Electric Machines I - Philadelphia University · 2018. 1. 22. · For Short Shunt Cumulatively Compound DC Generator 𝑇=𝐼𝐹 𝐹 𝐼𝐴=𝐼𝐹+𝐼𝐿 𝑇= 𝐴−𝐼𝐴

Electric Machines I DC Machines - DC Generators

1

Dr. Firas Obeidat

2

Table of contents

1 • Construction of Simple Loop Generator

2 • Working of Simple Loop Generator

3 • Types of DC Generators

4 • The Terminal Characteristic of a Separately Excited DC Generator

5 • The Terminal Characteristic of a Self Excited Shunt DC Generator

6 • The Terminal Characteristic of a Self Excited Series DC Generator

7 • The Terminal Characteristic of Cumulatively Compound DC Generator

8 • E.M.F. Equation of DC Generator

9 • Total Loss in a DC Generator

10 • Power Stages and Efficiency

11 • Voltage Regulation

12 • Uses of DC Generators

Dr. Firas Obeidat Faculty of Engineering Philadelphia University

3 Dr. Firas Obeidat Faculty of Engineering Philadelphia University

Construction of Simple Loop Generator

A single turn rectangular copper ABCD rotating about its own axis in a

magnetic field provided by either permanent magnet or electromagnet.

The two ends of the coil are joined to slip ring ‘a’ and ‘b’ which are

insulated from each other and from the central shaft. Two collecting

brushes press against the slip rings; their function is to collect the

current induced in the coil and to convey it to external load resistance.

The rotating coil is called the ‘armature’.


Working of Simple Loop Generator Imagine the coil to be rotating in clockwise

direction. As the coil assumes successive

positions in the field, the flux linked with it

changes. An emf is induced in it which is

proportional to the rate of change of flux

linkages (e=Ndϕ/dt).

When the plane of coil is in position 1, then

flux linked with the coil is maximum but rate

of change of flux linkage is minimum. Hence,

there is no induced emf in the coil.

As the coil continues rotating further, the rate

of change of flux linkages (and hence induced

emf in it) increases, till position 3 is reached

where θ=90o. The coil plane is horizontal

(parallel to the lines of flux). The flux linked

with the coil is minimum but rate of change of

flux linkage is maximum. Hence, maximum

emf is induced in the coil when in this position.

In the second half revolution, the direction of

the current flow is DCMLBA. Which is just

the reverse of the previous direction of flow.


Working of Simple Loop Generator In the next quarter revolution (from 90o

to 180o), the flux linked with the coil

gradually increases but the rate of change

of flux linkages decreases. Hence the

induced emf decreases gradually till in

position 5 of the coil, it reduces to zero

value.

In the first half revolution of the coil, no

emf is induced in it when in position 1,

maximum when in position 3 and no emf

when in position 5. In this half revolution,

the direction of the current flow is

ABMLCD. The current through the load

resistor R flows from M to L during the

first half revolution of the coil.

In the next half revolution (from 180o to

360o), the variations in the magnitude of

emf are similar to those in the first half

revolution. Its value is maximum when

the coil is in position 7 and minimum

when it in position 1.


Working of Simple Loop Generator

The current which is obtained from

such a simple generator reverses its

direction after every half

revolution, this current is known as

alternating current. To make the

flow of current unidirectional in

the external circuit, the slip rings

are replaced by split rings.

In the first half revolution segment

‘a’ is connected to brush 1 and

segment ‘b’ is connected to brush 2,

while in the second half revolution

segment ‘b’ is connected to brush 1

and segment ‘a’ is connected to

brush 2. In this case the current

will flow in the resistor from M to

L in the two halves of revolution.

The resulting current is

unidirectional but not continuous

like pure direct current.


Types of DC Generators

Generators are usually classified according to the way in which their fields

are excited

A. Separately Excited Generators: are those whose field magnets are

energized from an independent external source of DC current.

