Unit 1 Power System Stability
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1. The document discusses power system stability, including classifications of power system states as steady state, dynamic state, and transient state. 2. It describes synchronous machine swing equation and power angle equation, which relate the mechanical power input to the electrical power output of a generator through the power/torque angle. 3. An example calculation is shown to find the steady state power limit of a power system with a generator connected to an infinite bus through a transmission line.
Unit 1 Power System Stability
- 1. DEPARTMENT OF ELECTRICAL ENGINEERING JSPMS BHIVARABAISAWANTINSTITUTEOFTECHNOLOGYANDRESEARCH, WAGHOLI,PUNE A.Y. 2020-21 (SEM-I) Class: B.E. Subject: Power System Operation Control Unit No-1: Power System Stability Prepared by Prof. S. D. Gadekar [email protected] Mob. No-9130827661
- 2. Content • Introduction • Classification of Power System States • Power System Stability • Dynamics of Synchronous Machines • Synchronous Machines Swing Equation • Power Angle Equation and Power Angle Curve • Numerical of Steady State Stability Limit • Equal Area Criteria • Applications of equal area criteria Sudden change in mechanical input power Effect of clearing time on stability and Numerical Sudden Short circuit on one of parallel lines and Numerical • Numerical Solution of Swing Equation • Methods to Improve Steady State Stability Limit and Transient Stability Limit
- 3. Introduction to Power System Stability Modern power system is a complex non linear interconnected network. It consists of inter connected transmission lines, generating plants transformers and a variety of loads. With the increase in power demand nowadays some transmission lines are more loaded than their normal limits. With the increased loading of long transmission lines, the problem of transient stability has become a serious limiting factor.
- 4. Classification of Power System States The power system is a highly nonlinear system that operates in a constantly changing environment; loads, generator outputs and key operating parameters change continually . When subjected to a disturbance, the stability of the system depends on the initial operating condition as well as the nature of the disturbance. Stability of an electric power system is thus a property of the system motion around an equilibrium set, i.e., the initial operating condition. Steady state Dynamic state Transient state Power system states
- 5. POWER SYSTEM STATES STEADY STATE In an interconnected power system, the rotors of each synchronous machine in the system rotate at the same average electrical speed. The power delivered by the generator to the power system is equal to the mechanical power applied by the prime mover, neglecting losses. During steady state operation, the electrical power out balances the mechanical power in.
- 6. POWER SYSTEM STATES DYNAMIC STATE Dynamic instability is more probable than steady state stability. Small disturbances are continually occurring in a power system (variations in loadings, changes in turbine speeds, etc.) which are small enough not to cause the system to lose synchronism but do excite the system into the state of natural oscillations. In a dynamically unstable system, the oscillation amplitude is large and these persist for a long time (i.e., the system is under damped).
- 7. POWER SYSTEM STATES TRANSIENT STATE For a large disturbance, changes in angular differences may be so large as to cause the machines to fall out of step. This type of instability is known as transient stability and is a fast phenomenon usually occurring within 1sec for a generator close to the cause of disturbance.
- 8. POWER SYSTEM STABILITY STEADY STATE STABILITY DYNAMIC STABILITY TRANSIENT STABILITY VOLTAGE STABILITY Small- signal stability is the ability of the system to return to a normal operating state following a small disturbance Dynamic stability refers to the ability of a power system subject to a relatively small and sudden disturbance Transient stability is the ability of power system to maintain synchronism when it is suddenly subjected to a severe transient disturbance Voltage stability is concerned with the ability of a power system to maintain steady acceptable voltages at all buses
