PROJECT MENTOR:Dr.C.K.Panigrahi By:-
1) 2) 3)
4) 5) 6)
7)
8)
9) 10) 11)
ABSTRACT INTRODUCTION LITERATURE RIVIEW & MOTIVATION OF THE PRESENT WORK PROBLEM STATEMENT TRANSIENT STABILITY TRANSIENT INSTABILITY EFFECT OF TRANSIENT INSTABILITY ELEMENTARY VIEW OF TRANSIENT STABILITY SWING EQUATION EQUAL AREA CRITERION POINT-BY-POINT SOLUTION OF THE SWING EQUTION
12)
13)
14) 15)
16) 17) 18) 19)
20) 21)
MULTI-MACHINE SYSTEM , DIFFERENTIAL EQUATIONS METHODS OF SOLUTION OF SWING EQUATIONS MATHEMATICAL OVERVIEW MATHEMATICAL DERIVATION FOR COMPUTATIONAL ALGORITHM CASE STUDY MAT-LAB CODE MATLAB COMMAND WINDOW CASE STUDY OF A THREEMACHINE NINE-BUS SYSTEM SIMULINK REFERENCES
Power-system stability is a term applied to alternating-current electric power systems, denoting a condition in which the various synchronous machines of the system remain in synchronism, or "in step," with each other. Conversely, instability denotes a condition involving loss of synchronism, or falling "out of step." Occurrence of a fault in a power system causes transients. To stabilize the system load flow analysis is done. Actually in practice the fault generally occurs in the load side. As we controlling load side which will lead to complex problem in order avoid that we are controlling the generator side. A MATLAB simulation has been carried out to demonstrate the performance of the threemachine nine-bus system.
BACK GROUND
The classical model of a multi machine may be used to study the stability of a power system for a period of time during which the system dynamic response is dependent largely on the kinetic energy in the rotating masses. The classical three-machine nine-bus system is the simplest model used in studies of power system dynamics and requires of minimum amounts of data. Hence such studies can be connected in a relatively short time under minimum cost. Among various method of load flow calculation Newton Raphson method is chosen for calculation of load flow study. If the oscillatory response of a power system during the transient period following disturbance is damped and the system settles in a finite time to a new steady operating condition, we say the system is stable. If the system is not stable, it is considered unstable. This primitive definition of stability requires that the system oscillations should be damped. This condition is sometime called asymptotic stability and means that the system contains inherent forces that tend to reduce oscillation.
In recent years, energy, environment, right-of-way, and cost problems have delayed the construction of both generation facilities and new transmission lines, while the demand for electric power has continued to grow. This situation has necessitated a review of the traditional power system concepts and practices to achieve greater operating flexibility and better utilization of existing power systems. Transient stability of a transmission is a major area of research from several decades. Transient stability restores the system after fault clearance. Any unbalance between the generation and load initiates a transients that causes the rotors of the synchronous machines to “swing” because net accelerating torques are exerted on these rotors. If these net torques are sufficiently large to cause some of the rotors to swing far enough so that one or more machines “slip a pole” and synchronism is lost. So the calculation of transient stability should be needed. A system load flow analysis is required for it .The transient stability needs to be enhanced to optimize the load ability of a system, where the system can be loaded closer to its thermal limits.
Occurrence of fault may lead to instability in a system or the machine fall out of synchronism. Load flow study should be done to analyze the transient stability of the power system. If the system can’t sustain till the fault is cleared then the fault in-stabilize the whole system. If the oscillation in rotor angle around the final position go on increasing and the change in angular speed during transient condition go on increasing then system never come to its final position. The unbalanced condition or transient condition may leads to instability where the machines in the power system fall out of synchronism. Calculation of load flow equation by Newton Raphson method, Runge Kutta method, and decoupled method gives the rotor angle and initial condition.
Each generator operates at the same synchronous speed and frequency of 50 hertz while a delicate balance between the input mechanical power and output electrical power is maintained. Whenever generation is less than the actual consumer load, the system frequency falls. On the other hand, whenever the generation is more than the actual load, the system frequency rise. The generators are also interconnected with each other and with the loads they supply via high voltage transmission line. An important feature of the electric power system is that electricity has to be generated when it is needed because it cannot be efficiently stored. Hence using a sophisticated load forecasting procedure generators are scheduled for every hour in day to match the load. In addition, generators are also placed in active standby to provide electricity in times of emergency. This is referred as spinning reserved.
