Electrical Power and Machines
Lecture 5
<Dr Ahmed El-Shenawy>
<Dr Hadi El Helw>
DC Motor
theory of operation
• In a dc motor, the stator poles are supplied by dc excitation
current, which produces a dc magnetic field.
• The rotor is supplied by dc current through the brushes,
commutator and coils.
• The interaction of the magnetic field and rotor current generates a
force that drives the motor
2
IL
IF
IL
IF
Ia
Ia
RF
RF
Ra
Ra
Vt
Vt
Ea
Ea
Ea = KaφW
Generator
Vt = Ea – IaRa
Driving e.m.f
Motor
Back e.m.f
Vt = Ea + IaRa
Starting of DC motor
VS = Ea + Ia R
At Starting Ea is zero (back e.m.f = 0), so
Ia = VS/R , and since R is very small then Ia
is very big. Consequently a DC motor
should never be started at rated voltage
and an external resistance must be
added and acts as temporarily starting
resistance that is removed as soon as the
armature has attained its normal speed,
then
Ia = VS/(Ra+RS)
Ia
Ra
RF
VS
Ea
+
-
XF
RS>>>>Ra
Speed Regulation (SR): is a measure of the change in speed from no load to full
load
As Ia increases, Ea decreases thus N decreases
SR % = 100(NNL -NFL)/NNL = 100(WNL - WFL )/WNL
Series Motor: SR is high (variable speed motor)
Shunt Motor: SR is very low (constant speed motor)
P
CU
=I
2
a
R
a
+I
2
F
R
F
Power Flow Diagram
Pin = VSIL
Pd =TdWm= EaIa
P0 = TS Wm
=
Pr
P co
re
M
+P
ag
Efficiency “η”
η %= (P0 /Pin) ×100
Separately Excited and DC shunt Motors
IF = VF /RF
IL= Ia +IF
Ia = IL
Ia
IF =Vt /RF
Ra
Ra
RF
RF
Vt = Ea +IaRa
Vt = Ea +IaRa
VF
Ea
Ea
Separately Excited
Shunt
Td
Td = K Ia = Ka φp Ia
Ia
For a Shunt Motor
Ea = Vt – IaRa = Ka φP W
Then
W
Vt  I a Ra
I R
 WNL  a a
K a P
K a P
W
If Ra is very small then
WNL
W=W NL, consequently a
shunt motor is a constant
speed motor
Ia
Torque-Speed Characteristics
Td  K a P I a ..............I a 
W
Td
....................(1)
K a P
Vt
I R
 a a ..................(2)
K a P K a P
By substituting by (1) in (2) then
Td
Pd
K a PVt K a2 P2
Td 

W
Ra
Ra
0.5 WNL
W
Example
A 125 V shunt motor has an armature resistance of 0.2 Ω and a shunt field
resistance of 45 Ω. If the load current is 50 A, find both Ea and Pd
IL= Ia +IF
Solution
Ia
IF =Vt /RF
Ra
RF
Vt = Ea +IaRa
IF = Vt /RF = 125/45 = 2.78 A
Ea
Ia = IL-IF = 50 -2.78 = 47.22 A
Ea = Vt- IaRa = 125-47.22×0.2 = 115.55 V
Pd = Ea Ia = 115.55×47.22 = 5456.43 watts
Shunt
Series Motor
I a = IF = I L
RS
RF
Vt = Ea+ Ia (Ra +RF)
Ia
φP = KF IF = KF Ia
IF
IL
Ra
Ea = Ka φPW = Ka KF Ia W
Td = Ka φP Ia =
KaKFI2a
VS
……….(1)
+
-
Ea
W
Ea
V I R
 t a ..........( 2)
Ka K F I a Ka K F I a
•Speed of a series motor is inversely
Td
W
proportional to Ia
• A Series motor has a wide range of
speed variation
Ia
Ia 
Vt
K a K FW  R
Td  K a K F I a2 
K a K FVt 2
( K a K FW  R ) 2
As Td α 1/W2 the series motor has a very high starting torque and that
makes it suitable for electrical vehicles, electric traction applications and
drilling
Speed control
V I R
W t a
Ka K F I a
Td
• As R increases W decreases
• As φP increases W decreases (field control)
• Using a variable DC voltage source makes
W increases as Vt increases
Speed
Example
A series motor has a combined series and armature resistance of 0.85 Ω,
runs at 1000 rpm and draws 20A from a 250 V source, calculate:
1.The rotational speed if a 3.75 Ω resistor is connected in series with the
motor
2.The developed power and developed torque at both speeds
Solution
1. Ea1 = Vt –Ia (R +RS) = 250-20×0.85= 233 V,
Ea2 = 250-20×(0.85+3.75) = 158 V,
as Ea = KaφPW,
then (Ea1/Ea2) = (N1/N2)
then N2= 158×1000/233 = 678.1 rpm
or W2 = 678.11×2×3.14/60 = 71 rad /sec
2. Pd1= Ea1Ia = 233×20 = 4660 w, Pd2= Ea2Ia = 158×20= 3160 w
Td1= Pd1/ W1 = 4660/104.72 = 44.5 N.m
Td2 = Pd2/ W2 = 3160/71 = 44.5 N.m