Research Journal of Applied Sciences, Engineering and Technology 11(2): 190-197, 2015
DOI: 10.19026/rjaset.11.1707
ISSN: 2040-7459; e-ISSN: 2040-7467
© 2015 Maxwell Scientific Publication Corp.
Submitted: April 23, 2015
Accepted: June 14, 2015
Published: September 15, 2015
Research Article
Reduction of Losses at the Start-up of DC Motor Using the Voltage Amplitude
Variation Law
1
Eng. Abeer Hawatmeh, 1Mohammad AbuZalata and 2Ayman AL-Rawashdeh
1
Mechatronics Engineering Department,
2
Electical Engineering Department, Faculty of Engineering Technology, Al Balqa Applied University,
Amman, Jordan
Abstract: This study is concerned for exploring the possibilities of reducing losses in electric drives of heavy startup conditions. Under the latter refers to electric drives with large quantities of the total moment of inertia of the
rotating parts of the electric drives and machinery, in the presence of the moment of inertia at the starting process.
Keywords: DC motor, moment of inertia, startup losses, variation law
variations. In particular, the greatest attention is paid to
a law of linear rising voltage with the limitation by its
nominal value (Andreev and Sabinin, 1963; Keys,
1985). The starting up losses of a separately excited DC
motor will be studied.
INTRODUCTION
In Abuzalata et al. (2011) the relationships for
calculating the start-up losses in a separately
excited DC motor is derived using the dynamic
equations of the electric drive, One result of this
study is concluded that among other factors which
affect on the start-up losses is the variation of the
armature voltage, formed by an electric power
source during start-up process. In particular, it is shown
that, if the form of the voltage is as steps with equal
values and durations to power losses in the electric
drive when it starts without load is:
∆Α st = J
ω 20
2Κ
METHODOLOGY
The dynamic equations of the electric drive
represented by:
(Τm s + 1)ω (t ) = Va
Cе
(1)
Mm = J
where,
J : The total moment of inertia of all rotating parts
ω0 : No-load speed
K : The number of steps of the armature voltage
−
Ra
TL
C eCm
dω
− TL
dt
(2)
(3)
where, Eq. (2) is the equation of electrical equilibrium
in the rotor and (3) is the equation of mechanical
equilibrium. In these relations:
Tm
ω(t)
Va
Ra
Ce, Cm
: Electromechanical time constant –[s]
: The angular velocity of rotor –rad/s
: The applied armature voltage-[V]
: Resistance of the armature -[Ohm]
: Coefficients determined from structural data
of the motor
: Torque developed by motor-[N.m]
Tm
: The is moment of the (load)[N.m]
TL
J
: The total moment of inertia of all moving
parts and electric machinery-[kg/m2]
S = d/dt : The symbol of differentiation
Equation (1) shows that the losses in the armature
circuit of the motor are inversely proportional to the
number of steps of the armature voltage. It is assumed
that the acceleration in each step should be continued
until reaching the steady state speed. For this type of
voltage law relationships are achieved for determining
the energy losses and for the start-up of the electric
drive with load (Chilikin, 1965; Chilikin and Sandler,
1981; Andreev and Sabinin, 1963). However, to
minimize start-up losses, the step voltage control law is
not the best.
The aim of this study is to reduce the start-up
losses to show the advantages of other laws of voltage
Linear rising law with limited of nominal final
value can be represented as:
Corresponding Author: Eng. Abeer Hawatmeh, Mechatronics Engineering Department, Faculty of Engineering Technology, Al
Balqa Applied University, Amman, Jordan
This work is licensed under a Creative Commons Attribution 4.0 International License (URL: http://creativecommons.org/licenses/by/4.0/).
190
Res. J. Appl. Sci. Eng. Technol., 11(2): 190-197, 2015
In the time domain the Eq. (6) is represented as:
ω (t ) =
=
ω0
nTm
Κr
Се
t−
t
t

