Synchronous Generator Question

I have a tutorial question I'm a little stuck on, if anyone could give me some assistance it would be greatly appreciated

The Q:
A 200kVA, 660V, 50Hz, 4-pole star-connected synchronous generator has a synchronous reactance of 1ohm per phase. When the machine is operaing at lagging power factor, rated voltage and current, the torque angle [latex]\delta=\arctan(0.2)[/latex]. Calculate the power factor and torque.

What I've got so far:
[latex]V_ph=\frac{V_L}{sqrt{3}}=\frac{660}{sqrt{3}}=381.05[/latex]

[latex]S=sqrt{3}.V_L.I_L \Rightarrow I_L=\frac{S}{sqrt{3}.V_L}=\frac{200*10^3}{sqrt{3}.660}=174.95[/latex]

For star-connected [latex]I_L=I_ph[/latex]

[latex]I.X_s=174.95*1=174.95[/latex]

I've managed to get this far but I'm stuck on how to get Ef and phi.

Thanks for any help or ideas

This is not the solution to your problem but it maybe give you some ideas to solve yours
Please forgive me if the terminology isn't 100% correct but I translate it from my Dutch "Elektriciteit Deel III Elektrische machines" handbook from back in 1982!!

Given
2 pole, 3 fase, star-connected synchronous generator
Reactive power 12MVA
Uline = 6300V
Total Fe losses = PFe = 160kW
Excitation power = Pe = 44kW with 240V DC
Resistance/fase = 0.06Ohm
Reactance/fase = 0.9Ohm
Power factor = 0.8

Asked
Joule losses
Induced Eg/fase

Result
Usefull power Pu = Pr * cos phi = 12 * 0.8 = 9600kW
Current (fase & line) = Pu / sqrt(3) * Uline = 9600kW / sqrt(3) * 6300V = 880A
Joule losses = PJs = 3 * I² * R = 3 * 880² * 0.06 = 139.5kW
Total power applied to the generator
Pt = Pu + PFe + Pe + PJs
Pt = 9600 + 160 + 44 + 139.5 = 9943.5kW
Efficiency = 9600 / 9943.5 = 96.5%
Ohmic voltage losses = I * R = 880 * 0.06 = 52.8V
Inductive voltage losses = I * Xs = 880 * 0.9 = 792V
Voltage/phase = Uline / sqrt(3) = 6300 / sqrt(3) = 3630V
Induced Eg/fase see attached vector diagram Eg = 4191V

In your case you lack the power factor to split the inductive losses
But since you know three parameters of the triangle it can be solved
U = 381.05V
Inductive voltage losses = 174.95V
Torque angle = ATAN(0.2) = 11.3099°

Hope this helped...

synchronous generator torque angle, power factor, torque

The Q:
A 200kVA, 660V, 50Hz, 4-pole star-connected synchronous generator has a synchronous reactance of 1ohm per phase. When the machine is operating at lagging power factor, rated voltage and current, the torque angle . Calculate the power factor and torque.

Answer:

Voltage drop in the coil: V=(I)×(X)
Current (I) passing through the coil:
I=S÷(√3 V)=(200×〖10〗^3)/(1.732×660)=174.96 (amps)
As known X=1 ohms
V=(I)×(X)=174.96×1=174.96 (volts)
Computing ∠α,∠α is between vector E_f and vector (V)
sinα/V=sinδ/V

δ=tan^(-1)0.2=11.31°
α=sin^(-1)( sinδ/V_coil ×V)=sin^(-1)(sin11.31/174.96×660)=47.72°
Calculating ∠φ,
φ=90°-(α+δ)=90°-47.72°-11.31°=30.97°
Computing power factor pf
pf=cosφ=cos30.97°=0.86

Calculating torque T,
T=P/ω=(pf× S)/ω
ω=2π/60 ((120f))/n=2π/60 (120×50)/4=157 (rad/s)
T=(0.86×200×〖10〗^3)/157=1095.54 (N-m)=808.03 (F-lbs)