TRANSIENT STABILITY(cont.) • The power system is routinely subjected to a variety of disturbances. Even the act of switching on an appliance in the house can be regarded as a disturbance. • However, given the size of the system and the scale of the perturbation caused by the switching of an appliance in comparison to the size and capability of the interconnected system, the effects are not measurable. Large disturbance do occur on the system. These include severe lightning strikes, loss of transmission line carrying bulk power due to overloading.
• The ability of power system to survive the transition following a large disturbance and reach an acceptable operating condition is called transient stability.
TRANSIENT STABILITY(cont.) The physical phenomenon following a large disturbance can be described as follows:•Any disturbance in the system will cause the imbalance between the mechanical power input to the generator and electrical power output of the generator to be affected. As a result, some of the generators will tend to speed up and some will tend to slow down. • If, for a particular generator, this tendency is too great, it will no longer remain in synchronism with the rest of the system and will be automatically disconnected from the system. This phenomenon is referred to as a generator going out of step.
Transient instability refers to the condition where there is a disturbance on the system that causes a disruption in the synchronism or balance of the system. The disturbance can be a number of types of varying degrees of severity: • The opening of a transmission line increasing the XL of the system. • The occurrence of a fault decreasing voltage on the system. (The voltage at the fault goes to zero, decreasing all system voltages in the area.) • The loss of a generator disturbing the energy balance and requiring an increase in the angular separation as other generators adjust to make up the lost energy. • The loss of a large block of load in an exporting area.
EFFECT OF TRANSIENT INSTABILITY Acceleration or deceleration of these large generators causes severe mechanical stresses . Generators are also expensive. Damage to generators results in costly overhaul and long downtimes for repair. As a result, they are protected with equipment safety in mind. As soon as a generator begins to go out-of-step, sensor in the system sense the out-of-step condition and trip the generators. In addition, since the system is interconnected through transmission lines, the imbalance in the generator electrical output power and mechanical input power is reflected in a change in the flows of power on transmission lines.
As a result, there could be large oscillations in the flows on the transmission lines as generator try to overcome the imbalance and their output swing with respect to each other.
ELEMENTARY VIEW OF TRANSIENT STABILITY Consider the very simple power system of Fig. 2.2, consisting of a synchronous generator supplying power to a synchronous motor over a circuit composed of series inductive reactance X . Each of the synchronous machines may be represented, at least approximately, by a constant voltage source in series with a constant reactance. Thus the generator is represented by EG and XG ; and the motor, by EM and XM . Upon combining the machine reactance and the line reactance into a single reactance, we have an electric circuit consisting of two constantvoltage sources, EG and EM, connected through reactance X =X G + XL + XM .
ELEMENTARY VIEW OF TRANSIENT STABILITY (cont.)
Pa=((EG*EM)/x)*sin(δ) It is shown that the power transmitted from the generator to the motor depends upon the phase difference δ of the two voltages EG and EM . Since these voltages are generated by the flux produced by the field windings of the machines, their phase difference is the same as the electrical angle between the machine rotors.
The electromechanical equation describing the relative motion of the rotor load angle (δ) with respect to the stator field as a function of time is known as Swing equation .
