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7 DC machimes- DC motors

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Electric Machines I
DC Machines - DC Motors
Dr. Firas Obeidat
1
Table of Contents
1
• DC Motor Principle
2
• Types of DC Motors
3
• E.M.F. Equation of DC Motor
4
• Armature Torque of DC Motor
5
• Speed Regulation of DC Motor
6
• Total Losses in DC Motor
7
• Power Stages and Efficiency
8
2
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
DC Motor Principle
οƒ˜ Constructionally, there is no basic difference between DC generator and
DC motor. The DC machine can be used as generator or as motor.
οƒ˜ When field magnets are excited in multipolar DC motor, and its
armature conductors are supplied with current from the supply, they
experience a force tending to rotate the armature. Armature conductors
under N-pole are assumed to carry current downwards (crosses) and
those under S-poles to carry current upwards (dots).
οƒ˜ By applying Fleming’s left hand rule, each conductor experiences a force
F which tends to rotate the armature in anticlockwise direction. These
forces collectively produce a driving torque which sets the armature
rotating.
3
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Types of DC Motors
1- Separately Excited DC Motor
IF
+
𝐼𝐴 = 𝐼𝐿
RF
VF
𝑉𝑇 = 𝐸𝐴 + 𝐼𝐴 𝑅𝐴
𝑉𝐹 = 𝐼𝐹 𝑅𝐹
IA
LF
-
IL
+
+ RA
VT
EA
-
Where
IA: is the armature current
IL: is the load current
EA: is the internal generated voltage
VT: is the terminal voltage
IF: is the field current
VF: is the field voltage
RA: is the armature winding resistance
RF: is the field winding resistance
Ο•: is the flux
πœ”m: is the rotor angular speed
𝐸𝐴 = π‘˜Ο•πœ”π‘š
2- Shunt DC Motor
𝐼𝐿 = 𝐼𝐴 + 𝐼𝑓
𝑉𝑇 = 𝐸𝐴 + 𝐼𝐴 𝑅𝐴
IA
𝑉𝐹 = 𝐼𝐹 𝑅𝐹
+ RA RF
IF
EA
LF
-
𝐸𝐴 = π‘˜Ο•πœ”π‘š
IL
+
VT
4
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Types of DC Motors
Speed control of separately excited and shunt DC motors
Adjusting the field
resistance RF
• Increasing
RF causes
𝑰𝑭 = 𝑽𝑻 /𝑹𝑭 to decrease.
• Decreasing IF decreases Ο•.
• Decreasing Ο•, lowers 𝐸𝐴=π‘˜
Ο•πœ”π‘š .
• Decreasing EA increases
𝑰𝑨 = (𝑽𝑻 − 𝑬𝑨 )/𝑹𝑨.
• Increasing IA increases
Tind=π‘˜Ο•IA , with the
change in IA dominant
over the change in flux.
• Increasing Tind makes
Tind>Tload and the speed πœ”
m increases.
• Increasing πœ”m increases
𝐸𝐴=π‘˜Ο•πœ”π‘š.
• Increasing EA decreases
Adjusting the
terminal voltage
applied to the
armature
• An
increase
in
VT
increases 𝑰𝑨 = (𝑽𝑻 − 𝑬𝑨 )/𝑹𝑨 .
• Increasing IA increases
Tind=π‘˜Ο•IA
• Increasing Tind makes
Tind>Tload and the speed πœ”
m increases.
• Increasing πœ”m increases
𝐸𝐴=π‘˜Ο•πœ”π‘š.
• Increasing EA decreases
𝑰𝑨 = (𝑽𝑻 − 𝑬𝑨 )/𝑹𝑨 .
• Decreasing IA decreases
Tind until Tind=Tload at a
higher speed πœ”π‘š
IA.
• Decreasing IA decreases
Tind until Tind=Tload at a
higher speed πœ”π‘š.
Inserting a resistor
in series with the
armature circuit
• resistor is inserted in series
with the armature circuit,
the effect is to drastically
increase the slope of the
motor's
torquespeed
characteristic, making it
operate more slowly if
loaded.
• The insertion of a resistor is
a very wasteful method of
speed control, since the
losses in the inserted
resistor are very large. For
this reason, it is rarely
used. It will be found only
in applications in which the
motor spends almost all its
time operating at full speed
or in applications too
inexpensive to justify a
better form of speed
control.
5
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Types of DC Motors
3- The Permanent Magnet DC Motor
A permanent-magnet DC (PMDC) motor is a DC motor whose poles are made of permanent
magnets. Permanent-magnet dc motors offer a number of benefits compared with shunt dc motors
in some applications. Since these motors do not require an external field circuit, they do not have
the field circuit copper losses associated with shunt dc motors.
Because no field windings are required, they can be smaller than corresponding shunt DC motors.
they are especially common in smaller fractional- and sub fractional-horsepower sizes.
