ch. 3- d.c. generatorselearning.uokerbala.edu.iq/pluginfile.php/20807/mod...university of karbala dc...

University of Karbala DC Machine Electrical & Electronics Eng. Dep.

33

Ms.c. Haider M. Umran

CH. 3- D.C. GENERATORS

3.1. Generator Principle: An electrical generator is a machine which converts

mechanical energy (or power) into electrical energy (or

power).

The energy conversion is based on the principle of the

production of dynamically (or motion ally) induced

e.m.f.

Whenever a conductor cuts magnetic flux, dynamically

induced e.m.f. is produced in it according to Faraday’s

Laws of Electromagnetic Induction. This e.m.f. causes

a current to flow if the conductor circuit is closed.

Hence, two basic essential parts of an electrical

generator are (i) a magnetic field and (ii) a conductor or conductors which can so move as to

cut the flux.

3.2. Types of Generators

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

Generators may be divided into (a) separately-excited generators and (b) self-excited DC

generators.

(a) Separately-excited DC generator: The field winding are energized from an

independent external d.c. current source, as shown in Fig. 3.2.

(b) Self-excited DC Generators: The field windings are energized by the current produced by

the generators themselves. Due to residual magnetism, there is always present some flux in the

poles. When the armature is rotated, some e.m.f. and hence some induced current is produced

which is partly or fully passed through the field coils thereby strengthening the residual pole

flux.

There are three types of self-excited generators named according to the manner in which their

field coils (or windings) are connected to the armature.

Fig. 3.2 Separately-excited d.c generator.

F1

F2 A2

A1

Ea

Rf

If

Vt

Ra

IL

DC

Source

Ia = IL

+

-


34


(i) DC Shunt Generator: The field windings are connected across or in parallel

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

across them, as shown in Fig. 3.3.

Current and voltage relations can be expressed as:

Ia = IL + Ish

Ish =Vt

Rsh

The induced e.m.f is:

Ea= Vt + Ia. Ra + Vbrush

And Ea =Ф p N Z

60 a

In practice, Vbrush is neglected.

(ii) DC Series Generator: In this case, the field windings are joined in series with the

armature conductors, as shown in Fig. 3.4. As they carry full load current, they

consist of relatively few turns of thick wire or strips. Such generators are rarely

used except for special purposes i.e. as boosters etc.

Ea Vt

IL

Ia

Fig. 3.3: DC shunt generator.

F1

F2

A2

A1

Rsh

A1

A2

S2 S1

IL

Ia

Ise

Ea Vt

Fig. 3.4: DC series generator.


35


The relations between voltage and current can expressed as:

Ia = Ise= IL

Ise: Current through series winding.

E.m.f equation is:

Ea = Vt +Ia. Ra + Ia. Rse +Vbrush

Ea = Vt +Ia (Ra +Rse) +Vbrush

Where; Ea =Ф p N Z

60 a

(iii) DC Compound Generator: It is a combination of a few series and a few shunt

windings and can be either short-shunt or long-shunt, as shown in Fig. 3.5 and 3.6.

In a compound generator, the shunt field is stronger than the series field. When

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

compounded.

a) Long Shunt Compound Generator: -

Ia= Ise

And Ia= Ish+ IL

Ish =Vt

Rsh

Ea = Vt + Ia . Ra + Ia. Rse +Vbrush

Where; Rsh: Resistance of shunt field winding.

b) Short Shunt Compound Generator: -

Ia = Ise+ Ish

And Ise = IL

∴ Ia = IL+ Ish

∴ Ish =E− Ia.Ra

Rsh

Voltage equation is;

Ea = Vt+ Ia Ra +Ise Rse +Vbrush

Ise= IL

Ea = Vt + Ia Ra + IL Rse+ Vbrush

Neglecting Vbrush,

Ea = Vt + Ia Ra + IL Rse

Ea - Ia Ra = Vt + IL Rse

Fig. 3.5: Long shunt compound generator

Ea

Ish

Ia

Vt

Se.

Sh. A1

A2

Ise IL

E

Ish

Ia Vt

IL

Ise

Sh.

S

A1

A2

Fig. 3.6: Short shunt compound generators.


36


∴ Ish =Vt+ IL.Rse

Rsh

a) Cumulative and Differential Compound Generator:

When the field excitation is produced by a combination of shunt field winding and series field

winding as shown in Fig. 3.7, the shunt and series fields help each other, the compound

generator is termed cumulative compound.

