Electric Machinery Fundamentals (Solutions Manual) Part 4 ppsx

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E
A
2825

10.9
=

V
°

Problems 5-11

to

5-21

refer to a two-pole Y-connected synchronous generator rated at 470 kVA, 480 V, 60 Hz,
and 0.85 PF lagging. Its armature resistance

R
A
is 0.016

&
.

The core

losses of this

generator

at rated

conditions
are

7

kW,

and the friction and windage losses are 8 kW.

The open-circuit and short-circuit characteristics are
shown in Figure P5-2.

127

Note:

An electronic version of the saturated open circuit

characteristic can

be foun

in file

p52_occ.dat
,

and

an

electronic version of the air-gap characteristi

can

be

found

in

file

p52_ag_occ.dat
.

These

files

can

be

used

wit

MATLAB programs. Column 1

contains field current in

amps, and

column

contains

open-circuit

terminal

voltage in volts. An electronic version of th

short circuit

characteristic can

be found in file

p52_scc.dat
. Column

contains field current in amps, and

column

2 contains short-circuit

termina

current in amps.

5-11.

(a)

What is

the

saturated synchronous reactance of this generator at the rated

conditions?

(b)
What is the
unsaturated synchronous reactance

of

this generator?

(c)
Plot the saturated synchronous reactance

of

this
generator as

a

function of load.

S
OLUTION

(a)

The rated armature current for this generator is

S
I

I
=

=

47
=

0 kVA

565 A
=

3

3

(

)
480

V

A

L

T
V

The

field

current

required to produce this much

short-circuit current may

be read from the SCC.

It is

0.534

A
3
.

The open circuit voltage at 0.532 A is

880

V
4
, so the open-circuit phase voltage (=

508 V.

The approximate

saturated synchronous reactance

X

S
is

E
A
) is

880/

3

=

3
If

you have MATLAB available,

you can use the file

p52_scc.dat
and

the

interp1
function to look up

this
value as shown below.

Note that column 1 of

p52_scc
contains field

current,

and column 2 contains short-circuit
terminal current.

load p52_scc.dat
if = interp1(p52_scc(:,2),p52_scc(:,1),565)
if =
128

S

X

508 V

0.899
=

=

&

565 A

(b)

The unsaturated synchronous reactance

X

Su

is the ratio of the air-gap line to the SCC. This is a

straight

line, so we can determine its value by comparing

the

ratio of the air-gap voltage to the

short-circuit

current at any

given field

current. For example,

at

SCC

is 532 A.

I

F
= 0.50

A, the air-gap

line voltage is 1040

V, and

the

X

Su

(

)
1040 V

/

3

1.13
=

=

&

532 A

(c)

This

task

can best

be performed with MATLAB. The open-circuit characteristic is available in a file
called

p52_occ.dat
, and the short-circuit characteristic is available

in

a

file

called

p52_scc.dat
.
Each

of

these

files

are

organized

in two

columns, where the first column is field current and the second
column

is

either

open-circuit

terminal voltage or short-circuit current. A program to read these files and

calculate and

plot

X
S
is

shown

below.

% M-file: prob5_11c.m

%

M-file to calculate and

plot the

saturated

% synchronous reactance of a synchronous

% generator.

% Load the open-circuit characteristic.

It

is

in

% two

columns,

with the first column

being field

% current and the second

column being terminal

% voltage.
load p52.occ;

if_occ =

p52(:,1);

vt_occ =

p52(:,2);

% Load the short-circuit

characteristic. It is in

% two

columns,

with the first column

being field

% current and the second

column being line current

% (= armature current)

load p52.scc;
if_scc =

p52(:,1);
ia_scc =

p52(:,2);

% Calculate

Xs

if

= 0.001:0.02:1;

% Current steps

vt

= interp1(if_occ,vt_occ,If); % Terminal

voltage
ia

= interp1(if_scc,ia_scc,If); % Current

Xs =

(vt ./ sqrt(3))

./

ia;

0.534

4
If

you have MATLAB available,

you can use the file

p52_occ.dat
and

the

interp1
function to look up

this
value as shown below.

Note that column 1 of

p52_occ
contains field

current,

and column 2 contains open-circuit
terminal voltage.

load p52_occ.dat
vt = interp1(p52_occ(:,1),p52_occ(:,2),0.534)
vt =
880.400

129

% Plot the synchronous reactance
figure(1)
plot(If,Xs,'LineWidth',2.0);

title

('\bfSaturated Synchronous Reactance \itX_{s} \rm');

xlabel ('\bfField Current (A)');

ylabel ('\bf\itX_{s} \rm\bf(\Omega)');

grid on;

The

resulting plot is:

5-12.

(a)

What

are

the

rated

current and internal generated voltage of this generator?

(b)
What field current
does this

generator

require to operate

at the rated voltage,

current, and

power factor?

S
OLUTION

(a)

The rated line

and armature

current for this

generator

is

S
I

I
=

=

47
=

0 kVA

565 A
=

3

3

(

)
480

V

A L
T
V

The power factor is 0.85 lagging, so

I

A
565.3

31.8
=

A


°

.

The rated phase voltage

is

V
⎞

= 480 V /

3

= 277 V. The saturated

synchronous reactance at

rated

conditions

was

found

to be

0.450

&

in the previous
problem.

