Electric Machinery Fundamentals (Solutions Manual) Part 2 ppt

(a)

This

generator is Y-connected, so

L A
I

I
=

.

At rated conditions, the line and phase current
in this generator is


I

I






1000 kVA


251 A

=

=




P

=

=

at an

angle of –36.87
°

A

L

(
3

3

2300

)

V

L
V


=

=

. The internal generated voltage

of

the

machineThe phase voltage of this
machine

is

⎞

T

/
V

V

3

1328 V

is


=

+
E

V


+
I

I

A

A
⎞

A

S

A

R

jX


1328

0

(
0.15

)(


251

36.87

A
)

(
1.1

)(


251

36.87

A
)

E
A
=



°

+

&





°

j
+

&



°

E
A
1537

7.4

V
=



°


The input power

to this generator is equal to

the

output power plus losses.

The

rated output

power is

(

)(

)

OUT
P

1000 kVA

0.8

800 kW
=

=


2

(

)
=

=

2

(

)
&

=

CU

3

A

A
P

I

3
R

251 A

0.15

28.4 kW

F&
P


=

W
24 kW


co
P

re
=

18 kW


st
P

ray

=

(assumed 0)

=

+

+

+

+

=

IN

OUT
P

P

CU
P

F&W
P


core
P

stray

P

870.4 kW


η



OUT
P


100%




800 kW


100%

91.9%

=

⋅

=

⋅

=

IN
P

870.4 kW


(b)

If the generator

is loaded to rated kVA with

lagging loads,

the

phase

voltage

is

φ
V

=

1328

0


°

V

and


the

internal generated voltage is

E
A
1537

7.4

V
=

°

. Therefore, the phase voltage at no-
load would be

V
⎞

1537

0

V
=



°

.

The voltage

regulation would

be:


1537



1328

RV



100%

15.7%

=

⋅

=

1328


(c)

If the generator

is loaded to rated kVA with leading

loads, the

phase

voltage

is

φ
V

the internal

generated voltage

is



=

1328

0


°

V

and

=

+
E

V


+
I

I

A

A
⎞

A

S

A
R

jX

1328

0

(

)
0.

(
15

251

36.87

A

)

(
1.1

)(


251

36.87

A
)

E
A
=



°

+

&



°

j
+

&



°

E
A

1217

11.5
=



V
°


The voltage regulation

would be:


1217



1328

RV



100%

8.4%

=

⋅

=



1328


(d)

If

the

generator

is

loaded

to

rated

kVA

at

unity

power

factor,

the

phase

voltage

is

V
φ

=

1328

0


°
V

and the internal generated voltage

is


=

+
E

V


+
I

I

A

A
⎞

A

S

A
R

jX

115


1328

0

(
0.15

)(

)
251

0

A

(
1.1

)(

)
251

0

A

E
A
=



°

+

&



°

j
+

&



°

E
A
1393

11.4
=



V
°


The voltage regulation

would be:


1393



1328

RV



100%

4.9%

=

⋅

=

1328


(e)

For

this

problem,

we

will

assume

that

the

terminal

voltage

is

adjusted

to

2300

V

at

no

load
conditions, and see what happens to the voltage as load increases at 0.8 lagging, unity, and

0.8

leading
power factors. Note that the maximum current will be

251

A in any case.

A phasor

diagram

representing
the situation at

lagging

power factor is

shown

below:

E

A


⎝



™

⎝


I

A

By the Pythagorean

Theorem,




⎝

jX

I

V
⎞

S

A

I
R

AA



2

(

)
=

+


cos

+

sin

(

)

2

+

co
⎝

⎝


s



sin
⎝

2
⎝

A

A
E

V
⎞

R

A
I

S

A
X

I

S

A

X

I

A
R

S
I

2

(

)
=



cos

s


in

2

co
⎝

⎝




s

s


in
⎝

⎝

⎞

A

S
V

E

A
X

I

A

S
R

I

A

A
R

I

S

A
X

I


A phasor diagram representing the situation at

leading power

factor

is shown below:

E

A




jX

I

S

A

⎝

I

A

I
R

™

⎝

A

A

⎝


⎞
V


By the Pythagorean

Theorem,

2
(
=

+

cos



sin

)

(

)

2
+

co
⎝

⎝


s

+

sin
⎝

2
⎝

A
A
E

V
⎞

R

A

I

S A
X

I

S A
X

I

A
R

S
I

2

(

)
=



cos

s

+

in

2

co
⎝

⎝




s

s
+

in
⎝

⎝

⎞

A

S
V

E

A
X

I

A

S
R

I

A

A
R

I

S

A
X

I

A phasor diagram representing the situation at

unity power factor is shown

below:

E

A





™


I

A

V
⎞

116



jX

I

S

A



R

I

AA


By the Pythagorean

Theorem,


(
2

2
= +
)
2

A

S
E V
⎞

A
X I
2

(

)
=



2

⎞

A

S
V

E

A
X

I

The MATLAB

program

is shown below takes advantage of this

fact.