- 9. Transient stability is the ability of the power grid system to maintain synchronism when subjected to severe disturbances. Transient stability analysis is considered with large disturbances like : 1. Suddenly change in load. 2. Generation or transmission system configuration due to fault. 3. Switching. TRANSIENT STABILITY
- 10. It is the ability of the system to remain in synchronism when subjected to a disturbance. The rotor angle of a generator depends on the balance between the electromagnetic torque due to the generator electrical power output and mechanical torque due to the input mechanical power through a prime mover. Remaining in synchronism means that all the generators electromagnetic torque is exactly balanced by the mechanical torque. If in some generator the balance between electromagnetic and mechanical torque is disturbed, due to disturbances in the system, then this will lead to oscillations in the rotor angle. Rotor angle stability is further classified into small disturbance angle stability and large disturbance angle stability. Rotor Angle Stability
- 11. Voltage Stability Definition By IEEE: “Voltage stability refer to the ability of power system to maintain steady voltages at all buses in the system after being subjected to a disturbance from a given initial operating point. The system state enters the voltage instability region when a disturbance or an increase in load demand or alteration in system state results in an uncontrollable and continuous drop in system voltage.” Voltage Stability
- 12. Dynamics of Synchronous Machine The kinetic energy of the rotor at synchronous machine is, 𝐾. 𝐸 = 1 2 𝐽𝜔𝑠𝑚 2 × 10−6 𝑀𝐽 Where, J=rotor moment of inertia in 𝐾𝑔. 𝑚2 𝜔𝑠𝑚=rotor speed in radian (mechanical)/second But 𝜔𝑠 = ( 𝑃 2 ) 𝜔𝑠𝑚 rotor speed in radian (electrical)/second Where P is no of machine poles. 𝐾. 𝐸 = 1 2 𝐽 2 𝑃 2 𝜔𝑠 × 10−6 × 𝜔𝑠 𝐾. 𝐸 = 1 2 𝑀 𝜔𝑠 Where, 𝐌 = 𝑱 𝟐 𝑷 𝟐 𝝎 𝒔 × 𝟏𝟎−𝟔 𝒎𝒐𝒎𝒆𝒏𝒕 𝒐𝒇 𝒊𝒏𝒆𝒓𝒕𝒊𝒂 𝒊𝒏 𝑴𝑱. 𝑺/(𝒆𝒍𝒆𝒄𝒕 𝒓𝒂𝒅) GH= 𝐾. 𝐸 = 1 2 𝑀 𝜔𝑠 , Where G is machine rating in MVA and H is inertia constant in MJ/MVA 𝑀 = 2𝐺𝐻 𝜔𝑠 = 𝐺𝐻 𝜋𝑓 𝑀𝐽. 𝑆/(𝑒𝑙𝑒𝑐𝑡 𝑟𝑎𝑑) 𝑀 = 𝐺𝐻 180𝑓 𝑀𝐽. 𝑆/(𝑒𝑙𝑒𝑐𝑡 𝑑𝑒𝑔𝑟𝑒𝑒)
- 13. Synchronous Machine Swing Equation Under normal operating conditions, the relative position of the rotor axis and the resultant magnetic field axis is fixed. The angle between the two is known as the power angle or torque angle. During any disturbance, the rotor decelerates or accelerates with respect to the synchronously rotating air gap mmf, creating relative motion. The equation describing the relative motion is known as the swing equation, which is a non-linear second order differential equation that describes the swing of the rotor of synchronous machine. Generator𝑃𝑚 𝑇 𝑚 𝜔𝑠 𝑇𝑒 Motor𝑃𝑚 𝑇 𝑚 𝜔𝑠 𝑇𝑒 𝑃𝑒 𝑃𝑒
- 14. A synchronous generator is driven by a prime mover. The equation governing the rotor motion is given by: 𝐽 𝑑2 𝜃 𝑚 𝑑𝑡2 = 𝑇 𝑚 − 𝑇𝑒 = 𝑇𝑎…….1 where a) J is the total moment of inertia of the rotor mass in kgm2 b) Tm is the mechanical torque supplied by the prime mover in N-m c) Te is the electrical torque output of the alternator in N-m d) m is the angular position of the rotor in rad (Mechanical) Multiply equation 1 by 𝜔𝑠𝑚, then equation 1 becomes 𝐽𝜔𝑠𝑚 𝑑2 𝜃 𝑚 𝑑𝑡2 = 𝑃𝑚 − 𝑃𝑒 MW………2 𝐽 2 𝑃 2 𝜔𝑠 × 10−6 𝑑2 𝜃 𝑒 𝑑𝑡2 = 𝑃𝑚 − 𝑃𝑒 MW……….3 Where 𝜃𝑒 is the angle in radian (Electrical). Generator𝑃𝑚 𝑇 𝑚 𝜔𝑠 𝑇𝑒 𝑃𝑒
- 15. 𝐽 2 𝑃 2 𝜔𝑠 × 10−6 𝑑2 𝜃 𝑒 𝑑𝑡2 = 𝑃𝑚 − 𝑃𝑒 MW……….3 The above equation we can rewrite as, 𝐌 = 𝑱 𝟐 𝑷 𝟐 𝝎 𝒔 × 𝟏𝟎−𝟔 𝒎𝒐𝒎𝒆𝒏𝒕 𝒐𝒇 𝒊𝒏𝒆𝒓𝒕𝒊𝒂 𝒊𝒏 𝑴𝑱. 𝑺/(𝒆𝒍𝒆𝒄𝒕 𝒓𝒂𝒅) 𝑀 𝑑2 𝜃𝑒 𝑑𝑡2 = 𝑃𝑚 − 𝑃𝑒 MW … … … .4 𝛿 = 𝜃𝑒 − 𝜔𝑠 𝑇………5 Rotor angular displacement from synchronously rotating frame called as torque angle or power angle. 𝑑2 𝛿 𝑑𝑡2 = 𝑑2 𝜃 𝑒 𝑑𝑡2 ……...6 Thus equation 4 becomes 𝑀 𝑑2 𝛿 𝑑𝑡2 = 𝑃𝑚 − 𝑃𝑒………7 The above equation is called as swing equation of synchronous alternator.
