E16.1 the Input Power to the Dc Motor is Pin
-
Upload
iamthewalrus01 -
Category
Documents
-
view
219 -
download
0
Transcript of E16.1 the Input Power to the Dc Motor is Pin
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 1/35
CHAPTER 16
Exercises
E16.1 The input power to the dc motor is
in loss out source source P P I V P +==
Substituting values and solving for the source current we have 335074650220 +×=source I
A8.184=source I
Also we have
%76.91335074650
74650%100 =
+××
=×=in
out
P
P η
%35.4%1001150
11501200
%100regulationspeed
=×−
=
×
−
=−
−−
load full
load full load no
n
n n
E16.2 (a) The synchronous motor has zero starting torque and would not be
able to start a high-inertia load.
(b) The series-field dc motor shows the greatest amount of speed
variation with load in the normal operating range and thus has the poorest
speed regulation.
(c) The synchronous motor operates at fixed speed and has zero speed
regulation.
(d) The ac induction motor has the best combination of high starting
torque and low speed regulation.
(e) The series-field dc motor should not be operated without a load
because its speed becomes excessive.
E16.3 Repeating the calculations of Example 16.2, we have
(a) A4005.0
2
)0( ===+ A
T
A R
V
i N2440)3.0(2)0()0( ==+=+ A Bli f
m/s333.3)3.0(2
2===
Bl
V u T
(b) A667.6)3.0(2
4===
Bl
f i load A
540
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 2/35
V667.1)667.6(05.02 =−=−= A A T A I R V e
m/s778.2)3.0(2
667.1===
Bl
e u A
W11.11)778.2(4 === u f p load m
2
W222.2== R i p A R W33.13)667.6(2 === A T t i V p
%33.8333.13
11.11%100 ==×=
t
m
p
p η
(c) A333.3)3.0(2
2===
Bl
f i pull A
V167.2)333.3(05.02 =+=+= A A T A I R V e
m/s611.3)3.0(2
167.2===
Bl
e u A
W222.7)611.3(2 === u f p pull m W667.6)333.3(2 === A T t i V p
2 W5555.0== R i p A R
p %31.92
222.7
667.6%100 ==×=
m
t
p η
E16.4 Referring to Figure 16.15 we see that 125≅A E V for 2=F I A and
.1200=n Then for 1500=n , we have
V1561200
1500125 =×=A E
E16.5 Referring to Figure 16.15 we see that 145≅A E V for 5.2=F I A and
.1200=n Then for 1500=n , we have
V3.1811200
1500145 =×=A E
rad/s1.15760
2=×=
π ω n m
Nm49.471.157
74610=
×==
m
dev
dev
P T
ω
A15.413.181
74610=
×==
A
dev A E
P I
V6.193)15.41(3.03.181 =+=+= A A A T I R E V
541
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 3/35
E16.6 Ω2010
1010300=
×−=
−=
F
F F T adj I
I R V R
Because I F remains constant the value of K ϕ is the same value as in
Example 16.4, which is 2.228. Furthermore the loss torque also remains
constant at 11.54 Nm, and the developed torque remains at 261.5 Nm.Thus the armature current is still 117.4 A. Then we have
V4.292)4.117(065.0300 =−=−= A A T A I R V E
rad/s2.131228.2
4.292===
φω
K
E A m
rpm12532
60==
π ωm m n
Thus the motor speeds up when V T is increased.
E16.7Following Example 16.4, we have
A63010
240=
+=
+=
adj F
T F R R
V I
Referring to Figure 16.18 we see that 200≅A E V for 6=F I A and
.1200=n Thus we have
592.1)60/2(1200
200===
π ωφ
m
A E K
A3.164592.1
5.261===
φK
T I dev
A
V229.3)3.164(065.0240 =−=−= A A T A I R V E rad/s0.144
592.1
3.229===
φω
K
E A m
rpm13762
60==
π ωm m n
kW36== m out out T P ω
kW87.40)3.1646(240)( =+=+= A F T in I I V P
P %08.88%100 =×=
in
out
P η
E16.8 rad/s8.1776
127.125
3
113 ===
dev
dev m m T
T ωω
rpm16982
6033 ==
π ωm m n
W1066333 == m out out T P ω
542
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 4/35
E16.9 With R A = 0 and fixed V T , the shunt motor has constant speed
independent of the load torque. Thus we haverpm120012 == m m n n
rad/s7.12512 == m m ωω
W1508111 == m out out T P ω W3016222 == m out out T P ω
E16.10 Decreasing V T decreases the field current and therefore the flux ϕ . In
the linear portion of the magnetization curve, flux is proportional to the
field current. Thus reduction of V T leads to reduction of ϕ and according
to Equation 16.35, the speed remains constant. (Actually, some speed
variation will occur due to saturation effects.)
E16.11 The torque--speed relationship for the separately excited machine isgiven by Equation 16.27
)( m T A
dev K V R
K T φω
φ−=
which plots as a straight line in the T dev - ω m plane. A family of plots for
various values of V T is shown in Figure 16.27 in the book.
E16.12 The torque--speed relationship for the separately excited machine is
given by Equation 16.27
)( m T A
dev K V R K T φωφ −=
which plots as a straight line in the T dev - ω m plane. As the field current
is increased, the flux ϕ increases. A family of plots for various values of
I F and ϕ is shown in Figure 16.28 in the book.
