V. DC Machines
1
Introduction
DC machines are used in applications requiring a wide range
of speeds by means of various combinations of their field
windings
Types of DC machines:
•
•
•
•
Separately-excited
Shunt
Series
Compound
2
DC
Machines
Motoring
Generating
Mode
Mode
In Motoring Mode: Both armature
and field windings are excited by DC
In Generating Mode: Field winding
is excited by DC and rotor is rotated
externally by a prime mover coupled
to the shaft
3
1. Construction of DC Machines
Basic parts of a DC machine
4
Construction of DC Machines
Copper commutator segment
and carbon brushes are used
for:
(i) for mechanical rectification
of induced armature emfs
(ii) for taking stationary armature
terminals from a moving member
Elementary DC machine with commutator.
5
(a) Space distribution of air-gap flux density
in an elementary dc machine;
Average gives us a DC voltage, Ea
Ea = Kg φf ωr
Te = Kg φf Ιa
= Kd If Ιa
(b) waveform of voltage between brushes.
6
Electrical Analogy
(a) Space distribution of air-gap flux density
in an elementary dc machine;
(b) waveform of voltage between brushes.
7
2. Operation of a Two-Pole DC Machine
8
Space distribution of air-gap flux density, Bf
in an elementary dc machine;
One pole spans 180º electrical in space
B f = B peak sin(θ )
Mean air gap flux per pole: φavg / pole = Bavg A per
pole
= ∫0π B peak sin(θ )dA
Aper pole: surface area spanned
by a pole
= ∫0π B peak sin(θ )lrdθ
For a two pole DC machine, φavg / pole = 2 B peak lr
9
Space distribution of air-gap flux density, Bf
in an elementary dc machine;
One pole spans 180º electrical in space
B f = B peak sin(θ )
Flux linkage λa:
with α0= 0
ea =
λ a = Nφ avg / pole cos(α )
α : phase angle between the magnetic axes of
the rotor and the stator
λa = Nφavg / pole cos(ω r t )
dλ a
= −ω r Nφavg / pole sin(ω r t )
dt
For a two pole DC machine: E a =
2
π
α (t ) = ω r t + α 0
Ea =
1
π
π
∫0 ea ( t )dt
ω r Nφavg / pole
10
In general:
Kg: winding factor
φf: mean airgap flux per pole
ωr: shaft-speed in mechanical rad/sec
E a = K gφ f ω r
ω r = 2π
nr
nr: shaft-speed in revolutions per minute (rpm)
60
DC machines with number of poles > 2
f elec =
P nr Pnr
=
2 60 120
φavg / pole =
Ea =
P: number of poles
2
2 B peak lr
P
P 2
Nω rφavg / pole = K gφ f ω r
2π
A four-pole DC machine
11
Schematic representation of a DC machine
12
Typical magnetization curve of a DC machine
13
Torque expression in terms of mutual inductance
Te =
dM fa
1 2 dL f 1 2 dLa
+ ia
+ i f ia
if
2
dθ
2 dθ
dθ
Te = i f ia
dM fa
M fa = M̂ cos θ
dθ
ˆi i
Te = M
f a
Alternatively, electromagnetic torque Te can be derived from power conversion equations
Pmech = Pelec
Ea = K gφ f ω m
Teω m = Ea I a
Teω m = K gφ f ω m I a
Te = K gφ f I a
14
In a linear magnetic circuit
Te = K g K f I f I a
where
φf = K f If
15
Field-circuit connections of DC machines
(a) separately-excited
(c) shunt
(b) series
(d) compound
16
Separately-excited DC machine circuit in motoring mode
Pmech
=
Pout
+
P f &w
1444442
444443 1442
443 1442
443
internal electromechanical power
or gross output power
E a = K gφ f ω m
E a ≤ Vt
Kg =
p Ca
2π a
output power produced
friction and windage
p : number of poles
Ca : total number of conductors in armature winding
a : number of parallel paths through armature winding
Te produces rotation (Te and ωm are in the same direction)
Pmech > 0, Te > 0 and ωm > 0
17
Separately-excited DC machine circuit in generating mode
E a > Vt
Te and ωm are in the opposite direction
Pmech < 0, Te < 0 and ωm > 0
Generating mode
– Field excited by If (dc)
– Rotor is rotated by a mechanical prime-mover at ωm.
