A 60 Hz four-pole synchronous generator, connected to a power grid, has a synchronous reactance of 5 ohm. The line-line voltage of the stator winding connected in Wye is 15KV. The generator line current is 1000A at 0.9pf lagging. Determine:

(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

Explanation:

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answer: 5

explanation:

6-1