ΦT = Φsh + Φse

Where: Φsh = Flux produced by shunt.

Φse = Flux produced by series.

When the shunt and series field oppose each other, then the generator is differential compound

as shown in Fig. 3.8. As a result, the terminal voltage falls drastically with increasing load.

ΦT = Φsh - Φse

Ex.: A DC shunt generator has shunt field winding resistance of 100 Ω. It is supplying a load

of 5 Kw at a voltage of 250 V. if its armature resistance is 0.22 Ω. Calculate the induced e.m.f

of generator.

Sol.: - Ia = IL + Ish

Ish = Vt / Rsh = 250 v / 100 Ω = 2.5 A

IL= PL / Vt

= 5×103 / 250 = 20 A

Ia = IL + Ish

= 20 + 2.5 = 22.5 A

Ea = Vt + Ra.Ia

= 250 + 0.22 × 22.5 = 254.95 V.

A1

A2

A3

A4

F1

F1

F2

F1

Fig.3.7: Cumulative compound.

Fig. 3.8: Differential compound

F1

F2 A2

A1

A3

A4

G Vt

Ish IL Ia

F1

F2


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IL

Ex.: A 4 pole, compound DC generator long-shunt type, supply’s 100 A at a terminal voltage

of 500 V. If armature resistance is 0.02Ω, series field resistance 0.04 Ω and shunt field

resistance 100Ω, find the generated EMF. Take drop per brush as 1 V, Neglect armature

reaction.

Sol:

Ish = Vt / Rsh = 500 / 100 = 5 A

Ia = IL + Ish = 100 + 5 = 105 A

Voltage drop on series field windings =105×0.04= 4.2V

Armature voltage drop = 105 × 0.02=2.1 volt

Drop at brushes = 2 × 1= 2 V

Now,

E.m.f = V+ Ia. Ra +series drop+ brush drop

= 500 + 2.1 + 4.2 + 2 = 508.3 V.

Ex.: A 20 kW compound generator works on full load with a terminal voltage of 250 V. The

armature, series and shunt windings have resistances of 0.05Ω, 0.025Ω and 100 Ω respectively.

Calculate the total E.M.F generated in the armature when the machine is connected as short

shunt.

Sol.: Load current =P / V =20000/ 250 = 80 A

Voltage drop in the series windings = 80 × 0.025 = 2V

Voltage across shunt winding =Vt+ Voltage drop in the series windings

= 250 + 2 = 252 V.

Ish = Vt +IL Rse /Rsh

= 250 +80 ×0.025 / 100 =2.52 A

Ia = IL+Ish

= 80 + 2.52 = 82.52A

Ia.Ra = 82.52 × 0.05 = 4.13V

The generated E.m.f = Vt+ Ia. Ra +Voltage drop in the series windings

= 250 + 4.13 + 2 = 256.13 V.


38


3.3. Characteristics of D.C. Generators:

Following are the three most important characteristics or curves of a d.c. generator:

1. No-load saturation Characteristic (Eo/If):

It is also known as Magnetic Characteristic or Open-circuit Characteristic (O.C.C.). It

shows the relation between the no-load generated e.m.f. in armature, Eo and the field or exciting

current If at a given fixed speed. It is just the magnetization curve for the material of the

electromagnets. Its shape is practically the same for all generators whether separately-excited

or self-excited.

2. Internal or Total Characteristic (Eo/Ia):

It gives the relation between the e.m.f. Eo actually induces in the armature (after allowing for

the demagnetizing effect of armature reaction) and the armature current Ia.

3. External Characteristic (Vt/IL): It is also referred to as performance characteristic or sometimes voltage-regulating curve.

It gives relation between that terminal voltage Vt and the load current IL. This curve lies below

the internal characteristic because it takes into account the voltage drop over the armature

circuit resistance. The values of Vt are obtained by subtracting Ia .Ra from corresponding values

of Eo. This characteristic is of great importance in judging the suitability of a generator for a

particular purpose.

3.2.1 Characteristics of Separately Excited Generator:-

1. No-load Saturation or Open Circuit Characteristic (Eo/If):

The arrangement for obtaining the necessary data to plot this curve is shown in Fig. 3.9. The

exciting or field current If is obtained from an external independent d.c. source. It can be

varied from zero upwards by a potentiometer and its value read by an ammeter A connected

in the field circuit as shown.

