5.7 MATRIX REPRESENTATION OF ALL STATE EQUATIONS

 

It is in the form of

ẋ = Ax + BU + Fd

U – control vector

d – disturbance vector

x – state vector

Equation (5.18) is converted into standard state space model form.

i.e. ẋ = Ax + Bu is written as follows:

If x is made up of the sum of transient and steady state value, and U is also made up of transient and steady state value, then

. then substitute in the equation ẋ = Ax + BU + Fd

Therefore,    ẋ = Ax¹ + Ax_ss + BU¹ + BU_ss + Fd      (5.19)

If a sudden disturbance occurs and the system reaches the steady state value, then ẋ = 0

Therefore Ax_ss + BU_ss + BU¹ + Fd = 0

and the transient term is also zero under steady state. Hence ẋ = 0 ⇒ Ax_ss + BU_ss + Fd = 0

Example 5.1

Consider the block diagram model of LFC (load frequency control) given in Figure 5.11.

Make the following approximation

(1 + sT_g )(1 + sT_t ) = 1 + (T_g + T_t )s = 1 + sT_eq.

Solve for ∆ƒ(t) with parameter ∆P_D=0.01.

Solution:

Consider input ∆P_C (S) = 0, then there is only one input i.e., ∆P_D (S).

Redrawing the above block diagram as in Fig 5.12,

By equating the same order coefficients of s

 

 

Solving Equations (1) and (2).

Therefore  Δf(t) = −0.01537e^−0.05t + 0.00635e^35.92t

Example 5.2

Determine the automatic load frequency control (ALFC loop parameters for a control area having the following data:

Total rated area capacity = 2000 MW

Normal operating load = 1000 MW

Inertia constant H = 5

Regulation R = 2.4 H/MW

Solution:

Example 5.3

For an isolated single area, consider the following data:

Area capacity = 1000 MW

Normal operating load = 500 MW

Inertia constant = 5 sec

Speed Regulation R = 5 % = 2.5 Hz/p.u MW

Operating frequency f₀ = 50 Hz

The load decreases by 1 percent. For a decrease in frequency by 1 percent, find the gain and time constant of the power system represented by a first order transfer function. If it is an uncontrolled area then find out the change in frequency due to an increase of load by 75 MW. Find in what proportion (the increase in load demand is met by the increase in generation, and the decrease in load due to drop in frequency?

Solution:

Increase in generation APG,

Example 5.4

A control area had a total rated of capacity 1000 MW. The speed regulation ‘R’ for all the units in the area is 2 Hz/p.u MW. A one percent change in frequency causes a one percent change in load. If the system is operating at half of the rated capacity and the load increase by 2 percent,

1. find the static frequency drop.
2. if the speed governor loops were open what would be the frequency drop.

Solution:

1. R = 2Hz/p.u MW

   

2. Speed governor loop is open, then R → infinite

   

Example 5.5

An uncontrolled isolated power system has the following parameters:

Rated output=350 MW

Regulation=0.06 p.u

Inertia constant=5

Turbine time constant=0.6 sec

Governor time constant=0.3 sec

Nominal frequency=50 Hz

The load varies by 0.7 percent for a 1 percent change in frequency. Determine the steady state frequency deviation in Hz for a load change of 50MW.

We know that for the uncontrolled case,

k_g k_t = 1, and substituting for k_p, T_t, T_g, ΔP_D(S), R in above equation.

but

Now actual frequency of the system

f = f° + Δf|_{steady state} = 50 + (−0.429)

f = 49.571 Hz

Example 5.6

An uncontrolled single area has two alternators with speed governors. Generating unit ’1’ is rated 600MVA, with 5 percent speed regulation is p.u. and generating unit ‘2’ is rated at 400MVA, with 4 percent speed regulation in p.u. Under nominal operating conditions with ƒ° = 50 Hz, both the units share a load of 700MW. With an assumption that no frequency dependent load is connected to the system, find the new frequency and load share between each generators for an additional load of 85MW, if the initial share is 500MW and 200MW on alternators ‘1’ and ‘2’ respectively. Assume a base of 800MW.

Solution:

Example 5.7

Repeat Example 5.6 assuming frequency dependent load such that load increases by 1 percent for 1 percent increase in frequency.

Solution:

Example 5.8

Obtain the dynamic response of an uncontrolled isolated power system with the following loop parameters R = 2 Hz/p.u MW, k_p = 75 Hz/p.u MW, T_p = 20 sec, ∆P_D = 0.02 p.u.

Solution:

 

We know that the dynamic response of an uncontrolled isolated power system is given by

Based on Equation (1) the response can be obtained as shown below:

 

Example 5.9

A single area load frequency control system has the following parameters with free governor action:

Total capacity P_a = 1800 MW

Nominal operating load P_D = 900 MW

Nominal frequency f° = 50 Hz

Inertia constant H = 5 sec

Regulation R = 4 percent

k_p = 90 Hz/p.u MW

Given that T_p = 20 sec, T_g = T_t = 0.4 sec, 1 percent change in frequency causes 1 percent change is load.

1. Check whether the ALFC is stable or not.
2. Determine the location of three eigen values.

Solution:

1. From Fig 5.5
   

   Fig 5.15 Single area LFC with ∆P_C (s) = 0

   

   Therefore the characteristic equation is

   

   Using Routh’s array

   

   Since all the elements in the first column are > 0, the system is stable.

    

2. Finding the root of the equation 3.2s³ + 16.16s² + 20.8s + 46 = 0

   x₁ = −4.31, x₂ = −0.3672 ± j1.788

   The Eigen values are

   −4.31, −0.3672 + j1.788, −0.3672 − j1.788.

Example 5.10

A generator has a primary ALFC loop characterizer by R = 4 percent. The secondary loop is open. It is feeding a load consisting of three equal resistors ‘R’ connected in a star network. The frequency is 50 Hz. The resistances are decreased until the frequency drops to 49 Hz. Find by how many percent the resistance ‘R’ has to be decreased. Assume that the AVR loop keeps the terminal voltage constant.

Solution:

Per phase equivalent of the above data is shown below:

Resistance ‘R’ is decreased until the frequency drops to 49 Hz.

But we know that the power delivered to R is given by P = V²/R Watts and the change in frequency is ∆f = 1Hz.

With R = 4 percent, → 0.04 × 50 = 2 Hz/p.u MW

It shows that 2 Hz causes a 100 percent change in load

Therefore Δf(1 Hz) cause = = 50 percent change in load.

Let the new load be R_x = xR where ‘R’ is the given load.

x(0.5) = 1 − x ⇒ x = 0.667

Decrease in R is (100 – 66.67) = 33.33 percent

Resistance R has to be decreased by 33.33 percent for a frequency decrease of 1 Hz