E16.13
A14100
140
adj
=+
=+
=F
F F R R
V I
2601200
1000312 === A NL E V V
543
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 5/35
247065.0200260 =×−=−= A A A FL I R E V V
%26.5%100247
247260%100regulationvoltage =×
−=×
−=
FL
FL NL
V
V V
4.49247200out =×== FL L V I P kW
0.52)200(065.049400 22outdev =+=+= A A I R P P kW
7.10460
2=
π=ω m m n rad/sec 1.58
85.0
4.49
0.85out
in ===P
P kW
1.60.521.58devinlosses =−=−= P P P kW
5557.104
58100inin ==
ω=
m
P T nm 497
7.104
52000devdev ==
ω=
m
P T nm
Problems
P16.1 The principal parts of a rotating electrical machine include the stator,
rotor, shaft, field windings, and armature windings.
P16.2 Dc machines contain brushes and commutators which act as mechanical
switches that reverse the connections to the rotor conductors as needed
to maintain a constant output voltage polarity.
P16.3 The two principal types of three-phase motors are induction motors and
synchronous motors. Induction motors are far more common.
P16.4 The two types of windings found in electrical machines are field windings
and armature windings. Field windings establish the magnetic field in the
machine are not needed in permanent-magnet machines because the
magnets provide the field.
P16.5* Two disadvantages of dc motors compared to signal-phase ac induction
motors for a ventilation fan, which we can expect to operate most of the
time, are first that dc power is usually not readily available in a home andsecond that dc machines tend to require more frequent maintenance than
ac induction motors.
P16.6 Dc motors are advantageous in automotive applications because dc power
is available from the battery. Dc motors are advantageous when speed
544
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 6/35
and direction must be controlled; however, this advantage is rapidly being
lost to ac motors with electronic drives.
P16.7 The input power is the output power divided by the efficiency.
W407375.0 7465outin =×== ηP P
Solving Equation 16.1 for the line current, we have
A3.168.02203
4973
cos3 rms
inrms =
××==
θ V
P I
P16.8* %100regulationspeedload-full
load-fullload-no ×−
=n
n n
%100
1760
17601800=
−=
%27.2=
P16.9 sradian 3.18460
21760
60
2=×=×=
π π ω m m n
Nm2.203.184
7465outout =
×==
m
P T
ω
P16.10 θ cos3 rmsrmsload-fullin, I V P =
83.0354403 ×××= kW14.22=
outload-fullin,load-fullloss, P P P −=
746.02514.22 ×−=
kW49.3=
%100in
out ×=P
P η
%10014.22 746.025 ××=
%23.84=
θ cos3 rmsrmsload-noin, I V P =
30.05.64403 ×××=
kW49.1=
545
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 7/35
%100regulationspeedload-full
load-fullload-no ×−
=n
n n
%1001750
17501797×
−=
%69.2=
P16.11* At full load, we have:
sradian3.18360
21750
60
2=×=×=
π π ω m m n
Nm8.1013.183
74625outout =
×==
m
P T
ω
Starting with rated voltage, we have:Nm6.2032 outstart =×= T T
Starting with reduced line voltage, we estimate:2
start 440
2206.203
=T
Nm9.50=
P16.12 Rearranging Equation 16.9 we have 33.33120/)4(1000120/ === P n f s
Hz. Then, with a two-pole motor, we would have a speed of 2000 rpm
which is the highest speed attainable with a synchronous motor using thisfrequency. For six poles, we have a speed of 666.7 rpm, and for eight
poles we have a speed of 500 rpm.
P16.13 W52365440(14)0.8factorpowerin ==×= rms rms I V P
W48497465.6 =×=out P %6.92%100)/( =×=η in out P P
P16.14 First, we determine the value of the constant K .
( ) ( )
633
load2
load 102.487
6021000
74675.0 −×=
×
×===
π ωω m m
P T K
The equation for the torque--speed characteristic shown in Figure P16.10
is:
m T ω1.020 −=
546
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 8/35
Equating the motor torque to the load torque, we have:
m m K ωω 1.0202 −=
Solving, we find the equilibrium speed as:
sradian5.124=m ω (The negative root is extraneous.)
rpm11892
60=×=
π ωm m n
The torque is:( ) Nm55.75.1241.020 =−=T
The power is:
W1.940==
T P m ω
hp1.260=
P16.15 (a) At nearly zero speed, the torque required by the load is 40 Nm but
the motor produces only 20 Nm. Thus, the system will not start from a
standing start.
(b) To find the speeds for which the system can run at a constant speed,
we equate the load torque to the motor torque.
102020m
m 800
ω
ω −=+
Solving, we find two roots which are 97.251 =m ω rad/s and
0.1542 =m ω rad/s.
(c) If the speed becomes slightly less than the lower root, the load
torque exceeds the motor torque and the system slows to a stop. Thus,
the system is not stable when running at the lower root.
(d) If the speed becomes slightly less than the higher root, the loadtorque is less than the motor torque and the system speeds up. Thus, the
system is stable when running at the higher root.
