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AIM : To study how Induction Motor Operation is similar to Mechanical Clutch. To calculate Starting Time of a Drive with the given Parameters. To Explain the Stability of a Drive with the given parameters. Make a MATLAB script file which will plot speed-torque characteristics for frequency control method. OBJECTIVE : How…
Omkar Kudalkar
updated on 05 May 2021
AIM :
OBJECTIVE :
J=10 kg-m2,
T = 15 + 0.5wm
Tl=5+0.6ωm
Tm=(1+2ωm)
Tl= 3 √ωm
Obtain the equilibrium points and determine the stability.
SOFTWARE USED: MATLAB
NOtes Link: https://drive.google.com/file/d/1oOVelUmSYNNH0dfGUTeMiEt6p4d8QrDB/view?usp=sharing
STUDY_1:
INTRODUCTION:
Nikola Tesla invented induction motor with rotating Magnetic Field that made unit drives for machine feasible and made AC Motor transmission for economic necessity. The induction motor has 2 main parts Stator (stationary part) and Rotor (rotating part). The Stator is basically a 3 coil winding and 3-phase power input is given to it. The winding passes to slot of Stator made by stacking fin.
An Induction motor (also known as Asynchronous motor) is a commonly used AC electric Motor. In an Induction Motor, the Electric current in the rotor needed to produce torque which is obtained via electromagnetic induction from the rotating magnetic field of the stator winding. The rotor of induction motor can be Squirrel cage or Wound type rotor.
Induction motor is referred to as “Asynchronous motor” because they operate at a speed less than their synchronous speed.
Theory of Operation of Induction Motors:
In both Induction and Synchronous motors, the AC power supplied to the motor’s Stator creates Magnetic field that rotates in synchronism with AC Oscillations. Whereas Synchronous Motor’s Rotor turns at the same rate as the Stator field, an Induction motor’s Rotor rotates at a somewhat lower speed than the stator field. The Induction motor Stator’s Magnetic field is therefore changing or rotating relative to the Rotor. This induces an opposing current in the Induction Motor’s Rotor; its effect the Motor’s Secondary winding (closed loop conductor in rotor). The Rotating Magnetic flux induces current in the Winding of Rotor (close loop conductor), in the manner similar to current induced in a Transformer’s Secondary Winding(s).
The induced currents in the ROTOR winding in turn CREATE magnetic field in the ROTOR that react AGAINST the STATOR magnetic field. The direction of magnetic field created will be such as to oppose the change in current through the rotor winding (rotor winding will be the shaft of motor) by LENZ’s Law.
The cause of induced current in the rotor winding is the ROTATING STATOR’s MAGNETIC FIELD, so to OPPOSE the change in rotor winding currents the rotor will start to rotate in the direction of the rotating stator magnetic field. The ROTOR accelerates angularly until the magnitude of induced rotor current and TORQUE balances the applied MECHANICAL LOAD on the rotation of the ROTOR. Since the rotation at synchronous speed would result in no induced rotor current, AN INDUCTION MOTOR ALWAYS OPERATE SLIGHTLY SLOWER THAN SYNCHRONOUS SPEED.
For rotor current to be induced, the speed of the Physical rotor motor must be lower than that’s of Stator’s rotating magnetic field otherwise the magnetic field would not be moving relative to the rotor conductor and no current will be induced. As the Speed of the Rotor DROPs below Synchronous speed, the rotation rate of magnetic field in the rotor increases, inducing more current in the winding and creating more Torque. The ratio between the rotation rate of the Magnetic field induced in the rotor and the rotation rate of the stator’s rotating magnetic field is called “SLIP”. Under load, The Speed drops and the slip increases enough to create sufficient TORQUE to turn the load. For this reason induction motor are sometimes referred to as “Asynchronous Motors”.
Slip(s) is defined as the difference between Synchronous speed and operating speed, at the same frequency, expressed In RMP, or in % or ratio of synchronous speed. Thus
Whereas, Ns is Stator Electric Speed, Nr is Rotor Mechanical Speed.
