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Catégorie :Category: mViewer GX Creator App HP-Prime
Auteur Author: Mikinhomito
Type : Application
Page(s) : 63
Taille Size: 2.29 Mo MB
Mis en ligne Uploaded: 12/07/2019 - 15:21:05
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### Description

Chapter 8: DC Motors and Generators
Problems 8-1 to 8-12 refer to the following dc motor:
Prated = 30 hp I L,rated = 110 A
VT = 240 V N F = 2700 turns per pole
nrated = 1800 r/min N SE = 14 turns per pole
R A = 0.19 RF = 75
RS = 0.02 Radj = 100 to 400
Rotational losses = 3550 W at full load
Magnetization curve as shown in Figure P8-1

Note: An electronic version of this magnetization curve can be found in file
p81_mag.dat, which can be used with MATLAB programs. Column 1
contains field current in amps, and column 2 contains the internal generated
voltage EA in volts.
In Problems 8-1 through 8-7, assume that the motor described above can be connected in shunt. The equivalent
circuit of the shunt motor is shown in Figure P8-2.
214
8-1. If the resistor Radj is adjusted to 175 what is the rotational speed of the motor at no-load conditions?

SOLUTION At no-load conditions, E A VT 240 V . The field current is given by
VT 240 V 240 V
IF 0.960 A
Radj RF 175 75 250

From Figure P9-1, this field current would produce an internal generated voltage E Ao of 241 V at a speed
no of 1200 r/min. Therefore, the speed n with a voltage E A of 240 V would be
EA n
E Ao no
EA 240 V
n no 1200 r/min 1195 r/min
E Ao 241 V

8-2. Assuming no armature reaction, what is the speed of the motor at full load? What is the speed regulation
of the motor?
SOLUTION At full load, the armature current is
VT 240 V
IA IL IF IL 110 A 109 A

The internal generated voltage E A is

E A VT I A RA 240 V 109 A 0.19 219.3 V

The field current is the same as before, and there is no armature reaction, so E Ao is still 241 V at a speed
no of 1200 r/min. Therefore,
EA 219.3 V
n no 1200 r/min 1092 r/min
E Ao 241 V

The speed regulation is
nnl nfl 1195 r/min 1092 r/min
SR 100% 100% 9.4%
nfl 1092 r/min

215
8-3. If the motor is operating at full load and if its variable resistance Radj is increased to 250 , what is the
new speed of the motor? Compare the full-load speed of the motor with Radj = 175 to the full-load
speed with Radj = 250 . (Assume no armature reaction, as in the previous problem.)

SOLUTION If Radj is set to 250 , the field current is now

VT 240 V 240 V
IF 0.739 A
Radj RF 250 75 325

Since the motor is still at full load, E A is still 218.3 V. From the magnetization curve (Figure P8-1), the
new field current I F would produce a voltage E Ao of 212 V at a speed no of 1200 r/min. Therefore,
EA 218.3 V
n no 1200 r/min 1236 r/min
E Ao 212 V

Note that Radj has increased, and as a result the speed of the motor n increased.

8-4. Assume that the motor is operating at full load and that the variable resistor Radj is again 175 . If the
armature reaction is 1000 A turns at full load, what is the speed of the motor? How does it compare to the
result for Problem 8-2?
SOLUTION The field current is again 0.96 A, and the motor is again at full load conditions. However, this
time there is an armature reaction of 1200 A turns, and the effective field current is
AR 1000 A turns
IF* IF 0.96 A 0.59 A
NF 2700 turns

From Figure P9-1, this field current would produce an internal generated voltage E Ao of 185 V at a speed
no of 1200 r/min. The actual internal generated voltage E A at these conditions is

E A VT I A RA 240 V 109 A 0.19 219.3 V

Therefore, the speed n with a voltage of 240 V would be
EA 219.3 V
n no 1200 r/min 1423 r/min
E Ao 185 V

If all other conditions are the same, the motor with armature reaction runs at a higher speed than the motor
without armature reaction.
8-5. If Radj can be adjusted from 100 to 400 , what are the maximum and minimum no-load speeds possible
with this motor?
SOLUTION The minimum speed will occur when Radj = 100 , and the maximum speed will occur when
Radj = 400 . The field current when Radj = 100 is:

VT 240 V 240 V
IF 1.37 A
Radj RF 100 75 175

From Figure P9-1, this field current would produce an internal generated voltage E Ao of 271.5 V at a
speed no of 1200 r/min. Therefore, the speed n with a voltage of 240 V would be

216
EA n
E Ao no
EA 240 V
n no 1200 r/min 1061 r/min
E Ao 271.5 V

The field current when Radj = 400 is:

VT 240 V 240 V
IF 0.505 A
Radj RF 400 75 500

From Figure P8-1, this field current would produce an internal generated voltage E Ao of 167 V at a speed
no of 1200 r/min. Therefore, the speed n with a voltage of 240 V would be
EA n
E Ao no
EA 240 V
n no 1200 r/min 1725 r/min
E Ao 167 V

8-6. What is the starting current of this machine if it is started by connecting it directly to the power supply
VT ? How does this starting current compare to the full-load current of the motor?
SOLUTION The starting current of this machine (ignoring the small field current) is
VT 240 V
I L ,start 1260 A
RA 0.19
The rated current is 110 A, so the starting current is 11.5 times greater than the full-load current. This
much current is extremely likely to damage the motor.
8-7. Plot the torque-speed characteristic of this motor assuming no armature reaction, and again assuming a
full-load armature reaction of 1200 A turns. (Assume that the armature reaction increases linearly with
increases in armature current.)
SOLUTION This problem is best solved with MATLAB, since it involves calculating the torque-speed
values at many points. A MATLAB program to calculate and display both torque-speed characteristics is
shown below.

% M-file: prob8_7.m
% M-file to create a plot of the torque-speed curve of the
% the shunt dc motor with and without armature reaction.

% Get the magnetization curve. Note that this curve is
% defined for a speed of 1200 r/min.
if_values = p81_mag(:,1);
ea_values = p81_mag(:,2);
n_0 = 1200;

% First, initialize the values needed in this program.
v_t = 240; % Terminal voltage (V)
r_f = 75; % Field resistance (ohms)
r_adj = 175; % Adjustable resistance (ohms)
r_a = 0.19; % Armature resistance (ohms)
i_l = 0:1:110; % Line currents (A)

217
n_f = 2700; % Number of turns on field
f_ar0 = 1000; % Armature reaction @ 110 A (A-t/m)

% Calculate the armature current for each load.
i_a = i_l - v_t / (r_f + r_adj);

% Now calculate the internal generated voltage for
% each armature current.
e_a = v_t - i_a * r_a;

% Calculate the armature reaction MMF for each armature
% current.
f_ar = (i_a / 55) * f_ar0;

% Calculate the effective field current with and without
% armature reaction. Ther term i_f_ar is the fi...

### Archive contentsContenu de l'archive

Action(s) SizeTaille FileFichier
3.06 Ko KB readme.txt
3.65 Ko KB lisezmoi.txt
1.09 Ko KB RESAPPLI.hpprgm
2.27 Mo MB RESAPPLI 01_62.hpappdir.zip
26.46 Ko KB RESAPPLI 63.hpappdir.zip
164 octets bytes appslist.txt

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