P16.16 (a) For no load, we have:
( ) m m T ωωπ −== − 60100 2out
547
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 9/35
Two speeds satisfy this equation, π ωω 60and0 == m m .
(b) To find maximum torque, we have:
( )m
m d
dT ωπ
ω
260100 2out −== −
Solving, we find that the speed for maximum torque is:ω π 30=m
and the maximum torque isNm8.88maxout, =T
(c) outout T P m ω=
= ( ) 22 6010 m m ωωπ −−
To find the maximum output power, we have:
( )22out 3120100 m m m d
dP ωπωω
−== −
Solving, we find:ω π 40=m
(The root 0=ω corresponds to minimum output power.) Them
maximum power is:W9922maxout, =P
= hp13.3
(d) The starting torque is zero. The motor could be started with some
other source of mechanical power to get it moving.
P16.17* sradian4.12060
21150
60
2=×=×=
π π ω m m n
W18064.12015outout =×== m T P ω
hp42.2out =P
θ cosrmsrmsin I V 3P =
8.04.34403 ×××=
W2073=
outinloss P P P −=
W267=
548
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 10/35
%100in
out ×=P
P η
%1.87=
P16.18 W25383007463 =+×=+= loss out in P P P %18.88%100
2538
7463%100 =×
×=×=
in
out
P
P η
0.832582203
2538
3cos
rmsrms
in =××
==I V
P θ
Thus the power factor is 83.25%. Usually the power factor is lagging for
induction motors.
P16.19 The no-load speed of an induction motor is very close to the synchronous
speed given by Equation 16.9:
P
f n s =
120
in this case we have f = 60 Hz. Furthermore the number of poles must be
an even integer. Thus we have a 2-pole machine with a no-load speed of
approximately 3600 rpm. Because of mechanical losses, the actual no-
load speed will be slightly lower, perhaps 3590 rpm. In that case the
speed regulation is
%6.2
%1003500
35003590
%100regulationspeed
loadfull
loadfullloadno
=
×−
=
×−
=
−
−−
n
n n
P16.20* In steady-state with no load, we have u B e V A T l== and the current A i is
zero.
(a) If V T is doubled, the steady-state no-load speed is doubled.
(b) If the resistance is doubled, the steady-state no-load speed is not
changed. (However, it will take longer for the motor to achieve
this speed.)
(c) If B is doubled, the steady-state no-load speed is halved.
549
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 11/35
P16.21 Under starting conditions ( )0=u , we have ,,0 A T A A R V i u B e === l and
lB i f A = .
(a) If the source voltage V T is doubled, the starting force doubles.
(b) If R A is doubled, the starting force is halved.
(c) If the magnetic flux density B is doubled, the starting force is
doubled.
P16.22 The output power is W746hp1out ==P . Thus,
20746out ×=×== f u f P
N3.37=f
Solving Equation 16.11 for the current and substituting values, we have:
A6.7415.0
3.37=
×==
lB
f i A
V10205.01 =××== u B e A l
A A A T e R i V +=
1005.06.74 +×=
V73.13=
A T i V P =in
W1024=
%100in
out ×=P
P η
%1001024
746×=
%8.72=
P16.23* When the switch is closed, current flows toward the right through the
sliding bar. The force on the bar is given by:Blf ×= A i
Thus, the force is directed toward the bottom of the page. The starting
current and starting force are:A501.05starting, === A T A R V i
550
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 12/35
N75.483.175.050starting =××== lB i f A
Under no-load conditions in steady state, we have0=A i
u B e V A T l==
Thus, the steady-state speed is
75.03.1
5
×==
lB
V u T
sm13.5=
P16.24 In Problem 16.23, we found that the starting force is 48.75 N. Since
this is greater than the load force (10 N) applied to the bar, the machine
acts as a motor. In steady state, we have:lB i f A == 10
Solving, we find thatA26.10
3.175.0
10=
×==
lB
f i A
A A T A i R V e −=
26.101.05 ×−=V974.3=
The steady state velocity is:
sm076.475.03.1
974.3=
×==
lB
e u A
(a) The power supplied by the voltage source is:
A T i V P =in
26.105 ×=W28.51=
(b) The power absorbed by the resistance is:2
A A R i R P A =
( )226.101.0 ×=
W52.10=(c) The mechanical output power is:
fu P =out = 076.410 ×
= W76.40As a check, we note that outin P P P
A R += .
551
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 13/35
P16.25 In this case, the machine acts as a generator. We have:lB i f A =−= 10
Solving, we find that
A26.10
3.175.0
10−=
×
−=
l=
B
f i A
A A T A i R V e −=
( )26.101.05 −××=
V026.6=The steady-state velocity is
sm181.675.03.1
026.6=
×==
lB
e u A
(a) The power supplied to the voltage source is
A T i V P =out
= 26.105
× = W28.51(b) The power absorbed by the resistance is
2A A R i R P
A =
( )226.101.0 ×=
W52.10=(c) The mechanical input power is
fu P =in
181.610 ×=
W81.61=As a check, we note that outin P P P
A R += .
P16.26 (a) When the switch closes, current flows clockwise in the circuit. By the
right hand rule, the field created by the current in the rails points into
the page in the vicinity of the projectile. The force on the projectile is
given by f Bl ×= A i . Thus, the force is directed toward the right-hand
side of Figure P16.26.