- 16. Power Angle Equation and Power Angle Curve- 𝑀(𝑃𝑢) 𝑑2 𝛿 𝑑𝑡2 = 𝑃𝑚 − 𝑃𝑒 in Pu of machine rating as base Certain assumptions are usually made 1. Mechanical power input to the machine (Pm) remains constant during the period of electromechanical transient of interest. 2. Rotor speed changes are insignificant. 3. Effect of voltage regulating loop during the transient is ignored, as a consequence the generated machine emf remains constant. Generator𝑃𝑚 𝑇 𝑚 𝜔𝑠 𝑇𝑒 𝑃𝑒
- 17. 𝑃𝑒 = 𝐸′ |𝑉| 𝑋12 sin 𝛿 … … … . . This equation is called as power angle equation. The above equation shows that the power transmitted depends upon the transfer reactance and the angle between the two voltages. G1 𝐸′ < 𝛿 |𝑉| < 0° 𝑋12
- 18. 𝑃𝑒 = 𝐸′ |𝑉| 𝑋12 sin 𝛿 𝑃𝑒 = 𝑃𝑚𝑎𝑥 sin 𝛿 Where 𝑃𝑚𝑎𝑥 = 𝐸′ |𝑉| 𝑋12 The curve 𝑃𝑒 vs 𝛿 is known as the power angle curve. Steady State Stability limit occurs at an angular displacement of 90°. When a generator is suddenly short –circuited the current during the transient period is limited by its transient reactance 𝑋 𝑑 ′ . 𝐸′ = 𝑉𝑔 + 𝑗𝑋 𝑑 ′ 𝐼 𝑑
- 19. Power Angle Curve Generator𝑃𝑚 𝑇 𝑚 𝜔𝑠 𝑇𝑒 𝑃𝑒
- 20. Example-1 Find the steady state power limit of a power system consisting of a generator equivalent reactance 0.50 Pu connected to an infinite bus through a series reactance of 1.0 Pu. The terminal voltage of the generator is held at 1.2 Pu and the voltage of the infinite bus is 1.0 Pu. G L𝐸′ < 𝛿 𝑉 = |1| < 0° 𝑉𝑡 = 1.2 < 𝜃° 𝑋 𝑑 ′ = 0.5 𝑋 = 1.0 𝐸′ = 𝑉𝑡 + 𝑗𝑋 𝑑 ′ 𝐼 𝑑 = 𝟏. 𝟐 < 𝜽° + j0.5 * 𝑰 𝒅……….1 𝐼 𝑑 = 𝑉𝑡−𝑉 𝑗𝑋 = 𝟏.𝟐<𝜽°−|𝟏|<𝟎° 𝒋𝟏 ………..2 𝐼 𝑑
- 21. G L𝐸′ < 𝛿 𝑉 = |1| < 0° 𝑉𝑡 = 1.2 < 𝜃° 𝑋 𝑑 ′ = 0.5 𝑋 = 1.0 If we put equation 2 in equation 1 for 𝑰 𝒅 𝐸′ = 𝟏. 𝟐 < 𝜽° + 0.5j * 𝑉𝑡−𝑉 𝑗𝑋 𝐸′ = 𝟏. 𝟐 < 𝜽° + 0.5j * 𝟏.𝟐<𝜽°−|𝟏|<𝟎° 𝒋𝟏 𝐸′ = 𝟏. 𝟖 < 𝜽° − 𝟎. 𝟓 Separating Real and Imaginary Part 𝐸′ = (𝟏. 𝟖 cos 𝜽 − 𝟎. 𝟓)+j(1.8× sin 𝜃)……..3 𝐼 𝑑
- 22. G L𝐸′ < 𝛿 𝑉 = |1| < 0° 𝑉𝑡 = 1.2 < 𝜃° 𝑋 𝑑 ′ = 0.5 𝑋 = 1.0 𝐸′ = (𝟏. 𝟖 𝐜𝐨𝐬 𝜽 − 𝟎. 𝟓)+j(1.8× sin 𝜃)………3 Steady State Power Limit is reached when E has an angle of 𝜹 = 𝟗𝟎°, AS 𝑬′ < 𝜹 , here 𝜹 = 𝟗𝟎° This can be written as 𝑬′ (𝐜𝐨𝐬 𝟗𝟎 + 𝒋 𝐬𝐢𝐧 𝟗𝟎) Thus the real part will be zero. In the Equation 3 equate real part to zero. (𝟏. 𝟖 𝐜𝐨𝐬 𝜽 − 𝟎. 𝟓)=0 𝜃 =73.87° 𝐼 𝑑
- 23. G L𝐸′ < 𝛿 𝑉 = |1| < 0° 𝑉𝑡 = 1.2 < 𝜃° 𝑋 𝑑 ′ = 0.5 𝑋 = 1.0 𝐼 𝑑 = 𝑉𝑡−𝑉 𝑗𝑋 = 𝟏.𝟐<𝜽°−|𝟏|<𝟎° 𝒋𝟏 In above equation put 𝜃 =73.87° 𝐼 𝑑=1.152+j0.668 PU 𝐸′ = 𝑉𝑡 + 𝑗𝑋 𝑑 ′ 𝐼 𝑑 = 𝟏. 𝟐 < 𝜽° + j0.5 * 𝑰 𝒅……….1 Similarly In above equation put 𝜃 =73.87° & 𝐼 𝑑=1.152+j0.668 Amp 𝑬′ =1.728< 𝟗𝟎° Steady State Power Limit is given by 𝑃𝑚𝑎𝑥 = 𝐸′ |𝑉| 𝑋12 = 1.728∗1 0.5+1 =1.152 PU 𝐼 𝑑
- 24. G L𝐸′ < 𝛿 𝑉 = |1| < 0° 𝑉𝑡 = 1.2 < 𝜃° 𝑋 𝑑 ′ = 0.5 𝑋 = 1.0 If instead, the generator emf is held fixed at a value of 1.2 PU, the steady state power limit would be, 𝑃𝑚𝑎𝑥 = 𝐸′ |𝑉| 𝑋12 = 1.2∗1 0.5+1 =0.8 PU Conclusion- When 𝐸′ = 1.728 −−− −𝑃𝑚𝑎𝑥=1.152 PU And if 𝐸′ = 1.2 −−− −𝑃𝑚𝑎𝑥=0.8 PU It is observed that regulating the generator emf to hold the terminal generator voltage at 1.2 PU raises the Power Limit from 0.8 to 1.152 PU. 𝐼 𝑑
- 25. Equal Area Criteria If the system is unstable 𝛿 continues to increase indefinitely with time and the machine loses the synchronism. On the other hand if the system is stable 𝛿 performs oscillations whose amplitude decreases in actual practice because of damping terms. 𝜹 t 𝒅𝜹 𝒅𝒕 > 𝟎 Unstable Stable 𝒅𝜹 𝒅𝒕 = 𝟎
- 26. Consider the swing equation 𝑀 𝑑2 𝛿 𝑑𝑡2 = 𝑃𝑚 − 𝑃𝑒 𝑑2 𝛿 𝑑𝑡2 = 1 𝑀 𝑃𝑎 ……..1 Where 𝑃𝑎 is accelerating power and M= 𝐻 𝜋𝐹 𝑖𝑛 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑠𝑦𝑠𝑡𝑒𝑚 Now multiply both side by 2 𝑑𝛿 𝑑𝑡 to equation 1 2 𝑑𝛿 𝑑𝑡 𝑑2 𝛿 𝑑𝑡2 = 1 𝑀 𝑃𝑎 2 𝑑𝛿 𝑑𝑡 … … 2 Integrating both sides 𝑑𝛿 𝑑𝑡 2 = 2 𝑀 𝛿0 𝛿 𝑃𝑎 𝑑𝛿 … … . . 3 System is stable if 𝒅𝜹 𝒅𝒕 = 𝟎 𝛿0 𝛿 𝑃𝑎 𝑑𝛿 = 0
- 27. 𝛿0 𝛿 𝑃𝑎 𝑑𝛿 = 0 The condition of stability can therefore be stated as, the system is stable if the area under 𝑃𝑎 𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑛𝑔 𝐴𝑟𝑒𝑎 𝑉𝑠 𝛿curve reduces to zero at the same value of 𝛿. In other words, the positive (accelerating) area under 𝑃𝑎 𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑛𝑔 𝐴𝑟𝑒𝑎 𝑉𝑠 𝛿 curve must be equal to negative (decelerating) area and hence the name ‘equal area criteria’ of stability.