– As a result Ea and Ia are generated
18
3. Analysis of DC Generators
Separately-Excited DC Generator
also
Vt = E a − I a ra
where
E a = K gφ f ω m
Vt = I L R L
where
I L = Ia
19
Terminal V-I Characteristics
Terminal voltage (Vt) decreases slightly as load current increases
(due to IaRa voltage drop)
20
Terminal voltage characteristics of DC generators
Series generator is not used due to poor voltage regulation
21
Shunt DC Generator (Self-excited DC Generator)
– Initially the rotor is rotated by a mechanical prime-mover at ωm
while the switch (S) is open.
– Then the switch (S) is closed.
22
When the switch (S) is closed
Ea= (ra + rf) If
Load line of electrical circuit
Self-excitation uses the residual magnetization & saturation properties
of ferromagnetic materials.
– when S is closed Ea = Er and If = If0
– interdependent build-up of If and Ea continues
– comes to a stop at the intersection of the two curve
as shown in the figure below
23
Solving for the exciting current, If
E a = K gφf ω m
where
φf = K f I f
Ea = K d If ω m
where
Kd = K g K f
Integrating with the electrical circuit equations
K d i f ω m = ( La + Lf )
di f
+ (ra + rf ) i f
dt
Applying Laplace transformation we obtain
K d ω m I f ( s ) = ( La + Lf )sI f ( s ) + (ra + rf )I f ( s ) − ( La + Lf )I f0
So the time domain solution is given by
if ( t ) = I f0
⎛ r + r − Kdωm
− ⎜⎜ a f
La + Lf
e ⎝
⎞
⎟⎟ t
⎠
24
if ( t ) = I f0
⎛ r + r − Kdωm
− ⎜⎜ a f
La + Lf
e ⎝
⎞
⎟⎟ t
⎠
Let us consider the following 3 situations
(i) (ra + rf) > Kdωm
lim if ( t ) = 0
t −>∞
Two curves do not intersect.
(ii) (ra + rf) = Kdωm
if ( t ) = I f 0
Self excitation can just start
(iii) (ra + rf) < Kdωm
Generator can self-excite
25
Self-excited DC generator under load
Ia= If + IL
Vt= Ea - ra Ia = rf If
Ea= ra Ia + rf If
Ea= (ra + rf) If + ra IL
Load line of electrical circuit
26
Series DC Generator
also
Vt = E a − I a ( ra + rs )
where
E a = K gφ f ω m
Vt = I L R L
and
I L = Ia = I s
Not used in practical, due to poor voltage regulation
27
Compound DC Generators
(a) Short-shunt connected compound DC generator
also
Vt = E a − I a ra + I s rs
where
E a = K gφ f ω m
Vt = I L R L
and
IL = Is
and
Ia = If + Is
28
(b) Long-shunt connected compound DC generator
also
Vt = E a − I a (ra + rs )
where
E a = K gφ f ω m
Vt = I L R L
and
I L = Ia + If
and
Ia = Is
29
Types of Compounding
(i) Cumulatively-compounded DC generator (additive compounding)
Fd = Ff + Fs
{ {
{
field
mmf
shunt
field
mmf
series
field
mmf
for linear M.C. (or in the linear region of the magnetization curve, i.e. unsaturated magnetic circuits)
φ d = φf + φ s
(ii) Differentially-compounded DC generator (subtractive compounding)
Fd = Ff − Fs
for linear M.C. (or in the linear region of the magnetization curve, i.e. unsaturated magnetic circuits)
φ d = φf − φ s
Differentially-compounded generator is not used in practical, as it exhibits
poor voltage regulation
30
Terminal V-I Characteristics of Compound Generators
Above curves are for cumulatively-compounded generators
31
Magnetization curves for a 250-V 1200-r/min dc machine.
Also field-resistance lines for the discussion of self-excitation are shown
32
Examples
1.
