Electric Machinery Fundamentals (Solutions Manual) Part 10 ppt

(e)

I

A


(f)

co
P

nv


+

+

(g)

mech

core
P

P

stray
P

S
OLUTION


(a)

Since

this machine has 8 poles, it

rotates

at a speed of


120

120

(
50 Hz

)

n

f
e
=

=

1500 r/
=

min

P
m
4


If the output power is

100

kW, the output

torque is

(

)

⎮



ou
P

t
=

=



100,

000 W


637 N
=

m
⊕


load

⎤

m

(

)
1500 r/min

2


r



ad

1 min





(b)

The input power

is

1 r

60

s




P



OUT
P


100 kW
=

=



110 kW
=

IN

⎜

0.91

(c)

The mechanical

speed is


m
n

=

1500

r/min


(d)

The armature current

is




P

I

I
=

=



=



110 kW


156
=

A

A

L

(

)

3

PF

3

(

)
480 V

0.85

T
V


I

A
156

31.8
=



A
°

Therefore,

E
A
is


=


E

V



I

I

A

A
⎞

A

S

A
R

jX

(
277

0

V
)

(
0.08

)(

)
156

31.8

A

(
1.0

)(

)
156

31.8

A

E
A
=



°



&



°

j


&

°

E
A
375

21
=



.8


V
°


(e)

The magnitude of

the

armature current

is 375 A.

=



(f)

The power converted from electrical to

mechanical

form

is given

by the equation

conv IN C
P

P

U
P


2

(

)
=

=


2

(

)
&

=

CU

3

A

A
P

I

3

1
R

56 A

0.08

5.8 kW


=



=



=

conv IN C
P

P

U
11
P


0 kW

5.8 kW

104.2 kW


(g)

The mechanical, core, and stray losses are given by the equation


+
+ = 
=



=

mech core
P

P

stray
P

conv
P

OUT

1
P

04.2 kW

100 kW

4.2

kW


6-12.

The Y-connected synchronous motor

whose nameplate is shown

in Figure 6-21

has a per-unit

synchronous
reactance

of 0.90 and

a

per-unit resistance of 0.02.

(a)

What

is the rated input power of this motor?


(b)

What

is the magnitude of

E
A
at

rated conditions?

164

(c)

If the

input

power of

this motor is

10

MW,

what is the

maximum reactive

power

the motor

can
simultaneously supply?

Is it the

armature current or

the field current

that limits the reactive

power
output?


(d)

How much power does the field circuit

consume at

the

rated

conditions?


(e)

What

is the efficiency of this motor at

full load?


(f)

What is the output torque of the motor at the rated conditions?

Express

the

answer

both

in

newton-
meters

and in pound-feet.





S
OLUTION

The base quantities for this

motor are:


T
V

,base
=

6600 V

6600 V

3811 V
V

⎞

,base

=

=


3

=

=

,base ,base

14
A
L
I

I

04 A

=

=

(

)
=

(

)(

)

=

base

rated
S

P


3

P

T

L
V

F
I

3

6600 V

1404

A

1.0

16.05 MW


(a)

The rated input power

of

this motor is

(

)(
=

=



)(

)

=

IN

3

P
T

L

F
P

V

I

3

6600 V

1404

A

1.0

16.05 MW


(b)

At rated conditions,

⎞
V


1.0

0

pu
=

°

and

I
⎞

1.0

0

pu
=



°

, so

E
A
is given in per-unit

quantities

as

=


E

V



I

I

A

A
⎞

(

)
1

0

A

S

A

R

jX

(
0.02
)(
1.0

0

)

(
0.90
)(

)
1

0

E
A
=



°





°

j




°

E
A
1.33

42
=



.6


pu
°

The base phase voltage

of this motor is

6600 /

3

= 3810 V,

so

E
A
is

A
E

(

)
1.33

42.6
=





(

381
°

0 V
)

5067
=

42.6




V
°


(c)

From

the capability diagram, we

know that there are

two possible

constraints

on

the

maximum
reactive power—the maximum stator current and the

maximum rotor current.

