EAC - Stability (1).pptx glassgow scottland

Eas stability

1. Stability in Power System Swing Equation Equal Area Criterion Prof M Emad Farrag BEng, MSc, PhD, CEng, MIET, MIEEE, SFHEA Power Systems Operation and Protection

2. Outline Definition of Stability Dynamics of a synchronous machine S wing equation Steady-state Stability Theoretical steady-state stability limit Transient Stability Equal-area criterion Numerical solution of swing equation Stability Improvement Techniques 2

3. Definition of Stability S tability is the ability of a system to return back to its steady-state condition after being subjected to a disturbance. Stability for a synchronous machine in a power network refers to recovering its synchronous speed after having been subjected to a disturbance due to changes in the input or output power . Power system stability refers to maintaining the synchronism between the various parts of a power system. 3

4. Stability considerations are recognised a s major concerns in power system planning and studies . A stable power system must provide reliable power within the acceptable voltage and frequency range for its consumers. Instability ( loss of synchronism ) in part of an interconnected power system may lead to the propagation of the problem by cascading effect and may cause power outages or even complete blackout of the network. 4 Definition of Stability

5. Power System Stability denotes the ability of an electric power system, for a given initial operating condition, to regain a state of operating equilibrium after being subjected to a physical disturbance, with all system variables bounded so that the system integrity is preserved. 5 IEEE/CIGRE TF Report, "Definition and Classification of Power System Stability", IEEE Trans. on Power Systems, Vol. 19, pp. 1387-1401, August 2004 Definition of Stability

6. Outline Definition of Stability Dynamics of a synchronous machine S wing equation Steady-state Stability Theoretical steady-state stability limit Transient Stability Equal-area criterion Numerical solution of swing equation 6

7. In generator units (in steady-state and ignoring the losses): The mechanical power provided by the prime mover ( ) = electrical power produced by the generator( ). In mechanical term ( stability ): The accelerating torque (applied to the shaft by the prime mover) The decelerating torque (caused by the production of electric power) Stability Accelerating torque = Decelerating torque 7 Dynamics of a synchronous machine Generator Electrical system Mechanical system e , i , T , n,

8. Dynamics of a synchronous machine Consider a synchronous machine whose rotor is rotating with the synchronous speed ( ω s ). 8

9. Newton's second law for rotational dynamics: T a : net accelerating torque in N.m , J : rotor moment of inertia of all rotating masses attached to the rotor shaft in kg.m 2 α : rotor angular acceleration in rad/s 2 . 9 Dynamics of a synchronous machine

10. At equilibrium : The net torque is zero  The shaft rotates at a constant speed. The rotor angle (the angular position of the rotor measured in a reference frame that rotates at a synchronous speed ) is constant . 10 Dynamics of a synchronous machine

11. The rotor angular acceleration α is : ω : rotor angular velocity in rad/s  : angular displacement of the rotor with respect to a stationary axis fixed to the stator windings, in rad δ : angular displacement of the rotor with respect to a synchronously rotating axis. 11 Dynamics of a synchronous machine

12. 12 For a synchronous generator T a : net accelerating torque Dynamics of a synchronous machine

13. 13 For a synchronous generator T m : accelerating mechanical torque: the input mechanical power minus the mechanical losses due to friction . T e : retarding torque: electrical power on the shaft plus electrical losses due to iron and winding losses . Dynamics of a synchronous machine

14. 14 P a : the net accelerating power in W P m : the net mechanical power, i.e. the power supplied by the prime mover minus the mechanical losses, in W P e : the net electrical power, i.e. output electrical power plus electrical losses in core and windings, in W. M : the rotor angular momentum (or inertia constant ) expressed in J.s/rad Dynamics of a synchronous machine Multiply both side by

15. H constant : defined as the stored kinetic energy at synchronous speed based on the machine MVA rating: Typical values of the H constant 15   H constant (MJ/MVA) Turbine generators   3600 rpm 2.5 to 7 1800 rpm 6 to 9 Hydro generators 2 to 4 Synchronous condensers 1 to 1.25 Dynamics of a synchronous machine