PMDC motors are generally less expensive, smaller in size, simpler, and higher efficiency than
corresponding DC motors with separate electromagnetic fields. This makes them a good choice in
many DC motor applications. The armatures of PMDC motors are essentially identical to the
armatures of motors with separate field circuits, so their costs are similar too. However, the
elimination of separate electromagnets on the stator reduces the size of the stator, the cost of the
stator, and the losses in the field circuits.
PMDC motors also have disadvantages. Permanent magnets cannot produce as high a flux density
as an externally supplied shunt field, so a PMDC motor will have a lower induced torque Tind per
ampere of armature current IA than a shunt motor of the same size and construction. In addition,
PMDC motors run the risk of demagnetization.
A permanent-magnet DC motor is basically the same machine as a shunt dc motor, except that the
flux of a PMDC motor is fixed. Therefore, it is not possible to control the speed of a PMDC motor
by varying the field current or flux. The only methods of speed control available for a PMDC
motor are armature voltage control and armature resistance control.
The techniques to analyze a PMDC motor are basically the same as the techniques to analyze a
shunt dc motor with the field current held constant.
6
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Types of DC Motors
4- Series DC Motor
IA
𝐼𝐴 = 𝐼𝑆 = 𝐼𝐿
+ RA
EA
-
𝑉𝑇 = 𝐸𝐴 + 𝐼𝐴 (𝑅𝐴 + 𝑅𝑆 )
Is
Rs
IL
Ls +
VT
-
𝐸𝐴 = π‘˜Ο•πœ”π‘š
Speed control of series DC motors
1) Change the terminal voltage of the motor.
2) Insertion of a series resistor into the motor circuit, but this
technique is very wasteful of power and is used only for
intermittent periods during the start-up of some motors.
7
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Types of DC Motors
5- Compound DC Motor
For Long Shunt Cumulatively Compound DC Motor
𝐼𝐿 = 𝐼𝐴 + 𝐼𝐹
IA
𝑉𝑇 = 𝐸𝐴 + 𝐼𝐴 (𝑅𝐴 +𝑅𝑠 )
+
EA
-
𝑉𝑇 = 𝐼𝐹 𝑅𝐹
𝐸𝐴 = π‘˜Ο•πœ”π‘š
IL
RA
Rs
Ls
+
RF
IF
LF
VT
-
For Short Shunt Cumulatively Compound DC Motor
𝐼𝐿 = 𝐼𝐴 + 𝐼𝐹
IA
+ RA RF
IF
EA
LF
-
𝑉𝑇 = 𝐸𝐴 + 𝐼𝐴 𝑅𝐴 +𝐼𝐿 𝑅𝑠
𝐸𝐴 = π‘˜Ο•πœ”π‘š
IL
Rs
Ls
+
VT
-
Speed control of Cumulatively compound DC motors
1) Change the field resistance RF.
2) Change the armature voltage VA.
3) Change the armature resistance RA.
8
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Types of DC Motors
Example: A 220V DC shunt machine has an armature resistance of 0.5Ω. If
the full load armature current is 20A, find the induced emf when the machine
acts as (i) generator (ii) motor.
IA=20A
(i) As generator
𝐸𝐴 = 220 + 20 × 0.5 = 230𝑉
+ RA RF
IF
EA
LF
-
-
IA=20A
+
EA
-
𝑉𝑇 = 𝐸𝐴 + 𝐼𝐴 𝑅𝐴
𝐸𝐴 = 𝑉𝑇 − 𝐼𝐴 𝑅𝐴
𝐸𝐴 = 220 − 20 × 0.5 = 210𝑉
RA RF
IF
LF
IL
+
-
VT=220V
(ii) As motor
VT=220V
𝐸𝐴 = 𝑉𝑇 + 𝐼𝐴 𝑅𝐴
IL
+
9
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Types of DC Motors
Example: A 440V shunt DC motor has an armature resistance of 0.8Ω and
field resistance of 200Ω. find the back emf when giving an output of 7.46kW at
85% efficiency.
π‘ƒπ‘œπ‘’π‘‘
IA
IL
πœ‚=
× 100%
𝑃𝑖𝑛
𝑃𝑖𝑛 =
7.46π‘˜
= 8.7765π‘˜π‘Š
0.85
+
VT
+ RA
EA
-
7.46π‘˜
85% =
× 100%
𝑃𝑖𝑛
RF
IF
LF
-
𝑃𝑖𝑛 = 8.7765π‘˜ = 𝑉𝑇 𝐼𝐿 = 440 × πΌπΏ
8.7765π‘˜
= 19.9𝐴
440
440
𝐼𝐹 =
= 2.2𝐴
200
𝐼𝐿 =
𝐼𝐴 = 𝐼𝐿 − 𝐼𝐹 = 19.9 − 2.2 = 17.7𝐴
𝐸𝐴 = 𝑉𝑇 − 𝐼𝐴 × π‘…π΄ = 440 − 17.7 × 0.8 = 425.84𝑉
10
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Types of DC Motors
Example: A 25kW, 250V DC shunt machine has an armature and field
resistances of 0.06Ω and 100Ω respectively. Determine the total armature
power developed when working (i) as generator delivering 25kW output and
IA
IL
(ii) as motor taking 25kW input.
