Compare the full-load speed of the motor with Radj = 175 Ω to the full-load speed with R = 250 adj Ω.. If the armature reaction is 1200 A⋅turns at full load, what is the speed of the mot
Trang 1Note: Figure P9-2 shows incorrect values for RA and RF in the first printing of this
book The correct values are given in the text, but shown incorrectly on the figure This will be corrected at the second printing
9-1 If the resistor Radj is adjusted to 175 Ω what is the rotational speed of the motor at no-load conditions?
SOLUTION At no-load conditions, E A=V T =240 V The field current is given by
adj
240 V 240 V
0.873 A
175 100 250
T F
F
V I
From Figure P9-1, this field current would produce an internal generated voltage E Ao of 271 V at a speed
o
n of 1200 r/min Therefore, the speed n with a voltage E of 240 V would be A
A
Ao o
240 V
1200 r/min 1063 r/min
271 V
A o Ao
E
E
9-2 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
adj
55 A 0.87 A 54.13 A
T
F
V
+ The internal generated voltage E is A
240 V 54.13 A 0.40 218.3 V
The field current is the same as before, and there is no armature reaction, so E Ao is still 271 V at a speed
o
n of 1200 r/min Therefore,
218.3 V
1200 r/min 967 r/min
271 V
A o Ao
E
E
The speed regulation is
nl fl fl
1063 r/min 967 r/min
967 r/min
n
Trang 2216
9-3 If the motor is operating at full load and if its variable resistance R is increased to 250 adj Ω, what is the
new speed of the motor? Compare the full-load speed of the motor with Radj = 175 Ω to the full-load speed with R = 250 adj Ω (Assume no armature reaction, as in the previous problem.)
SOLUTION If R is set to 250 adj Ω, the field current is now
adj
240 V 240 V
0.686 A
250 100 325
T F
F
V I
Since the motor is still at full load, E is still 218.3 V From the magnetization curve (Figure P9-1), the A
new field current I would produce a voltage F E Ao of 247 V at a speed no of 1200 r/min Therefore,
218.3 V
1200 r/min 1061 r/min
247 V
A o Ao
E
E
Note that R has increased, and as a result the speed of the motor n increased adj
9-4 Assume that the motor is operating at full load and that the variable resistor Radj is again 175 Ω If the
armature reaction is 1200 A⋅turns at full load, what is the speed of the motor? How does it compare to the result for Problem 9-2?
SOLUTION The field current is again 0.87 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
2700 turns
F
N
⋅
From Figure P9-1, this field current would produce an internal generated voltage E Ao of 181 V at a speed
o
n of 1200 r/min The actual internal generated voltage EA at these conditions is
240 V 54.13 A 0.40 218.3 V
Therefore, the speed n with a voltage of 240 V would be
218.3 V
1200 r/min 1447 r/min
181 V
A o Ao
E
E
If all other conditions are the same, the motor with armature reaction runs at a higher speed than the motor without armature reaction
9-5 If Radj can be adjusted from 100 to 400 Ω, what are the maximum and minimum no-load speeds possible
with this motor?
