Chapter 9: DC Motors and Generators

Problems 9-1 to 9-12 refer to the following dc motor:

Prated = 15 hp I L ,rated = 55 A
VT = 240 V N F = 2700 turns per pole
nrated = 1200 r/min N SE = 27 turns per pole
RA = 0.40 Ω RF = 100 Ω
RS = 0.04 Ω Radj = 100 to 400 Ω
Rotational losses = 1800 W at full load. Magnetization curve as shown in Figure P9-1.
Note: 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 = VT = 240 V . The field current is given by

VT 240 V 240 V
IF = = = = 0.873 A
Radj + RF 175 Ω + 100 Ω 250 Ω

From Figure P9-1, this field current would produce an internal generated voltage E Ao of 271 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 ) = 1063 r/min
E Ao 271 V

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
VT
IA = IL − IF = IL − = 55 A − 0.87 A = 54.13 A
Radj + RF

The internal generated voltage E A is

E A = VT − I A R A = 240 V − (54.13 A )( 0.40 Ω ) = 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
no of 1200 r/min. Therefore,
Note: An electronic version of this magnetization curve can be found in file EA 218.3 V
p91_mag.dat, which can be used with MATLAB programs. Column 1 n= no = (1200 r/min ) = 967 r/min
E Ao 271 V
contains field current in amps, and column 2 contains the internal generated
voltage EA in volts. The speed regulation is
In Problems 9-1 through 9-7, assume that the motor described above can be connected in shunt. The equivalent
nnl − nfl 1063 r/min − 967 r/min
circuit of the shunt motor is shown in Figure P9-2. SR = × 100% = × 100% = 9.9%
nfl 967 r/min

214 215

9-3. If the motor is operating at full load and if its variable resistance Radj is increased to 250 Ω, what is the EA n
=
new speed of the motor? Compare the full-load speed of the motor with Radj = 175 Ω to the full-load speed E Ao no
with Radj = 250 Ω. (Assume no armature reaction, as in the previous problem.) n=
EA
no =
240 V
(1200 r/min ) = 1004 r/min
E Ao 287 V
SOLUTION If Radj is set to 250 Ω, the field current is now
The field current when Radj = 400 Ω is:
VT 240 V 240 V
IF = = = = 0.686 A VT 240 V 240 V
Radj + RF 250 Ω + 100 Ω 325 Ω IF = = = = 0.480 A
Radj + RF 400 Ω + 100 Ω 500 Ω
Since the motor is still at full load, E A is still 218.3 V. From the magnetization curve (Figure P9-1), the
new field current I F would produce a voltage E Ao of 247 V at a speed no of 1200 r/min. Therefore, From Figure P9-1, this field current would produce an internal generated voltage E Ao of 199 V at a speed
no of 1200 r/min. Therefore, the speed n with a voltage of 240 V would be
EA 218.3 V
n= no = (1200 r/min ) = 1061 r/min EA n
E Ao 247 V
=
E Ao no
Note that Radj has increased, and as a result the speed of the motor n increased.
EA 240 V
9-4. Assume that the motor is operating at full load and that the variable resistor Radj is again 175 Ω. If the
n= no = (1200 r/min ) = 1447 r/min
E Ao 199 V
armature reaction is 1200 A⋅turns at full load, what is the speed of the motor? How does it compare to the
9-6. What is the starting current of this machine if it is started by connecting it directly to the power supply VT ?
result for Problem 9-2?
How does this starting current compare to the full-load current of the motor?
SOLUTION The field current is again 0.87 A, and the motor is again at full load conditions. However, this
SOLUTION The starting current of this machine (ignoring the small field current) is
time there is an armature reaction of 1200 A⋅turns, and the effective field current is
VT 240 V
AR 1200 A ⋅ turns I L,start = = = 600 A
I F* = I F − = 0.87 A − = 0.426 A RA 0.40 Ω
NF 2700 turns
The rated current is 55 A, so the starting current is 10.9 times greater than the full-load current. This much
From Figure P9-1, this field current would produce an internal generated voltage E Ao of 181 V at a speed
current is extremely likely to damage the motor.
no of 1200 r/min. The actual internal generated voltage E A at these conditions is
9-7. Plot the torque-speed characteristic of this motor assuming no armature reaction, and again assuming a
E A = VT − I A R A = 240 V − (54.13 A )( 0.40 Ω ) = 218.3 V full-load armature reaction of 1200 A⋅turns.

Therefore, the speed n with a voltage of 240 V would be 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
EA 218.3 V
n= no = (1200 r/min ) = 1447 r/min below.
E Ao 181 V
% M-file: prob9_7.m
If all other conditions are the same, the motor with armature reaction runs at a higher speed than the motor % M-file to create a plot of the torque-speed curve of the
without armature reaction. % the shunt dc motor with and 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
% Get the magnetization curve. Note that this curve is
with this motor? % defined for a speed of 1200 r/min.
load p91_mag.dat
SOLUTION The minimum speed will occur when Radj = 100 Ω, and the maximum speed will occur when if_values = p91_mag(:,1);
Radj = 400 Ω. The field current when Radj = 100 Ω is: ea_values = p91_mag(:,2);
n_0 = 1200;
VT 240 V 240 V
IF = = = = 1.20 A % First, initialize the values needed in this program.
Radj + RF 100 Ω + 100 Ω 200 Ω
v_t = 240; % Terminal voltage (V)
From Figure P9-1, this field current would produce an internal generated voltage E Ao of 287 V at a speed r_f = 100; % Field resistance (ohms)
r_adj = 175; % Adjustable resistance (ohms)
no of 1200 r/min. Therefore, the speed n with a voltage of 240 V would be r_a = 0.40; % Armature resistance (ohms)
i_l = 0:1:55; % Line currents (A)
n_f = 2700; % Number of turns on field
216 217
 f_ar0 = 1200; % Armature reaction @ 55 A (A-t/m) The resulting plot is shown below:
Shunt DC Motor Torque-Speed Characteristic
% Calculate the armature current for each load. 1250
i_a = i_l - v_t / (r_f + r_adj);
1200
% Now calculate the internal generated voltage for No armature reaction
% each armature current. With armature reaction
1150
e_a = v_t - i_a * r_a;
1100
% Calculate the armature reaction MMF for each armature
% current.

nm (r/min)
f_ar = (i_a / 55) * f_ar0; 1050

% Calculate the effective field current with and without 1000

% armature reaction. Ther term i_f_ar is the field current

% with armature reaction, and the term i_f_noar is the 950
% field current without armature reaction.
i_f_ar = v_t / (r_f + r_adj) - f_ar / n_f; 900
i_f_noar = v_t / (r_f + r_adj);
850
% Calculate the resulting internal generated voltage at
% 1200 r/min by interpolating the motor's magnetization 800
0 20 40 60 80 100 120
% curve. τ (N-m)
ind
e_a0_ar = interp1(if_values,ea_values,i_f_ar);
e_a0_noar = interp1(if_values,ea_values,i_f_noar);
For Problems 9-8 and 9-9, the shunt dc motor is reconnected separately excited, as shown in Figure P9-3. It has a
% Calculate the resulting speed from Equation (9-13). fixed field voltage V F of 240 V and an armature voltage V A that can be varied from 120 to 240 V.
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)'); Note: Figure P9-3 shows incorrect values for RA and RF in the first printing of this
ylabel('\bf\itn_{m} \rm\bf(r/min)'); book. The correct values are given in the text, but shown incorrectly on the
title ('\bfShunt DC Motor Torque-Speed Characteristic'); figure. This will be corrected at the second printing.
legend('No armature reaction','With armature reaction');
axis([ 0 125 800 1250]); 9-8. What is the no-load speed of this separately excited motor when Radj = 175 Ω and (a) V A = 120 V, (b) V A
grid on; = 180 V, (c) V A = 240 V?
hold off;
SOLUTION At no-load conditions, E A = VA . The field current is given by

VF 240 V 240 V
IF = = = = 0.873 A
Radj + RF 175 Ω + 100 Ω 275 Ω

From Figure P9-1, this field current would produce an internal generated voltage E Ao of 271 V at a speed
no of 1200 r/min. Therefore, the speed n with a voltage of 240 V would be

218 219

EA n EA n
= =
E Ao no E Ao no
EA EA 120 V
n= no n= no = (1200 r/min ) = 502 r/min
E Ao E Ao 287 V

(a) If V A = 120 V, then E A = 120 V, and 9-10. If the motor is connected cumulatively compounded as shown in Figure P9-4 and if Radj = 175 Ω, what is
120 V its no-load speed? What is its full-load speed? What is its speed regulation? Calculate and plot the torque-
n= (1200 r/min ) = 531 r/min speed characteristic for this motor. (Neglect armature effects in this problem.)
271 V
(a) If V A = 180 V, then E A = 180 V, and
180 V
n= (1200 r/min ) = 797 r/min
271 V
(a) If V A = 240 V, then E A = 240 V, and
240 V
n= (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 A and Radj ?