M(d2 δ /dt2) =Pt - PU
or , M(d2 δ /dt2) =Pt - Pm sin(δ)
M = inertia constant Pt = Shaft power input corrected for rotational loss Pu = Pm sin(δ)=electric power output corrected for rotational losses Pm= amplitude of power angle curve δ = rotor angle with respect to a synchronously rotating referance
M(d2 δ /dt2) =Pt – PU or, 2((d2δ /dt2)* (d δ /dt))=2(Pa /M)(dδ /dt) or, d/dt((dδ/dt) ^2)=2(Pa /M)(dδ /dt) or, d(dδ /dt) ^2)=2(Pa /M)(d δ )
or, (lim. 0 , t)∫ d((dδ /dt) ^2)= (lim. δo,δ)∫ 2(Pa /M)(dδ )
or, ((dδ /dt) ^2) )= (lim. δo,δ)∫ 2(Pa /M)(dδ ) or, dδ /dt=√((lim. δo,δ)
∫ 2(Pa /M)(dδ ))
Under equilibrium dδ /dt =0; (lim. δo ,δ) ∫ 2(Pa /M)(d δ ))=0 ; Or, (lim. δo , δ) ∫( Pt – PU ) (d δ ) = 0 ; • (lim. δo,δ1)∫ Pa (d δ ) =A1 (positive area); • (lim. δ1,δ2) ∫ Pa (d δ) =A2 (negative area);
EQUAL AREA CRITERION (cont.) A 1= Area of acceleration A2= Area of deceleration If the area of acceleration is larger than the area of deceleration, i.e., A 1 > A2. The generator load angle will then cross the point δm, beyond which the electrical power will be less than the mechanical power forcing the accelerating power to be positive.
The generator will therefore start accelerating before is slows down completely and will eventually become unstable. If, on the other hand, A1< A2, i.e., the decelerating area is larger than the accelerating area, the machine will decelerated completely before accelerating again. The rotor inertia will force the subsequent acceleration and deceleration areas to be smaller than the first ones and the machine will eventually attain the steady state. If the two areas are equal, i.e., A 1 = A 2 , then the accelerating area is equal to decelerating area and this is defines the boundary of the stability limit.
M (d2 δ /dt2) =Pa
Where , δ= displacement angle of rotor w.r.t to a reference axis rotating at normal speed. M = Inertia constant of machine Pa = accelerating power t = time The solution of this equation gives δ as a function of t . A graph
of the solution is known as a swing curve. Inspection of the swing curves of all the machines of a system will show whether the machines will remain in synchronism after a disturbance. In a multi machines system , the output and hence the accelerating power of each machine depend upon the angular positions –and also upon the angular speeds-of all the machines of the system.
i.
M1(d2 δ1 /dt2) =Pi1 – PU1 (δ1 , δ2 , δ3 , dδ1 /dt , dδ2 /dt , dδ3 /dt )
ii.
M2(d2 δ2 /dt2) =Pi2 – PU2 (δ1 , δ2 , δ3 , dδ1 /dt , dδ2 /dt , dδ3 /dt )
iii.
M3(d2 δ3 /dt2) =Pi3 – PU3 (δ1 , δ2 , δ3 , dδ1 /dt , dδ2 /dt , dδ3 /dt
)
Formal solution of such a set of equations is not feasible even the simplest case.so we go for point to point solution which is most feasible and widely used way of solving the swing equations. such solution which are also called step by step solutions ,are applicable to the numerical solution of all sorts of differential equations. Good accuracy can be attained and the computational are simple. In a point by point solution one or more of the variables are assumed either to be constant or to vary according to assumed laws throughout a short interval of time ∆t , so that as a result of the assumptions made the equations can be solved for the changes in the other 2variables during the same time interval. Then, from the values of the other variables at the end of the interval, new value can be calculated for the variables which were assumed constant . These new value are then used in the next time interval.
In applying the point by point method to the solution of swing equations, it is customary to assume that the accelerating power is constant during each time constant during each time interval, although it has different values in different variables. The point by point solution of swing curve consists of two processes which are carried out alternately:
The first process is the computational of the angular positions, and perhaps also of the angular speeds at the end of the time interval from a knowledge of the positions and speeds at the beginning of the interval and the accelerating power assumed constant for interval. The second process is the computational of the accelerating power of each machine from the angular position of all machines of the system .The second process requires knowledge of network solution.