SOLUTION The minimum speed will occur when Radj = 100 Ω, and the maximum speed will occur when
adj
R = 400 Ω The field current when R = 100 adj Ω is:
adj
240 V 240 V
1.20 A
100 100 200
T F
F
V I
From Figure P9-1, this field current would produce an internal generated voltage E Ao of 287 V at a speed
o
n of 1200 r/min Therefore, the speed n with a voltage of 240 V would be
Trang 3Ao o
240 V
1200 r/min 1004 r/min
287 V
A o Ao
E
E
The field current when R = 400 adj Ω is:
adj
240 V 240 V
0.480 A
400 100 500
T F
F
V I
From Figure P9-1, this field current would produce an internal generated voltage E Ao of 199 V at a speed
o
n of 1200 r/min Therefore, the speed n with a voltage of 240 V would be
A
Ao o
240 V
1200 r/min 1447 r/min
199 V
A o Ao
E
E
9-6 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
,start
240 V
600 A 0.40
T L
A
V I
R
Ω The rated current is 55 A, so the starting current is 10.9 times greater than the full-load current This much current is extremely likely to damage the motor
9-7 Plot the torque-speed characteristic of this motor assuming no armature reaction, and again assuming a
full-load armature reaction of 1200 A⋅turns
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: prob9_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
load p91_mag.dat
if_values = p91_mag(:,1);
ea_values = p91_mag(:,2);
n_0 = 1200;
% First, initialize the values needed in this program
v_t = 240; % Terminal voltage (V)
r_f = 100; % Field resistance (ohms)
r_adj = 175; % Adjustable resistance (ohms)
r_a = 0.40; % Armature resistance (ohms)
i_l = 0:1:55; % Line currents (A)
n_f = 2700; % Number of turns on field
Trang 4218
f_ar0 = 1200; % Armature reaction @ 55 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 field current
% with armature reaction, and the term i_f_noar is the
% field current without armature reaction
i_f_ar = v_t / (r_f + r_adj) - f_ar / n_f;
i_f_noar = v_t / (r_f + r_adj);
% Calculate the resulting internal generated voltage at
% 1200 r/min by interpolating the motor's magnetization
% curve
e_a0_ar = interp1(if_values,ea_values,i_f_ar);
e_a0_noar = interp1(if_values,ea_values,i_f_noar);
% Calculate the resulting speed from Equation (9-13)
n_ar = ( e_a / e_a0_ar ) * n_0;
n_noar = ( e_a / e_a0_noar ) * n_0;
% Calculate the induced torque corresponding to each
% speed from Equations (8-55) and (8-56)
t_ind_ar = e_a * i_a / (n_ar * 2 * pi / 60);
t_ind_noar = e_a * i_a / (n_noar * 2 * pi / 60);
% Plot the torque-speed curves
figure(1);
plot(t_ind_noar,n_noar,'b-','LineWidth',2.0);
hold on;
plot(t_ind_ar,n_ar,'k ','LineWidth',2.0);
xlabel('\bf\tau_{ind} (N-m)');
ylabel('\bf\itn_{m} \rm\bf(r/min)');
title ('\bfShunt DC Motor Torque-Speed Characteristic');
legend('No armature reaction','With armature reaction');
axis([ 0 125 800 1250]);
grid on;
hold off;
Trang 5The resulting plot is shown below:
800 850 900 950 1000 1050 1100 1150 1200 1250
τind (N-m)
n m
Shunt DC Motor Torque-Speed Characteristic
No armature reaction With armature reaction
For Problems 9-8 and 9-9, the shunt dc motor is reconnected separately excited, as shown in Figure P9-3 It has a fixed field voltage VF of 240 V and an armature voltage VA that can be varied from 120 to 240 V
Note: Figure P9-3 shows incorrect values for RA and RF in the first printing of this
book The correct values are given in the text, but shown incorrectly on the figure This will be corrected at the second printing
9-8 What is the no-load speed of this separately excited motor when R = 175 adj Ω and (a) V = 120 V, (b) A V A
= 180 V, (c) V = 240 V? A
SOLUTION At no-load conditions, E A= The field current is given by V A
adj
240 V 240 V
0.873 A
175 100 275
F F
F
V I
From Figure P9-1, this field current would produce an internal generated voltage E Ao of 271 V at a speed
o
n of 1200 r/min Therefore, the speed n with a voltage of 240 V would be
Trang 6220
A
Ao o
A o Ao
E
E
=
(a) If V = 120 V, then A E = 120 V, and A
120 V
1200 r/min 531 r/min
271 V
(a) If V = 180 V, then A E = 180 V, and A
180 V
1200 r/min 797 r/min
271 V
(a) If V = 240 V, then A E = 240 V, and A
240 V
1200 r/min 1063 r/min
271 V
9-9 For the separately excited motor of Problem 9-8:
(a) What is the maximum no-load speed attainable by varying both V and A R ? adj
(b) What is the minimum no-load speed attainable by varying both V and A R ? adj
SOLUTION
when R = 400 adj Ω is:
adj
240 V 240 V
0.48 A
400 100 500
T F
F
V I
From Figure P9-1, this field current would produce an internal generated voltage E Ao of 199 V at a speed
o
n of 1200 r/min At no-load conditions, the maximum internal generated voltage E A= = 240 V V A
Therefore, the speed n with a voltage of 240 V would be
A
Ao o
240 V
1200 r/min 1447 r/min
199 V
A o Ao
E
E
(b) The minimum speed will occur with the minimum V and the minimum A R The field current when adj
adj
R = 100 Ω is:
adj
240 V 240 V
1.2 A
100 100 200
T F
F
V I
From Figure P9-1, this field current would produce an internal generated voltage E Ao of 287 V at a speed
o
n of 1200 r/min At no-load conditions, the minimum internal generated voltage E A= = 120 V V A
Therefore, the speed n with a voltage of 120 V would be
Trang 7Ao o
120 V
1200 r/min 502 r/min
287 V
A o Ao
E
E
9-10 If the motor is connected cumulatively compounded as shown in Figure P9-4 and if Radj = 175 Ω, what is
its no-load speed? What is its full-load speed? What is its speed regulation? Calculate and plot the torque-speed characteristic for this motor (Neglect armature effects in this problem.)
