Synchronous Generator Question

[Timmymna]

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

 

[mcs51mc]

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...

 

[garyluo]

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)