VI. DC Machines - Hacettepe Universityusezen/eem473/dc_machines-2p.pdf3 Construction of DC Machines...
Transcript of VI. DC Machines - Hacettepe Universityusezen/eem473/dc_machines-2p.pdf3 Construction of DC Machines...
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VI. DC Machines
Introduction
• Separately-excited• Shunt• Series• Compound
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:
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DC Machines
Motoring
Mode
Generating
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
Basic parts of a DC machine
1. Construction of DC Machines
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Construction of DC Machines
Elementary DC machine with commutator.
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
(a) Space distribution of air-gap flux density in an elementary dc machine;
(b) waveform of voltage between brushes.
Average gives us a DC voltage, Ea
Ea = Kg f r
Te = Kg f a= Kd If a
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(a) Space distribution of air-gap flux density in an elementary dc machine;
(b) waveform of voltage between brushes.
Electrical Analogy
2. Operation of a Two-Pole DC Machine
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Space distribution of air-gap flux density, Bfin an elementary dc machine;
One pole spans 180º electrical in space )sin(peakf BB
Mean air gap flux per pole: poleperavgpoleavg AB /
0 )sin( dABpeak
0 )sin( rdBpeak
Aper pole: surface area spanned by a pole
rBpeakpoleavg 2/ For a two pole DC machine,
Space distribution of air-gap flux density, Bfin an elementary dc machine;
One pole spans 180º electrical in space
)sin(peakf BB
Flux linkage a: )cos(/ poleavga N : phase angle between the magnetic axes of the rotor and the stator
0)( tt rwith 0= 0 )cos(/ tN rpoleavga
)sin(/ tNdt
de rpoleavgr
aa
0 )(1
dtteE aa
poleavgra NE /2
For a two pole DC machine:
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rfga KE In general: Kg: winding factorf: mean airgap flux per poler: shaft-speed in mechanical rad/sec
602 r
rn nr: shaft-speed in revolutions per minute (rpm)
DC machines with number of poles > 2
120602rr
elecPnnP
f P: number of poles
rBP peakpoleavg 22
/
rfgpoleavgra KNP
E
/2
2
A four-pole DC machine
Schematic representation of a DC machine
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Typical magnetization curve of a DC machine
Torque expression in terms of mutual inductance
d
dMii
d
dLi
d
dLiT fa
afa
af
fe 22
2
1
2
1
d
dMiiT faafe cosMM fa
afe iiMT ˆ
Alternatively, electromagnetic torque Te can be derived from power conversion equations
elecmech PP
aame IET mfga KE
amfgme IKT
afge IKT
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In a linear magnetic circuit
affge IIKKT fff IKwhere
(a) separately-excited (b) series
(c) shunt (d) compound
Field-circuit connections of DC machines
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Separately-excited DC machine circuit in motoring mode
mfga KE 2 a
CpK a
g
p : number of polesCa : total number of conductors in armature windinga : number of parallel paths through armature winding
ta VE
Te produces rotation (Te and m are in the same direction) 0and 0 ,0 memech ωTP
windage and frictionproducedpower output
poweroutput grossor power hanicalelectromec internal
wfoutmech PPP &
Separately-excited DC machine circuit in generating mode
ta VE
Te and m are in the opposite direction 0 and 0 ,0 memech ωTP
Generating mode– Field excited by If (dc)– Rotor is rotated by a mechanical prime-mover at m.– As a result Ea and Ia are generated
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Separately-Excited DC Generator
rIEV aaat mfga KE where
LLt RIV also aL II where
3. Analysis of DC Generators
Terminal V-I Characteristics
Terminal voltage (Vt) decreases slightly as load current increases (due to IaRa voltage drop)
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Terminal voltage characteristics of DC generators
Series generator is not used due to poor voltage regulation
Shunt DC Generator (Self-excited DC Generator)
– Initially the rotor is rotated by a mechanical prime-mover at mwhile the switch (S) is open.
– Then the switch (S) is closed.
