Session: Summer 2008 Page: 1of 8 Question Number: 1 (a) A single winding machine cannot generate starting torque. During starting the switch connects the starting winding via the capacitor. The capacitor causes phase lead (a time shift) between the starting winding current and the main winding current. The starting winding is aligned at 90 to the main winding so the combination of main winding and starting winding creates a field which rotates from one axis to the other. This rotating field generates torque to start the machine. Once the machine is rotating the starter winding may be switched out and the main winding will continue to provide a pulsating torque. Diagram Description [6 ] [2 ]
Session: Summer 2008 Page: 2of 8 Question Number: 1 (b) Total Electricity Cost for 2 years operation with standard motor: 8kW Electricit y Cost (standard) = 2 years*365 days* 24 hours * * 0.15 / kwh= 24,73 0.85 Total Electricity Cost for 2 years with high efficiency motor: 8kW Electricit y Cost (Efficient) = 2 years*365 days* 24 hours * * 0.15 / kwh= 23,360 0.9 Saving in electricity with high efficiency motor = 24,734-23,360 = 1,374 Since the 2 year saving is greater than the 500 price difference between standard and high efficiency motor we should choose the high efficiency motor. (c) Fuse Characterisitic Time (t) (Log Scale) Typical Melting Time Curves Slow Blow General Purpose Fuse Rating Current (I) [6 ] Typical Application requiring a slow blow fuse: A Motor which draws high inrush during starting. Typical Application requiring a Fast Blow fuse: A Semiconductor load such as a rectifier. [1 mark] [1 mark]
Session: Summer 2008 Page: 3of 8 Question Number: 1 (d) The four main functions with the appropriate components are as follows: (1) Isolation Lockable MCB or Isolator (2) Short-circuit protection A magnetic coil in the MCB which activates on high current, opening the supply to the load (3) Overload protection A bi-metallic strip in the MCB which bends when heated OR a Bi-metallic thermal overload relay (Motor) (4) Power switching Electromagnetic contactor (e) N S = (f 60)/p = (10 60)/3 = 200 revs/min. N R = 200 4% = 192 revs/min. One problem of running the motor at such a low speed is that the temperature of the motor can rise as there may not be enough air being drawn along the cooling fins of the motor. A solution is to use a separate cooling fan to assist in the motor cooling. Total Question 1 [40 marks]
Session: Summer 2008 Page: 4of 8 Question Number: 2 (a) (b) (c) V a = E a + R i a a E = V R 2πN 60Ea Ea = kaφ kaφ = 60 2πN Rotationallosses= E i P P nm a a a a nm i a a = 200 20 1= 180V 60 180 = = 0.859Vs / A 2π 2000 = 0 ( no load) Rotationallosses= 180V 1= 180W Factors contributing to rotational losses: Iron losses Windage Friction [5 marks] [3 marks] (d) If the field current was reduced then the the flux φ would be reduced and the machine would have to rotate faster in order to produce the same back emf E a. Therefore the machine would speed up. [6 marks] Total Q2 [30 marks]
Session: Summer 2008 Page: 5of 8 Question Number: 3 (a) The peak load power requirement is 10 kw [2 marks] (b) The average load power requirement is 4kW 5s+ 10kW 5s+ 2kW 10s = 4.5kW 20s [6 marks] (c) The root mean square load power is (d) 4 2 2 5+ 10 5+ 2 20 2 10 = 5.57kW Peak Power requirement requires that the motor rating be greater than the peak load power / (1+service factor) = 10kW / (1.2) = 8.33kW The rms power requirement requires that the motor rating be greater than rms load power = 5.57kW [10 marks] (e) In this case the peak power requirement dominates and an 8.33kW motor is required. 2π N 60P 60 10kW P= T T = = = 47. 7Nm 60 2πN 2π 2000 [4 marks] Total Q3 [30 marks]
Session: Summer 2008 Page: 6of 8 Question Number: 4 (a) Q4 (a) H1 is the motor overload tripped indicator and H2 is the motor run indicator. S1 is the stop push-button and S2 is the start push-button. (b) The utilisation categories cover the main application areas foe alternating current (AC-Categories) and direct current (DC-Categories). For normal operation of contactors they define the making and breaking conditions at rated operational current (Ie) and at rated operational voltage (Ue). They depend on (1) the nature of the load to be switched (resistive, squirrel cage motor, etc), (2) the conditions under which making and breaking occurs (motor started or stalled, reversing direction of operation, etc.). The operating rate is less than 30 starts per hour and standard motors are designed for 6 starts per hour. Applications AC-1: this applies to all loads supplied by alternating current at a power factor greater than 0.95, for example resistive loads. Also a low number of cycles/hour operation. AC-2: starting and inching of asynchronous 3-phase slip-ring motors. AC-3: controlling the full load breaking current for squirrel cage motors, and breaking with motor running. AC-4: breaking the stalled rotor current for a squirrel cage motor.
Session: Summer 2008 Page: 7of 8 Question Number: 4 (cont.) (c) (i) AC Input Smoothing Capacitor 3-Phase Motor Rectifier (AC to DC) Inverter (DC to AC) Figure 1 Block diagram of frequency inverter Figure 2 More detail on frequency inverter The rectifier converts the supply voltage, single-phase or three-phase A.C. at 50 hertz, to a D.C. value. The D.C. has ripples which the filter capacitor smoothes out to give a constant D.C. value (straight line). This signal is the fed into an inverter that converts the D.C. back to an A.C. signal at whatever frequency is desired, depending on the required motor speed. To switch the current from D.C. to A.C. IGBT s are used. Insulated Gate Bipolar Transistor (IGBT) is a special transistor which is controlled by a very low voltage, hence there is no relation between the energy necessary for control and the current switched. Each time the device is switched puts stress on the semiconductor, because of the dv/dt relation due to the high speed switching. To minimise this stress, inductances and switching circuits comprised of resistors, capacitors and diodes are employed. In order to achieve a constant torque whatever the speed, it is necessary to maintain a constant flux. This means that the voltage and frequency must evolve simultaneously and in the same proportion. This is achieved using the Pulse Width Modulation (PWM) technique, as the width of the pulse is increased, the effective voltage is increased. [9 marks]
Session: Summer 2008 Page: 8of 8 Question Number: 4 (cont.) (c) (ii) Advantages of Inverter control over DOL Range of speeds available Soft start whereby speed ramps up slowly reducing sudden mechanical jerks Soft stop available Regenerative braking possible Direction is easily reversed using software Overload current value is programmable Disadvantages of Inverter control over DOL At low speeds, cooling may not be sufficient At high speeds, mechanical problems with bearings, fixings coming loose Cost [5 marks] Total Q4 [30 marks]