Induction Motor Torque - Speed Characteristics and Stead State Stability of a Drive

AIM :

1. To explain about how Induction motor Operation is similar to clutch.

2. Calculate the starting time of a drive with the following parameters:

3. Obtain the equillibrium points and determine the stability for a drive.

EXPLANATION :

STUDY -1 :

Induction Motor Operation :

Induction motor has two main parts, Stator and Rotor.Stator is basically 3 coil winding and 3-phase AC power input is given to it. The Winding passes through slots of the stator, which are made by thin and highly permiable steel laminations inside a steel or cast-iron frame. When the 3-phase current passes through windings and it produces a Rotating magentic field. This Rotating Magnetic field is causes the rotor to turn. 3 coils are connected 120 degree apart. A wire carrying current produces magnetic field arounded. Rotational Speed of the magnetic field is known as Synchronous Speed(Ns).

Fig : Key Components of AC Induction Motor

Assume Conductor is placed in centre of 3 windings. Magnetic field will vary, emf will be induced in the loop. An EMF will produce current in the loop. Electro Magnetic force is produced on the loop and the loop was start to rotate. The same phenomenon occurs in the Induction motor as well. Instead of loop squirrel cage rotor is used in Induction motor.

3-phase AC power input is pasing through the stator produces magnetic field and current will be induced in the bars of the squirrel cage. So the rotor starts to rotate. Thats why it is Induction motor, Electricity is induced on the rotor with help of Electro magnetic induction rather direct connection.

Fig : Rotating Magnetic field with Squirrel cage rotor

Both Magnetic field and Rotor are rotating. When the rotor and Magnetic field both are rotating in a same speed, magentic field will never cut the loop. Thus, there will not be any induced EMF and Current. This translates zero force of the rotor part and the rotor will gradually slow down. As it slows down, magnetic field will cut the rotor loop, so induced EMF and current rise again and the rotor will then speed up. Rotor will never be catch up the speed of the magnetic field. Rotor will rotates in a specific speed which is slightly less than the synchronous speed(Ns). Difference between the Synchronous Speed(Ns) and Rotor Speed(Nr) is known as Slip.

Slip = (Ns-Nr)/Ns

Ns - Synchronous Speed,

Nr - Rotor or Motor Speed,

Sometimes, Ns-Nr is called Slip speed

Obviously Rotor or Motor speed is

Nr = Ns(1-s).....eqn(a)

It may be kept in mind revolving flux is rotating synchronously, relative to the stator(i.e., stationary space) but at slip speed relative to the motor.

Induction Motor do not require permanent magnets, Brushes, Commutator rings, Position sensors like other machine mechanical parts.Induction motor are also Self Started. The main advantage is Induction motor Speed can control easily by controlling input power of Frequency.

Clutch Operation :

A Clutch is a mechanical device which engage and disengage power transmission especially from driving shaft to driven shaft.In simplest application Clutch connect and disconnect two rotating shafts.

A Electromagnetic clutch is a clutch (a mechanism for transmitting rotation) that is engaged and disengaged by an electromagnetic actuator. In this type of clutch flwheel consists of winding. The current supplied to the winding from battery or dynamo.

AC Induction motor is coupled to an eddy current clutch. Eddy Current Clutch, also known generally as Magnetic drive, eddy current clutch or magnetic clutch, is a electro mechanical variable speed drive that uses a constatnt speed motor as the input. The eddy current clutch is installed between the motor and driven equipment.

Fig : Eddy Current clutch used in AC Induction Motor

Eddy current clutches are driven at a constant speed by an AC induction motor. The AC motor interacts with an electromagnetic clutch that controls both the torque and speed that is transferred to the driven load. The clutch itself is installed between the constant speed AC motor and the equipment being driven. The clutch controller governs the amount of power that the clutch receives (which is separate from the AC motor power). Included as part of the magnetic rotor is a direct-current (DC) excitation coil that is regulated by the clutch controller. The magnetic rotor combined with the DC excitation coil serves as the electromagnetic coupling that varies the output speed and torque of the clutch’s output rotor and shaft.

Fig : Eddy Current Clutch Working Operation

A rotating drum (sometimes called a ring) is coupled directly to the Drive shaft (shaft of the motor) and rotates at a constant speed. A Rotor (contains Electromagnetic coils) coupled directly to the Driven Shaft. The Speed of the clutch can be controlled by varying the amount of direct current applied to the rotor. Since the Output Speed(ω2) is determine by the amount of slip between the rotor and drum.

Slip = (ω1-ω2)/ω1

ω1 - Drive Shaft Speed(Input Speed)

ω2 - Driven Shaft Speed (Output Speed)

Therefore, Output speed is

ω2 = ω1(1-S) .... eqn(b)

From Eqn (a) and Eqn (b), we know that Clutch and Induction motor has similarities in both operation.