B. Self Excited Generators: are those whose field magnets are energized by

current produced by the generators themselves. There are three types of

self excited generators named according to the manner in which their

field coils are connected to the armature.

i. Shunt Wound: the field windings are connected across or in parallel

with the armature conductors and have the full voltage of the generator

applied across them.

ii. Series Wound: the field windings are joined in series with the armature

conductors

iii.Compound Wound: it is a combination of a few series and a few shunt

windings and can be either short-shunt or long-shunt. In compound

generator, the shunt field is stronger than the series field. When series

field aids the shunt field, generator is said to be commutatively-

compound. In series field oppose the shunt field, the generator is said to

be differentially compounded.



Separately Excited Generators Shunt Wound Generators Series Wound Generators

Short Shunt Generators Long Shunt Generators




The Terminal Characteristic of a Separately Excited DC Generator

𝐼𝐴 = 𝐼𝐿

𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴 𝑅𝐴

𝑉𝐹 = 𝐼𝐹 𝑅𝐹

For Separately Excited DC Generator

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

ϕ: is the flux

𝜔m: is the rotor angular speed

𝐸𝐴 = 𝑘ϕ𝜔𝑚

The terminal voltage can be controlled by:

1. Change the speed of rotation: If 𝜔

increases, then 𝐸𝐴=𝑘ϕ𝜔𝑚 increases, so

𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴 𝑅𝐴 increases as well.

2. Change the field current. If RF is

decreased. then the field current

increases (𝑉𝐹 = 𝐼𝐹 𝑅𝐹 ). Therefore, the

flux in the machine increases. As the

flux rises, 𝐸𝐴=𝑘ϕ𝜔𝑚 must rise too, so

𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴 𝑅𝐴 increases.

EA

-

+ RA

IA IL

-

+

VT

RF

LF

-

+

VF

IF


The Terminal Characteristic of a Self Excited Shunt DC Generator

𝐼𝐴 = 𝐼𝐹 + 𝐼𝐿

𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴 𝑅𝐴

𝑉𝑇 = 𝐼𝐹 𝑅𝐹

For Self Excited Shunt DC Generator

The terminal voltage can be controlled by:

1. Change the speed of rotation: If 𝜔

increases, then 𝐸𝐴=𝑘ϕ𝜔𝑚 increases, so


2. Change the field current. If RF is

decreased. then the field current

increases (𝑉𝐹 = 𝐼𝐹 𝑅𝐹 ). Therefore, the

flux in the machine increases. As the

flux rises, 𝐸𝐴=𝑘ϕ𝜔𝑚 must rise too, so

𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴 𝑅𝐴 increases.

EA

-

+ RA

IA IL

-

+

VT

RF

LF

IF



The Terminal Characteristic of a Self Excited Series DC Generator

𝐼𝐴 = 𝐼𝑠 = 𝐼𝐿

𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴(𝑅𝐴+𝑅𝑠)

For Self Excited Series DC Generator

At no load, there is no field current, so VT is

reduced to a small level given by the

residual flux in the machine. As the load

increases, the field current rises, so EA rises

rapidly The IA(RA+ Rs) drop goes up too,

but at first the increase in EA goes up more

rapidly than the IA(RA+ Rs) drop rises, so

VT increases. After a while, the machine

approaches saturation, and EA becomes

almost constant. At that point, the resistive

drop is the predominant effect, and VT

starts to fall.


EA

-

+ RA

IA IL

-

+

VT

Rs Ls

Is


The Terminal Characteristic of Cumulatively Compound DC Generator


𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴(𝑅𝐴+𝑅𝑠)

For Long Shunt Cumulatively Compound DC Generator


EA

-

+ RA

IA IL

-

+

VT

Rs Ls RF

LF

IF

For Short Shunt Cumulatively Compound DC Generator

𝑉𝑇 = 𝐼𝐹 𝑅𝐹


𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴𝑅𝐴 −𝐼𝐿𝑅𝑠

𝐸𝐴 = 𝑘ϕ𝜔𝑚 EA

-

+ RA

IA IL

-

+

VT

Rs LsRF

LF

IF

The terminal voltage Cumulatively Compound DC Generator can be controlled by:

1. Change the speed of rotation: If 𝜔 increases, then 𝐸𝐴=𝑘ϕ𝜔𝑚 increases, so


2. Change the field current. If RF is decreased. then the field current increases

(𝑉𝐹 = 𝐼𝐹 𝑅𝐹). Therefore, the flux in the machine increases. As the flux rises, 𝐸𝐴=𝑘ϕ

𝜔𝑚 must rise too, so 𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴 𝑅𝐴 increases.


