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Transcript
Direct – Current Motor Characteristics and Applications • Straight Shunt Motor – Essentially a constant speed motor • Compound or Stabilized – Shunt Motors – Has both shunt and series field windings – Series field generates mmf in the same direction as the shunt field mmf. ECE 441 1 Circuit Diagram of a Compound Motor ECE 441 2 Differential Connection of Fields • Both the series and shunt fields must provide fluxes that are additive. • If the series field is reversed with respect to the shunt field, the net flux decreases, and the speed increases. • The time constant of the series field is such that the current increases faster than the shunt field current. ECE 441 3 Differential Connection of Fields • If the series field is reversed, – The motor will start in the wrong direction – Depending upon the load and the structure of the series field, the motor could • slow down and stop, tripping the breaker • slow down, stop, reverse direction, and accelerate • slow down, stop, reverse direction, slow down, stop, reverse direction, etc. until a breaker trips ECE 441 4 Reversing the Direction of Compound Motors • Reverse either the armature current or reverse both the series and shunt fields. – If only one field is reversed, a “differential” connection results! – The field mmfs will be reduced, resulting in excessive speed! ECE 441 5 Reversing the Armature Current ECE 441 6 Using NEMA standard terminal markings ECE 441 7 Series Motor • Series field – Heavy windings – Must conduct the armature current • Potentially dangerous problem if the shaft load is removed! ECE 441 8 Field winding is in series with the armature ECE 441 9 More Details • When shaft load is removed, TD>Tload – Motor speed increases – cemf increases – armature current decreases – series field flux decreases ECE 441 10 Reversing the Direction of a Series Motor • Reverse the current in the armatureinterpole-compensating branch • Reverse the current in the series field windings ECE 441 11 Reversing the Armature Current ECE 441 12 Using NMEA standard Terminal Markings ECE 441 13 Using NEMA standard terminal markings Reversing the series field ECE 441 14 Effect of Magnetic Saturation on DC Motor Performance • Pole flux is not directly proportional to the applied mmf due to magnetic saturation • Net mmf is made up of the following components, as applicable – – – – – Fnet = Ff + Fs - Fd Fnet = net mmf (A-t/pole) Ff = shunt field mmf (NfIf)(A-t/pole) Fs = series field mmf (NsIa)(A-t/pole) Fd = equivalent demagnetizing mmf due to armature reaction (A-t)/pole ECE 441 15 Effect of Magnetic Saturation on DC Motor Performance • Note that Fd is not exactly proportional to the armature current, but is assumed to be. • If a compensating winding is used, Fd = 0. ECE 441 16 Developed Torque and Speed TD B p I a k M VT I a Racir n p kG p 0 Racir Ra RIP RCW Rs ECE 441 17 Defining Parameters • • • • Racir = resistance of armature circuit (Ω) Ra = resistance of armature windings (Ω ) RIP = resistance of interpole windings (Ω) RCW = resistance of compensating windings (Ω) • Rs = resistance of series field winding (Ω) • Bp = air-gap flux density (T) • Φp = pole flux (Wb) ECE 441 18 Solve Problems with Proportions TD1 [ B p I a ]1 TD 2 [ B p I a ]2 VT I a Racir n1 p kG 1 , 0 n2 VT I a Racir k p G 2 p Bp A Bp n1 VT I a Racir n2 Bp V I R 1 T a acir 2 ECE 441 19 Example 11.1 • A 240-V, 40-hp, 1150 r/min stabilizedshunt motor, operating at rated conditions, has an efficiency at rated load of 90.2%. The motor parameters are • Ra = 