- 28. Applications of Equal Area Criteria i) Sudden change in mechanical Input Consider the transient model of a single machine tied to infinite bus bar as shown in fig. Generator𝑃𝑚 𝑇 𝑚 𝜔𝑠 𝑇𝑒 𝑃𝑒 G ∞ 𝐸′ < 𝛿 |𝑉| < 0° 𝑋 𝑑 ′ 𝑋 𝑒 𝑃𝑚 The electrical power transmitted is given by 𝑃𝑒 = 𝐸′ 𝑉 𝑋 𝑑 ′ + 𝑋 𝑒 sin 𝛿 = 𝑃𝑚𝑎𝑥 sin 𝛿
- 29. 𝑷 𝒆 𝜹 𝑷 𝒎𝟎 𝑷 𝒎𝟏 𝜹 𝟎 𝜹 𝟏 𝜹 𝟐 Under Steady State operating condition 𝑷 𝒎𝟎 = 𝑷 𝒆𝟎 = 𝑷 𝒎𝒂𝒙 𝒔𝒊𝒏 𝜹 𝟎 The initial operating point is given by point “a”. As mechanical input power is increased from 𝑷 𝒎𝟎 to 𝑷 𝒎𝟏, the new operating point will be given by point "b“. 𝒂 𝒃 𝒄 Generator𝑃𝑚 𝑇 𝑚 𝜔𝑠 𝑇𝑒 𝑃𝑒 𝑨 𝟏 𝑨 𝟐 A1 --- Pm1>Pe----Rotor Accelerates A2 --- Pm1<Pe----Rotor decelerates
- 30. 𝑷 𝒆 𝜹 𝑷 𝒎𝟎 𝑷 𝒎𝟏 𝜹 𝟎 𝜹 𝟏 𝜹 𝟐 𝒂 𝒃 𝒄 𝑨 𝟏 𝑨 𝟐 By equal area criteria the system is stable under sudden change in input mechanical power only when, 𝛿0 𝛿2 𝑃𝑎 𝑑𝛿 = 0 𝑨 𝟏 = 𝑨 𝟐
- 31. Applications of Equal Area Criteria ii) Effect of clearing time on stability A generator is connected to an infinite bus bar through two parallel lines. Assume that input power Pm is constant and the machine is delivering power to the system with a power angle 𝛿0. A temporary three phase fault occurs at the sending end of the bus 1. ∞ |𝑉| < 0°G 𝐸′ < 𝛿 𝑃𝑚 Bus 1 Bus 2 Three Phase Fault During the fault at sending end of the line, no power is transmitted to the infinite bus and 𝑃𝑒 is zero.
- 32. 𝑷 𝒆 𝜹 𝑷 𝒎 𝜹 𝟎 𝜹 𝟏 𝜹 𝒎𝒂𝒙 𝒂 𝑨 𝟏 𝑨 𝟐 𝒃 𝒄 𝒅 𝒆 𝒇 𝒈 A1 --- Pm>Pe----Rotor Accelerates Here Pe is zero due to three phase fault A2 --- Pm<Pe----Rotor decelerates 𝟎 ∏𝜹 𝟐
- 33. 𝑷 𝒆 𝜹 𝑷 𝒎 𝜹 𝟎 𝜹 𝒄 𝜹 𝒎𝒂𝒙 𝒂 𝒃 𝒄 𝒅 𝒆 𝒇 𝑨 𝟐 𝑨 𝟏 Critical Clearing Angle “𝜹 𝒄" - The critical clearing angle is reached when any further increase in 𝛿1 causes the area A2 representing decelerating energy to become less than the area representing accelerating energy. Critical Clearing Time “t 𝒄" – The maximum allowable value of the clearing time is called the critical clearing time.
