Electric Machinery Fundamentals (Solutions Manual) Part 8 pps

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The

resulting plot is

shown below

260

Synchronous Motor V-Curve

250

240

230

220

210

200
350 400 450 500 550 600 650 700
E (V)
A

6-3.

A 2300-V 1000-hp 0.8-PF leading 60-Hz two-pole

Y-connected

synchronous

motor

has

a

synchronous
reactance of

2.8

&

and an armature resistance of 0.4

&
.

At 60 Hz,

its friction and

windage losses are 24
kW, and

its core losses

are

18 kW.

The field circuit

has

a

dc

voltage

of

200 V,

and the maximum

I

F

is 10

A.

The open-circuit characteristic of this motor is shown in

Figure

P6-1. Answer the following

questions
about the motor, assuming that it is

being supplied by

an infinite bus.

(a)
How much field current would be required to make this machine

operate

at

unity

power

factor

when
supplying full

load?

(b)

What

is the motor’s efficiency at full

load and

unity power factor?

(c)

If

the

field current were increased by 5

percent, what would

the new value

of

the armature current be?
What

would the new power

factor be? How much reactive power is

being consumed

or

supplied

by the
motor?

(d)

What

is

the

maximum

torque

this

machine

is theoretically

capable of supplying at unity power factor?
At 0.8 PF leading?

Note:

An electronic version of this open circuit

characteristic can

be

found

in

fil

p61_occ.dat
,

which

can

be

used

with

MATLAB

programs.

Column

contains

field

current

in

amps, and column 2 contains

open-circuit termina

voltage in

volts.

152

S
OLUTION

(a)

At full

load, the

input power to the motor is

IN
P

=
OUT
P

+
m
P
ech

+
co
P

re

+
CU
P

We can’t know the copper

losses until the armature current

is

known, so

we will find

the

input

power

and
armature current

ignoring that term, and then correct

the

input power after we know

it.

(

)(

)

IN
P

1000 hp

746 W/hp

24 kW

18 kW

78
=

+

8 kW
+

=

Therefore,

the

line and phase

current at unity power

factor

is

P

I

I
=

=

=

788 kW

198
=

A

A L
(

)

3

PF

3

(

)
2300 V

1.0

T

V

The copper

losses

due to a current

of 198 A are

2

(

)
=

=

2

(

)
&

=

CU

3

A

A
P

I

3

19
R

8 A

0.4

47.0 kW

Therefore,

a

better
estimate of the input

power at full load

is

(

)(

)

IN
P

1000 hp

746 W/hp

24 kW

18 kW

47
=

+

kW

83
+

5

kW
+

=

and

a

better
estimate of the line

and phase current

at unity

power factor is

153

P

I

I
=

=

=

835 kW

210
=

A

A L
(

)

3

PF

3

(

)
2300 V

1.0

T
V

The phasor

diagram of this

motor operating

a

unity power factor

is shown

below:

A
I

V
⎞

A
E

R

I

A

A

jX

I

S

A

The phase voltage of this

motor

is 2300 /

3

= 1328 V.

The required internal generated voltage

is

=


E

V


I

I

A
A
⎞

A S A
R

jX

1328

0

V

(

)
0.4

(

210

0

A
)

(
2.8

)(

210

0

A

)

E

A

=

°



&

°

j


&

°

E
A
1376

25.3
=



V

°

This

internal

generated

voltage corresponds to a terminal voltage of

would require a field current of 4.6 A.

(b)

The motor’s efficiency at full load

and unity

power factor is

⎜

OUT
P

100%
=

⋅

746 k
=

W

100%
⋅

89.3%
=

3

(

)
1376

=

2383 V

. This voltage

IN
P

835 kW

(c)

To

solve this

problem, we will temporarily ignore the effects of the armature

resistance

R
A

.

If
R
A

is

ignored, then

E

A

sin

™

is

directly proportional to

the power supplied by the

motor. Since the power

supplied by the motor does

not

change when

I

F

is changed,

this quantity will

be a constant.

If the field current

is increased by 5%,

then the new field current will

be

4.83

A,

and

the

new

value of

the

open-circuit terminal voltage will be 2450 V. The new value of

Therefore, the new torque

angle

™
will

be

E
A

will be 2450 V

/

3

=

1415 V.

sin

E

1
A
sin

sin

1376 V
™

™

sin

(

)
1 1
=

=

25.