Examples

Example: A shunt DC generator delivers 450A at 230V and the resistance of

the shunt field and armature are 50Ω and 0.3 Ω respectively. Calculate emf.

𝐼𝑓 =230

50= 4.6𝐴

𝐼𝐴 = 𝐼𝐹 + 𝐼𝐿 = 4.6 + 450 = 454.6𝐴

𝐸𝐴 = 𝑉𝑇 + 𝐼𝐴𝑅𝐴 = 230 + 454.6 × 0.3 = 243.6V

Example: A long shunt compound DC generator delivers a load current of

50A at 500V and has armature, series field and shunt field resistances of

0.05Ω, 0.03Ω and 250Ω respectively. Calculate the generated voltage and the

armature current. Allow 1V per brush for contact drop.

𝐼𝐹 =500

250= 2𝐴

𝐼𝐴 = 𝐼𝐹 + 𝐼𝐿 = 2 + 50 = 52𝐴

Voltage drop across series winding=𝐼A𝑅s=52×0.03=1.56V

Armature voltage drop=𝐼A𝑅A=52×0.05=2.6V

Drop at brushes=2×1=2V

𝐸𝐴 = 𝑉𝑇 + 𝐼𝐴𝑅𝐴 + 𝑠𝑒𝑟𝑖𝑒𝑠 𝑑𝑟𝑜𝑝 + 𝑏𝑟𝑢𝑠ℎ𝑒𝑠 𝑑𝑟𝑜𝑝 = 500 + 2.6 + 1.56 + 2 = 506.16V

EA

-

+ RA

IA IL=450A

-

+ VT =

230V

RF

LF

IF

EA

-

+ RA

IA

-

+Rs Ls RF

LF

IF

VT=

500V

IL=50A


Examples

Example: A short shunt compound DC generator delivers a load current of


0.05Ω, 0.3Ω and 200Ω respectively. Calculate the induced emf and the

armature current. Allow 1V per brush for contact drop.

𝐼𝐹 =229

200= 1.145𝐴

Armature voltage drop=𝐼𝐴𝑅𝐴 = 31.145 × 0.05 = 1.56V

Voltage drop across series winding=𝐼L𝑅s

=30 ×0.3=9V

Drop at brushes=2×1=2V

𝐸𝐴 = 𝑉𝑇 + 𝐼𝐴𝑅𝐴 + 𝑠𝑒𝑟𝑖𝑒𝑠 𝑑𝑟𝑜𝑝 + 𝑏𝑟𝑢𝑠ℎ𝑒𝑠 𝑑𝑟𝑜𝑝 = 220 + 9 + 1.56 + 2 = 232.56V

Voltage across shunt winding=220 + 9=229V

EA

-

+ RA

IA

-

+Rs LsRF

LF

IF

VT =

220V

IL=30A


Examples

Example: A long shunt compound DC generator delivers a load current of


0.032Ω, 0.015Ω and 92Ω respectively. Calculate (i) induced emf (ii) total

power generated and (iii) distribution of this power.

EA

-

+ RA

IA

-

+Rs Ls RF

LF

IF

VT =

230V

IL=150A

𝐼𝐹 =230

92= 2.5𝐴

𝐼𝐴 = 𝐼𝐹 + 𝐼𝐿 = 2.5 + 150 = 152.5𝐴

Voltage drop across series winding=

𝐼A𝑅s=152.5 ×0.015=2.2875V

Armature voltage drop=𝐼A𝑅A=152.5 ×0.032=4.88V

Total power generated by the armature=𝐸𝐴𝐼𝐴=237.1675×152.5=36168.04375W

𝐸𝐴 = 𝑉𝑇 + 𝐼𝐴𝑅𝐴 + 𝐼A𝑅s = 230 + 2.2875 + 4.88 = 237.1675V

(i)