Result :

When the direct current is applied to the rotor, the spinning electromagnetic induce eddy currents into the metal drum. The induce eddy currents form magnetic poles inside the drum.The magnetic fields of the rotor and drum attracted to eachother, and clutch turns in the same direction of the motor.The main advanatge of eddy current clutch is that no mechanical connection between the rotor and drum. Since there is no mechanical connection, there is no friction to produce excessive heat and there is no wear as is the case with mechanical clutches.

Fig : Torque-Speed Characteristics of Induction Motor

The Speed of the clutch can be controlled by varying the amount of direct current applied to the rotor. Since the Output Speed is determined by the amount of slip between the rotor and drum.When load is added, the slip will become greater and causing a decrease in speed. This can be compensated by increasing the amount of direct current applied to the motor. In Stabe region the clutch is fully enganged to transfer 100% efficient as there is no related slip and in Unstable region Clutch is disengaged.

STUDY - 2 :

Given :

Moment of Inertia, J = 10 kg-m^2,

Motor Torque, Tm = 15+0.5*ωm

Load Torque, Tl = 5+0.6ωm

To find :

To Calculate Starting Time of drive with given parameters

Solution :

By using Motor Torque and Load Torque equation, Constant Speed (equilibrium points) will be calculated.

For finding equilibrium point or Constant Speed(ωm),

Motor Torque(Tm) = Load Torque(Tl)

15+(0.5*ωm) = 5+(0.6*ωm)

10=0.1*ωm

ωm = 100 rad/s.

Using Another Method for finding Equilibrium Point or Constant Speed (ωm) is Verified by using MATLAB Program

clear all
close all
clc

w = linspace(0,200)
Tm = 15+(0.5*w);
Tl = 5+(0.6*w); 
plot(w,Tm)
hold on
plot(w,Tl)
xlabel("Speed")
ylabel("Torque")
title("Calculating Equilibrium point")
legend('Motor Torque','Load Torque')
grid on

Explanation for above MATLAB Progaram,

Step-1 : Giving Input of Speed(w), Motor Torque Equation(Tm) and Load Torque Equation(Tl)

Step-2 : For finding Constant Speed(ωm) or Equilibrium point, Use "plot" command to plot Speed(w) with Motor Torque(Tm) and Speed(w) with Load Torque(Tl)

Step-3 : Use "hold on" command between two "plot" commands to plot the resuts in Single plot window.

Step-4 : Use "xlabel","ylabel","title","legend" and "grid on" Command for Proper Identification of output results.

Step-5 : Run the Program to get the Output Plot Results.

From the about Plot result, Verified that Equilibrium Point or Constant Speed (ωm) is 100 rad/s.

After finding Constant Speed(ωm), to find the Starting time to reach constant angular speed. For finding Starting Time of drive Solve the Steady State Stability Equation which is related with Newton second law(F = m*a),

Fig (A) : Torque-Speed Characteristic Stability Analysis

Here, Force(F) is reffered as Torque(T) which is Motor Torque(Tm) and Load Torque(Tl), In Linear motion consider Mass(m) and here rotational motion so considered Moment of Inertia (J) and linear Acceleration(a) is dv/dt and in angular acceleration is dw/dt.

Steady State Stability equation of induction motor is

Tm - Tl = J * (dw/dt)...............eqn(*)

Substitute the Given input parameters in above eqn(*),

(15+(0.5*ωm))-(5+(0.6*ωm)) = 10 * (dw/dt)

15+0.5ωm-5-0.6ωm = 10*(dw/dt)

10-0.1ωm = 10*(dw/dt)

dw/dt = (10-(0.1*ωm)) / 10

dw/dt = 1-(0.01*ωm)................eqn(a)

Rearranging the above equation in terms of change in time(dt) with respect to change in speed(dw) for finding the Starting time of drive,

dt/dw = 1/(1-(0.01*ωm))

By reffering fig(A) Torque-Speed Characteristic Stability Analysis, when motor reaching the steady state speed slip will occur. Here Assuming 5% of slip is occuring at Constat Speed(ωm)

Therefore, Integrating the above differential equation with 0 to 95 rad/s boundary limits, we get the starting time of drive,

∫(dt/dw) = ∫( 1/(1-(0.01*ωm)))

t = C-100*ln(ωm-100)

Applying 0 to 95 boundary limits,

t = (-100*ln(95-100)) - (-100*ln(0-100))

t = (100*ln(5)) - (100*ln(100))

t = (160.94)-(460.52)

t = 299.58 sec.

