**LOAD-FREQUENCY DYNAMICS OF SINGLE-AREA AND TWO AREA POWER SYSTEMS**

**AIM:**

To become familiar with the modeling and analysis of load-frequency and tie-line flow dynamics of a power system with load-frequency controller (LFC) under different control modes and to design improved controllers to obtain the best system response.

**OBJECTIVES:**

1. To study the time response (both steady state and transient) of area frequency deviation and transient power output change of regulating generator following a small load change in a single-area power system with the regulating generator under “free governor action”, for different operating conditions and different system parameters.

2. To study the time response (both steady state and transient) of area frequency deviation and turbine power output change of regulating generator following a small load change in a single- area power system provided with an integral frequency controller, to study the effect of changing the gain of the controller and to select the best gain for the controller to obtain the best response.

3. To analyze the time response of area frequency deviations and net interchange deviation following a small load change in one of the areas in an inter connected two-area power system under different control modes, to study the effect of changes in controller parameters on the response and to select the optimal set of parameters for the controller to obtain the best response under different operating conditions.

**SOFTWARE REQUIRED:**

‘LOAD FREQUENCY CONTROL’ module of AU Power Lab or equivalent

**EXERCISES**

1. It is proposed to simulate using the software available the load-frequency dynamics of a single-area power system whose data are given below:

Rated capacity of the area = 2000 MW

Normal operating load = 1000 MW

Nominal frequency = 50 Hz

Inertia constant of the area = 5.0 s

Speed regulation (governor droop)

of all regulating generators = 4 percent

Governor time constant = 0.08 s

Turbine time constant = 0.3 s

Assume linear load–frequency characteristics which means the connected system load increases by one percent if the system frequency increases by one percent. The area has a governor control but not a load-frequency controller. The area is subjected to a load increase of 20 MW.

(a) Simulate the load-frequency dynamics of this area using available software and check the following:

(i) Steady – state frequency deviation Dfs in Hz. Compare it with the hand-calculated value using “Area Frequency Response Coefficient” (AFRC).

(ii) Plot the time response of frequency deviation Df in Hz and change in turbine power DPT in p.u MW upto 20 sec. What is value of the peak overshoot in Df?

(b) Repeat the simulation with the following changes in operating condition, plot the time

response of Df and compare the steady-state error and peak overshoot.

(i) Speed regulation = 3 percent

(ii) Normal operating load = 1500 MW

Hz

**MANUAL CALCULATION: A)LOAD FREQ SINGLE AREA- PROBLEM 1(a)**

*Steady State frequency deviation Δfs=-(M/*

*b) Hz*

*where*

*b=Area Frequency Response Coefficient(AFRC)=D+(1/R) Hz/p.u.M.W.*

*M is in p.u.MW =20/2000=0.01 p.u.MW/Hz*

*D =20/2000=0.01 p.u.MW/Hz*

*1/R=1000/2000=0.5 p.u.MW/Hz*

*b=0.01+0.5=0.51 p.u MW/Hz*

*Δfs = -(M/*

*b)Hz = -(0.01)/(0.51)*

*Δfs = -0.0196 Hz*

**MANUAL CALCULATION: B)LOAD FREQ SINGLE AREA- PROBLEM 1(b)**

*Steady State frequency deviation Δfs=-(M/*

*b) Hz*

*where*

*b=Area Frequency Response Coefficient(AFRC)=D+(1/R) Hz/p.u.M.W.*

*M is in p.u.MW =20/2000=0.01 p.u.MW/Hz*

*D =30/2000=0.015 p.u.MW/Hz*

*1/R=1333/2000=0.666 p.u.MW/Hz*

*b=0.015+0.666=0.681 p.u MW/Hz*

*Δfs = -(M/*

*b)Hz = -(0.01)/(0.681)*

*Δfs = -0.01467 Hz***COMPARISON OF A AND B:**

D | R | Δfs Hz | Δf peak Hz |

0.01 | 4% | -0.0196 | -0.0256 |

0.015 | 3% | -0.0155 | -0.0215 |

**2.**Assume that the single-area power system given in exercise 9.5.1 is provided with a load frequency controller (an integral controller) whose gain KI can be tuned.

(a) Carryout the simulation for the same disturbance of load change of 20 MW for different values of KI, obtain the time response Df for each case, critically compare these responses and comment on their suitability for practical application.

9-13

**(Hint: For choosing different values of K**

**I**

**, first set the governor and turbine time**

**constants to zero and determine analytically the value of integral gain K**

**I,cr**

**to have**

**critical damping on the response**D

**f (t). Choose the range of K**

**I**

**to include**

**K**

**I,cr**

**as 0**£

**K**

**I**£

**( K**

**I,cr**

**+ 1.0 ) )**

(b) From the investigations made in (a) above, choose the best value of KI which gives an “optimal” response Df (t) with regard to peak overshoot, settling time, steady-state error and Mean Sum- Squared-Error (MSSE).

**3**It is proposed to simulate the load frequency dynamics of a two-area power system. Both the areas are identical and has the system parameters given in exercise 9.5.1. Assume that the tie-line has a capacity of Pmax 1-2 = 200 MW and is operating at a power angle of ( d01- d20 ) = 300. Assume that both the areas

**do not**have load –frequency controller. Area 2 is subjected to a load increase of 20 MW.

(a) Simulate the load-frequency dynamics of this system using available software and

check the following:(i). Steady-State frequency deviation Dfs in Hz and tie-line flow deviation, DP12,S inp.u. MW. Compare them with hand-calculated values using AFRC’s

(ii) Compare result Dfs with that obtained in single area simulation in exercise 1

(a), and comment on the support received from area 1 and the advantages of

interconnecting with neighbouring areas .

(iii) Plot the time responses, Df1(t), Df2 (t), DPT1(t) , DPT2(t) and DP12(t). Comment

on the peak overshoot of Df1, and Df2.

**MANUAL CALCULATION:TWO AREA LOAD FREQUENCY CONTROL PROGRAM OUTPUT**

*Δfs = -(M1+M2)/(*

*b1+*

*b2)*

*M1 =0;M2=0.01 p.u .MW/Hz*

*b1 =*

*b2=0.51 p.u .MW/Hz (identical areas)*

*Δfs = - (0.01)/(0.51+0.51)= - 0.0098 Hz*

*ΔP*

_{1-2s}= - ΔP_{1-2s}= (*b1M2-*

*b2M1)/ (*

*b1+*

*b2) p.u .MW*

*= (0.51*0.01)/(0.51+0.51)*

*= 10 MW.*

Comparison of manual calculation and simulated values.

MANUALY CALCULATED VALUES | SIMULATED VALUES |

Δfs = - 0.0098 Hz | |

ΔP _{1-2s}=10MW | |

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