ELEC 3105 Basic EM and Power Engineering Rotating DC Motor PART 2 Electrical.
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Transcript of ELEC 3105 Basic EM and Power Engineering Rotating DC Motor PART 2 Electrical.
![Page 1: ELEC 3105 Basic EM and Power Engineering Rotating DC Motor PART 2 Electrical.](https://reader033.fdocuments.in/reader033/viewer/2022052915/5a4d1add7f8b9ab059975853/html5/thumbnails/1.jpg)
ELEC 3105 Basic EM and Power Engineering
Rotating DC MotorPART 2 Electrical
![Page 2: ELEC 3105 Basic EM and Power Engineering Rotating DC Motor PART 2 Electrical.](https://reader033.fdocuments.in/reader033/viewer/2022052915/5a4d1add7f8b9ab059975853/html5/thumbnails/2.jpg)
Motor / Generator Action
BrVemf
2
Equivalent circuit
emfVbat
V
R
terminalv
Expression of Vemf
Loop
Slide extracted from linear motor and modified for loop motor.
I
emfbatVIRV
BrRrB
Vbat
22
rBI2
![Page 3: ELEC 3105 Basic EM and Power Engineering Rotating DC Motor PART 2 Electrical.](https://reader033.fdocuments.in/reader033/viewer/2022052915/5a4d1add7f8b9ab059975853/html5/thumbnails/3.jpg)
Motor / Generator Action
BrRrB
Vbat 22
Linear relation betweenspeed and torque
rBVbat
loadno 2
RVrB bat2
0
Current flows in a direction to charge the battery.
Motor
Slide extracted from linear motor and modified for loop motor.
Stall torque
GeneratorLink
![Page 4: ELEC 3105 Basic EM and Power Engineering Rotating DC Motor PART 2 Electrical.](https://reader033.fdocuments.in/reader033/viewer/2022052915/5a4d1add7f8b9ab059975853/html5/thumbnails/4.jpg)
Electrical Equivalent
V
fV
aI π ππΏπ
E
ππ ,π πππ£
πΏ π
π π
E = motor voltageRa = armature resistanceLa = armature inductanceV = Applied motor voltageIa = armature current
= magnetic fluxRf = field resistanceLf = field inductanceVf = Field voltageIf = field current
![Page 5: ELEC 3105 Basic EM and Power Engineering Rotating DC Motor PART 2 Electrical.](https://reader033.fdocuments.in/reader033/viewer/2022052915/5a4d1add7f8b9ab059975853/html5/thumbnails/5.jpg)
Electrical Equivalent
V
fV
aI π ππΏπ
E
ππ ,π πππ£
πΏ π
π π
In steady state operation: FIELD SIDE
πΌ π=π π
π πΞ¦=
2ππΌ ππreluctance
Magnetic circuit
![Page 6: ELEC 3105 Basic EM and Power Engineering Rotating DC Motor PART 2 Electrical.](https://reader033.fdocuments.in/reader033/viewer/2022052915/5a4d1add7f8b9ab059975853/html5/thumbnails/6.jpg)
Electrical Equivalent
V
fV
aI π ππΏπ
E
ππ ,π πππ£
πΏ π
π π
In steady state operation: ARMATURE SIDE
πΈ=ππΈΞ¦ππ
Motor constant
Back emf
![Page 7: ELEC 3105 Basic EM and Power Engineering Rotating DC Motor PART 2 Electrical.](https://reader033.fdocuments.in/reader033/viewer/2022052915/5a4d1add7f8b9ab059975853/html5/thumbnails/7.jpg)
Motor / Generator Action
BrVemf
2
Equivalent circuit
emfVbat
V
R
terminalv
Expression of Vemf
Loop
Slide extracted from linear motor and modified for loop motor.
