This is a complicated question but I will attempt to be as detailed as possible without over killing it.
First, let us try to understand H and what it actual means.
If we have a mass of inertia J, rotating at a speed of ωr itstores a kinetic energy Ws=0.5* ωr2/S and its momentum is m0=J ωr .
H, the inertia constant with dimensions of time, is the ratio of the stored energy (joules) at normal speed to the rating (VA) to the machine:
H= ωr/S = 0.5*J* ωr2/S = 0.5* m0* ωr2/S ≈ 20*J*n2/S (n is speed in frequency in r/s)
So, imagine a generator connected to the grid is at full load. If its load is suddenly thrown off (think disconnected from grid), it will accelerate at the rate (in rad/s2) of ωr /2H.
H represents one-half of the time taken to bring the rotating parts of the machine from rest up to rated speed with a torque corresponding to the rating.
Now, if you go back to your laws of rotation you will see the following:
Id2θ/dt2 = T
Where I is the moment of inertia and T is the torque.
Now this law also applies to the motion of a synchronous machine.
For a machine the net torque is the sum of all the torque acting on the machine, including shaft torque due to the prime mover, torque due to rotational losses (friction, windage, core losses etc) and the electro-mechanical torque. In our world of power engineers it comes out to:
Md2δ/dt2 = PM -PE
This is the swing equation. It is a differential equation that governs the motion of each machine of a system.
Where M is the inertia constant in MJ per electrical degrees.
Pm = mechanical power input
Pe= electrical power output
PM -PE is also replaced with Pa (acceleration)
δ=displacement angle of rotor with respect to a reference axis rotating at normal speed.
Replace PM -PE with Pa
Md2δ/dt2 = Pa
So if we divide out by M and integrate
dδ/dt = ω=ω0 + Pat/M
and
δ = δ0 + ω0 t + Pat2/(2M)
M is related to H by the following M= (MVA of machine)*H/(180*f)
The simplest case is one machine connected to an infinite bus, with damping neglected.
Md2δ/dt2 = PM -PE sin(δ)
Let’s work through a very simple problem and it will become clearer.
A 60 Hz machine with a H=2.7 Mj per MVA is operating at steady state with input and output of 1 per unit and an angular displacement of 45° with respect to an infinite bus. A fault occurs. Assume that the input remains constant and the output is given by:
PE = δ/90° or in radians 2δ/π
Plot the oscillation with respect to time for H=2.7. Compare it to a H with 2.6.
Answer:
Now we have our equation for one generator connected to an infinite bus:
Md2δ/dt2 = PM -PE
Md2δ/dt2 = 1 - 2δ/π
We know the initial conditions (i.e. at t=0):
δ =45° in radians π/4
dδ/dt=0 (basically steady state)
To solve for this we must move things around and integrate .
I am going to skip all that math and just give you the answer:
δ = π/2 - π/4 * cos(√(2/( π*M)*t)
Just to double check this at time 0 seconds I get:
π/2 - π/4*cos(0) which is equal π/4 or 45° which is what I expect.
At time 0.05 i get 46.38°.
At time 0.1 i get 50.46°
You get the point. I keep going until I have enough points to plot.
Now if I was to plot this with a time interval step of 0.05 seconds I would get the blue line below for H=2.7.
If I was to change H to 2.6 you can already see the difference. It oscillates much more and gets to its destination much quicker. The reason why it oscillates forever because we ignored damping. In reality this problem gets much more complicated as we begin to loosen assumptions and take other considerations. Much, Much more complicated. I can not stress that enough. This is just the starting point to really understanding this problem and understand how "H" effects stability.