A 240kW, 240V, 600 rpm separately excited DC generator has an armature
resistance, ra = 0.01Ω and a field resistance rf = 30Ω. The field winding is
supplied from a DC source of Vf = 100V. A variable resistance R is connected
in series with the field winding to adjust field current If. The magnetization
curve of the generator at 600 rpm is given below:
If (A)
1
1.5
2
2.5
3
4
5
6
Ea (V)
165
200
230
250
260
285
300
310
If DC generator is delivering rated voltage and is driven at 600 rpm determine:
a) Induced armature emf, Ea
b) The internal electromagnetic power produced (gross power)
c) The internal electromagnetic torque
d) The applied torque if rotational loss is Prot = 10kW
e) Efficiency of generator
f) Voltage regulation
33
2.
A shunt DC generator has a magnetization curve at nr = 1500 rpm as shown
below. The armature resistance ra = 0.2 Ω, and field total resistance rf = 100 Ω.
a) Find the terminal voltage Vt and field current If of the generator when it
delivers 50A to a resistive load
b) Find Vt and If when the load is disconnected (i.e. no-load)
34
Solution:
a)
b)
NB. Neglected raIf voltage drops
35
3.
The magnetization curve of a DC shunt generator at 1500 rpm is given below,
where the armature resistance ra = 0.2 Ω, and field total resistance rf = 100 Ω,
the total friction & windage loss at 1500 rpm is 400W.
a) Find no-load terminal voltage at 1500 rpm
b) For the self-excitation to take place
(i) Find the highest value of the total shunt field resistance at 1500 rpm
(ii) The minimum speed for rf = 100Ω.
c) Find terminal voltage Vt, efficiency η and mechanical torque applied to
the shaft when Ia = 60A at 1500 rpm.
36
Solution:
a)
b) (i)
b) (ii)
37
4. Analysis of DC Motors
DC motors are adjustable speed motors. A wide range of torquespeed characteristics (Te-ωm) is obtainable depending on the motor
types given below:
•
•
•
•
Series DC motor
Separately-excited DC motor
Shunt DC motor
Compound DC motor
38
DC Motors Overview
(a) Series DC Motors
39
(b) Separately-excited DC Motors
40
(c) Shunt DC Motors
41
(d) Compound DC Motors
42
DC Motors
(a) Series DC Motors
The back e.m.f:
E a = K gφ f ω m
Electromagnetic torque:
Te = K gφ f I a
Terminal voltage equation:
Vt = E a + I a ( ra + rs )
Ia = I s
Assuming linear equation:
φ f = K f Is
43
Te = K gφ f I a
K φ f = K f Is
Te = K g K f I s I a
K Ia = I s
Te = K d I a2
K gφ f = K d I a
ωm =
V − I a (ra + rs )
Ea
= t
K gφ f
K gφ f
K E a = K gφ f ω m , Vt = E a + I a ( ra + rs )
ωm =
Vt − I a (ra + rs )
KI a
K K gφ f = KI a
E a = K d I a ω m = Vt − I a (ra + rs )
Ia =
Te =
K E a = K gφ f ω m
Vt
K d ω m + (ra + rs )
K d Vt 2
[K d ω m + (ra + rs )]2
K Te = K d I a2
44
Te =
K d Vt 2
[K d ω m + (ra + rs )]
2
thus
Te ∝
1
ω m2
Note that: A series DC motor should never run no load!
Te → 0
⇒
ωm = ∞
overspeeding!
45
(b) Separately-excited DC Motors
The back e.m.f:
E a = K gφ f ω m
Electromagnetic torque:
Te = K gφ f I a
Terminal voltage equation:
Vt = E a + I a ra
Assuming linear equation:
φf = K f If
46
Vt = E a + I a ra
Vt = K g φ f ω m +
Te ra
K gφ f
Vt
Te ra
= ωm +
K gφ f
K gφ f
)2
Vt
ra
−
K gφ f
K gφ f
)2
(
ωm =
(
K E a = K gφ f ω m ,
Te
Te = K gφ f I a
ω m = ω 0 − K l Te
No load (i.e. Te = 0) speed: ω 0 =
Slope:
Kl =
ra
(K gφ f )2
Vt
K gφ f
very small!
