Let's analyze and solve the problem step by step. --- ## Given Data: - **Type:** Three-phase, 4.17 kV, 60 Hz, wye-connected, cylindrical-rotor synchronous generator - **Leakage Reactance per phase:** \( X_s = 2.2\, \Omega \) - **Winding Resistance per phase:** \( R_s = .15\, \Omega \) - **Generator Power:** \( P_{load} = 950\, \text{kVA} \) - **Power factor:** .85 lagging - **Line-to-line Voltage:** \( V_{L} = 4.17\, \text{kV} = 417\, \text{V} \) --- ## Step 1: Convert line-to-line voltage to phase voltage Since the generator is wye-connected: \[ V_{ph} = \frac{V_{L}}{\sqrt{3}} = \frac{417\, V}{\sqrt{3}} \approx 241\, V \] --- ## Step 2: Calculate the secondary (phase) load current Total apparent power: \[ S_{load} = 950\, \text{kVA} \] Power factor: \[ \cos \phi = .85 \] Reactive power: \[ Q_{load} = S \sin \phi = 950 \times 10^3 \times \sin(\cos^{-1} .85) \] Calculate \(\sin \phi\): \[ \sin \phi = \sqrt{1 - \cos^2 \phi} = \sqrt{1 - .85^2} = \sqrt{1 - .7225} \approx \sqrt{.2775} \approx .527 \] Reactive power: \[ Q_{load} = 950 \times 10^3 \times .527 \approx 500,150\, \text{VAR} \] Active power: \[ P_{load} = S \cos \phi = 950 \times 10^3 \times .85 \approx 807,500\, \text{W} \] Calculate phase current: \[ I_{ph} = \frac{S_{load}}{3 V_{ph}} = \frac{950,000}{3 \times 241} \approx \frac{950,000}{723} \approx 131.3\, \text{A} \] --- ## Step 3: Determine the power factor angle \[ \phi = \cos^{-1}(.85) \approx 31.8^\circ \] --- ## Step 4: Find the internal generated emf \(E_{ph}\) The generator's internal emf per phase is related by the per phase voltage, the current, and the reactance: \[ V_{ph} = |E_{ph} - j X_s I_{ph}| \] But we need to consider the **air-gap voltage** (which is \(E_{ph}\)), and the load current phase angle. --- ## Step 5: Write the per-phase voltage equation The phasor diagram: \[ V_{ph} = E_{ph} - j X_s I_{ph} \] Expressed in complex form: \[ V_{ph} = E_{ph} \angle ^\circ \] \[ I_{ph} = I \angle -\phi = 131.3 \angle -31.8^\circ \] The reactive term: \[ j X_s I_{ph} = X_s I_{ph} \angle 90^\circ - \phi = 2.2 \times 131.3 \angle 90^\circ - 31.8^\circ = 288.9 \angle 58.2^\circ \] Expressed as a complex number: \[ j X_s I_{ph} = 288.9 (\cos 58.2^\circ + j \sin 58.2^\circ) \approx 288.9 (.529 + j .849) \approx 153 + j 245.4 \] --- ## Step 6: Calculate the internal emf \(E_{ph}\) Using the phasor form: \[ V_{ph} = E_{ph} - j X_s I_{ph} \] So, \[ E_{ph} = V_{ph} + j X_s I_{ph} \] In rectangular form: \[ V_{ph} = 241\, V \angle ^\circ = 241 + j \] Adding: \[ E_{ph} = (241 + 153) + j 245.4 = 2563 + j 245.4 \] Magnitude: \[ |E_{ph}| = \sqrt{2563^2 + 245.4^2} \approx \sqrt{6,568,369 + 60,213} \approx \sqrt{6,628,582} \approx 2574\, V \] --- ## **Final results:** ### **Induced air-gap voltage (per phase):** \[ \boxed{ E_{ph} \approx 2574\, \text{V} } \] ### **Induced line-to-line voltage:** \[ E_{L} = \sqrt{3} \times E_{ph} \approx 1.732 \times 2574 \approx 4465\, \text{V} \] --- ## **Summary:** | Quantity | Value | |---|---| | **Air-gap (internal emf)** | **2574 V per phase** | | **Line-to-line air-gap voltage** | **4465 V** | --- This completes the analysis. The exact voltage regulation would require comparing the no-load emf to the terminal voltage, but this provides the core calculations for the induced voltage and the internal emf.