(b) What is the minimum no-load speed attainable by varying both V A and Radj ?
Note: Figure P9-4 shows incorrect values for RA + RS and RF in the first printing of
SOLUTION this book. The correct values are given in the text, but shown incorrectly on
the figure. This will be corrected at the second printing.
(a) The maximum speed will occur with the maximum V A and the maximum Radj . The field current
SOLUTION At no-load conditions, E A = VT = 240 V . The field current is given by
when Radj = 400 Ω is:
VF 240 V 240 V
VT 240 V 240 V IF = = = = 0.873 A
IF = = = = 0.48 A Radj + RF 175 Ω + 100 Ω 275 Ω
Radj + RF 400 Ω + 100 Ω 500 Ω
From Figure P9-1, this field current would produce an internal generated voltage E Ao of 271 V at a speed
From Figure P9-1, this field current would produce an internal generated voltage E Ao of 199 V at a speed
no of 1200 r/min. Therefore, the speed n with a voltage of 240 V would be
no of 1200 r/min. At no-load conditions, the maximum internal generated voltage E A = V A = 240 V.
Therefore, the speed n with a voltage of 240 V would be EA n
=
E Ao no
EA n
= EA 240 V
E Ao no n= no = (1200 r/min ) = 1063 r/min
E Ao 271 V
EA 240 V
n= no = (1200 r/min ) = 1447 r/min
E Ao 199 V At full load conditions, the armature current is

(b) The minimum speed will occur with the minimum V A and the minimum Radj . The field current when VT
IA = IL − IF = IL − = 55 A − 0.87 A = 54.13 A
Radj + RF
Radj = 100 Ω is:
The internal generated voltage E A is
VT 240 V 240 V
IF = = = = 1.2 A
Radj + RF 100 Ω + 100 Ω 200 Ω E A = VT − I A ( RA + RS ) = 240 V − (54.13 A )(0.44 Ω) = 216.2 V

From Figure P9-1, this field current would produce an internal generated voltage E Ao of 287 V at a speed The equivalent field current is
no of 1200 r/min. At no-load conditions, the minimum internal generated voltage E A = V A = 120 V. N SE 27 turns
Therefore, the speed n with a voltage of 120 V would be I F* = I F + I A = 0.873 A + (54.13 A ) = 1.41 A
NF 2700 turns

220 221
 From Figure P9-1, this field current would produce an internal generated voltage E Ao of 290 V at a speed t_ind = e_a .* i_a ./ (n * 2 * pi / 60);
no of 1200 r/min. Therefore,
% Plot the torque-speed curves
EA 216.2 V figure(1);
n= no = (1200 r/min ) = 895 r/min plot(t_ind,n,'b-','LineWidth',2.0);
E Ao 290 V
xlabel('\bf\tau_{ind} (N-m)');
The speed regulation is ylabel('\bf\itn_{m} \rm\bf(r/min)');
title ('\bfCumulatively-Compounded DC Motor Torque-Speed
nnl − nfl 1063 r/min − 895 r/min Characteristic');
SR = × 100% = × 100% = 18.8%
nfl 895 r/min axis([0 125 800 1250]);
grid on;
The torque-speed characteristic can best be plotted with a MATLAB program. An appropriate program is
shown below. The resulting plot 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);
Compare this torque-speed curve to that of the shunt motor in Problem 9-7. (Both curves are plotted on the
% Now calculate the internal generated voltage for same scale to facilitate comparison.)
% each armature current. 9-11. The motor is connected cumulatively compounded and is operating at full load. What will the new speed of
e_a = v_t - i_a * r_a; the motor be if Radj is increased to 250 Ω? How does the new speed compared to the full-load speed
% Calculate the effective field current for each armature calculated in Problem 9-10?
% current.
SOLUTION If Radj is increased to 250 Ω, the field current is given by
i_f = v_t / (r_f + r_adj) + (n_se / n_f) * i_a;
VT 240 V 240 V
% Calculate the resulting internal generated voltage at IF = = = = 0.686 A
% 1200 r/min by interpolating the motor's magnetization Radj + RF 250 Ω + 100 Ω 350 Ω
% curve.
e_a0 = interp1(if_values,ea_values,i_f); At full load conditions, the armature current is
I A = I L − I F = 55 A − 0.686 A = 54.3 A
% Calculate the resulting speed from Equation (9-13).
n = ( e_a ./ e_a0 ) * n_0; The internal generated voltage E A is
% Calculate the induced torque corresponding to each E A = VT − I A ( RA + RS ) = 240 V − (54.3 A )(0.44 Ω ) = 216.1 V
% speed from Equations (8-55) and (8-56).
222 223

The equivalent field current is N SE 27 turns

I F* = I F − I A = 0.873 A − (40 A ) = 0.473 A
N SE 27 turns NF 2700 turns
I F* = I F + I A = 0.686 A + (54.3 A ) = 1.23 A
NF 2700 turns From Figure P9-1, this field current would produce an internal generated voltage E Ao of 197 V at a speed
From Figure P9-1, this field current would produce an internal generated voltage E Ao of 288 V at a speed no of 1200 r/min. Therefore,
no of 1200 r/min. Therefore, EA 227.4 V
n= no = (1200 r/min ) = 1385 r/min
EA 216.1 V E Ao 197 V
n= no = (1200 r/min ) = 900 r/min
E Ao 288 V At I A = 60A, the internal generated voltage E A is
The new full-load speed is higher than the full-load speed in Problem 9-10. E A = VT − I A ( RA + RS ) = 240 V − ( 60 A )( 0.44 Ω ) = 213.6 V
9-12. The motor is now connected differentially compounded.
The equivalent field current is
(a) If Radj = 175 Ω, what is the no-load speed of the motor?
N SE 27 turns
(b) What is the motor’s speed when the armature current reaches 20 A? 40 A? 60 A?
I F* = I F − I A = 0.873 A − (60 A ) = 0.273 A
NF 2700 turns
(c) Calculate and plot the torque-speed characteristic curve of this motor. From Figure P9-1, this field current would produce an internal generated voltage E Ao of 121 V at a speed
SOLUTION no of 1200 r/min. Therefore,
At no-load conditions, E A = VT = 240 V . The field current is given by EA 213.6 V
(a) n= no = (1200 r/min ) = 2118 r/min
E Ao 121 V
VF 240 V 240 V
IF = = = = 0.873 A
Radj + RF 175 Ω + 100 Ω 275 Ω (c) The torque-speed characteristic can best be plotted with a MATLAB program. An appropriate
program is shown below.
From Figure P9-1, this field current would produce an internal generated voltage E Ao of 271 V at a speed
no of 1200 r/min. Therefore, the speed n with a voltage of 240 V would be % M-file: prob9_12.m
% M-file to create a plot of the torque-speed curve of the
EA n % a differentially compounded dc motor withwithout
= % armature reaction.
E Ao no
EA 240 V % Get the magnetization curve. Note that this curve is
n= no = (1200 r/min ) = 1063 r/min % defined for a speed of 1200 r/min.
E Ao 271 V
load p91_mag.dat
(b) At I A = 20A, the internal generated voltage E A is if_values = p91_mag(:,1);
ea_values = p91_mag(:,2);
E A = VT − I A ( RA + RS ) = 240 V − ( 20 A )(0.44 Ω ) = 231.2 V n_0 = 1200;

The equivalent field current is % First, initialize the values needed in this program.
v_t = 240; % Terminal voltage (V)
N SE 27 turns
I F* = I F − I A = 0.873 A − (20 A ) = 0.673 A r_f = 100; % Field resistance (ohms)
NF 2700 turns r_adj = 175; % Adjustable resistance (ohms)
r_a = 0.44; % Armature + series resistance (ohms)
From Figure P9-1, this field current would produce an internal generated voltage E Ao of 245 V at a speed i_l = 0:50; % Line currents (A)
no of 1200 r/min. Therefore, n_f = 2700; % Number of turns on shunt field
n_se = 27; % Number of turns on series field
EA 231.2 V
n= no = (1200 r/min ) = 1132 r/min
E Ao 245 V % Calculate the armature current for each load.
i_a = i_l - v_t / (r_f + r_adj);
At I A = 40A, the internal generated voltage E A is
% Now calculate the internal generated voltage for
E A = VT − I A ( RA + RS ) = 240 V − ( 40 A )(0.44 Ω ) = 222.4 V % each armature current.
e_a = v_t - i_a * r_a;
The equivalent field current is
% Calculate the effective field current for each armature
224 225
 % current. in Figure P9-5. The core losses are 200 W, and the mechanical losses are 240 W at full load. Assume that
i_f = v_t / (r_f + r_adj) - (n_se / n_f) * i_a; the mechanical losses vary as the cube of the speed of the motor and that the core losses are constant.
% 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).
t_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 ('\bfDifferentially-Compounded DC Motor Torque-Speed
Characteristic');
axis([0 100 800 1600]);
grid on;
The resulting plot is shown below:

Note: An electronic version of this magnetization curve can be found in file

p95_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.
(a) What is the efficiency of the motor at full load?
(b) What are the speed and efficiency of the motor if it is operating at an armature current of 35 A?
(c) Plot the torque-speed characteristic for this motor.

Compare this torque-speed curve to that of the shunt motor in Problem 9-7 and the cumulatively- SOLUTION
compounded motor in Problem 9-10. (Note that this plot has a larger vertical scale to accommodate the
(a) The output power of this motor at full load is
speed runaway of the differentially-compounded motor.)
POUT = ( 7.5 hp )( 746 W/hp ) = 5595 W
9-13. A 7.5-hp 120-V series dc motor has an armature resistance of 0.2 Ω and a series field resistance of 0.16 Ω.
At full load, the current input is 58 A, and the rated speed is 1050 r/min. Its magnetization curve is shown The input power is