Now, the voltage equation of a d.c. generator is, 𝐄𝐨 =Ф 𝐩 𝐍 𝐙

𝟔𝟎 𝐚 volt

Fig. 3.9: Separately excited gen. with no-load.

A1

A2

A

F2

F1

D.c

supply Rheosta

t

V


39


When current If is increased from its initial small value, the flux Φ changed and hence induced

e.m.f. Eo increase directly along the poles are unsaturated. This is represented by the straight

portion OA in Fig. 3.10. But as the flux density increases, the poles become saturated, so a

greater increase in If is required to produce a given small increase in voltage than on the lower

part of the curve.

2. Load Saturation Curve (Vt/If):

The curve showing relation between the terminal voltage Vt and field current If when the

generator is loaded, is known as Load Saturation Curve.

The curve can be deduced from the no-load saturation curve provided the values of armature

reaction and armature resistance are known. While considering this curve, account is taken of

the demagnetizing effect of armature reaction and the voltage drop in armature which are

practically absent under no-load conditions.

Fig. 3.11: Load saturation curve.

The no-load saturation curve of Fig. 3.10 has been repeated in Fig. 3.11 on a base of field amp-

turns (and not current) and it is seen that at no-load, the field amp-turns required for rated no-

load voltage are given by Oa. Under load conditions, the voltage will decrease due to

E0

If

Increasing A Constant

Open circuit Characteristic

Saturation

O

Fig. 3.10: Magnetization ch.cs. for constant speed.


40


demagnetizing effect of armature reaction. This decrease can be made up by suitably increasing

the field amp-turns. Let ac represent the equivalent demagnetizing amp-turns per pole. Then,

it means that in order to generate the same e.m.f. on load as at no-load, the field amp-turns/pole

must be increased by an amount ac = bd. The point d lies on the curve LS which shows relation

between the voltage E generated under load conditions and the field amp-turns. The curve LS

is practically parallel to curve Ob. The terminal voltage V will be less than this generated

voltage E by an amount (Ia Ra) where Ra is the resistance of the armature circuit. From point

d, a vertical line de = Ia Ra is drawn. The point e lies on the full-load saturation curve for the

generator. Similarly, other points are obtained in the same manner and the full-load saturation

curve Mp is drawn. The right-angled triangle bde is known as drop reaction triangle. Load

saturation curve for half-load can be obtained by joining the mid-points of such lines as mn and

bd etc. In the case of self-excited generators, load saturation curves are obtained in a similar

way.

3. Internal and External Characteristics:

Let us consider a separately-excited generator giving its rated no-load voltage of Eo for a certain

constant field current. If there were no armature reaction and armature voltage drop, then this

voltage would have remained constant as shown in Fig. 3.11. By the dotted horizontal line I.

But when the generator is loaded, the voltage falls due to these two causes, thereby giving

slightly dropping characteristics.

If we subtract from Eo the values of voltage drops due to armature reaction for different loads,

then we get the value of E, the e.m.f. actually induced in the armature under load conditions.

Curve II is plotted in this way and is known as the internal characteristic. The straight line Oa

represents the Ia Ra drops corresponding to different armature currents. If we subtract from Eo

the armature drop Ia Ra, we get terminal voltage Vt. Curve III represents the external

characteristic and is obtained by subtracting ordinates the line Oa from those of curve II.

Fig. 3.11: Load characteristics curve.


41


3.2.2: Characteristics of Self- Excited DC Shunt Generator:-

1. No-load Curve Characteristic (Eo / If):

This curve shows the relation between the generated e.m.f. at no-load (Eo) and the field

current (If).

The field or exciting current If is varied by rheostat and its value read on the ammeter (A). The

generator is derived at constant speed by the prime mover and the generated e.m.f. on no-load

is measured by the voltmeter connected across the armature as shown Fig. 3.12. Due to residual

magnetism in the poles, some e.m.f. (OA) is generated even when If = 0. Hence, the curve

shown in Fig. 3.13; starts a little way up. The generated e.m.f. is directly proportional to the

exciting current. However, at high flux densities, where μ is small, iron path reluctance

becomes appreciable and straight relation between E and If after point B, saturation of poles

starts.

