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Catégorie :Category: mViewer GX Creator App HP-Prime
Auteur Author: pauloreto
Type : Application
Page(s) : 10
Taille Size: 595.83 Ko KB
Mis en ligne Uploaded: 17/10/2020 - 22:48:29
Uploadeur Uploader: pauloreto (Profil)
Téléchargements Downloads: 1
Visibilité Visibility: Archive publique
Shortlink : http://ti-pla.net/a2648546

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Chapter 5: Synchronous Motors

5-1. A 480-V, 60 Hz, 400-hp 0.8-PF-leading eight-pole -connected synchronous motor has a synchronous
reactance of 0.6  and negligible armature resistance. Ignore its friction, windage, and core losses for the
purposes of this problem. Assume that E A is directly proportional to the field current I F (in other
words, assume that the motor operates in the linear part of the magnetization curve), and that E A = 480
V when I F = 4 A.
(a) What is the speed of this motor?
(b) If this motor is initially supplying 400 hp at 0.8 PF lagging, what are the magnitudes and angles
of E A and I A ?
(c) How much torque is this motor producing? What is the torque angle  ? How near is this value
to the maximum possible induced torque of the motor for this field current setting?

(d) If E A is increased by 30 percent, what is the new magnitude of the armature current? What is
the motor’s new power factor?
(e) Calculate and plot the motor’s V-curve for this load condition.
SOLUTION
(a) The speed of this motor is given by
120 f se 120  60 Hz 
nm    900 r/min
P 8
(b) If losses are being ignored, the output power is equal to the input power, so the input power will be
PIN   400 hp  746 W/hp   298.4 kW

This situation is shown in the phasor diagram below:
V

IA
jX I
S A


EA

The line current flow under these circumstances is
P 298.4 kW
IL    449 A
3 VT PF 3  480 V  0.8

Because the motor is -connected, the corresponding phase current is I A  449 / 3  259 A . The angle
of the current is  cos 1  0.80  36.87 , so I A  259  36.87 A . The internal generated voltage E A
is
E A  V  jX S I A

E A   4800 V   j  0.6   259  36.87 A   406  17.8 V


123
(c) This motor has 6 poles and an electrical frequency of 60 Hz, so its rotation speed is nm = 1200
r/min. The induced torque is
POUT 298.4 kW
 ind    3166 N  m
m 1 min  2 rad 
 900 r/min    
 60 s  1 r 
The maximum possible induced torque for the motor at this field setting is the maximum possible power
divided by m

3V E A 3  480 V  406 V 
 ind,max    10,340 N  m
m X S  1 min  2 rad 
 900 r/min      0.6  
 60 s  1 r 
The current operating torque is about 1/3 of the maximum possible torque.
(d) If the magnitude of the internal generated voltage E A is increased by 30%, the new torque angle
can be found from the fact that E A sin   P  constant .

E A 2  1.30 E A1  1.30  406 V   487.2 V

 E A1   406 V 
 2  sin 1  sin 1   sin 1  sin  17.8    14.8
 E A2   487.2 V 
The new armature current is
V  E A 2 4800 V  487.2  14.8 V
I A2    208  4.1 A
jX S j 0.6 

The magnitude of the armature current is 208 A, and the power factor is cos (-24.1) = 0.913 lagging.
(e) A MATLAB program to calculate and plot the motor’s V-curve is shown below:

% M-file: prob5_1e.m
% M-file create a plot of armature current versus Ea
% for the synchronous motor of Problem 5-1.

% Initialize values
Ea = (0.90:0.01:1.70)*406; % Magnitude of Ea volts
Ear = 406; % Reference Ea
deltar = -17.8 * pi/180; % Reference torque angle
Xs = 0.6; % Synchronous reactance (ohms)
Vp = 480; % Phase voltage at 0 degrees
Ear = Ear * (cos(deltar) + j * sin(deltar));

% Calculate delta2
delta2 = asin ( abs(Ear) ./ abs(Ea) .* sin(deltar) );

% Calculate the phasor Ea
Ea = Ea .* (cos(delta2) + j .* sin(delta2));

% Calculate Ia
Ia = ( Vp - Ea ) / ( j * Xs);

% Plot the v-curve

124
figure(1);
plot(abs(Ea),abs(Ia),'b','Linewidth',2.0);
xlabel('bfitE_{A}rmbf (V)');
ylabel('bfitI_{A}rmbf (A)');
title ('bfSynchronous Motor V-Curve');
grid on;
The resulting plot is shown below




5-2. Assume that the motor of Problem 5-1 is operating at rated conditions.
(a) What are the magnitudes and angles of E A and I A , and I F ?

(b) Suppose the load is removed from the motor. What are the magnitudes and angles of E A and I A
now?
SOLUTION
(a) The line current flow at rated conditions is
P 298.4 kW
IL    449 A
3 VT PF 3  480 V  0.8

Because the motor is -connected, the corresponding phase current is I A  449 / 3  259 A . The angle
of the current is cos 1  0.80   36.87 , so I A  25936.87 A . The internal generated voltage E A is

E A  V  jX S I A

E A   4800 V   j  0.6   25936.87 A   587  12.2 V

The field current is directly proportional to E A , with = 480 V when I F = 4 A. Since the real E A is
587 V, the required field current is
E A2 IF 2

E A1 I F1

125
E A2 587 V
IF 2  I F1   4 A   4.89 A
E A1 480 V

(b) When the load is removed from the motor the magnitude of E A remains unchanged but the torque
angle goes to   0 . The resulting armature current is
V  E A 4800 V  5870
IA    178.390 A
jX S j 0.6 
5-3. A 230-V, 50 Hz, two-pole synchronous motor draws 40 A from the line at unity power factor and full
load. Assuming that the motor is lossless, answer the following questions:
(a) What is the output torque of this motor? Express the answer both in newton-meters and in pound-
feet.
(b) What must be done to change the power factor to 0.85 leading? Explain your answer, using
phasor diagrams.
(c) What will the magnitude of the line current be if the power factor is adjusted to 0.85 leading?
SOLUTION
(a) If this motor is assumed lossless, then the input power is equal to the output power. The input
power to this motor is
PIN  3VT I L cos   3  230 V  40 A 1.0   15.93 kW

The rotational speed of the motor is
120 f se 120  50 Hz 
nm    1500 r/min
P 4
The output torque would be
POUT 15.93 kW
 LOAD    101.4 N  m
m  1 min  2 rad 
1500 r/min    
 60 s  1 r 
In English units,
7.04 POUT  7.04 15.93 kW 
 LOAD    74.8 lb  ft
nm 1500 r/min 
(b) To change the motor’s power factor to 0.8 leading, its field current must be increased. Since the
power supplied to the load is independent of the field current level, an increase in field current increases
E A while keeping the distance E A sin  constant. This increase in E A changes the angle of the
current I A , eventually causing it to reach a power factor of 0.8 leading.




126
P




}
I A2
I A1 V




...

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