226 227

PIN = VT I L = (120 V )(58 A ) = 6960 W

% M-file: prob9_13.m
Therefore the efficiency is % M-file to create a plot of the torque-speed curve of the
% the series dc motor in Problem 9-13.
P 5595 W
η = OUT × 100% = × 100% = 80.4%
PIN 6960 W % Get the magnetization curve. Note that this curve is
% defined for a speed of 1200 r/min.
(b) If the armature current is 35 A, then the input power to the motor will be load p95_mag.dat
if_values = p95_mag(:,1);
PIN = VT I L = (120 V )( 35 A ) = 4200 W ea_values = p95_mag(:,2);
n_0 = 1200;
The internal generated voltage at this condition is
E A2 = VT − I A ( RA + RS ) = 120 V − ( 35 A )(0.20 Ω + 0.16 Ω ) = 107.4 V % First, initialize the values needed in this program.
v_t = 120; % Terminal voltage (V)
and the internal generated voltage at rated conditions is r_a = 0.36; % Armature + field resistance (ohms)
i_a = 9:1:58; % Armature (line) currents (A)
E A1 = VT − I A ( RA + RS ) = 120 V − (58 A )(0.20 Ω + 0.16 Ω ) = 99.1 V
% Calculate the internal generate voltage e_a.
The final speed is given by the equation e_a = v_t - i_a * r_a;
E A2 K φ2 ω 2 E Ao ,2 n2
= = % Calculate the resulting internal generated voltage at
E A1 K φ2 ω 2 E Ao ,1 n1 % 1200 r/min by interpolating the motor's magnetization
% curve. Note that the field current is the same as the
since the ratio E Ao ,2 / E Ao ,1 is the same as the ratio φ 2 / φ1 . Therefore, the final speed is % armature current for this motor.
e_a0 = interp1(if_values,ea_values,i_a,'spline');
E A 2 E Ao ,1
n2 = n1
E A1 E Ao ,2 % Calculate the motor's speed, using the known fact that
% the motor runs at 1050 r/min at a current of 58 A. We
From Figure P9-5, the internal generated voltage E Ao,2 for a current of 35 A and a speed of no = 1200 % know that
r/min is E Ao,2 = 115 V, and the internal generated voltage E Ao,1 for a current of 58 A and a speed of no = %
% Ea2 K' phi2 n2 Eao2 n2
1200 r/min is E Ao ,1 = 134 V. % ----- = ------------ = ----------
% Ea1 K' phi1 n1 Eao1 n1
E A2 E Ao ,1 107.4 V 134 V
n2 = n1 = (1050 r/min ) = 1326 r/min %
E A1 E Ao ,2 99.1 V 115 V % Ea2 Eao1
% ==> n2 = ----- ------ n1
The power converted from electrical to mechanical form is % Ea1 Eao2
%
Pconv = E A I A = (107.4 V )( 35 A ) = 3759 W % where Ea0 is the internal generated voltage at 1200 r/min
% for a given field current.
The core losses in the motor are 200 W, and the mechanical losses in the motor are 240 W at a speed of %
1050 r/min. The mechanical losses in the motor scale proportionally to the cube of the rotational speedm % Speed will be calculated by reference to full load speed
so the mechanical losses at 1326 r/min are % and current.
3 3 n1 = 1050; % 1050 r/min at full load
n2 1326 r/min
Pmech = (240 W ) = (240 W ) = 483 W Eao1 = interp1(if_values,ea_values,58,'spline');
n1 1050 r/min Ea1 = v_t - 58 * r_a;

Therefore, the output power is % Get speed

Eao2 = interp1(if_values,ea_values,i_a,'spline');
POUT = Pconv − Pmech − Pcore = 3759 W − 483 W − 200 W = 3076 W n = (e_a./Ea1) .* (Eao1 ./ Eao2) * n1;
and the efficiency is
% Calculate the induced torque corresponding to each
POUT 3076 W % speed from Equations (8-55) and (8-56).
η= × 100% = × 100% = 73.2% t_ind = e_a .* i_a ./ (n * 2 * pi / 60);
PIN 4200 W

(c) A MATLAB program to plot the torque-speed characteristic of this motor is shown below: % Plot the torque-speed curve
228 229
 figure(1); important that interpolation be done using smooth curves, so be sure to specify the 'spline' option in
plot(t_ind,n,'b-','LineWidth',2.0); the MATLAB interp1 function:
hold on;
xlabel('\bf\tau_{ind} (N-m)'); load prob9_14_mag.dat;
ylabel('\bf\itn_{m} \rm\bf(r/min)'); mmf_values = prob9_14_mag(:,1);
title ('\bfSeries DC Motor Torque-Speed Characteristic'); ea_values = prob9_14_mag(:,2);
grid on; ...
hold off; Eao = interp1(mmf_values,ea_values,mmf,'spline')
The resulting torque-speed characteristic is shown below: (a) Since full load corresponds to 76 A, this calculation must be performed for armature currents of 25.3
A, 50.7 A, 76 A, and 101.3 A.
If I A = 23.3 A, then

E A = VT − I A ( RA + RS ) = 240 V − ( 25.3 A )(0.09 Ω + 0.06 Ω ) = 236.2 V

The magnetomotive force is F = NI A = ( 33 turns)( 25.3 A ) = 835 A ⋅ turns , which produces a voltage E Ao
of 134 V at no = 900 r/min. Therefore the speed of the motor at these conditions is

EA 236.2 V
n= no = (900 r/min ) = 1586 r/min
E Ao 134 V

The power converted from electrical to mechanical form is

Pconv = E A I A = ( 236.2 V )( 25.3 A ) = 5976 W

Since the rotational losses are ignored, this is also the output power of the motor. The induced torque is
Pconv 5976 W
τ ind = = = 36 N ⋅ m
ωm 2π rad 1 min
(1586 r/min )
1r 60 s
If I A = 50.7 A, then

9-14. A 20-hp 240-V 76-A 900 r/min series motor has a field winding of 33 turns per pole. Its armature E A = VT − I A ( RA + RS ) = 240 V − (50.7 A )( 0.09 Ω + 0.06 Ω ) = 232.4 V
resistance is 0.09 Ω, and its field resistance is 0.06 Ω. The magnetization curve expressed in terms of
magnetomotive force versus EA at 900 r/min is given by the following table: The magnetomotive force is F = NI A = ( 33 turns )(50.7 A ) = 1672 A ⋅ turns , which produces a voltage E Ao
of 197 V at no = 900 r/min. Therefore the speed of the motor at these conditions is
EA , V 95 150 188 212 229 243
EA 232.4 V
F, A ⋅ turns 500 1000 1500 2000 2500 3000 n= no = (900 r/min ) = 1062 r/min
E Ao 197 V
Note: An electronic version of this magnetization curve can be found in file The power converted from electrical to mechanical form is
prob9_14_mag.dat, which can be used with MATLAB programs. Column
1 contains magnetomotive force in ampere-turns, and column 2 contains the Pconv = E A I A = ( 232.4 V )(50.7 A ) = 11, 780 W
internal generated voltage EA in volts.
Since the rotational losses are ignored, this is also the output power of the motor. The induced torque is
Armature reaction is negligible in this machine.
Pconv 11,780 W
(a) Compute the motor’s torque, speed, and output power at 33, 67, 100, and 133 percent of full-load τ ind = = = 106 N ⋅ m
ωm 2π rad 1 min
armature current. (Neglect rotational losses.) (1062 r/min )
1r 60 s
(b) Plot the terminal characteristic of this machine.
If I A = 76 A, then
SOLUTION Note that this magnetization curve has been stored in a file called prob9_14_mag.dat. The
first column of the file is an array of mmf_values, and the second column is an array of ea_values. E A = VT − I A ( RA + RS ) = 240 V − (76 A )(0.09 Ω + 0.06 Ω ) = 228.6 V
These values are valid at a speed no = 900 r/min. Because the data in the file is relatively sparse, it is

230 231

The magnetomotive force is F = NI A = (33 turns)(76 A ) = 2508 A ⋅ turns , which produces a voltage E Ao f = n_s * i_a;
of 229 V at no = 900 r/min. Therefore the speed of the motor at these conditions is % Calculate the internal generate voltage e_a.
e_a = v_t - i_a * r_a;
EA 228.6 V
n= no = (900 r/min ) = 899 r/min
E Ao 229 V % Calculate the resulting internal generated voltage at
% 900 r/min by interpolating the motor's magnetization
The power converted from electrical to mechanical form is % curve. Specify cubic spline interpolation to provide
Pconv = E A I A = ( 228.6 V )( 76 A ) = 17,370 W % good results with this sparse magnetization curve.
e_a0 = interp1(mmf_values,ea_values,f,'spline');
Since the rotational losses are ignored, this is also the output power of the motor. The induced torque is
% Calculate the motor's speed from Equation (9-13).
Pconv 17,370 W n = (e_a ./ e_a0) * n_0;
τ ind = = = 185 N ⋅ m
ωm
(899 r/min ) 2π rad 1 min
% Calculate the induced torque corresponding to each
1r 60 s
% speed from Equations (8-55) and (8-56).
If I A = 101.3 A, then t_ind = e_a .* i_a ./ (n * 2 * pi / 60);

E A = VT − I A ( RA + RS ) = 240 V − (101.3 A )( 0.09 Ω + 0.06 Ω ) = 224.8 V % Plot the torque-speed curve

figure(1);
The magnetomotive force is F = NI A = (33 turns)(101.3 A ) = 3343 A ⋅ turns , which produces a voltage plot(t_ind,n,'b-','LineWidth',2.0);
hold on;
E Ao of 252 V at no = 900 r/min. Therefore the speed of the motor at these conditions is
xlabel('\bf\tau_{ind} (N-m)');
EA 224.8 V ylabel('\bf\itn_{m} \rm\bf(r/min)');
n= no = (900 r/min ) = 803 r/min title ('\bfSeries DC Motor Torque-Speed Characteristic');
E Ao 252 V %axis([ 0 700 0 5000]);
The power converted from electrical to mechanical form is grid on;
hold off;
Pconv = E A I A = ( 224.8 V )(101.3 A ) = 22,770 W
The resulting torque-speed characteristic is shown below:
Since the rotational losses are ignored, this is also the output power of the motor. The induced torque is
Pconv 22,770 W
τ ind = = = 271 N ⋅ m
ωm 2π rad 1 min
(803 r/min )
1r 60 s
(b) A MATLAB program to plot the torque-speed characteristic of this motor is shown below:

% M-file: series_ts_curve.m
% M-file to create a plot of the torque-speed curve of the
% the series dc motor in Problem 9-14.