−


V −
Vn
V
t − n + n e Tm =
t − Tm 1 − e Tm  =
Се n Се n
 Се nTm


ω0
n
+
ω0
n
e
−
t
Tm
(7)
Assuming that the transient process comes to an
−
t
end when ( e Tm ≈ 0 ) for time t = 5Tm, let tr >> 5Tm. Let tr
= 20Tm, as shown in Fig. 2, during 5Tm, the component
Fig. 1: The terminal armature voltage
ω0
n
−
e
t
Tm
ω0
n
tends to zero and Eq. (7) remain as:
−
e
t
Tm
ω (t ) =
ω0
nTm
ω (t ) = ω 0 −
V n ⟨ Vr
Κ rt
Va = 
 Vn
(4)
n
ω (t ) = ω − ω0 + ω0 1 − e
0

n
Vn
tr
Va ⟨ Vn
Va = Vn
(Τm s + 1)ω (s ) = Va , Or ω (s ) = Va
С e (Tm s + 1)
Се
−
t
Tm
n 
t
ω0 − Tm

e
 = ω0 −
n


(8)
During t = tr + 5Tm the transient process is over and
speed ω (t) = ω0.
Thus, if the time of linear ramp voltage tr = nTm,
then the transient process of the speed is ttp = nTm + 5
Tm (n + 5)Tm from the achieved velocity law, we find
the behavior of the armature current vs. time.
Using the equation of mechanical equilibrium Cm Ia
= J dω/dt; then Ia = J/Cm dω/dt; by differentiating the
Eq. (7) the equation below is achieved:
Using Laplace transformation of the dynamic
equation of start up without load, the result is:
t
t
−
ω −
ω 
dω ω 0
=
− 0 e Tm = 0 1 − e Tm
dt nTm nTm
nTm 
(5)




For the armature current in this part we have:
For the linear rising part vs. time of the voltage:
Va (s ) =
n
ω0
Va = Vr
Relation (4) corresponds to the graph in Fig. 1:
where, Kr = tga = Vn/tr Slope (0, 1, 2, 3, 4…), tr = nTm
(tr- voltage rising time)
K r = tgα =
ω 0 if t = t , t = nT then
s
m
Or, in other words, the speed is different from the
ideal no load speed with the value ∆ω = ω0/n, but at the
end of the linear ramp voltage the transient process of
the speed continues, then for the part of the transient
process when tr<t<∞ we have the following equation:
Fig. 2: Chart of the voltage changing
 Κ rt
Va = 
 Vr
t−
Vn
Κ
= r
nTm s 2 s 2
t
−
J ω 0 
Tm
Ia =
1− e
C m nTm 




(9)
Then Eq. (5) becomes:
After a period of time equals 5Tm at the origin the
Κ rTm
Κr
Κr
=
−
ω (s ) =
С e (Tm s + 1)s 2 С е S 2 С е s (Tm s + 1)
(6)
e
191
−
t
Tm
≈ 0 and the Eq. (9) becomes:
Res. J. Appl. Sci. Eng. Technol., 11(2): 190-197, 2015
Fig. 3: The plot of the starting current
Fig. 4: Charts of voltage, speed and armature current simulation results
Ia =
J ω0
J
=
Сm nTm Сm
ω0
n
JRa
Сe Сm
=
(10)
Vn
nRa
The third region; t > nTm
t
V −
J ω0 − Tm
Ia =
e
= n e Tm
nRa
Cm nTm
 Vn 
2
∫0 I Ra dt =  nR a 
5T m
Thus, for the armature current, as well as for the
speed, we can distinguish three regions of the transient
process:
The first region; 0 ≤ t ≤ 5Tm
=
The second region; 5Tm< t ≤ nTm
=
The corresponding current schedule is shown in
Fig. 3. Charts show voltage, speed and armature current
for the transient process which is presented in Fig. 4.
Starting losses will also consist of three components;
In the first period:
For the latter part of the transient process:
t
=
1 − ∆Α e1 =
t
−

