Guard ring sizing for 3-unit string at 20 kV

The guard ring adds a capacitance $C_x$ between the ring (at line potential) and the link pin at the top of the line-end unit. This capacitance is chosen so that the two units nearest the line share the voltage equally.

 Tower ─[C]─ A ─[C]─ B ─[C]─ Line (20 kV)
             |       |  \
          C/5       C/5  Cx (ring to pin B)
             |       |
           earth   earth

Units: 1 (tower end), 2, 3 (line end), with voltages $V_1, V_2, V_3$. The shunt capacitance is $C/5 = 0.2C$, and the condition is $V_2 = V_3$.

Junction A (no guard ring capacitance)

$$\omega CV_2 = \omega CV_1 + \omega(0.2C)V_1 \;\Rightarrow\; V_2 = 1.2V_1$$

Junction B (guard ring capacitance $C_x$ to line)

The voltage of B to earth is $V_1+V_2$, and to the line it is $V_3$:

$$\omega CV_3 + \omega C_xV_3 = \omega CV_2 + \omega(0.2C)(V_1+V_2)$$

With $V_3 = V_2$:

$$C_x V_2 = 0.2C(V_1 + V_2) \;\Rightarrow\; C_x = 0.2C\left(\frac{1}{1.2} + 1\right) = 0.3667C$$

(i) Voltage on the line-end unit

$$V = V_1 + V_2 + V_3 = V_1 + 1.2V_1 + 1.2V_1 = 3.4V_1 = 20\ \text{kV}$$ $$V_1 = 5.882\ \text{kV},\qquad V_2 = V_3 = 7.059\ \text{kV}$$

String efficiency $= 20/(3\times 7.059) = 94.44\ \%$.

(ii) Capacitance between ring and pin

$$C_x = 0.3667C \;(= 11C/30)$$

Answer: (i) The line-end unit carries 7.06 kV (top unit 5.88 kV, middle unit 7.06 kV). (ii) The ring-to-pin capacitance is $0.367C$.