ENERGY CONVERSION ONE (Course 25741) CHAPTER NINE ….continued DC MOTORS AND GENERATORS

TERMINAL CHARCATERISTIC of a SHUNT DC MOTOR If a shunt dc motor has compensating windings so that flux is constant regardless of load, & motor’s speed & armature current known at any one value of load, then speed at any other load can be determined if I A is known at that load EXAMPLE-1: A 50 hp, 250 V, 1200 r/min dc shunt motor with compensating windings, R A =0.06 Ω (including brushes, comp. windings, & inter-poles). The field cct. has a total resistance of R adj + R F of 50 Ω which develops a no load speed of 1200 r/min. There are 1200 turns per pole on shunt field winding

TERMINAL CHARCATERISTIC of a SHUNT DC MOTOR (a)Find speed of this motor when its input current is 100 A (b) Find speed of this motor when its input current is 200 A (c) Find the speed of this motor when its input current is 300 A (d) plot torque-speed characteristic of this motor

TERMINAL CHARCATERISTIC of a SHUNT DC MOTOR EXAMPLE-1;SOLUTION: E A =K’φn, since I F is constant (V T & R F const.) & since there are no A.R.  φ would be constant  relationships between speeds & internal generated voltages of motor at 2 different load conditions is: E A2 /E A1 =[K’φn 2 ]/[K’φn 1 ] constant K’ cancels, also φ canceled  n 2 = E A2 /E A1. n 1 at no load I A is zero so E A1 =V T =250 V While speed n 1 =1200 r/min If internal generated voltage at any other load is determined, motor speed at the load can be determined

TERMINAL CHARCATERISTIC of a SHUNT DC MOTOR (a) if I L =100 A then armature current : I A =I L -I F =I L - V T / R F =100 – 250/ 50=95 A  E A =V T -R A I A =250 –(95)(0.06)=244.3 V Resulting speed: n 2 = E A2 /E A1x n 1 =244.3/250 x1200=1173 r/min (b) I L =200 A  I A = /50=195 A E A =250-(95)(0.06)=238.3 V n 2 = 238.3/250 x1200=1144 r/min

TERMINAL CHARCATERISTIC of a SHUNT DC MOTOR (c) if I L =300 A, then I A =I L -I F = /50=295 A E A = 250-(295)(0.06)=232.3 V n 2 =232.3/250 x 1200=1115 r/min (d) torque versus speed At no load induced torque is zero P conv =E A I A =T ind ω From this equation T ind = E A I A / ω I L =100 A  T ind =(244.3)(95)/ [1173x1/60x2π]=190 N.m. I L =200 A  T ind = 238.3x195/[1144x1/60x2π]=388 N.m. I L =300 A  T ind =587 N.m.

TERMINAL CHARCATERISTIC of a SHUNT DC MOTOR Torque – speed characteristic of motor

NONLINEAR ANALYSIS of a SHUNT DC MOTOR flux φ & E A of a dc machine is a nonlinear function of mmf  anything that changes mmf cause a nonlinear effect on E A mmf should be used to determine E A & mmf determined based on field current and A.R. magnetization curve is a direct plot of E A versus I F for a given speed ω 0  effect of variation in field current can be determined directly from its magnetization curve

NONLINEAR ANALYSIS of a SHUNT DC MOTOR If a machine has armature reaction, its flux will be reduced with each increase in load. The total mmf in a shunt dc motor is the field circuit mmf less the mmf due to armature reaction (AR): F net = N F I F – F AR magnetization curves are expressed as plots of E A vs field current, normally an equivalent field current is defined that would produce the same output voltage as the combination of all the mmf in the machine The equivalent field current :

NONLINEAR ANALYSIS of a SHUNT DC MOTOR one other effect must be considered when non linear analysis is used to determine E A of a dc motor The magnetization curves for a machine are drawn for a particular speed, usually the rated speed of the machine How can the effects of a given field current be determined if the motor is turning at other than rated speed? The equation for the induced voltage in a dc machine when speed is expressed as rev/min: E A = K’φn, For a given effective field current, the flux in the machine is fixed, so the E A is related to speed by

NONLINEAR ANALYSIS of a SHUNT DC MOTOR EXAMPLE-2: A 50 hP, 250V, 1200r/min DC shunt motor without compensating windings has an armature resistance (including the brushes and interpoles) of 0.06 . Its field circuit has a total resistance R adj + R F of 50 , which produces a no-load speed of 1200 r/min. There are 1200 turns per pole on the shunt field winding, and the armature reaction produces a demagnetising mmf of 840 A.turns at a load current of 200A. magnetization curve of this machine is shown next

NONLINEAR ANALYSIS of a SHUNT DC MOTOR Magnetization Curve of typical 250 V dc motor Example -2

NONLINEAR ANALYSIS of a SHUNT DC MOTOR (a) Find the speed of this motor when its input current is 200A. (b) This motor is essentially identical to the one in Example 1, except for the absence of compensating windings. How does its speed compare to that of the previous motor at a load current of 200A? (c) calculate & plot torque-speed characteristic of motor

NONLINEAR ANALYSIS of a SHUNT DC MOTOR (a) I L =200 A  I A =I L -V T /R F = /50=195 A E A = x0.06=238.3 V at this current A.R. demagnetizing mmf is 840 A.T.s so effective shunt field current is: I F *= I F - F AR /N F = /1200 =4.3 A from magnetization curve, effective field current produce E A0 of 233 V at a speed of n 0 =1200 r/min However since voltage is V, shows actual operating speed of motor is: E A /E A0 n 0 =238.3/233 x 1200 =1227 r/min