1 − e Tm
R
∫0 a 

2 5T
m
2

 dt


In the second period:
 Vn
∫5T  nR a
m
nTm
∆Α e2 =
192
2
 V

 Ra dt =  n
 nR a

2

 Ra (nTm − 5Tm )

Res. J. Appl. Sci. Eng. Technol., 11(2): 190-197, 2015
Table 1: Comparison, energy levels of the control voltage and time ramp voltage
Number of the voltage steps
3
4
n- for linear law
10
15
The ratio of losses ∆ Astep / ∆ Alinear
1.85
2.009
5Tm
∫
0
6
25
2.17
2
2t
−
 Vn 

 Ra e Tm dt
 nR a 
If the last option ∆Ast = J
It is useful to sum the first two components, so:
2
 V 
∆Αe1 + ∆Αe3 =  n  Ra
 nR a 
2
5Tm
5Tm
∫
0

 Ra

 V
=  n
 nR a
t
 5Tm
−

 R я t + 2Tm e Tm
 0


0
t

−
1 − e Tm


2
5Tm

dt =


2t
−2
0
Tm − Tm
e
2
2
 V

 Ra (5Tm − 2Tm + Tm ) =  n
 nRa

2
 v 
∆Α e1 + ∆Α e 2 =  n  Ra 4Tm
 nR a 
 V
=  n
 nR a
5Tm
0




∆Α = ∆Α e1 + ∆Α e2 + ∆Α e 3 =  Vn
 nR
 a
, then the linear growth
ttp = nTm + 5Tm
It is useful to compare, when K ≥ 2 and assume K
= 2. Then:
2

 Ra 4Tm

ttp = 5Tm ∗ 2 = nTm + 5Tm `
From Eq. (13) if n = 5, then ∆Ast =
Then the summation of all starting losses is:
=
8
35
2.25
∆Ast = [n -1]. How is the difference between these
losses? By comparing them with each other is necessary
for consideration of ways to set the same start time of
transient process. As previously noted, starting with a
step function ttp = 5Tm * K, where K- the number of
start-up stages of the armature voltage. When the ramp
of armature voltage is limited with the nominal voltage:
2
2t 
−

 + e Tm dt =




t
2t
2t
−
−
−

1 − 2e Tm + e Tm + e Tm


V
=  n
 nRa
∫
7
30
2.217
equals 20In. If it is necessary to starting current is equal
In, you should install tr = 20Tm.
It is interesting the losses comparison between a
linear start-up and a step-laws method.
In the third period:
∆Αe3 =
5
20
2.1
up ∆Ast =