% Get the magnetization curve. Note that this curve is

% defined for a speed of 900 r/min.
load prob9_14_mag.dat
mmf_values = prob9_14_mag(:,1);
ea_values = prob9_14_mag(:,2);
n_0 = 900;

% First, initialize the values needed in this program.

v_t = 240; % Terminal voltage (V)
r_a = 0.15; % Armature + field resistance (ohms)
i_a = 15:1:76; % Armature (line) currents (A)
n_s = 33; % Number of series turns on field
9-15. A 300-hp 440-V 560-A, 863 r/min shunt dc motor has been tested, and the following data were taken:
% Calculate the MMF for each load Blocked-rotor test:
232 233
 V A = 16.3 V exclusive of brushes VF = 440 V RF = 200 Ω nm = 1200 r/min
I A = 500 A I F = 8.86 A Radj = 0 to 300 Ω, currently set to 120 Ω
This motor has compensating windings and interpoles. The magnetization curve for this motor at 1200 r/min is
No-load operation: shown in Figure P9-6.
V A = 16.3 V including brushes I F = 8.76 A
I A = 231
. A n = 863 r/min
What is this motor’s efficiency at the rated conditions? [Note: Assume that (1) the brush voltage drop is 2
V; (2) the core loss is to be determined at an armature voltage equal to the armature voltage under full load;
and (3) stray load losses are 1 percent of full load.]
SOLUTION The armature resistance of this motor is
VA,br 16.3 V
RA = = = 0.0326 Ω
I A,br 500 A

Under no-load conditions, the core and mechanical losses taken together (that is, the rotational losses) of
this motor are equal to the product of the internal generated voltage E A and the armature current I A , since
this is no output power from the motor at no-load conditions. Therefore, the rotational losses at rated speed
can be found as
E A = VA − Vbrush − I A RA = 442 V − 2 V − ( 23.1 A )(0.0326 Ω ) = 439.2 V

Prot = Pconv = E A I A = ( 439.2 V )( 23.1 A ) = 10.15 kW

The input power to the motor at full load is

PIN = VT I L = ( 440 V )(560 A ) = 246.4 kW

The output power from the motor at full load is

POUT = PIN − PCU − Prot − Pbrush − Pstray

The copper losses are

PCU = I A2 RA + VF I F = (560 A ) ( 0.0326 Ω ) + ( 440 V )(8.86 A ) = 14.1 kW

2

Note: An electronic version of this magnetization curve can be found in file

The brush losses are p96_mag.dat, which can be used with MATLAB programs. Column 1
contains field current in amps, and column 2 contains the internal generated
Pbrush = Vbrush I A = ( 2 V )(560 A ) = 1120 W
voltage EA in volts.
Therefore, 9-16. The motor described above is connected in shunt.
POUT = PIN − PCU − Prot − Pbrush − Pstray (a) What is the no-load speed of this motor when Radj = 120 Ω?
POUT = 246.4 kW − 14.1 kW − 10.15 kW − 1.12 kW − 2.46 kW = 218.6 kW
(b) What is its full-load speed?
The motor’s efficiency at full load is
(c) Under no-load conditions, what range of possible speeds can be achieved by adjusting Radj ?
P 218.6 kW
η = OUT × 100% = × 100% = 88.7%
PIN 246.4 kW SOLUTION Note that this magnetization curve has been stored in a file called p96_mag.dat. The first
Problems 9-16 to 9-19 refer to a 240-V 100-A dc motor which has both shunt and series windings. Its column of the file is an array of ia_values, and the second column is an array of ea_values. These
characteristics are values are valid at a speed no = 1200 r/min. These values can be used with the MATLAB interp1
RA = 0.14 Ω N F = 1500 turns function to look up an internal generated voltage as follows:
RS = 0.04 Ω N SE = 12 turns load p96_mag.dat;
234 235

if_values = p96_mag(:,1); (a) At full load, I A = I L − I F = 100 A − 0.75 A = 99.25 A , and

ea_values = p96_mag(:,2);
... E A = VT − I A ( RA + RS ) = 240 V − (99.25 A )(0.14 Ω + 0.05 Ω ) = 221.1 V
Ea = interp1(if_values,ea_values,if,'spline')
The actual field current will be
(a) If Radj = 120 Ω, the total field resistance is 320 Ω, and the resulting field current is
VT 240 V
IF = = = 0.75 A
IF =
VT
=
240 V
= 0.75 A RF + Radj 200 Ω + 120 Ω
RF + Radj 200 Ω + 120 Ω
and the effective field current will be
This field current would produce a voltage E Ao of 256 V at a speed of no = 1200 r/min. The actual E A is N SE 12 turns
240 V, so the actual speed will be I F* = I F + I A = 0.75 A + (99.25 A ) = 1.54 A
NF 1500 turns
E 240 V
n = A no = (1200 r/min ) = 1125 r/min This field current would produce a voltage E Ao of 290 V at a speed of no = 1200 r/min. The actual E A
E Ao 256 V
is 240 V, so the actual speed at full load will be
(b) At full load, I A = I L − I F = 100 A − 0.75 A = 99.25 A , and
EA 221.1 V
n= no = (1200 r/min ) = 915 r/min
E A = VT − I A R A = 240 V − (99.25 A )( 0.14 Ω ) = 226.1 V E Ao 290 V

Therefore, the speed at full load will be (b) A MATLAB program to calculate the torque-speed characteristic of this motor is shown below:
EA 226.1 V
n= no = (1200 r/min ) = 1060 r/min % M-file: prob9_17.m
% M-file to create a plot of the torque-speed curve of the
E Ao 256 V
% a cumulatively compounded dc motor.
(c) If Radj is maximum at no-load conditions, the total resistance is 500 Ω, and
% Get the magnetization curve.
VT 240 V load p96_mag.dat;
IF = = = 0.48 A if_values = p96_mag(:,1);
RF + Radj 200 Ω + 300 Ω
ea_values = p96_mag(:,2);
This field current would produce a voltage E Ao of 200 V at a speed of no = 1200 r/min. The actual E A is n_0 = 1200;
240 V, so the actual speed will be % First, initialize the values needed in this program.
EA 240 V v_t = 240; % Terminal voltage (V)
n= no = (1200 r/min ) = 1440 r/min r_f = 200; % Field resistance (ohms)
E Ao 200 V r_adj = 120; % Adjustable resistance (ohms)
r_a = 0.19; % Armature + series resistance (ohms)
If Radj is minimum at no-load conditions, the total resistance is 200 Ω, and i_l = 0:2:100; % Line currents (A)
n_f = 1500; % Number of turns on shunt field
VT 240 V
IF = = = 1.2 A n_se = 12; % Number of turns on series field
RF + Radj 200 Ω + 0 Ω
% Calculate the armature current for each load.
This field current would produce a voltage E Ao of 287 V at a speed of no = 1200 r/min. The actual E A is i_a = i_l - v_t / (r_f + r_adj);
240 V, so the actual speed will be
% Now calculate the internal generated voltage for
EA 240 V
n= no = (1200 r/min ) = 1004 r/min % each armature current.
e_a = v_t - i_a * r_a;
E Ao 287 V

9-17. This machine is now connected as a cumulatively compounded dc motor with Radj = 120 Ω. % Calculate the effective field current for each armature
% current.
(a) What is the full-load speed of this motor? i_f = v_t / (r_f + r_adj) + (n_se / n_f) * i_a;
(b) Plot the torque-speed characteristic for this motor.
% Calculate the resulting internal generated voltage at
(c) What is its speed regulation? % 1800 r/min by interpolating the motor's magnetization
% curve.
SOLUTION e_a0 = interp1(if_values,ea_values,i_f);

236 237
 % a differentially compounded dc motor.
% Calculate the resulting speed from Equation (9-13).
n = ( e_a ./ e_a0 ) * n_0; % Get the magnetization curve.
load p96_mag.dat;
% Calculate the induced torque corresponding to each if_values = p96_mag(:,1);
% speed from Equations (8-55) and (8-56). ea_values = p96_mag(:,2);
t_ind = e_a .* i_a ./ (n * 2 * pi / 60); n_0 = 1200;

% Plot the torque-speed curves % First, initialize the values needed in this program.
figure(1); v_t = 240; % Terminal voltage (V)
plot(t_ind,n,'b-','LineWidth',2.0); r_f = 200; % Field resistance (ohms)
xlabel('\bf\tau_{ind} (N-m)'); r_adj = 120; % Adjustable resistance (ohms)
ylabel('\bf\itn_{m} \rm\bf(r/min)'); r_a = 0.19; % Armature + series resistance (ohms)
title ('\bfCumulatively-Compounded DC Motor Torque-Speed i_l = 0:2:40; % Line currents (A)
Characteristic'); n_f = 1500; % Number of turns on shunt field
axis([0 200 900 1600]); n_se = 12; % Number of turns on series field
grid on;
% Calculate the armature current for each load.
The resulting torque-speed characteristic is shown below: 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

% 1800 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).
t_ind = e_a .* i_a ./ (n * 2 * pi / 60);

% Plot the torque-speed curves

figure(1);
plot(t_ind,n,'b-','LineWidth',2.0);
(c) The no-load speed of this machine is the same as the no-load speed of the corresponding shunt dc xlabel('\bf\tau_{ind} (N-m)');
motor with Radj = 120 Ω, which is 1125 r/min. The speed regulation of this motor is thus ylabel('\bf\itn_{m} \rm\bf(r/min)');
title ('\bfDifferentially-Compounded DC Motor Torque-Speed
nnl − nfl 1125 r/min - 915 r/min Characteristic');
SR = × 100% = × 100% = 23.0%
nfl 915 r/min axis([0 200 900 1600]);
grid on;
9-18. The motor is reconnected differentially compounded with Radj = 120 Ω. Derive the shape of its torque-
speed characteristic.
SOLUTION A MATLAB program to calculate the torque-speed characteristic of this motor is shown below:

% M-file: prob9_18.m
% M-file to create a plot of the torque-speed curve of the
238 239

The resulting torque-speed characteristic is shown below:

% 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. (Note that the magnetization curve is defined
% in terms of shunt field current, so we will have to
% translate the series field current into an equivalent
% shunt field current.
i_f = (n_se / n_f) * i_a;

% Calculate the resulting internal generated voltage at

% 1800 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).
t_ind = e_a .* i_a ./ (n * 2 * pi / 60);

% Plot the torque-speed curves

This curve is plotted on the same scale as the torque-speed curve in Problem 6-17. Compare the two figure(1);
curves. plot(t_ind,n,'b-','LineWidth',2.0);
xlabel('\bf\tau_{ind} (N-m)');
9-19. A series motor is now constructed from this machine by leaving the shunt field out entirely. Derive the ylabel('\bf\itn_{m} \rm\bf(r/min)');
torque-speed characteristic of the resulting motor. title ('\bfSeries DC Motor Torque-Speed Characteristic');
grid on;
SOLUTION This motor will have extremely high speeds, since there are only a few series turns, and the flux
in the motor will be very small. A MATLAB program to calculate the torque-speed characteristic of this The resulting torque-speed characteristic is shown below:
motor is shown below:

% M-file: prob9_19.m
% M-file to create a plot of the torque-speed curve of the
% a series dc motor. This motor was formed by removing
% the shunt field from the cumulatively-compounded machine
% if Problem 9-17.

