Induction Motor Characteristic-I
TARGET>>
1.How induction motor operation is similar to the clutch?
2.Calculate the starting time of a drive with the following parameters:
J=10 kg-m2,
Tm = 15 + 0.5wm
Tl=5+0.6ωm
3.A drive has following equations for motor and load torques:
Tm=(1+2ωm)
Tl= 3 √ωm
Obtain the equillibrium points and determine the stability
SOLUTION>>
1. Similarity between Induction motor operation and Clutch operation---
Induction Motor Operation- >When the motor is excited with a three-phase supply,three-phase stator winding produces a rotating magnetic field with 120 displacements at a constant magnitude which rotates at synchronous speed. This changing magnetic field cuts the rotor conductors and induces a current in them according to the principle of Faraday’s laws of electromagnetic induction. As these rotor conductors are shorted, the current starts to flow through these conductors.
>In the presence of the magnetic field of the stator, rotor conductors are placed, and therefore, according to the Lorenz force principle, a mechanical force acts on the rotor conductor. Thus, all the rotor conductors force, i.e., the sum of the mechanical forces produces torque in the rotor which tends to move it in the same direction of the rotating magnetic field.
>This rotor conductor’s rotation can also be explained by Lenz’s law which tells that the induced currents in the rotor oppose the cause for its production, here this opposition is rotating magnetic field. This result the rotor starts rotating in the same direction of the stator rotating magnetic field.
>If the rotor speed more than the stator speed, then no current will induce in the rotor because the reason for rotor rotation is the relative speed of the rotor and stator magnetic fields. This stator and the rotor field difference are called slip. Stator speed or Synchronous speed is marked with yellow colour in the picture stated below.
If Stator speed is Wms and and Rotor speed is Wmr. Then Slip(s)= [(Ns - Nr)/ Ns]. Where 0 < s < 1 Nr = (1 - s) * Ns ......(i) Percent of Slip = [(Ns - Nr)/ Ns] *100
Clutch Operation- >A clutch is the mechanical device that transfers all power from the source into the load of a vehicle. Without a properly operating clutch, power transfer and gear shifting would be very difficult. >The clutch is located between the engine flywheel and the transmission. The first section of this system starts at the flywheel. Connected to the flywheel is the pressure plate, with the clutch-friction disc between the two items. On the out side of the pressure plate will be the clutch control unit, or the throwout bearing.The pressure plate assembly is secured to the flywheel via bolts connecting the cover stamping to the flywheel.
>During engagement, the pressure plate assembly clamps the disc assembly against the flywheel, transmitting engine power to the transmission. >During disengagement, power flow is interrupted when the pressure plate no longer clamps the disc against the flywheel. Instead, the pressure plate lifts away from the flywheel, creating a gap large enough for the disc to disengage from the flywheel.
The clutch connects the driving shaft and driven shaft each other and transfer equal ammount of torque but they does not spin in same speed because of slip.
As we can see speed in the both shaft is not same. Driving shaft speed = W1 and Driven shaft speed = W2. Slip(s) = [(W1 - W2)/ W1] Where 0 < s < 1 W2 = (1 - s) * W1 ........(ii) We can say both eqn (i) & (ii) is similar . By this we can say Induction motor operation is similar to the Clutch operation.
2.Calculating the starting time of a Induction motor drive---
Starting time - The duration of time when large current, which flows during the starting of an induction motor. Normally, the starting current of an induction motor is six to eight time of full load current.
OR
The time from starting torque point to maximum or peak torque point after which motor is engaging with load is called as starting time.
To Calculate the starting time of induction motor-
Parameters Given- Moment of Inertia of motor and combined load system(J)= 10 kg-m^2 Motor Torque (Tm)= 15 + 0.5Wm Load Torque(Tl)= 5 + 0.6Wm
A motor generally drives the load or machine with some transmission system. While the motor always rotates, the load may rotate or may undergo a translational motion.
Motor –load system can be described by the fundamental equation- Tm - Tl = d/dt(J*Wm) or, Tm - Tl =J*(dWm/dt) + Wm*(dJ/dt) As here to calculate starting time of drive the Moment of Inertia J constant so, (dJ/dt)=0. Now, Tm - Tl = J*(dWm/dt) ..............(i) In steady state (dWm/dt)=0 so, Tm - Tl = 0 or, Tm = Tl or, 15 + 0.5Wm = 5 + 0.6Wm or, Wm = 100 rad/sec The torque component J*(dWm/dt) is called the dynamic torque component because it is present only during the transient operations like starting, braking, speed reversal, acceleration or deacceleration. Now, (15 + 0.5Wm) - (5+ 0.6Wm) = 10*(dWm/dt) ...........( putting all the values of parameters given in eq (i). ) or, (10 - 0.1Wm) = 10*(dWm/dt) or, dt = [1/(1 - 0.01Wm)]dWm or, ∫dt = ∫[1/(1 - 0.01Wm)]dWm or t = -100*ln(1-0.01Wm) + C Lets consider slip of the motor is 6% to the steady state speed of the motor will be {100 - (100*0.07)} = 94 rad/sec . So, to calculate starting time we need to calculate the time for the motor to rotate from 0rad/sec to 94rad/sec. For this we have consider 0 rad/sec as lower limit and 94 rad/sec as upper limit in the time interval of 0 to t. Now, ∫dt ( 0 to t) = ∫[1/(1 - 0.01Wm)]dWm ( 0 to 94) So, t - 0= -100*ln(1-.94) - 0 So, t = -100*(-2.8134) So, t = 281.34 sec So the starting time of the drive with a 6% of slip is 281.34 sec.
3.Obtaining Equilibrium point and stability of Induction motor drive---
Stability- Stability of a Motor is defined as its capacity to drive a load. If a motor has enough power and is able to drive a load then it is called a stable motor drive.
Given motor & load torque equations are below- Tm=(1+2Wm) Tl= 3*√Wm In case of steady state (dWm/dt)= 0 so, Tm = Tl Then, (1+2Wm) = 3*√Wm so, 2Wm - 3*√Wm +1 = 0 .......(i) Comparing eq-(i) with standard quadratic eq [ ax^2 + bx + c = 0], We get a=2 b=-3 and c=1. √Wm= [-(-3) ± √{(-3)^2 - 4*2*1}]/ 2*2 Wm= 1(considering positive sign) √Wm= 0.5(considering negative sign) So, Wm = 1 or 0.25 The stability condition of an induction motor is - [(dTl/dWm)-(dTm/dWm)] > 0 Now [(3*√Wm /dWm)- {(1+2Wm)/dWm}] > 0 or, [(3/2√Wm) - 2] > 0 or, 3/Wm > 4 For Wm = 1 3 ≯ 4(not satisfied the stability condition) but for Wm = 0.25 12 > 4 (satisfied the stability condition)
The stability is achieved at a equilibrium point where Wm=0.25 rad/sec.