We will have to

check

each
one separately,

and limit the reactive power

to the lesser of the two limits.


The stator

apparent power limit defines a maximum safe stator current.

This

limit

is

the

same

as

the

rated
input

power

for

this

motor,

since

the

motor is rated at unity power factor. Therefore, the stator apparent



165

power

limit

is

16.05

MVA. If the input

power is 10 MW, then

the maximum reactive power that still
protects the stator current is


2 2
Q

S

P
=



(

)
16.

=

05 MVA

(
10

MW


)
2 2

12.6 M
=


VAR


Now we must determine the rotor current limit. The per-unit power supplied to the motor

is

10

MW

/

16.05

MW = 0.623.

The maximum

E
A
is 5067 V

or

1.33 pu,

so

with

E
A
set to maximum

and the motor
consuming 10 MW, the torque

angle (ignoring

armature resistance) is

(

)

(

)

sin

1

1


S
X

P

sin



0.90

0.
=

=

623





24.9
=

°

™

3

⎞

A
V

E

(

)
1.0

(
1.

)
33


At rated voltage

and 10 MW

of

power supplied,

the

armature current

will be




⎞



A
V

E

I


1

0


°

1


.33


2
4.





°
9

0.663

20.2

pu

=

=

=



°

+
A
0.90

A S
R

jX

j


In

actual amps, this

current is

A
I

(

)
1404

A

(
0.663

20.2

)

931

20
=



.2

A
°

=



°


The

reactive power

supplied

at the conditions of

maximum
E
A
and

10 MW power is

3

sin

3
(

)
3811 V

(
=

=

931 A
)
sin

(
20.2

)

3.68 MV
°

AR
=
⎞

⎝

A
Q

V

I




Therefore,

the

field

current limit

occurs before

the stator current limit
for these conditions, and the
maximum reactive power that

the

motor can supply is 3.68 MVAR under these conditions.


(d)

At rated conditions, the field circuit consumes

(

)
=

=

(

)

=

field

F

F
P

V

12
I


5 V

5.2

A

650 W


(e)

The efficiency of this

motor

at

full load is

(

)(

)

⎜

OUT
P


100%
=

⋅

2100
=

0 hp

746 W/hp


100%
⋅

97.6%
=


IN
P

16.05 MW


(f)

The output

torque in SI and

English

units

is

(

)(

)

⎮

OUT
P




2
=

=

1000 hp

746 W/hp

124,

700 N
=

m
⊕


load
⎤

m

(

)
1200 r/min


1 min

2


rad

60 s

1 r

⎮

load


5252

P


5252

(

)


21000

hp



(

)
1200 r/min

91,

910 lb

ft

=

=

=

⊕

m
n


6-13.

A

440-V

three-phase

Y-connected synchronous motor has a synchronous reactance of 1.5

&
per phase.

The

field

current

has been

adjusted so that the torque angle

™
is 28
°
when the power supplied

by

the
generator is 90 kW.

(a)

What

is the magnitude of the internal generated voltage

E

A
in this machine?

(b)

What

are the magnitude and angle of the armature current

in the

machine?

What

is

the

motor’s

power
factor?


(c)

If the field current remains

constant, what is

the

absolute maximum power this motor could

supply?



166



S
OLUTION


(a)

The power supplied to the motor is

90 kW. This

power is

give by the equation

P

A
V

E

sin
=

3

⎞

™

X

S

so the magnitude of

E
A
is

(

)
1.

(
5

90

kW
)

E



S
X

P




&




=

=

=

3

sin

3
(

)
254 V

sin

28
⎞

™

377 V


V
A

°


(b)

The armature current

in

this machine is given by




⎞



A
V

E

254

0

V


°


I


3


77


28


°





129

24

A

=

=

=



°

A
1.5

jX

S
j


The power factor of the motor is

PF =

cos 24º =

0.914

leading.


(c)

The maximum power that

the

motor

could supply at this field current

3

3
(

)
254 V

(
377 V
)

P



⎞

A
V

E




=

=

=

max
X

S

1.5

&

191.5 kW



6-14.