(i) As generator
→ 𝐼𝐿 =
π‘ƒπ‘œπ‘’π‘‘ 25000
=
= 100𝐴
𝑉𝑇
250
+
VT
π‘ƒπ‘œπ‘’π‘‘ = 𝐼𝐿 𝑉𝑇
+ RA RF
IF
EA
LF
-
-
250
𝐼𝐹 =
= 2.5𝐴
𝐼𝐴 = 𝐼𝐿 + 𝐼𝐹 = 100 + 2.5 = 102.5𝐴
100
𝐸𝐴 = 𝑉𝑇 + 𝐼𝐴 𝑅𝐴 = 250 + 0.06 × 102.5 = 256.15
𝑃𝐴 = 𝐸𝐴 𝐼𝐴 = 256.15 × 102.5 = 26255.375π‘Š
(ii) As motor
𝑃𝑖𝑛 = 𝐼𝐿 𝑉𝑇
250
𝐼𝐹 =
= 2.5𝐴
𝐼𝐴 = 𝐼𝐿 − 𝐼𝐹 = 100 + 2.5 = 97.5𝐴
100
𝐸𝐴 = 𝑉𝑇 − 𝐼𝐴 𝑅𝐴 = 250 − 97.5 × 0.06 = 244.15V
IA
+
EA
-
IL
RA RF
IF
LF
+
VT
𝑃𝑖𝑛 25000
→ 𝐼𝐿 =
=
= 100𝐴
𝑉𝑇
250
-
𝑃𝐴 = 𝐸𝐴 𝐼𝐴 = 244.15 × 97.5 = 23804.625π‘Š
11
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
E.M.F. Equation of DC Motor
Let
Ο•: flux/pole in weber.
Z: total number of armature conductors
Z=number of slots × number of conductors/slot
A: number of parallel paths in armature
N: armature rotation in rpm
E: emf induced in any parallel path in armature
Generated emf EA=emf generated in any one of the parallel paths
Average emf generated/conductor=dΟ•/dt volt
Flux cut/conductor in one revolution dΟ•=Ο•P Wb
Number of revolutions /second=N/60
Time for one revolution dt=60/N second
E.M.F. generated/conductor= dΟ•/dt= Ο•PN/60 volt
12
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
E.M.F. Equation of DC Motor
For simplex wave-wound motor
Number of parallel paths=2
Number of conductors (in series) in one path=Z/2
𝐸. 𝑀. 𝐹. π‘”π‘’π‘›π‘’π‘Ÿπ‘Žπ‘‘π‘’π‘‘/π‘π‘Žπ‘‘β„Ž(𝐸𝐴 ) =
πœ™π‘ƒπ‘ 𝑍 πœ™π‘ƒπ‘π‘
× =
π‘£π‘œπ‘™π‘‘
60
2
120
For simplex lap-wound motor
Number of parallel paths=P
Number of conductors (in series) in one path=Z/P
πœ™π‘ƒπ‘ 𝑍 πœ™π‘π‘
𝐸. 𝑀. 𝐹. π‘”π‘’π‘›π‘’π‘Ÿπ‘Žπ‘‘π‘’π‘‘/π‘π‘Žπ‘‘β„Ž(𝐸𝐴 ) =
× =
π‘£π‘œπ‘™π‘‘
60
𝑃
60
In general
where
πœ™π‘π‘ 𝑃
A=2 for simplex wave-winding
𝐸𝐴 =
× π‘£π‘œπ‘™π‘‘
60
𝐴
A=P for simplex lap-winding
1 2π𝑁
𝑃
𝑍𝑃
2π𝑁
Where πœ”π‘š = 60
𝐸𝐴 =
×
× πœ™π‘ × =
πœ™πœ”π‘š π‘£π‘œπ‘™π‘‘
2π
60
𝐴 2π𝐴
For a given DC machine Z,P and A are constant
𝑍𝑃
Where π‘˜ = 2π𝐴
𝐸𝐴 = π‘˜πœ™πœ”π‘š π‘£π‘œπ‘™π‘‘
13
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
E.M.F. Equation of DC Motor
Example: A 4-pole, 32 conductor, lap-wound DC shunt generator with
terminal voltage of 200V delivering 12A to the load has RA=2 Ω and field
circuit resistance of 200Ω. It is driven at 1000rpm. Calculate the flux per pole
in the machine. If the machine has to be run as a motor with the same
terminal voltage and drawing 5A from the mains, maintaining the same
magnetic field, find the speed of the machine.