The Power loss Equation is given below:
Pr : Pcr : Pm = 1 : S : (1-s)
Where , Pr = Power applied to the Rotor
Pcr = Rotor losses
Pm = Mechanical power
There will be losses are acting during the transmission of power and variation in rotational of winding leads to losses of speed in motor. The input current is processes to stator and rotor rotates the connected shaft by the magnetic field of winding and outputs are obtained by it. During that operation, there will be losses in speed of motor from the stator to rotor of the motor
The Speed losses in a motor are given below,
Where,
= Rotor Speed.
= Stator Synchronous Speed.
s = Slip
CLUTCH:
A Clutch is a Mechanical device that is used for transferring power from 1 part of an engine to another. A typical example is an Automobile. The reciprocating action of piston inside the Engine produces a POWER. The power generated (provided) by the engine is transmitted via a Clutch to the wheel through Gear to a Transmitting shaft. The gears are of various diameters and hence they could control the speed of rotation. The Clutch works by Engaging and disengaging the gear from the transmitting shaft.
ELECTROMAGNETIC CLUTCH:
The main components of the EM clutch are a Coil shell, an Armature, Rotor, and Hub. The Armature plate is lined with with friction coating. The coil is placed behind the rotor. When the clutch is active the electric circuit Energizes the coil, it generates a magnetic field. The Rotor portion of the clutch gets magnetized. When the magnetic field exceeds the air gap between the rotor and armature it then pulls the armature towards the rotor. The frictional force generated at the contact surface transfer the Torque. Engagement time depends on the strength of magnetic fields, inertia, and air gap.
When voltage is removed from the coil, the contact is gone. In most design a spring is used to hold back the armature to provide an air gap when current is removed.
EDDY CURRENT CLUTCH:
An eddy current clutch & an Induction Machine are identical in working principle. The input shaft is directly coupled to the motor and variable speed output shaft is connected to load. A metallic drum is connected to the driving shaft and a field coil system is mounted on output shaft. The Slip ring used to excite the field system. The load is initially at rest when the motor reaches full speed; DC current is passed through the electromagnets. The resultant magnetic field produced an eddy current in the metallic drum. The magnetic flux produces torques that brings the load to Desired Speed.
Loss = Input Power – Output Power
Input power = T.
Output Power = T.
Loss = T. T. ………………………….(1)
Where , T = Torque of input and output shaft
speed at driving shaft.
speed at driven shaft.
S = slip
The Clutch losses are given below,
Substituting the value of in equation (1)
Loss = T. T.
Loss = (1-s) . T
The output from the rotor of the induction motor is connected to clutch. Eddy current clutches are so named because they induce Eddy current into a metal, cylinder or drum. One part of the clutch contains slip rings and a winding. The rotor is constructed in such a way that when the winding is excited with direct current magnetic pole pieces are formed. The rotor in motor is connected to driving motor input of clutch. The rotor is the input of the clutch and is connected to AC induction motor. The rotor is mounted between metallic drum field winding of clutch that form output shaft of clutch. When rotor transfers the force to clutch, A current is applied to driving motor (rotor) directly, the spinning electromagnets induce Eddy current into Metal drum. The magnetic field of the rotor and drum are attracted to each other, and the clutch turns in same direction as the motor.
Once it reaches the high level of TORQUE to obtain the constant speed the clutch is engaged to the output shaft of the driving vehicle. The main advantage of Eddy current Clutch is that there is no mechanical connection between rotor and drum. Since there is no mechanical connection, there is no friction to produce excessive Heat and there is no wear as is in the case of Mechanical Clutch. A Clutch Engaged and disengaged are based on the maximum speed of rotor obtain from motor to clutch. The speed of the clutch can be controlled by varying the amount of direct current applied to the armature or clutch rotor. Since the output speed is determined by the amount of slip between the rotor and drum, when the Load is added, the Slip will become greater, causing a decrease in Speed. This can be compensated by increasing the amount of direct current applied to the clutch rotor.Many Eddy current clutch circuit contain a Speed sensing device that will automatically increase or decrease the DC excitation current when load is added or removed.