This starting time is also verified by using "ode45" solver in MATLAB Program to solve the differential equation of eqn(a),

Main Program :

clear all
close all
clc

t_span = [0,500];
w_i = [0,0];
[t,w] = ode45(@startingtime,t_span,w_i);
plot(t,w)
xlabel("time(sec)")
ylabel("Speed(rad/s)")
grid on

Function Program :

function [dw_dt] = startingtime(t,w)
dw_dt = 1-(0.01*w);
end

Explanation for above MATLAB Program,

Step-1 : Giving inputs of t_span(time range),w_i(initial speed).

Step-2 : use "ode45" solver to solve the differential equation.

Syntax : [t,y] = ode45(odefun,tspan,y0),

where tspan = [t0 tf], integrates the system of differential equations y′=f(t,y) from t0 to tf with initial conditions y0. Each row in the solution array y corresponds to a value returned in column vector t.

Step-3 : By referring above "ode45" syntax, calling a function by giving a name as startingtime to integrates the system of differential equations

Step-4 : After Solving the differential equation using "ode45" solver, Use "plot" command to plot the output results of (t,w).

Step-5 : Use "xlabel","ylabel" and "grid on" command for proper identification of output results.

Step-6 : Run the Program and get the Output results.

From the above plot result, know that Steady state is achieved in time range of 299-300 secs.

Result :

For Given Input Parameters of Motor Torque(Tm),Load Torque(Tl), Moment of Inertia(J) ,Starting Time of drive is obtained at 300 sec.

STUDY - 3 :

Given :

A drive has following equations for motor and load torque,

Tm = (1+2ωm)

Tl = 3√ωm

To Find :

Obtain the equlibrium points and determine the stability.

Solution :

1. Finding Equilibrium Point :

For Obtaining Equilibrium point, Refer STUDY-2 and same steps to be follwed

Motor Torque(Tm) = Load Torque(Tl)

MATLAB Program for Obtaining Equilibrium Point :

clear all
close all
clc

w = linspace(0,2);
Tm = 1+(2*w);
Tl = 3*sqrt(w); 
plot(w,Tm)
hold on
plot(w,Tl)
xlabel("Speed(rad/s)")
ylabel("Torque(N-m)")
title("Calculating Equilibrium point")
legend('Motor Torque','Load Torque')
grid on

Explanation for above MATLAB Progaram,

Step-1 : Giving Input of Speed(w), Motor Torque Equation(Tm) and Load Torque Equation(Tl)

Step-2 : For finding Constant Speed(ωm) or Equilibrium point, Use "plot" command to plot Speed(w) with Motor Torque(Tm) and Speed(w) with Load Torque(Tl)

Step-3 : Use "hold on" command between two "plot" commands to plot the resuts in Single plot window.

Step-4 : Use "xlabel","ylabel","title","legend" and "grid on" Command for Proper Identification of output results.

Step-5 : Run the Program to get the Output Plot Results.

From the above plot we obtain two equilibrium points, ωm = 0.24 & 1.01 rad/s.

Determining the Stability :

Motor is going to drive the load sufficiently is known as Stability. That is motor has enough power to drive the load. If motor does not have enough power or load is running in other region of Characteristic is known as Unstable operation and the motor is never going to satisfy the load. For Determining the Stability, it should satisfy the condition for Stability.

Fig : Torque-Speed Characteristic Stability Analysis of a drive

Stability theory tells that, if change in Load torque(dTl) vs Change in Positive Speed(dωm) and it difference with change in motor torque(dTm) vs same Change in Positive Speed(dωm) should be greater than zero for motor drive system to become stable.

Condition for Stability,

(dTl/dωm) - (dTm/dωm) > 0.............eqn(a)

Condition - (i)

if dωm>0, Tl > Tm

while Speed is increasing, Load torque(Tl) will be greater than Motor Torque(Tm)

Condition - (ii)

if dωm<0, Tl < Tm

while Speed is decreasing, Load torque(Tl) will be lesser than Motor Torque(Tm)

Substituting the Given Load Torque(Tl) and Motor Torque(Tm) equation in eqn(a) condition for Stability.

(dTl/dωm) - (dTm/dωm) > 0

[d(3√ωm)/dωm] - [d(1+2ωm)/dωm] > 0

By differentiating the above equation, we get

[3/(2√ωm)] - [0+2] > 0

[3/(2√ωm)] - 2 > 0 .......... eqn(b)

Case(i) - substitute ωm = 0.24 rad/s in above eqn (b)

[3/(2√0.24)] -2 > 0

[3/0.9797] - 2 > 0

3.062 - 2 > 0

1.062 > 0

It satisfied the above condition.

Case(ii) - substitute ωm = 1.01 rad/s in above eqn (b)

[3/(2√1.01)] -2 > 0

[3/2.01] - 2 > 0

1.5 - 2 > 0

-0.5 > 0

It not satisfied the above condition.

Result :

Case(i) satisfies the Condition for Stability, So ωm = 0.24 rad/s is the stability of drive(motor has going to drive the load sufficiently)