(ii)

(iii) Power lost in armature=𝐼𝐴2𝑅𝐴=152.52×0.032=744.2W

Power dissipated in shunt winding=𝑉𝑇𝐼𝐹=230×2.5=575W

Power dissipated in series winding=𝐼𝐴2𝑅𝑠=152.52×0.015=348.84375W

Power delivered to the load=𝑉𝑇𝐼𝐿=230×150=34500W

Total power generated by the armature=744.2 + 575 + 348.843 + 34500=36168.04375W


E.M.F. Equation of DC Generator

Let

ϕ: 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=dϕ/dt volt

Flux cut/conductor in one revolution dϕ=ϕP Wb

Number of revolutions /second=N/60

Time for one revolution dt=60/N second

E.M.F. generated/conductor= dϕ/dt= ϕPN/60 volt



For simplex wave-wound generator

Number of parallel paths=2

Number of conductors (in series) in one path=Z/2

𝐸. 𝑀. 𝐹. 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑/𝑝𝑎𝑡ℎ(𝐸𝐴) =𝜙𝑃𝑁

60×

𝑍

2=

𝜙𝑃𝑍𝑁

120𝑣𝑜𝑙𝑡

For simplex lap-wound generator

Number of parallel paths=P

Number of conductors (in series) in one path=Z/P

𝐸. 𝑀. 𝐹. 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑/𝑝𝑎𝑡ℎ(𝐸𝐴) =𝜙𝑃𝑁

60×

𝑍

𝑃=

𝜙𝑍𝑁

60𝑣𝑜𝑙𝑡

In general

𝐸𝐴 =𝜙𝑍𝑁

60×

𝑃

𝐴𝑣𝑜𝑙𝑡

where

A=2 for simplex wave-winding

A=P for simplex lap-winding

𝐸𝐴 =1

2π×

2π𝑁

60× 𝜙𝑍 ×

𝑃

𝐴=

𝑍𝑃

2π𝐴𝜙𝜔𝑚 𝑣𝑜𝑙𝑡 Where 𝜔𝑚 =

2π𝑁

60

For a given DC machine Z,P and A are constant

𝐸𝐴 = 𝑘𝜙𝜔𝑚 𝑣𝑜𝑙𝑡 Where 𝑘 =𝑍𝑃

2π𝐴



Example: A four pole generator, having wave wound armature winding has 51

slots, each slot containing 20 conductor. What will be the voltage generated in

the machine when driven at 1500 rpm assuming the flux per pole to be

7mWb?


60×

𝑃

𝐴=

7 × 10−3 × 51 × 20 × 1500

60×

4

2= 357𝑣𝑜𝑙𝑡

Example: An 8 pole Dc generator has 500 armature conductors, and a useful

flux of 0.05Wb per pole. what will be the emf generated if it is lap-connected

and runs at 1200 rpm? What must be the speed at which it is to be driven

produce the same emf if it is wave-wound?


60×

𝑃

𝐴=

0.05 × 500 × 1200

60×

8

8= 500𝑣𝑜𝑙𝑡

With lap-wound, P=A=8

With wave-wound, P=8, A=2


60×

𝑃

𝐴=

0.05 × 500 × 𝑁

60×

8

2= 500 → 𝑁 = 300𝑟𝑝𝑚



Example: A four pole lap-connected armature of a DC shunt generator is

required to supply the loads connected in parallel:

(a) 5kW Geyser at 250 V and (b) 2.5kW lighting load also at 250V.

The generator has an armature resistance 0.2Ω and a field resistance of 250Ω.