Therefore the operation of the Induction Motor and the clutch are similar to each other, By increasing the Motor speed the direct current is obtained by the rotor and it transfers the current to drum and field winding thus they engaged with motor and clutch also rotates as the same direction of a motor. After the Speed changes the disengagement is appeared and the Clutch is released for fraction of seconds and engaged while it reaches the constant speed.
STUDY_2
The duration of time when a large current, which flow during the starting of an induction motor. Normally, the starting current of an Induction Motor is 6 to 8 times of full load current.
The time from starting point to maximum or peak torque point after which motor is engaging with load is called as starting time.
Motor starting time is the period from when the electrical supply is connected to the motor to when the motor accelerates to full speed. The length of the starting period is dependent on the combination of the motor and mechanical load, and it can be anything from a fraction of a second to 30 seconds or longer.
High levels of current are required during the start-up period, and they can have detrimental effects on the electrical supply system and other equipment connected to it. The duration of starting transients depends on the load characteristics and how long it takes the motor to run up to speed.
The figure below illustrates what happens during motor starting. During the starting period a current significantly larger than the motor’s normal full load running current is drawn, the magnetic fields within the motor and back EMF increase, and the mechanical load accelerates. The start-up current can be as high as five to eight times the full load current.
Large Motor Over current protection is normally set to trip prior or to Locked Rotor Withstand Time (LRWT) provided by Motor Manufacturer, after the calculation motor starting time is determined by the motor designer based on the heating the rotor part during Locked Rotor Condition, where motor continuously requires large value of in rush current.
At the time of Starting Induction motor drives high value of current (Motor is a constant impedance device during the starting condition) current that are very closed to the motor’s Locked Rotor value & remain at this value for the time required to start the motor.
The Capability to calculate Motor starting Time for large induction motor is important in order to evaluate the relative strength of the power system.
Parameters given as follows:
Moment of inertia of motor and combined Load System (J) = 10 Kg-m2
Motor Torque (Tm) = 15 + 0.5 Wm
Load Torque (Tl) = 5 + 0.6 Wm
A Motor generally drives the load or machine with some transmission system. While the motor always rotates, a load may rotates or may undergo Translation motion.
Motor Torque or Load Torque relation is defined as,
For drives with constant inertia (dJ / dt) = 0
Therefore,
Where , is the Synchronous speed of the motor
Tm = Tl
15 + 0.5 Wm = 5 + 0.6 Wm
15 - 5 = 0.6 Wm - 0.5 Wm
10 = 0.1Wm
Wm = 100 rad/sec ………………………Steady State Speed
Therefore,
15 + 0.5Wm = 5 + 0.6. Wm(dWmdt)
(dWmdt)=(10-0.1Wm)10
1–0.01Wm
dt=(1(1-0.01Wm))dWm
Integrating the above equation w.r.t Time (t)
The Integral limit should be considered from initial condition of motor to the end of transit state after which the motor attains its steady speed. So we consider Steady speed. So we consider Wm1 = 0 & Wm2= 95% of Wm
Wm = 100 rad/sec.
Wm1 = 0 rad/sec.
Wm2 = 95 rad/sec.
Therefore,
∫950dt=∫950[11-0.01ωm].dWm
t=ln[1-0.01ωm]lim(950)+C
At initial condition, when t = 0, Wm = 0 , C = 0
Applying integral limits,
t=ln[1-0.01(95)]-ln[1-0.01(0)]
t=ln[1-0.95]-ln[1]
t=-2.99-0
t=+2.99
Wm = 100 rad/sec ………………………Steady State Speed
(2.99*100)=299 sec
Therefore, starting time for given Induction motor is t = 3s
MATLAB SCRIPT:
LINK:https://drive.google.com/file/d/1_nCuMO6Z0XqRygV51lDTcLAbPYb4tZp_/view?usp=sharing
clear all
close all
clc
% Claculating the Equilibrium points
W = linspace(0,160);
Tm = 15 +(0.5*(W));
Tl = 5 +(0.6*(W));
%Plotting the Omega and Motor Torque that represented as Blue colour
plot(W,Tm,'b')
title('Determining Equilibrium points')
legend('Motor Torque','Load Torque')
xlabel('Omega (rad/sec)')
ylabel('Torque (N-m)')
hold on
% Plotting the Omega and load torque that represent as Red colour
plot(W,Tl,'r')
grid on
RESULTS:
CODE EXPLANED:
The Output value is W = 100.2 rad/sec as we derived by the equation.
