Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN: 9780133923605
Author: Robert L. Boylestad
Publisher: PEARSON
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Number 14 letter a. 

12 A separatel otor turns at ng. c. Compare the brak 200 r/min to the in voltage required 1500 r/min. At 100 r/min. dissipated in resis 16 The armature of a 2: tor has a diameter o length of 235 mm. C a. The approximate knowing that irc kg/m³ b. The kinetic ener when it turns at Advanced level nd make a erator ef- insulation class: H d polarity weight: 3400 kg external diameter of the frame: 915 m c. The total kineti parts at a speed the windings an the J calculated length of frame: 1260 mm re volt- . Calculate the total losses and efficiency at full-load. ary the exciting current if the shunt field canse 20 percent of the total losses. c. Calculate the value of the armature resistance as well as the counter-emf. 17 If we reduce the n practical shunt me speed increases, b Explain why, bea of the iron under a shunt tes. ing a 18 The speed of a se ing temperature, increases. Expla of ro- d. If we wish to attain a speed of 1100 r/min what should be the approximate exciting na- current? Industrial Application 14 We wish to stop a 120 hp, 240 V, 400 r/min A permanent ma cobalt-samarium motor by using the dynamic braking circuit shown in Fig. 17. If the nominal armature current is 400 A, calculate the following: a. The value of the braking resistor R if we want to limit the maximum braking [M che magnetism per I ture. The motor 2500 r/min whe in an ambient te the speed if the where the ambi current to 125 percent of its nominal value pa has decelerated to 200 r/min, 50 r/min, 0 r/min IC A. 15 a. The motor in Problem 14 is now stopped Answers to Pr Calculate the new braking resistor R so the maximum braking current is 500 A. Note: The following rounded off to an a are preceded by the 1S 9a. 221 V 9b. 13 800 W 10b. 1 85 O
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Transcribed Image Text:12 A separatel otor turns at ng. c. Compare the brak 200 r/min to the in voltage required 1500 r/min. At 100 r/min. dissipated in resis 16 The armature of a 2: tor has a diameter o length of 235 mm. C a. The approximate knowing that irc kg/m³ b. The kinetic ener when it turns at Advanced level nd make a erator ef- insulation class: H d polarity weight: 3400 kg external diameter of the frame: 915 m c. The total kineti parts at a speed the windings an the J calculated length of frame: 1260 mm re volt- . Calculate the total losses and efficiency at full-load. ary the exciting current if the shunt field canse 20 percent of the total losses. c. Calculate the value of the armature resistance as well as the counter-emf. 17 If we reduce the n practical shunt me speed increases, b Explain why, bea of the iron under a shunt tes. ing a 18 The speed of a se ing temperature, increases. Expla of ro- d. If we wish to attain a speed of 1100 r/min what should be the approximate exciting na- current? Industrial Application 14 We wish to stop a 120 hp, 240 V, 400 r/min A permanent ma cobalt-samarium motor by using the dynamic braking circuit shown in Fig. 17. If the nominal armature current is 400 A, calculate the following: a. The value of the braking resistor R if we want to limit the maximum braking [M che magnetism per I ture. The motor 2500 r/min whe in an ambient te the speed if the where the ambi current to 125 percent of its nominal value pa has decelerated to 200 r/min, 50 r/min, 0 r/min IC A. 15 a. The motor in Problem 14 is now stopped Answers to Pr Calculate the new braking resistor R so the maximum braking current is 500 A. Note: The following rounded off to an a are preceded by the 1S 9a. 221 V 9b. 13 800 W 10b. 1 85 O
50 Speed +, DIRECT-CURRENT MOTORS 110 75 dyna R. 25 plugging 0. Figure 17b Armature on open circuit generating a voltage E.. Figure 18otor Speed versus tin ors to vinulo E, E R reversing the a minals of the L- Under nor 41 rent I is give Figure 17c Dynamic braking. where R, is t In practice, resistor R is chosen so that the initial braking current is about twice the rated motor cur- rent. The initial braking torque is then twice the nor- mal torque of the motor. As the motor slows down, the gradual decrease in E, produces a corresponding decrease in I,. Consequently, the braking torque becomes smaller and smaller, finally becoming zero when the arma- ture ceases to turn. The speed drops quickly at first reverse the t acting on th The so-calle longer cour supply volt an enormo greater tha current wo
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Transcribed Image Text:50 Speed +, DIRECT-CURRENT MOTORS 110 75 dyna R. 25 plugging 0. Figure 17b Armature on open circuit generating a voltage E.. Figure 18otor Speed versus tin ors to vinulo E, E R reversing the a minals of the L- Under nor 41 rent I is give Figure 17c Dynamic braking. where R, is t In practice, resistor R is chosen so that the initial braking current is about twice the rated motor cur- rent. The initial braking torque is then twice the nor- mal torque of the motor. As the motor slows down, the gradual decrease in E, produces a corresponding decrease in I,. Consequently, the braking torque becomes smaller and smaller, finally becoming zero when the arma- ture ceases to turn. The speed drops quickly at first reverse the t acting on th The so-calle longer cour supply volt an enormo greater tha current wo
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