Let's solve the problem step by step: --- ## **Given Data** - **Load:** 300 kVA, .8 p.f. (lagging), 11 kV (line-to-line) - **Transformer ratio:** 6:1 (assume HV:LV, so 66 kV/11 kV) - **Transmission line resistance:** 2 Ω/phase, reactance: 3 Ω/phase - **Primary side (Y-connected) winding:** 5 Ω, 10 Ω (assume R = 5 Ω, X = 10 Ω) - **Secondary side (Δ-connected) winding:** 1.5 Ω, 3 Ω (assume R = 1.5 Ω, X = 3 Ω) - **Required:** Voltage and power factor at the sending end of the transmission line. --- ## **Step 1: Find Load Current** ### **Load Apparent Power** \[ S = 300\,\text{kVA} = 3,000,000\,\text{VA} \] ### **Load Voltage (Line-to-line)** \[ V_{L} = 11,000\,\text{V} \] ### **Load Current** \[ I_L = \frac{S}{\sqrt{3}V_L} = \frac{3,000,000}{\sqrt{3} \times 11,000} = \frac{3,000,000}{19,052.56} \approx 157.5\,\text{A} \] --- ## **Step 2: Calculate Per-phase Load Current and Power Factor Angle** \[ \cos \phi = .8 \implies \phi = \cos^{-1}(.8) = 36.87^\circ \] \[ I_{phase} = I_L = 157.5\,\text{A} \quad (\text{since load is star/Y}) \] --- ## **Step 3: Load Per-phase Complex Power** \[ S_{ph} = \frac{S}{3} = \frac{3,000,000}{3} = 1,000,000\,\text{VA per phase} \] But we already have the current, so let's move on. --- ## **Step 4: Equivalent Impedances (Referred to Secondary Side)** ### **Transformer Secondary (Δ-Connected):** - \( R_2 = 1.5\,\Omega \), \( X_2 = 3\,\Omega \) ### **Transmission Line:** - \( R_{line} = 2\,\Omega \), \( X_{line} = 3\,\Omega \) ### **Transformer Primary (Y-Connected, Referred to Secondary Side)** Transformer ratio = 6:1 Impedances referred from primary to secondary side: \[ Z_{1,\text{sec}} = Z_1 \left(\frac{V_2}{V_1}\right)^2 = Z_1 \left(\frac{1}{6}\right)^2 = \frac{Z_1}{36} \] \[ R_{1,\text{sec}} = \frac{5}{36} = .139\,\Omega \] \[ X_{1,\text{sec}} = \frac{10}{36} = .278\,\Omega \] --- ## **Step 5: Total Per-phase Impedance** \[ Z_{total} = R_{1,\text{sec}} + jX_{1,\text{sec}} + R_{line} + jX_{line} + R_2 + jX_2 \] \[ = (.139 + 2 + 1.5) + j(.278 + 3 + 3) \] \[ = 3.639 + j6.278 \] --- ## **Step 6: Sending End Voltage Calculation** ### **Load Phase Voltage** \[ V_{ph,load} = \frac{V_L}{\sqrt{3}} = \frac{11,000}{\sqrt{3}} = 6,351\,\text{V} \] ### **Load Phase Current (with angle)** \[ I_{ph} = 157.5 \angle -36.87^\circ \,\text{A} \] ### **Total Voltage Drop** \[ \text{Voltage drop per phase:} \quad \Delta V = I_{ph} \times Z_{total} \] \[ Z_{total} = 3.639 + j6.278 = 7.22 \angle \theta_Z \] Where, \[ \theta_Z = \tan^{-1}\left(\frac{6.278}{3.639}\right) = \tan^{-1}(1.726) = 60.47^\circ \] \[ I_{ph} \times Z_{total} = 157.5 \angle -36.87^\circ \times 7.22 \angle 60.47^\circ = (157.5 \times 7.22) \angle (60.47 - 36.87) = 1,137.15 \angle 23.6^\circ\,\text{V} \] --- ### **Sending End Phase Voltage** \[ V_{ph,sending} = V_{ph,load} + \Delta V \] Add these two as phasors: First, convert to rectangular form: \[ \Delta V = 1,137.15 \cos 23.6^\circ + j\,1,137.15\sin 23.6^\circ = 1,044.7 + j456.\,\text{V} \] \[ V_{ph,load} = 6,351 \angle ^\circ = 6,351 + j \] \[ V_{ph,sending} = (6,351 + 1,044.7) + j ( + 456.) = 7,395.7 + j456. \] \[ |V_{ph,sending}| = \sqrt{(7,395.7)^2 + (456.)^2} = \sqrt{54,708,732 + 207,936} = \sqrt{54,916,668} = 7,411\,\text{V} \] \[ \theta_{V_{ph,sending}} = \tan^{-1}\left( \frac{456.}{7,395.7} \right) = \tan^{-1}(.0616) = 3.53^\circ \] --- ### **Sending End Line-to-Line Voltage** \[ V_{L,sending} = \sqrt{3} \times V_{ph,sending} = \sqrt{3} \times 7,411 \approx 12,841\,\text{V} \] --- ## **Step 7: Sending End Power Factor** The sending end voltage leads the load voltage by \(3.53^\circ\). The phase difference between the sending end voltage and current: \[ \text{Angle between } V_{ph,sending} \text{ and } I_{ph} = 3.53^\circ + 36.87^\circ = 40.4^\circ \] So, the sending end power factor is: \[ \cos(40.4^\circ) = .76 \quad \text{(lagging)} \] --- ## **Final Answers** ### **1. Sending End Voltage** \[ \boxed{12,841\,\text{V (line-to-line)}} \] ### **2. Sending End Power Factor** \[ \boxed{.76 \text{ (lagging)}} \] --- ### **Summary Table** | Parameter | Value | |------------------------|--------------| | Sending end voltage | 12,841 V | | Sending end power factor | .76 (lagging) | --- **If you have any specific step you'd like explained in more detail, let me know!**