Note: Figure P9-4 shows incorrect values for RA + RS and RF in the first printing of
this book The correct values are given in the text, but shown incorrectly on the figure This will be corrected at the second printing
SOLUTION At no-load conditions, 240 E A=V T = V The field current is given by
adj
240 V 240 V
0.873 A
175 100 275
F F
F
V I
From Figure P9-1, this field current would produce an internal generated voltage E Ao of 271 V at a speed
o
n of 1200 r/min Therefore, the speed n with a voltage of 240 V would be
A
Ao o
240 V
1200 r/min 1063 r/min
271 V
A o Ao
E
E
At full load conditions, the armature current is
adj
55 A 0.87 A 54.13 A
T
F
V
+ The internal generated voltage E is A
( ) 240 V 54.13 A 0.44 ( )( ) 216.2 V
The equivalent field current is
0.873 A 54.13 A 1.41 A
2700 turns
F
N
N
Trang 8222
From Figure P9-1, this field current would produce an internal generated voltage E Ao of 290 V at a speed
o
n of 1200 r/min Therefore,
216.2 V
1200 r/min 895 r/min
290 V
A o Ao
E
E
The speed regulation is
nl fl fl
1063 r/min 895 r/min
895 r/min
n
The torque-speed characteristic can best be plotted with a MATLAB program An appropriate program is shown below
% M-file: prob9_10.m
% M-file to create a plot of the torque-speed curve of the
% a cumulatively compounded dc motor without
% armature reaction
% Get the magnetization curve Note that this curve is
% defined for a speed of 1200 r/min
load p91_mag.dat
if_values = p91_mag(:,1);
ea_values = p91_mag(:,2);
n_0 = 1200;
% First, initialize the values needed in this program
v_t = 240; % Terminal voltage (V)
r_f = 100; % Field resistance (ohms)
r_adj = 175; % Adjustable resistance (ohms)
r_a = 0.44; % Armature + series resistance (ohms)
i_l = 0:55; % Line currents (A)
n_f = 2700; % Number of turns on shunt field
n_se = 27; % Number of turns on series field
% 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 effective field current for each armature
% current
i_f = v_t / (r_f + r_adj) + (n_se / n_f) * i_a;
% Calculate the resulting internal generated voltage at
% 1200 r/min by interpolating the motor's magnetization
% curve
e_a0 = interp1(if_values,ea_values,i_f);
% Calculate the resulting speed from Equation (9-13)
n = ( e_a / e_a0 ) * n_0;
% Calculate the induced torque corresponding to each
% speed from Equations (8-55) and (8-56)
Trang 9t_ind = e_a * i_a / (n * 2 * pi / 60);
% Plot the torque-speed curves
figure(1);
plot(t_ind,n,'b-','LineWidth',2.0);
xlabel('\bf\tau_{ind} (N-m)');
ylabel('\bf\itn_{m} \rm\bf(r/min)');
title ('\bfCumulatively-Compounded DC Motor Torque-Speed
Characteristic');
axis([0 125 800 1250]);
grid on;
The resulting plot is shown below:
Compare this torque-speed curve to that of the shunt motor in Problem 9-7 (Both curves are plotted on the same scale to facilitate comparison.)