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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
Solving for the exciting current, If
mga KE f fff IK
mda IKE f fgd KKK
where
where
Integrating with the electrical circuit equations
fff
ff irrdt
diLLiK aamd
Applying Laplace transformation we obtain
f0ffffff )()()( ILLsIrrssILLsIK aaamd
So the time domain solution is given by
tLL
Krr
a
mda
eIti
f
f
f0f )(
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Let us consider the following 3 situations
tLL
Krr
a
mda
eIti
f
f
f0f )(
(iii) (ra + rf) < Kdm
(ii) (ra + rf) = Kdm
(i) (ra + rf) > Kdm
0)(lim f
tit
Two curves do not intersect.
0ff )( Iti
Self excitation can just start
Generator can self-excite
Self-excited DC generator under load
Ea= ra Ia + rf If
Load line of electrical circuit
Ia= If + IL Vt= Ea - ra Ia = rf If
Ea= (ra + rf) If + ra IL
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Series DC Generator
)( saaat rrIEV mfga KE where
LLt RIV also saL III and
Not used in practical, due to poor voltage regulation
Compound DC Generators
ssaaat rIrIEV mfga KE where
LLt RIV also sL II and
(a) Short-shunt connected compound DC generator
sf IIIa and
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(b) Long-shunt connected compound DC generator
saaat rrIEV mfga KE where
LLt RIV also fIII aL and sIIa and
Types of Compounding
(i) Cumulatively-compounded DC generator (additive compounding)
mmf field
series
mmf field
shunt
f
mmffield
sd FFF
for linear M.C. (or in the linear region of the magnetization curve, i.e. unsaturated magnetic circuits)
sd f
(ii) Differentially-compounded DC generator (subtractive compounding)
sd FFF f
for linear M.C. (or in the linear region of the magnetization curve, i.e. unsaturated magnetic circuits)
sd f
Differentially-compounded generator is not used in practical, as it exhibits poor voltage regulation
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Terminal V-I Characteristics of Compound Generators
Above curves are for cumulatively-compounded generators
Magnetization curves for a 250-V 1200-r/min dc machine. Also field-resistance lines for the discussion of self-excitation are shown
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Examples
1. A 240kW, 240V, 600 rpm separately excited DC generator has an armature resistance, ra = 0.01 and a field resistance rf = 30. 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:
310300285260250230200165Ea (V)
65432.521.51If (A)
If DC generator is delivering rated load 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
2. A shunt DC generator has a magnetization curve at nr = 1500 rpm as shown below. The armature resistance ra = 0.2 , and field total resistance rf = 100 .
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)
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Solution:
a) b)
NB. Neglected raIf voltage drops
3. The magnetization curve of a DC shunt generator at 1500 rpm is given below, where the armature resistance ra = 0.5 , and field total resistance rf = 100 , 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 = 100.
c) Find terminal voltage Vt, efficiency and mechanical torque applied to
the shaft when Ia = 60A at 1500 rpm.
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Solution:
a) b) (i)
b) (ii)
• Series DC motor• Separately-excited DC motor• Shunt DC motor• Compound DC motor
DC motors are adjustable speed motors. A wide range of torque-speed characteristics (Te-m) is obtainable depending on the motor types given below:
4. Analysis of DC Motors
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(a) Series DC Motors
DC Motors Overview
(b) Separately-excited DC Motors
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(c) Shunt DC Motors
(d) Compound DC Motors
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(a) Series DC Motors
DC Motors
)( saaat rrIEV
mfga KE
sa II
The back e.m.f:
Electromagnetic torque: afge IKT
Terminal voltage equation:
Assuming linear equation: sff IK
)( saaat rrIEV fg
saat
fg
am K
rrIV
K
E
sa II
afge IKT sff IK
2ade IKT
adfg IKK
asfge IIKKT
, mfga KE
ad
saatm IK
rrIV adfg IKK
saatmada rrIVIKE mfga KE
samd
ta rrK
VI
22
samd
tde
rrK
VKT
2 ade IKT
23
22
samd
tde
rrK
VKT
2
1
meT
thus
Note that: A series DC motor should never run no load!
m 0 eT overspeeding!
(b) Separately-excited DC Motors
aaat rIEV
mfga KE The back e.m.f:
Electromagnetic torque: afge IKT
Terminal voltage equation:
Assuming linear equation: fff IK
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, mfga KE
aaat rIEV
fg
aemfgt K
rTKV
afge IKT
2fg
aem
fg
t
K
rT
K
V
efg
a
fg
tm T
K
r
K
V2
elm TK 0
No load (i.e. Te = 0) speed: fg
t
K
V
0
Slope: 2fg
al
K
rK
very small!
