Electrical electronics technology

As we know the torque equation of induction motor is T = K1 * φ * sE2 * R2/( R2² + (sX2)²) So , if s=0 then T=0 Let start the curve from point 0. Let's say, motor is running near to synchronous speed. The term  sX2 = 0 negligible Therefore T =  K1 * φ * sE2 /R2 T > s [ other is let say constant] So for low value of slip, torque slip curve is straight line. as a load increases sleep increases. As the load further increases it the touches Tmax point when s= R2/ X2 called pull out torque. Any further increases in load i.e slip increases R2 becomes negligible  Therefore T > (1/s) Hence after pull out aur breakdown torque the curve will be rectangular hyperbola. Hance we can see that any further increases in load motor slow down and eventually stop. If we provide some electrical protection for overload it will trip circuit. From figure it is also clear that by increasing the value of rotor resistance we can get maximum torque at the s = 1 or zero speed Effect of change i...

Since we know the motor running torque is T = K1 * φ * sE2 * R2/( R2² + X2²) Where φ  - flux per status pole I₂ - rotor current E2 rotor emf per phase. So if S = 0 then T = 0 Let's start from right hand side of the curve from point 0. Let's say, motor is running near to synchronous speed. The term sX2 approximately equals to zero negligible. Therefore, T = K1 * s * φ * E2 /R2 Assuming stator flux per pole, rotor EMF per phase, rotor resistance per phase is constant. T ⇒ s  So for low value of sleep torque slip service straight line. As the load increases sleep increases. As the load further increases it touches the maximum torque point when S = R2/X2 called pull- out torque or breakdown torque. Any further increase is in load that is slip increases R2 becomes negligible.  So,  T ⇒ 1/s Hence after pull out torque or breakdown torque the curve will be rectangular hyperbola. Hence we can see that any further increases in load motor slow down and eventually stopped. If w...

When we connect three phase applied to stator winding then rotating flux will set up due to this flux rotor start rotating but it will never catch up the synchronous speed of flux Ns=120 f/p. if it did so then no relative speed between flux and rotor then no EMF will induces, no current and so no torque to maintain rotation. So motor will always rotate less than synchronous speed of magnetic flux Ns = 120 f/p. The difference between synchronous speed and actual speed Nr is known as slip.      Slip = Ns - Nr.  Unit revolution per second. Express percentage of synchronous speed  % slip S = ( (Ns -  Nr) /NS) * 100 % One more term we call it as slip speed Slip speed = sNs = Ns- Nr  when rotor is stationary then frequency of rotor current is same as supply frequency. But when rotor start rotating then frequency depend upon relative speed between the two rotating magnetic flux and rotor. Let's say at any sleep Speed the frequency of rotor current is f' . Ns ...

The torque in induction motor is proportional to the product of flux per stator pole and the rotor current one more term we will take in account is power factor of the rotor.   T →  φ  *  I₂  * cos∅₂  Where φ  - flux per status pole I₂ - rotor current cos∅₂ - rotor power factor ∅₂ phase angle  rotor current and rotorEMF. The induced rotor EMF E2 is proportional to flux per stator φ. Therefore T = K1 * E₂ *  I₂ * cos φ2  Where K1 is another constant. This is torque at standstill condition.   Under running condition induced EMF is T = K1 * Er *  Ir * cos ∅2 ---------------------- (1) Where Er = rotor induced EMF per phase under running condition. Ir = rotor current per phase under running condition. φ = flux per stator pole. rotor current, Ir = Er / Zr = sE2/  √( R2² + (sX2)²)  ----------- (2) power factor, cos∅2 = R2 / √( R2² + (sX2)²)   ----------------(3) So the rotor torque is,  After putting Ir from ...

When we supply three phase input to the rotor of the motor then voltage is induced in the rotar e2 and current I to start flowing. The frequency of rotar and voltage of rotor at stand still is equal to supply frequency. Because relative velocity is maximum at a stand still when s= 1 between status flux and rotar. Let at stand still S =1 , E2= stand still rotar induced EMF per phase f2 = rotar current frequency at stand still X2 = stand still rotor induced reactance per phase. when rotor starts rotating then relative velocity between rotor and stator flux decreases then induced voltage E2  start decrease. When  rotor speed catches rotating flux speed than relative velocity becomes zero that is S= 0 then induced voltage then become zero. let's say for a slip 's' the rotor induced EMF will be s times the induced emf at a stand still. Therefore under running condition Rotar induced EMF running condition Er = s E2 Frequency of induced EMF at running condition fr = sf2 Since freq...

The torque in induction motor is proportional to the product of flux per stator pole and the rotor current one more term we will take in account is power factor of the rotor.   T →  φ  *  I₂  * cos∅₂  Where φ  - flux per status pole I₂ - rotor current cos∅₂ - rotor power factor ∅₂ phase angle  rotor current and rotorEMF. The induced rotor EMF E2 is proportional to flux per status φ. Therefore T = K1 * E₂ *  I₂ * cos φ2  Where K1 is another constant. From the above equation it is clear that when φ2 increases cos φ2 decreases and vice versa. Due to the revolving stator flux EMF is induced in rotor conductor, and this EMI is also sinusoidal. When rotor is non inductive. φ₂=0 in this case rotor current is in phase with rotor EMF. So the instantaneous value of torque is product of instantaneous value of flux and current. It is seen that torque is always positive. When rotor is inductive load. In inductive load I₂ lags behind E₂ bY an angle φ2...