NONLINEAR ANALYSIS of a SHUNT DC MOTOR (b) at 200 A load in Example 1, motor’s speed was n=1144 r/min in this example motor’s speed is 1227 r/min Note: speed of motor with A.R. is higher than speed of motor with no A.R. This relative increase in speed is due to flux weakening in machine with A.R. (c) for this item, (since there is no information about F AR in other load currents) it is assumed that F AR varies linearly with load current, & using MATLAB speed-torque characteristic obtained

NONLINEAR ANALYSIS of a SHUNT DC MOTOR Torque-speed characteristic of motor with armature reaction

SPEED CONTROL of SHUNT DC MOTOR Two common methods: 1- Adjusting the field resistance R F (and thus the field flux) 2- Adjusting the terminal voltage applied to the armature Less common method: 3- Inserting a resistor in series with the armature circuit

SPEED CONTROL of SHUNT DC MOTOR Changing the Field Resistance If the field resistance increases, field current decreases (I F ↓ = V T /R F ↑), and as the field current decreases, flux decreases as well. A decrease in flux causes an instantaneous decrease in the internal generated voltage E A ↓ (=Kφ↓ω), which causes a large increase in the machine’s armature current since

SPEED CONTROL of SHUNT DC MOTOR Induced torque in a motor is given by  ind =KφI A since flux in machine decreases while current I A increases, which way does the induced torque change? Look at this example:  armature current flow is I A =(250V-245V)/ 0.25Ω= 20A

SPEED CONTROL of SHUNT DC MOTOR What happens in this motor if there is a 1% decrease in flux? If the flux decrease by 1%, then E A must decrease by 1% too, because E A = Kφω Therefore, E A will drop to: E A2 = 0.99 E A1 = 0.99 (245) = V armature current must then rise to I A = ( )/0.25 = 29.8 A Thus, a 1% decrease in flux produced a 49% increase in armature current

SPEED CONTROL of SHUNT DC MOTOR back to original discussion, the increase in current predominates over the decrease in flux so,  ind >  load, the motor speeds up However, as the motor speeds up, E A rises, causing I A to fall. Thus, induced torque  ind drops too, and finally  ind equals  load at a higher steady-sate speed than originally Summarizing behaviour: 1- increasing R F causes I F (=V T /R F ) to decrease 2- decreasing I F decrease φ

SPEED CONTROL of SHUNT DC MOTOR 3 – Decreasing φ lowers E A (=Kφω) 4 - Decreasing E A increases I A (=V T -E A )/R A 5- increasing I A increases T ind (=KφI A ), with change in I A dominant over change in flux 6-increasing T ind makes T ind >T load & speed ω increases 7-increase in ω, increases E A = Kφω again 8-increasing E A decreases I A 9-Decreasing I A decreases T ind until T ind =T load at a higher ω Effect of increasing R F on O/P characteristic of a shunt motor shown in next figure

SPEED CONTROL of SHUNT DC MOTOR Effect of R F speed control on a shunt motor’s torque-speed (over motor’s normal operating range)

SPEED CONTROL of SHUNT DC MOTOR Effect of R F speed control on a shunt motor’s torque-speed (over entire range from no-load to stall conditions)

SPEED CONTROL of SHUNT DC MOTOR According to equation of speed presented before: (a) no-load speed is proportional to reciprocal of flux in motor (b) while slope of the curve is proportional to reciprocal of flux squared Therefore a decrease in flux causes slope of torque- speed to become steeper over this range, an increase in field resistance increases motor’s speed For motors operating between no-load & full-load conditions, an increase in R F may reliably be expected to increase operating speed

SPEED CONTROL of SHUNT DC MOTOR In previous figure (b) terminal characteristic of motor over full range from no-load to stall shown In figure can see at very slow speeds, an increase in field resistance will actually decrease speed of motor This effect occurs because at very low speeds, the increase in armature current caused by decrease in E A not enough to compensate for decrease in flux in induced torque field resistance Some small dc motors used for control purposes operate at speeds close to stall conditions For these motors, an increase in R F might have no effect or it might even decrease speed of motor Since the results are not predictable, field resistance speed control should not be used in these types of dc motors. Instead, the armature voltage method should be employed

SPEED CONTROL of SHUNT DC MOTOR CHANGING ARMATURE VOLTAGE This method involves changing the voltage applied to the armature of the motor without changing the voltage applied to the field If the voltage V A is increased, then the I A must rise [ I A = (V A ↑ -E A )/R A ]. As I A increases, the induced torque  ind =KφI A ↑ increases, making  ind >  load, and the speed of the motor increases But, as the speed increases, the E A (=Kφω↑) increases, causing the armature current to decrease This decrease in I A decreases the induced torque, causing  ind =  load at a higher rotational speed

SPEED CONTROL of SHUNT DC MOTOR Effect of armature voltage speed control

SPEED CONTROL of SHUNT DC MOTOR Inserting a Resistor in Series with the Armature Circuit If a resistor is inserted in series with the armature circuit, the effect is to drastically increase the slope of the motor’s torque-speed characteristic, making it operate more slowly if loaded. This fact can be seen from the speed equation: The insertion of a resistor is a very wasteful method of speed control, since the losses in the inserted resistor are very large. For this reason, it is rarely used