NO LOAD TEST:1. No load power factor (CosФ0) = W0/ V0I0Where W0= No load power in watts V0= No load voltage in volts. I0= No load current in amps2.Working component current (Iw) = I0CosФ0amps3.Magnetizing component current (Im) = I0SinФ0amps4. No load resistance R0= V0/ Iwohm5. No load reactance X0= V0/ ImohmBLOCKED ROTOR TEST:6.Motor equivalent impedance referred to stator Zsc = (Vsc / Isc) ohm7.Motor equivalent resistance referred to stator Rsc = ZscCosФsc ohm= Wsc / Isc2ohm8.Power factor CosФsc = Wsc / VscIsc9.Motor equivalent reactance referred to stator Xsc = √Zsc 2 – Rsc2ohm10.Rotor resistance referred to stator R 2’= Rsc – R 1ohm11.Rotor reactance referred to stator X2’= Xsc / 2 = X1ohmWhere R1= Rac = 1.6 *RdcR 1= stator resistance X1= stator reactance12.Magnetizing reactance Xm = 2(Xo –X1– X2’/2)13.Slip = (Ns-N)/NsNs = synchronous speed in rpm N = speed of the motor in rpmTHEORY:Theequivalent circuit of a single phase induction motor can be developed by usingdouble fieldrevolving theory. By using the equivalent circuit the performance of the single phase inductionmotor can be obtained.The single phase induction motor can be visualized to be made of single stator winding and twoimaginary rotors. The developing torques of the induction motor is forwardtorque and backwardtorque.When the single phase induction motor is running in the direction of forwardrevolving field at aslip S, then the rotor currents induced by the forward field has frequencysf. The rotor mmfrotates at slip speed with respect to the rotor but at synchronous speed withrespect to the stator.The resultant forward stator flux and the rotor flux produce a forward air gap flux. This fluxinduces the voltage in rotor. Thus due to the forward flux, the rotor circuitreferred to stator hasan impedance of R2’ /2s + jX2’/2.The backward flux induces a current in the rotor at a frequency (2-s)f. thecorresponding rotormmf rotates in the air gap at synchronous speed in the backwarddirection. The resultantbackward stator flux and the rotor flux produce a backward airgapflux. This flux induces thevoltage in rotor. Thus due to backward flux the rotor circuitreferred to stator has an impedance ofR2’/2(2-s )+ jX2’/2
An induction motor is simply an electric transformer whose magnetic circuit is separated by anair gap into two relatively movable portions, one carrying the primary and the other thesecondary winding. Alternating current supplied to the primary winding induces an opposingcurrent in the secondary winding, when later is short circuited or closed through an externalimpedance. Relative motion between the primary and secondary ie, stator and rotor is producedby the electromagnetic forces corresponding to the power thus transferred across the air gap byinduction.Equivalent load resistance (RL = R 2’ (1/S -1) in ohm 1 1V 0 R L R2 1 1 s sSwinburne’s test theory The field coil and the armature windings are connected in shunt or parallel across the powersource. The armature winding consists of relatively few turns of heavy gauge wire. The voltageacross two windings is the same but the armature draws considerably more current than the fieldcoil. Torque is caused by the interaction of the current carrying armature winding with themagnetic field produced by the field coil. If the DC line voltage is constant, the armature voltageand the field strength will be constant. The speed regulation is quite good; the speed is a functionof armature current and is not precisely constant. As the armature rotates within the magneticfield, an EMF is induced in its wining. This EMF is in the direction opposite to the source EMFand is called the counter EMF (CEMF), which varies with rotational speed. Finally, the currentflow through the armature winding is a result of the difference between source EMF and CEMF.When the load increases, the motor tends to slow down and less CEMF is induced, which in turnincreases the armature current providing more torque for the increased load.Motor speed is increased by inserting resistance into the field coil circuit, which weakens themagnetic field. Therefore, the speed can be increased from “basic” or full-load, full-field value tosome maximum speed set by the electrical and mechanical limitations of the motor. The powerdifference between the motor input and the output is dissipated in form of heat and constitutes tothe losses of the machine. These losses increase with load, since the motor heats up as it deliversmechanical power.
D.C. Shunt MotorH.P Voltage Current Speed Field current Excitation current Winding type D.C. Shunt MotorH.P / KW Voltage Current Speed Field current Excitation current Winding typeratingSumpners testThe O.C and S.C. tests give us the equivalent circuit parameters but cannot give heatinginformation under various load conditions. The Sumpner’s test gives heating information also. InO.C. test, there is no load on the transformer while in S.C. test also only fractional load getsapplied. In all in O.C. and S.C. tests, the loading conditions are absent. Hence the results areinaccurate. In sumpner’s test, actual loading conditions are simulated hence the results obtainedare much more accurate. Thus Sumpner’s test is much improved method of predeterminingregulation and efficiency thanScott testWith the help of Scott connection it is possible to obtain 2 – phase supply which is required forfurnaces or even three phase load can be driven from the available 2 – phase supply source.No load and Blocked rotor test theoyAn induction motor is simply an electric transformer whose magnetic circuit is separated by anair gap into two relatively movable portions, one carrying the primary and the other thesecondary winding. Alternating current supplied to the primary winding induces an opposingcurrent in the secondary winding, when later is short circuited or closed through an externalimpedance. Relative motion between the primary and secondary ie, stator and rotor is producedby the electromagnetic forces corresponding to the power thus transferred across the air gap byinduction
FORMULAE:Torque T = (S1 S2) x R x 9.81 NmInput Power Pi = VI Watts 2 NTOutput Power Pm= ------------ Watts 60 Output PowerEfficiency % = -------------------- x 100% 1. Ammeter, Voltmeter readings, speed and spring balance readings are noted under no load condition. 2. The load is then added to the motor gradually and for each load, voltmeter, ammeter, spring balance readings and speed of the motor are noted. 3. The motor is then brought to no load condition and field rheostat to minimum position, then DPST switch is opened.PRECAUTIONS:DC shunt motor should be started and stopped under no load condition.Field rheostat should be kept in the minimum position.Brake drum should be cooled with water when it is under load.