16. Swing Equation Swing Equation δ in electrical radians δ in electrical degrees 16 M : rotor angular momentum M=J ω

17. Outline Definition of Stability Dynamics of a synchronous machine S wing equation Steady-state Stability Theoretical steady-state stability limit Transient Stability Equal-area criterion Numerical solution of swing equation 17

18. Steady-state stability: the ability of a network to recover from slow and small gradual changes in the operating conditions. Small disturbances can be due to small changes in the consumption of electrical loads or in the input mechanical power on the prime mover . Steady-state stability limit : the maximum real power that can be transmitted without loss of synchronism . 18 Steady-state stability

19. Power transfer in a two-bus system A synchronous machine is connected to an infinite bus through a transmission line with a total reactance of X L. The voltage at the infinite bus (V) is constant and considered to be the phasor reference . The real (active) power transmitted to the infinite bus ( P e ) is: 19 Steady-state stability

20. Power angle characteristics   δ Z s

21. R a is neglected Power angle characteristics

22. Power-angle curve The transmitted electrical power to infinite bus as a function of power angle ( δ ), called the power-angle curve 22 Steady-state stability

23. Stable Vs Equilibrium 23

24. Power transfer in a two-bus system C onsider a synchronous generator operating at steady-state conditions delivering active electrical power P e0 to the load. Since the system is in steady-state condition the net accelerating power P a is zero and the rotor rotates at synchronous speed . Therefore the input mechanical power on prime mover ( P m ) equals the output electrical power: 24 Steady-state stability

25. Power transfer in a two-bus system For any given transmitted power ( P e0 ) less than P e , max we can find two corresponding power angles ,  and  ' =180   , corresponding to operating points A and A' respectively, on the power-angle curve 25 Steady-state stability ? ?

26. Unstable equilibrium point Case 1- A generator is operating at power angle  ' > 90° (point A '). Due to a small disturbance, the output electrical power decreases to P e1 < P e0 . The rotor tends to increase δ to  ' 1 (point B') to compensate for the unbalance in the o/p & i /p power. The net accelerating power is positive, accelerates the rotor and increase the power angle . This will cause the rotor to accelerate indefinitely, leading to loss of synchronism and therefore loss of stability . 26  ' P e1 < P e0 P a >0 Accelerate the rotor

27. Stable equilibrium point Case 2- the generator is operating at power angle  < 90 ° (point A). Due to a small disturbance, the output electrical power decreases to P e1 < P e0 . Consider that the input mechanical power on the prime mover ( P m ) remains constant due to the slow response of the governor . The rotor tends to decrease the power angle to the new value of  1 (point B) in order to compensate the unbalance in the output and input power . Under this situation the net accelerating power is no longer zero. 27  '

28. Stable equilibrium point Case 2- This positive accelerating power tends to accelerate the rotor and increase the power angle. This will cause the rotor to oscillate around point B. In reality, because of the line resistance and damper windings the oscillations will decay and the rotor will fix the power angle at the new value of  1 . 28 Steady-state stability

29. Stable equilibrium point Case 3- Starting again from the steady-state condition with power angle  , assume that due to a small disturbance, the output electrical power increases to P e2 > P e0 . The rotor tends to increase the power angle to the new value of  2 (point C) in order to compensate the unbalance in the output and input power. Under this situation the net accelerating power is no longer zero: 29  '

30. Stable equilibrium point Case 3- The negative accelerating power tends to decelerate the rotor and decrease the power angle. This will cause the rotor to oscillate around point C. In reality because of line resistance and damper windings the oscillations will decay and the rotor will fix the power angle at the new value of  2 . For a power angle  < 90°, it is observed that small disturbances do not cause the machine to lose its synchronism, and that it remains stable. This point is said to be a stable equilibrium point . 30

31. T heoretical steady-state stability limit If the slope of power-angle characteristic at any working point is positive the system is said to be stable . Contrarily , if the slope is negative the system is unstable . The maximum power that can be transmitted without losing synchronism occurs at  = 90° and is called theoretical steady-state stability limit : 31 Steady-state stability Stable

