# 50Hz N S Load All Actions A1  A weight is balanced at equilibrium position of a spring and produce a Simple Harmonic Motion with acceleration Which.

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50Hz N S Load

All Actions A1  A weight is balanced at equilibrium position of a spring and produce a Simple Harmonic Motion with acceleration Which is proportional to Displacement of x= ASinwt.  The Harmonic motion has the Damping ratio which tends to stop the Oscillation but energy flow from electrical side of M3 protect the damping and makes the system undamped free Vibration means continuous oscillation occurs..

 Where it is balancing the Periodic motion of spring into Circular Motion. It is going up and down Which Creates a full revolution and creates a full Rotation.  The time period is same as the periodic motion and circumference of the circle is equal to the path traveled by the weight of spring. A2

A3  The circle taking a periodic motion and converting circular motion which is exactly equal to the time period of spring and expand of spring is equal to the circumference of circular motion.  At steady state condition spring takes the motion from circle come from electrical force.

A4  A4. Here the circumference of the circular motion is increased by increasing its radius which time period is exactly equal to A3. That means A5 experiences a Huge distance travelled on its surface within the same Time period of A1.

A5  A5. Is the circle which surface is connected to A4 surface and produce increased Rotation that time period can be much smaller than A1

S N e1= dФ/dt f=PN/120=5 Hz N=300 rmp ω= 5 rps A6 M1

A6 At M1(e1) the coil of Rotor is rotating within a magnetic field and producing an e.m.f in rotor circuit by Faraday’s law of electro-magnetic induction e1=dФ/dt and the frequency is f=PN/120 Hz, N∞f depends upon the rotation of A5.

A7 M2 e1 e2

A7 (At M2 e1 e2)Then the e1 is transmitted to Rotor of 2 nd Machine which taking the e1 at Rotor and producing a rotating magnetic field in the air gap and produces e2 in the the stator by the same law and the frequency is same as the rotor frequency of 1 st Machine.

A8  (At System board) The system is taking the stator emf e2 from the 2 nd Machine and rectified it into dc then inverts it into ac with 50Hz from the pulse width modulation in the system and provides the amount energy(e3) generated in 2 nd machine with having 50Hz frequency.

A9  (M3 e3 e4) Here the e3 is applied on the stator of 3 rd Machine and produce a rotating magnetic field in air gap which cuts fluxes on the conductor of Rotor and generates emf in rotor. At first it stores energy as a energy storing device and after getting the sufficient amount of energy it starts moving with a angular velocity which is increasing and trying to catch up the angular velocity of rotating magnetic field. That means a force is developing in rotor which tends to rotate faster to catch up synchronous speed which is Ns=120f/p. At Ns rotor rotating at synchronous speed with slip s=0.  Then no emf is generating in rotor and no more force to rotate faster.  Rotor has the tendency to rotate slower but the continuous supply of frequency will produce the continuous synchronous speed means the continuous rotating magnetic field which is cause of producing emf e4

A10  (M4 e5 e6) the rotor of M3 and M4 are coupled together. That means the energy is flowing from M3 to M4 to be magnetize enough and it is also rotating with Ns which producing a rated emf e6 on the stator or as output.

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