% Get the magnetization curve.

load p96_mag.dat;
if_values = p96_mag(:,1);
ea_values = p96_mag(:,2);
n_0 = 1200;

% First, initialize the values needed in this program.

v_t = 240; % Terminal voltage (V)
r_a = 0.19; % Armature + series resistance (ohms)
i_l = 20:1:45; % Line currents (A)
n_f = 1500; % Number of turns on shunt field
n_se = 12; % Number of turns on series field

% Calculate the armature current for each load.

i_a = i_l;

240 241
 The extreme speeds in this characteristic are due to the very light flux in the machine. To make a practical 9-21. A 15-hp 120-V 1800 r/min shunt dc motor has a full-load armature current of 60 A when operating at rated
series motor out of this machine, it would be necessary to include 20 to 30 series turns instead of 12. conditions. The armature resistance of the motor is RA = 0.15 Ω, and the field resistance RF is 80 Ω.
9-20. An automatic starter circuit is to be designed for a shunt motor rated at 15 hp, 240 V, and 60 A. The The adjustable resistance in the field circuit Radj may be varied over the range from 0 to 200 Ω and is
armature resistance of the motor is 0.15 Ω, and the shunt field resistance is 40 Ω. The motor is to start currently set to 90 Ω. Armature reaction may be ignored in this machine. The magnetization curve for this
with no more than 250 percent of its rated armature current, and as soon as the current falls to rated value, motor, taken at a speed of 1800 r/min, is given in tabular form below:
a starting resistor stage is to be cut out. How many stages of starting resistance are needed, and how big
should each one be? EA , V 5 78 95 112 118 126

SOLUTION The rated line current of this motor is 60 A, and the rated armature current is I A = I L − I F = 60 IF , A 0.00 0.80 1.00 1.28 1.44 2.88
A – 6 A = 54 A. The maximum desired starting current is (2.5)(54 A) = 135 A. Therefore, the total initial
Note: An electronic version of this magnetization curve can be found in file
starting resistance must be
prob9_21_mag.dat, which can be used with MATLAB programs. Column
240 V 1 contains field current in amps, and column 2 contains the internal generated
R A + Rstart,1 = = 1.778 Ω voltage EA in volts.
135 A
Rstart,1 = 1.778 Ω − 0.15 Ω = 1.628 Ω (a) What is the speed of this motor when it is running at the rated conditions specified above?
The current will fall to rated value when E A rises to (b) The output power from the motor is 7.5 hp at rated conditions. What is the output torque of the motor?
E A = 240 V − (1.778 Ω )(54 A ) = 144 V
(c) What are the copper losses and rotational losses in the motor at full load (ignore stray losses)?
At that time, we want to cut out enough resistance to get the current back up to 135 A. Therefore, (d) What is the efficiency of the motor at full load?
240 V − 144 V (e) If the motor is now unloaded with no changes in terminal voltage or Radj , what is the no-load speed of
R A + Rstart,2 = = 0.711 Ω
135 A the motor?
Rstart,2 = 0.711 Ω − 0.15 Ω = 0.561 Ω
(f) Suppose that the motor is running at the no-load conditions described in part (e). What would happen
With this resistance in the circuit, the current will fall to rated value when E A rises to
to the motor if its field circuit were to open? Ignoring armature reaction, what would the final steady-
E A = 240 V − ( 0.711 Ω )(54 A ) = 201.6 V state speed of the motor be under those conditions?
At that time, we want to cut out enough resistance to get the current back up to 185 A. Therefore, (g) What range of no-load speeds is possible in this motor, given the range of field resistance adjustments
available with Radj ?
240 V − 201.6 V
R A + Rstart,3 = = 0.284 Ω
135 A SOLUTION
Rstart,3 = 0.284 Ω − 0.15 Ω = 0.134 Ω
(a) If Radj = 90 Ω, the total field resistance is 170 Ω, and the resulting field current is
With this resistance in the circuit, the current will fall to rated value when E A rises to
E A = 240 V − ( 0.284 Ω )(54 A ) = 224.7 V VT 230 V
IF = = = 1.35 A
RF + Radj 90 Ω + 80 Ω
If the resistance is cut out when E A reaches 228,6 V, the resulting current is
This field current would produce a voltage E Ao of 221 V at a speed of no = 1800 r/min. The actual E A is
240 V − 224.7 V
IA = = 102 A < 135 A ,
0.15 Ω E A = VT − I A RA = 230 V − ( 60 A )( 0.15 Ω ) = 221 V
so there are only three stages of starting resistance. The three stages of starting resistance can be found so the actual speed will be
from the resistance in the circuit at each state during starting.
EA 221 V
Rstart,1 = R1 + R2 + R3 = 1.628 Ω n= no = (1800 r/min ) = 1800 r/min
E Ao 221 V
Rstart,2 = R2 + R3 = 0.561 Ω
Rstart,3 = R3 = 0.134 Ω (b) The output power is 7.5 hp and the output speed is 1800 r/min at rated conditions, therefore, the
torque is
Therefore, the starting resistances are
τ out =
Pout
=
(15 hp)(746 W/hp) = 59.4 N ⋅ m
R1 = 1.067 Ω
R2 = 0.427 Ω
ωm
(1800 r/min ) 2π rad 1 min
1r 60 s
R3 = 0.134 Ω (c) The copper losses are

242 243

PCU = I A2 RA + VF I F = ( 60 A ) ( 0.15 Ω ) + ( 230 V )(1.35 A ) = 851 W

2
EA 230 V
n= no = (1800 r/min ) = 1711 r/min
E Ao 242 V
The power converted from electrical to mechanical form is
9-22. The magnetization curve for a separately excited dc generator is shown in Figure P9-7. The generator is
Pconv = E A I A = ( 221 V )( 60 A ) = 13,260 W rated at 6 kW, 120 V, 50 A, and 1800 r/min and is shown in Figure P9-8. Its field circuit is rated at 5A.
The output power is The following data are known about the machine:

POUT = (15 hp )( 746 W/hp ) = 11,190 W

Therefore, the rotational losses are

Prot = Pconv − POUT = 13,260 W − 11,190 W = 2070 W
(d) The input power to this motor is
PIN = VT ( I A + I F ) = ( 230 V )(60 A + 1.35 A ) = 14,100 W

Therefore, the efficiency is

POUT 11,190 W
η= × 100% = × 100% = 79.4%
PIN 14,100 W

(e) The no-load E A will be 230 V, so the no-load speed will be

EA 230 V
n= no = (1800 r/min ) = 1873 r/min
E Ao 221 V

(f) If the field circuit opens, the field current would go to zero Ÿ φ drops to φ res Ÿ E A ↓ Ÿ I A ↑Ÿ
τ ind ↑ Ÿ n↑ to a very high speed. If I F = 0 A, E Ao = 8.5 V at 1800 r/min, so
EA 230 V
n= no = (1800 r/min ) = 48,700 r/min
E Ao 8.5 V

(In reality, the motor speed would be limited by rotational losses, or else the motor will destroy itself first.)
(g) The maximum value of Radj = 200 Ω, so

VT 230 V
IF = = = 0.821 A
RF + Radj 200 Ω + 80 Ω

This field current would produce a voltage E Ao of 153 V at a speed of no = 1800 r/min. The actual E A is
230 V, so the actual speed will be
EA 230 V
n= no = (1800 r/min ) = 2706 r/min
E Ao 153 V

The minimum value of Radj = 0 Ω, so

VT 230 V
IF = = = 2.875 A Note: An electronic version of this magnetization curve can be found in file
RF + Radj 0 Ω + 80 Ω
p97_mag.dat, which can be used with MATLAB programs. Column 1
This field current would produce a voltage E Ao of about 242 V at a speed of no = 1800 r/min. The actual contains field current in amps, and column 2 contains the internal generated
voltage EA in volts.
E A is 230 V, so the actual speed will be

244 245
 EA n
=
E Ao no

n 2000 r/min
EA = E Ao = (129 V ) = 143 V
no 1800 r/min

The minimum possible field current occurs when Radj = 30 Ω. The current is

VF 120 V
I F ,max = = = 2.22 A
RF + Radj 24 Ω + 30 Ω

From the magnetization curve, the voltage E Ao at 1800 r/min is 87.4 V. Since the actual speed is 1500
R A = 0.18 Ω VF = 120 V
r/min, the maximum no-load voltage is
Radj = 0 to 30 Ω RF = 24 Ω
EA n
N F = 1000 turns per pole =
E Ao no
Answer the following questions about this generator, assuming no armature reaction.
n 1500 r/min
(a) If this generator is operating at no load, what is the range of voltage adjustments that can be achieved EA = E Ao = (87.4 V ) = 72.8 V
by changing Radj ? no 1800 r/min

9-23. If the armature current of the generator in Problem 9-22 is 50 A, the speed of the generator is 1700 r/min,
(b) If the field rheostat is allowed to vary from 0 to 30 Ω and the generator’s speed is allowed to vary from
and the terminal voltage is 106 V, how much field current must be flowing in the generator?
1500 to 2000 r/min, what are the maximum and minimum no-load voltages in the generator?
SOLUTION The internal generated voltage of this generator is
SOLUTION
E A = VT + I A RA = 106 V + (50 A )( 0.18 Ω ) = 115 V
(a) If the generator is operating with no load at 1800 r/min, then the terminal voltage will equal the
internal generated voltage E A . The maximum possible field current occurs when Radj = 0 Ω. The current at a speed of 1700 r/min. This corresponds to an E Ao at 1800 r/min of
is EA n
=
VF 120 V E Ao no
I F ,max = = =5A
RF + Radj 24 Ω + 0 Ω no 1800 r/min
E Ao = EA = (115 V ) = 121.8 V
n 1700 r/min
From the magnetization curve, the voltage E Ao at 1800 r/min is 129 V. Since the actual speed is 1800
r/min, the maximum no-load voltage is 129 V. From the magnetization curve, this value of E Ao requires a field current of 4.2 A.