The armature has 120 conductors in the slots and runs at 1000 rpm. Allowing

1V per brush for contact drops, find

(1) Flux per pole, (2) armature current per parallel path


60×

𝑃

𝐴=

𝜙 × 120 × 1000

60×

4

4= 258.2𝑣𝑜𝑙𝑡 → 𝜙 = 129.1𝑚𝑊𝑏

With lap-wound, P=A=4

𝐼𝐿 =5000 + 2500

250= 30𝐴

𝐼𝐹 =250

250= 1𝐴

𝐼𝐴 = 𝐼𝐿 + 𝐼𝐹 = 30 + 1 = 31𝐴

𝐸𝐴 = 𝑉𝑇 + 𝐼𝐴𝑅𝐴 + 𝑏𝑟𝑢𝑠ℎ𝑒𝑠 𝑑𝑟𝑜𝑝 = 250 + 31 × 0.2 + 2 × 1 = 258.2V

EA

-

+

RA=0.2ΩIA IL

-

+

LF

IF

VT =

250V

RF=250Ω

5k

W G

eyser

2.5

kW

ligh

ting

(2)

(1)

Armature current per parallel path=31/4=7.75A



Example: A separately excited DC generator, when running at 1000 rpm

supplied 200A at 125V. What will be the load current when the speed drops to

800 rpm if IF is unchanged? Given that the armature resistance 0.04Ω and

brush drop 2V.

𝐸𝐴2 = 𝑉𝑇2 + 𝐼𝐴2𝑅𝐴 + 𝑏𝑟𝑢𝑠ℎ𝑒𝑠 𝑑𝑟𝑜𝑝

𝐸𝐴1 = 𝑉𝑇1 + 𝐼𝐴𝑅𝐴 + 𝑏𝑟𝑢𝑠ℎ𝑒𝑠 𝑑𝑟𝑜𝑝 = 125 + 200 × 0.04 + 2 = 135V

𝑁𝐴1 = 1000𝑟𝑝𝑚

𝐸𝐴2 = 𝐸𝐴1

𝑁𝐴2

𝑁𝐴1= 135

800

1000= 108𝑉

𝑅𝑙𝑜𝑎𝑑 =125

200= 0.625Ω

108 = 𝐼𝐴2 × 0.625 + 𝐼𝐴2 × 0.04 + 2

𝑉𝑇2 = 𝐼𝐴2𝑅𝑙𝑜𝑎𝑑

𝐼𝐴2 =108 − 2

0.625 + 0.04= 159.4A

𝑉𝑇2 = 𝐼𝐴2𝑅𝑙𝑜𝑎𝑑 = 159.4 × 0.625 = 99.6𝑉

EA2

-

+ RA

IL=159.4A

-

+RF

LF

-

+

VF

IF

VT

2 =99.6

V

EA1

-

+ RA

IL=200A

-

+RF

LF

-

+

VF

IF

VT

1 =125V


Total Loss in a DC Generator

(A) Copper Losses

(i) Armature copper losses=Ia2Ra

This loss is about 30-40% of full load losses.

(ii) Field copper loss:

In case of shunt generator, field copper losses=IF2RF

In case of shunt generator, field copper losses=IL2Rs


(iii) The loss due to brush contact resistance.

(B) Magnetic (Iron or Core) Losses

(i) Hysteresis Loss, 𝑾𝒉 ∝ 𝑩𝒎𝒂𝒙𝟏.𝟔𝒇

(ii) Eddy Current Loss, 𝑾𝒆 ∝ 𝑩𝒎𝒂𝒙𝟐𝒇𝟐

These losses are practically constant for shunt and compound wound

generators, because in their case, field current is approximately constant.


(C) Mechanical Losses

(i) Friction Loss at bearing and commutator. (ii) Air Friction or Windage Loss of rotating armature



Total Loss in a DC Generator

Tota

l L

oss

es

Copper Losses

Armature Cu Loss

Shunt Cu Loss

Series Cu Loss

Iron Losses

Hysteresis Loss

Eddy Current Loss

Mechanical Losses

Friction Loss

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).