Slightly dropping ωm with load
47
(c) Shunt DC Motors
The back e.m.f:
E a = K gφ f ω m
Electromagnetic torque:
Te = K gφ f I a
Terminal voltage equation:
Vt = E a + I a ra
Assuming linear equation:
φf = K f If
48
Vt = E a + I a ra
Vt = K g φ f ω m +
Te ra
K gφ f
Vt
Te ra
= ωm +
K gφ f
K gφ f
)2
Vt
ra
−
K gφ f
K gφ f
)2
(
ωm =
(
K E a = K gφ f ω m ,
Te
Te = K gφ f I a
ω m = ω 0 − K l Te
Same as separately excited motor
No load (i.e. Te = 0) speed: ω 0 =
Slope:
Kl =
ra
(K gφ f )2
Vt
K gφ f
very small!
Slightly dropping ωm with load
49
ω m = ω 0 − K l Te
Note that: In the shunt DC motors, if suddenly the field terminals are
disconnected from the power, supply while the motor was running,
overspeeding problem will occur
E a = K gφ f ω m
so
φ f → 0 ⇒ ωm → ∞
Ea is momentarily constant, but φf will decrease rapidly.
overspeeding!
50
Motor Speed Control Methods
(a) Controlling separately-excited DC motors
Shaft speed can be controlled by
i.
Changing the terminal voltage
ii. Changing the field current (magnetic flux)
51
i.
Changing the terminal voltage
ω m = ω 0 − K l Te
Te = K gφ f I a
ω0 =
Vt
K gφ f
Vt ↓ ⇒ ω 0 ↓ ,
Vt = E a + I a ra
Te ↓
52
ii.
Changing the field curerent
ω m = ω 0 − K l Te
Te = K gφ f I a
φf = K f I f
ω0 =
Vt
K gφ f
(linear magnetic circuit)
I f ↓ ⇒ φf ↓ , ω 0 ↑,
Te ↓
53
Ex1:
A separately excited DC motor drives the load at nr = 1150 rpm.
a)
Find the gross output power (electromechanical power output)
produced by the dc motor.
b) If the speed control is to be achieved by armature voltage control and
the new operating condition is given by:
nr = 1150 rpm and Te = 30 Nm
find the new terminal voltage V′t while φf is kept constant.
54
(b) Controlling series DC motors
Shaft speed can be controlled by
i.
Adding a series resistance
ii. Adding a parallel field diverter resistance
iii. Using a potential divider at the input (i.e. changes the
effective terminal voltage)
55
i.
Adding a series resistance
For the same Te produced
Ea drops, Ia stays the same
E a = Vt − I a ( ra + rs + rt )
For the same Te,
φf is constant
but ωm drops since Ea = Kg φf ωm..
New value of the motor speed, ωm is
given by
ωm =
Ea
K gφ f
rt ↑ ⇒
K Te = K gφ f I a
K E a = K gφ f ω m
Ea ↓, ωm ↓
56
ii.
Adding a parallel field diverter resistance
When we add the diverter resistance
Is drops i.e. Is < Ia.
Ea remains constant,
For the same Te produced, Ia increases
Ia =
Vt − E a
ra + ( rs || rd )
K rs || rd < rs
When series field flux drops, the motor
speed ωm = Ea / Kg φf should rise, while
driving the same load.
ωm =
rd ↓ ⇒
Ea
K gφ f
I s ↓,
K E a = K gφ f ω m
K φ f = K f Is
I a ↑,
φf ↓ ,
ωm ↑
57
iii.
Using a potential divider
Let us apply Thévenin theorem to the
right of V′t
VTh =
R2
Vt
R1 + R2
RTh = R1 || R2
This system like the speed control by
adding series resistance as explained in
section (i) where rt ≡ RTh and Vt ≡ VTh.
For the same Te produced
Ea drops rapidly, Ia stays the same
E a = VTh − I a ( ra + rs + RTh )
For the same Te,
φf is constant
but ωm drops rapidly since Ea = Kg φf ωm..
New value of the motor speed, ωm is
given by
Ea
K Te = K gφ f I a
ωm =
K gφ f
K E =K φ ω
a
Vt′ ↓ ⇒
g
f
m
E a ↓↓ , ω m ↓ ↓
If the load increases, Te and Ia increases and Ea decreases, thus motor speed ωm drops down more.
58