To calculate the maximum power that a synchronous generator can supply, we use the power-angle equation for a synchronous generator:

$P_{\text{max}} = \frac{V_t \cdot E_g}{X_s}$

Where:

• $P_{\text{max}}$ = Maximum power (in watts or kW)
• $V_t$ = Terminal voltage (in volts, V)
• $E_g$ = Internal air-gap voltage (in volts, V)
• $X_s$ = Synchronous reactance (in ohms, $\Omega$)

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Step 1: Identify given values

• $V_t = 13.2 \, \text{kV} = 13,200 \, \text{V}$
• $E_g = 17.5 \, \text{kV} = 17,500 \, \text{V}$
• $X_s = 2 \, \Omega$

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Step 2: Calculate maximum power

Apply the power formula:
$P_{\text{max}} = \frac{V_t \cdot E_g}{X_s}$ $P_{\text{max}} = \frac{13,200 \, \text{V} \cdot 17,500 \, \text{V}}{2 \, \Omega}$ $P_{\text{max}} = \frac{231,000,000 \, \text{V}^2}{2 \, \Omega}$ $P_{\text{max}} = 115,500,000 \, \text{W} = 115.5 \, \text{MW}$

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Final Answer

The maximum power that the synchronous generator can supply is:
$P_{\text{max}} = 115.5 \, \text{MW}$

Would you like a step-by-step explanation or details on any part of the process?

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5 Related Questions

1. What is the effect of synchronous reactance on the maximum power a generator can supply?
2. How does the power angle impact the real power output of a synchronous generator?
3. Can you calculate the current flowing through the generator under maximum power conditions?
4. How does increasing terminal voltage affect the maximum power of the generator?
5. What is the physical meaning of "internal air-gap voltage" in synchronous machines?

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Quick Tip

Reducing synchronous reactance ($X_s$) increases the maximum power that a generator can deliver. This is why modern generator designs aim to minimize synchronous reactance.