IA=13A
(i) As generator
πœ™π‘π‘ 𝑃
A=P
×
60
𝐴
226 × 60
πœ™=
= 0.42375𝑀𝑏
1000 × 32
𝐸𝐴 = 226 =
for lap-winding
𝐼𝐴 = 𝐼𝐿 − 𝐼𝐹 = 5 + 1 = 4𝐴
πœ™π‘π‘ 𝑃
𝐸𝐴 = 𝑉𝑇 − 𝐼𝐴 𝑅𝐴 = 200 + 4 × 2 = 192 =
× π‘‰
60
𝐴
192 × 60
𝑁=
= 850π‘Ÿπ‘π‘š
0.42375 × 32
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Action as Generator
IA=4A
+
EA
-
IL=5A
+
RA RF
IF=1A
LF
-
Action as Motor
VT=200V
(ii) As motor
+
EA
-
VT=200V
200
𝐼𝐹 =
= 1𝐴
𝐼𝐴 = 𝐼𝐿 + 𝐼𝐹 = 12 + 1 = 13𝐴
200
𝐸𝐴 = 𝑉𝑇 + 𝐼𝐴 𝑅𝐴 = 200 + 13 × 2 = 226𝑉
IL=12A
+
RA RF
IF=1A
LF
-
14
Armature and Shaft Torque of DC Motor
Armature Torque of DC Motor
Let Ta be the torque developed by the armature of a motor, then the power
developed
π‘π‘œπ‘€π‘’π‘Ÿ π‘‘π‘’π‘£π‘’π‘™π‘œπ‘π‘’π‘‘ = π‘‡π‘Ž × 2πœ‹π‘/60 π‘€π‘Žπ‘‘π‘‘
The electrical power converted into mechanical power in the armature=EAIA
Equating the above two equations yields
π‘‡π‘Ž × 2πœ‹π‘ = 𝐸𝐴 𝐼𝐴
π‘‡π‘Ž =
𝐸𝐴 𝐼𝐴
60 𝐸𝐴 𝐼𝐴
𝐸𝐴 𝐼𝐴
=
= 9.55
N. m
2πœ‹π‘/60 2πœ‹ 𝑁
𝑁
𝐸𝐴 =
πœ™π‘π‘ 𝑃
× π‘£π‘œπ‘™π‘‘
60
𝐴
Or
9.55
𝑃
𝑃
π‘‡π‘Ž =
πœ™π‘πΌπ΄ × = 0.159πœ™π‘πΌπ΄ × N. m
60
𝐴
𝐴
Shaft Torque of DC Motor
π‘‡π‘ β„Ž
π‘€π‘œπ‘‘π‘œπ‘Ÿ π‘œπ‘’π‘‘π‘π‘’π‘‘
= 9.55
N. m
𝑁
(π‘‡π‘ β„Ž − π‘‡π‘Ž ) is known as lost torque and is due iron and friction losses of the motor
15
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Armature Torque of DC Motor
Example: A DC motor takes an armature current of 110A at 480V. The
armature circuit resistance is 0.2Ω. The machine has 6 poles and the armature
is lap-connected with 864 conductors. The flux per pole is 0.05wb. Calculate
the speed and the gross torque developed by the armature.
𝐸𝐴 = 𝑉𝑇 − 𝐼𝐴 𝑅𝐴 = 480 − 110 × 0.2 = 458V
πœ™π‘π‘ 𝑃 0.05 × 864 × π‘
× =
= 458
60
𝐴
60
𝐸𝐴 𝐼𝐴
458 × 110
π‘‡π‘Ž = 9.55
= 9.55
≈ 756N. m
𝑁
636
𝐸𝐴 =
𝑁 = 636 π‘Ÿπ‘π‘š
π‘‡π‘Ž = 0.159 × πœ™ × π‘ × πΌπ΄ = 0.159 × 0.05 × 864 × 110 ≈ 756N. m
Or
Example: Determine armature torque and motor speed of 220V, 4-pole series
motor with 800 conductors wave connected supplying a load by taking 45A from
the mains. The flux per pole is 25mwb and its armature circuit resistance is 0.6Ω.
𝑃
4
π‘‡π‘Ž = 0.159πœ™π‘πΌπ΄ × = 0.159 × 0.025 × 800 × 45 × = 286.2N. m
𝐴
2
𝐸𝐴 = 𝑉𝑇 − 𝐼𝐴 𝑅𝐴 = 220 − 45 × 0.6 = 193V
𝐸𝐴 =
πœ™π‘π‘ 𝑃 0.025 × 800 × π‘ 4
× =
× = 193
60
𝐴
60
2
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
𝑁 = 579 π‘Ÿπ‘π‘š
16
Armature Torque of DC Motor
Example: A 220V DC shunt motor runs at 500 rpm when the armature
current is 50A. Calculate the speed if the torque id doubled. Given that
RA=0.2Ω.
In shunt DC motor the flux is constant.
