Question

4-pole, 50-Hz synchronous generator has a rating of 200 MVA, 0.8 power factor lagging. The kinetic energy of the machine at synchronous speed is 1000 MJ. The machine is running steadily at synchronous speed and delivering 50 MW to a load at a power angle of 8 electrical degrees. The load is suddenly removed. Determine the acceleration of the rotor in rpm/S. If the acceleration computed for the generator is constant for a period of 8 cycles, determine the following at the end of 8th cycle. (i) the value of the power angle in mechanical degrees (ii) the speed of rotor in rpm.

Answer 1

please vote up
p = Pole 4. given data -50HZ F = Frequency Machine rabing in mun S 200 IVA power factor = 0.8 Jogging R.E. stored in the whor at synchronous speed is = 10oomJ K. - S. H. Н. stored enery in nou mega al Joules machine rating in MVA H = 1000 m] 200 MUA = 5 MI - 5 MYA 6.H mw er & m. arf 6H mj-sec ms 6H Tf 18 of elect, degree M 1000 180X80 10 1885 I MJ-sec ğ elect. degree

the accelerating power = mechanical power input- Electreal power output Pa Pri - Pe the clood is suddenly removed that is electreal power output is zeno then Pe to at power at steady state before disturbance then Pirio = per = somw (Pa 50) Swing equation need JË Pa d²d Pmo - le Pme dta Mechanical power not change m d²d dta 450 elec, deyne

dele & Omech mach Oolec & 11 450 X mech. degree 152 d+२ uso x et =225 mech. degree 4 5२ a = 225 X 60 2x180 Ypm Sec = 37.5 rpm sec T - I f ㅗ So 0.02 see period time it=85 (1 0.16 See a) 니 &+2 change in dood angle Dohou af -$#450X-102 A5 5.76 ele-degrees 48 - (5. 76)x] 2.88 mech.-deynee power anyle 이 So + Ad 872 10 + Ad = or +2.88

Power angle 6.88 mech. degree I (ii) Synchronous honous speed NS lauf 120x50 u Mss Isoo rpm Since 37.5 rpm is there fore. speed at them of 8 cyeles 37.5 =lSoot H7.16 = 1506 7pm