I
emfbatVIRV
BrRrB
Vbat
22
rBI2
![Page 8: ELEC 3105 Basic EM and Power Engineering Rotating DC Motor PART 2 Electrical.](https://reader033.fdocuments.in/reader033/viewer/2022052915/5a4d1add7f8b9ab059975853/html5/thumbnails/8.jpg)
Electrical Equivalent
V
fV
aI π ππΏπ
E
ππ ,π πππ£
πΏ π
π π
In steady state operation: ARMATURE SIDE
πΈ=ππΈΞ¦ππ
Motor constant
Back emf
![Page 9: ELEC 3105 Basic EM and Power Engineering Rotating DC Motor PART 2 Electrical.](https://reader033.fdocuments.in/reader033/viewer/2022052915/5a4d1add7f8b9ab059975853/html5/thumbnails/9.jpg)
Electrical Equivalent
V
fV
aI π ππΏπ
E
ππ ,π πππ£
πΏ π
π π
In steady state operation: ARMATURE SIDE
π πππ£=ππΞ¦ πΌπ
Motor constant
Developed torque
ππΈ=ππ Same motor constant in emf and developed torque
![Page 10: ELEC 3105 Basic EM and Power Engineering Rotating DC Motor PART 2 Electrical.](https://reader033.fdocuments.in/reader033/viewer/2022052915/5a4d1add7f8b9ab059975853/html5/thumbnails/10.jpg)
Motor / Generator Action
BrVemf
2
Equivalent circuit
emfVbat
V
R
terminalv
Expression of Vemf
Loop
Slide extracted from linear motor and modified for loop motor.
I
emfbatVIRV
BrRrB
Vbat
22
rBI2
![Page 11: ELEC 3105 Basic EM and Power Engineering Rotating DC Motor PART 2 Electrical.](https://reader033.fdocuments.in/reader033/viewer/2022052915/5a4d1add7f8b9ab059975853/html5/thumbnails/11.jpg)
Electrical Equivalent
V
fV
aI π ππΏπ
E
ππ ,π πππ£
πΏ π
π π
In steady state operation: ARMATURE SIDE
π πππ£=ππΞ¦ πΌ π
Motor constant
Developed torque
ππΈ=ππ Same motor constant in emf and developed torque
πΈ=ππΈΞ¦ππ
Motor constant
Back emf
![Page 12: ELEC 3105 Basic EM and Power Engineering Rotating DC Motor PART 2 Electrical.](https://reader033.fdocuments.in/reader033/viewer/2022052915/5a4d1add7f8b9ab059975853/html5/thumbnails/12.jpg)
Electrical Equivalent
V
fV
aI π ππΏπ
E
ππ ,π πππ£
πΏ π
π π
Power flow: ARMATURE SIDE; Conservation of energy
π=πΌππ π+πΈKVL
POWER π πΌ π= πΌπ2 π π+πΌ ππΈ
Power in
Armature copper loss
Power developed
![Page 13: ELEC 3105 Basic EM and Power Engineering Rotating DC Motor PART 2 Electrical.](https://reader033.fdocuments.in/reader033/viewer/2022052915/5a4d1add7f8b9ab059975853/html5/thumbnails/13.jpg)
Motor / Generator Action
BrVemf
2
Equivalent circuit
emfVbat
V
R
terminalv
Expression of Vemf
Loop
Slide extracted from linear motor and modified for loop motor.