32. Theoretical steady-state stability limit Theoretical maximum power cannot be achieved in practice. Limiting factors T hermal limit: thermal rating of conductors , The maximum allowable voltage drop (i.e. 5 percent : V / E >95%) Stability considerations. The third limitation imposes a minimum steady-state stability security margin ( M ) which is defined as : For example for a security margin of 50%, the maximum allowable transmittable power, P limit , considering this margin is P e , max /2, i.e. power angle must be less than 30°. 32 Steady-state stability

33. Outline Definition of Stability Dynamics of a synchronous machine Swing equation Steady-state Stability Theoretical steady-state stability limit Transient Stability Equal-area criterion Numerical solution of swing equation 33

34. Transient stability A stable system operating under steady-state conditions may become unstable when exposed to more severe transient disturbances. Transient disturbances: S hort-circuit (faults) Important changes in the circuit configuration after faulted transmission lines are removed or added L oss of a considerable part of generating plants Sudden load changes Switching operations. 34

35. Transient stability A system is said to be transiently stable if, after being impacted by a severe disturbance, it eventually reaches its initial or a different stable operating state following some transient oscillations. 35

36. Swing Curve( 𝝳, t) The swing curve is the variation of the power angle as a function of time (  = f( t )), (the solution of the swing equation). If the power angle tends to increase without limit, the system is said to be unstable . In a stable system , the power angle after disturbance occurrence undergoes oscillation with respect to time 36

37. Swing Curve( 𝝳, t) 37 The criterion for transient stability is the existence of a time for which:

38. Equal-area criterion The equal-area criterion is another way to present the transient stability criterion i.e. the system is stable if at an instant the slope of the swing curve becomes zero . The stability criterion implies that for an instant with power angle  s the expression becomes zero, which yields the following criterion for transient stability: 38 and

39. Equal-area criterion 39 As the integral is the net area under the curve it can be concluded that the system is stable if the net signed area of the region bounded by the graph of P a , the   axis, and the vertical lines  =  and  =  s is zero . In other words, positive area under the P a   curve which accelerates the rotor should be equal to the negative area which decelerates it.

40. 40 Equal-area criterion and power-angle curve decelerating area accelerating area P e = 0 Fault cleared here

41. 41 EAC- Change in mechanical input

42. 42 Athletics & end-line

43. 43 Athletics & end-line

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47. Disturbance due to faults Complete interruption of power delivery Partial loss of power delivery Pre-fault During fault Post-fault 47 Equal-area criterion

48. Equal-area criterion Disturbance due to faults Complete interruption of power delivery Assume that at t =0 a three-phase fault occurs at point F As the voltage at this point becomes zero, the electrical power output P e is also suddenly reduced to zero However, the mechanical power input P m remains unchanged This results in a positive net accelerating power P a = P m  P e = P m The accelerating power will accelerate the rotor and increase the power angle 48 t =0 V=0 P e =0

49. Equal-area criterion Disturbance due to faults Complete interruption of power delivery If it is assumed that the fault will be cleared by itself (temporary fault) or by the action of a circuit breaker at time t c ( clearing time ), the rotor angular acceleration from t =0 to t = t c can be found from the swing equation 49 t = t c

50. Equal-area criterion Complete interruption of power delivery During the period spanned by t =0 to t = t c , the rotor accelerates from δ=δ to δ= δ c , corresponding to points B and C . The accelerating area ( A 1 ) is the area under the power-angle curve: 50 t =0 t = t c power-angle curve

51. Equal-area criterion At t = t c when the fault is cleared at angle δ c , P e increases suddenly to a value corresponding to δ c , (point D ) P m < P e  negative net accelerating power  the rotor decelerates while the power angle increases due to inertia 51 decelerating area accelerating area Complete interruption of power delivery

52. Equal-area criterion The rotor recovers its synchronous speed at point E The decelerating area ( A 2 ) corresponds to: 52 decelerating area accelerating area Complete interruption of power delivery

53. Equal-area criterion Complete interruption of power delivery The stability criterion implies equality between the accelerating area ( A 1 ) and the decelerating area ( A 2 ). 53 Departing from point E, the power angle decreases again to be followed by oscillations. Neglecting damping, the rotor should oscillate infinitively between δ and δ 1 . In practice, however, these oscillations will decay and eventually die away because of the presence of damping.