(b) Equating the energy stored in the capacitor to the kinetic energy inthe projectile we have
2212
21 mu CV =
Solving for the velocity and substituting values, we have
m/s5773103
)10(1010003
2462
=×
××== −
−
m
CV u
552
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 14/35
(c) The velocity attained will be less than the value computed in part (b)
because of air resistance, friction, energy lost in the resistance of the
circuit. Furthermore, some of the energy may remain stored in the
capacitor or in the magnetic field at the instant that the projectile
leaves the end of the rails.
P16.27* Using the right-hand rule we see that in Figure 16.10, the north pole of
the rotor is at the top of the rotor. Because the north rotor pole is
attracted to the south stator pole, the torque is counterclockwise, as
indicated in the figure.
In Figure 16.11, the north rotor poles are in the upper right-hand and
lower left-hand portions of the rotor. South poles appear in the upper
left-hand and lower right-hand parts of the rotor. Because the northrotor poles are attracted to the south stator poles, the torque is
counterclockwise, as indicated in the figure.
P16.28 The magnetic field in the yoke is nearly constant in magnitude and
direction. Thus, voltages that result in eddy currents are not induced in
the yoke, and laminations are not necessary.
On the other hand as the rotor turns, the field alternates in direction
through the rotor material. This induces voltages that could cause eddy
currents and large power losses if the rotor was not laminated.
P16.29 Equation 16.15 states
m A K E φω=
With constant field current, the magnetic flux is constant. Therefore,
the back emf E
φ
A is proportional to machine speed ω (or equivalently tom
m n ). Thus, we have
( )rpmm n ( )VA E 600 120
1200 240
1500 300
553
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 15/35
P16.30* Converting the speed of 1200 rpm to angular velocity, we have
π π π
ω 4060
21200
60
2=×=×= m m n
Solving Equation 16.15 for the machine constant φK and substituting
values, we have
π π ωφ
6
40
240===
m
A E K
Then we use Equation 16.16 to compute the torque
Nm10.19106
dev =×==π
φ A I K T
W2400devdev == T P m ω
The voltage applied to the armature circuit is
A A A T E I R V +=
240103.1+×=
V253=
P16.31 π π π
ω 4060
21200
60
211 =×=×= m m n
π π π
ω 5060
21500
60
222 =×=×= m m n
592.15
40
200
1
====π π ω
φm
A E K
Nm75.2350
7465
2
dev
dev=
×==
π ωm
P T
A92.14592.1
75.23dev ===φK
T I A
P16.32 Under no-load conditions we haveV47925.0480 =×−=− A A T =A I R V E
π π π
4060
21200
60
2==m n ω =m
812.3
40
479==
π ωm
A =φE
K
Nm906.1=== A loss dev I K T T φ
Then with the load applied:91.5150906.1 =+=+= load loss dev T T T
A62.13812.3
91.51==
φK =
T I A dev
554
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 16/35
V8.452262.13480 =×−=− A A T =A I R V E
rpm1134479
8.4521200
,
, ===−
−−
load full A
load full A load no load E −full
E n n
%82.5
%100regulation
=
×speed−
=−
−−
load full
load full load no
n
n n
P16.33* The voltage induced in each armature conductor is given byu B e A = l
where is the length of the conductor,m3.0=l T 1=B is the flux
density, and u is the linear velocity of the conductor which is given by the
product of the number of revolutions per second 20=60m n and the
circumference of the rotor.
sm41.0220 π π =××=u
Thus, we haveV770.343.01 =××== π u B e A l
The machine is designed to have 240≅A E (because in a good design A A I R
is small and A T E V ≅ ). Thus, we need to have
6477.3
240≅==
A
A
e
E N
armature conductors connected in series.
P16.34 (a) The magnetic intensity in the air gap is
mA10500105.12
32502
2
2 33
gap
×=××××
== −l
F NI H
The flux density is
= T 6283.010500104 370 =×××= −π µ H B
(b) The force exerted on each armature conductor is given byB I f A A l=
where is the length of an armature conductor. Thus, wehave
m5.0=A l
N425.96283.05.030 =××=f
555
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 17/35
P16.35 V168102.1180 =×−=−= A A T A I R V E
π π π
4060
21200
60
2=×=ω ×= m m n
337.1
40
168===
π ω
φ
m
A E K
N37.1310337.1dev =×== A I K T φ
hp2.252W1680devdev === m T P ω 2
F
F A A R
V R I P 2
heat +=
150
1802.110
22 +×=
W336=
P16.36* Because the field current is constant so is K ϕ . Then because thedeveloped torque is constant, I A is constant. For V T = 200 V, we have
V15010520011 =×−=−= A A T A I R V E
and for V T = 250 V, we haveV20010525022 =×−=−= A A T A I R V E
Then we have
rpm1600150
2001200
1
212 ==
A
A
E =
E n n
P16.37 For a permanent-magnet motor there are no field losses and K ϕ is
constant. Under no-load conditions, we haveV23317240 =×−=− A A T =A I R V E
π π π
5060
21500
60
2==m n ω =m
483.150
233==
π ωm
A =φE
K
Nm438.1=== A loss dev I K T T φ
Under loaded conditions, we have
rad/s1.136
60
21300
60
2==
π π m n =ωm
=A =m K E φω 201.9 V
A444.57
9.201240=
−=
−
A
A T
R =A
E V I
=in W1307=A T I V P
W7.1951.136438.1 =×== m loss loss T P ω
556
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 18/35
W4.903=−=−= loss A A loss dev out P I E P P P
%1.69%100 =×in
out
P =
P η
P16.38 Under full-load in a well designed machine, I A should be much larger thanI F because all of the field power I F V T is converted to heat while the
output power is part of the armature power I A V T . For high efficiency,
the field power must be small compared to the output power. Perhaps an
acceptable ratio would be I A /I F = 20.