% M-file: prob5_4e.m

%

M-file to calculate and

plot the

terminal voltage

% of a synchronous

generator as a

function of load

% for

power

factors of 0.8 lagging, 1.0,

and 0.8 leading.


% Define

values for this

generator

EA

= 1328;

% Internal gen voltage

I = 0:2.51:251;

% Current values (A)
R = 0.15;

% R (ohms)

X = 1.10;

% XS (ohms)

% Calculate

the

voltage for

the lagging PF case

VP_lag = sqrt( EA^2 - (X.*I.*0.8

- R.*I.*0.6).^2

)

- R.*I.*0.8 -

X.*I.*0.6;
VT_lag = VP_lag .* sqrt(3);


% Calculate

the voltage for

the leading

PF case

VP_lead = sqrt( EA^2 - (X.*I.*0.8

+ R.*I.*0.6).^2

)

- R.*I.*0.8 +

X.*I.*0.6;
VT_lead = VP_lead .* sqrt(3);


% Calculate

the voltage for

the unity PF case

VP_unity

= sqrt( EA^2

- (X.*I).^2

);
VT_unity

= VP_unity .* sqrt(3);


% Plot the terminal voltage

versus load
plot(I,abs(VT_lag),'b-','LineWidth',2.0);
hold on;

plot(I,abs(VT_unity),'k ','LineWidth',2.0);
plot(I,abs(VT_lead),'r ','LineWidth',2.0);
title

('\bfTerminal Voltage

Versus Load');
xlabel ('\bfLoad (A)');

ylabel ('\bfTerminal Voltage (V)');

legend('0.8

PF

lagging','1.0 PF','0.8 PF

leading');

axis([0 260

1500 2500]);

grid on;

hold off;

The

resulting plot is

shown below:













117



5-5.

Assume that the field current of the generator in Problem 5-2 has been adjusted so

that

it

supplies

rated

voltage when loaded with rated current at unity power factor.

(You

may

ignore

the

effects

of

answering these

questions.)

R
A

when

(a)

What

is the torque

angle

™
of the generator when supplying

rated current at unity power factor?


(b)

When this

generator

is running at full

load with

unity power factor, how

close is it to

the

static

stability
limit of the

machine?




S
OLUTION


(a)


The

torque

™
angle can be

found

by calculating

E

A

:


=

+
E

V


+
I

I

A
A
⎞

A S A
R

jX

1328

0

(
0.15

)(

)
251

0

A

(
1.1

)(

)
251

0

A

E

A

=



°

+

&



°

j
+

&



°

E
A
1393

11.4
=



V
°

Thus the

torque angle

™
= 11.4
°
.

(b)

The static stability limit

occurs

at

™

=

°

90


.

This generator is a very long way

from

that

limit.

If we

ignore the internal resistance of the generator, the output

power will be given

by


P

A
V

E

sin
=

3

⎞

™

X

S


and the output power is proportional to

sin

™

.

Since

sin

11.4
°=

0.198

, and

sin

90

1.
°=

stability limit

is about 5 times

the

current output power of the generator.


00

, the static



5-6.

A 480-V 400-kVA 0.85-PF-lagging 50-Hz four-pole



-connected generator is driven by a 500-hp diesel
engine

and

is

used as a standby or emergency generator. This machine can also be paralleled with the
normal power supply (a very

large power system) if

desired.

(a)

What

are the

conditions required

for

paralleling the emergency generator with the existing

power
system? What

is the generator’s rate of shaft rotation after paralleling occurs?

118

(b)

If

the

generator

is

connected

to

the power system and

is initially floating on

the line, sketch the resulting
magnetic fields and

phasor

diagram.

(c)

The governor setting

on the diesel is now increased. Show

both by means of house diagrams and

by
means of phasor

diagrams what happens to the generator.

How

much reactive power

does the generator
supply now?


(d)

With the diesel generator now supplying

real power to the

power

system,

what

happens

to the generator as
its field current is increased and

decreased? Show this behavior

both

with phasor diagrams

and

with
house diagrams.




S
OLUTION


(a)

To

parallel

this generator to the large power system, the required conditions

are:


1.

The generator must have

the

same voltage
as the power

system.




2.

The

phase sequence
of the oncoming generator must be the

same as

the phase sequence of the
power system.




3.

The

frequency
of the oncoming

generator

should

be

slightly
higher than the frequency of the
running system.


4.

The circuit breaker connecting the two systems together should be shut when

the

above conditions
are met

and the generator is

in phase
with the power

system.


