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Module PE5 Problem 1 Recall that the speed of a DC motor is given by w Vt R a T . A particular separately excited DC motor 2 k k has k 1.44 , Ra 0.86 . When it is operating at 150 rad/sec, the armature current is I a 40 A . The terminal voltage Vt is held constant under all conditions. (a) Compute Vt (b) Compute the no-load speed in rad/sec Solution to problem 1 (a) Ea kwm 1.44 150 216V Vt E a I a ra 216 40 0.86 250 .4V (b) "No load" speed occurs when T=0 (No load !), therefore, w w Vt RT V a 2 t when T=0 k k k 250 .4 173 .9rad / sec 1.44 Problem 2 A separately excited DC motor runs at 100 rpm supplying rated torque when the applied voltage is 50 volts and the armature current is 10A. The armature resistance is Ra = 1 ohm. A thyristor- controlled chopper circuit is used to decrease the applied voltage to 40 volts, at constant field flux, to drive a load at 75% of the rated torque. The conducting time T on of this circuit is fixed at 0.2 seconds. Compute the speed of the motor at 40 volts and the blocking time T off of the chopper circuit. Solution to problem 2 T1 k I a1 T2 0.75T1 k I a 2 0.75(k I a1 ) I a 2 0.75I a1 (0.75)(10 A) 7.5 Amps E a1 Vt1 I a1 Ra 50 (10)(1) 40Volts E a 2 Vt 2 I a 2 Ra 40 (7.5)(1) 32.5Volts E a1 k 1 , E a 2 k 2 E a1 1 Ea 2 2 2 Ea 2 32.5 1 (100) 81.25rpm E a1 40 VLoad , DC Ton 0.2 40 Toff 0.05 sec onds Ton Toff V 0.2 Toff 50 Problem 3 A separately excited D.C motor is rated for a full load torque of 100Nm. For an applied voltage of 240V and for flux ( ) per pole of 30 mWb, the no-load and full load speeds of the motor are 1595 r.p.m and 1195 r.p.m respectively. Find a) b) i) The constant k ii) The armature resistance Ra If the applied voltage and the flux per pole are reduced to 200V and 20mWb respectively, Find i) The corresponding no load speed ii) The corresponding full load speed Solution to problem 3 Vt RT a 2 . From the problem, we have: k (k ) 100 Nm Vt 240V 30mWb nnoload 1595rpm n fullload 1195rpm a) From the text, we know that T fullload At no load, 1595rev 1 min 2rad 167.03rad / sec min 60 sec rev V Vt 240 167.03 t k 47.9 k (167.03) (167.03)(0.03) T 0and noload At full load, TFL 100Nm 2 125.1rad / sec 60 V RT (k ) 2 125.1 t a FL2 Ra k (k ) TFL FL (1195) Vt 125.1 k (( 47.9)(0.03)) 2 240 Ra 125.1 0.865 100 (47.9)(0.03) b) Vt = 200 V, = 0.02 Webers. V 200 60 noload t 204.08rad / sec n NL 204.08 1949.8rpm k (49)(0.02) 2 V RT 0.865(100) fullload t a 2 204.08 88.265rad / sec k (k ) [( 49)(0.02)] 2 60 n FL 88.265 843.3rpm 2 Problem 5 An engineer wants to construct a half-wave rectifier for speed control of a DC motor using a thyristor, but there is no gate-pulse circuit available. The implication is that he/she can supply no gate pulse. Without a gate pulse circuit, can the engineer just use the thyristor like a diode, so that at least the circuit can supply one DC voltage level to make the motor run at a single speed? Indicate yes or no, and identify the feature(s) of a thyristor that support your answer. Solution to problem 5 Answer 1: No, it cannot be done, because without the gate pulse, the thyristor will not be able to turn on and it will appear as open-circuit at all times. Answer 2: Yes, it can work if the applied AC voltage has a peak that is significantly higher than the forward breakover voltage of the thyristor. Problem 6 A square wave voltage waveform having a period of 2 seconds is applied through a half wave thyristorcontrolled rectifier circuit to the armature terminals of a separately excited DC motor driving It is known that the DC motor has k=1.44 (a constant under all conditions), and Ra=0.86 ohms. When Ia=40 amperes, the speed is 150 rad/sec. The source voltage and the thyristor firing time is shown in the figure below. (a) Draw this circuit, including the rectifier and the motor. (b) What is the effective DC voltage at the motor armature terminals under this condition ? (c) What is the peak value of the square wave under this condition ? Vsource Time (sec) 0.5 1 2 Fire thyristor Solution to problem 6 a) Ia V(t) _ Ra Ea b) Ea = k = (1.44)(150) = 216, Vt = 216 + IaRa = 250.4 Volts c) 2 Vdceff 1 1 1 1 V (t )dt V (t )dt V 250.4 V 1001.6Volts T0 2 0.5 4