Slightly dropping m with load
(c) Shunt DC Motors
aaat rIEV
mfga KE The back e.m.f:
Electromagnetic torque: afge IKT
Terminal voltage equation:
Assuming linear equation: fff IK
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, mfga KE
aaat rIEV
fg
aemfgt K
rTKV
afge IKT
2fg
aem
fg
t
K
rT
K
V
efg
a
fg
tm T
K
r
K
V2
elm TK 0
No load (i.e. Te = 0) speed: fg
t
K
V
0
Slope: 2fg
al
K
rK
very small!
Slightly dropping m with load
Same as separately excited motor
elm TK 0
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
m 0 f
mfga KE Ea is momentarily constant, but f will decrease rapidly.
so overspeeding!
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Motor Speed Control Methods
Shaft speed can be controlled byi. Changing the terminal voltageii. Changing the field current (magnetic flux)
(a) Controlling separately-excited DC motors
afge IKT
aaat rIEV
i. Changing the terminal voltage
elm TK 0fg
t
K
V
0
et TV , 0
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afge IKT
ii. Changing the field current
elm TK 0fg
t
K
V
0
eTI , , 0ff
(linear magnetic circuit)fff IK
Shaft speed can be controlled byi. Adding a series resistanceii. Adding a parallel field diverter resistanceiii. Using a potential divider at the input (i.e. changes the
effective terminal voltage)
(b) Controlling series DC motors
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i. Adding a series resistance
)( tsaata rrrIVE
fg
am K
E
afge IKT
Ea drops, Ia stays the same
For the same Te, f is constant
but m drops since Ea = Kg f m..
New value of the motor speed, m is given by
mfga KE
, r mt aE
For the same Te produced
ii. Adding a parallel field diverter resistance
fg
am K
E
sff IK
Is drops i.e. Is < Ia.
Ea remains constant,
When series field flux drops, the motor speed m = Ea / Kg f should rise, while driving the same load.
mfga KE
, , , r mfd as II
For the same Te produced, Ia increases
)||( dsa
ata rrr
EVI
sds rrr ||
When we add the diverter resistance
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iii. Using a potential divider
This system like the speed control by adding series resistance as explained in section (i) where rt RTh and Vt VTh.
tTh VRR
RV
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2
Let us apply Thévenin theorem to the right of Vt
21 || RRRTh
)( ThsaaTha RrrIVE
fg
am K
E
afge IKT
Ea drops rapidly, Ia stays the same
For the same Te, f is constant
but m drops rapidly since Ea = Kg f m..
mfga KE
, V mt aE
For the same Te produced
New value of the motor speed, m is given by
If the load increases, Te and Ia increases and Ea decreases, thus motor speed m drops down more.
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 = 1000 rpm and Te = 30 Nm
find the new terminal voltage Vt while f is kept constant.
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Soln:
a) Gross output power is given by
For Ia = 36 A
Thus
V11237036125 .aaTa rIVE
aame IETP conv
kW03436112conv . aame IETP
Note that induced torque is found to be Te = 33.4 Nm
b) New gross output power is found to be
kW1436010002302222conv2 ./ aame IETP
As Ea = Kg f m and f is constant
V 1971121150
10001
1
22 . a
r
ra E
n
nE
2
1
2
1
2
1 r
r
m
m
a
a
n
n
E
E
A332197
143
2
22 .
..
k
E
PI
a
conva
Then, new armature current is found to be
Consequently, the new terminal voltage is given by
V10937033219722 ...aaat rIEV
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Ex2: A 50 hp, 250 V, 1200 rpm DC shunt motor with compensating windings has an armature resistance (including the brushes, compensating windings, and interpoles) of 0.06 . Its field circuit has a total resistance Radj + RF of 50 , which produces a no-load speed of 1200 rpm. The shunt field winding has 1200 turns per pole.
a) Find the motor speed when its input current is 100 A.
b) Find the motor speed when its input current is 200 A.
c) Find the motor speed when its input current is 300 A.
d) Plot the motor torque-speed characteristic.
