Let's proceed with the step-by-step calculation for the elements of the P matrix, including the effect of the shield. --- # Step-by-step solution ## Given Data: - Conductor, \( d = 1.345'' \) - Shield wire diameter, \( d_s = .385'' \) - Phase spacing, \( D = 15\, \textfeet} \- Conductor at tower, \( h = 50\, \textfeet} \) - Shield wire height at tower, \( h_s = 70\, \text{feet} \) - Shield wire Sag, \( S_s = 15\, \text{feet} \) --- ## 1. **Calculate the geometric mean radius (GMR) of the conductor \( GMR_c \)** \[ GMR_c = d \times e^{-.25} \approx d \times .7788 \] \[ GMR_c = 1.345 \times .7788 \approx 1.046 \text{ inches} \] --- ## 2. **Calculate the GMR of the shield wire \( GMR_s \)** \[ GMR_s = d_s \times e^{-.25} \approx .385 \times .7788 \approx .3 \text{ inches} \] --- ## 3. **Calculate the geometric factors for the P matrix** ### a. **Calculate the average height of the conductor and shield wire** Since the shield wire is often supported with sag, the effective height is: \[ h_{eff,s} = h_s - S_s/2 = 70 - 15/2 = 70 - 7.5 = 62.5\, \text{feet} \] The conductor height: \[ h_{conductor} = 50\, \text{feet} \] ### b. **Calculate the distance between the conductor and the shield wire** \[ D_{s} = \sqrt{(D)^2 + (h_{eff,s} - h)^2} \] \[ D_s = \sqrt{(15)^2 + (62.5 - 50)^2} = \sqrt{225 + 156.25} = \sqrt{381.25} \approx 19.52\, \text{feet} \] --- ## 4. **Calculate the potential coefficients \( P_{ii} \) and \( P_{ij} \)** ### a. **Self potential coefficient \( P_{ii} \)** \[ P_{ii} = \ln \left( \frac{D_{ref}}{GMR} \right) \] Where \( D_{ref} \) is the reference distance, typically taken as 1 foot for these calculations. - For the conductor: \[ P_{cc} = \ln \left( \frac{1}{GMR_c} \right) = \ln \left( \frac{1}{1.046} \right) = -.045 \] - For the shield wire: \[ P_{ss} = \ln \left( \frac{1}{GMR_s} \right) = \ln \left( \frac{1}{.3} \right) = 1.204 \] ### b. **Mutual potential coefficient \( P_{cs} \) (conductor-shield wire)** \[ P_{cs} = \ln \left( \frac{D_s}{\sqrt{GMR_c \times GMR_s}} \right) \] Calculate the denominator: \[ \sqrt{1.046 \times .3} \approx \sqrt{.3138} \approx .56 \] Then: \[ P_{cs} = \ln \left( \frac{19.52}{.56} \right) = \ln(34.86) \approx 3.55 \] --- ## 5. **Construct the P matrix** \[ \boxed{ P = \begin{bmatrix} P_{cc} & P_{cs} \\ P_{cs} & P_{ss} \end{bmatrix} = \begin{bmatrix} -.045 & 3.55 \\ 3.55 & 1.204 \end{bmatrix} } \] --- ## 6. **Inclusion of shield wire effect** The above approach accounts for the shield wire's influence via the mutual potential coefficient \( P_{cs} \). The effect of the shield wire is embedded in the off-diagonal terms. --- # **Final P matrix:** \[ \boxed{ \begin{bmatrix} -.045 & 3.55 \\ 3.55 & 1.204 \end{bmatrix} } \] --- # **Summary:** - The diagonal elements represent self-potential coefficients. - The off-diagonal elements account for mutual coupling, including the effect of the shield wire. - Heights and sag influence the mutual distances, affecting the off-diagonal terms. Let me know if you'd like further steps such as calculating the entire P matrix or specific effects!