A 250 MVA, 25 kV, 3 phase synchronous generator has a synchronous reactance X, of 1.6 pu. It is delivering the rated output of 250 MVA, 25 kV at the generator terminals at a power factor of 1.0. (a) Calculate the per unit base impedance of the generator ZB and the synchronous reactance X, in Ohm. (b) Calculate the phase current magnitude. (c) Calculate the per phase voltage drop E – V across the synchronous reactance.

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**Title: Understanding 3 Phase Synchronous Generator Fault Current Calculations**

**Introduction:**

This study exercise explores the characteristics and calculations related to a 3 phase synchronous generator with a rated output of 250 MVA and 25 kV. These assessments focus on the generator's impedance, reactance, and behavior during fault conditions.

**Details:**

1. **System Specifications:**
   - **Generator Rating:** 250 MVA, 25 kV, 3 phase
   - **Synchronous Reactance (Xs):** 1.6 pu
   - **Power Factor:** 1.0

2. **Calculations:**

   - **(a)** Determine the per unit base impedance \( Z_B \) and the synchronous reactance \( X_s \) in ohms.
   
   - **(b)** Calculate the magnitude of the phase current.
   
   - **(c)** Calculate the per phase voltage drop \( E - V \) across the synchronous reactance.
   
   - **(d)** Find the line to neutral no load voltage \( E \).
   
   - **(e)** Consider a scenario with a 3 phase fault near the generator terminals. Assuming constant field current and voltage \( E \), calculate the steady state fault current magnitude.

   - **(f)** Discuss the initial fault current due to the 3 phase fault, which is larger in transient situations due to decreased effective synchronous reactance. The transient synchronous reactance \( X'_d \) is 0.23 pu. Calculate the initial fault current using this transient synchronous reactance. This current is critical for the design of circuit breakers as it represents the maximum current that they must interrupt.

**Conclusion:**

This exercise provides insights into calculating and understanding the electrical parameters and behaviors of synchronous generators, particularly under fault conditions. It highlights the importance of accurate calculations in ensuring system stability and safety.
Transcribed Image Text:**Title: Understanding 3 Phase Synchronous Generator Fault Current Calculations** **Introduction:** This study exercise explores the characteristics and calculations related to a 3 phase synchronous generator with a rated output of 250 MVA and 25 kV. These assessments focus on the generator's impedance, reactance, and behavior during fault conditions. **Details:** 1. **System Specifications:** - **Generator Rating:** 250 MVA, 25 kV, 3 phase - **Synchronous Reactance (Xs):** 1.6 pu - **Power Factor:** 1.0 2. **Calculations:** - **(a)** Determine the per unit base impedance \( Z_B \) and the synchronous reactance \( X_s \) in ohms. - **(b)** Calculate the magnitude of the phase current. - **(c)** Calculate the per phase voltage drop \( E - V \) across the synchronous reactance. - **(d)** Find the line to neutral no load voltage \( E \). - **(e)** Consider a scenario with a 3 phase fault near the generator terminals. Assuming constant field current and voltage \( E \), calculate the steady state fault current magnitude. - **(f)** Discuss the initial fault current due to the 3 phase fault, which is larger in transient situations due to decreased effective synchronous reactance. The transient synchronous reactance \( X'_d \) is 0.23 pu. Calculate the initial fault current using this transient synchronous reactance. This current is critical for the design of circuit breakers as it represents the maximum current that they must interrupt. **Conclusion:** This exercise provides insights into calculating and understanding the electrical parameters and behaviors of synchronous generators, particularly under fault conditions. It highlights the importance of accurate calculations in ensuring system stability and safety.
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