- 34. Critical Clearing Angle “𝜹 𝒄" – Applying equal area criteria to the figure as shown. A1=A2 Using integration we can find out the areas. 𝛿0 𝛿 𝑐 𝑃𝑚 𝑑𝛿 = 𝛿 𝑐 𝛿 𝑚𝑎𝑥 (𝑃𝑚𝑎𝑥sin 𝛿 − 𝑃𝑚)𝑑𝛿 cos 𝛿 𝑐 = 𝑃 𝑚 𝑃 𝑚𝑎𝑥 𝛿 𝑚𝑎𝑥 − 𝛿0 + cos 𝛿 𝑚𝑎𝑥
- 35. Critical Clearing Time “t 𝒄" – For this case the electrical power 𝑃𝑒 during fault is zero, the swing equation becomes 𝐻 𝜋𝑓0 𝑑2 𝛿 𝑑𝑡2 = 𝑃𝑚 𝑑2 𝛿 𝑑𝑡2 = 𝜋𝑓0 𝐻 𝑃𝑚 Integrating both side w.r.t. time 𝑑𝛿 𝑑𝑡 = 𝜋𝑓0 𝐻 𝑃𝑚 0 𝑡 𝑑𝑡 = 𝜋𝑓0 𝐻 𝑃𝑚 𝑡 Again Integrating both side w.r.t. time and applying limits of 𝛿0 𝑡𝑜 𝛿 𝑐 𝛿 𝑐 − 𝛿0 = 𝜋𝑓0 2𝐻 𝑃 𝑚 𝑡 𝑐 2 𝑡 𝑐 = 2𝐻 𝛿 𝑐 − 𝛿0 𝜋𝑓0 𝑃𝑚
- 36. Example-2 A 60 Hz synchronous generator having inertia constant H=5 MJ/MVA and direct axis transient reactance 𝑋 𝑑 ′ = 0.3 𝑃𝑈 is connected to an infinite bus through a purely reactive circuit as shown in fig. Reactance are marked on the diagram on a common system base. The generator is delivering real power 𝑃𝑒 = 0.8 𝑃𝑈 and 𝑄 = 0.074 𝑃𝑈 to the infinite bus at a voltage of V=1 PU. a. A temporary three phase fault occurs at the sending end of the line at point F. When the fault is cleared both lines are intact. Determine the critical clearing angle and the critical fault clearing time. ∞ 𝑉 = 1 < 0° G 𝐸′ < 𝛿 𝑃𝑚 Bus 1 Bus 2 Three Phase Fault 𝑿 𝒅 ′ =0.3 𝐗 𝐭 = 𝟎. 𝟐 𝐗 𝐋𝟏 = 𝟎. 𝟑 𝐗 𝐋𝟐 = 𝟎. 𝟑
- 37. The Internal Voltage of Generator is given by 𝐸′ = 𝑉 + 𝑗𝑋12 𝐼 … … . . 1 The transfer voltage of Internal Voltage of Generator and the infinite bus is given as, 𝑋12 = 𝑋 𝑑 ′ + 𝑋𝑡 + 𝑋 𝐿1 ||𝑋 𝐿2 𝑋12 = 0.3 + 0.2 + 0.3 2 𝑋12 = 0.65 𝑃𝑈 … … . . 2 The total complex power delivered by generator is given by 𝑆 = 𝑉 × 𝐼∗ 𝐼 = 𝑆∗ 𝑉∗ (𝑆 = 𝑃𝑒 + 𝑗Q , Where𝑃𝑒=𝑃𝑚 = 0.8 𝑃𝑈 𝑎𝑛𝑑 𝑄 = 0.074 𝑃𝑈, 𝑆𝑜 𝑆 = 0.8 + 𝑗0.074 𝑃𝑈) 𝐼 = (0.8 + 𝑗0.074)∗ (1 < 0)∗ = (0.8 − 𝑗0.074) 1 = 0.8 − 𝑗0.074 𝑃𝑈 … . . 3 Using equation 1,2 and 3 𝐸′ = 1.0 + 𝑗0.65 × 0.8 − 𝑗0.074 = 𝟏. 𝟏𝟕 < 𝟐𝟔. 𝟑𝟖𝟕 𝑷𝑼
- 38. Case-1 A temporary three phase fault occurs at the sending end of the line at point F. When the fault is cleared both lines are intact. Determine the critical clearing angle and the critical fault clearing time. Since both lines are intact when the fault is cleared, the power angle equation before and after the fault is 𝑃𝑒 = |𝐸′ ||𝑉| 𝑋12 sin 𝛿 = 𝑃𝑚𝑎𝑥 sin 𝛿 𝑃𝑒 = 1.17 × 1 0.65 sin 𝛿 = 1.8 sin 𝛿 The initial operating power angle is 𝛿0 and corresponding 𝑃𝑒=𝑃𝑚 = 0.8 0.8 = 1.8sin 𝛿0 𝜹 𝟎 = 𝟐𝟔. 𝟑𝟖° = 𝟎. 𝟒𝟔𝟎𝟓𝟓 𝒓𝒂𝒅𝒊𝒂𝒏 𝜹 𝒎𝒂𝒙 = 𝟏𝟖𝟎° − 𝜹 𝟎 = 𝟏𝟓𝟑. 𝟔𝟏𝟐 ° = 𝟐. 𝟔𝟖𝟏 𝒓𝒂𝒅 Critical Clearing Angle- cos 𝛿 𝑐 = 𝑃𝑚 𝑃𝑚𝑎𝑥 𝛿 𝑚𝑎𝑥 − 𝛿0 + cos 𝛿 𝑚𝑎𝑥 cos 𝛿 𝑐 = 0.8 1.8 2.681 − 0.46055 + cos 153.61 cos 𝛿 𝑐 = 0.09106 𝜹 𝒄 = 𝟖𝟒. 𝟕𝟕𝟓° = 𝟏. 𝟒𝟖 𝒓𝒂𝒅𝒊𝒂𝒏
- 39. Critical Clearing Time- 𝑡 𝑐 = 2𝐻 𝛿 𝑐 − 𝛿0 𝜋𝑓0 𝑃𝑚 𝑡 𝑐 = 2 × 5 × 1.48 − 0.46055 𝜋 × 60 × 0.8 𝑡 𝑐 = 0.26 𝑆𝑒𝑐𝑜𝑛𝑑𝑠