P16.39* (a) V7.4291031.0440 =×−=−= A A T A I R V E
π π π
ω =m 5060
21500
60
211 =×=×m n
736.250 7.4291
== π m
A = ω φ E K
N8.281103736.2dev =×== A I K T φ
hp33.59kW26.441devdev === m T P ω 2 =R kW061.1=A A R I P
A
outdevrot P P P −=
hp9.330kW96.6746.05026.44 ==×−=
(b) Since we are assuming that the rotational power loss is
proportional to speed, we can write:
W31.4469601
rot m
m
m P ω ω
ω =×=
Also, we have
m A I K P ωφ=dev
m
A
A T
R
E V K ωφ
−=
m A
m T
R
K V K ω
φωφ
−=
Substituting values, we have23
dev 86.741004.12 m m P ωω −×=
With no load, we have 0out =P , and
rotdev P P =
12 m m m ωωω 31.4486.741004. 23 =−×
Solving, we find
557
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 19/35
rpm1530 or 24.160 == m m n ω
(The root 0=ω is extraneous.)m
P16.40 For operation at 1000 rpm, we have
rad/s7.104602 == π ω m m n
146.17.104
120==
m
A =E
K φ ω
The torque-speed relationship is given by Equation 16.27:
( )
−=−=60
2dev
π φ
φφω
φm T
A m T
A
n K V R
K K V
R
K T
Substituting values, we obtain
m
m
n T
n T
03438.076.6860
2146.1240
4
146.1
dev
dev
−=
−=π
A sketch of the torque-speed characteristic is:
P16.41 (a) The field current is
A5.1160
240
adj
==+
=R R
V I
F
T F
From the magnetization curve shown in Figure P16.41, we find that
V200≅A E with A5.1=F I and rpm1000=m n . Neglecting lossesat no load, we have 0=A I and V240== T A V E . Since A E is
proportional to speed, the no-load speed is
rpm1200rpm1000200
240load-no =×=n
(b) When the speed drops by 6%, we have
558
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 20/35
1200 rpm112894.0 =×=m n
sradian1.11860
2=×
π 1128=ωm
240 V6.22594.0 =×=A E
A6.91.5 6.225240 =−=−A
A T R =A E V I
W2166devout === A A I E P P
Nm34.18devdevload ===
m
P T T
ω
W3605.1240 =×==F F T I V P 2 =R W2.138=A A I R P
A
P16.42* (a) The field current is
A0.1240240
adj
==+ R R
= V I F F
T
From the magnetization curve shown in Figure P16.41, we find that
V165=A E with A0.1=F I and rpm1000=m n . Neglecting losses
at no load, we have 0=A I and V240== T A V E . Since E A is
proportional to speed, the no-load speed is:
rpm1455rpm1000165
240load-no =×=n
(b) When the speed drops by 6%, we have
rpm136794.01455 =×=m n
sradian2.14360
21367 =×
π =ωm
240 V6.22594.0 =×=A E
A6.95.1
6.225240=
−=
−
A
A T
R =A
E V I
W2166devout === A A I E P P
Nm13.15devdevload ===
m
P T T
ω
W2400.1240 =×== T F F I V P W2.1382 == A A R I R P A
559
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 21/35
P16.43 A5.0400
200
adj
==+
=R R
V I
F
T F
A7.05.02.1 =−=−= F L A I I I
V7.1997.05.0200 =×−=−= A A T A I R V E
Since 0out =P , we have W8.139devrot === A A I E P P
P16.44 A165.210
200
adj
=+
=+
=R R
V I
F
T F
From Figure 16.19, we find that V320ref, =A E at rpm1200ref, =m n . Next,
we determine the value of the machine constant φK .
546.2
60
21200
320
ref,
ref, =×
==π ω
φm
A E K
We assume that the rotational power loss is proportional to speed, which
is equivalent to assuming constant torque.
Nm96.76021200
1000
ref,
rot
rot=
×==
π ωm
P
T
Nm96.20796.7200rotoutdev =+=+= T T T
A68.81546.2
96.207dev ===φK
T I A
V06.193=−= A A T A I R V E
rpm724.0320
06.1931200
ref,ref, =×=×=
A
A m m E
E n n
sradian81.7560
2=×=
π ω m m n
kW16.15outout == m T P ω
( )
in kW54.19=+ F A T = I I V P
%6.77%100in
==P out=
P η
560
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 22/35
P16.45* (a) A0.1200
200
adj
==+
=R R
V I
F
T F
We are given that W50rot =P and V175ref, =A E at
rpm1200ref, =m n . Next, we determine the value of the machine
constant φK .
