X'd = 20% @ 200MVA, H=4.0 @ 200MVA

Added by Kathy S.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Yatu Khandelwal

Step 1

First, let's calculate the Thevenin impedance of the 138 kV system. We are given that the Thevenin impedance is j0.20 per unit on a 100 MVA base. Since the base is 100 MVA, we can convert the impedance to the actual value in ohms by multiplying it by the base MVA. Show more…

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A 200 MW generator is connected to a 138 kV system with two transmission lines as shown below. With the strongest line out of service, the 138 kV system Thevenin impedance is j0.20 per unit on a 100 MVA base. Following closing into a close-in three-phase fault on the strongest line, determine the critical clearing time for this fault condition. (20 pts.) To be conservative, assume the system equivalent voltage is at 1.0 / 0 deg. Vpu. Assume the power flow is 200 MW and OMvar at the equivalent system voltage. GSU: 20 kV to 138 kV, Impedance = 9% @ 200 MVA Generator: 20 kV, 200 MW, Xd = 20% @ 200 MVA, H = 4.0 @ 200 MVA GSU

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