- 40. Applications of Equal Area Criteria iii) Sudden Short circuit on one of parallel lines Case 1-Short circuit at one end of line A generator is connected to an infinite bus bar through two parallel lines. Assume that input power Pm is constant and generator is operating at 𝛿0. A temporary three phase fault occurs at the generator end of the line. Thus at the instant of fault CB-0 operates and isolates the generator during fault, Power flow during the fault is zero. After 𝑡 𝑐 time the circuit breaker at faulted lines operates and isolates the faulted line and power flow restores through healthy line. ∞ 𝑉 < 0° G 𝐸′ < 𝛿 Bus 1 Bus 2 𝑿 𝒅 ′ 𝐗 𝐋𝟏 𝐗 𝐋𝟐 CB1 CB2 CB0 CB3 CB4 CB5𝑃𝑚 The electrical power delivered by generator before fault occurs is given by 𝑃𝑒1 = 𝐸′ 𝑉 𝑋 𝑑 ′ + 𝑋𝑙1||𝑋𝑙2 sin 𝛿
- 41. 𝑷 𝒆 𝜹 𝑷 𝒎 𝜹 𝟎 𝜹 𝒄 𝜹 𝒎𝒂𝒙 𝒂 𝑨 𝟏 𝑨 𝟐 𝒃 𝒄 𝒅 𝒈 𝟎 ∏𝜹 𝟐 ∞ 𝑉 < 0° G 𝐸′ < 𝛿 Bus 1 Bus 2 Three Phase Fault 𝑿 𝒅 ′ 𝐗 𝐋𝟏 𝐗 𝐋𝟐 CB1 CB3 CB5𝑃𝑚 CB0 𝑷 𝒆𝑰 𝐷𝑢𝑟𝑖𝑛𝑔 𝛿0 𝑡𝑜 𝛿1 A1 --- Pm>Pe----Rotor Accelerates Here Pe is zero due to three phase fault 𝑷 𝒆𝑰𝑰𝑰 𝒆 f 𝒉 CB0 CB4CB2 𝐷𝑢𝑟𝑖𝑛𝑔 𝛿1 𝑡𝑜 𝛿2 A2 --- Pm<Pe----Rotor decelerates 𝑃𝑒𝐼𝐼=0 𝑃𝑒𝐼𝐼𝐼 = 𝐸′ 𝑉 𝑋 𝑑 ′ + 𝑋𝑙1 sin 𝛿
- 42. Applications of Equal Area Criteria iii) Sudden Short circuit on one of parallel lines Case 2-Short circuit away from line ends A generator is connected to an infinite bus bar through two parallel lines. Assume that input power Pm is constant and generator is operating at 𝛿0. When the fault occurs away from line ends, there is some power flow during the fault. After 𝑡 𝑐 time the circuit breaker at faulted lines operates and isolates the faulted line and power flow restores through healthy line. The electrical power delivered by generator before fault occurs is given by 𝑃𝑒1 = 𝐸′ 𝑉 𝑋 𝑑 ′ + 𝑋𝑙1||𝑋𝑙2 sin 𝛿
- 43. 𝑷 𝒆 𝜹 𝑷 𝒎 𝜹 𝟎 𝜹 𝒄 𝜹 𝒎𝒂𝒙 𝒂 𝑨 𝟏 𝑨 𝟐 𝒃 𝒄 𝒅 𝒈 𝟎 ∏𝜹 𝟐 ∞ 𝑉 < 0° G 𝐸′ < 𝛿 Bus 1 Bus 2 Three Phase Fault 𝑿 𝒅 ′ 𝐗 𝐋𝟏 𝐗 𝐋𝟐 CB1 CB3 CB5𝑃𝑚 CB0 𝑷 𝒆𝑰 𝐷𝑢𝑟𝑖𝑛𝑔 𝛿0 𝑡𝑜 𝛿1 A1 --- Pm>Pe----Rotor Accelerates Here Pe is not zero during the fault. 𝑷 𝒆𝑰𝑰𝑰 𝒆 f 𝒉 CB4CB2 𝐷𝑢𝑟𝑖𝑛𝑔 𝛿1 𝑡𝑜 𝛿2 A2 --- Pm<Pe----Rotor decelerates 𝑃𝑒𝐼𝐼= 𝐸′ 𝑉 𝑋 𝐼𝐼 sin 𝛿 𝑃𝑒𝐼𝐼𝐼 = 𝐸′ 𝑉 𝑋 𝑑 ′ + 𝑋𝑙1 sin 𝛿 𝑷 𝒆𝑰𝑰
- 44. Applications of Equal Area Criteria iii) Sudden Short circuit on one of parallel lines Case 3-Reclosure A generator is connected to an infinite bus bar through two parallel lines. Assume that input power Pm is constant and generator is operating at 𝛿0. When the fault occurs away from line ends, there is some power flow during the fault. After 𝑡 𝑐 time the circuit breaker at faulted lines operates and isolates the faulted line and power flow restores through healthy line. Now if the circuit breaker of line 2 are reclosed successfully, i. e. the fault was a transient one and therefore vanished, The power flow is again restored using both lines. The electrical power delivered by generator before fault occurs is given by 𝑃𝑒1 = 𝐸′ 𝑉 𝑋 𝑑 ′ + 𝑋𝑙1||𝑋𝑙2 sin 𝛿
- 45. 𝑷 𝒆 𝜹 𝑷 𝒎 𝜹 𝟎 𝜹 𝒄 𝜹 𝒎𝒂𝒙 𝒂 𝑨 𝟏 𝑨 𝟐 𝒃 𝒄 𝒅 𝒈 𝟎 ∏𝜹 𝟐 ∞ 𝑉 < 0° G 𝐸′ < 𝛿 Bus 1 Bus 2 Three Phase Fault 𝑿 𝒅 ′ 𝐗 𝐋𝟏 𝐗 𝐋𝟐 CB1 CB3 CB5𝑃𝑚 CB0 𝑷 𝒆𝑰 𝐷𝑢𝑟𝑖𝑛𝑔 𝛿0 𝑡𝑜 𝛿1 A1 --- Pm>Pe----Rotor Accelerates Here Pe is not zero during the fault. 𝑷 𝒆𝑰𝑰𝑰 𝒆 f 𝒉 CB4CB2 𝐷𝑢𝑟𝑖𝑛𝑔 𝛿1 𝑡𝑜 𝛿2 A2 --- Pm<Pe----Rotor decelerates 𝑃𝑒𝐼𝐼= 𝐸′ 𝑉 𝑋 𝐼𝐼 sin 𝛿 𝑃𝑒𝐼𝐼𝐼 = 𝐸′ 𝑉 𝑋 𝑑 ′ + 𝑋𝑙1 sin 𝛿 𝑃𝑒𝐼𝑉 = 𝑃𝑒𝐼 𝑷 𝒆𝑰𝑰 CB2 CB4 𝜹 𝒄𝒓