Assumptions:The only assumption is that accelerating power is assumed as constant throughout the time interval ∆t and has the value computed for the beginning of the interval. M (d2 δ /dt2) = Pa Integrating both sides: dδ /dt= ω = ωo +(Pat/M); ……………(1) δ = δo+ tωo +(Pat2/2M); …………...(2) Dividing the time interval t in n intervals From equations (1) &(2) ωn = ωn-1 +(Pa(n-1) ∆t /M); .…………..(3) δn = δn-1+ ∆tωn-1+(Pa(n-1)(∆t )2/2M); …….....(4) The increments of speed and angle during the nth interval are: ∆ ωn = ωn - ωn-1=(Pa(n-1) ∆t /M); ……………(5) ∆ δn = δn -δn-1 =∆tωn-1+(Pa(n-1)(∆t )2/2M);….(6)
Mathematical derivation for computational algorithm
The equations 3,4,5&6 are suitable for point-by-point calculation. the preceding interval can be calculated as ωn-1 = ωn-2 +(Pa(n-2) ∆t /M); .…….(7) 2 δn-1 = δn-2+ ∆t ωn-2 +(Pa(n-2)(∆t ) /2M); ……..(8)
Subtract eq.7 from 4: (δn - δn-1 )= (δn-1 -δn-2 )+ ∆t(ωn-1 -ωn-2 ) +(Pa(n-1)(∆t )2/2M)-(Pa(n-2)(∆t )2/2M) ; …………….(9) but ; δn - δn-1 = ∆δn ; or, δn-1 - δn-2 = ∆δn-1 ; or, ωn -1 - ωn-2 = ∆ωn-1 ; From equation (5) ∆ ωn-1 = (Pa(n-2) ∆t /M); ………….(10) using eq. (9) &(10) ∆ δn = ∆δn-1 + ∆t Pa(n-2) ∆t /M +(Pa(n-1) - Pa(n-2)) (∆t )2/2M;
Or,
∆ δn = ∆δn-1 +(Pa(n-1) + Pa(n-2)) (∆t )2/2M;
…… (final eq. for point by point soln.)
Case:-
A synchronous machine performing oscillation of small amplitude with respect
to an infinite bus ,its power output may be assumed to be directly proportional to its angular displacement from the infinite bus .considering a 50 cycle machine for which H = 2.7 Mj-per Mva. And which is operating in the steady state with input and output of 1.00 pu. and angular displacement of 45o (elec.) with respect to an infinite bus. Upon occurrence of a fault ,we assume that the input remains constant and that output is given by
Pu=δ/90o
may be great.
even though the amplitude of oscillation
The initial conditions are :δ = (pi/4) ; dδ/dt = 0;
The period of oscillation is given by :T = (2*pi)/sqrt.(2/(pi*M)); or, = pi*sqrt.(2*pi*M);
The inertial constant is :-
M = H /(pi*f) =2.7/(pi*50) = .017188(unit power sec.2 per elec.rad) or ,M = H/(180*f) =0.30micro (unit power sec.2 per elec.deg.)
Code:-
clc ,clear all, close all; Problem='Point by solution of swing equation by Method 1 ' Assumptions=':the synchronous machine performing oscillation of small amplitude with
respect to an infinite bus ,its power output may be assumed to be directly proportional to its angular displacement from infinite bus under steady state input and output of 1pu. input remain constant and output is given by =del/90(degree) '
data ='required' are=':-1. step size for numerical computation 2. range of time of operation 3. inertial constant(H) in mega 4. frequency of operation 5. initial displacement delta' s=1; while s<3 delT=input('enter step size') lim=input('enter range of time') H=input('enter inertial constant in mega') f=input('enter frequency') del=input('enter initial displacement')
Mat-lab code cont.(1)
Pu=del/90; Pa=1-Pu; M=(H/(180*f)); t = [0:delT:lim]; [m,n] = size(t); x=0; Wn=0; l=1; delWn=(delT/M)*Pa; while x
Mat-lab code cont.(2) Pa=1-Pu; x=x+1; l=l+1; end s=s+1; if s==2 output1='output data of first data input' time1=transpose(t) electrical_opower1=transpose(ele) accelrating_power1=transpose(acc) change_angular_velocity_deg_sec1=transpose(Dwn) angular_velocity_deg_sec1=transpose(wn) change_in_power_angle_degree1=transpose(Dd) power_angle_degree1=transpose(d)
