BATAAN PENINSULA STATE UNIVERSITY
COLLEGE OF ENGINEERING & ARCHITECTURE
DC MACHINES Lab Simulation # _____
DC SERIES MOTOR
I. OBJECTIVES
1. To study the characteristics of a series dc motor without armature reaction
2. To calculate the efficiency of the motor and its power stages.
II. INTRODUCTORY LESSON
1. The series winding and armature are
connected in series.
2. The field winding carries the same
current as the armature winding.
3. It is called a universal motor. It is
universal in the sense that it will run
equally well using either an ac or a dc
voltage source.
II. PROCEDURE
1. Working Problem: 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 rpm. Its magnetization curve is shown on next page. The core
losses are 200 W, and the mechanical losses are 240 W at full load. Assume that the
mechanical losses vary as the cube of the speed of the motor and that the core losses
are constant.
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.
SERIES DC MOTOR
2. Ask your laboratory instructor for a copy of “MagDCSeries.dat.” This data is the field
current (If) and the generated voltage EA of a motor of Figure 1.
3. Refer to circuit of Figure 1.
1
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IL
Po = 7.5 hp RA = 0.2 Ω
0.16 +
IA RS = 0.16 Ω NSE = 27 turns per pole
+
NRated = 1200 rpm
M EA VA
120 V RS = 0.04 Ω
-
Radj = 100 to 400 Ω rotational loss = 240 W
-
4. To compute for (a) the efficiency of the motor at full load, answer or calculate the
following:
a. Output power Po = hp x 746 = _________ W
b. Input power Pin = VT IL = _____________ W
c. Efficiency = Po/Pin = _______________ %
5. To compute for the speed and efficiency of the motor with an armature current I A = 35
A, calculate the following:
a. Pin = VT IL = _________ W
b. The internal generated voltage at this condition:
EA1 = VT – IA(RA + RS)= _______________ V
c. The internal generated voltage at rated condition (full load, IA = 58 A)
EA1 = VT – IA(RA + RS)= _______________ V
d. The final speed given by equation:
𝐸𝐴2 𝐸𝐴01
𝑛2 = 𝑛
𝐸𝐴1 𝐸𝐴02 1
EAO1 = ________ V (obtained from the graph of IA = 58 A)
EAO2 = ________ V (obtained from the graph with IA = 35 A)
n1 (given)
6. To compute for the efficiency at rated condition, calculate the following:
a. Armature power developed PA:
PA = EA IA = ___________ W
SERIES DC MOTOR
b. Mechanical Power PMech:
𝑛2 3
𝑃𝑀𝑒𝑐ℎ = ( ) 𝑃𝑚𝑒𝑐ℎ 𝐿𝑜𝑠𝑠𝑒𝑠 = ___________𝑊
𝑛1
c. Output power Po: 2
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Po = PA – Pmech – Pcore = __________ W
d. Efficiency:
η = Po/Pin = ________ %
7. Plot the torque-Speed characteristics of this series motor by compiling program5.
8. Complete the Summary box.
Summary Box
4. a. b. c.
5. a. b. c. d.
6. a. b. c. d.
III. CONCLUSION
In this simulation experiment, we can note that:
1. The speed decreases as the torque increases.
SERIES DC MOTOR
3
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% M-file: program5.m
% M-file to create a plot of the torque-speed curve of the
% the series dc motor in Problem 9-13.
% Get the magnetization curve. Note that this curve is
% defined for a speed of 1200 r/min.
h0=figure('Units', 'normalized',...
'Color', [1 1 1],...
'HitTest', 'off',...
'Position', [0.01 0.05 0.8 0.7],...
'Name', 'Series DC Motor Torque-Speed Characteristics',...
'NumberTitle','off');
load p95_mag.dat
if_values = p95_mag(:,1);
ea_values = p95_mag(:,2);
n_0 = 1200;
% First, initialize the values needed in this program.
v_t = 120; % Terminal voltage (V)
r_a = 0.36; % Armature + field resistance (ohms)
i_a = 9:1:58; % Armature (line) currents (A)
% Calculate the internal generate voltage e_a.
e_a = v_t - i_a * r_a;
% Calculate the resulting internal generated voltage at
% 1200 r/min by interpolating the motor's magnetization
% curve. Note that the field current is the same as the
% armature current for this motor.
e_a0 = interp1(if_values,ea_values,i_a,'spline');
% Calculate the motor's speed, using the known fact that
% the motor runs at 1050 r/min at a current of 58 A. We
% know that
%
% Ea2 K' phi2 n2 Eao2 n2
% ----- = ------------ = ----------
% Ea1 K' phi1 n1 Eao1 n1
%
% Ea2 Eao1
% ==> n2 = ----- ------ n1
% Ea1 Eao2
%
% where Ea0 is the internal generated voltage at 1200 r/min
% for a given field current.
%
% Speed will be calculated by reference to full load speed
% and current.
n1 = 1050; % 1050 r/min at full load
Eao1 = interp1(if_values,ea_values,58,'spline');
Ea1 = v_t - 58 * r_a;
% Get speed
Eao2 = interp1(if_values,ea_values,i_a,'spline');
n = (e_a./Ea1) .* (Eao1 ./ Eao2) * n1;
% 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 curve
SERIES DC MOTOR
figure(1);
plot(t_ind,n,'b-','LineWidth',2.0);
hold on;
xlabel('\bf\tau_{ind} (N-m)');
ylabel('\bf\itn_{m} \rm\bf(r/min)');
title ('\bfSeries DC Motor Torque-Speed Characteristic');
grid on;
hold off;
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5
SERIES DC MOTOR
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