electrical electronics knowledge topics

Induction motor star delta starter.

Music: Good time 02

and speed.

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 in supply voltage is on torque and speed.

T = K1 * φ * sE2 * R2/( R2² + (sX2)²)

Since ,φ > E2

And E2 > V

Therefore T > V²

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 we provide some electrical protection for overload it will trip the circuit.

From figure it is also clear that by increasing the value of rotor resistance we can get maximum torque at s= 1 or zero speed.

To start a motor successfully, the load torque always be less than motor torque. If motor start successfully and when due to some reason load torque becomes more than motor torque than motor will stop. So to run motor successfully motor torque must be greater than load torque all the time.