Therefore, the internal generated voltage

is

=

+

+

A

A
E

V
⎞

A
I

S

A
R

jX

I

277

0

(

)
0.

(
016

565.3

31.8

A

)

(

)
0.899

(

565.3

31.8

A
)

E
A
=

°

+

&



°

j
+

&



°

E
A
509

30.5
=

V
°

(b)

This

internal generated voltage corresponds to

a

no-load

terminal

voltage

of

(

)
3

509

=
881 V. From the open-circuit characteristic, the required

field current would be 0.535 A.

5-13.

What is

the

voltage regulation of this generator at the rated current

and power factor?

S
OLUTION

The voltage regulation

is

130

V

V




VR

,nl ,fl
10
T T
0%

881
=

⋅

480
=

100%

83.5
⋅

%
=

T
V

,fl
480

5-14.

If

this generator

is

operating at the

rated conditions

and the load is suddenly removed, what will the
terminal voltage be?

S
OLUTION

From the above calculations,

T
V

will be 881 V.

5-15.

What are the electrical losses in

this generator at rated conditions?

S
OLUTION

The electrical losses are

2

(

)
=

=

2

(

)
&

=

CU

3

A

A

P

I

3
R

565 A

0.016

15.3 kW

5-16.

If

this

machine is operating

at

rated conditions, what input torque must be applied to the shaft of

this
generator?

Express your

answer both in newton-meters and in pound-feet.

S
OLUTION

To get

the

applied torque, we must know

the

input power. The input

power to this generator

is
equal to the output

power plus losses. The rated output

power and the losses are

(

)(

)

OUT
P

470 kVA

0.85

400 kW
=

=

2

(

)
=

=

2

(

)
&

=

CU

3

A

A
P

I

3
R

565 A

0.016

15.3 kW

F&
P

=

W
8 kW

co
P

re
=

7 kW

st
P

ray

=

(assumed 0)

=

+

+

+

+

=

IN

OUT
P

P

CU
P

F&W
P

Therefore, the applied

torque

is

core

P

stray
P

430.3 kW

⎮

IN
P
=

=

430.3 kW

2280 N
=

m
⊕

APP

⎤

m
(

)
1800 r/min

2

r


ad

1

min

7.04

P

1 r

60 s

7.04

(

430.3

kW
)

or

⎮

APP

1800 r/min

1680 lb

ft

=

=

=

⊕

m
n

5-17.

What

is the torque

angle

™
of this generator at rated conditions?

S
OLUTION

From the calculations

in Problem 5-12,

™
= 30.5
°
.

5-18.

Assume that the generator field current is adjusted to supply

480

V

under

rated

conditions.

What is

the

static

stability limit of this generator? (

Note:
You may ignore

R
A
to make this calculation easier.) How

close

is the full-load condition of this

generator to the static

stability limit?

S
OLUTION

At rated conditions,

E
A
509

30.5
=

V
°

.

Therefore, the

static stability

limit is

3

(

)
3

277 V

(

)
509

V

P

⎞

A
V

E

=

=

=

MAX
X

S

0.899

&

471 kW

The

full-load

rated

power

of

this

generator

is

reasonably

close

to

the

static

stability

limit.

Normal
generators

would have

more margin than

this.

131

5-19.

Assume that the generator field current is adjusted

to supply 480 V

under

rated

conditions.

Plot

the

power

supplied by the generator as a function of the torque angle

™
. (
Note:
You may ignore

calculation easier.)

R
A
to make this

S
OLUTION

We will

again ignore

R

A
to make

this calculation easier.

The power supplied

by the generator is

3

⎞
V

E

P

sin
A

3
(

)
277 V

(
509

V

)

(

)

sin

471 kW

sin
™

™

™

=

=

=

S
X
G

0.899

&

The power supplied

as a function of the torque angle

™

may be plotted using a simple

MATLAB program:

% M-file: prob5_19.m

% M-file

to

plot the power output

of

a

% synchronous generator as a function of

% the

torque angle.

% Calculate

Xs

delta

= (0:1:90);

%

Torque

angle

(deg)
Pout = 561 .* sin(delta * pi/180); %

Pout

% Plot the output power
figure(1)

plot(delta,Pout,'LineWidth',2.0);

title

('\bfOutput power vs torque

angle \delta');

xlabel ('\bfTorque

angle

\delta (deg)');

ylabel ('\bf\itP_{OUT} \rm\bf(kW)');

grid on;

The

resulting plot is:

5-20.

Assume that the generator’s field current is

adjusted so

that

the

generator

supplies

rated

voltage at

the

rated
load current

and power factor.

If the field current

and the magnitude

of the load

current are held

constant,
how

will

the terminal

voltage

change

as

the load power factor varies from 0.85 PF lagging to 0.85 PF

132

leading?

Make

a

plot

of the terminal voltage versus the

impedance angle of the load being supplied by this
generator.

S
OLUTION
If the field current is held constant, then the magnitude of

E

A

will be constant, although its

angle

™
will vary.

Also, the magnitude of the armature current

is constant.

Since

we

also

know

R
A
,

X

S
,

and the current angle

⎝
, we know enough to find the phase

voltage

⎞
V
, and therefore the terminal voltage

T
V

.

At lagging

power factors,