9-11 The motor is connected cumulatively compounded and is operating at full load What will the new speed of
the motor be if R is increased to 250 adj Ω? How does the new speed compared to the full-load speed calculated in Problem 9-10?
SOLUTION If R is increased to 250 adj Ω, the field current is given by
adj
240 V 240 V
0.686 A
250 100 350
T F
F
V I
At full load conditions, the armature current is
55 A 0.686 A 54.3 A
The internal generated voltage E is A
( ) 240 V 54.3 A 0.44 ( )( ) 216.1 V
Trang 10224
The equivalent field current is
0.686 A 54.3 A 1.23 A
2700 turns
F
N
N
From Figure P9-1, this field current would produce an internal generated voltage E Ao of 288 V at a speed
o
n of 1200 r/min Therefore,
216.1 V
1200 r/min 900 r/min
288 V
A o Ao
E
E
The new full-load speed is higher than the full-load speed in Problem 9-10
9-12 The motor is now connected differentially compounded
(a) If Radj = 175 Ω, what is the no-load speed of the motor?
(b) What is the motor’s speed when the armature current reaches 20 A? 40 A? 60 A?
(c) Calculate and plot the torque-speed characteristic curve of this motor
SOLUTION
(a) At no-load conditions, E A=V T =240 V The field current is given by
adj
240 V 240 V
0.873 A
175 100 275
F F
F
V I
From Figure P9-1, this field current would produce an internal generated voltage E Ao of 271 V at a speed
o
n of 1200 r/min Therefore, the speed n with a voltage of 240 V would be
A
Ao o
240 V
1200 r/min 1063 r/min
271 V
A o Ao
E
E
(b) At I = 20A, the internal generated voltage A E is A
( ) 240 V 20 A 0.44 ( )( ) 231.2 V
The equivalent field current is
2700 turns
F
N
N
From Figure P9-1, this field current would produce an internal generated voltage E Ao of 245 V at a speed
o
n of 1200 r/min Therefore,
231.2 V
1200 r/min 1132 r/min
245 V
A o Ao
E
E
At I = 40A, the internal generated voltage A E is A
( ) 240 V 40 A 0.44 ( )( ) 222.4 V
The equivalent field current is
Trang 11( )
2700 turns
F
N
N
From Figure P9-1, this field current would produce an internal generated voltage E Ao of 197 V at a speed
o
n of 1200 r/min Therefore,
227.4 V
1200 r/min 1385 r/min
197 V
A o Ao
E
E
At I = 60A, the internal generated voltage A E is A
( ) 240 V 60 A 0.44 ( )( ) 213.6 V
The equivalent field current is
2700 turns
F
N
N
From Figure P9-1, this field current would produce an internal generated voltage E Ao of 121 V at a speed
o
n of 1200 r/min Therefore,
213.6 V
1200 r/min 2118 r/min
121 V
A o Ao
E
E
(c) The torque-speed characteristic can best be plotted with a MATLAB program An appropriate
program is shown below
% M-file: prob9_12.m
% M-file to create a plot of the torque-speed curve of the
% a differentially compounded dc motor withwithout
% armature reaction
% Get the magnetization curve Note that this curve is
% defined for a speed of 1200 r/min
load p91_mag.dat
if_values = p91_mag(:,1);
ea_values = p91_mag(:,2);
n_0 = 1200;
% First, initialize the values needed in this program
v_t = 240; % Terminal voltage (V)
r_f = 100; % Field resistance (ohms)
r_adj = 175; % Adjustable resistance (ohms)
r_a = 0.44; % Armature + series resistance (ohms)
i_l = 0:50; % Line currents (A)
n_f = 2700; % Number of turns on shunt field
n_se = 27; % Number of turns on series field
% 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 effective field current for each armature