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6/17/2015
1
ET 332a
Dc Motors, Generators and Energy Conversion Devices
1Lesson 22 332a.pptx
Learning Objectives
After this presentation you will be able to:
Explain how a self-excited dc generator operates Explain how voltage builds in self-excited dc
generators Use magnetization curves and dc machine
formulas to find an unloaded generator’s terminal voltage
Identify the voltage characteristics of compound generators.
2Lesson 22 332a.pptx
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Self-Excited Generator Connection
Lesson 22 332a.pptx 3
With field switch open only residual flux can produce Ea
nΦr
+
If
Racir
Rf
VT
Shunt field is connected in parallel with armature
Prime mover provides mechanical power
Residual magnetism in pole pieces causes induced voltage
Voltage Buildup in Self-Excited Generators
Lesson 22 332a.pptx 4
nΦr
+
If
Racir
Rf
VT
1.) Residual magnetism causesEa0 with If=0
Ea0
2.) Close switch and parallelarmature and field Ea=VT
3.) Ea0/Rf = If1 causes Ea to increase due to field flux increase
Ea1
4.) Ea increases to Ea1 causing If2
If2
5.) Process repeats until operating point reached
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Self-Excited Generator Graphical Analysis
Lesson 22 332a.pptx 5
Operating pointoccurs whenmagnetizing curveand field R linesintersect
Ea
If
Rf
Magnetizing curve
Operating Point
Ea0
Ea1
If1
Field Resistance and Voltage Buildup
Lesson 22 332a.pptx 6
Analytic methods
Reluctance is nonlinear w.r.t. If
In field circuit so
Gff
Gpa kIN
nknE ⋅
⋅⋅=⋅Φ⋅=R
rheof RRR +=aT EV =and
Equate field and armature circuit equations
Gff
rheoff kIN
n)RR(I ⋅
⋅⋅=+⋅R
Non-linear equation with respect to (w.r.t.) If. Difficult to solve.
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Lesson 22 332a.pptx 7
Field Resistance and Voltage Buildup
R4 > R3 > R2 > R1
R4 is critical resistanceEa will not build up dueto excessive field resistance
Ea
If
R4R3 R2
R1
rheof RRR +=
Voltage Buildup and Speed
Lesson 22 332a.pptx 8
Speed of rotation effects the voltage buildup. Lower speed - lower Ea
Adjusting the rheostat controlsterminal voltage. Critical resistanceprevents voltage buildup at all speeds
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Self-Excited Generator Example
Lesson 22 332a.pptx 9
Example 22-1: Given a generators magnetization curve (Fig 12-4, p443, text). Self-excited generator ratings: 125 Vdc, 50 kW, 1750 rpm. No-load voltage with rheostat shorted 156 V
Determine:
a.) field circuit resistanceb.) field rheostat setting for no-load voltage of 140 Vdcc.) armature voltage if rheostat is set at 14.23 ohmsd.) rheostat setting that will give critical resistancee.) armature voltage at 80% rated speed and rheostat
shortedf.) rheostat setting for no-load voltage of 140 Vdc at 1750 rpm if field is
separately excited from 120 Vdc source
Example 22-1 Solution (1)
Lesson 22 332a.pptx 10
Part a.) Field circuit resistance (rheostat shorted)
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 70
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Magnetization CurveField R Line
Field Current If
Ind
uce
d V
olta
ge
Ea
1564.7
For voltage of 156 Vdc, graph gives current of 4.7 A
Use Ohm’s Law to find the valueof field resistance
Answer
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Lesson 22 332a.pptx 11
Example 22-1 Solution (2)Part b.) Rheostat setting to give Ea = 140 Vdc
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 70
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Magnetization CurveField Current If
Ind
uce
d V
olta
ge
Ea
140
3.2
Field current value from graph is 3.2 A
Subtract off field resistance to findrheostat setting
Rrheo
Ea
If
Rf−=
140 V
33.2 Ω
3.2 A
Answer
Lesson 22 332a.pptx 12
Example 22-1 Solution (3)Part c.) Find value of Ea when field rheostat increased to:
Rrheo 14.23 ohm⋅=
R f R rheo+ 47.421 ohm=Total field circuit resistance is:
Graph this equation: Ea = 47.421∙If
The value of Ea is where the above equation intersects the magnetizationcurve.
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Lesson 22 332a.pptx 13
Example 22-1 Solution (4)Part d.) Find the critical resistance value.
Critical resistance has the same value as the slope of the linear part of the generator’s magnetization curve. Select two points and compute the slope
fc
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 70
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Magnetization CurveCritical Resistance Line
Example Self-Excited Generator
Field Current If
Induced V
oltage E
a
A 51.1I V 90E f11a ==
A 50.0I V 30E f2a2 ==
Plot the line Ea = 59.4∙If This is the critical resistance line
Rcrit
Ans
Lesson 22 332a.pptx 14
Example 22-1 Solution (5)
Part e.) Armature voltage at 80% rated speed and rheostat shorted out.
Multiply the rated speed Ea
curve data by 0.80 to get new curve then plot the line Ea = 33.2∙If line. The intersection is the solution
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 70
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Magnetization Curve80% Speed EaField Resistance Line
Field Current If
Induced V
oltage E
a 115
Ea = 115 V Ans
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Lesson 22 332a.pptx 15
Example 22-1 Solution (6)Part f.) Find the field rheostat for separately excited operation for a 120 V dc source if Ea = 140 V dc
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 70
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Magnetization Curve
Example Self-Excited Generator
Field Current If
Induced V
oltage E
a
If =3.2 A
Separately excitedsource voltage andRf from part a.)
Rrheo
Vf
IfRf−=
R rheo 4.309 ohm=
33.191 Ω
3.2 A
120 V
Ans
Terminal Voltage Characteristic
Lesson 22 332a.pptx 16
Rated
Rated
VT
Ia
Increasing load increasesIa. IaR drop.
Increase load until breakdown occurs
Breakdown
Constant IRegion
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Compound Generator Connections
Lesson 22 332a.pptx 17
Short shunt connection Avoids series field drop
Long shunt connectionPreferred in calculations
Compound Generator Characteristics
Lesson 22 332a.pptx 18
Loading effects on induced voltage
Ff = shunt field mmfFs = series field mmfFd = demagnetizing mmf
Gdsf
a kE ⋅
−+=
R
FFF
Series field flux related to load. Increasing loadincreases field flux which increase Ea.
OverCompounded
FlatCompounded
UnderCompounded
ShuntGenerator
DifferentiallyCompounded
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ET 332a
Dc Motors, Generators and Energy Conversion Devices
Lesson 22 332a.pptx 19