2. Load Characteristics of DC Shunt Generator:-

A. Internal Characteristic (E/Ia):

Ideally the induced e.m.f. is not dependent on the load current IL or armature current Ia. but

as load current increased, the armature current Ia increases to supply load demand. As Ia

increased armature flux increases. The effect of flux produced by armature on main flux

produced by the field winding is called armature reaction. Due to the armature reaction, main

flux distorted. Hence lesser flux gets linked with the armature conductor and this reduces the

induced e.m.f as shown Fig. 3.14.

Drop due to Armature

reaction

O

E

IL

Fig.3.14: Internal characteristics curve.

Eo

Q

O

B

Q

If

Fig.3.13. Load characteristics curve

A

Eo

G Vt

Ish IL Ia

V

G

Fig.3.12. DC shunt generator

cteristics curve

F1

F2

A


42


B. External Characteristic (Vt/IL):

For D.C generator (E = Vt - Ia Ra) neglected other drops. So as load current IL increases,

Ia increases. Thus will increase drop Ia Ra and terminal voltage Vt = E - Ia Ra decreases. The

value of armature resistance Ra is very small; the drop in terminal voltage as IL changes from

no load to full load is very small. Hence shunt generator is called constant voltage generator.

If the load resistance is reduced beyond point a break down point i.e. load IL is increased beyond

P then it increases momentarily. This is very large current and generator gets overloaded. Due

to a large current the armature reaction is high and drop Ia Ra decreases from P to Q, rather than

increasing. Thus on curve aqr, the voltage goes on reducing rapidly and point r becomes zero

as shown Fig. 3.15.

3.2.2.1 Critical Field Resistance in DC Shunt Generator:

Now, connect the field windings back to the armature and run the machine as a shunt

generator. Due to residual magnetism in the poles, some e.m.f. and hence current, would be

generated. This current while passing through the field coils will strengthen the magnetism of

the poles (provided field coils are properly connected as regards polarity). This will increase

the pole flux which will further increase the generated e.m.f. Increased e.m.f. means more

current which further increases the flux and so on. This mutual reinforcement of e.m.f. and flux

proceeds on till equilibrium is reached at some point like P (Fig. 3.16). The point lies on the

resistance line OA of the field winding. Let R be the resistance of the field winding. Line OA

is drawn such that its slope equals the field winding resistance i.e. every point on this curve is

such that volt/ampere = R. The voltage OL corresponding to point P represents the maximum

voltage to which the machine will build up with R as field resistance. OB represents smaller

resistance and the corresponding voltage OM is slightly greater than OL. If field resistance is

increased, then slope of the resistance line increased, and hence the maximum voltage to which

the generator will build up at a given speed, decreases. If R is increased so much that the

resistance line does not cut the O.C.C. at all (like OT), then obviously the machine will fail to

excite i.e. there will be no ‘build up’ of the voltage. If the resistance line just lies along the

slope, then with that value of field resistance, the machine will just excite. The value of the

Small drop Ia.Ra

O

Vt

IL

Fig.3.15: External characteristics curve.

Eo

a

P Q

q r

Break down

Drooping nature


43


resistance represented by the tangent to the curve, is known as critical resistance Rc for a given

speed.

Steps to Find Critical Resistance Rc:-

1. O.C.C. is plotted from the given data. As shown in Fig. 3.16.

2. To draw the line of field resistance Rf, consider an equation of line (y = m.x). Where

y = Eo, x = If and m= slope = Rf. one point of the line is (0, 0), second point (Eo, If) If

(If) is any value from the given data by the above eq. we determine the new value of Eo.

The second point on the line is (Eo, If) and draw the line passing through (0, 0) and

(Eo , If), OA line.

3. Draw a tangent line to (O.C.C) i.e. tan θ, is the critical resistance of the field resistance.

Slope of tangent line is 𝐑𝐜 =∆𝐄

∆𝐈𝐟=

𝐃𝐄

𝐂𝐃= 𝐭𝐚𝐧 𝛉.

We know that E α N. As speed decreased the induced e.m.f. decreases, we gate (O.C.C) below

the (O.C.C) just tangential to normal field resistance line. As shown in Fig.3.17.

