Let's solve each question step-by-step. --- ## **Question 1** **A 3-phase transmission line 10 km long delivers 2,500 kVA at 600 V. The power factor of the load is 0.80 lagging. If the impedance of each load is (0.3 + j0.4) ohm/km, the sending end voltage between lines is nearest to:** **Options:** - a. 8,021 V - b. 8,210 V - c. 8,201 V --- ### **Step 1: Find Load Current** Given: - Apparent Power, \( S = 2500 \) kVA \( = 2,500,000 \) VA - Line-to-line Voltage (Receiving), \( V_r = 600 \) V (line-to-line) - Power Factor = 0.8 lagging For a 3-phase system: \[ I_L = \frac{S}{\sqrt{3} \cdot V_{r}} \] \[ I_L = \frac{2,500,000}{\sqrt{3} \cdot 600} \] \[ I_L = \frac{2,500,000}{1,039.23} \] \[ I_L \approx 2,406.1 \text{ A} \] --- ### **Step 2: Find Total Line Impedance** Given impedance per phase per km: \( Z = 0.3 + j0.4 \) \(\Omega\)/km Line length = 10 km Total per phase impedance: \[ Z_{total} = (0.3 + j0.4) \times 10 = (3 + j4) \; \Omega \] --- ### **Step 3: Receiving End Phase Voltage** \[ V_{r(phase)} = \frac{V_r}{\sqrt{3}} = \frac{600}{1.732} \approx 346.4 \text{ V} \] --- ### **Step 4: Receiving End Phase Current (Phasor Form)** Power factor = 0.8 lagging So, \(\cos \phi = 0.8\), \(\sin \phi = 0.6\) \[ I_{r(phase)} = \frac{2,500,000}{3 \times 346.4} = \frac{2,500,000}{1,039.2} \approx 2,406.1 \text{ A} \] Phasor current (lagging): \[ I_r = 2,406.1 \angle -\cos^{-1}(0.8) = 2,406.1 \angle -36.87^\circ \text{ A} \] --- ### **Step 5: Voltage Drop Across the Line** \[ \Delta V = I_r \cdot Z_{total} \] First, write \( I_r \) in rectangular form: \[ I_r = 2,406.1 (\cos(-36.87^\circ) + j\sin(-36.87^\circ)) \] \[ I_r = 2,406.1 \times (0.8 - j0.6) = 1,924.88 - j1,443.66 \text{ A} \] Now multiply by \( Z_{total} = 3 + j4 \): \[ \Delta V = (1,924.88 - j1,443.66) \times (3 + j4) \] Let’s expand: - \( (a + jb) \times (c + jd) = (ac - bd) + j(ad + bc) \) So, - \( a = 1,924.88 \) - \( b = -1,443.66 \) - \( c = 3 \) - \( d = 4 \) Calculate: - Real part: \( 1,924.88 \times 3 - (-1,443.66) \times 4 = 5,774.64 + 5,774.64 = 11,549.28 \) - Imaginary part: \( 1,924.88 \times 4 + (-1,443.66) \times 3 = 7,699.52 - 4,330.98 = 3,368.54 \) \[ \Delta V = 11,549.28 + j3,368.54 \text{ V} \] Magnitude: \[ |\Delta V| = \sqrt{(11,549.28)^2 + (3,368.54)^2} \] \[ = \sqrt{133,376,879 + 11,351,173} \] \[ = \sqrt{144,728,052} \] \[ \approx 12,032.8 \text{ V} \] But this is per phase. Remember, the per-phase voltage drop is added to the per-phase voltage. --- ### **Step 6: Sending End Phase Voltage** \[ V_{s(phase)} = V_{r(phase)} + \Delta V \] But voltage drop is a phasor addition. So, let's do it in phasor form: - \( V_{r(phase)} = 346.4 \angle 0^\circ \) V - \( \Delta V = 12,032.8 \angle \theta \), where \( \theta = \arctan(\frac{3,368.54}{11,549.28}) = \arctan(0.2916) \approx 16.3^\circ \) Add the two phasors: \[ V_{s(phase)} = 346.4 \angle 0^\circ + 12,032.8 \angle 16.3^\circ \] Let's convert both to rectangular and add: - \( V_{r(phase)} = 346.4 + j0 \) - \( \Delta V = 12,032.8 \times \cos(16.3^\circ) + j12,032.8 \times \sin(16.3^\circ) \) - \( = 11,548.5 + j3,377.8 \) Add: \[ V_{s(phase)} = (346.4 + 11,548.5) + j(0 + 3,377.8) = 11,894.9 + j3,377.8 \] Magnitude: \[ = \sqrt{(11,894.9)^2 + (3,377.8)^2} = \sqrt{141,502,276 + 11,410,537} = \sqrt{152,912,813} \approx 12,366.2 \text{ V} \] --- ### **Step 7: Find Line-to-Line Sending End Voltage** \[ V_{s(line)} = V_{s(phase)} \times \sqrt{3} = 12,366.2 \times 1.732 = 21,418.5 \text{ V} \] Wait, but this value looks much higher than the given options (8,021 V, 8,210 V, 8,201 V). Let's check for a possible mistake. #### **Check the initial data:** - The load voltage is only 600 V (quite low). - Line current calculated is 2,406.1 A (very high, but makes sense for 2,500 kVA at 600 V). #### **But the voltage drop per phase is 12,032.8 V, much higher than the load voltage.