Synchronizing
Synchronizing
(OP)
1. Does anyone have some simple formulas for determining the amount of current that will flow right at generator breaker closeing based on the voltage difference between a generator and the bus. Assume the phase angle difference is 0 degrees. Assume an infinite bus. I am thinking it is Delta volts divided by the subtransient reactance of the generator.
2. Does anyone have a simple formula to estimate the amount of power that will flow based on phase angle difference or slip rate of a generator synchronizing to a bus? Assume the voltage of the the generator and bus are exactly the same. The only variable is the slip rate.
I am trying to set a synch check relay and I am thinking to set it wide as long as I don't exceed the nameplate ratings of the generator, as a check device that should be OK.
2. Does anyone have a simple formula to estimate the amount of power that will flow based on phase angle difference or slip rate of a generator synchronizing to a bus? Assume the voltage of the the generator and bus are exactly the same. The only variable is the slip rate.
I am trying to set a synch check relay and I am thinking to set it wide as long as I don't exceed the nameplate ratings of the generator, as a check device that should be OK.






RE: Synchronizing
If the machine slips a pole while you're standing there, you won't be worrying about current magnitude in the future.
Get some good references for your type of generator and follow the guidelines for synch check settings. This will also be heavily influenced by the speed control capability of the prime mover and governor.
RE: Synchronizing
I hope you are talking about momentary paralleling, less than 0.1 sec.
Rafiq Bulsara
http://www.srengineersct.com
RE: Synchronizing
at first I thoght the increase in MW at synch would be propertional to the delta speed change as a function of the governor response. which would be a part of the answer IF, the unit was synched on the governor.
But it is possible to sych with a constain valve opening (LL or SVBP) and thus the increase in power would be the change in inertai of slowing the speed at synch. this would be a short duration bump in current since the steam flow to maintain speed would still be at no load.
RE: Synchronizing
Out of phase s more serious. There will be high transient torques and currents as the unit pulls into sync. Possible results may range from tripped circuit breakers to sheared or exploded couplings, damaged Generator Step Up transformers, blown diodes, cracked foundations or destroyed generators.
The governor has no effect, the damage is done by the inertia of the rotor as it abruptly tries to alter its angular position.
I have seen successful closings at 30 degrees phase angle error on a mix of 600 kW and 350 kW diesel sets. This was really the edge of the envelope. I would suggest 15 degrees max and as little as 5 degrees if you are able to control that precisely.
A couple of caveats that may upset the best planned protection if not considered.
1> Beware of slow responding breakers. More an issue with old existing plants. Evaluate the possibility of the unit going too far out of sync between the permission for a breaker close and the actual closing of the breaker.
2> Closing frequency; out of sync implies off frequency. Avoid closing when the frequency is below grid frequency. This may lead to reverse power or no load trips.
The best manual sync is with the synchroscope at twelve o-clock and rotating slowly clockwise.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Synchronizing
thread238-233765: Synchronizing Torque
----------------------------------
If we learn from our mistakes I'm getting a great education!
RE: Synchronizing
% Assume the phase angle difference is 120 degrees
tetaG = 120; %deg generator angle
Vg0=1; % pu % voltage gen
angrG = tetaG * pi/180; % angle in radian
Vg= Vg0 * exp (j * angrG); % vectorial tension
Xg = 0.20; % pu - generator impedance at Nb and Vb base
%3phase fault at generator terminal
Icc3 = Vg / Xg;
Icc3 = abs(Icc3);
Icc3 = Icc3 * Ib13;
tetaS = 0; % system angle
Vs0=1; %pu
angrS = tetaS * pi/180;
Vs= Vs0 * exp (j * angrS); % vectorial tension
Xs=0.03; %pu system impedance at Nb and Vb base
% I- resultant corrente due to 120 deg out-of-phase sincronization
I= (Vg-Vs)/ (Xg + Xs) ;% pu
I= abs(I); % pu
Ib = Nb/(1.732*Vb); % Ibase
I = I * Ib; % Ampere
RE: Synchronizing
So to reiterate:
1. Does anyone have some simple formulas for determining the amount of current that will flow right at generator breaker closeing based on the voltage difference between a generator and the bus. Assume the phase angle difference is 0 degrees. Assume an infinite bus. I am thinking it is Delta volts divided by the subtransient reactance of the generator.
2. Does anyone have a simple formula to estimate the amount of power that will flow based on phase angle difference or slip rate of a generator synchronizing to a bus? Assume the voltage of the the generator and bus are exactly the same. The only variable is the slip rate.
RE: Synchronizing
2. Initially there will be enough energy to accelerate or de-cellerate the rotor to synchronous speed. Then the power will depend on the phase angle which is controlled by the throttle setting. Typically very little until the governor is adjusted to pick up the load.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Synchronizing
answer 1.
% I- resultant corrente due to 120 deg out-of-phase sincronization
I= (Vg-Vs)/ (Xg + Xs) ;% pu
RE: Synchronizing
thanks for the info but why 120 deg out of phase? you mean in phase but with VG-VS right?
RE: Synchronizing
No! Vg-Vs are in complex form.
See my comments before using Matlab
tetaG = 120; %deg generator angle
Vg0=1; % pu % voltage gen
angrG = tetaG * pi/180; % angle in radian
Vg = Vg0 * exp (j * angrG); % vectorial tension
tetaS = 0; % system angle
Vs0=1; %pu
angrS = tetaS * pi/180;
Vs= Vs0 * exp (j * angrS); % vectorial tension
RE: Synchronizing
It is all in complex form.
See my comments before using Matlab
tetaG = 120; %deg generator angle
Vg0=1; % pu % voltage gen
angrG = tetaG * pi/180; % angle in radian
Vg = Vg0 * exp (j * angrG); % vectorial tension
tetaS = 0; % system angle
Vs0=1; %pu
angrS = tetaS * pi/180;
Vs= Vs0 * exp (j * angrS); % vectorial tension