Utilisation of Electrical Energy Introduction

Last time … Assignment – Transformers contd. Open & short circuit test recap. Efficiency calculations. Measured efficiency. Maximum efficiency. Assignment working time. 10/01/2019 UEE

Introduction - objectives Induction motor introduction. Poles, slip & rpm. Wiring options. Equivalent circuit of induction motor. No-load power factor and branch current. Equivalent impedance, Zeq. Assignment 4 (induction motors) working time. 10/01/2019 UEE

Diagram of induction motor 10/01/2019 UEE

Induction motors Three-phase squirrel-cage induction motors. Rugged, reliable, no electrical connection to rotor. Windings on stator induce a rotating magnetic field in the rotor. Causes the rotor to spin. With no load, the rotor spins at the stator’s magnetic field rotation frequency. 10/01/2019 UEE

Stator, rotor and poles 10/01/2019 UEE

Key to abbreviations Supply frequency – f, Hz. p - number of pole pairs (2 poles = 1 pole pair). ns - synchronous speed, rev.s-1 or rpm. nr - rotor speed, rev.s-1 or rpm. s – slip. Note: convert all speeds to rev.s-1. when calculating. 10/01/2019 UEE

Example 1 A 6-pole, 3-phase induction motor is connected to a 300 Hz supply. Determine the speed of rotation of the magnetic field produced by the stator. Speed of rotation, = ns = supply frequency/ number of pole pairs. = 300 / (6/2) = 100 rev.s-1 or 6000 rpm. 10/01/2019 UEE

Adding a load to the motor In reality, the rotor spins at a lower speed than the stator frequency nr. The difference between stator magnetic field speed and rotor speed is known as slip, s. Slip, s = (ns – nr) / ns. Often expressed as a percentage. Slip %, s = (ns – nr) / ns x 100% 10/01/2019 UEE

Example 2 A 6-pole, 3-phase induction motor runs at 970 rpm at a certain load. If the stator is connected to a 50 Hz supply, find the percentage slip at this load. ns = f / p = 50 / 3 = 16.667 rev.s-1. nr = 970 / 60 = 16.166 rev.s-1. s = (ns – nr) / ns = (16.667 - 16.166) / 16.667 = 0.03 = 3% slip. 10/01/2019 UEE

Motor wiring configurations Can be star or delta wired. Delta is the usual configuration for normal running. Star is suitable for starting. In delta, full supply voltage appears across the 3 windings, L1, L2 and L3. In star, supply voltage / 3 appears across the windings. 10/01/2019 UEE

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Example 3 A 3 phase induction motor is connected to a 750 V supply. What is the voltage across the windings L1 and L2 when: A) It is star connected. B) It is delta connected. For star connections, voltage across windings = supply voltage / 3 = 750 / 1.73 = 433V. For delta connection, voltage across windings = supply voltage = 750 V. 10/01/2019 UEE

Induction motor equivalent circuit To all intents and purposes, the induction motor can be thought of as a rotating transformer. The equivalent circuit for a transformer can be used to analyse an induction motor with a little alteration to represent the load. 10/01/2019 UEE

Key to symbols Vs = voltage across stator. Rc = core losses. Xm = magnetising losses. Rs = resistance of stator. Rr = resistance of rotor. Xs = impedance of stator. Xr = impedance of rotor. 10/01/2019 UEE

Simplified induction motor circuit 10/01/2019 UEE

No-load current calculations When there is no load, all current flows through the Rc and Xm components only. Supply voltage is across Rc and Xm . Ic = Vs / Rc Im = Vs / Xm These are the real and reactive parts of the total branch impedance. Reactive part is inductive therefore lagging. 10/01/2019 UEE

Power factor calculations 10/01/2019 UEE

Branch current, Io 10/01/2019 UEE

Example 4 A star-connected three phase induction motor is supplied with 500 V. Rc is 75W and Xm is 25W. Calculate the no-load power factor and Io. In star connections, Vs = supply voltage / 3 = 500 / 3 = 288.7 V. Io = Ic + Im 10/01/2019 UEE

Example 4 continued Ic = Vs / Rc = 288.7 / 75 = 3.85 A. Im = Vs / Xm = 288.7 / 25 = 11.55 A. Power factor q = tan-1 Im / Ic = tan-1 (11.55 /3.85) = tan-1 (3) = 71.6 degrees lagging. Power factor = cos 71.6 degrees = 0.316 lag. Io = (Ic2 + Im2) = (3.852 + 11.552) = 12.2 A. 10/01/2019 UEE

Full-load calculations These calculations ignore the magnetising branch current initially. This simplifies the model of the motor and leaves only series resistances and reactances which form the total impedance, Zeq. It can be shown that the real part of the load Rr varies with the rotor speed but the reactance Xr remains constant. A variable resistor represents the rotor speed. 10/01/2019 UEE

Full-load induction motor model 10/01/2019 UEE

Complex notation for Zeq When adding the resistances and reactances, it is sometimes easier to use complex notation. Zeq is the resultant of series impedances. Ohmic part is the real part of the impedance. Reactive part is the imaginary part of the impedance. 10/01/2019 UEE

Zeq calculations Zeq = equivalent impedance. Zeq = (Rs + R`r/s) + j(Xs + Xr). Where s = slip. 10/01/2019 UEE

Example 5 An induction motor has the following parameters: Rs = 0.6W, R`r = 0.4W, Xs = Xr = 0.6W, slip s = 0.04. Zeq = (0.6 + 0.4/0.04) + j(0.6 + 0.6) = 10.6 + j1.2 W. To be continued…. 10/01/2019 UEE

Summary Induction motor introduction. Poles, slip & rpm. Wiring options. Equivalent circuit of induction motor. No-load power factor and branch current. Equivalent impedance, Zeq. Assignment (induction motors) working time. 10/01/2019 UEE