π‘‡π‘Ž1
𝑃
= 0.159πœ™π‘πΌπ΄1 ×
𝐴
π‘‡π‘Ž1
𝐼𝐴1
=
2π‘‡π‘Ž1 𝐼𝐴2
→
π‘‡π‘Ž2 = 2π‘‡π‘Ž1 = 0.159πœ™π‘πΌπ΄2 ×
1 50
=
2 𝐼𝐴2
𝑃
𝐴
→ 𝐼𝐴2 = 2 × 50 = 100𝐴
𝐸𝐴1 = 𝑉𝑇 − 𝐼𝐴1 𝑅𝐴 = 220 − 50 × 0.2 = 210V
𝐸𝐴2 = 𝑉𝑇 − 𝐼𝐴2 𝑅𝐴 = 220 − 100 × 0.2 = 200V
𝐸𝐴1 =
πœ™π‘π‘1 𝑃
×
60
𝐴
𝐸𝐴1 𝑁1
=
→
𝐸𝐴1 𝑁2
𝐸𝐴2 =
210 500
=
200
𝑁2
πœ™π‘π‘2 𝑃
×
60
𝐴
→
𝑁2 =
500 × 200
= 476π‘Ÿπ‘π‘š
210
17
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Armature Torque of DC Motor
Example: A 500V, 37.3kW, 1000rpm DC shunt motor has on full load an
efficiency of 90%. Determine (i) full load line current (ii) full load armature
torque (neglect iron and friction losses).
(i) π‘€π‘œπ‘‘π‘œπ‘Ÿ 𝑖𝑛𝑝𝑒𝑑 =
π‘ƒπ‘œπ‘’π‘‘ 37300
=
= 41444π‘Š
πœ‚
0.9
41444
𝐹𝑒𝑙𝑙 π‘™π‘œπ‘Žπ‘‘ 𝑙𝑖𝑛𝑒 π‘π‘’π‘Ÿπ‘Ÿπ‘’π‘›π‘‘ =
= 82.9𝐴
500
(ii) π‘‡π‘Ž = 9.55
π‘‡π‘Ž = 9.55
𝐸𝐴 𝐼𝐴
π‘œπ‘’π‘‘π‘π‘’π‘‘
≈ 9.55
= N. m
𝑁
𝑁
(neglect iron and friction losses)
37300
= 356N. m
1000
Example: Determine the torque established by the armature of a four-poles
DC motor having 774 conductors, two paths in parallel, 24 milli-webers of
pole-flux and the armature current is 50A.
π‘‡π‘Ž = 0.159 × πœ™ × π‘ × πΌπ΄ = 0.159 × 0.024 × 774 × 50 × 4/2 = 295.36N. m
18
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Speed Regulation of DC Motor
The speed regulation refers to the change in speed of a motor with change in
applied load torque, other conditions remaining constant.
The Speed Regulation (SR) is defined as the change in speed when the load on
the motor is reduced from rated value to zero, expressed as percent of the
rated load speed.
𝑆𝑅 =
𝑁𝑁𝐿 − 𝑁𝐹𝐿
× 100%
𝑁𝐹𝐿
Where
NNL: noload speed
NFL: full load speed
Example: A 4-pole series motor has 944 wave connected armature conductors.
At a certain load, the flux per pole is 34.6 mWb and the total mechanical
torque developed is 209 N.m. calculate the line current taken by the motor and
the speed at which it will run with an applied voltage of 500V. Total motor
resistance is 3 ohm.
𝑃
4
π‘‡π‘Ž = 209 = 0.159 × πœ™ × π‘ × πΌπ΄ × = 0.159 × 0.0346 × 944 × πΌπ΄ ×
𝐴
2
𝐼𝐴 = 20.1𝐴
𝐸𝐴 = 𝑉𝑇 − 𝐼𝐴 𝑅𝐴 = 500 − 20.1 × 3 = 439.7V
πœ™π‘π‘ 𝑃 0.0346 × 944 × π‘ 4
𝐸𝐴 = 439.7 =
× =
×
60
𝐴
60
2
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
→
𝑁 = 403.8π‘Ÿπ‘π‘š
19
Speed Regulation of DC Motor
Example: A 230V DC shunt motor has an armature resistance of 0.5Ω and
field resistance of 115Ω. At no load, the speed is 1200rpm and the armature
current 2.5A. On application of rated load, the speed drops to 1120 rpm.
Determine the speed regulation, line current and power input when the motor
delivers rated voltage.
