Respuesta :
Answer:
(a) δ = [tex]27.27[/tex]°
(b)
[tex]I_{a} =194.3<-7.96[/tex] [tex]A[/tex]
[tex]PF = 0.99[/tex] [tex]lagging[/tex]
[tex]Q=27.96[/tex] [tex]kVAR[/tex]
(c)
[tex]E_{a}=463.9<21.93[/tex]° [tex]V[/tex]
δ = [tex]21.93[/tex]°
Given Information:
Internal generated voltage = Ea= 420 V
Voltage = VLL= 600 V
Output Power = P = 200 kW
Synchronous Reactance = Xs= 1 Ω
Required Information:
Torque angle = δ = ?
Generator current = Ia = ?
Power factor = PF = ?
Reactive power = Q = ?
Excitation voltage at 0.9 PF lagging = Ea = ?
New torque angle = δ = ?
Solution:
For a wye (Y) connected generator the relationship between terminal voltage and phase voltage is [tex]V_{ph}=V_{T} / \sqrt{3}[/tex]
[tex]V_{ph}=600/\sqrt{3}[/tex]
[tex]V_{ph}=346.4[/tex] [tex]V[/tex]
(a) Torque angle = δ = ?
δ = [tex]sin^{-1} (X_{s}P/3V_{ph} E_{a} )[/tex]
δ = [tex]sin^{-1} (1*200,000/3*346.4*420)[/tex]
δ = [tex]27.27[/tex]°
(b) Generator current = Ia = ?
[tex]I_{a} =(E_{a} -V_{ph}) /X_{s}[/tex]
[tex]I_{a} =(420<27.27-346.4<0)/j1[/tex] (complex notation)
[tex]I_{a} =194.3<-7.96[/tex] [tex]A[/tex]
(b) Power factor = PF = ?
PF is cosine of the angle between [tex]V_{ph}[/tex] and [tex]I_{a}[/tex]
[tex]PF=cos(0-(-7.96)) = 0.99[/tex] [tex]lagging[/tex]
(b) Reactive power = Q = ?
[tex]Q=3V_{ph}I_{a} sin(7.96)[/tex]
[tex]Q=3*346.4*194.3*sin(7.96)[/tex]
[tex]Q=27.96[/tex] [tex]kVAR[/tex]
(c) Excitation voltage at 0.9 PF lagging = Ea = ?
[tex]I_{a} =\frac{P}{3V_{ph}cos(0.9)}[/tex]
[tex]I_{a} =\frac{200,000}{3*346.4*cos(0.9)}[/tex]
[tex]I_{a}=192.5<-25.84[/tex]° [tex]A[/tex] ( [tex]cos^{-1}(0.9)=25.84[/tex]° )
[tex]E_{a}=V_{ph} + I_{a}X_{s}[/tex]
[tex]E_{a}=346.4<0 + (192.5<-25.84)(1j)[/tex]
[tex]E_{a}=463.9<21.93[/tex]° [tex]V[/tex]
(c) New torque angle = δ = ?
Torque angle is the angle of internal generated voltage Ea
δ = [tex]21.93[/tex]°
Decreasing the PF from 0.99 lagging to 0.9 lagging will increase the Ea from 420 to 463.9 and decreases the torque angle from 27.27 to 21.93