Power Stages and Efficiency

Mechanical Efficiency

𝜂𝑚 =𝑇𝑜𝑡𝑎𝑙 𝑤𝑎𝑡𝑡𝑠 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑖𝑛 𝑎𝑟𝑚𝑎𝑡𝑢𝑟𝑒

𝑀𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑝𝑜𝑤𝑒𝑟 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑× 100% =

𝐸𝐴𝐼𝐴

𝑂𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑑𝑟𝑖𝑣𝑖𝑛𝑔 𝑒𝑛𝑔𝑖𝑛𝑒× 100%

Electrical Efficiency

𝜂𝑒 =𝑊𝑎𝑡𝑡𝑠 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 𝑖𝑛 𝑙𝑜𝑎𝑑 𝑐𝑖𝑟𝑢𝑖𝑡

𝑇𝑜𝑡𝑎𝑙 𝑤𝑎𝑡𝑡𝑠 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑖𝑛 𝑎𝑟𝑚𝑎𝑡𝑢𝑟𝑒× 100% =

𝑉𝐼𝐿

𝐸𝐴𝐼𝐴× 100%

Overall or Commercial Efficiency

𝜂𝑐 = 𝜂𝑚 × 𝜂𝑒 =𝑊𝑎𝑡𝑡𝑠 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 𝑖𝑛 𝑙𝑜𝑎𝑑 𝑐𝑖𝑟𝑢𝑖𝑡

𝑀𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑝𝑜𝑤𝑒𝑟 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑× 100% =

𝑉𝐼𝐿

𝑂𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑑𝑟𝑖𝑣𝑖𝑛𝑔 𝑒𝑛𝑔𝑖𝑛𝑒× 100%



Example: A shunt generator delivers 195A at terminal voltage of 250V. The

armature resistance and shunt field resistance are 0.02Ω and 50Ω respectively.

The iron and friction losses equal 950W. Find

(a) emf generated (b) Cu losses (c) output of the prime motor

(d) commercial, mechanical and electrical efficiencies.

(a) 𝐼𝑓 =250

50= 5𝐴

𝐼𝐴 = 𝐼𝐹 + 𝐼𝐿 = 5 + 195 = 200𝐴

𝐸𝐴 = 𝑉𝑇 + 𝐼𝐴𝑅𝐴 = 250 + 200 × 0.02 = 254V

(b) 𝐴𝑟𝑚𝑎𝑡𝑢𝑟𝑒 𝐶𝑢 𝑙𝑜𝑠𝑠 = 𝐼𝐴2𝑅𝐴 = 2002 × 0.02 = 800𝑊

𝑆ℎ𝑢𝑛𝑡 𝐶𝑢 𝑙𝑜𝑠𝑠 = 𝐼𝑓2𝑅𝑓 = 52 × 50 = 1250𝑊

𝑇𝑜𝑡𝑎𝑙 𝐶𝑢 𝑙𝑜𝑠𝑠 = 800 + 1250 = 2050𝑊

(c) Stray losses=950W

Total losses=950+2050=3000W

𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑜𝑢𝑡𝑝𝑢𝑡 = 𝑉𝐼𝐿 = 250 × 195 = 48750𝑊

𝑂𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑟𝑖𝑚𝑒 𝑚𝑜𝑡𝑜𝑟 = 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑖𝑛𝑝𝑢𝑡



𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑖𝑛𝑝𝑢𝑡 = 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑜𝑢𝑡𝑝𝑢𝑡 + 𝑡𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠𝑒𝑠 = 48750 + 3000 = 51750𝑊

𝑂𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑟𝑖𝑚𝑒 𝑚𝑜𝑡𝑜𝑟 = 51750𝑊

𝜂𝑚 =𝐸𝐴𝐼𝐴

𝑂𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑑𝑟𝑖𝑣𝑖𝑛𝑔 𝑒𝑛𝑔𝑖𝑛𝑒× 100% =

50800

51750× 100% = 98.2%

𝜂𝑒 =𝑉𝐼𝐿

𝐸𝐴𝐼𝐴=

48750

50800× 100% = 95.9%

𝜂𝑐 =𝑉𝐼𝐿

𝑂𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑑𝑟𝑖𝑣𝑖𝑛𝑔 𝑒𝑛𝑔𝑖𝑛𝑒× 100% =

48750

51750× 100% = 94.2%

(c) 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑝𝑜𝑤𝑒𝑟(𝐸𝐴𝐼𝐴) = 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑖𝑛𝑝𝑢𝑡 − 𝑠𝑡𝑟𝑎𝑦 𝑙𝑜𝑠𝑠

𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑝𝑜𝑤𝑒𝑟(𝐸𝐴𝐼𝐴) = 51750 − 950 = 50800𝑊



Example: A shunt generator has a full load current of 196 A at 220V. The stray

lassos are 720W and the shunt field coil resistance is 55Ω. If it has full load

efficiency of 88%, find the armature resistance.