𝑁𝑁𝐿 − 𝑁𝐹𝐿
1200 − 1120
𝑆𝑅 =
× 100% =
× 100% = 7.1%
𝑁𝐹𝐿
1120
𝑁1 = 1200π‘Ÿπ‘π‘š
𝐸𝐴1 = 𝑉𝑇 − 𝐼𝐴1 𝑅𝐴 = 230 − 2.5 × 0.5 = 228.75V
𝑁2 = 1120π‘Ÿπ‘π‘š
𝐸𝐴2 = 𝑉𝑇 − 𝐼𝐴2 𝑅𝐴 = 230 − 𝐼𝐴2 × 0.5
𝐸𝐴1 𝑁1
=
𝐸𝐴2 𝑁2
→
228.75 1200
=
𝐸𝐴2
1120
𝐸𝐴2 = 213.5 = 230 − 𝐼𝐴2 × 0.5
230
𝐼𝐿 = 𝐼𝐴2 + 𝐼𝐹 = 33 +
= 35𝐴
115
→ 𝐸𝐴2 =
→
1120 × 228.75
= 213.5𝑉
1200
𝐼𝐴2 =
230 − 213.5
= 33𝐴
0.5
𝑃𝑖𝑛 = 𝐼𝐿 𝑉𝑇 = 35 × 230 = 8050W
20
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Total Loss in a DC Motor
Armature Cu Loss
Total Losses
Copper Losses
Shunt Cu Loss
Series Cu Loss
Hysteresis Loss
Iron Losses
Eddy Current Loss
Friction Loss
Mechanical Losses
Air Friction or Windage Loss
Stray Losses
Iron and mechanical losses are collectively known as Stray (Rotational) losses.
Constant or Standing Losses
Field Cu losses is constant for shunt and compound generators. Stray losses
and shunt Cu loss are constant in their case. These losses are together known
as Constant or Standing Losses (Wc).
21
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Power Stages and Efficiency
Mechanical Efficiency
πœ‚π‘š =
π‘€π‘œπ‘‘π‘œπ‘Ÿ 𝑂𝑒𝑑𝑝𝑒𝑑 𝑖𝑛 π‘Šπ‘Žπ‘‘π‘‘π‘ 
π‘€π‘œπ‘‘π‘œπ‘Ÿ 𝑂𝑒𝑑𝑝𝑒𝑑 𝑖𝑛 π‘Šπ‘Žπ‘‘π‘‘π‘ 
× 100% =
× 100%
π·π‘Ÿπ‘–π‘£π‘–π‘›π‘” π‘ƒπ‘œπ‘€π‘’π‘Ÿ 𝑖𝑛 π΄π‘Ÿπ‘šπ‘Žπ‘‘π‘’π‘Ÿπ‘’ 𝑖𝑛 π‘Šπ‘Žπ‘‘π‘‘π‘ 
𝐸𝐴 𝐼𝐴
Electrical Efficiency
πœ‚π‘’ =
π·π‘Ÿπ‘–π‘£π‘–π‘›π‘” π‘ƒπ‘œπ‘€π‘’π‘Ÿ 𝑖𝑛 π΄π‘Ÿπ‘šπ‘Žπ‘‘π‘’π‘Ÿπ‘’ 𝑖𝑛 π‘Šπ‘Žπ‘‘π‘‘π‘ 
𝐸𝐴 𝐼𝐴
× 100% =
× 100%
π‘€π‘œπ‘‘π‘œπ‘Ÿ 𝐼𝑛𝑝𝑒𝑑 𝑖𝑛 π‘Šπ‘Žπ‘‘π‘‘π‘ 
𝑉𝑇 𝐼𝐿
Overall or Commercial Efficiency
πœ‚π‘ = πœ‚π‘š × πœ‚π‘’ =
π‘€π‘œπ‘‘π‘œπ‘Ÿ 𝑂𝑒𝑑𝑝𝑒𝑑 𝑖𝑛 π‘Šπ‘Žπ‘‘π‘‘π‘ 
π‘€π‘œπ‘‘π‘œπ‘Ÿ 𝑂𝑒𝑑𝑝𝑒𝑑 𝑖𝑛 π‘Šπ‘Žπ‘‘π‘‘π‘ 
× 100% =
× 100%
π‘€π‘œπ‘‘π‘œπ‘Ÿ 𝐼𝑛𝑝𝑒𝑑 𝑖𝑛 π‘Šπ‘Žπ‘‘π‘‘π‘ 
𝑉𝑇 𝐼𝐿
22
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Power Stages and Efficiency
Example: The armature winding of a 4-pole, 250V DC shunt motor is lap
connected. There are 120 slots, each slot containing 8 conductors. The flux per
pole is 20mWb and current taken by the motor is 25A. The resistance of
armature and field circuit are 0.1Ω and 125Ω respectively. If the rotational
losses amount to be 810W. Find
(i) gross torque (Ta) (ii) useful torque (Tsh) (iii) efficiency.