A

460-V,

200-kVA, 0.80-PF-leading,

400-Hz,

six-pole, Y-connected

synchronous motor has negligible
armature resistance and a synchronous reactance of 0.50 per unit.

Ignore all

losses.

(a)

What

is the speed of

rotation of this motor?


(b)

What

is the output

torque of this motor at the rated conditions?


(c)

What

is the internal generated voltage

of

this motor at the rated conditions?

(d)

With the field current remaining at the value present in the

motor

in

part

(c)
, what is the maximum
possible output power

from

the machine?




S
OLUTION


(a)

The

speed of rotation of this

motor

is

120

120
(

)
400 Hz

n



f

e
=

=

=

sync
P

8000 r/min

6


(b)

Since

all losses are ignored,

=

=

(

)
⋅

=

(

)

=

.

The output


torque of this

motor is

IN ,rated OUT ,rated
P

P



rated
PF
S



200 kVA

0.8

160 kW



⎮

OUT
P
=

=



160 kW


191 N
=

m
⊕

load
⎤

m

(

)
8000 r/min


1 min

2


rad

60 s

1

r


(c)

The phase

voltage of this motor is 460 V /

3

=

266

V.

The

rated armature current of

this motor is




P

I

I
=

=



=



160 kW


251
=

A

A L
(

)


3

PF

3

(

)
460 V

0.80

T
V




Therefore,

I

A
251

36.87
=




2


A
°

.

The base impedance of this

motor

is


(

)
2

Z

base

3
V
⎞

,base






3

266 V


1.06

=

=

=

&

S
base

200,

000 VA




167

so the actual synchronous reactance

is

(

)
0.50 pu

(
1.06

)

0.53

X

S
=

&

=

&

. The internal generated voltage
of

this machine at

rated conditions

is given

by


=


E

V

I

A
S
⎞

jX

A


266

0

V

(

)
0.53

(
251

36.87

A
)

362

17.1

V

E
A

=



°

j


&



°

=





°


(d)

The maximum power that

the

motor

could supply at these conditions is

3

3
(
266 V
)(
362 V
)

P



⎞

A
V

E

=

=

=

MAX
X

S

0.53

&

545 kW



6-15.

A 100-hp 440-V

0.8-PF-leading


-connected

synchronous motor has an armature resistance of 0.22

&
and a

synchronous reactance of 3.0

&
.

Its

efficiency at full load

is 89

percent.

(a)

What is

the

input

power to the motor at

rated conditions?


(b)

What is the line current of the motor at rated conditions?

What

is

the

phase

current

of

the

motor

at
rated conditions?


(c)

What

is the reactive power

consumed by

or

supplied by the motor at

rated conditions?


(d)

What

is the internal generated voltage

E

A
of

this motor at rated conditions?


(e)

What are the stator

copper losses in

the

motor at

rated

conditions?


(f)


What

is

P
conv
at

rated conditions?


(g
) If

E

A
is decreased by

10 percent, how much reactive power will be consumed by or supplied by

the
motor?




S
OLUTION


(a)

The input power

to the motor at rated conditions is

(

)(

)

P



OUT
P

100 h
=

=

p

746

W/hp


83.8 kW
=


IN

⎜

0.89


(b)

The

line

current

to the motor

at rated conditions is


I



P




83.8 kW
=

=



137 A
=

L
3

PF

3

(

)
440

V

(
0.8
)

T
V


The

phase

current

to the motor at rated conditions is


I

L


137

A

79.4

A

I
⎞

=

=

=

3

3


(c)

The reactive power supplied by

this motor to the power

system at

rated conditions

is

(

)(
=

=

)

°

=

rated

3

⎞

A

sin
Q

V

I

⎝

3

440 V

79.4

A

sin

36.87

62.9

kVAR


(d)

The internal

generated voltage at rated conditions is

=


E

V



I

I

A

A
⎞

A

S

A
R

jX


440

0

V

(
0.22

)(

)
79.4

36.87

A

(
3.0

)(

)
79.4

36.87


A

E
A
=



°



&



j
°



&



°

E
A
603

19

=



.5


V
°


(e)

The stator

copper losses at

rated conditions

are



168