Critical Speed (Nc) is the speed at which machine just excites for the given field circuit

resistance. 𝐄𝐨𝟏

𝐄𝐨𝟐=

𝐍𝟏

𝐍𝟐

∴ 𝐄𝐨𝟐 =𝐍𝟐

𝐍𝟏 . 𝐄𝐨𝟏

E.m.f

A

If

C

O

Critical Voltage EC

Normal max; excitation voltage

O.C.C at Normal speed B

Normal field resistance line

(Rsh)

(Rsh) increased, slope increases,

Max; generate voltage decreases.

(B)

Line of Rc

E

D

Fig.3.16: No-load saturation curve.


44


Steps to Find Critical Speed Nc Graphically at Different Speeds:-

1. Drawn O.C.C for the given speed N1.

2. Draw a line tangential to this O.C.C, OA.

3. Draw a line representing the given Rsh, OP.

4. Select any field current value, point R.

5. Draw vertical line from R to intersect OA at S and OP at T.

6. The critical speed Nc is, 𝐑𝐓

𝐑𝐒=

𝐍𝐂

𝐍𝟏

∴ 𝐍𝐂 =𝐑𝐓

𝐑𝐒 . 𝐍𝟏

Ex.: The magnetization curve of a d.c. shunt generator at 1500 r.p.m. is:

If (A): 0 0.4 0.8 1.2 1.6 2 2.4 2.8 3

Eo (V): 6 60 120 172.5 202.5 221 231 237 240

For this generator find (i) no load e.m.f. for a total shunt field resistance of 100 Ω (ii) the critical

field resistance at 1500 r.p.m. and (iii) the magnetization curve at 1200 r.p.m. and therefrom

the open-circuit voltage for a field resistance of 100 Ω.

(b) A long shunt, compound generator fitted with inter-poles is commutatively-compounded.

With the supply terminals unchanged, the machine is now run as compound motor. Is the motor

differentially or cumulatively compounded?

Sol.:

The magnetization curve at 1500 r.p.m. is plotted in Fig. from the given data. The 100 Ω

resistance line OA is obtained by joining the origin (0, 0) with the point (1A, 100 V).

The voltage corresponding to point A is 227.5 V. Hence, no-load voltage to which the generator

will build-up is 227.5 V. The tangent OT represents the critical resistance at 1500 r.p.m.

Considering point B, Rc = 225/1.5 = 150 Ω.

T

R

S

A P Eo

O.C.C at Nc

O.C.C

at N1

Line for given

Rsh at Nc

Fig.3.17 Determine Critical speeds.


45


For 1200 r.p.m., the induced voltages for different field currents would be (1200/1500) = 0.8

of those for 1500 r.p.m. The values of these voltages are tabulated below:

If (A): 0 0.4 0.8 1.2 1.6 2 2.4 2.8 3

Eo (V): 4.8 48 96 138 162 176.8 184.8 189.6 192

The new magnetization curve is also plotted in Fig. 28.9. The 100 Ω line cuts the curve at

point C which corresponds to an induced voltage of 166 V.

Ex.: The O.C.C of a separately excited DC generator driven at 1000 r.p.m. is as follows:

Field current: 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6

𝑬. 𝑴. 𝑭. 𝒗𝒐𝒍𝒕𝒔: 30 55 75 90 100 110 115 120

If the machine is connected as shunt generator and driven at 1000 r.p.m. and has a field

resistance of 100 Ω, find (a) open circuit voltage and exciting current (b) the critical resistance

and (c) resistance to induce 115 volts on open circuit.

Sol.: The O.C.C. has been plotted in Fig. below. The shunt resistance line OA is drawn as usual.

Draw a line for Rsh = 100 Ω, the slope of the line is 100.

Eo = Rsh × If, if exciting current (If = 1A) from the table; Eo = 100 V.

The slope of tangent line is critical resistance Rc:


46


Slope of tangent line (Rc) =CD

FG=

30

0.2= 150Ω.

C. Line OB represents shunt resistance for getting 115 V on open circuit. From the table

the field current If = 1.4A at E=115.

∴ Field resistance 𝑅𝑓 =𝐸

𝐼𝑓=

115

1.4= 82.1 Ω.

Ex.: The magnetization characteristic for a 4-pole, 110V and 1000 r.p.m. shunt generator as

follows:

𝐼𝑓: 0 0.5 1 1.5 2 2.5 3 𝐴

𝐸𝑜: 5 50 85 102 112 116 120 𝑉

Armature is lap connected with 144 conductors; field resistance is 45 Ω. Determine

i. Voltage the machine will build up at no-load.

ii. The critical resistance.

iii. The speed at which the machine just excite.

iv. Residual flux per pole.

