e16.1 the input power to the dc motor is pin

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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



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



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



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



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



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



%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



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



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



%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



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



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



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



(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



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



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



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



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



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



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



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



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



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



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



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



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



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



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



sradian4.134dev

devm ==

T

P ω

rpm1284=m n


sradian7.12560

21200

60

2m =×=×=

π π ω m n

( ) V5.267=+−= A A F T A I R R V E


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


sradian7.12560

21200

60

2m =×=×=

π π ω m n

( ) V5.267=+−= A A F T A I R R V E


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



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



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



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



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



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



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

e16.1 the input power to the dc motor is pin

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