(i) gross torque (Ta)
𝑉𝑇 250
𝐼𝐹 =
=
= 2𝐴
𝑅𝐹 125
𝐼𝐴 = 𝐼𝐿 − 𝐼𝐹 = 25 − 2 = 23𝐴
𝐸𝐴 = 𝑉𝑇 − 𝐼𝐴 𝑅𝐴 = 250 − 23 × 0.1 = 247.7𝑉
𝐸𝐴 = 247.7 =
π‘‡π‘Ž = 9.55
πœ™π‘π‘ 𝑃 0.02 × (120 × 8) × π‘ 4
× =
×
60
𝐴
60
4
→
𝑁 = 773π‘Ÿπ‘π‘š
𝐸𝐴 𝐼𝐴
247.7 × 23
= 9.55
= 70.4N. m
𝑁
773
23
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Power Stages and Efficiency
(ii) useful torque (Tsh)
Driving power in armature = 𝐸𝐴 𝐼𝐴 = 247.7 × 23 = 5697.1π‘Š
Optput power = 𝐸𝐴 𝐼𝐴 − π‘Ÿπ‘œπ‘‘π‘Žπ‘‘π‘–π‘œπ‘›π‘Žπ‘™ π‘™π‘œπ‘ π‘ π‘’π‘  = 5697.1 − 810 = 4887.1π‘Š
π‘‡π‘ β„Ž = 9.55
π‘€π‘œπ‘‘π‘œπ‘Ÿ π‘œπ‘’π‘‘π‘π‘’π‘‘
4887.1
= 9.55
= 60.4N. m
𝑁
773
(iii) efficiency
πœ‚π‘š =
π‘€π‘œπ‘‘π‘œπ‘Ÿ 𝑂𝑒𝑑𝑝𝑒𝑑 𝑖𝑛 π‘Šπ‘Žπ‘‘π‘‘π‘ 
4887.1
× 100% =
× 100% = 85.8%
𝐸𝐴 𝐼𝐴
5697.1
πœ‚π‘’ =
𝐸𝐴 𝐼𝐴
5697.1
× 100% =
× 100% = 91%
𝑉𝑇 𝐼𝐿
250 × 25
πœ‚π‘ =
4887.1
× 100% = 78.2%
250 × 25
24
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Power Stages and Efficiency
Example: A 20hp (14.92kW), 230V, 1150rpm, 4poles DC shunt motor has a
total of 620 conductors arranged in two parallel paths and yielding an
armature circuit resistance of 0.2Ω. When it delivers rated power at rated
speed, it draws a line current of 74.8A and a field current of 3A. Calculate
(i) the flux per pole (ii) armature torque (iii) the rotational losses (iv) total
losses expressed as a percentage of power.
(i) the flux per pole
𝐼𝐴 = 𝐼𝐿 − 𝐼𝐹 = 74.8 − 3 = 71.8𝐴
𝐸𝐴 = 𝑉𝑇 − 𝐼𝐴 𝑅𝐴 = 230 − 71.8 × 0.2 = 215.64𝑉
πœ™π‘π‘ 𝑃 πœ™ × 620 × 1150 4
× =
×
60
𝐴
60
2
(ii) armature torque
𝐸𝐴 = 215.64 =
→
πœ™ = 9π‘šπ‘Šπ‘
𝐸𝐴 𝐼𝐴
215.64 × 71.8
= 9.55
= 128.6N. m
𝑁
1150
(iii) the rotational losses
π‘‡π‘Ž = 9.55
Rotational losses=EAIA – output power =215.64×71.8 - 14920= 562.952W
(iv) total losses expressed as a percentage of power
Total losses=input power (VTIL)-output power = 230×74.8 - 14920= 17204 - 14920= 2284W
total losses expressed as a percentage of power=2284/17204=13.3%
25
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Power Stages and Efficiency
Example: A 7.46kW, 250V shunt motor takes a line current of 5A when
running light. Calculate the efficiency as a motor when delivering full load
output, if the armature resistance are 0.5Ω and 250 Ω respectively.