393.1
60
21200
175
ref,
ref, =×
==π ω
φm
A E K
We assume that the rotational power loss is proportional to speed,
which is equivalent to assuming constant torque.
Nm3980.0
60
21200
50ref,
rotrot =
×==
π ωm
P T
At no load, we have 0out =T and rotdev T T = .
A2857.0393.1
3980.0dev ===φK
T I A
V71.199=−= A A T A I R V E
rpm1369175
71.1991200ref,
ref,load-no, =×=×=A
A m m E
E n n
sradian4.14360
2load-no,load-no, =×=
π ω m m n
(b) We have
m
A
m T
R A
K V I = ω
φω393.1200 −=
−
m A I K T ωφ 940.16.278dev −== ( ) m m m T P ωωω 940.16.278devdev −==
The plots are:
561
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 23/35
P16.46 (a) I A doubles, and the speed remains constant.
(b) I A doubles, and the speed remains constant.
(c) The field current is cut in half, the speed remains constant, and I A
doubles.
(d) The speed increases to 2400 rpm and I A remains constant.
P16.47 A4.4100440
adj
==+
=R R
V I F F
T
A6.45=−= F L A I I I
V7.43705.06.45440 =×−=−= A A T A R I V E
kW96.19dev == A A I E P
562
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 24/35
Nm8.158
602
1200
19960
m
devdev =
×==
π ω
P T
kW22in == L T I V P
%4.81%10022000
74624%100
in
out
=××
=×= P
P η
P16.48* We have
A
m T
A
A T
R A
K V
R
E V I
φω−=
−=
A A I E P =dev
m
A
m T K R
K V P φω
φω−=dev
Substituting values, we obtain7465 × 283333.067.166 m m ωω −=
Solving, we find the two roots and the corresponding armature currents
as
1 A4.21 and 3.174 1 == A m I ω for which %2.87=η
2 A141 and 67.25 2 == A m I ω for which %5.13=η
The first solution is more likely to fall within the rating because the
efficiency for the second solution is very low.
P16.49 (a) W4660in == L T I V P (b) W300== F T F I V P
(c) A8.21=−= F L A I I I
W1.1902 == A A R I R P A
(d) outinrot P P P P P A R F −−−=
= W440
(e) %0.80%100in
out =×=P
P η
P16.50 (a) V1.204=−= A A T A I R V E
052.2)60/2(950
1.204===
π ω ϕ
m
A E K
W2684in == A T I V P
W23387463out =×=P
563
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 25/35
W5.1932 == A A R I R P A
W5.152outinrot =−−=A R P P P P
W5.2490rotoutdev =+= P P P
Nm03.25602950
5.2490
m
dev
dev=
×==
π ω
P
T
(b) Nm533.1
602
950
5.152rotrot =
×==
π ωm
P T
A7470.0rotload-no, ==
φK
T I A
V0.219load-no, =−= A A T A I R V E
srad74.106load-no,load-no, == φ
ωK
E A m
rpm1019load-no, =m n
P16.51* The magnetization curve is a plot of E A versus the field current I F at a
stated speed. Because a permanent magnet motor does not have field
current, the concept of a magnetization curve does not apply to it.
P16.52 (a) With a locked rotor, 0and0 == A m E ω . Then we have
Ω== 6.02012
locked,A
T
I =A V R
(b) and (c)( ) A A A T A A I I R V I E P −==dev
A A T A
I R V I
dP 20 −==
ddev
A
T
R A
V I =
2
A
T
R
V
4
2
maxdev, =P ( )VT V ( )Wmaxdev,P
10 41.67
12 60
14 81.66
564
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 26/35
P16.53 In Problem 16.52, we determined that Ω= 6.0A R . We have the
following equations:2
dev m T K T ω= (1)
(Because load torque is given to be proportional to the square of speed.)
A I K T φ=dev (2)
A
m T
R A
K V I = (3)
φω−
Using the given data rpm800=m n (or equivalently, ),77.83=m ω
A5.3=A I , and V12=T V , we can find the values of the constants:
=φ 11817.0K
=T 61093.58 −×K
Then, we set the right hand sides of Equations (1) and (2) equal and use
Equation (3) to substitute for I A resulting in:
A
m T m R T
K V K
φωφ
−=2K ω
Substituting values and solving for the speed gives the answers which
are:( )VT V ( )sradianm ω ( )rpmm n
10 71.63 694
12 83.78 800
14 95.42 911
(We have discarded the extraneous negative roots of the quadraticequation.)
