vi. dc machines - ee.hacettepe.edu.trusezen/eem473/dc_machines-1p.pdfconstruction of dc machines...
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VI. DC Machines
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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
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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
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(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
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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,
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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
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 rpoleavgra
a
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
elecPnnPf P: number of poles
rBP peakpoleavg 22
/
rfgpoleavgra KNPE
/2
2
A four-pole DC machine
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Schematic representation of a DC machine
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Typical magnetization curve of a DC machine
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Torque expression in terms of mutual inductance
ddM
iiddLi
ddL
iT faaf
aa
ffe 22
21
21
ddM
iiT 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
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(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) 0 and 0 ,0 memech ωTP
windage and frictionproducedpower output
poweroutput grossor power hanicalelectromec internal
wfoutmech PPP &
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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
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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
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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
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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 irrdtdiLLiK 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
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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
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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
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Types of Compounding
(i) Cumulatively-compounded DC generator (additive compounding)
mmf field
series
mmf field
shuntf
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
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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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Examples1. 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 torqued) The applied torque if rotational loss is Prot = 10kWe) Efficiency of generatorf) Voltage regulation
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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
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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 rpmb) 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)
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• 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
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(b) Separately-excited DC Motors
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(c) Shunt DC Motors
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(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
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)( saaat rrIEV fg
saat
fg
am K
rrIVK
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
rrKVK
T
2 ade IKT
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22
samd
tde
rrKVK
T
2
1 m
eT
thus
Note that: A series DC motor should never run no load!
m 0 eT overspeeding!
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(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
KrT
KV
efg
a
fg
tm T
Kr
KV
2
elm TK 0
No load (i.e. Te = 0) speed: fg
t
KV
0
Slope: 2fg
al
Kr
K
very small!
Slightly dropping m with load
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(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
![Page 49: VI. DC Machines - ee.hacettepe.edu.trusezen/eem473/dc_machines-1p.pdfConstruction of DC Machines Elementary DC machine with commutator. Copper commutator segment and carbon brushes](https://reader030.vdocuments.site/reader030/viewer/2022020113/5acfe7287f8b9a1d328d999c/html5/thumbnails/49.jpg)
, mfga KE
aaat rIEV
fg
aemfgt K
rTKV
afge IKT
2fg
aem
fg
t
KrT
KV
efg
a
fg
tm T
Kr
KV
2
elm TK 0
No load (i.e. Te = 0) speed: fg
t
KV
0
Slope: 2fg
al
Kr
K
very small!
Slightly dropping m with load
Same as separately excited motor