I
emfbatVIRV
BrRrB
Vbat
22
rBI2
![Page 14: ELEC 3105 Basic EM and Power Engineering Rotating DC Motor PART 2 Electrical.](https://reader033.fdocuments.in/reader033/viewer/2022052915/5a4d1add7f8b9ab059975853/html5/thumbnails/14.jpg)
Electrical Equivalent
V
fV
aI π ππΏπ
E
ππ ,π πππ£
πΏ π
π π
Power flow: ARMATURE SIDE; Conservation of energy
ππππ£= πΌππΈ=ππππππ£
Power developed
π πππ π π ππ
ππππ£
Electrical Mechanicalcopper
π πΌ π= πΌπ2 π π+πΌ ππΈ
![Page 15: ELEC 3105 Basic EM and Power Engineering Rotating DC Motor PART 2 Electrical.](https://reader033.fdocuments.in/reader033/viewer/2022052915/5a4d1add7f8b9ab059975853/html5/thumbnails/15.jpg)
Electrical Equivalent
V
fV
aI π ππΏπ
E
ππ ,π πππ£
πΏ π
π π
Power flow: ARMATURE SIDE πππ’π‘=ππππ£βππππ‘
π πππ π π ππ
ππππ£
Electrical Mechanical
ππππ‘
πππ’π‘
Rotational lossCopper
![Page 16: ELEC 3105 Basic EM and Power Engineering Rotating DC Motor PART 2 Electrical.](https://reader033.fdocuments.in/reader033/viewer/2022052915/5a4d1add7f8b9ab059975853/html5/thumbnails/16.jpg)
Electrical Equivalent
V
fV
aI π ππΏπ
E
ππ ,π πππ£
πΏ π
π π
Motor sequence πππ’π‘=ππππ£βππππ‘
πππ π πππ£
Speed of rotation limiting loop
πΈπΌπ ππππ£
![Page 17: ELEC 3105 Basic EM and Power Engineering Rotating DC Motor PART 2 Electrical.](https://reader033.fdocuments.in/reader033/viewer/2022052915/5a4d1add7f8b9ab059975853/html5/thumbnails/17.jpg)
Shunt Connected Field
V
aIπ ππΏπ
E
ππ ,π πππ£πΏ π
π πR
π πππ£=ππΞ¦ πΌπ
π=πΌππ π+πΈ
πΈ=ππΈΞ¦ππ
π πππ£=πΞ¦ππ π
β π2Ξ¦2
π πππ
Developed torque Rotation rate
![Page 18: ELEC 3105 Basic EM and Power Engineering Rotating DC Motor PART 2 Electrical.](https://reader033.fdocuments.in/reader033/viewer/2022052915/5a4d1add7f8b9ab059975853/html5/thumbnails/18.jpg)
Shunt Connected Field
V
aIπ ππΏπ
E
ππ ,π πππ£πΏ π
π πR
π πππ£=πΞ¦ππ π
β π2Ξ¦2
π πππ
Similar type of graph
![Page 19: ELEC 3105 Basic EM and Power Engineering Rotating DC Motor PART 2 Electrical.](https://reader033.fdocuments.in/reader033/viewer/2022052915/5a4d1add7f8b9ab059975853/html5/thumbnails/19.jpg)
Shunt Connected Field
V
aIπ ππΏπ
E
ππ ,π πππ£πΏ π
π πR
π πππ£=πΞ¦ππ π
β π2Ξ¦2
π πππ
π πππ£
ππ
πΎΞ¦ππ π
ππΞ¦
MotorGenerator
Force motor to spin backwards
Generator
Force motor to spin to fast
![Page 20: ELEC 3105 Basic EM and Power Engineering Rotating DC Motor PART 2 Electrical.](https://reader033.fdocuments.in/reader033/viewer/2022052915/5a4d1add7f8b9ab059975853/html5/thumbnails/20.jpg)
Series Connected Field
V
aIπ ππΏπ
E
ππ ,π πππ£
πΏ ππ π
Universal motor design: works for D.C. and for A.C.
π πππ£=ππΞ¦ πΌ π
π=πΌ π(π ΒΏΒΏπ+π π )+πΈ ΒΏ
πΈ=ππΈΞ¦ππ
π πππ£=πβ² πΌπ2 Since Ξ¦β πΌπ
![Page 21: ELEC 3105 Basic EM and Power Engineering Rotating DC Motor PART 2 Electrical.](https://reader033.fdocuments.in/reader033/viewer/2022052915/5a4d1add7f8b9ab059975853/html5/thumbnails/21.jpg)
Series Connected Field
V
aIπ ππΏπ
E
ππ ,π πππ£
πΏ ππ π
Universal motor design: works for D.C. and for A.C.