MATLAB SCRIPT:
LINK:https://drive.google.com/file/d/17N_7WhQzyrSsWJPHbqj1rbIIgf29Yr-7/view?usp=sharing
clear all
close all
clc
T_p = [0 299];
ic = [0 0];
% Using ode45 solver we obtain W for each time span and initial condition
[t,W] = ode45(@omega,T_p,ic);
plot(t,W)
title('W-dot (dw/dt)')
xlabel('time (sec)')
ylabel('Omega (rad/sec)')
grid on
% function program
% here we consider W_dot = 1 - 0.01*(W) and Equate it in main program for
% differentiation as solver ode45
function [W_dot] = omega(t,W)
W_dot = 1 - 0.01*(W);
end
RESULT:
CODE EXPLANED:
STUDY_3:
To Calculate the stability and equilibrium points of the drive with the following load torque and motor torque equation:
Given data:
Tm=(1+2ωm)
Tl= 3 √ωm
To FIND :
The stability and equilibrium points of the drive
Solution:
General conditions:
condition for stability:
dTldωm-dTmdωm>0
If dωm>0,Tl>Tm
dωm<0,Tl<Tm
Where,
`T_m` = Torque of Motor.
Tl = Load Torque.
`ω_m` = Speed of motor in RMP
AT STEADY –STATE STABILITY
Tm=Tl
Equation both the value,
1+2ωm = 3 √ωm
By solving this equation we obtain two equilibrium points ωm = 0.25 and 1
MATLAB SCRIPT:
LINK: https://drive.google.com/file/d/14jLyeGm7el7HvBCYA4ZTUwOVaSaDql2s/view?usp=sharing
clear all
close all
clc
% Here Tm and Tl value are taken from study-3 Given equations.
W = linspace(0,2);
Tm = 1+(2*W);
Tl = 3 *sqrt(W);
% Plotting the Omega and motor torque that represented as red colour
plot(W,Tm,'r')
title ('Determining equilibrium points')
xlabel('Omega(rad/sec)')
ylabel('Torque(N-m)')
hold on
% plotting the omega and load torque that represented as blue colour
plot(W,Tl,'b')
legend('motor torque', 'load torque')
grid on
RESULT:
CODE EXPLANATION:
Tl = (3)/(2).
Tl = 3
Hence Tl= 3 and Tm = 2 are obtained in motor speed of ωm = 0.25, according to the conditions of stability the value of load torque is greater than value of the motor torque that system is considered as a stable condition.
Whereas by applying the speed of ωm = 0.25 in the differential equation of load torque and motor torque. So the value = 0.25is the stability of a drive.
Tl = (3)/(2).
Tl = 1.5
Hence Tl = 1.5 and Tm= 2 are obtained in the motor speed of ωm = 1, according to condition of stability the value of load torque is lesser than value of the motor torque, such system is considered as Unstable Condition.
So the value ωm = 1 is not stability point of the drive.
STUDY_4:
4. Make a MATLAB script file which will plot speed-torque characteristics for frequency control method.
INTRODUCTION:
The three-phase induction motor operates in some ways like a transformer. In the transformer, AC voltage is applied to the primary, which creates AC flux in the core. That flux links the secondary and induces a voltage of the same frequency, but with a voltage that depends on the transformer turns ratio.
The three-phase induction motor has voltage applied to the stator that creates a rotating flux wave. As that wave sweeps by the rotor bars, voltages are induced; however, the frequency of the voltage is determined by the slip of the motor. It turns out that the similarity extends to the equivalent circuits of the two devices.