P16.54* Under no load conditions, we have:V35.125.05.06.12 =×−=− A A T =A I R V E
rot W175.6dev === A A I E P P
Nm1011.55
60
21070
175.6 3
m
rotrot
−×=×
==π ω
P T
3dev
102.110−
×== A I
T
K φ
With the load applied, we have:sradian99.48 and 950 m == ωm n
V96.10m == φωK E A
A274.3=−
=A
A T A R
E V I
565
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 27/35
W88.35dev == A A I E P
W014.5175.61170
950rot =×=P
W87.30rotdevout =−= P P P
W25.41in == A T I V P %83.74%100
in
out =×=P
P η
P16.55 (a) According to Equation 16.34 for a series connected motor, we
have:
( )22
devm F F A
T F
KK R R
V KK T
ω++=
However, in this problem we have 0=+ F A R R , so we have:
2
2
devm F
T
KK V T
ω=
Since the rotational losses are negligible, we have devout T T = . Thus,
torque is inversely proportional to speed squared so we can write:2
1
2
out2
out1
=
m
m
n
n
T
T
2
2
1200300
100
= m n
Solving, we find the speed as:rpm6932 =m n
(b) In theory, the no-load speed is infinite. Of course, rotational
losses will limit the speed to a finite value. However, the speed can
become high enough to damage the machine unless protective
circuits remove electrical power when the load is removed.
P16.56 We have:
in W440020220 =×=A T = I V P ( ) V180=+− A A F = T A I R R V E
W360020180dev =×== A A I E P
W34501503600rotdevout =−=−= P P P
P %41.78%100
in
out =×=P
η
566
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 28/35
P16.57* For A40=A I , we have:
sradian25.9460
2900
60
2m =×=×=
π π ω m n
( ) V196=+−= A A F T A I R R V E
Rearranging Equation 16.30 and substituting values, we have:31099.51
25.9440
196 −×=×
==m A
A F I
E KK
ω
For A20=A I , we have:
sradian0.200201099.51
2083m =××
== −A F
A
I KK
E ω
rpm1910=m n
P16.58 For A40=A I , we have:
sradian25.94602900
602
m =×=×= π π ω m n
( ) V196=+−= A A F T A I R R V E
Rearranging Equation 16.30 and substituting values, we have:
31099.5125.9440
196 −×=×
==m A
A
I F
E KK
ω
dev = W7840=A A I E P
out = W7440rotdev =− P P P
Nm18.83m
devdev =
ω
=P
T
Nm94.78m
outout =
ω=
P T
Nm24.4outdevrot =−= T T T
(Since we assume that P rot is proportional to speed, T rot is constant with
speed.)
Now when the output torque is reduced by a factor of 2, we have:
Nm71.4324.42
94.78dev =+=T
From Equation 16.31, we have:
A00.29dev ==F
A KK
T I
( ) V6.202=+−= A A F T A I R R V E
W5875dev == A A I E P
567
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 29/35
sradian4.134dev
devm ==
T
P ω
rpm1284=m n
P16.59 For A25=A I , we have:
sradian7.12560
21200
60
2m =×=×=
π π ω m n
( ) V5.267=+−= A A F T A I R R V E
Rearranging Equation 16.30 and substituting values, we have:
31015.857.12525
5.267 −×=×
==m A
A F I
E KK
ω
For A10=A I , we have:
sradian0.323101015.85
275
3m =
××==ω −
A F
A
I KK
E
rpm3084=m n
P16.60 For A25=A I , we have:
sradian7.12560
21200
60
2m =×=×=
π π ω m n
( ) V5.267=+−= A A F T A I R R V E
Rearranging Equation 16.30 and substituting values, we have:
31012.857.12525
5.267 −×=×
==m A
A
I F
E KK
ω
dev = W6688=A A I E P
out W6338rotdev =−= P P P
Nm 21.53m
dev =ωdev =P
T
Nm42.50m
out =ωout =P
T
rot Nm784.2outdev =−= T T T
(Since we assume that P rot is proportional to speed, T rot is constant with
speed.)
Now when the output torque is increased by a factor of 2, we have:Nm6.103784.242.502dev =+×=T
From Equation 16.31, we have:
568
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 30/35
A88.34dev ==F
A KK
T I
( ) V6.262=+−= A A F T A I R R V E
W9158dev == A A I E P
sradian39.88dev
devm ==ω
T P
rpm1.844=m n
P16.61 1. Higher power-to-mass ratio.
2. Higher starting torque.
3. Slows down for higher torque loads. Suitable for variable torque
loads.
4. Speed can be higher.
P16.62 A universal motor would be a poor choice for a clock because the speed is
variable with the output torque required. The clock would be very
inaccurate.
A universal motor would be a poor choice for a furnace fan because it
would have too short a service life because of brush and commutator
wear.
A universal motor would be a good choice for a coffee grinder because ofits higher power-to-mass ratio and small amount of time in use.
P16.63 Any ac motor that contains brushes and a commutator is most likely a
universal motor.
P16.64 Three methods to control the speed of dc motors and the types of
motors for which each is practical are:
1. Vary the voltage supplied to the armature circuit while holdingthe field constant. (Separately excited motors and permanent-
magnet motors.)
2. Vary the field current while holding the armature supply voltage
constant. (Shunt connected and separately excited motors.)
569
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 31/35
3. Insert resistance in series with the armature circuit. (Shunt
connected, separately excited, permanent magnet, and series-
connected motors.)
P16.65* See Figures 16.26, 16.27 and 16.28 in the book.
P16.66 For this shunt-connected machine, we have
A T V E = (Because 0=A R .)
F F I K =φ (Because we assume operation on the linear portion of
the magnetization curve.)
adjR R
V I
F
T F +=
m A K E φω=
From these equations, we obtain the following expression for speed:
F
F m KK
R R adj+=ω
Thus, speed (either ω orm m n ) is proportional to the resistance adjR R F + .
