Load EqualisationIn some drive application load torque fluctuates widely within short intervals of time. Examples of loads are, 1) In pressing machine 2) Electrical hammer 3) Steel rolling mills 4) Reciprocating pumps In such drives if motor is required to supply peak torque demanded by load following problems may arrives, 1) Motor with high rating should be use 2) Motor will draw a pulsed current from the supply, it gives rise to the voltage fluctuations 3) Reduces System Stability In above mentioned problems of fluctuations of loads are overcome by mounting a flywheel on motor shaft in a non-reversible drives. During high load period load torque will be much higher than motor torque. Deacceleration occur producing large dynamic torque component. Because of Deacceleration motor speed falls during light load period .The motor torque exceeds the load torque acceleration speed is brought back to original value before the next high load period. It show that peak torque required from the motor has much smaller value than peak load torque hence a motor with small rating than peak load can be used. Fluctuations in motor torque and speed are also reduced since power drawn from the source fluctuates very little. This is called load equalization. The moment of inertia of the fly wheel required for load equalization is calculated as follows. Assuming a linear motor speed-torque curve in the region of interest (drooping characteristic AC) It is the time required for the motor speed to change by mentioned constant at rated value Tr ω (¿ ¿ m 0−ω mr ) ¿ . From fundamental torque equation & equation (2) τm dT +T =T l dt when motor torque is .ω T (1) Tr ω m=ω m 0−¿ (¿ ¿ m 0−ω mr ) Where.Rated torque in N-m Differentiating equation (1) and multiplying both sides by J gives ω (¿ ¿ m 0−ω mr ) dT /T dt r Jd ωm =−J ¿ dt ¿− τ m dT (2) dt ω (¿ ¿ m 0−ω mr )/T r τ m=J ¿ Where Term τm (3) is defined as a mechanical time constant of the motor.Rated speed in rad/sec . ωm 0 .load speed in rad/sec ω mr Tr .no. T= T min . ( T min=T ¿ 1−e −−t l τm )+T max e −−t l τm From equation (4) th τ m= log [ ( T l h−T min ) T l h−T max ] (6) From equation (3) & (6) ω ¿ m 0−ω (¿ m )(7) th J= ∗T r /¿ ( T l h−T min ) log T l h−T max [ [ ] ] Also From equation (5) (5) tl . When operating in steady-state. For high load period (0<=t<= −th τm th −−t h τm min e th ) the solution of above equation is ( 4) is the motor torque at t=0 which is also the instant when heavy load is applied and motor torque at the end of heavy load period is T max . a cycle of which consists of one high load Tlh period with torque T¿ and duration tl and duration ( T max=T l h 1−e T min Where Tlh )+T and one light load period with torque .Consider now a periodic load torque. motor torque at the end of the cycle will be the same as at the beginning of the cycle Hence at t’= We get. Kg−m 2 (10) Where W is weight of flywheel (kg) and R is radius (m). (6) or(7) further J =W R2 .tl τm= log [ ( T max −T ¿ ) T min −T ¿ ] ( 8) From equation (6) & (3) ω ( ¿ ¿ m 0−ω mr)(9) tl J= ∗T r /¿ ( T max−T ¿ ) log T min−T ¿ [ ] Moment of inertia of flywheel required can be calculated either from Eq. .