In developing the equivalent circuit for the induction motor, we can recall the equivalent circuit of the transformer. The primary circuit contained inductances to account for the leakage and mutual fluxes and resistances to account for the resistance of the primary winding and the core losses. The stator of the induction motor is essentially the same; there are mutual and leakage fluxes, winding resistance, and core losses due to hysteresis and eddy currents.
Figure 2 shows the equivalent circuit of one phase of the stator of the induction motor. It is assumed the windings are connected in wye, so the voltage applied to the circuit is a line-to-neutral voltage. The elements Rs and Xs are the stator winding resistance and leakage reactance, and Xm is the magnetizing reactance. This circuit is essentially the same as the primary circuit of a transformer. The only difference is we have not included a core loss resistance. The core losses are often accounted for separately and are thus not represented in the equivalent circuit.
Looking at the stator circuit of Figure 2, I1 is the current entering the winding. As already discussed, a significant current Im, is required to establish the magnetic field. The remaining current, I2, is the load portion of the stator current. The MMF of I2 will exactly cancel the MMF of the rotor current. In phasor notation, we can write,
V1 = E1 + I1(Rs +jXs)(1)"> V1 = E1 + I1(Rs +jXs)(1)V1 = E1 + I1(Rs +jXs) ....................Eqn(1)
Where E1 is the EMF induced in the stator coil by the mutual flux. We need to add the rotor to the equivalent circuit.
Looking at figure 3, as the stator flux sweeps by the rotor conductor, a voltage and current will be induced. If the rotor is not allowed to turn (blocked rotor), then the voltage and current induced in the rotor will have the same frequency as the stator.
We have an unusual form of a transformer, in which the flux rotates around the rotor conductors. In the case of the transformer, we refer quantities from one side to the other using the turns ratio. Because most induction motors have squirrel cage rotors, it’s not easy to determine the number of turns on the rotor. Fortunately, we can avoid the problem by always working with quantities referred to the stator.
Figure 4 shows what’s happening when the rotor is blocked. As the stator field sweeps by the rotor conductors, a blocked rotor voltage, EBR (equal to E1), is induced. Since the coils are shorted, current flows through the resistance and leakage reactance of the rotor coils.
In figure 4, Rr is the resistance of one phase of the rotor winding and Xr is the leakage reactance of the rotor when stator frequency currents flow in the rotor, which only happens when the rotor is stationary (slip=1.0). Of course, both Rr and Xr are referred to the stator by an appropriate turns ratio.
For the induction motor to be of use to us, it must turn, which means the slip is less than 1.0. If the rotor is moving, two things happen:
Replacing E1 by sE1, and Xr by sXr in Figure 4 yields the circuit shown in Figure 5, which is valid at any value of slip. In order to connect the rotor circuit of Figure 5 to the stator circuit of Figure 2, we must account for the different frequencies.
Just as we referred impedances by the turns ratio, we can refer them by the frequency. From the circuit of Figure 5, we can write
sE1 + I2(Rr + jsXr)(2)"> sE1 + I2(Rr + jsXr)(2)sE1 + I2(Rr + jsXr) .....................................(2)
Dividing equation 7-8 by s yields
E1+I2(Rrs+jXr)(3)"> E1+I2(Rrs+jXr)(3)E1+I2(Rrs+jXr) .........................(3)
Equation 3 can be represented by the circuit of Figure 6, which is the rotor equivalent circuit referred to the stator both by turns ratio and by frequency.
This circuit can be connected to the stator equivalent circuit, but it is instructive to split the resistance into two separate components. For convenience, we can write:
Rrs=Rr+Rrs−Rr(4)"> Rrs=Rr+Rrs−Rr(4)Rrs=Rr+Rrs−Rr ...................(4)
Combining the last two terms on the right side of equation 4 yields
Rrs=Rr+Rr(1−ss)(5)"> Rrs=Rr+Rr(1−ss)(5)Rrs=Rr+Rr(1−ss) ...................(5)
Replacing the resistive element in Figure 6 by the two resistive elements on the right-hand side of equation 5 yields the rotor equivalent circuit shown in Figure 7. The reasons for this manipulation will be discussed shortly.