- 46. Example-2 A 60 Hz synchronous generator having inertia constant H=5 MJ/MVA and direct axis transient reactance 𝑋 𝑑 ′ = 0.3 𝑃𝑈 is connected to an infinite bus through a purely reactive circuit as shown in fig. Reactance are marked on the diagram on a common system base. The generator is delivering real power 𝑃𝑒 = 0.8 𝑃𝑈 and 𝑄 = 0.074 𝑃𝑈 to the infinite bus at a voltage of V=1 PU. a. A temporary three phase fault occurs at the sending end of the line at point F. When the fault is cleared both lines are intact. Determine the critical clearing angle and the critical fault clearing time. b. A three phase fault occurs at the middle of one of the lines the fault is cleared, and the faulted line is isolated. Determine the critical clearing angle. ∞ 𝑉 = 1 < 0° G 𝐸′ < 𝛿 𝑃𝑚 Bus 1 Bus 2 Three Phase Fault 𝑿 𝒅 ′ =0.3 𝐗 𝐭 = 𝟎. 𝟐 𝐗 𝐋𝟏 = 𝟎. 𝟑 𝐗 𝐋𝟐 = 𝟎. 𝟑
- 47. Critical Clearing Angle- cos 𝛿 𝑐 = 𝑃𝑚 𝛿 𝑚𝑎𝑥 − 𝛿0 − 𝑃2𝑚𝑎𝑥 cos 𝛿0 + 𝑃3𝑚𝑎𝑥 cos 𝛿 𝑚𝑎𝑥 𝑃3𝑚𝑎𝑥 − 𝑃2𝑚𝑎𝑥
- 48. Step 1-Before Fault occurs The Internal Voltage of Generator is given by 𝐸′ = 𝑉 + 𝑗𝑋12 𝐼 … … . . 1 The transfer voltage of Internal Voltage of Generator and the infinite bus is given as, 𝑋1 = 𝑋 𝑑 ′ + 𝑋𝑡 + 𝑋 𝐿1 ||𝑋 𝐿2 𝑋1 = 0.3 + 0.2 + 0.3 2 𝑋1 = 0.65 𝑃𝑈 … … . . 2 The total complex power delivered by generator is given by 𝑆 = 𝑉 × 𝐼∗ 𝐼 = 𝑆∗ 𝑉∗ (𝑆 = 𝑃𝑒 + 𝑗Q , Where𝑃𝑒=𝑃𝑚 = 0.8 𝑃𝑈 𝑎𝑛𝑑 𝑄 = 0.074 𝑃𝑈, 𝑆𝑜 𝑆 = 0.8 + 𝑗0.074 𝑃𝑈) 𝐼 = (0.8 + 𝑗0.074)∗ (1 < 0)∗ = (0.8 − 𝑗0.074) 1 = 0.8 − 𝑗0.074 𝑃𝑈 … . . 3 Using equation 1,2 and 3 𝐸′ = 1.0 + 𝑗0.65 × 0.8 − 𝑗0.074 = 𝟏. 𝟏𝟕 < 𝟐𝟔. 𝟑𝟖𝟕 𝑷𝑼
- 49. PeI = |E′ ||V| X12 sin δ = Pmax sin δ PeI = 1.17 × 1 0.65 sin δ = 1.8 sin δ 𝑃1𝑚𝑎𝑥 = 1.8 𝑃𝑈 The initial operating power angle is δ0 and corresponding Pe=Pm = 0.8 0.8 = 1.8sin δ0 𝛅 𝟎 = 𝟐𝟔. 𝟑𝟖° = 𝟎. 𝟒𝟔𝟎𝟓𝟓 𝐫𝐚𝐝𝐢𝐚𝐧
- 50. Step 2-During Fault During fault the generator delivers some amount of power to infinite bus. As 𝑃𝑒𝐼𝐼 ≠ 0 The equivalent circuit can be draw by referring above mentioned single diagram, ∞ 𝐕 = 𝟏 < 𝟎° G 𝐄′ < 𝛅 Three Phase Fault 𝑿 𝒅 ′ + 𝐗 𝐭 = 0.3 + 0.2 𝐗 𝐋𝟏 = 𝟎. 𝟑 𝐗 𝐋𝟐 = 𝟎. 𝟑
- 51. 𝐕 = 𝟏 < 𝟎° 𝐄′ < 𝛅 Three Phase Fault X=0.5 𝐗 𝐋𝟏 = 𝟎. 𝟑 G ∞ 𝐗 𝟑 = 𝟎. 𝟑 𝑿 𝟏=0.5 Convert this star in to Delta 1 2 3 N 1 2 3 N 𝑋12 = 𝑋1 𝑋2 + 𝑋2 𝑋3 + (𝑋3 𝑋1) 𝑋2 𝑋12 = 0.5 ∗ 0.15 + 0.15 ∗ 0.3 + (0.3 ∗ 0.5) 0.15 𝑋12 = 1.8 𝑃𝑈 Similarly 𝑋13 = 0.8 𝑃𝑈 𝑎𝑛𝑑𝑋23 = 0.54 𝑃𝑈
- 52. 𝐕 = 𝟏 < 𝟎° 𝐄′ < 𝛅 Three Phase Fault 𝐗 𝟏𝟐 = 𝟏. 𝟖 𝐗𝟏𝟑=𝟎.𝟖 G ∞ 𝐗𝐋=𝟎.𝟏𝟓 𝐗𝟐𝟑=𝟎.𝟓𝟒 PeII = |E′||V| 𝑇𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 sin δ = 𝑃2𝑚𝑎𝑥 sin δ PeII = 1.17 ∗ 1 1.8 sin δ P2max = 0.65 𝑃𝑈
- 53. Step 3-After Fault When the fault is cleared, the faulted line is isolated from the system as shown below. The transfer reactance in step 3 is given as, 𝑋3 = 𝑋 𝑑 ′ + 𝑋𝑡 + 𝑋 𝐿1 =0.3+0.2+0.3 =0.8 PU PeIII = |E′ ||V| 𝑇𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 sin δ = 𝑃3𝑚𝑎𝑥 sin δ PeIII = 1.17 ∗ 1 0.8 sin δ 𝑷 𝟑𝒎𝒂𝒙 = 𝟏. 𝟒𝟔𝟐 𝑷𝑼 𝛿 𝑚𝑎𝑥 = 180° − 𝛿0 𝛿0=sin−1 𝑃 𝑚 𝑃3𝑚𝑎𝑥 = sin−1 0.8 1.462 =33.17° 𝜹 𝒎𝒂𝒙=146.83° = 𝟐. 𝟓𝟔𝟐𝟖 𝒓𝒂𝒅𝒊𝒂𝒏