![Page 50: VI. DC Machines - ee.hacettepe.edu.trusezen/eem473/dc_machines-1p.pdfConstruction of DC Machines Elementary DC machine with commutator. Copper commutator segment and carbon brushes](https://reader030.vdocuments.site/reader030/viewer/2022020113/5acfe7287f8b9a1d328d999c/html5/thumbnails/50.jpg)
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!
![Page 51: VI. DC Machines - ee.hacettepe.edu.trusezen/eem473/dc_machines-1p.pdfConstruction of DC Machines Elementary DC machine with commutator. Copper commutator segment and carbon brushes](https://reader030.vdocuments.site/reader030/viewer/2022020113/5acfe7287f8b9a1d328d999c/html5/thumbnails/51.jpg)
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
![Page 52: VI. DC Machines - ee.hacettepe.edu.trusezen/eem473/dc_machines-1p.pdfConstruction of DC Machines Elementary DC machine with commutator. Copper commutator segment and carbon brushes](https://reader030.vdocuments.site/reader030/viewer/2022020113/5acfe7287f8b9a1d328d999c/html5/thumbnails/52.jpg)
afge IKT
aaat rIEV
i. Changing the terminal voltage
elm TK 0fg
t
KV
0
et TV , 0
![Page 53: VI. DC Machines - ee.hacettepe.edu.trusezen/eem473/dc_machines-1p.pdfConstruction of DC Machines Elementary DC machine with commutator. Copper commutator segment and carbon brushes](https://reader030.vdocuments.site/reader030/viewer/2022020113/5acfe7287f8b9a1d328d999c/html5/thumbnails/53.jpg)
afge IKT
ii. Changing the field current
elm TK 0fg
t
KV
0
eTI , , 0ff
(linear magnetic circuit)fff IK
![Page 54: VI. DC Machines - ee.hacettepe.edu.trusezen/eem473/dc_machines-1p.pdfConstruction of DC Machines Elementary DC machine with commutator. Copper commutator segment and carbon brushes](https://reader030.vdocuments.site/reader030/viewer/2022020113/5acfe7287f8b9a1d328d999c/html5/thumbnails/54.jpg)
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
![Page 55: VI. DC Machines - ee.hacettepe.edu.trusezen/eem473/dc_machines-1p.pdfConstruction of DC Machines Elementary DC machine with commutator. Copper commutator segment and carbon brushes](https://reader030.vdocuments.site/reader030/viewer/2022020113/5acfe7287f8b9a1d328d999c/html5/thumbnails/55.jpg)
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
![Page 56: VI. DC Machines - ee.hacettepe.edu.trusezen/eem473/dc_machines-1p.pdfConstruction of DC Machines Elementary DC machine with commutator. Copper commutator segment and carbon brushes](https://reader030.vdocuments.site/reader030/viewer/2022020113/5acfe7287f8b9a1d328d999c/html5/thumbnails/56.jpg)
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
![Page 57: VI. DC Machines - ee.hacettepe.edu.trusezen/eem473/dc_machines-1p.pdfConstruction of DC Machines Elementary DC machine with commutator. Copper commutator segment and carbon brushes](https://reader030.vdocuments.site/reader030/viewer/2022020113/5acfe7287f8b9a1d328d999c/html5/thumbnails/57.jpg)
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
RV21
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.
![Page 58: VI. DC Machines - ee.hacettepe.edu.trusezen/eem473/dc_machines-1p.pdfConstruction of DC Machines Elementary DC machine with commutator. Copper commutator segment and carbon brushes](https://reader030.vdocuments.site/reader030/viewer/2022020113/5acfe7287f8b9a1d328d999c/html5/thumbnails/58.jpg)
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 Nmfind the new terminal voltage Vt while f is kept constant.
![Page 59: VI. DC Machines - ee.hacettepe.edu.trusezen/eem473/dc_machines-1p.pdfConstruction of DC Machines Elementary DC machine with commutator. Copper commutator segment and carbon brushes](https://reader030.vdocuments.site/reader030/viewer/2022020113/5acfe7287f8b9a1d328d999c/html5/thumbnails/59.jpg)
Soln:
a) Gross output power is given by
For Ia = 36 A
Thus
V 11237036125 .aaTa rIVE
aame IETP conv
kW 03436112conv . 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
![Page 60: VI. DC Machines - ee.hacettepe.edu.trusezen/eem473/dc_machines-1p.pdfConstruction of DC Machines Elementary DC machine with commutator. Copper commutator segment and carbon brushes](https://reader030.vdocuments.site/reader030/viewer/2022020113/5acfe7287f8b9a1d328d999c/html5/thumbnails/60.jpg)
As Ea = Kg f m and f is constant
V 19711211501000
11
22 . a
r
ra E
nnE
2
1
2
1
2
1 r
r
m
m
a
ann
EE
A332197
143
2
22 .
..
k
EPI
a
conva
Then, new armature current is found to be
Consequently, the new terminal voltage is given by
V10937033219722 ...aaat rIEV
![Page 61: VI. DC Machines - ee.hacettepe.edu.trusezen/eem473/dc_machines-1p.pdfConstruction of DC Machines Elementary DC machine with commutator. Copper commutator segment and carbon brushes](https://reader030.vdocuments.site/reader030/viewer/2022020113/5acfe7287f8b9a1d328d999c/html5/thumbnails/61.jpg)
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.
