Induction-motor2 plot Speed-Torque characteristics for frequency control method

(1) T∝φI2*cosθ2

where T=torque produced by induction motor

φ=flux for producing induced emf

I2=induced rotor current

cosθ2=power factor of rotor circuit

φ∝E1(stator emf)

K(transformation ratio)=E2/E1

E2=φ

I2=sE2/Z2 (where Z2 on rotor side)

now Z2=sqrt(R2^2+(sX2)^2)

I2=sE2/sqrt(R2^2+(sX2)^2)

cosθ2=R2/sqrt(R2^2+(sX2)^2)

T∝E2*sE2/sqrt(R2^2+(sX2)^2)*R2/sqrt(R2^2+(sX2)^2)

T=K*sE2^2*(R2/sqrt(R2^2+(sX2)^2)

K=3/(2*pi*ns)

ns=sychronous speed (rps)

so from this further final torque value comes out to be stated below:

this is the torque value for 3 phase induction motor

Induction Motor circuit equivalent diagram stated below:

Theory:The current that flows in the rotor bars will create a flux around them.This flux interacts with the stator flux to generate the necessary torque in the rotor bar.Without the generator of motor would not rotates.The energy stored in the rotor flux is represented in the equivalent circuit in the form of reactance. Current flows to the stator resistance is less than the rotor resistance .Slip occurs when the force on the shaft is more.

I1,Is=stator current

I2=rotor current

s=slip

X1=stator leakage reactance

X2=rotor circuit reactance

Xm=magnteising reactance

(2) Given: 3 phase ,50Hz,X1&X2=0.1,R1&R2=0.2ohm with half rated voltage and frequency.

Solution and formula used

V1=Vpeak/sqrt(2);(where V1=Vrms)

V2=V1/2;

ns2=ns1/2; (where ω=2*pi*N) & (ns=N/60) (ns in rps and N in rpm)

T∝(V)^2/ns; rest of the things are kept constant because they will get rid off because of having same value .

T1/T2=(V2/V1)^2*(ns1/ns2);

T1/T2=0.5

Therefore the starting torque ratio of half voltage and frequency comes out to be 0.5

(3)

clear all
close all
clc
V=230/sqrt(3)
no.of_phase=3;
poles=4;
R1=0.095;
R2=0.23;
X1=0.680;
X2=0.672;
Xm=18.7;
Zequiv=1i*Xm*(R1+1i*X1)/(R1+1i*(X1+Xm));
Req=real(Zequiv);
Ieq=imag(Zequiv);
Veq=abs(V*1i*Xm/(R1+1i*(X1+Xm)))
Vf=linspace(10,50,4)    %(Vf=variable frequency)
for i=1:length(Vf)
    omega_ms=120*Vf(i)/poles;
    omega=4*pi*Vf(i)/poles;
    for t=1:200
        s(t)=t/200
        rpm(t)=omega_ms*(1-s(t));
        PR2=abs(Veq/(Zequiv+1i*X2+R2/s(t)));
         T(t)=no.of_phase*PR2^2*R2/(s(t)*omega);
         
    end
    
    

     plot(rpm,T,\'linewidth\',2)
    hold on
end
    axis([0 1500,0 600])
    xlabel(\'motor speed(rpm)\')
    ylabel(\'torque(N-M)\')
     grid on
     legend(\'10HZ\',\'20HZ\',\'30HZ\',\'40HZ\')
    

 

GRAPH:

(4)(i) BLDC motor works on the principle similar to that of conventional DC motor i.e lorentz force law which states that whenever a current carrying conductor placed in a magnetic field it experiences the force.As a consequence of reaction force,the magnet will experience an equal and opposite force .In case of BLDC motor,the current carrying conductor is stationary while the permanent magnet moves .

(ii)When the stator coils are electrically switched by a supply source ,it becomes the electromagnet and starts producing the uniform field in the air gap.Though the source of supply is DC,switching makes to generate an AC voltage waveform with trapezoidal shape .Due to force of interaction between electromagnet stator and permanent magnet rotor ,the rotor continues to rotates.

(iii) IT has no mechanical commutator and associated problems ,high efficiency due to the use of permanent magnet rotor ,high speed of operation even in loaded and unloaded conditions due to the absence of brushes that limits the speed along with the smaller motor geomatry and lighter in weight than brushed type DC and induction AC and lastly less electromagnetic intereference.And quiet operation due to absence of brushes.