

Power Grid Kinetic Energy and Inertia 

Here we discuss the kinetic energy that is stored in the power grid and the inertia of the power grid.
For this level of analysis we shall ignore transient effects such as occur due to short circuits
and switching operations. In other words we consider that the measured frequency is ubiquitous, i.e.,
nearly the same everywhere in the interconnected power system.

Terminology 

The power industry expresses the kinetic energy of the grid as follows: 

H = E0 / S0 
Kinetic Energy Factor of the Grid.
Generally expressed as grid energy (E0)
per grid size (S0) at nominal frequency (F0)
Typically H ≈ 5 (in seconds = MWsec/ MVA) approximately and is not constant under all conditions. 
E0 
Grid Kinetic Energy in MWseconds in our discussion 
P0 
Grid Size 
F0 
Nominal Frequency 
E 
Grid Kinetic Energy expressed in MWseconds 
E = I ω^2 / 2 
The kinetic energy of a rotating object (a flywheel for
example) 
I 
Moment of Inertia ( in our discussion, H serves as a proxy for
I ) 
ω = 2 π F 
Angular Velocity in Radians per Second 


By manipulation we get the following: 

E = H * (F / F0) ^ 2 * S0 

In other words, the kinetic energy of the grid varies as the square of the frequency

Example 1 

What is the approximate kinetic energy of a 200,000 MVA grid operating at nominal frequency? 

E = (5) (200,000) = 1,000,000 MWseconds 
Example 2 

Given a loss of 1000 MW of generation on the grid of the previous example, how many seconds will it take for the frequency to drop
from its nominal value of 60 Hz down to 59.5 Hz? For this example, assume that
the corresponding loss of apparent power is 1250 MVA. Also assume that there is no corrective action taken either by the
system operator or by the governors on all the generators and that the loads do not change (decrease) with the frequency
drop. This is the fastest that the frequency will decrease to 59.5 Hz.


The change in kinetic energy, starting after the loss of generation, is calculated as follows: 

ΔE = H * (F / F0) ^ 2 * (S0 –
1250)  H * (S0 – 1250) = H * (S0 –
1250) * [(F / F0)
^ 2 – 1] 

ΔE = 993,750 * [(59.5/60) ^ 2 – 1] = 16,493 MWseconds


T = (16,493 MWseconds) / (1000 MW) = 16.493 seconds


In the real world the frequency would not drop this quickly because of automatic
generator governor action and the decrease in load as the frequency drops.