Sol.: The O.C.C. as shown in Fig., OA represents the 45Ω line which is drawn according to eq.

Eo = Rsh × If.

Drawing a horizontal line from 60V on Y-axis and vertical line to If = 1.1A on X-axis, intersect

at point C.

i. The voltage to which machine will build up = OM = 118 V.

ii. OT is tangent to the initial part of the O.C.C. It represents critical resistance.

Slope of tangent line (Rc) =NC

HR=

30

0.3= 100 Ω

iii. From any point on OT, at point B, drop the perpendicular BD on X-axis.

C

F G If

Eo

D


47


CD

BD=

NC

N1 or

50

110=

NC

1000 ∴ Nc = 454.54 𝑟. 𝑝. 𝑚

iv. As given in the table, induced e.m.f. due to

residual flux (i.e. when there is no exciting current)

is 5 V.

Eo =Z N ϕ P

60 a , 5 =

144×𝛷×1000×4

60 ×4

∴ Φ = 2.08 𝑚𝑊𝑏.

3.2.2.2 Voltage Build-up of a DC Shunt Generator:

Before loading a shunt generator, it is allowed to build up its voltage. Usually, there is always

present some residual magnetism in the poles, hence a small e.m.f. is produced initially.

This e.m.f. circulates a small current in the field circuit which increases the pole flux (provided

field circuit is properly connected to armature, otherwise this current may wipe off the residual

magnetism).

When flux is increased, generated e.m.f. is increased which further increases the flux and so

on. As shown in Fig. 28.17, Oa is the induced e.m.f. due to residual magnetism which appears

across the field circuit and causes a field current Ob to flow. This current aids residual flux and

hence produces, a larger induced e.m.f. Oc. In turn, this increased e.m.f. Oc causes an even

larger current Od which creates more flux for a still larger e.m.f. and so on.

Now, the generated e.m.f. in the armature has (a) to supply the ohmic drop If Rsh in the winding

and (b) to overcome the opposing self-induced e.m.f. in the field coil i.e. L. (d If /dt) because

field coils have appreciable self-inductance.

Ea = If . Rsh + L.d If

dt V.

If (and so long as), the generated e.m.f. is in excess of the ohm drop If Rsh, energy would

continue being stored in the pole fields. For example, as shown in Fig. 3.18, corresponding to

field current OA, the generated e.m.f. is AC. Out of this, AB goes to supply ohm drop If Rsh

and BC goes to overcome self-induced e.m.f. in the coil. Corresponding to If = OF, whole of

the generated e.m.f. is used to overcome the ohm drop.

None is left to overcome L. dIf /dt. Hence no energy is stored in the pole fields. Consequently,

there is no further increase in pole flux and the generated e.m.f. With the given shunt field

resistance represented by line OP, the maximum voltage to which the machine will build up is

OE. If resistance is decreased, it will built up to a somewhat higher voltage. OR represents the

Eo

If

S

N

R

C

H


48


resistance known as critical resistance. If shunt field resistance is greater than this value, the

generator will fail to excite.

Fig. 3.18: Voltage build-up of a DC shunt generator.

Conditions Required for Voltage Build-up of DC Shunt Generator:

1. There must be some residual magnetism in the generator poles.

2. For the given direction of rotation, the shunt field coils should be correctly connected to the

armature.

3. If excited on open circuit no-load, its shunt field resistance should be less than the critical

resistance (which can be found from its O.C.C.).

4. The rotational speed of generator should be greater than the critical speed. This is given by

internal characteristic.

3.2.3. Load Characteristics of DC Series Generator:-

In this type of generator, Ise = Ia = IL

As load current IL is increases, Ise increases. The flux Φ is directly proportional to Ise. So flux

increases. The induced e.m.f. E is proportional to flux hence induced e.m.f. also increases.

The internal characteristics (E/Ise) are of increasing nature.

Ea

If

T


49


As Ia is increases, armature reaction increases but its effect is negligible compared to increase

in E. For high load current, saturation occurs and flux remains constant. In such case, due to

armature reaction E starts decreasing as shown in the Fig. 3.19.

As Ia = IL increases, the drop in armature and field winding increases Ia (Ra +Rse), Where

Vt= E- Ia (Ra +Rse), thus the external c/s; as shown under the internal c/s due to drop Ia (Ra +Rse).