Let's solve the problem by breaking it down into clear steps to answer all parts, including the induced voltage, air-gap voltage, and voltage regulation. --- ## Given Data: - **Generator Type:** Three-phase, wye-connected, cylindrical-rotor synchronous generator - **Line-to-line Voltage (\(V_L\)):** \(4.17\, \text{kV} = 4170\, \text{V}\) - **Frequency:** \(60\, \text{Hz}\) - **Leakage Reactance per Phase (\(X_s\)):** \(2.2\, \Omega\) - **Winding Resistance per Phase (\(R_s\)):** \(0.15\, \Omega\) - **Load Power (\(S_{load}\)):** \(950\, \text{kVA}\) - **Power Factor (pf):** \(0.85\) lagging --- ## Step 1: Convert Line-to-Line Voltage to Phase Voltage Since the generator is wye-connected: \[ V_{ph} = \frac{V_L}{\sqrt{3}} = \frac{4170}{\sqrt{3}} \approx 2410\, \text{V} \] --- ## Step 2: Calculate the Load Current Total apparent power: \[ S_{load} = 950\, \text{kVA} = 950 \times 10^3\, \text{VA} \] Active Power: \[ P_{load} = S_{load} \times \text{pf} = 950 \times 10^3 \times 0.85 = 807500\, \text{W} \] Reactive Power: \[ Q_{load} = S_{load} \times \sin(\phi) \] where \[ \phi = \cos^{-1}(0.85) \approx 31.8^\circ \] \[ \sin(\phi) \approx \sqrt{1 - (0.85)^2} \approx 0.527 \] Calculating \(Q_{load}\): \[ Q_{load} = 950 \times 10^3 \times 0.527 \approx 500150\, \text{VAR} \] Phase current: \[ I_{ph} = \frac{S_{load}}{3 V_{ph}} = \frac{950 \times 10^3}{3 \times 2410} \approx 131.3\, \text{A} \] --- ## Step 3: Determine the Power Factor Angle \[ \phi = \cos^{-1}(0.85) \approx 31.8^\circ \] --- ## Step 4: Calculate the Internal Generated EMF \(E_{ph}\) The voltage equation for the generator is: \[ V_{ph} = E_{ph} - jX_s I_{ph} \] 1. Calculate the reactive component: \[ jX_s I_{ph} = j(2.2)(131.3) = j288.86 \] 2. Convert phase current to complex form: \[ I_{ph} = 131.3 \angle -31.8^\circ = 131.3 (\cos(-31.8^\circ) + j\sin(-31.8^\circ) \approx 131.3 (0.85 - j0.527) \approx 111.6 - j69.2 \] 3. Calculate \(V_{ph}\): \[ V_{ph} = 2410 + j(-288.86) \approx 2410 - j288.86 \] 4. Rearranging gives: \[ E_{ph} = V_{ph} + jX_s I_{ph} = (2410 + 288.86) + j(-288.86 + 288.86) \approx 2698.86 + j0 \] 5. Magnitude of \(E_{ph}\): \[ |E_{ph}| = 2698.86\, \text{V} \] --- ## Step 5: Calculate the Air-gap Voltage The air-gap voltage is equal to the internal generated emf: ### **Air-gap Voltage:** \[ \boxed{E_{ph} \approx 2698.86\, \text{V}} \] ### **Line-to-Line Air-gap Voltage:** \[ E_L = \sqrt{3} \times E_{ph} \approx \sqrt{3} \times 2698.86 \approx 4675.54\, \text{V} \] --- ## Step 6: Calculate Voltage Regulation Voltage regulation is calculated as: \[ \text{Voltage Regulation} = \frac{E_{ph} - V_{ph}}{V_{ph}} \times 100\% \] Substituting values: \[ \text{Voltage Regulation} = \frac{2698.86 - 2410}{2410} \times 100\% \approx \frac{288.86}{2410} \times 100 \approx 12.0\% \] --- ## Final Summary: | Quantity | Value | |--------------------------------|-------------------| | **Induced Air-gap Voltage (per phase)** | **2698.86 V** | | **Line-to-line Air-gap Voltage** | **4675.54 V** | | **Voltage Regulation** | **12.0%** | This provides a full analysis of the induced voltage, air-gap voltage, and voltage regulation.