54. Equal-area criterion Complete interruption of power delivery In order to find the critical clearing time ( t c , critical ), angle δ 1 can be allowed to reach the maximum allowable power angle  max = 180    max = 180   54

55. 55 Equal-area criterion accelerating area decelerating area Complete interruption of power delivery P e =0 stability A 1 =A 2

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58. Equal-area criterion Disturbance due to faults b) Partial loss of power delivery If a fault occurs on one line at some distance from the line sending end, some power, smaller than the transmitted power during pre-fault conditions, can be transmitted during the fault . Consider a synchronous generator is delivering power to an infinite bus through two parallel lines: 58

59. Equal-area criterion b ) Partial loss of power delivery Pre-fault During fault Post-fault 59

60. Equal-area criterion b ) Partial loss of power delivery Pre-fault: Power angle  delivering electrical power P e0 .. The equivalent reactance between the generator and the infinite bus is given by while the power-angle characteristic is expressed as: 60

61. Equal-area criterion b) Partial loss of power delivery During fault: a symmetrical, fault occurs at point F on line 2. The transmitted electrical power at during-fault condition : 61

62. b) Partial loss of power delivery During fault : P e,max2 < P e,max1 . Initially , the system is working at steady-state condition (point A). When the fault occurs, the output electrical power decreases suddenly to P e1 (point B). 62 Equal-area criterion

63. b) Partial loss of power delivery During fault : 63 The input mechanical power remains constant  P ositive net accelerating power  R otor accelerates and the power angle increases Equal-area criterion Fault is isolated by the action of circuit breaker at time t c ( clearing time ). The rotor angular acceleration from t =0 to t = t c can be found from the swing equation

64. b) Partial loss of power delivery During fault : 64 During time period t = 0 to t = t c , the rotor accelerates from δ=δ to δ= δ c , corresponding to points B and C. The accelerating area ( A 1 ) can be calculated from the area under the power-angle curve: Equal-area criterion

65. b) Partial loss of power delivery Post-fault: After a few cycle the fault is detected and the faulty line is isolated by opening breakers B 1 and B 2 . The power-angle characteristic is : where P e,max3 is less than P e,max1 but greater than P e,max2 . 65 Equal-area criterion

66. Equal-area criterion b) Partial loss of power delivery Post-fault: At t = t c fault is cleared at angle δ c , P e increases suddenly to a value corresponding to δ c (point D ). P m < P e , post-fault ,  Negative net accelerating power The rotor decelerates while the power angle increases due to inertia . 66 The relative angular velocity decreases until the rotor acquires its synchronous speed at point E corresponding to power angle δ 1 . The decelerating area ( A 2 );

67. Equal-area criterion b) Partial loss of power delivery The stability criterion implies that A 1 = A 2 . The critical clearing angle  c, critical can be found by using the stability criterion condition A 1 = A 2 and allowing for the maximum power angle δ 1 =  max =   sin -1 ( P m / P e,max3 ) . Any further increase in power angle beyond  max will produce a positive net accelerating power which will accelerate the rotor and cause instability and loss of synchronism. The critical clearing angle can be expressed as: 67

68. Equal-area criterion, Example A synchronous generator is initially delivering active power of 1 pu to the infinite bus at V =1 pu through two 150-km transmission lines with reactances of X L1 = X L2 = 0.6 pu . The transient reactance of the generator is X ' d = 0.2 pu and the internal transient voltage E '=1.4 pu . The transformer reactance is X T = 0.1 pu . A three-phase fault occurs on line 2 at a 50-km distance from generator’s side. Determine the critical clearing angle for preserving the stability of the system. 68

69. 69 Equal-area criterion, Example

70. 70 Equal-area criterion, Example

71. 71 Equal-area criterion, Example

72. 72 Equal-area criterion, Example

73. 73 In this case, it is impossible to find the critical clearing time with analytical methods . Numerical methods should be for solving the swing equation and calculating the critical clearing time .   Equal-area criterion, Example

74. 74 Equal-area criterion, Example

75. Outline Definition of Stability Dynamics of a synchronous machine S wing equation Steady-state Stability Theoretical steady-state stability limit Transient Stability Equal-area criterion Numerical solution of swing equation 75