- 54. Critical Clearing Angle- cos 𝛿 𝑐 = 𝑃𝑚 𝛿 𝑚𝑎𝑥 − 𝛿0 − 𝑃2𝑚𝑎𝑥 cos 𝛿0 + 𝑃3𝑚𝑎𝑥 cos 𝛿 𝑚𝑎𝑥 𝑃3𝑚𝑎𝑥 − 𝑃2𝑚𝑎𝑥 cos 𝛿 𝑐 = 0.8 ∗ 2.5628 − 0.4605 − 0.65 ∗ cos 26.38 + 1.4625 ∗ cos 146.83 146.83 − 0.65 cos 𝛿 𝑐 = −0.15356 𝛿 𝑐 = 98.83°
- 55. Example-3 A 50 Hz, Four Pole turbo generator rated 100 MVA, 11kV has an inertia constant of 8.0 MJ/MVA. a. Find the stored energy in the rotor at synchronous speed. b. If the mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW, find the rotor acceleration, neglecting mechanical and electrical losses. c. If the acceleration calculated in part (b) is maintained for 10 cycles, find the changes in torque angle and rotor speed in revolutions per minute at the end of this period. a. Stored Kinetic Energy- GH= 𝐾. 𝐸 = 1 2 𝑀 𝜔𝑠 , Where G is machine rating in MVA and H is inertia constant in MJ/MVA Stored Energy=100*8 =800 MJ
- 56. Example-3 A 50 Hz, Four Pole turbo generator rated 100 MVA, 11kV has an inertia constant of 8.0 MJ/MVA. a. Find the stored energy in the rotor at synchronous speed. b. If the mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW, find the rotor acceleration, neglecting mechanical and electrical losses. c. If the acceleration calculated in part (b) is maintained for 10 cycles, find the changes in torque angle and rotor speed in revolutions per minute at the end of this period. Generator𝑃𝑚 𝑇 𝑚 𝜔𝑠 𝑇𝑒 𝑃𝑒 b. Swing Equation is given as, 𝑀 𝑑2 𝛿 𝑑𝑡2 = 𝑃𝑚 − 𝑃𝑒 Where M= 𝐺𝐻 180𝐹 = 800 180∗50 = 4 45 𝑀𝐽 − 𝑆/𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙𝐷𝑒𝑔𝑟𝑒𝑒 4 45 𝑑2 𝛿 𝑑𝑡2 = 80 − 30 𝑑2 𝛿 𝑑𝑡2 = 337.5 𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙𝐷𝑒𝑔𝑟𝑒𝑒/𝑆2
- 57. Methods to Improve Steady State and Transient Stability Limit G1 𝐸′ < 𝛿 |𝑉| < 0° 𝑋12 The Power Angle equation is given as, 𝑃𝑒 = 𝐸′ |𝑉| 𝑋12 sin 𝛿 Steady State Stability limit occurs at 𝛿 = 90° 1. The Steady State Stability limit can be increased by reducing 𝑋12 and increasing either or both 𝐸′ 𝑎𝑛𝑑 |𝑉|. 2. If the transmission lines are of sufficiently high reactance, the stability limit can be increased by using two parallel lines which incidentally also increases the reliability of the system.
- 58. Methods to Improve Steady State and Transient Stability Limit The Power Angle equation is given as, 𝑃𝑒 = 𝐸′ |𝑉| 𝑋12 sin 𝛿 1. Series capacitor are sometimes employed in lines to get better voltage regulation and to raise the stability limit by decreasing the line reactance. 2. Some time excitation voltages and quick excitation system are also employed to improve the stability limit. 3. Use of high speed reclosing circuit breaker.
- 59. Numerical Solution of Swing Equation • The critical clearing time can not be obtained from equal area criteria and we have to make this calculation numerically through swing equation. • If more than one machine is connected to the infinite bus bar then solving the swing equation will become very complicated task. • The point by point method for one machine tied to infinite bus bar gives the solution of the swing equation . • This method can be applied to every machine of a multi machine system.