Finally, by combining the rotor equivalent circuit of Figure 7 with the stator equivalent circuit of Figure 2, we obtain the steady-state equivalent circuit for one phase of a wye-connected induction motor, as shown in Figure 8.
Locking again at the rotor part of the equivalent circuit in Figure 8, the resistor Rr represents the resistance of the rotor winding. The power used by it is the power lost in the resistive heating of the rotor winding. The additional resistive element on the right end is a function of the slip and the rotor resistance. It arises from the necessity to transform the rotor circuit not only by turns ratio but also by frequency. The power consumed in this element is the developed power of the machine.
Developed power is the power converted from electrical form to mechanical form and includes the load power plus mechanical losses such as friction and windage.
Subtracting the mechanical losses from the developed power would yield the shaft power, which is the actual power delivered to the load. Developed torque and shaft torque can be calculated from the developed and shaft power, respectively.
Locking again at the rotor part of the equivalent circuit in Figure 8, the resistor Rr represents the resistance of the rotor winding. The power used by it is the power lost in the resistive heating of the rotor winding. The additional resistive element on the right end is a function of the slip and the rotor resistance. It arises from the necessity to transform the rotor circuit not only by turns ratio but also by frequency. The power consumed in this element is the developed power of the machine.
Developed power is the power converted from electrical form to mechanical form and includes the load power plus mechanical losses such as friction and windage.
Subtracting the mechanical losses from the developed power would yield the shaft power, which is the actual power delivered to the load. Developed torque and shaft torque can be calculated from the developed and shaft power, respectively.
MATLAB CODE:
LINK:https://drive.google.com/file/d/1zTIA6T6-qHFOI_bMY_45aHP315CiLscs/view?usp=sharing
clear all
close all
clc
% Inputs
% Voltage per Phase
V =230/sqrt(3);
% Number of Phase
Ph = 3;
% Number of Poles
P = 4;
% Stator Resistance and Inductance
R1 = 0.095;
X1 = 0.680;
% Rotor resistance and inductance
R2 = 0.23;
X2 = 0.672;
% Magnetizing Inductance
Xm = 18.7;
% Frequency Range from 50Hz To 150Hz in ther interval of 20 Hz
Fz = [50:20:150];
% For continously varying frequency 1 unit
i = length(Fz)
% Loop
for x = 1:i
% Motor Speed calculation
Ws = (4*pi*Fz(x))/P; % FOR ANGULAR SPEED
Ns = (120*Fz(x))/P; % FOR SYNCHRONOUS SPEED
for n = 1:200 % FOR VARYING SLIP
s = n/200; % SLIP CALCULATION
N(n) = Ns*(1-s) % FOR ROTOR SPEED OR ACTUAL SPEED CALCULATION
Zeq = (R1+R2*(1-s)/s)+(X1+X2); % FOR EQUIVALENT IMPEDENCE CALCULATION
T(n) = 3*((V^2)*(R2/s))/(Ws*((R1+(R2/s)^2))+((X1+X2)^2)) % FOR TORQUE
end
% To Display Maximum Speed of Motor at different Frequencies
Curve = ['The Maximum Speed of Motor at ', num2str(Fz(x)), ' Hz is '];
disp(Curve)
disp(N(1))
% PLOTTING FOR TORQUE AND SPEED CURVE
plot(N,T);
hold on
grid on
title('Speed Torque Charateristics for Frequency Controlled Induction Motor');
xlabel('Speed of Motor in (RMP)');
ylabel('Torque of Motor in (N-m)');
end
CODE EXPLANED:
RESULT's:
CONCLUSION:
REFERENCE:
UPLOADED SOURCE GDRIVE LINK:https://drive.google.com/drive/folders/17zyPMMTcBN4pWqzu5s7ux8uPV5R4IKee?usp=sharing
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