![Page 62: VI. DC Machines - ee.hacettepe.edu.trusezen/eem473/dc_machines-1p.pdfConstruction of DC Machines Elementary DC machine with commutator. Copper commutator segment and carbon brushes](https://reader030.vdocuments.site/reader030/viewer/2022020113/5acfe7287f8b9a1d328d999c/html5/thumbnails/62.jpg)
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 is250100 9550
TA L F L
F
VI I I I AR
Therefore 250 95 0.06 244.3A T A AE V I R V
and the resulting motor speed is:
22 1
1
244.31200250
1173A
A
En n mE
rp
![Page 63: VI. DC Machines - ee.hacettepe.edu.trusezen/eem473/dc_machines-1p.pdfConstruction of DC Machines Elementary DC machine with commutator. Copper commutator segment and carbon brushes](https://reader030.vdocuments.site/reader030/viewer/2022020113/5acfe7287f8b9a1d328d999c/html5/thumbnails/63.jpg)
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 195 3882 1144 / 60ind N m
-
2323 295 5872 1115 / 60ind N m
-
The torque-speedcharacteristic of the motor is:
![Page 64: VI. DC Machines - ee.hacettepe.edu.trusezen/eem473/dc_machines-1p.pdfConstruction of DC Machines Elementary DC machine with commutator. Copper commutator segment and carbon brushes](https://reader030.vdocuments.site/reader030/viewer/2022020113/5acfe7287f8b9a1d328d999c/html5/thumbnails/64.jpg)
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 isconnected 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.
![Page 65: VI. DC Machines - ee.hacettepe.edu.trusezen/eem473/dc_machines-1p.pdfConstruction of DC Machines Elementary DC machine with commutator. Copper commutator segment and carbon brushes](https://reader030.vdocuments.site/reader030/viewer/2022020113/5acfe7287f8b9a1d328d999c/html5/thumbnails/65.jpg)
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 1250126 120
41.67A L FI I I A
1 1 250 120 0.03 246.4A T A AE V I R V
2250 550FI A
![Page 66: VI. DC Machines - ee.hacettepe.edu.trusezen/eem473/dc_machines-1p.pdfConstruction of DC Machines Elementary DC machine with commutator. Copper commutator segment and carbon brushes](https://reader030.vdocuments.site/reader030/viewer/2022020113/5acfe7287f8b9a1d328d999c/html5/thumbnails/66.jpg)
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 nE K n
Since the armature current is assumed constant, EA1 = EA2 and, therefore
1 12
2
nn
1 1 1 12
2 2
2681103 1187250
A
A
n E nn rpmE
![Page 67: VI. DC Machines - ee.hacettepe.edu.trusezen/eem473/dc_machines-1p.pdfConstruction of DC Machines Elementary DC machine with commutator. Copper commutator segment and carbon brushes](https://reader030.vdocuments.site/reader030/viewer/2022020113/5acfe7287f8b9a1d328d999c/html5/thumbnails/67.jpg)
and since the flux is constant
Since the both the torque and the flux are constants, the armature current IA isalso constant. Then
c) For a separately excited motor, the initial generated voltage is
Since1 1 1A T A AE V I R
2 2 2 2 2
1 1 1 1 1
A
A
E K nE K n
2 12
1
A
A
E nnE
2 22 1
1 1
200 120 0.031103 879250 120 0.03
T A A
T A A
V I Rn n rpmV I R
separately excited
![Page 68: VI. DC Machines - ee.hacettepe.edu.trusezen/eem473/dc_machines-1p.pdfConstruction of DC Machines Elementary DC machine with commutator. Copper commutator segment and carbon brushes](https://reader030.vdocuments.site/reader030/viewer/2022020113/5acfe7287f8b9a1d328d999c/html5/thumbnails/68.jpg)
Ex4: A 250-V series dc motor with compensating windings. and a total seriesresistance 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
![Page 69: VI. DC Machines - ee.hacettepe.edu.trusezen/eem473/dc_machines-1p.pdfConstruction of DC Machines Elementary DC machine with commutator. Copper commutator segment and carbon brushes](https://reader030.vdocuments.site/reader030/viewer/2022020113/5acfe7287f8b9a1d328d999c/html5/thumbnails/69.jpg)
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 thecorrect speed of the motor.
rpm 3690120080246
00
ra
ar n
EEn
![Page 70: VI. DC Machines - ee.hacettepe.edu.trusezen/eem473/dc_machines-1p.pdfConstruction of DC Machines Elementary DC machine with commutator. Copper commutator segment and carbon brushes](https://reader030.vdocuments.site/reader030/viewer/2022020113/5acfe7287f8b9a1d328d999c/html5/thumbnails/70.jpg)
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 thatPconv = Ea Ia = Tem. Therefore,
Nm 8316036902
50246.
/
m
aae
IET
%.% 4981005025050246100100100
tt
aa
tt
me
in
outIVIE
IVT
PP