After paralleling, the generator’s shaft will

be rotating at


120

120
(

)
50 Hz

n



f

e
=

=

1500 r/
=

min

P
m
4


(b)

The magnetic field and phasor diagrams immediately after paralleling are shown below:


BB

E

S R A

I

jX
I

S A

B

net
A
V
⎞


(c)

When

the

governor

setpoints

on

the generator are increased, the emergency generator begins

to

supply

more

power to the loads, as shown below:


f

e





E

A


jX

I

S

A

I

P
P

P
1

sys

A

V
⎞

2

P

G


Note that as the load

increased with

E

A

reactive power.


constant, the generator began to consume a small amount of

(d)

With

the generator

now

supplying power to

the system,

an increase in field current increases the
reactive power supplied to the

loads, and a decrease in field current decreases the reactive

power supplied to
the loads.


119

V
T

I

A1



E

A1





E

A2
E
A3




QQ QQ
3

G

2

1




Q
sys



I

A2
I



⎞
V

jX
I

S A
A3

V
T






I

A2




A2
EE

A1









jX

I

S A

Q Q Q
G

1

2


Q
sys


I

A1
V
⎞


5-7.

A

13.8-kV

10-MVA

0.8-PF-lagging

60-Hz

two-pole

Y-connected

steam-turbine

generator

has

a
synchronous reactance of 12

&

per phase and

an armature

resistance of 1.5

&
per phase.

This

generator

is
operating in

parallel

with a large power

system

(infinite bus).

(a)

What

is the magnitude of

E

A
at

rated conditions?


(b)

What

is the torque

angle of the generator at

rated conditions?


(c)

If the field current is constant, what is the maximum power possible out of this

generator?

How

much
reserve power

or

torque does this

generator

have at full load?


(d)

At the absolute maximum power possible,

how much

reactive power will this

generator

be

supplying or
consuming? Sketch the corresponding phasor diagram.

(Assume

I

F
is still unchanged.)




S
OLUTION


(a)

The

phase voltage

of

this

generator at rated conditions is


V

13,

800 V

7967 V

⎞

=

=

3


The armature current per

phase

at rated conditions is




S


10,

000,
=

=

000 VA

418 A
=


A
I

3

3

(

)
13,800

V

T
V


Therefore, the internal generated voltage

at

rated

conditions

is


=

+

+

A

A
E

V
⎞

A
I

S

A
R

jX

I


7967

0

(

)
1.5

(
418

36.87

A
)

(
12.0

)(
418

36.87

A

)

E
A
=



°

+

&



°

j
+

&





°

E
A
12,

040

17.6

V
=

°



120

The magnitude of

E

A
is 12,040 V.


(b)

The torque angle of the generator

at rated

conditions is

™
= 17.6
°
.


(c)

Ignoring
R
A
, the

maximum output

power
of the generator

is given by

3

3
(

)
7967 V

(
12,040 V

)

P

⎞

A
V

E




=

=

=

MAX
X

S

12

&

24.0 MW



The power at maximum load is 8 MW, so the maximum output power is three times the

full

load output
power.


(d)

The phasor diagram at these conditions is

shown below:

E

A




jX
I

S A



I

A

R
I

A A
V
⎞


Under these conditions,

the

armature current

is


A
E

V


I


⎞

12,

040

90

V


°

-
7967

0

V


°





1194

40.6

A

=

=

=



°

A
1.5

12.0
+

+

&

A S
R

jX

j

The reactive

power produced by the generator at this

point is

3

sin

3
(

)
7967 V

(
119
=

=
⎞

⎝

A
Q

V

I

4

A
)

sin

(
0

40.6

)

18
°

.6


MVA
°

R
=



The generator is actually consuming

reactive power at this time.


5-8.

A 480-V,

100-kW, two-pole, three-phase, 60-Hz

synchronous generator’s

prime

mover

has

a

no-load speed of

3630 r/min and a full-load speed of 3570 r/min. It is operating

in

parallel

with

a 480-V,

75-kW,

four- pole,

60-Hz synchronous generator whose prime mover has a no-load speed of 1800 r/min and a

full-load speed
of 1785

r/min.

The loads supplied

by the two generators consist of 100 kW at 0.85 PF

lagging.

(a)

Calculate

the

speed droops of generator 1 and

generator

2.


(b)

Find

the operating frequency of

the

power system.


(c)

Find

the power being

supplied by

each of the generators in this

system.


(d)

If

V
T
is

460

V, what must the generator’s

operators

do to correct

for the low

terminal

voltage?




S
OLUTION

The no-load frequency

of

generator 1

corresponds to a frequency

of

(

)
3630 r/min

(
2
)

f



m
n

P
=

=

=

nl1

120

120

60.5 Hz



The full-load frequency of generator

1 corresponds

to a frequency of