 Ra [4Tm + nTm − 5Tm ] =

2
V n2
JR a
[n − 1] = J ω20 [n − 1]
2
n Ra C еC m
n
(11)
If n >>1, then ∆A ≈ J / n
(12)
=
.
at step start-
and ramp voltage. This shows
that the efficiency of the second method ÷ =
.
1.56 times the step in starting.
For comparison, a larger number of energy levels
of the control voltage and the corresponding time ramp
voltage are made in the Table 1.
The table shows that, compared with step-start
performance increases linearly with an increase in the
number of stages and consequently the rise time. But it
hardly makes sense to take the number of steps above
start four to eight. The equivalent rise time is it (15 35) Tm and inrush current is equal (1.2 – 0.6)
Ir. Therefore we can say that when you start the motor
with a voltage ramp the starting losses are about 2 times
lower than that for step change.
2
2
Finally ∆Αs = Jω 0 (n − 1)
n2
∗ (13)
Motor starting in the presence of the moment of
Inertia: Electric drive starting conditions deteriorate in
the presence of the electric moment of inertia. We
assume that such a reactive torque that comes with the
beginning of the movement and that for the sake of
simplicity ML = const (Andreev and Sabinin, 1963).
Then the dynamic equation of the electric drive of
the type:
where n = tr/Tm;
where n-number of Tm, that fit into the length of time of
the armature voltage ramp.
From these expressions, in addition to Eq. (11) and
(12), it is very important the equation (10). It allows
selecting the time of the ramp of the control action
depending on the desired acceleration current (see DC
section in Fig. 3). If at the start up the current exceeds
the nominal start up current, say ten times, then by
selecting tr = nTm any value can be limited. For one of
the above selected motors the starting rush current
(Τm s + 1)ω (t ) =
193
Va −
Ra
ML
Сm
, Can be converted to
Ce
Res. J. Appl. Sci. Eng. Technol., 11(2): 190-197, 2015
(Τm s + 1)ω (t ) = Va − I L Ra And:
delay on the time axis is equal to the rise time of the
armature current to a value Ia = IL. But in this case
significantly increases the difficulties of the analytical
solutions (15). Therefore, trying to stay within the
framework of analytical solutions, we adopt an
approximate solution in the form of (15) and the error
estimate assumptions simulation on a PC, knowing that
it is small enough. The equation of mechanical
equilibrium at a constant load torque becomes:
Ce
(Τm s + 1)ω (t ) = Va − ∆VL
(14)
Ce
The difference (Va - ∆ VL) is equal to the armature
voltage, which determines the rate of motor
acceleration and its rate at steady state. In particular, if
the voltage rise to a value, the rate is established not on
the level Vr and where, then expression (7) for speed at
the site will change linearly increasing voltage to the
form:
ω (t ) =
ωc
nTm
t−
ωc
n
+
ωc
n
e
−
t
Tm
Cm I a = J
Ia =
(15)
dω
+ Cm I L
dt
J dω
or we get dω from
+ IL
dt
С m dt
t
−

ω −t
ω 
dω ω L
=
− L e Tm = L 1 − e Tm 
dt nTm nTm
nTm 

Strictly speaking, the solution (15) did not
accurately reflect the behavior of a variable ω (t), since
the latter should not start from the origin and have some
For the armature current view of (18) we have:
Fig. 5: starter armature current
Fig. 6: Charts of voltage, speed and armature current simulation result
194
(16)
(17)
(18)
Res. J. Appl. Sci. Eng. Technol., 11(2): 190-197, 2015
Ia =
Second period:
t
−

J ωL 
1 − e Tm  + I L


С m nTm 

2
nTm

V
∆Α e 2 = ∫  L + I L  Ra dt =
nR
a

5Tm 
Substituting in this expression instead of:
 VL2

2V I
Ra + L L Ra + I L2 Ra dt =
2 2
R
nR
a
a

5Tm
nTm
∫  n
U
ωс = c
Cm
t
And T = JRa we obtain I = VL 1 − e − Tm  + I
m
a
 L
CeСm
nR a 

For the latter part of the transition process, when
we have the following changes in the current law:
VL
e
nR a
Ia =
t
−
Tm
=
nTm
nTm
V L2 Ra nTm 2V L I L
t +
Ra t + I L2 Ra t
=
2 2
nRa
n Ra 5 T m
5 Tm
5 Tm
=
VL2
(nTm − 5Tm ) + 2vL I L (nTm − 5Tm ) + I L2 Ra (nTm − 5Tm )
n
n 2 Ra
The third period:
+ IL
nTm + 5Tm
∫
∆Α э3 =
nTm
The corresponding plot is shown in Fig. 5 the
current charts conduct voltage, speed and armature
current for the transition process is presented in Fig. 6.
+
i =1
2
a
V 2 R 5Tm
V 2R
= L2 2a t − 2 L2 2a
n Ra
n Ra 0
=
V L2 5Tm
2
n Ra
+
0
−2
5T m
0
+
2VL I L
5Tm + I L2 Ra 5Tm
n
V L2 Tm 8V L I L Tm 2V L I L (nTm − 5Tm )
+
+
+
n
n
2n 2 Ra
10VL I LTm
+ 5 I L2 RATm + I L2 Ra (nTm − 5Tm ) + 5I L2 RaTm =
n
V 2T (n − 1) 2VL I LTm (n + 4) 2
= L m2
+
+ I L RaTm (n + 5) =
n
n Ra
2


