A cylindrical rotor synchronous generator has steady state synchronous reactance of 0.7 pu and sub transient 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 sub transient internal emf (rounded off to 2 decimal places) is______ pu.

Correct Answer: The magnitude of the subtransient internal emf is 1.02 pu (or 1.0198 pu, rounded off to two decimal places: 1.02).

Step-by-Step Explanation:

1. Given Parameters:
Terminal voltage under steady-state condition, $V_t=1+j0\text{pu}$
Steady-state synchronous reactance, $X_d=0.7\text{pu}$
Subtransient reactance, $X_d^{″}=0.2\text{pu}$
Steady-state internal emf (excitation voltage), $E=1+j0.7\text{pu}$

2. Finding the Initial Load Current ($I$):
The steady-state internal voltage equation of a synchronous generator is given by:

$E=V_t+jI{X}_d$

Rearranging the equation to solve for the steady-state armature current $I$:

$jI{X}_d=E-V_t$

$I=\frac{E-V_t}{j{X}_d}$

Substitute the given values:

$I=\frac{(1+j0.7)-(1+j0)}{j0.7}=\frac{j0.7}{j0.7}=1+j0\text{pu}$

3. Calculating the Subtransient Internal EMF ($E^{″}$):
Immediately following the fault, the internal subtransient voltage $E^{″}$ before the fault is given by using the subtransient reactance $X_d^{″}$:

$E^{″}=V_t+jI{X}_d^{″}$

Substituting $V_t=1+j0$, $I=1+j0$, and $X_d^{″}=0.2$:

$E^{″}=(1+j0)+j(1+j0)\left(0.2\right)=1+j0.2\text{pu}$

4. Calculating the Magnitude of $E^{″}$:

$\left|E^{″}\right|=\sqrt{1^2+{0.2}^2}=\sqrt{1+0.04}=\sqrt{1.04}\approx 1.0198\text{pu}$

Rounding off to 2 decimal places, the magnitude of the subtransient internal emf is 1.02 pu.