3

24.6


°

=



°

2 1
E
A2
1415 V

Therefore, the new armature

current will

be

⎞



A
V

E

I

1328

0

V

°

141



5

-
25

.3

V
°

214.5

3.5

A

=

=

=

°

A
0.4

2.8
+

+

&

A S
R

jX

j

The

new current is about the same as before, but the phase angle has become positive.

The

new

power
factor is

cos 3.5
°
= 0.998 leading, and the reactive power

supplied by the motor is

3

sin

3

(
2300

V

)(
21
=

=

4.5 A
)
sin

(
3.5

)

52.2 kVAR
°

=

T L
Q

V

I

⎝

(d)

The maximum torque possible

at unity

power factor (ignoring

the

effects

of

3

3
(
1328 V
)

(
1376 V
)

R
A

) is:

⎮

⎞

A
V

E

=

=

ind,max

1 min

2

rad

5193 N

m
=

⊕

⎤

m
S
X

(

)
3600 r/min



(

)
2.8

&

60 s

1 r

154

If we are ignoring the resistance of the motor, then the input power would be 788

kW

(note

that

copper
losses

are ignored!). At a power

factor

of

0.8 leading, the current

flow will be

P

I

I
=

=

=

788 kW

247
=

A

A L
(

)

3

PF

3

(

)
2300 V

0.8

T
V

so

I

A
247

36.8
=

7

A
°

.

The

internal generated

voltage at 0.8 PF leading (ignoring copper losses) is

=


E

V


I

I

A
A
⎞

A S A
R

jX

1328

0

V

(

)
2.8

(
247

36.87

A
)

E

A

=

j
°



&

°

E
A
1829

17.6
=



V
°

Therefore,

the

maximum torque at

a

power factor of 0.8 leading is

3
3
(
1328 V
)

(
1829 V
)

⎮

⎞

A
V E

=

=

ind,max

1 min 2 rad
6093 N
=
m
⊕

⎤

m
S
X

(

)
3600 r/min



(

)
2.8
&

60 s 1 r

6-4.

Plot the V-curves

(

I

A
versus

I

F
)

for the synchronous

motor of

Problem

6-3 at no-load, half-load,

and full-
load

conditions.

(Note

that an

electronic version of the open-circuit

characteristics

in Figure P6-1

is
available at the book’s Web site. It may simplify the

calculations

required by

this

problem.

Also,

you

may
assume that

R
A
is negligible

for

this calculation.)

S
OLUTION

The

input power

at no-load, half-load and full-load

conditions is given below.

Note that we

are

assuming that

R
A
is

negligible in

each case.

IN
P

,nl
24

kW

18 kW
=

+

42 kW
=

(

)(

)

IN
P

,half

500 hp

746 W/hp

24

kW

18 kW

3
=

+

73

kW
+

=

(

)(

)

IN
P

,full

1000 hp

746 W/hp

24 kW

18 kW

78
=

+

8 kW
+

=

If the power factor

is adjusted to

unity,

then armature currents will be

I

P

42 kW
=

=

10.5 A
=

A
,nl

3

PF

3

(

)
2300 V

(
1.0
)

T
V

I

P

373 kW
=

=

93.6 A
=

A,fl

3

PF

3

(

)
2300 V

(
1.0
)

T
V

I

P

788 kW
=

=

198 A
=

A,fl

3

PF

3

(

)
2300 V

(
1.0
)

T
V

The corresponding

internal

generated

voltages at unity power

factor are:

=


E

V

I

A

S
⎞

jX

A

(

)(

)

E
A,nl
1328

0

V
=

j
°

2.8


10.5
&

0

A

°

1328.
=

3

1.27



V
°

(

)(

)
E

,h
A

alf

1328

0

V
=

j
°

1.5


93.6
&

0

A

°

1354
=

11.2



V
°

(

)(

)

E
A,full
1328

0

V
=

j
°

2.8


198
&

0

A

°

1439
=

22.

7


V
°

These

values

of

E

A

and

™

at

unity

power

factor

can

serve

as

reference

points

in

calculating

the

synchronous motor V-curves. The MATLAB program to solve this problem

is shown below:

155

% M-file: prob6_4.m

% M-file

create

a plot of armature current versus

field

%

current

for the synchronous motor of Problem 6-4

at

%

no-load, half-load, and

full-load.

