A three-phase, 318.75kVA, 2300-Volt, alternator has an armature resistance of 0.35/phase and a synchronous reactance of 1.2/phase. Determine the no-load lineto- line generated voltage and the voltage regulation at:

The complete question asks us to: Determine the no-load line-to-line generated voltage and the voltage regulation at Full-load kVA, 0.8 power factor lagging, rated voltage, and voltage regulation.

The percentage of the voltage regulation from the power factor lagging and rated voltage is 6.12%

How do you determine the no-load line-to-line generated voltage and voltage regulation?

From the given information, we have the following parameters:

3-Phase alternator with 318.75 kVA, 2300-Volt
Armature resistance = 0.35 Ω/phase
Synchronous reactance = 1.2 Ω/phase

The Full load current ([tex]\mathbf{I_{fl}}[/tex]) can be determined by using the expression:

318.75 × 10³ = √3 ×2300 × [tex]\mathbf{I_{fl}}[/tex]

[tex]\mathbf{I_{fl}= \dfrac{318.75 \times 10^3}{\sqrt{3} \times 2300}}[/tex]

[tex]\mathbf{I_{fl}= 80.01 \ amperes}[/tex]

Now, the line-to-line generated voltage no-load (Emf) and the voltage regulation can be computed as follows:

Power factor = 0.8 lagging
Power factor angle = cos⁻¹(0.8) = 36.87°
Current [tex]\mathbf{I_{fl}}[/tex] = 80.01 ∠ -36.87°

For alternator:

[tex]\mathbf{E_f(phase) = V_t (phase) + I_a(R_a+js)}[/tex]

[tex]\mathbf{E_f(phase) =\dfrac{2300}{\sqrt{3}}+(80.01 < -36.87)(0.35+j1.2)}[/tex]

[tex]\mathbf{E_f(phase) =1327.90 + 80.01 +60j}[/tex]

[tex]\mathbf{E_f(phase) =1409.18 < 2.44^0}[/tex]

However, the open-circuit voltage is:

[tex]\mathbf{E_f (line) = \sqrt{3}\times 1409.18 < 2.44^0}[/tex]

[tex]\mathbf{E_f (line) = (2400.78 < 2.44^0) volt}[/tex]

Voltage Regulation (V.R) = [tex]\mathbf{\dfrac{E_f-V_t}{V_t}}[/tex]

Voltage Regulation (V.R) = [tex]\mathbf{=\dfrac{1409.18 - 1327.90}{1327.90}\times 100}[/tex]

Voltage Regulation (V.R%) = 6.12%

Learn more about voltage regulation here:

https://brainly.com/question/15875284

#SPJ9