To achieve rpm1200=m n , we need
( ) 5.112800
12002550adj =+=+R R F
Ω= 5.62adjR
The lowest speed that can be achieved is for 0adj=
R .rpm3.533
75
50800min, =×=m n
P16.67 Neglecting rotational losses, the no-load speed of a PM motor is
proportional to average applied voltage. To achieve a no-load speed of
1000 rpm, the applied voltage must be:
V06.7121700
1000avg =×=V
588.012
06.7on
==T
T
P16.68* Equation 16.34 gives the developed torque of the series motor.
( )22
devm F F A
T F
KK R R
V KK T
ω++=
Substituting 0=+ F A R R and solving for speed, we have:
570
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 32/35
devT KK
V
F
T m =ω
Thus, speed for a constant torque load is proportional to the applied
voltage. To achieve a speed of 1000 rpm, the average applied voltage
must be:V33.3350
1500
1000=×=T V
667.050
33.33on ==T
T
P16.69 We have m T load K T T ω==dev and 0=+ F A R R . Substituting this into
Equation 16.34 and solving for speed, we obtain:
3
2
F T
T m
KK K
V =ω
Thus, speed is proportional to the 2/3 power of the applied voltage.
V21.27501500
10002/3
=×
=T V
5443.050
21.27on ==T
T
P16.70 (a) A165.210
200
adj
=+
=+
=R R
V I
F
T F
For this field current at a speed of 1200 rpm, from the magnetizationcurve we find E A = 320 V. Thus we have
546.2)60/2(1200
320==
π ωm
A =φE
K
Nm958.7)60/2(1200
1000===
π ωm
rot rot
P T
A66.81546.2
958.7200=
+=
+=
φφ K =
T T
K
T I A rot load dev
(b) With zero speed, we have E A = 0. Thus the initial armature current is
A2353085.0
200==
A
T
R =A
V I
Nm59902353546.2startdev, =×== A I K T φ
571
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 33/35
Thus the starting current is 28.8 times larger than the steady-
state value. This is why additional resistance is usually placed in series
with the armature to start the shunt dc motor.
(c) For a starting current of 200 A, we requireΩ1
200== T
added +A
V R R
Ω915.0085.01 =−=added R
P16.71* Operating at 1400 rpm without added resistance, we have:
W36652560
21400devdev =××===
π ωT E I P m A A
A ( ) A F A T I R R V E +−=
[ ]( ) ( )[ ] 36651.075 =−=+− A A A F A T A I I I R R V I
=A A55.52I (The other root is unrealistically large for a practical
machine.) Then solving Equation 16.31 for KK F , we have
322
10056.955.52
25 −×===A
dev
I F
T KK
After adding series resistance, we have:
( )22
devm F F A added
T F
KK R R R
V KK T
ω+++=
Substituting values and solving for the added resistance, we obtain:
added 379.0=R Ω
P16.72 From highest to lowest voltage regulation the generators shown in Figure
16.30 are:
1. Differential compound connected
2. Shunt connected
3. Separately excited
4. Cumulative compound connected
P16.73 (a) To increase the load voltage of a separately-excited generator,increase V F , reduce R adj, or increase the shaft speed.
(a) To increase the load voltage of a shunt connected generator, reduce
R adj or increase the shaft speed.
P16.74 1. Cumulative long-shunt compound connected.
572
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 34/35
2. Cumulative short-shunt compound connected.
3. Differential long-shunt compound connected.
4. Differential short-shunt compound connected.
P16.75 Voltage regulation is zero for a fully compensated cumulative compoundconnected dc generator.
P16.76
A5.14060
150
adj
=+
=+
=F
F F R R
V I
5.2661000
1300205 === A NL E V V
5.251105.15.266 =×−=−= A A A FL I R E V V
%96.5%1005.251 5.2515.266%100regulationvoltage =×−=×−=FL
FL NL
V V V
515.25.25110out =×== FL L V I P kW
665.2)10(5.12515 22outdev =+=+= A A I R P P kW
1.13660
2==
π ω m m n rad/sec 144.3
80.0
515.2
0.80out
in ===P
P kW
479.0665.2144.3devinlosses =−=−= P P P kW
1.231.136
3144inin ===
m
P T
ω nm 6.19
1.136
2665devdev ===
m
P T
ω nm
P16.77* %667.6%100150
150160%100regulationvoltage =×
−=×
−=
FL
FL NL
V
V V
Ω=== 5.720
150
L
L L I
V R Ω=
−=
−= 5.0
20
150160
FL
FL NL A I
V V R
573
8/14/2019 E16.1 the Input Power to the Dc Motor is Pin
http://slidepdf.com/reader/full/e161-the-input-power-to-the-dc-motor-is-pin 35/35
1.15760
2==
π ω m m n rad/s 10.19
1.157
20150=
×===
m
A A
m
dev dev
I E P T
ω ω Nm
P16.78* In Problem P16.77, we determined that Ω= 5.7L R and .5.0 Ω=A R
1201500
1200150 === A NL E V V 15
5.75.0
120=
+=
+=
L A
A L R R
E I A
5.112== L L L I R V V 1800dev == A A I E P W