Explain "Ferranti-Effect" with respect to transmission lines. The figure below shows a nominal - T model of a power line; Show that Is = Vr + Ir (1 + 2Y/2).

2. (a) (i) State three classifications of overhead power transmission lines.

• Short transmission lines (typically up to 80 km)
• Medium transmission lines (typically between 80 km and 240 km)
• Long transmission lines (typically over 240 km)

2. (a) (ii) Explain "Ferranti-Effect" with respect to transmission lines. The Ferranti effect is a phenomenon in long or medium length AC transmission lines where the receiving end voltage ($V_r$) becomes greater than the sending end voltage ($V_s$) under light load or no-load conditions. This occurs due to the charging current flowing through the line's shunt capacitance and the line's series inductance, causing a voltage rise along the line.

2. (b) The figure below shows a nominal - T model of a power line; Show that $I_s = V_r Y + I_r (1 + ZY/2)$.

Step 1: Identify the voltage at the shunt admittance. From the nominal T-model diagram, the voltage across the shunt admittance $Y$ (let's call it $V_c$) can be expressed in terms of the receiving end voltage $V_r$ and the current $I_r$ flowing through the second half of the series impedance $Z/2$. $V_c = V_r + I_r \frac{Z}{2}$

Step 2: Calculate the current through the shunt admittance. The current flowing through the shunt admittance $Y$ (let's call it $I_c$) is given by Ohm's law: $I_c = V_c Y$ Substitute the expression for $V_c$ from Step 1: $I_c = \left(V_r + I_r \frac{Z}{2}\right) Y$ $I_c = V_r Y + I_r \frac{ZY}{2}$

Step 3: Apply Kirchhoff's Current Law (KCL) at the central node. The current flowing into the central node from the sending end side (which is $I_s$) splits into the current through the shunt admittance ($I_c$) and the current flowing towards the receiving end ($I_r$). $I_s = I_c + I_r$

Step 4: Substitute the expression for $I_c$ into the KCL equation. $I_s = \left(V_r Y + I_r \frac{ZY}{2}\right) + I_r$ Factor out $I_r$: $I_s = V_r Y + I_r \left(\frac{ZY}{2} + 1\right)$ Rearranging the terms: $I_s = V_r Y + I_r \left(1 + \frac{ZY}{2}\right)$ This matches the expression to be shown. $\boxed{I_s = V_r Y + I_r \left(1 + \frac{ZY}{2}\right)}$

2. (c) A 200km long, three phase overhead line has a resistance of $48.7 \Omega$ per phase, inductive reactance of $80.2 \Omega$ per phase and capacitance (line to Neutral) power factor of 0.9 lagging. This question is incomplete. To solve for any specific parameter of the transmission line (e.g., sending end voltage, current, or efficiency), the following information is missing:

• The value of the capacitance (line to neutral). The phrase "capacitance (line to Neutral) power factor of 0.9 lagging" is grammatically incorrect and does not provide a capacitance value. It is likely intended to state a capacitance value and a receiving end power factor.
• The receiving end voltage ($V_r$) or receiving end power ($P_r$).
• The specific quantity to be calculated (e.g., sending end voltage, sending end current, voltage regulation, efficiency).