Synchronous Machine Swing Equation
11-23-2016, 08:54 PM (This post was last modified: 11-29-2016 11:30 AM by Ángel Martin.)
Post: #1
 Ángel Martin Senior Member Posts: 1,046 Joined: Dec 2013
Synchronous Machine Swing Equation
Synchronous Machine Swing Equation. [SWING ]
From the author’s Engineering Collection, included in the ETSII6 module.

With this program you can model the transients of a synchronous machine (either as generator or as motor) when a sudden load change is imposed to the system. Typically this is done looking at the electrical angle "x" between a point on the rotor and the synchronous reference frame.

Using this variable, the basic swing equation expressed in system Torque terms (Mm = mechanical torque, kd = damping constant, and Pe = electrical power) has the form:

(J/2p) d2x/dt2 = Mm + Kd dx/dt + Pe sin x / [w1 + (1/2p) dx/dt ] (1)

where 2p is the number of poles in the machine and w1 is the synchronous rotation speed. Note that the last term cannot be simplified for oscillations of large amplitude; i.e. sin x is not equivalent to the angle x, and dx/dt is not null. This introduces some complexity but produces more accurate results.

The program uses the 4th-degree Runge-Kutta formula to solve numerically this second-order differential equation. The routine is called for each time instant, and the results are stored in an X-memory data file (named “SWING”) as data pairs (t, x). The stabilization time and the maximum value of the oscillation will occur at tmax, determined using the control systems theory – according to the general expression form of a transfer function it follows:

tmax = pi / sqrt [ (2p Pe /J w1) – (2p Kd /2J)^2 ]
tstb = 2pi J / 2p Kd

whereas the stationary final value can be easily obtained from the equation (1) above, zeroing the derivatives of the angle:

x finl = asin (Mn w1 / Pe)

Finally, the program uses a step size for the increments as h/2 = tmax /60

Example.

Characterize the oscillations when a 28-pole, 600 kW synchronous generator rotating at 22 rad/s experiences a sudden change of the load given by Mn = 10 kN.m. Assume a dampening constant Kd = 200 N.m/rad.s, and angular momentum of inertia J = 444 kg/m^2

Results: The time for maximum angle is tm = 0,107755 s
And the stabilization angle: x fnl = 0,375424

The table below summarizes the values for each instant of time where the calculations have been made:

Code:
t (s)        angle(rad)    |   t (s)       angle(rad)   |   t (s)       angle(rad) ---------------------------|----------------------------|--------------------------- 0,035918     0,172011      |   0,215510    0,214867     |   0,395102    0,382101 0,172011     0,482708      |   0,251429    0,252188     |   0,431021    0,311437 0,107755     0,634696      |   0,287347    0,386210     |   0,466939    0,310819 0,143674     0,546672      |   0,323266    0,480650     |   0,502858    0,364791 0,179592     0,343725      |   0,359184    0,467154     |   0,538776    0,415264

Note that the program will generate data pair results until the SWING file is full, presenting the "END OF FILE" message. At that juncture you can enlarge the file (using RESZFL) and continue getting more results if so desired.

Attached File(s)