Impedance traingle of rotor circuit of induction motor. Let E2 = rotar EMF per phase at stand still  R2 = rotar resistance per phase X2 = rota reactance per per phase at stand still . Z2 = √( R2² + X2²) rotar impedance per phase at stand Still. I2 = E2 / √( R2² + X2²)  = rotar current cos∅2 = R2 / √( R2² + X2²) = power factor  T = K1 E₂ R2/( R2² + X2²)  = rotor torque = K1 * E2 / √( R2² + X2²) * E2 * R2 / √( R2² + X2²) = K1 E₂² R2 /(R2² + X2²) if we increase the resistance of rotor circuit then we can improve the power factor. Torque in squirrel cage induction motor :- The resistance of squirrel cage rotor is fixed and small as compared to reactance. Reactance of rotor is high as frequency of a current is equals to supply frequency. Hance at starting power factor is poor. starting current is although very high but lags by very large angle behind E2. So the torque is only 1.5 times the full load torque but starting current is 5 to 7 times the full load current. Hence ...

when we connect three phase applied to stator winding then rotating flux will set up due to this flux rotor start rotating but it will never catch up the synchronous speed of flux Ns=120 f/p. if it did so then no relative speed between flux and rotor then no EMF will induces, no current and so no torque to maintain rotation. So motor will always rotate less than synchronous speed of magnetic flux Ns = 120 f/p. The difference between synchronous speed and actual speed Nr is known as slip.      Slip = Ns - Nr.  Unit revolution per second. Express percentage of synchronous speed  % slip S = ( (Ns -  Nr) /NS) * 100 % One more term we call it as slip speed Slip speed = sNs = Ns- Nr  when rotor is stationary then frequency of rotor current is same as supply frequency. But when rotor start rotating then frequency depend upon relative speed between the two rotating magnetic flux and rotor. Let's say at any sleep Speed the frequency of rotor current is f' . Ns ...

Induction motor has a stationary part known as stator containing status winding and a rotor having bar in squirrel gauge induction motor and three phase winding in slipping induction motor. When three phase winding of stator fed by three phase supply then a rotating magnetic field  of constant magnitude 1.5 times  of maximum flux due to any phases is generated. Rotating at synchronous speed NS = 120 f/p.  This flux passage through the air gap and cut the rotor bar as that they are still stationery. Due to the relative velocity between the rotating flux and the stationary rotar conductor, EMF is induced in the rotor conductor according to faraday's law of electromagnetic induction. The the frequency of the induced EMF is same as supply frequency, magnitude is proportional to the relative velocity between the two rotating flux and rotor conductor. And direction is given by Fleming's right hand rule. since the rotor conductors are short circuit by end ring so current start f...

  In AC induction motor when we start it, at starting it will take 5-7 times of motor rated current and it should come down in 4-5 second and will take no load current 40% - 50% of motor rated current. Then we can now give it on load. Now suppose if there is jam in mechanical coupling side, motor shaft then what would happen. In this case if we start the motor then it take it's starting current 5-6 times of rated current and because it is jam either from mechanical coupling,gearbox or motor shaft problem  Motor can't start rotating and motor will take  current 5-6 times and  will not come down to it no load current as the delay time of EOCR end it will trip in stall fault. So if it trip in stall fault then it is mechanically stuck up, it happens at start. If any queries then do comment. Regards.

If 3 phase Induction motor was in running condition and trip during running in unbalance current.  Probable cause may be :- ​1. Winding resistance unbalance. ​2. Winding inductance unbalance. ​3. Motor shaft play. ​4. Motor vibration may increase. 5. Motor power cable burn somewhere/ carbonized.  May be other cause too. 1. First this is check the motor resistance and inductance by motor checker.  Procedure to check the motor Resistance:-  Disconnect all three motor phase connection from motor terminal , and take out the three lead from stud. Then take reading of resistance of R- phase,Y- phase, B -phase by connecting checker lead to motor phase lead one by one. Resistance should be balance. Procedure to check the motor inductance:-  Disconnect all three motor phase connection from motor terminal , and take out the three lead from stud. Then take reading of inductance of R- phase,Y- phase, B -phase by connecting checker lead to motor phase lead one by one. Induct...

  Well " insulation resistance"  this is very basic term in electrical world.  If some motor, switchgear, transformer trip in Earth fault then we need to check the insulation resistance of the equipment with megger.  For LT equipment we use 500volt megger. Our equipment is generally of 415 volt ac. So the minimum IR value formula is  IR min. = KV +1 Mega ohm SO if the voltage is 415 vac then min. Value of IR  will be IR  = 0.415 +1 Mega ohm       = 1.415 Mega ohm. If voltage level is 11kv then min. Value of insulation resistance will be IR min. = 11kv + 1 Mega oh               = 12 Mega ohm. If any queries please comment. Thanks.