The minimum possible field current occurs when Radj = 30 Ω. The current is 9-24. Assuming that the generator in Problem 9-22 has an armature reaction at full load equivalent to 400
A⋅turns of magnetomotive force, what will the terminal voltage of the generator be when I F = 5 A, nm =
VF 120 V
I F ,max = = = 2.22 A 1700 r/min, and I A = 50 A?
RF + Radj 24 Ω + 30 Ω
SOLUTION When I F is 5 A and the armature current is 50 A, the magnetomotive force in the generator is
From the magnetization curve, the voltage E Ao at 1800 r/min is 87.4 V. Since the actual speed is 1800
r/min, the minimum no-load voltage is 87 V. Fnet = NI F − FAR = (1000 turns )(5 A ) − 400 A ⋅ turns = 4600 A ⋅ turns

(b) The maximum voltage will occur at the highest current and speed, and the minimum voltage will or I F * = Fnet / N F = 4600 A ⋅ turns / 1000 turns = 4.6 A
occur at the lowest current and speed. The maximum possible field current occurs when Radj = 0 Ω. The
The equivalent internal generated voltage E Ao of the generator at 1800 r/min would be 126 V. The actual
current is
voltage at 1700 r/min would be
VF 120 V
I F ,max = = =5A n 1700 r/min
RF + Radj 24 Ω + 0 Ω EA = E Ao = (126 V ) = 119 V
no 1800 r/min
From the magnetization curve, the voltage E Ao at 1800 r/min is 129 V. Since the actual speed is 2000 Therefore, the terminal voltage would be
r/min, the maximum no-load voltage is
VT = E A − I A RA = 119 V − (50 A )(0.18 Ω ) = 110 V

246 247

9-25. The machine in Problem 9-22 is reconnected as a shunt generator and is shown in Figure P9-9. The shunt (b) At an armature current of 20 A, the internal voltage drop in the armature resistance is
field resistor Radj is adjusted to 10 Ω, and the generator’s speed is 1800 r/min. (20 A )(0.18 Ω) = 3.6 V .
As shown in the figure below, there is a difference of 3.6 V between E A and
VT at a terminal voltage of about 106 V.

(a) What is the no-load terminal voltage of the generator?

(b) Assuming no armature reaction, what is the terminal voltage of the generator with an armature current
of 20 A? 40 A?
(c) Assuming an armature reaction equal to 200 A⋅turns at full load, what is the terminal voltage of the
generator with an armature current of 20 A? 40 A?
(d) Calculate and plot the terminal characteristics of this generator with and without armature reaction.

SOLUTION
(a) The total field resistance of this generator is 34 Ω, and the no-load terminal voltage can be found A MATLAB program to locate the position where the triangle exactly fits between the E A and VT lines is
from the intersection of the resistance line with the magnetization curve for this generator. The shown below. This program created the plot shown above. Note that there are actually two places where
magnetization curve and the field resistance line are plotted below. As you can see, they intersect at a the difference between the E A and VT lines is 3.6 volts, but the low-voltage one of them is unstable. The
terminal voltage of 112 V.
code shown in bold face below prevents the program from reporting that first (unstable) point.

% M-file: prob9_25b.m
% M-file to create a plot of the magnetization curve and the
% field current curve of a shunt dc generator, determining
% the point where the difference between them is 3.6 V.

% Get the magnetization curve. This file contains the

% three variables if_values, ea_values, and n_0.
clear all
load p97_mag.dat;
if_values = p97_mag(:,1);
ea_values = p97_mag(:,2);
n_0 = 1800;

% First, initialize the values needed in this program.

r_f = 24; % Field resistance (ohms)
r_adj = 10; % Adjustable resistance (ohms)
r_a = 0.19; % Armature + series resistance (ohms)
i_f = 0:0.02:6; % Field current (A)
n = 1800; % Generator speed (r/min)

% Calculate Ea versus If
Ea = interp1(if_values,ea_values,i_f);

248 249
% Calculate Vt versus If
Vt = (r_f + r_adj) * i_f;

% Find the point where the difference between the two

% lines is 3.6 V. This will be the point where the line
% line "Ea - Vt - 3.6" goes negative. That will be a
% close enough estimate of Vt.
diff = Ea - Vt - 3.6;

% This code prevents us from reporting the first (unstable)

% location satisfying the criterion.
was_pos = 0;
for ii = 1:length(i_f);
if diff(ii) > 0
was_pos = 1;
end
if ( diff(ii) < 0 & was_pos == 1 )
break;
end;
end;

% We have the intersection. Tell user.

disp (['Ea = ' num2str(Ea(ii)) ' V']);
disp (['Vt = ' num2str(Vt(ii)) ' V']);
(c) The rated current of this generated is 50 A, so 20 A is 40% of full load. If the full load armature
disp (['If = ' num2str(i_f(ii)) ' A']);
reaction is 200 A⋅turns, and if the armature reaction is assumed to change linearly with armature current,
% Plot the curves then the armature reaction will be 80 A⋅turns. The figure below shows that a triangle consisting of 3.6 V
figure(1); and (80 A⋅turns)/(1000 turns) = 0.08 A fits exactly between the E A and VT lines at a terminal voltage of
plot(i_f,Ea,'b-','LineWidth',2.0); 103 V.
hold on;
plot(i_f,Vt,'k--','LineWidth',2.0);
% Plot intersections
plot([i_f(ii) i_f(ii)], [0 Ea(ii)], 'k-');
plot([0 i_f(ii)], [Vt(ii) Vt(ii)],'k-');
plot([0 i_f(ii)], [Ea(ii) Ea(ii)],'k-');
xlabel('\bf\itI_{F} \rm\bf(A)');
ylabel('\bf\itE_{A} \rm\bf or \itV_{T}');
title ('\bfPlot of \itE_{A} \rm\bf and \itV_{T} \rm\bf vs field
current');
axis ([0 5 0 150]);
set(gca,'YTick',[0 10 20 30 40 50 60 70 80 90 100 110 120 130 140
150]')
set(gca,'XTick',[0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0]')
legend ('Ea line','Vt line',4);
hold off;
grid on;
At an armature current of 40 A, the internal voltage drop in the armature resistance is
(40 A )(0.18 Ω) = 7.2 V . As shown in the figure below, there is a difference of 7.2 V between E A and
VT at a terminal voltage of about 98 V.

250 251

% Calculate Ea versus If
Ea = interp1(if_values,ea_values,i_f);

% Calculate Vt versus If
Vt = (r_f + r_adj) * i_f;

% Find the point where the difference between the two

% lines is exactly equal to i_a*r_a. This will be the
% point where the line line "Ea - Vt - i_a*r_a" goes
% negative.
i_a = 0:1:50;
for jj = 1:length(i_a)

% Get the voltage difference

diff = Ea - Vt - i_a(jj)*r_a;

% This code prevents us from reporting the first (unstable)

% location satisfying the criterion.
was_pos = 0;
for ii = 1:length(i_f);
if diff(ii) > 0
was_pos = 1;
end
if ( diff(ii) < 0 & was_pos == 1 )
break;
end;
end;

The rated current of this generated is 50 A, so 40 A is 80% of full load. If the full load armature reaction % Save terminal voltage at this point
v_t(jj) = Vt(ii);
is 200 A⋅turns, and if the armature reaction is assumed to change linearly with armature current, then the
i_l(jj) = i_a(jj) - v_t(jj) / ( r_f + r_adj);
armature reaction will be 160 A⋅turns. There is no point where a triangle consisting of 3.6 V and (80
A⋅turns)/(1000 turns) = 0.16 A fits exactly between the E A and VT lines, so this is not a stable operating end;
condition.
% Plot the terminal characteristic
(c) A MATLAB program to calculate the terminal characteristic of this generator without armature figure(1);
reaction is shown below: plot(i_l,v_t,'b-','LineWidth',2.0);
xlabel('\bf\itI_{L} \rm\bf(A)');
% M-file: prob9_25d.m ylabel('\bf\itV_{T} \rm\bf(V)');
% M-file to calculate the terminal characteristic of a shunt title ('\bfTerminal Characteristic of a Shunt DC Generator');
% dc generator without armature reaction. hold off;
axis( [ 0 50 0 120]);
% Get the magnetization curve. This file contains the grid on;
% three variables if_values, ea_values, and n_0.
load p97_mag.dat;
if_values = p97_mag(:,1);
ea_values = p97_mag(:,2);
n_0 = 1800;

% First, initialize the values needed in this program.

r_f = 24; % Field resistance (ohms)
r_adj = 10; % Adjustable resistance (ohms)
r_a = 0.18; % Armature + series resistance (ohms)
i_f = 0:0.005:6; % Field current (A)
n = 1800; % Generator speed (r/min)
252 253
 The resulting terminal characteristic is shown below: % the line "Ea_ar - Vt - i_a*r_a" goes negative.
i_a = 0:1:37;
for jj = 1:length(i_a)

% Calculate the equivalent field current due to armature

% reaction.
i_ar = (i_a(jj) / 50) * 200 / n_f;

% Calculate the Ea values modified by armature reaction

Ea_ar = interp1(if_values,ea_values,i_f - i_ar);

% Get the voltage difference

diff = Ea_ar - Vt - i_a(jj)*r_a;

% This code prevents us from reporting the first (unstable)

% location satisfying the criterion.
was_pos = 0;
for ii = 1:length(i_f);
if diff(ii) > 0
was_pos = 1;
end
if ( diff(ii) < 0 & was_pos == 1 )
break;
end;
A MATLAB program to calculate the terminal characteristic of this generator with armature reaction is end;
shown below:
% Save terminal voltage at this point
% M-file: prob9_25d2.m v_t(jj) = Vt(ii);
% M-file to calculate the terminal characteristic of a shunt i_l(jj) = i_a(jj) - v_t(jj) / ( r_f + r_adj);
% dc generator with armature reaction.
end;
% Get the magnetization curve. This file contains the
% three variables if_values, ea_values, and n_0. % Plot the terminal characteristic
clear all figure(1);
load p97_mag.dat; plot(i_l,v_t,'b-','LineWidth',2.0);
if_values = p97_mag(:,1); xlabel('\bf\itI_{L} \rm\bf(A)');
ea_values = p97_mag(:,2); ylabel('\bf\itV_{T} \rm\bf(V)');
n_0 = 1800; title ('\bfTerminal Characteristic of a Shunt DC Generator w/AR');
hold off;
% First, initialize the values needed in this program. axis([ 0 50 0 120]);
r_f = 24; % Field resistance (ohms) grid on;
r_adj = 10; % Adjustable resistance (ohms)
r_a = 0.18; % Armature + series resistance (ohms)
i_f = 0:0.005:6; % Field current (A)
n = 1800; % Generator speed (r/min)
n_f = 1000; % Number of field turns

% Calculate Ea versus If
Ea = interp1(if_values,ea_values,i_f);

% Calculate Vt versus If
Vt = (r_f + r_adj) * i_f;

% Find the point where the difference between the Ea

% armature reaction line and the Vt line is exactly
% equal to i_a*r_a. This will be the point where
254 255

The resulting terminal characteristic is shown below: If Radj decreases to 5 Ω, the total field resistance becomes 29 Ω, and the terminal voltage line gets
shallower. The new point where the distance between the E A and VT curves is exactly 4.5 V corresponds
to a terminal voltage of 115 V, as shown below.

