b. The braking power [kW] when the motor We wish to stop a 120 hp, 240 V, 400 r/min 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 current to 125 percent of its nominal value 14 has decelerated to 200 r/min 50 r/min.

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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
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
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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