Let's solve each question step by step as they appear in your message. --- ### **Question 1** #### **Given:** - Impedance per phase: \( Z = (1 + j3) \) Ω - Load: 20 MW at unity power factor - Efficiency: 92% #### **To Find:** - Sending-end voltage #### **Solution:** **Step 1: Calculate Load Current** For a 3-phase system: \[ P_L = \sqrt{3} V_{r} I_{L} \cos\phi \] Given unity power factor, \( \cos\phi = 1 \). Let \( V_r \) = receiving end line voltage (not given), so let's assume per-phase value for now. But we need more data (either voltage or current) to continue. Let's see what we can do using efficiency. **Step 2: Calculate Sending-end Power** Efficiency \( \eta = \frac{\text{Receiving End Power}}{\text{Sending End Power}} \) \[ \eta = \frac{P_L}{P_S} \implies P_S = \frac{P_L}{\eta} \] \[ P_S = \frac{20,000,000}{0.92} = 21,739,130.43 \text{ W} \] Losses = \( P_S - P_L = 1,739,130.43 \) W **Step 3: Calculate Line Current** \[ P_L = \sqrt{3} V_{r} I_{L} \implies I_{L} = \frac{P_L}{\sqrt{3} V_r} \] But \( V_r \) is not given, so we have to assume something is missing in the question. **Please provide the receiving end voltage or more data for this question.** --- ### **Question 2** #### **Given:** - Impedance per phase: \( Z = 5 + j10 \) Ω - Input (sending) line voltage: \( V_s = 12.8 \) kV - Receiving end load: \( P_L = 1,200 \) kW, \( I_L = 60 \) A, **lagging** power factor (\( \cos\phi \) unknown, but can be calculated) #### **To Find:** - Load voltage between lines (\( V_r \)) #### **Solution:** **Step 1: Calculate Power Factor** \[ P_L = \sqrt{3} V_{r} I_L \cos\phi \] \[ \cos\phi = \frac{P_L}{\sqrt{3} V_{r} I_L} \] But \( V_r \) is unknown. Alternatively, find phase current: \[ I_L = 60 \, \text{A} \] \[ V_{r} = \text{unknown} \] **Step 2: Per-Phase Calculations** Per-phase load: \[ P_{ph} = \frac{P_L}{3} = \frac{1200\,kW}{3} = 400\,kW \] Line current is given, so: \[ V_{r(phase)} = \frac{V_{r(line)}}{\sqrt{3}} \] But let's use the per-phase equivalent circuit: \[ V_{s(phase)} = V_{r(phase)} + I_L Z \] But \( V_{s(line)} = 12.8\,kV \implies V_{s(phase)} = \frac{12,800}{\sqrt{3}} = 7,389\,V \) \[ I_L = 60\,A \] \[ Z = 5 + j10 = 11.18 \angle 63.4^\circ \, \Omega \] Let power factor angle \( \phi \): \[ P = \sqrt{3} V_{r(line)} I_L \cos\phi \] \[ \cos\phi = \frac{1200 \times 10^3}{\sqrt{3} V_{r(line)} \times 60} \] Let \( x = V_{r(line)} \): \[ \cos\phi = \frac{1200000}{\sqrt{3} \times x \times 60} \] But also, draw the phasor diagram: \[ V_{s(phase)} = V_{r(phase)} + I_L Z \] Let - \( I_L = 60 \) A, angle \( -\phi \) (since lagging) - \( Z = 5 + j10 \) Ω So, \[ I_L Z = 60 \times (5 + j10) = (300 + j600) = 670.8 \angle 63.4^\circ \text{ V} \] Let the receiving end voltage per phase be \( V_{r(phase)} \), at angle 0°. \[ V_{s(phase)} = V_{r(phase)} + 670.8 \angle (\text{current lags voltage by } \phi) \] But \( I_L Z \) will have a phase of \( 63.4^\circ - \phi \). So, \[ V_{s(phase)} = V_{r(phase)} + 670.8 \angle (63.4^\circ - \phi) \] But \( V_{s(phase)} = 7,389 \) V. Let's work with \[ V_{r(phase)} = x \] \[ \cos\phi = \frac{1200000}{\sqrt{3} \times x \times 60} \] \[ \cos\phi = \frac{1200000}{103.923 x} \] \[ \cos\phi = \frac{11,548}{x} \] Also, \[ V_{s(phase)} = x + 670.8 \angle (63.4^\circ - \phi) \] Write in rectangular form: \( I_L Z = (300 + j600) \) V Since current lags by \(\phi\), so: \[ I_L = 60 \angle -\phi \] \[ I_L Z = 60 \angle -\phi \times 11.18 \angle 63.4^\circ = 670.8 \angle (63.4^\circ - \phi) \] So, write \[ V_{s(phase)} = x + 670.8 \cos(63.4^\circ - \phi) + j670.8 \sin(63.4^\circ - \phi) \] But \( V_{s(phase)} = 7,389 \) V. Set up the magnitude relation: \[ 7,389^2 = x^2 + [670.8]^2 + 2x \cdot 670.8 \cos(63.4^\circ - \phi) \] But we have two unknowns: \( x \) and \(\phi \). Recall: \[ \cos\phi = \frac{11,548}{x} \] Let’s try to solve numerically. Let’s guess \( x \) around 11,000 V (for 12.8 kV sending, expect something lower at the receiving end). Try \( x = 11,000 \): \[ \cos\phi = \frac{11,548}{11,000} = 1.05 \] Too high; not possible. Try \( x = 12,000 \): \[ \cos\phi = \frac{11,548}{12,000} = 0.962 \] Now, let's get \(\phi\): \[ \phi = \arccos(0.962) = 15.7^\circ \] Now \( 63.4^\circ - \phi = 63.4^\circ - 15.7^\circ = 47.7^\circ \) So, \[ \cos(47.7^\circ) = 0.671 \] \[ 7,389^2 = x^2 + 670.8^2 + 2x \cdot 670.8 \cdot 0.671 \] \[ 54,597,721 = x^2 + 450,971 + 899.8x \] \[ x^2 + 899.8x + 450,971 - 54,597,721 = 0 \] \[ x^2 + 899.8x - 54,146,750 = 0 \] Solve quadratic: \[ x = \frac{ -899.8 \pm \sqrt{899.8^2 + 4 \times 54,146,750} }{2} \] \[ = \frac{ -899.8 \pm \sqrt{809,640 + 216,587,000} }{2} \] \[ = \frac{ -899.8 \pm \sqrt{217,396,640} }{2} \] \[ = \frac{ -899.8 \pm 14,743 }{2} \] Take the positive root: \[ x = \frac{ -899.8 + 14,743 }{2} \approx \frac{13,843}{2} = 6,921.5 \] That can't be, as it's lower than phase voltage. Let’s try with \( x = 11,500 \): \[ \cos\phi = \frac{11,548}{11,500} = 1.004 \] Still too high. Try \( x = 12,000 \) (as above), \(\cos\phi = 0.962\), \(\phi = 15.7^\circ\) Let’s try \( x = 13,000 \): \[ \cos\phi = \frac{11,548}{13,000} = 0.888 \] \[ \phi = \arccos(0.888) = 27.3^\circ \] So \( 63.4^\circ - 27.3^\circ = 36.1^\circ \) \[ \cos(36.1^\circ) = 0.807 \] \[ 7,389^2 = 13,000^2 + 670.8^2 + 2 \times 13,000 \times 670.8 \times 0.807 \] \[ = 169,000,000 + 450,971 + 14,070 \times 0.807 = 14,070 \times 0.807 = 11,355 \] Wait, that's wrong. The correct calculation is \( 2 \times 13,000 \times 670.8 \times 0.807 \): \[ 2 \times 13,000 \times 670.8 \times 0.807 = 2 \times 13,000 \times 541.47 = 2 \times 7,038,990 = 14,077,980 \] Wait, but \( 13,000 \times 670.8 = 8,720,400 \), times 2 = 17,440,800, times 0.807 = 14,082,626 So, total: \[ 7,389^2 = 13,000^2 + 450,971 + 14,082,626 \] \[ 54,597,721 = 169,000,000 + 450,971 + 14,082,626 = 183,533,597 \] But 54 million = 183 million, so this is not matching. Therefore, perhaps the question is ambiguous or data is inconsistent. **If you want, I can make a reasonable assumption and move on.** --- ### **Question 3** #### **Given:** - Length: 10 km - Load: 2,500 kVA at 600 V, 3-phase - Power factor: 0.80 lagging - Impedance per phase per km: \( 0.3 + j0.4 \) Ω/km - To find: Nearest sending end voltage (choose from 8,021V, 8,210V, 8,201V) #### **Solution:** **Step 1: Total Line Impedance per phase** \[ Z_{total} = 10 \times (0.3 + j0.4) = (3 + j4) \, \Omega \] **Step 2: Load Current** \[ \text{Apparent Power} = S = 2,500 \text{ kVA} = 2,500,000 \text{ VA} \] \[ I_L = \frac{S}{\sqrt{3} V} \] \[ I_L = \frac{2,500,000}{\sqrt{3} \times 600} = \frac{2,500,000}{1,039.23} = 2,406 \text{ A} \] **Step 3: Per-phase Equivalent** - Per-phase load voltage: \( V_{r,ph} = \frac{600}{\sqrt{3}} = 346.4 \) V - Load power factor \( \phi = \arccos(0.80) = 36.87^\circ \) - Load current lags voltage by \( 36.87^\circ \): \[ I_{ph} = 2,406 \angle -36.87^\circ \text{ A} \] **Step 4: Sending End Per-Phase Voltage** \[ V_{s,ph} = V_{r,ph} + I_{ph} Z_{total} \] \[ Z_{total} = 5 \angle 53.13^\circ