0.0680 Ω RIP = 0.0198 Ω Rs = 0.00911 Ω Rshunt = 99.5 Ω • Turns/pole series - ½ shunt - 1231 ECE 441 20 Example 11.1 (continued) • The circuit diagram and magnetization curve are shown on the next slide. Determine (a) the armature current when operating at rated conditions; (b) the resistance and power rating of an external resistance required in series with the shunt field in order to operate at 125% rated speed. Assume the shaft load is adjusted to a value that limits armature current to 115% of rated current. ECE 441 21 ECE 441 22 Solution for Armature Current P 40 746 P VT IT IT VT 0.902 240 IT 137.84 A VT 240 If 2.4121A R f 99.5 I a IT I f 137.84 2.41 135.43 A ECE 441 23 Solution for External Resistance • The series field of a compound motor is designed to be approximately equal and opposite to the equivalent demagnetizing mmf of armature reaction. Therefore, the net flux is due to the shunt field alone. Fnet Ff N f I f 1231 2.412 2969.2 A t / pole ECE 441 24 net mmf = 0.70 T ECE 441 25 Racir Ra RIP Rs Racir 0.0680 0.0198 0.0091 0.0969 Bp n1 VT I a Racir n2 Bp V I R 1 T a acir 2 n1 [VT I a Racir ]2 B p 2 B p1 n2 [VT I a Racir ]1 Bp 2 Bp 2 1150 240 1.15 135.43 0.0969 0.70 1.25 1150 240 135.43 0.0969 0.56T ECE 441 26 Ff = 2.3 X 1000 = 2300 A-t/pole ECE 441 27 Ff 2300 Ff N f I f I f N f 1231 I f 1.87 A VT VT If Rx Rf R f Rx Rf 240 Rx 99.5 28.8 187 PRx I 2f Rx (1.87) 2 28.8 100.7W ECE 441 28 Linear Approximations • If the magnetization curve is not available – rough approximation obtained by assuming magnetization effects are negligible – Do not use approximations if the motor is operating under heavy overload or locked rotor conditions. • If the net mmf is to be reduced below its rated value, approximation using the linear assumption is OK. ECE 441 29 Approximate Equations for Torque and Speed [Fnet I a ]1 TD1 [ B p I a ]1 TD 2 [ B p I a ]2 [Fnet I a ]2 Bp n1 VT I a Racir n2 Bp 1 VT I a Racir 2 Fnet n1 VT I a Racir , 0 n2 Fnet 1 VT I a Racir 2 ECE 441 30 For the Series Motor If the range of operation is in the unsaturated region, and armature reaction effects are either negligible or compensated for, TD1 [Fnet I a ]1 TD , series Fnet I a N s I a TD 2 [Fnet I a ]2 TD , series I 2 a The developed torque is proportional to the square of the armature current. ECE 441 31 Example 11.2 • Example 11.1 is re-solved using the linear approximation, and the solution is compared to the results obtained in Example 11.1. ECE 441 32 VT 240 I f1 2.412 A R f 99.5 I a1 IT I f 137.84 2.412 135.43 A Fnet Ff N f I f 1231 2.412 2969.2 A t / pole Fnet n1 VT I a Racir n2 Fnet V I R 1 T a acir 2 n1 [VT I a Racir ]2 Fnet 2 Fnet1 n2 [VT I a Racir ]1 Fnet 2 Fnet 2 1150 240 1.15 135.43 0.0969 2969.2 1.25 1150 240 135.43 0.0969 2354.8 A t / pole ECE 441 33 Fnet 2354.8 If 1.91A Nf 1231 VT VT If Rx Rf R f Rx If 240 Rx 99.5 26.15 1.91 From Example 11.1, the value of resistance was determined to be 28.8 Ω ECE 441 34 Calculate the Percent Error %error Ractual Rapprox Ractual 100% 28.8 26.15 %error 100% 28.8 %error 9.2% This lower value of resistance would cause a slightly higher field current, and therefore, a speed slightly lower than 1437.5 r/min. ECE 441 35 Comparison of Steady – State Operating Characteristics of DC Motors • The steady-state operating characteristics of typical shunt, compound, and series motors of the same torque and speed ratings are shown on the next slide. ECE 441 36 ECE 441 37 Comparisons (continued) • Shunt Motor – relatively constant speed from no-load to full-load – does not have high starting torque – essentially constant flux – torque