9-26. If the machine in Problem 9-25 is running at 1800 r/min with a field resistance Radj = 10 Ω and an
armature current of 25 A, what will the resulting terminal voltage be? If the field resistor decreases to 5 Ω
while the armature current remains 25 A, what will the new terminal voltage be? (Assume no armature
reaction.)
SOLUTION If I A = 25 A, then I A RA = ( 25 A )( 0.18 Ω ) = 4.5 V. The point where the distance between the
Note that decreasing the field resistance of the shunt generator increases the terminal voltage.
E A and VT curves is exactly 4.5 V corresponds to a terminal voltage of 104 V, as shown below.
9-27. A 120-V 50-A cumulatively compounded dc generator has the following characteristics:
R A + RS = 0.21 Ω N F = 1000 turns
RF = 20 Ω N SE = 20 turns
Radj = 0 to 30 Ω, set to 10 Ω nm = 1800 r/min

The machine has the magnetization curve shown in Figure P9-7. Its equivalent circuit is shown in Figure
P9-10. Answer the following questions about this machine, assuming no armature reaction.

(a) If the generator is operating at no load, what is its terminal voltage?

(b) If the generator has an armature current of 20 A, what is its terminal voltage?

256 257
(c) If the generator has an armature current of 40 A, what is its terminal voltage'?
(d) Calculate and plot the terminal characteristic of this machine.

SOLUTION
(a) The total field resistance of this generator is 30 Ω, and the no-load terminal voltage can be found
from the intersection of the resistance line with the magnetization curve for this generator. The
magnetization curve and the field resistance line are plotted below. As you can see, they intersect at a
terminal voltage of 121 V.

(c) If the armature current is 40 A, then the effective field current contribution from the armature current
N SE 15
IA = (40 A ) = 0.6 A
NF 1000
and the I A (R A + RS ) voltage drop is I A (RA + RS ) = (80 A )(0.20 Ω ) = 8 V . The location where the
N SE
triangle formed by I A and I A R A exactly fits between the E A and VT lines corresponds to a terminal
NF
(b) If the armature current is 20 A, then the effective field current contribution from the armature current voltage of 116 V, as shown below.

N SE 20
IA = (20 A ) = 0.4 A
NF 1000

and the I A ( R A + RS ) voltage drop is I A ( R A + RS ) = ( 20 A ) ( 0.21 Ω ) = 4.2 V . The location where the
N
triangle formed by SE I A and I A RA exactly fits between the E A and VT lines corresponds to a terminal
NF
voltage of 120 V, as shown below.

258 259

i_a = 20;
Ea_a = interp1(if_values,ea_values,i_f + i_a * n_se/n_f);

% Find the point where the difference between the

% enhanced Ea line and the Vt line is 4 V. This will
% be the point where the line "Ea_a - Vt - 4" goes
% negative.
diff = Ea_a - Vt - 4;

% This code prevents us from reporting the first (unstable)

% location satisfying the criterion.
was_pos = 0;
for ii = 1:length(i_f);
if diff(ii) > 0
was_pos = 1;
end
if ( diff(ii) < 0 & was_pos == 1 )
break;
end;
end;

% We have the intersection. Tell user.

disp (['Ea_a = ' num2str(Ea_a(ii)) ' V']);
A MATLAB program to locate the position where the triangle exactly fits between the E A and VT lines is disp (['Ea = ' num2str(Ea(ii)) ' V']);
disp (['Vt = ' num2str(Vt(ii)) ' V']);
shown below. This program created the plot shown above. disp (['If = ' num2str(i_f(ii)) ' A']);
disp (['If_a = ' num2str(i_f(ii)+ i_a * n_se/n_f) ' A']);
% M-file: prob9_27b.m
% M-file to create a plot of the magnetization curve and the
% Plot the curves
% field current curve of a cumulatively-compounded dc generator
figure(1);
% when the armature current is 20 A.
plot(i_f,Ea,'b-','LineWidth',2.0);
hold on;
% Get the magnetization curve. This file contains the
plot(i_f,Vt,'k--','LineWidth',2.0);
% three variables if_values, ea_values, and n_0.
clear all
% Plot intersections
load p97_mag.dat;
plot([i_f(ii) i_f(ii)], [0 Vt(ii)], 'k-');
if_values = p97_mag(:,1);
plot([0 i_f(ii)], [Vt(ii) Vt(ii)],'k-');
ea_values = p97_mag(:,2);
plot([0 i_f(ii)+i_a*n_se/n_f], [Ea_a(ii) Ea_a(ii)],'k-');
n_0 = 1800;
% Plot compounding triangle
% First, initialize the values needed in this program.
plot([i_f(ii) i_f(ii)+i_a*n_se/n_f],[Vt(ii) Vt(ii)],'b-');
r_f = 20; % Field resistance (ohms)
plot([i_f(ii) i_f(ii)+i_a*n_se/n_f],[Vt(ii) Ea_a(ii)],'b-');
r_adj = 10; % Adjustable resistance (ohms)
plot([i_f(ii)+i_a*n_se/n_f i_f(ii)+i_a*n_se/n_f],[Vt(ii)
r_a = 0.21; % Armature + series resistance (ohms)
Ea_a(ii)],'b-');
i_f = 0:0.02:6; % Field current (A)
n = 1800; % Generator speed (r/min)
xlabel('\bf\itI_{F} \rm\bf(A)');
n_f = 1000; % Shunt field turns
ylabel('\bf\itE_{A} \rm\bf or \itE_{A} \rm\bf(V)');
n_se = 20; % Series field turns
title ('\bfPlot of \itE_{A} \rm\bf and \itV_{T} \rm\bf vs field
current');
% Calculate Ea versus If
axis ([0 5 0 150]);
Ea = interp1(if_values,ea_values,i_f);
set(gca,'YTick',[0 10 20 30 40 50 60 70 80 90 100 110 120 130 140
150]')
% Calculate Vt versus If
set(gca,'XTick',[0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0]')
Vt = (r_f + r_adj) * i_f;
legend ('Ea line','Vt line',4);
hold off;
% Calculate the Ea values modified by mmf due to the
grid on;
% armature current
260 261
(d) A MATLAB program to calculate and plot the terminal characteristic of this generator is shown
below. % Save terminal voltage at this point
v_t(jj) = Vt(ii);
% M-file: prob9_27d.m i_l(jj) = i_a(jj) - v_t(jj) / ( r_f + r_adj);
% M-file to calculate the terminal characteristic of a
% cumulatively compounded dc generator without armature end;
% reaction.
% Plot the terminal characteristic
% Get the magnetization curve. This file contains the figure(1);
% three variables if_values, ea_values, and n_0. plot(i_l,v_t,'b-','LineWidth',2.0);
clear all xlabel('\bf\itI_{L} \rm\bf(A)');
load p97_mag.dat; ylabel('\bf\itV_{T} \rm\bf(V)');
if_values = p97_mag(:,1); string = ['\bfTerminal Characteristic of a Cumulatively ' ...
ea_values = p97_mag(:,2); 'Compounded DC Generator'];
n_0 = 1800; title (string);
hold off;
% First, initialize the values needed in this program. axis([ 0 50 0 130]);
r_f = 20; % Field resistance (ohms) grid on;
r_adj = 10; % Adjustable resistance (ohms) The resulting terminal characteristic is shown below. Compare it to the terminal characteristics of the
r_a = 0.21; % Armature + series resistance (ohms)
shunt dc generators in Problem 9-25 (d).
i_f = 0:0.02:6; % Field current (A)
n = 1800; % Generator speed (r/min)
n_f = 1000; % Shunt field turns
n_se = 20; % Series field turns

% Calculate Ea versus If
Ea = interp1(if_values,ea_values,i_f);

% Calculate Vt versus If
Vt = (r_f + r_adj) * i_f;

% Find the point where the difference between the two

% lines is exactly equal to i_a*r_a. This will be the
% point where the line line "Ea - Vt - i_a*r_a" goes
% negative.
i_a = 0:1:50;
for jj = 1:length(i_a)

% Calculate the Ea values modified by mmf due to the

% armature current
Ea_a = interp1(if_values,ea_values,i_f + i_a(jj)*n_se/n_f);

% Get the voltage difference

diff = Ea_a - Vt - i_a(jj)*r_a;
9-28. If the machine described in Problem 9-27 is reconnected as a differentially compounded dc generator, what
% This code prevents us from reporting the first (unstable)
% location satisfying the criterion. will its terminal characteristic look like? Derive it in the same fashion as in Problem 9-27.
was_pos = 0; SOLUTION A MATLAB program to calculate and plot the terminal characteristic of this generator is shown
for ii = 1:length(i_f); below.
if diff(ii) > 0
was_pos = 1; % M-file: prob9_28.m
end % M-file to calculate the terminal characteristic of a
if ( diff(ii) < 0 & was_pos == 1 ) % differentially compounded dc generator without armature
break; % reaction.
end;
end; % Get the magnetization curve. This file contains the
262 263