% First,

initialize the field current values (21 values

% in the

range

3.8-5.8 A)

If =

2.5:0.1:8;

% Get

the OCC

load p61_occ.dat;
if_values =

p61_occ(:,1);
vt_values =

p61_occ(:,2);

% Now

initialize all other values

Xs

= 1.5;

% Synchronous

reactance

Vp

= 1328;

% Phase voltage

% The

following

values of Ea and delta are for

unity

% power factor. They

will serve as reference values

% when calculating

the V-curves.

d_nl = -1.27 *

pi/180;

% delta at

no-load

d_half =

-11.2

* pi/180;

% delta at

half-load
d_full =

-22.7

* pi/180;

% delta

at

full-load
Ea_nl

= 1328.3;

% Ea at no-load
Ea_half = 1354;

% Ea at half-load
Ea_full = 1439;

% Ea at full-load

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% Calculate

the actual Ea corresponding

to each level

% of field current

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Ea =

interp1(if_values,vt_values,If) /

sqrt(3);

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% Calculate

the armature

currents

associated with

% each value of Ea

for the no-load case.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%

First, calculate delta.

delta

= asin (

Ea_nl ./ Ea .* sin(d_nl)

);

% Calculate

the phasor Ea

Ea2

= Ea .* (cos(delta)

+

j .* sin(delta));

% Now

calculate Ia

Ia_nl

= ( Vp - Ea2

) / (j *

Xs);

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% Calculate

the armature

currents

associated with

% each value of Ea

for the half-load

case.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%

First, calculate delta.

delta

= asin (

Ea_half ./ Ea .* sin(d_half)

);

% Calculate

the phasor Ea

Ea2

= Ea .* (cos(delta)

+

j .* sin(delta));

156

% Now

calculate Ia

Ia_half = (

Vp - Ea2 ) /

(j

* Xs);

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%

Calculate the armature currents associated

with

% each value of Ea

for the full-load

case.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%

First, calculate delta.

delta

= asin (

Ea_full ./ Ea .* sin(d_full)

);

% Calculate

the phasor Ea

Ea2

= Ea .* (cos(delta)

+

j .* sin(delta));

% Now

calculate Ia

Ia_full = (

Vp - Ea2 ) /

(j

* Xs);

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% Plot the v-curves

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

plot(If,abs(Ia_nl),'k-','Linewidth',2.0);

hold on;

plot(If,abs(Ia_half),'b ','Linewidth',2.0);
plot(If,abs(Ia_full),'r:','Linewidth',2.0);
xlabel('\bfField Current

(A)');
ylabel('\bfArmature Current

(A)');

title

('\bfSynchronous Motor V-Curve');

grid on;

The resulting

plot

is

shown

below. The flattening

visible to the right of the V-curves is due to magnetic
saturation in the machine.

6-5.

If a 60-Hz synchronous motor

is to be

operated at

50 Hz, will its synchronous reactance be the

same as at

60

Hz, or will

it change?

(Hint:

Think about the derivation

of

X

S
.)

157

S
OLUTION

The

synchronous

reactance

represents the effects of the armature reaction

voltage

E
stat
and

the
armature self-inductance. The

armature reaction voltage is caused by

the armature

magnetic field

B

S
, and
the

amount of voltage is directly proportional to the speed with which the magnetic field sweeps

over

the

stator surface.

The

higher the frequency, the faster

B

S

sweeps over the stator, and

the

higher the armature

reaction voltage

E
stat
is. Therefore, the armature reaction

voltage

is

directly

proportional

to

frequency.

Similarly, the reactance of the armature self-inductance is directly

proportional

to

frequency,

so

the

total
synchronous reactance

X

S
is

directly

proportional

to

frequency.

If the frequency is

changed from 60 Hz

to

50

Hz, the synchronous

reactance

will be

decreased

by a factor

of 5/6.

6-6.

A 480-V

100-kW 0.85-PF leading 50-Hz six-pole Y-connected synchronous motor has a

synchronous
reactance of 1.5

&

and

a

negligible armature

resistance.

The

rotational losses are also to be ignored. This
motor

is

to

be

operated

over a continuous

range of speeds from 300 to

1000 r/min, where the speed changes
are

to

be

accomplished by

controlling the system frequency

with a solid-state

drive.

(a)

Over what range

must

the

input frequency be

varied to provide

this speed control range?

(b)

How large is

E

A
at the motor’s rated conditions?

(c)

What

is the maximum

power the motor can

produce at

the

rated conditions?

(d)

What

is the largest

E

A
could

be at

300 r/min?

(e)

Assuming

that the applied voltage
V
⎞

is derated by the same amount as

E

A
, what is the maximum
power the motor could supply at 300

r/min?

(f)

How does the power capability of a synchronous

motor

relate to its speed?

S
OLUTION

(a)

A

speed of 300 r/min

corresponds to a frequency

of

(