 + I L  Ra dt =
∆Αe1 = ∫ I Ra dt = ∫


0
0

2
t
t
−
 V L2
V2
V2 − 
 2 2 − 2 2 L 2 e Tm + 2 L 2 e Tm 
5Tm
n Ra
n Ra
 n Ra

= ∫ 
Ra dt
t
−
0 
VL I L
VL I L Tm

+ I L2
+ 2V nR − 2 nR e

a
a


V
 L
 nRa
5Tm
5Tm
VL I l Ra
nR a
Tm − Tm
e
2
+
First period:
−2
2t
−
VL2Tm 10VL I LTm
+
+ I L2 Ra 5Tm
n
2n 2 Ra
3
3,5V L2Tm V L2 (nTm − 5Tm )
∆Α e = ∑ ∆Α ie =
+
n 2 Ra
n 2 Ra
i =1
3
5Tm
VL2
n 2 Ra
=
∆Α e st = ∑ ∆Α ei .
2t
=
0
To calculate the energy losses during acceleration
the engine is useful to consider separately three regions
of acceleration and then summarize them in the form of
T −
− m e Tm
2
∫
 V 2 − T2t

2V I
2
 2 L 2 e m Ra + L L Ra + I L Ra dt =
n
R
nR
a
a


5Tm
RESULTS AND DISCUSSION
2

 V − Tt
 L e m + I L  Ra dt =

 nRя
t
−

1 − e Tm


− Tme
−
t
5Tm
Tm
0
−
t
5Tm
Tm
+ I L2 Ra t
=
0
V L2 Tm
n Ra
+
V L2 Tm
n 2 Ra ∗ 2
∆Α e st =
J ω L2 (n − 1) 2ω L M L (n + 4 )
+
+ I L2 R a t пп
n2
n
(19)
In order to have an understanding of the
components of losses in (19), we take a numerical
example: P = 50 kW, J = 20 kgm2, Ir = 220a, Ra = 0.05
ohm, Ce = Cm = 1.4, Tm = 0, 51s, Ic = 200A, Mc = 280
N.ωL, = 154.9 rad/s.
Let n = 15 Then
=
0
+
VL2Tm (n − 1) 2VL I LTm (n + 4) 2
+
+ I L RaTm (n + 5) =
n
n 2 Ra
Then
V2
+ 2L 2
n Ra
5Tm
V L2 Tm (3.5 + n − 5 + 0.5) 2VL I L (2nTm − 10Tm + 10Tm + 8Tm )
+
n
n 2 Ra
+ I L2 Ra (5Tm + nTm − 5Tm + 5Tm ) =
2VL I L Ra 5Tm
t −
nRa
0
− Tm e
2
=
2V L I L 5Tm
n
∆Α e1 =
V I T
− 2 L L m + I L2 Ra 5Tm
n
∆Α e3
195
Joules [j];
2 ∗154,9 ∗ 280(19)
= 109875 j;
15
= 200 2 ∗ 0,05 ∗ 0,51 ∗ 20 = 20400 j;
∆Αe2 =
3.5VL2Tm 8VL I LTm
=
+
+ I L2 Ra 5Tm
n 2 Ra
n
20 ∗154,9 2 (14 )
= 29859
15 2
Res. J. Appl. Sci. Eng. Technol., 11(2): 190-197, 2015
450000
= Jω L2 n 2 − 2 Jω L n 2 + 2 Jω L n + 2ω L M L n 3
400000
− 2ω L M L n 3 + 8ω L M L n 2 + I L2 R a Tm n 4 = 0
= − J ω L n + 2 J ω L + 8ω L M L n + I L2 R a Tm n 3 = 0
350000
300000
Finally I L2 Ra Tm n 3 + n(8ω L M L − Jω L ) + 2 Jω L = 0 ;
This equation can be found analytically or by
selection of "n". We use the latter:
250000
200000
150000
100000
Let n = 24,
50000
2
then ∆Αe1 = 20 *154,9 * 23 = 19161 j,
2
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
0
24
2 * 154,9 * 280 * 28
∆Αe 2 =
= 101201 j;
24
Fig. 7: Plotted to determine the optimum energy losses when
the load torque, equal to half the nominal value
∆Αe3 = 200 2 * 0,05 * 0,51* 29 = 29580 j;
300000
n = 26
250000
20 * 154,9 2 * 25
= 17747
26 2
2 *154 ,9 * 280 * 30
∆Α e 2 =
= 100089 j;
26
∆Α e1 =
200000
150000
∆Αe3 = 200 2 * 0,05 * 0,51* 31 = 31620 j;
100000
n = 28
50000
20 *154,9 2 * 27
= 16526 j
28 2
