A three-phase, 60 Hz, 600 VL-L, wye-connected synchronous generator is delivering 200 kW at rated voltage. The excitation voltage Ear is 420 V per phase, and the synchronous reactance xs-1.0 Ω. (a) Calculate torque angle δ. (b) Find generator current, power factor, and reactive generation QT (c) For the same terminal voltage and active power generation P, determine the excitation voltage Ear to achieve a power factor of 0.9 lagging. What is the new value of the torque angle δ in this case?

(a) δ = °

(b)

(c)

°

δ = °

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

(a) Torque angle = δ = ?

δ =

δ =

δ = °

(b) Generator current = Ia = ?

  (complex notation)

(b) Power factor = PF = ?

PF is cosine of the angle between and

(b) Reactive power = Q = ?

(c) Excitation voltage at 0.9 PF lagging = Ea  = ?

°   ( ° )

°

(c) New torque angle = δ = ?

Torque angle is the angle of internal generated voltage Ea

δ = °

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

x+2 because it just doubles hon so the pattern can actually tell you the expression.

so for example,

if x=2 then 2+2 is 4

if x=4 then 4+4 is 8

and so on and so forth.

i really hope that this makes sense and you out in a way hon: )

can you make it more obvious. i can't see it properly