π πππ£=πβ² [ ππ π+π π+πβ²ππ ]
2
π πππ£
ππ
1ππ
2
![Page 22: ELEC 3105 Basic EM and Power Engineering Rotating DC Motor PART 2 Electrical.](https://reader033.fdocuments.in/reader033/viewer/2022052915/5a4d1add7f8b9ab059975853/html5/thumbnails/22.jpg)
Maximum Power Transfer
V
fV
aI π ππΏπ
E
ππ ,π πππ£
πΏ π
π ππ=πΌ ππ π+πΈ
Power developed in the motor ππππ£=πΈ πΌπ
ππππ£=πΈ (π βπΈ )/π π
ππππ£=πΈπ βπΈ2
π π
Find maximum with respect to the motor voltage
![Page 23: ELEC 3105 Basic EM and Power Engineering Rotating DC Motor PART 2 Electrical.](https://reader033.fdocuments.in/reader033/viewer/2022052915/5a4d1add7f8b9ab059975853/html5/thumbnails/23.jpg)
Maximum Power Transfer
V
fV
aI π ππΏπ
E
ππ ,π πππ£
πΏ π
π ππ=πΌ ππ π+πΈ
For extremes of a function, take derivatives and set to zero
ππππ£=πππ₯πππ’π hπ€ πππΈ=π2
ππππ£ πππ₯ = π£2
4 π π
![Page 24: ELEC 3105 Basic EM and Power Engineering Rotating DC Motor PART 2 Electrical.](https://reader033.fdocuments.in/reader033/viewer/2022052915/5a4d1add7f8b9ab059975853/html5/thumbnails/24.jpg)
Calculation example
V
fV
aI π ππΏπ
E
ππ ,π πππ£
πΏ π
π π
A 120 volt dc motor has an armature resistance of 0.70 Ξ©. At no-load, it requires 1.1 A armature current and runs at 1000 rpm. Find the output power and torque at 952 rpm output speed. Assume constant flux.
π πππ£=ππΞ¦ πΌππ=πΌππ π+πΈ
πΈ=ππΈΞ¦ππ
Solution provided in class
![Page 25: ELEC 3105 Basic EM and Power Engineering Rotating DC Motor PART 2 Electrical.](https://reader033.fdocuments.in/reader033/viewer/2022052915/5a4d1add7f8b9ab059975853/html5/thumbnails/25.jpg)
Calculation example
V
fV
aI π ππΏπ
E
ππ ,π πππ£
πΏ π
π π
A permanent magnet dc motor has the following information: 50 hp, 200 V, 200 A, 1200 rpm and armature resistance of 0.05 Ξ©. Determine the output power if the voltage is lowered to 150 V and the current is 200 A. Assume rotational losses are proportional to speed. Determine the rotational loss, armature resistance, no-load rpm, machine constant, efficiency?
π πππ£=ππΞ¦ πΌππ=πΌππ π+πΈ
πΈ=ππΈΞ¦ππ
Solution provided in class
![Page 26: ELEC 3105 Basic EM and Power Engineering Rotating DC Motor PART 2 Electrical.](https://reader033.fdocuments.in/reader033/viewer/2022052915/5a4d1add7f8b9ab059975853/html5/thumbnails/26.jpg)
Calculation example
V
fV
aI π ππΏπ
E
ππ ,π πππ£
πΏ π
π π
An 80 V dc motor has constant field flux, separately excited, and a nameplate speed of 1150 rpm with 710 W output power. The nameplate armature current is 10 A and the no-load current is 0.5 A. Assume constant rotational losses.
π πππ£=ππΞ¦ πΌππ=πΌππ π+πΈ
πΈ=ππΈΞ¦ππ
Solution provided in class