{"id":4379,"date":"2025-03-18T13:15:39","date_gmt":"2025-03-18T07:45:39","guid":{"rendered":"https:\/\/study.madeeasy.in\/?p=4379"},"modified":"2025-07-16T14:19:48","modified_gmt":"2025-07-16T08:49:48","slug":"sudden-change-in-mechanical-input","status":"publish","type":"post","link":"https:\/\/www.madeeasy.in\/study\/ee\/power-generation-concepts\/sudden-change-in-mechanical-input","title":{"rendered":"Sudden Change in Mechanical Input"},"content":{"rendered":"

Applications of Equal Area Criterion<\/h2>\n

Below figure shows the transient model of a single machine connected to infinite bus bar.<\/p>\n

\"Mechanical<\/p>\n

Let the mechanical input to the rotor be suddenly increased to Pm1<\/sub>. Due to the accelerating power\u00a0 Pa<\/sub> = (Pm1<\/sub>-Pe<\/sub>) rotor accelerates and its speed increases (\u03c9 > \u03c9s) as a result \u03b4 also increases as shown below in figure.<\/p>\n

\"Curve\"<\/p>\n

At angle \u03b41, Pa<\/sub> = (Pm1<\/sub>-Pe<\/sub>) = Pmax Sin \u03b41 = 0 (at point b), but the rotor angle continues to increase as \u03c9 > \u03c9s. Now, Pa<\/sub> becomes negative (decelerating) hence, the rotor speed starts reducing but, the angle continues to increase till at angle \u03b42, \u03c9 = \u03c9s (at point C). At C, the decelerating area A2<\/sub> equals the accelerating area A1<\/sub>, i.e.,<\/p>\n

\"Rotor\"<\/p>\n

Since the rotor is deaccelerating, the speed reduces below \u03c9s and the rotor angle begins to reduce. The state point now traverses the Pe<\/sub> – \u03b4 curve in opposite direction as shown below in figure (power angle curve). The system oscillates about the new steady-state point b (\u03b4 = \u03b41<\/sub>) with angle upto \u03b40<\/sub> and \u03b42<\/sub> on the two sides. When the oscillations decay out, the system settles to the new steady state where<\/p>\n

Pm1<\/sub>-Pe<\/sub> = Pmax<\/sub> Sin \u03b41<\/sub><\/p>\n

Now, using figure (power angle curve), the two areas are<\/p>\n

\"Power<\/p>\n

Now, any further increase in 1 m P means that the area available for A2<\/sub> is less than A1<\/sub>, so that the excess kinetic energy causes \u03b4 to increase beyond point c and the decelerating power changes over to accelerating power\u00a0 and hence, the system becomes unstable. The system will remain stable even though the rotor oscillate beyond \u03b4 = 90\u00b0, as long as the equal area criterion is met. The oscillations for \u03b4 = 90\u00b0 is shown below in figure which is a stable system as A2<\/sub> (decelerating area) is greater than A1<\/sub> (accelerating area).<\/p>\n

\"Transient\"<\/p>\n

Note:<\/strong><\/p>\n