Solution:
At no load, the armature current is zero and therefore Ea = VT = 250 V.
a) Since the input current is 100 A, the armature current is
250100 95
50T
A L F LF
VI I I I A
R
Therefore 250 95 0.06 244.3A T A AE V I R V
and the resulting motor speed is:
22 1
1
244.31200
2501173A
A
En n m
Erp
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b) Similar computations for the input current of 200 A lead to n2 = 1144 rpm.
c) Similar computations for the input current of 300 A lead to n2 = 1115 rpm.
d) At no load, the torque is zero. The induced torque is A Aind
E I
For the input current of 100 A:2443 95
1902 1173 / 60ind N m
-
For the input current of 200 A:
For the input current of 300 A:
2383 195388
2 1144 / 60ind N m
-
2323 295587
2 1115 / 60ind N m
-
The torque-speed
characteristic of the motor is:
Ex3: A 100 hp, 250 V, 1200 rpm DC shunt motor with an armature resistance of 0.03 and a field resistance of 41.67 . The motor has compensating windings, so armature reactance can be ignored. Mechanical and core losses may be ignored also. The motor is driving a load with a line current of 126 A and an initial speed of 1103 rpm. Assuming that the armature current is constant and the magnetization curve is
a) What is the motor speed if the field
resistance is increased to 50 ?
b) Calculate the motor speed as a function
of the field resistance, assuming a
constant-current load.
c) Assuming that the motor next is
connected as a separately excited and
is initially running with VA = 250 V,
IA = 120 A and at n = 1103 rpm while
supplying a constant-torque load,
estimate the motor speed if VA is
reduced to 200 V.
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Soln:
a) For the given initial line current of 126 A, the initial armature current will be
Therefore, the initial generated voltage for the shunt motor will be
After the field resistance is increased to 50 Ω, the new field current will be
shunt
1 1 1
250126 120
41.67A L FI I I A
1 1 250 120 0.03 246.4A T A AE V I R V
2
2505
50FI A
The values of EA on the magnetization curve are directly proportional to the flux. Therefore, the ratio of internal generated voltages equals to the ratio of the fluxes within the machine. From the magnetization curve, at IF = 5A, EA1 = 250V, and at IF = 6A,
EA1 = 268V. Thus:
b) A speed vs. RF characteristic is shown on the
right
The ratio of the two internal generated voltages is
2 2 2 2 2
1 1 1 1 1
A
A
E K n
E K n
Since the armature current is assumed constant, EA1 = EA2 and, therefore
1 12
2
nn
1 1 1 12
2 2
2681103 1187
250A
A
n E nn rpm
E
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and since the flux is constant
Since the both the torque and the flux are constants, the armature current IA is
also constant. Then
c) For a separately excited motor, the initial generated voltage is
Since
1 1 1A T A AE V I R
2 2 2 2 2
1 1 1 1 1
A
A
E K n
E K n
2 12
1
A
A
E nn
E
2 22 1
1 1
200 120 0.031103 879
250 120 0.03T A A
T A A
V I Rn n rpm
V I R
separately excited
Ex4: A 250-V series dc motor with compensating windings. and a total series
resistance ra + rs of 0.08 . The series field consists of 25 turns per pole.
with the magnetization curve (at 1200 rpm) shown below.
a) Find the speed and induced torque of
this motor for when its armature current
is 50 A.
b) Calculate the efficiency of the motor
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Soln:
a) To analyze the behavior of a series motor with saturation. pick points along the operating curve and find the torque and speed for each point. Notice that the magnetization curve is given in units of magnetomotive force (ampere-turns) versus Ea for a speed of 1200 r/min. so calculated Ea values must be compared to the equivalent values at 1200 r /min to deternine the actual motor speed.
For Ia = 50 A
Since Ia = Is = 50 A. the magnetomotive force is
V 24608.050250 saaTa rrIVE
turns-A12505025 ss NIF
From the magnetization curve at F = 1250 A-turns. Ea0 = 80 V. To get the
correct speed of the motor.
rpm 3690120080
2460
0 r
a
ar n
E
En
b) Efficiency is given by the ratio of the output and input power values. Thus,
To find the induced torque supplied by the motor at that speed, recall that
Pconv = Ea Ia = Tem. Therefore,
Nm 8316036902
50246.
/
m
aae
IET
%.% 49810050250
50246100100100
tt
aa
tt
me
in
out
IV
IE
IV
T
P
P