Mat-lab code cont.(3) else s==3 output2='output data of second data input' time2=transpose(t) electrical_opower2=transpose(ele) accelrating_power2=transpose(acc) change_angular_velocity_deg_sec2=transpose(Dwn) angular_velocity_deg_sec2=transpose(wn) change_in_power_angle_degree2=transpose(Dd) power_angle_degree2=transpose(d) end end plot(time1,power_angle_degree1,time2,power_angle_degree2) xlabel('time(sec.)') ylabel('displacement angle del(electrical degree)') title('Swing curve solution by point by point solution ') grid
MATLAB Command Window Problem = Point by solution of swing equation by Method 1 Assumptions = :the synchronous machine performing oscillation of small amplitude with respect to an infinite bus ,its power output may be assumed to be directly proportional to its angular displacement from infinite bus under steady state input and output of 1pu. input remain constant and output is given by =del/90(degree) data required are = :1. step size for numerical computation 2. range of time of operation 3. inertial constant(H) in mega 4. frequency of operation 5. initial displacement delta enter step size.05 delT = 0.0500 enter range of time1 lim = 1 enter inertial constant in mega2.7 H = 2.7000 enter frequency50 f = 50 enter initial displacement45 del = 45 enter step size0.04 delT = 0.0400 enter range of time1 lim = 1 enter inertial constant in mega2.7 H = 2.7000 enter frequency50 f= 50 enter initial displacement45 del = 45
output1 = output data of first data input time1 = 0.0000 0.0500 0.1000 0.1500 0.2000 0.2500 0.3000 0.3500 0.4000 0.4500 0.5000 0.5500 0.6000 0.6500 0.7000 0.7500 0.8000 0.8500 0.9000 0.9500 1.0000
electrical_ opower1 = 0.5000 0.5000 0.5463 0.6389 0.7735 0.9415
1.1305 1.3250 1.5073 1.6596 1.7648 1.8090 1.7824 1.6809 1.5069 1.2699 0.9860 0.6770 0.3694 0.0916 -0.1277
acceleratin g_ power1 =
change angular Velocity degsec1 =
Angular velocity_ deg_sec1 =
Change in power angle_ degree1 =
power_ angle_degree1 =
0.5000 0.5000 0.4537 0.3611 0.2265 0.0585 -0.1305 -0.3250 -0.5073 -0.6596 -0.7648 -0.8090 -0.7824 -0.6809 -0.5069 -0.2699 0.0140 0.3230 0.6306 0.9084 1.1277
83.3333 83.3333 75.6173 60.1852 37.7515 9.7451 -21.7568 -54.1611 -84.5508 -109.9256 -127.4716 -134.8393 -130.4041 -113.4838 -84.4890 -44.9865 2.3391 53.8301 105.1045 151.3947 187.9529
0.0000 83.3333 166.6667 242.2839 302.4691 340.2206 349.9657 328.2089 274.0478 189.4971 79.5715 -47.9001 -182.7394 -313.1435 -426.6273 -511.1164 -556.1029 -553.7638 -499.9337 -394.8292 -243.4345
0.0000 4.1667 8.3333 12.1142 15.1235 17.0110 17.4983 16.4104 13.7024 9.4749 3.9786 -2.3950 -9.1370 -15.6572 -21.3314 -25.5558 -27.8051 -27.6882 -24.9967 -19.7415 -12.1717
45.0000 45.0000 49.1667 57.5000 69.6142 84.7377 101.7487 119.2470 135.6574 149.3598 158.8347 162.8132 160.4182 151.2813 135.6241 114.2927 88.7369 60.9317 33.2436 8.2469 -11.4946
output2 = output data of second data input time2 =
Electrical opower2 =
Accelerating power2 =
change_ angular Velocity degsec2 =
0 0.0400 0.0800 0.1200 0.1600 0.2000 0.2400 0.2800 0.3200 0.3600 0.4000 0.4400 0.4800 0.5200 0.5600 0.6000 0.6400 0.6800 0.7200 0.7600 0.8000 0.8400 0.8800 0.9200 0.9600 1.0000
0.5000 0.5000 0.5296 0.5889 0.6760 0.7875 0.9182 1.0615 1.2096 1.3541 1.4862 1.5973 1.6795 1.7264 1.7330 1.6966 1.6167 1.4955 1.3378 1.1508 0.9437 0.7277 0.5150 0.3184 0.1506 0.0232
0.5000 0.5000 0.4704 0.4111 0.3240 0.2125 0.0818 -0.0615 -0.2096 -0.3541 -0.4862 -0.5973 -0.6795 -0.7264 -0.7330 -0.6966 -0.6167 -0.4955 -0.3378 -0.1508 0.0563 0.2723 0.4850 0.6816 0.8494 0.9768
66.6667 66.6667 62.7160 54.8148 43.1971 28.3310 10.9052 -8.1995 -27.9505 -47.2156 -64.8243 -79.6351 -90.6044 -96.8546 -97.7357 -92.8773 -82.2270 -66.0730 -45.0462 -20.1040 7.5076 36.3105 64.6686 90.8749 113.2490 130.2380
