Question

A 50 Hz, 275 kV line of length 400 km has the following parameters: Resistance, $R = 0.035 \,\Omega/\text{km}$
Inductance, $L = 1 \,\text{mH}/\text{km}$
Capacitance, $C = 0.01 \,\mu\text{F}/\text{km}$
The line is represented by the nominal-$\pi$ model. With the magnitudes of the sending end and the receiving end voltages of the line (denoted by $V_S$ and $V_R$, respectively) maintained at 275 kV, the phase angle difference $(\theta)$ between $V_S$ and $V_R$ required for maximum possible active power to be delivered to the receiving end, in degree is \underline{\hspace{2cm}} (Round off to 2 decimal places).

Hint:

For transmission line power transfer, the angle $\theta$ of the $A$ parameter of the ABCD matrix gives the phase angle required for maximum active power delivery when $V_S=V_R$.

Answer 1

Step 1: Compute total series impedance.
\[ R' = 0.035 \times 400 = 14 \,\Omega \] \[ X_L = \omega L' = 2 \pi \times 50 \times (1 \times 10^{-3}) \times 400 = 125.66 \,\Omega \] \[ Z = R' + j X_L = 14 + j125.66 \]

Step 2: Compute shunt admittance.
\[ C' = 0.01 \times 10^{-6} \times 400 = 4 \times 10^{-6} \,\text{F} \] \[ Y = j \omega C' = j (2 \pi \times 50 \times 4 \times 10^{-6}) = j 0.0012566 \,\text{S} \]

Step 3: Nominal-$\pi$ parameters.
\[ A = D = 1 + \frac{YZ}{2} \] \[ B = Z, C = Y \left(1 + \frac{YZ}{4}\right) \] Since $YZ \ll 1$, approximate: \[ A \approx 1 + \frac{YZ}{2} \]

Step 4: Maximum power transfer angle.
The maximum power transfer occurs when $P = \frac{V_S V_R}{|B|} \sin \theta$, hence maximum when $\theta = \angle A$. \[ \theta \approx \tan^{-1}\left(\frac{\Im(A)}{\Re(A)}\right) = \tan^{-1}\left(\frac{\Im(1+YZ/2)}{\Re(1+YZ/2)}\right) \] \[ YZ = (j0.0012566)(14+j125.66) = -0.1578 + j0.0176 \] \[ YZ/2 = -0.0789 + j0.0088 \] \[ A \approx 0.9211 + j0.0088 \] \[ \theta = \tan^{-1}\left(\frac{0.0088}{0.9211}\right) \approx 0.00956 \,\text{rad} = 2.07^\circ \] % Final Answer \[ \boxed{2.07^\circ} \]