To solve for the sending end voltage ($V_s$) in this 3-phase transmission line problem, we can proceed through a systematic calculation using the given parameters.

Step-by-Step Solution

1. Identify Given Parameters:

   • Line resistance, $R = 5 \ \Omega$ per wire
   • Line reactance, $X = 12 \ \Omega$ per wire
   • Power transmitted to the load, $P = 1000 \ \text{kW}$
   • Load voltage (line-to-line), $V_L = 13,200 \ \text{V}$
   • Power factor, $\text{pf} = 0.8$ lagging
   • Frequency, $f = 60 \ \text{Hz}$

2. Calculate the Load Current:

   The power transmitted in a 3-phase system is given by: $P = \sqrt{3} \times V_L \times I \times \text{pf}$ Rearranging to solve for the load current $I$: $I = \frac{P}{\sqrt{3} \times V_L \times \text{pf}}$ Substituting values: $I = \frac{1000 \times 10^3}{\sqrt{3} \times 13200 \times 0.8}$ $I \approx 54.7 \ \text{A}$

3. Calculate the Total Line Impedance:

   Since the resistance and reactance are given per wire, for a 3-phase system, the impedance per phase is: $Z = R + jX = 5 + j12 \ \Omega$

   Calculate the magnitude of $Z$: $|Z| = \sqrt{R^2 + X^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \ \Omega$

4. Calculate the Voltage Drop in the Line:

   The voltage drop across the line impedance is given by: $V_{\text{drop}} = I \times Z = 54.7 \times (5 + j12)$ Expanding this: $V_{\text{drop}} = 54.7 \times 5 + j(54.7 \times 12)$ $V_{\text{drop}} = 273.5 + j656.4$ This voltage drop has both real and imaginary components, and its magnitude is: $|V_{\text{drop}}| = \sqrt{273.5^2 + 656.4^2} \approx 710 \ \text{V}$

5. Calculate the Receiving End Phase Voltage:

   The line-to-line voltage at the receiving end is given as 13,200 V, so the phase voltage at the receiving end ($V_R$) is: $V_R = \frac{V_L}{\sqrt{3}} = \frac{13200}{\sqrt{3}} \approx 7621 \ \text{V}$

6. Determine the Sending End Voltage:

   Using the voltage drop and the receiving end voltage, the sending end voltage ($V_s$) is: $V_s = V_R + V_{\text{drop}}$ Since $V_{\text{drop}}$ has both real and imaginary parts, we must add them vectorially: $V_s = 7621 + (273.5 + j656.4)$

   Converting $V_s$ to polar form gives us the magnitude of the sending end voltage.

   After calculation: $|V_s| \approx 7621 + 710 \approx 8331 \ \text{V (phase)}$ Converting back to line-to-line voltage: $V_{s (\text{line-to-line})} = \sqrt{3} \times 8331 \approx 14,430 \ \text{V}$

Final Answer:

The sending end voltage $V_s$ is approximately 14,430 V (line-to-line).

Would you like further clarification on any of these steps?

Related Questions

1. How would the solution change if the power factor were leading instead of lagging?
2. How does the frequency affect the impedance in AC circuits?
3. What is the impact of increasing the line length on the sending end voltage?
4. How do you determine the phase angle of the sending end voltage relative to the receiving end voltage?
5. What are the consequences of high resistance and reactance in transmission lines?

Tip

For accurate results in AC circuit analysis, always use vector notation to handle phase relationships in voltages and currents.