2 *154,9 * 280 * 32
∆Αe2 =
= 99136 j;
28
∆Αe3 = 200 2 * 0,05 * 0,51* 33 = 33660 j;
∆Αe1 =
160
13
5
130
85
110
54
38
32
8
16
4
0
Fig. 8: Plotted to determine the decreasing values of load
torque, the optimum energy loss shifts toward larger
"n"
n = 30
We investigate (19) on the optimum (minimum
losses) for "n". Increase the length of the linear growth
of the armature voltage up to n = 20. Then:
∆Αe1 =
20 *154,92 * 29
= 15462
30 2
∆Αe 2 =
2 *154,9 * 280 * 34
= 98309
30
j;
∆Αe3 = 2002 * 0,05 * 0,51* 35 = 35700 j;
20 ∗154 ,9 (19 )
= 22794 j;
20 2
2 ∗154,9 ∗ 280(24)
∆Α e2 =
= 104093 j;
20
∆Α e3 = 200 2 ∗ 0,05 ∗ 0,51 ∗ 25 = 25500 j;
2
∆Α e1 =
That is, when n = 26, ∑∆A = 149456j with n = 28
∑∆A = 149322j, with n = 30 ∑∆A = 149471j.
Additionally in Fig. 7 plotted to determine the
optimum energy-loss program using excel, at the
moment the load is equal to the nominal value. From it
also shows that the minimum loss occurs when "n" =
28.
From the graph in Fig. 8 shows that with
decreasing values of load torque, the optimum energy
loss shifts toward larger "n" (n opt = 48).
We see that as n increases the first component of
losses (19) decreases, decreases slightly and the second
component and the third " RЯ (n + 5)Tm is growing as
increasing the acceleration time.
It follows that the function (19) has an extremum
on (n). Find the value of (n) providing for (19)
minimum:
CONCLUSION
•
d∆Α e Jω L2 n 2 − 2nJω L (n − 1)
=
dn
n4
2ω L M L n − 2ω L M L (n + 4 )
+
+ I L2 Ra Tm = 0
n2
•
196
One of the important factors affecting the value of
start-up losses is the variation of armature voltage,
formed by an electric transformer.
A relation (10), allowing to select parameters of the
law ramp voltage to provide the selected value of
the current acceleration.
Res. J. Appl. Sci. Eng. Technol., 11(2): 190-197, 2015
•
Andreev, V.P. and Y.A. Sabinin, 1963. Osnovy
Elektroprivoda. Fundamentals of electric drives.
Gosenergoizdat, Moscow-Leningrad, pp: 674-675.
Chilikin, M.G., 1965. Obshchiy Kurs Elektroprivoda
(General Course of Electric Drive). Energiya Press,
Moscow.
Chilikin, M.G. and A. Sandler, 1981. The Total Rate of
Electric. Energoatomizdat, Moscow.
Keys, V.I., 1985. Electric Theory. Energoatomizdat,
Moscow.
Linear law of increase effective armature voltage
step, since approximately 2-times reduced start-up
losses.
REFERENCES
Abuzalata, M., O. Barbarawi, M. Awad, S. Alqsos and
S. Asad, 2011. A new approach for evaluation of
the energy losses when starting-up separately
excited DC motors. Int. J. Elec. Power Eng., 5:
157-160.
197