𝑆ℎ𝑢𝑛𝑡 𝐶𝑢 𝑙𝑜𝑠𝑠 = 𝐼𝑓𝑉 = 4 × 220 = 880𝑊

Constant losses=Shunt Cu losses+stray losses=880+720=1600W

𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑜𝑢𝑡𝑝𝑢𝑡 = 𝑉𝐼𝐿 = 220 × 196 = 43120𝑊


𝐸𝐴𝐼𝐴× 100% = 88% → 𝐸𝐴𝐼𝐴 = 43120 ÷ 0.88 = 49000𝑊

𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠𝑒𝑠 = 49000 − 43120 = 5880𝑊

𝐼𝑓 = 220 ÷ 55 = 4𝐴

𝐼𝐴 = 𝐼𝐿 + 𝐼𝑓 = 195 + 4 = 199𝐴

Total losses=Armature losses + Constant losses=𝐼𝐴2𝑅𝐴+1600=5880

𝐼𝐴2𝑅𝐴 = 5880 − 1600 = 4280𝑊

𝑅𝐴 = 4280 ÷ 1992 = 0.108Ω


Voltage Regulation

Example: A 4-pole shunt DC generator is delivering 20A to a load of 10Ω. If

the armature resistance is 0.5 Ω and the shunt field resistance is 50 Ω,

calculate the induced emf and the efficiency of the machine. Allow a drop of

1V per brush.

𝑇𝑒𝑟𝑚𝑖𝑛𝑎𝑙 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 = 𝐼𝐿𝑅 = 20 × 10 = 200𝑉


𝐸𝐴𝐼𝐴× 100% =

200 × 20

214 × 24× 100% = 77.9%

𝐼𝑓 = 200 ÷ 50 = 4𝐴 𝐼𝐴 = 𝐼𝐿 + 𝐼𝑓 = 20 + 4 = 24𝐴

𝐼𝐴𝑅𝐴 = 24 × 0.5 = 12𝑉

𝐸𝐴 = 𝐼𝐴𝑅𝐴 + 𝑉 + 𝑏𝑟𝑢𝑠ℎ 𝑑𝑟𝑜𝑝 = 12 + 200 + 2 = 214𝑉

The voltage regulation (VR) is defined as the difference between the no-load

terminal voltage (VNL) to full load terminal voltage (VFL) and is expressed as

a percentage of full load terminal voltage. It is therefore can be expressed as,

𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑅𝑒𝑔𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑉𝑅 =𝑉𝑁𝐿 − 𝑉𝐹𝐿

𝑉𝐹𝐿× 100% =

𝐸𝐴 − 𝑉𝐹𝐿

𝑉𝐹𝐿× 100%

𝑉𝑅 =𝐸𝐴 − 𝑉𝐹𝐿

𝑉𝐹𝐿× 100% =

214 − 200

200× 100% = 7%


Uses of DC Generators

• Shunt generators with field regulators are used for ordinary lighting and power supply purposes. They are also used for charging batteries because their terminal voltages are almost constant.

Shunt Generators

• Series generators are used as boosters in a certain types of distribution systems particularly in railway service.

Series Generators

• The cumulatively compound generator is the most used DC generator because its external characteristics can be adjusted for compensating the voltage drop in the line resistance. Cumulatively compound generators are used for motor driving which require DC supply at constant voltage, for lamp loads and for heavy power service such as electric railways.

• The differential compound DC generator has an external characteristic similar to that of shunt generator but with large demagnetization armature reaction. Differential compound DC generators re widely used in arc welding where larger voltage drop is desirable with increase in current.

Compound Generators

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Documents

Electric Machines I - Philadelphia University · 2018. 1. 22. · For Short Shunt Cumulatively Compound DC Generator 𝑇=𝐼𝐹 𝐹 𝐼𝐴=𝐼𝐹+𝐼𝐿 𝑇= 𝐴−𝐼𝐴