varies linearly with armature current – speed regulation around 5% ECE 441 38 Relatively Constant Speed Linear Torque ECE 441 39 Comparisons (continued) • Compound Motor – Higher torque, lower speed than shunt motor – speed regulation between 15 and 25% – used with loads requiring high starting torques or have pulsating loads • smoothes out the energy required by the pulsating load, lowering the demand on the electrical supply ECE 441 40 Lower Speed at Higher Torque Higher Torque above base speed than Shunt motor ECE 441 41 Comparisons (continued) • Series Motor – high starting torque – wide speed range – REMOVING THE LOAD CAUSES IT TO RUN AWAY! • CONNECT LOAD BY GEARS OR SOLID COUPLING – NO BELT DRIVES! ECE 441 42 Wide Speed Range High Starting Torque ECE 441 43 Dynamic Braking, Plugging, and Jogging • Dynamic Braking is the deceleration of the motor by converting the energy stored in the moving masses into electrical energy and dissipating it as heat via resistors. Also called resistive braking. ECE 441 44 Dynamic Braking (continued) • Disconnect the armature from the electrical supply lines and connect across a suitable resistor while maintaining the field at full strength. • The motor behaves as a generator, feeding current to the resistor, dissipating heat. ECE 441 45 Dynamic Braking (continued) • Choose the resistance for current between 150 and 300% of rated current. • The armature current is in a direction to oppose the armature motion, producing a negative, or, counter-torque, slowing down the load. ECE 441 46 Compound Motor Example Normal Operation Dynamic Braking ECE 441 47 Normal Operation Closed Open ECE 441 48 Dynamic - Braking Open Closed ECE 441 49 Regenerative Braking • Convert energy of overhauling loads into electrical energy and pumps it back into the electrical system. • The overhauling load drives a DC motor faster than normal, causing the cemf to become greater than the supply voltage and results in generator action. • Trains, elevators, hybrid automobiles ECE 441 50 Plugging • The electrical reversal of a motor before it stops • Reverse the voltage applied to the armature • Current in the series and shunt fields is not reversed • Insert resistance in series with the armature to limit the current ECE 441 51 Normal Operation ECE 441 52 Plugging ECE 441 53 Jogging • • • • Very brief application of power to a motor Fraction of a revolution Used for positioning the load Place resistance in series with the armature to limit the current ECE 441 54 Example 11.7 • A 240-V, compensated shunt motor driving a 910 lb-ft torque load is running at 1150 r/min. The efficiency of the motor at this load is 94.0%. The combined armature, compensating winding, and interpole resistance is 0.00707Ω, and the resistance of the shunt field is 52.6Ω. Determine the resistance of a dynamic-braking resistor that will be capable of developing 500 lb-ft of braking torque at a speed of 1000 r/min. Assume windage and friction at 1000r/min are essentially the same as at 1150 r/min. ECE 441 55 Circuit for Dynamic Braking ECE 441 56 T n 910 1150 Pshaft 199.257 hp 5252 5252 Pshaft 199.257 746 Pin 158134W 0.940 158134 Pin VT IT IT 658.89 A 240 VT 240 If 4.56 A R f 52.6 I a IT I f 658.89 4.56 654.33 A VT Ea1 I a1 Racir Ea1 240 654.33 0.00707 Ea1 235.37V ECE 441 57 I T T1 [ B p I a ]1 I a1 I a 2 a1 2 T2 [ B p I a ]2 I a 2 T1 654.33 500 Ia2 359.52 A 910 Ea1 [n p kG ]1 n1 n2 Ea 2 Ea1 Ea 2 [n p kG ]2 n2 n1 1000 Ea 2 235.37 204.67V 1150 Ea 2 I a 2 ( Racir RDB ) RDB RDB RDB Ea 2 I a 2 Racir Ia2 204.67 359.52 0.00707 0.562 359.52 0.562 ECE 441 58