A cylindrical rotor synchronous generator has steady state synchronous reactance of 0.7 pu and subtransient reactance of 0.2 pu. It is operating at (1+ j0) pu terminal voltage with an internal emf of (1 + j0.7) pu. Following a three-phase solid short circuit fault at the terminal of the generator, the magnitude of the subtransient internal emf (rounded off to 2 decimal places) is ________ pu.Correct answer is '1.02'. Can you explain this answer?

Question

Answer

Given:

Steady state synchronous reactance (Xs) = 0.7 pu

Subtransient reactance (X'd) = 0.2 pu

Terminal voltage = 1 + j0 pu

Internal emf = 1 + j0.7 pu

Calculation:

During the fault, the fault current causes a sudden drop in the terminal voltage of the generator, leading to a sudden change in the internal emf.

The subtransient internal emf is given by E' = E - jX'dI, where E is the internal emf before the fault and I is the fault current.

The fault current can be calculated using the Thevenin equivalent of the generator. The Thevenin voltage is equal to the terminal voltage, and the Thevenin impedance is equal to the synchronous reactance Xs.

The fault current is given by I = E_th/Xs, where E_th is the Thevenin voltage.

Substituting the given values, we get I = (1 + j0)/(0.7) = 1.429 - j0.714 pu.

Substituting the values of E, X'd, and I, we get E' = (1 + j0.7) - j(0.2)(1.429 - j0.714) = 1.02 + j0.226 pu.

Rounding off to two decimal places, we get E' = 1.02 pu.

Therefore, the magnitude of the subtransient internal emf is 1.02 pu.

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