** This suggests that the line losses are enormous, and the voltage drop is huge compared to the receiving end voltage, which makes sense with such a high current and such a high impedance. But the options suggest a sending voltage just above 8,000 V. #### **Let’s check the per-phase calculation again:** - \( V_{r(phase)} = 346.4 \) V - \( I_{ph} = 2,406.1 \angle -36.87^\circ \) A - \( Z_{total} = 3 + j4 = 5 \angle 53.13^\circ \) - \( I_{ph} \times Z_{total} = 2,406.1 \times 5 = 12,030.5 \) - Angle: \( -36.87^\circ + 53.13^\circ = 16.26^\circ \) - So, \( \Delta V = 12,030.5 \angle 16.26^\circ \) Adding to \( V_{r(phase)} = 346.4 \angle 0^\circ \) Now, let's add as vectors: - Real: \( 346.4 + 12,030.5 \times \cos(16.26^\circ) \) - Imaginary: \( 0 + 12,030.5 \times \sin(16.26^\circ) \) - \( \cos(16.26^\circ) = 0.9584 \), \( \sin(16.26^\circ) = 0.2800 \) - Real: \( 346.4 + 11,532.4 = 11,878.8 \) - Imaginary: \( 0 + 3,368.5 = 3,368.5 \) - Magnitude: \( \sqrt{(11,878.8)^2 + (3,368.5)^2} = \sqrt{141,079,085 + 11,353,703} = \sqrt{152,432,788} = 12,349.8 \) V (matches before) Line-to-line: \( 12,349.8 \times 1.732 = 21,403 \) V But the options are all ~8,200 V. #### **Is the impedance per phase or per line?** - **If the impedance is per line:** For a star connection, the per-phase impedance is the same as per line. - **If the impedance given is for the total line:** (which is more likely given such a low voltage) Alternatively, perhaps the voltage should be 6,000 V, not 600 V, for a transmission line. Let's recalculate **assuming a typo and the voltage is actually 6,000 V**. #### **Recalculate for 6,000 V:** - \( V_r = 6,000 \) V - \( V_{r(phase)} = 6,000 / 1.732 = 3,464.1 \) V - \( I_L = 2,500,000 / (1.732 \times 6,000) = 2,500,000 / 10,392 = 240.6 \) A Now, \( I_{ph} = 240.6 \angle -36.87^\circ \) A Voltage drop per phase: - \( Z_{total} = 3 + j4 = 5 \angle 53.13^\circ \) - \( I_{ph} \times Z_{total} = 240.6 \times 5 = 1,203 \) - Angle: \( -36.87^\circ + 53.13^\circ = 16.26^\circ \) - So, \( \Delta V = 1,203 \angle 16.26^\circ \) Add to \( 3,464.1 \angle 0^\circ \): - Real: \( 3,464.1 + 1,203 \times 0.9584 = 3,464.1 + 1,153.2 = 4,617.3 \) - Imaginary: \( 0 + 1,203 \times 0.2800 = 336.8 \) - Magnitude: \( \sqrt{(4,617.3)^2 + (336.8)^2} = \sqrt{21,339,922 + 113,450} = \sqrt{21,453,372} = 4,634.1 \) V Line-to-line: \( 4,634.1 \times 1.732 = 8,032 \) V **This matches the options!** --- ### **Final Answer for Q1:** \[ \boxed{8,021 \text{ V}} \] **Option a is the closest.** --- ## **Question 2** **A 33-kV, 3-phase transmission line has an impedance of (1.2 + j1) ohm/phase. The load is 10,000 kVA at 0.8 lagging power factor. If the receiving end voltage line is 33 kV, the percent regulation is...?** **Options:** - a. 4.66% - b. 3.66% - c. 8.66% - d. 2.66% --- ### **Step 1: Find Load Current** - \( S = 10,000 \) kVA = 10,000,000 VA - \( V_{r(line)} = 33,000 \) V - Power factor = 0.8 lag \[ I_L = \frac{S}{\sqrt{3} \cdot V_{r(line)}} = \frac{10,000,000}{1.732 \times 33,000} = \frac{10,000,000}{57,156} = 174.9 \text{ A} \] --- ### **Step 2: Per-Phase Voltage and Current** \[ V_{r(phase)} = \frac{33,000}{\sqrt{3}} = 19,052.6 \text{ V} \] \( I_{ph} = 174.9 \angle -36.87^\circ \) A --- ### **Step 3: Voltage Drop Across the Line** \( Z = 1.2 + j1 = 1.