angular velocity deg_sec2 =
change_in_po wer_angle_de gree2 =
power angle degree2 =
0 66.6667 133.3333 196.0494 250.8642 294.0613 322.3923 333.2975 325.0980 297.1475 249.9319 185.1076 105.4725 14.8681 -81.9865 -179.7223 -272.5995 -354.8266 -420.8996 -465.9458 -486.0499 -478.5423 -442.2318 -377.5632 -286.6883 -173.4393
0.0000 2.6667 5.3333 7.8420 10.0346 11.7625 12.8957 13.3319 13.0039 11.8859 9.9973 7.4043 4.2189 0.5947 -3.2795 -7.1889 -10.9040 -14.1931 -16.8360 -18.6378 -19.4420 -19.1417 -17.6893 -15.1025 -11.4675 -6.9376
45.0000 45.0000 47.6667 53.0000 60.8420 70.8765 82.6390 95.5347 108.8666 121.8705 133.7564 143.7537 151.1580 155.3769 155.9716 152.6921 145.5033 134.5993 120.4062 103.5702 84.9324 65.4904 46.3487 28.6594 13.5569 2.0894
The graphical output of mat-lab command The output curve of two step size:1) ∆t=0.04 2) ∆t=0.05 H=2.7; Frequency=50Hz; Initial displacement angle=45o
The assumptions for a multi-machine system: 1. Mechanical power input is constant. 2. Damping or asynchronous power is negligible. 3. Constant-voltage-behind-transient-reactance model for the synchronous machines is valid. 4. The mechanical rotor angle of a machine coincides with the angle of the voltage behind the transient reactance. 5. Passive impedances represent loads.
This model is useful for stability analysis but is limited to the study of transients for only the “first swing” or for periods on the order of one second. Assumptions 2 are improved upon somewhat by assuming a linear damping characteristic. A damping torque (or power) D is frequently added to the inertia torque (or power) in the swing equation. The damping coefficient D includes the various damping torque coefficients, both mechanical and electrical. This represents turbine damping, generator electrical damping, and the damping effect of electrical loads.
I. II. III. IV. V. VI.
Simulink is advanced software which is increasingly being used as a basic building block in many areas of research. As such, it holds a great potential in the area of power system example to demonstrate the features and scope of Simulink –based model for transient stability analysis. The stability of power systems continues to be major concern in system operation. Modern electrical power systems have grown to a large generating units and extra high voltage tie-lines, etc. The transient stability is a function of both operating conditions and disturbances. Thus the analysis of transient stability is complicated. Simulink is an interactive environment for modeling, analyzing and simulating a wide variety of dynamic systems.
The key features of Simulink are:
Interactive simulations with live display; A comprehensive block library for creating linear, non linear, discrete or hybrid multi- input/output systems; Seven integration methods for fixed step, variable step and stiff systems; Unlimited hierarchical model structure; Scalar and vector connections; Mask facility for creating custom blocks and block libraries;
I.J. NAGRATH and D.P.KOTHARI, ―Power system Engineering (Tata Mc -Graw-Hill, New York, 1964) RAMNARAYAN PATEL, T. S. BHATTI and D. P. KOTHARI Centre for Energy Studies, Indian Institute of Technology, Hauz Khas, New Delhi, India “MATLAB/ Simulink-based transient stability analysis of a multi-machine power system ”September 1,2003. Edward Wilson Kimbark ,-Power System Stability volume-1( IEEE Press; Wiley Inter-science , New York,1948) bing.com; google.co.in library.nu