% three variables if_values, ea_values, and n_0. plot(i_l,v_t,'b-','LineWidth',2.0);

clear all xlabel('\bf\itI_{L} \rm\bf(A)');
load p97_mag.dat; ylabel('\bf\itV_{T} \rm\bf(V)');
if_values = p97_mag(:,1); string = ['\bfTerminal Characteristic of a Cumulatively ' ...
ea_values = p97_mag(:,2); 'Compounded DC Generator'];
n_0 = 1800; title (string);
hold off;
% First, initialize the values needed in this program. axis([ 0 50 0 120]);
r_f = 20; % Field resistance (ohms) grid on;
r_adj = 10; % Adjustable resistance (ohms)
r_a = 0.21; % Armature + series resistance (ohms) The resulting terminal characteristic is shown below. Compare it to the terminal characteristics of the
i_f = 0:0.02:6; % Field current (A) cumulatively compounded dc generator in Problem 9-28 and the shunt dc generators in Problem 9-25 (d).
n = 1800; % Generator speed (r/min)
n_f = 1000; % Shunt field turns
n_se = 20; % Series field turns

% Calculate Ea versus If
Ea = interp1(if_values,ea_values,i_f);

% Calculate Vt versus If
Vt = (r_f + r_adj) * i_f;

% Find the point where the difference between the two

% lines is exactly equal to i_a*r_a. This will be the
% point where the line line "Ea - Vt - i_a*r_a" goes
% negative.
i_a = 0:1:26;
for jj = 1:length(i_a)

% Calculate the Ea values modified by mmf due to the

% armature current
Ea_a = interp1(if_values,ea_values,i_f - i_a(jj)*n_se/n_f);

% Get the voltage difference

diff = Ea_a - Vt - i_a(jj)*r_a;

% This code prevents us from reporting the first (unstable) 9-29. A cumulatively compounded dc generator is operating properly as a flat-compounded dc generator. The
% location satisfying the criterion. machine is then shut down, and its shunt field connections are reversed.
was_pos = 0; (a) If this generator is turned in the same direction as before, will an output voltage be built up at its
for ii = 1:length(i_f); terminals? Why or why not?
if diff(ii) > 0
was_pos = 1; (b) Will the voltage build up for rotation in the opposite direction? Why or why not?
end
(c) For the direction of rotation in which a voltage builds up, will the generator be cumulatively or
if ( diff(ii) < 0 & was_pos == 1 )
break; differentially compounded?
end; SOLUTION
end;
(a) The output voltage will not build up, because the residual flux now induces a voltage in the opposite
% Save terminal voltage at this point direction, which causes a field current to flow that tends to further reduce the residual flux.
v_t(jj) = Vt(ii);
i_l(jj) = i_a(jj) - v_t(jj) / ( r_f + r_adj); (b) If the motor rotates in the opposite direction, the voltage will build up, because the reversal in voltage
due to the change in direction of rotation causes the voltage to produce a field current that increases the
end; residual flux, starting a positive feedback chain.
(c) The generator will now be differentially compounded.
% Plot the terminal characteristic
figure(1);
264 265
9-30. A three-phase synchronous machine is mechanically connected to a shunt dc machine, forming a motor- and the reactive power supplied by the ac machine to the ac power system is
generator set, as shown in Figure P9-11. The dc machine is connected to a dc power system supplying a
constant 240 V, and the ac machine is connected to a 480-V 60-Hz infinite bus. QAC = S sin θ = (50 kVA ) sin cos −1 (0.8) = 30 kvar

The power out of the dc motor is thus 40 kW. This is also the power converted from electrical to
mechanical form in the dc machine, since all other losses are neglected. Therefore,
Pconv = E A I A = (VT − I A R A ) I A = 40 kW

VT I A − I A 2 R A − 40 kW = 0
The base resistance of the dc machine is

VT ,base 2 ( 230 V )
2

Rbase,dc = = = 1.058 Ω
Pbase 50 kW

Therefore, the actual armature resistance is

R A = (0.04)(1.058 Ω ) = 0.0423 Ω

Continuing to solve the equation for Pconv , we get

The dc machine has four poles and is rated at 50 kW and 240 V. It has a per-unit armature resistance of
0.04. The ac machine has four poles and is Y-connected. It is rated at 50 kVA, 480 V, and 0.8 PF, and its 0.0423 I A 2 − 230 I A + 40, 000 = 0
saturated synchronous reactance is 2.0 Ω per phase. I A2 − 5434.8 I A + 945180 = 0
All losses except the dc machine’s armature resistance may be neglected in this problem. Assume that the I A = 179.9 A
magnetization curves of both machines are linear.
and E A = 222.4 V.
(a) Initially, the ac machine is supplying 50 kVA at 0.8 PF lagging to the ac power system.
1. How much power is being supplied to the dc motor from the dc power system? Therefore, the power into the dc machine is VT I A = 41.38 kW , while the power converted from electrical
to mechanical form (which is equal to the output power) is E A I A = ( 222.4 V )(179.9 A ) = 40 kW . The
2. How large is the internal generated voltage E A of the dc machine?
internal generated voltage E A of the dc machine is 222.4 V.
3. How large is the internal generated voltage E A of the ac machine?
The armature current in the ac machine is
(b) The field current in the ac machine is now increased by 5 percent. What effect does this change have
S 50 kVA
on the real power supplied by the motor-generator set? On the reactive power supplied by the motor- IA = = = 60.1 A
generator set? Calculate the real and reactive power supplied or consumed by the ac machine under 3 Vφ 3 ( 480 V )
these conditions. Sketch the ac machine’s phasor diagram before and after the change in field current. I A = 60.1∠ − 36.87° A
(c) Starting from the conditions in part (b), the field current in the dc machine is now decreased by 1 Therefore, the internal generated voltage E A of the ac machine is
percent. What effect does this change have on the real power supplied by the motor-generator set? On
the reactive power supplied by the motor-generator set? Calculate the real and reactive power supplied E A = Vφ + jX S I A
or consumed by the ac machine under these conditions. Sketch the ac machine’s phasor diagram before
E A = 277∠0° V + j ( 2.0 Ω )( 60.1∠ − 36.87° A ) = 362∠15.4° V
and after the change in the dc machine’s field current.
(d) From the above results, answer the following questions: (b) When the field current of the ac machine is increased by 5%, it has no effect on the real power
supplied by the motor-generator set. This fact is true because P = τω , and the speed is constant (since the
1. How can the real power flow through an ac-dc motor-generator set be controlled? MG set is tied to an infinite bus). With the speed unchanged, the dc machine’s torque is unchanged, so the
2. How can the reactive power supplied or consumed by the ac machine be controlled without total power supplied to the ac machine’s shaft is unchanged.
affecting the real power flow? If the field current is increased by 5% and the OCC of the ac machine is linear, E A increases to
SOLUTION E A′ = (1.05)( 262 V ) = 380 V
(a) The power supplied by the ac machine to the ac power system is
The new torque angle δ can be found from the fact that since the terminal voltage and power of the ac
PAC = S cos θ = (50 kVA )( 0.8) = 40 kW machine are constant, the quantity E A sinδ must be constant.

266 267

E A sin δ = E A′ sin δ ′ The phasor diagram of the ac machine before and after the change in dc machine field current is shown
EA 362 V below.
δ ′ = sin −1 sin δ = sin −1 sin15.4° = 14.7° E A2
E A′ 380 V

Therefore, the armature current will be E A1

E A − Vφ 380∠14.7° V − 277∠0° V
IA = = = 66.1∠ − 43.2° A jX I
jX S j 2.0 Ω S A
I A2 Vφ
The resulting reactive power is
I A1
Q = 3 VT I L sin θ = 3 ( 480 V )( 66.1 A ) sin 43.2° = 37.6 kvar
(d) The real power flow through an ac-dc motor-generator set can be controlled by adjusting the field
The reactive power supplied to the ac power system will be 37.6 kvar, compared to 30 kvar before the ac current of the dc machine. (Note that changes in power flow also have some effect on the reactive power of
machine field current was increased. The phasor diagram illustrating this change is shown below. the ac machine: in this problem, Q dropped from 35 kvar to 30 kvar when the real power flow was
E A1 E A2 adjusted.)
The reactive power flow in the ac machine of the MG set can be adjusted by adjusting the ac machine’s
field current. This adjustment has basically no effect on the real power flow through the MG set.

I A1 Vφ jX I
I A2 S A

(c) If the dc field current is decreased by 1%, the dc machine’s flux will decrease by 1%. The internal
generated voltage in the dc machine is given by the equation E A = K φ ω , and ω is held constant by the
infinite bus attached to the ac machine. Therefore, E A on the dc machine will decrease to (0.99)(222.4 V)
= 220.2 V. The resulting armature current is
VT − E A 230 V − 220.2 V
I A,dc = = = 231.7 A
RA 0.0423 Ω
The power into the dc motor is now (230 V)(231.7 A) = 53.3 kW, and the power converted from electrical
to mechanical form in the dc machine is (220.2 V)(231.7 A) = 51 kW. This is also the output power of the
dc machine, the input power of the ac machine, and the output power of the ac machine, since losses are
being neglected.
The torque angle of the ac machine now can be found from the equation
3Vφ E A
Pac = sin δ
XS
P X
δ = sin ac S = sin −1
−1 (51 kW )( 2.0 Ω ) = 18.9°
3Vφ E A 3 ( 277 V )( 380 V )

The new E A of this machine is thus 380∠18.9° V , and the resulting armature current is
E A − Vφ 380∠18.9° V − 277∠0° V
IA = = = 74.0∠ − 33.8° A
jX S j 2.0 Ω

The real and reactive powers are now

P = 3 VT I L cos θ = 3 ( 480 V )(74.0 A ) cos 33.8° = 51 kW
Q = 3 VT I L sin θ = 3 ( 480 V )( 74.0 A ) sin 33.8° = 34.2 kvar

268 269