1- When loaded lightly
𝑃𝑖𝑛 = 𝑉𝑇 𝐼𝐿 = 250 × 5 = 1250π‘Š
𝑉𝑇 250
𝐼𝐴 = 𝐼𝐿 − 𝐼𝐹 = 5 − 1 = 4𝐴
𝐼𝐹 =
=
= 1𝐴
𝑅𝐹 250
Field Cu loss = 𝑉𝑇 𝐼𝐹 = 250 × 1 = 250W
Armature Cu loss = 𝐼𝐴 2 𝑅𝐴 = 42 × 0.5 = 8W
Iron and friction losses=input power-Armature Cu loss-Filed Cu loss
Iron and friction losses=1250-250-8=992W
Iron and friction losses is constant
2- at full load
𝑉𝑇 𝐼𝐴 = 𝐼𝐴 2 𝑅𝐴 + 𝐸𝐴 𝐼𝐴
𝑉𝑇 𝐼𝐴 = 250𝐼𝐴
𝐸𝐴 𝐼𝐴 = π‘œπ‘’π‘‘π‘π‘’π‘‘ π‘π‘œπ‘€π‘’π‘Ÿ + π‘–π‘Ÿπ‘œπ‘› π‘™π‘œπ‘ π‘  + π‘π‘Ÿπ‘’π‘ β„Žπ‘’π‘  π‘™π‘œπ‘ π‘  = 7460 + 992 + 0 = 8452π‘Š
250𝐼𝐴 = 𝐼𝐴 2 × 0.5 + 8452
26
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Power Stages and Efficiency
250𝐼𝐴 = 𝐼𝐴 2 × 0.5 + 8452
Rearrange the above equation yields
0.5𝐼𝐴 2 − 250𝐼𝐴 + 8452 = 0
−𝑏 ± 𝑏2 − 4π‘Žπ‘ 250 ± (−250)2 −4 × 0.5 × 8452
𝐼𝐴 =
=
= 36.5 π‘œπ‘Ÿ 463.5
2π‘Ž
2 × 0.5
𝐼𝐴 = 36.5𝐴
𝐼𝐿 = 𝐼𝐴 + 𝐼𝐹 = 36.5 − 1 = 37.5𝐴
𝑃𝑖𝑛 = 𝑉𝑇 𝐼𝐿 = 250 × 37.5 = 9375π‘Š
π‘ƒπ‘œπ‘’π‘‘ = 7460π‘Š
πœ‚πΉπΏ
π‘œπ‘’π‘‘π‘π‘’π‘‘ π‘π‘œπ‘€π‘’π‘Ÿ
7460
=
× 100% =
× 100% = 79.6%
𝑖𝑛𝑝𝑒𝑑 π‘π‘œπ‘€π‘’π‘Ÿ
9375
27
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Power Stages and Efficiency
Example: A 6-pole, 500V, wave connected shunt motor has 1200 armature
conductors and useful flux/pole of 20mWb. The armature and field resistances
are 0.5Ω and 250Ω respectively. What will be the speed and torque developed
by the motor when it draws 20A from the supply mains? If magnetic and
mechanical losses amount to 900W, find (i) output in kW (ii) useful torque
(Tsh) (iii) efficiency (πœ‚π‘ ) at this load.
𝐼𝐹 =
𝑉𝑇 500
=
= 2𝐴
𝑅𝐹 250
𝐼𝐴 = 𝐼𝐿 − 𝐼𝐹 = 20 − 2 = 18𝐴
𝐸𝐴 = 𝑉𝑇 − 𝐼𝐴 𝑅𝐴 = 500 − 18 × 0.5 = 491𝑉
πœ™π‘π‘ 𝑃 0.02 × 1200 × π‘ 6
𝐸𝐴 = 491 =
× =
×
60
𝐴
60
2
𝐸𝐴 𝐼𝐴
491 × 18
π‘‡π‘Ž = 9.55
= 9.55
= 205.9N. m
𝑁
410
(i) output in kW
→
𝑁 = 410π‘Ÿπ‘π‘š
Field Cu loss = 𝑉𝑇 𝐼𝐹 = 500 × 2 = 1000W
Armature Cu loss = 𝐼𝐴 2 𝑅𝐴 = 182 × 0.5 = 162W
Iron and friction losses=900W
28
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Power Stages and Efficiency
Total losses=armature Cu loss + field Cu loss + iron and friction loss
Total losses=162+1000+900=2062W
Motor input power=VTIL=500×20=10000W
Motor output power=motor input power-motor losses=10000-2062=7938W
(ii) useful torque (Tsh)
π‘‡π‘ β„Ž = 9.55
π‘€π‘œπ‘‘π‘œπ‘Ÿ π‘œπ‘’π‘‘π‘π‘’π‘‘
7938
= 9.55
= 184.9N. m
𝑁
410
(iii) Efficiency (πœ‚π‘ ) at this load
π‘œπ‘’π‘‘π‘π‘’π‘‘ π‘π‘œπ‘€π‘’π‘Ÿ
7938
πœ‚π‘ =
× 100% =
× 100% = 79.38%
𝑖𝑛𝑝𝑒𝑑 π‘π‘œπ‘€π‘’π‘Ÿ
10000
29
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Characteristics and Applications of DC Motors
Type of Motor
Shunt Motor
Series Motor
Cumulative
Compound
Motor
Characteristics
Applications
Approximately constant
speed
Adjustable speed
Medium starting torque (up
to 1.5 F.L. torque)
For driving constant speed line
shafting
Lathes
Centrifugal pumps
Machine tools
Blowers and fans
Reciprocating pumps
Variable speed
Adjustable variable speed
High starting torque
For traction work i.e. Electric
locomotives
Rapid transit systems
Trolley, Cars etc
Cranes and hoists
conveyers
Variable speed
Adjustable variable speed
High starting torque
For intermittent high torque loads
For shears and punches
Elevators
Conveyers
Heavy planers
Rolling mills, Ice machine, printing
presses, Air compressors
30
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
31
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