If self-excited series generator, O.C.C. cannot be obtained. The O.C.C. can be obtained in

separately -excited the field winding.

3.2.4. Load Characteristic of DC Compound Generator:-

In a compound generator, both series and shunt excitation are combined; we shall discuss

the characteristics of cumulatively compounded generator. It may be noted that external

characteristics of long and short shunt compound generators are almost identical, as shown in

Fig. (3.4-5). Compound generator can be classified in to:-

a. Under-compounded generator:-

In cumulatively compounded, ΦT = Φsh +Φse. As load current IL is increases, Ia increases hence

Ise also increases and flux increasing more. The induced e.m.f. increases and terminal voltage

also increases. But voltage drop and armature reaction drop also increases, hence there is drop

in the terminal voltage. As shown in Fig. (3.20)

b. Over-compound generator:-

If series winding turns are so adjusted that with the increase in load current the terminal voltage

increases. In such a case, as the load current increases, the series field e.m.f. increases and tends

to increase the flux and hence the generated voltage. The increase in generated voltage is

greater than the Ia.Ra drop so that instead of decreasing, the terminal voltage increases. As

shown in Fig. (3.20).

V

Ise = Ia = IL

O.C.C. characteristics

Internal c/s

External c/s Voltage

due to

residual flux

Armature

reaction

effect Ia (Ra +Rse)

Fig.3.19: Characteristics of DC series generator.


50


c. Level Compound generator (flat).

If series winding turns are so adjusted that with the increase in load current, the terminal voltage

substantially remains constant. The series winding of such a machine has lesser number of turns

than the one in over-compounded machine and, therefore, does not increase the flux as much

for a given load current. Consequently, the full-load voltage is nearly equal to the no-load

voltage. As shown in Fig. (3.20).

Fig.3.20: Compound generator C/S.

IL

VT

Eo

Differentially compounded

Full load

Flat or level Under

Over


51


3.3 Generator Power Losses:- In DC generators, as in most electrical devices, certain forces act to decrease the

efficiency. These forces, as they affect the armature, are considered as losses and may be

defined as follows:

1. Copper loss or I2.R in the winding

2. Iron loss or core Losses.

4. Mechanical Losses.

5- Brush contact losses.

1. Copper Losses: These losses are taking place due to the current flow in a winding. There

are various copper losses can be given by:

Armature copper loss = Ia2.Ra

Shunt field copper loss = I2sh.Rsh

Series field copper loss = I2se.Rse

The brushes contact resistances usually included in copper losses, copper losses are about 30

to 40% of full-load losses.

2. Iron or Core Losses: These losses are also called magnetic losses. Its include hysteresis loss

and eddy current loss in the core. Booths eddy currents and hysteresis losses total up to about

20 to 30% of full-load, as it explained in Ch.1

a. Mechanical Losses: These losses consist of:-

i. Mechanical friction losses

ii. Wind age, Air-friction or wind resists at the shaft.

Not. The Magnetic and Mechanical losses together called stray losses. For shunt and

compound DC generators where field current is constant, loss is constant; field copper losses

are also constant. Stray losses along with constant field copper losses are called constant losses

(Wc). While the armature current is dependent on the load, thus armature copper losses are

called variable losses (Pcu).

∴ Total losses = Constant losses + Variable losses.

Total Losses

Copper Losses

Iron

Losses

Mechanical

Losses

Brush Contact Losses

Armature Cu Loss

Shunt Cu Loss Series Cu Loss

Hysteresis

Eddy Current

Friction Wind age


52


3.4 Power Stages energy transformation in D.C Generator:- Various power stages in the case of a DC generator are shown below:

Armature Cupper loss = Ia2 . Ra

Note: (power developed in armature). Pg = Eg. Ia Or Pg = Pin − (Pmech. + Pin)

In the case of shunt generators: Field Cupper Loss =Ish2 . Rsh

In the case of series generator: Field Cupper Loss= Ise2 . Rse

3.5 Efficiency of D.C Generator: - The efficiencies for DC generator are divided to:

1. Mechanical Efficiency: ηm =B

A=

Eg.Ia

Pin× 100%

2. Electrical Efficiency: ηele. =C

B=

VL. IL

Eg.Ia× 100%

3. Total Efficiency: η =Pout

Pin= ηm. ηe =

C

A=

VL. IL

Pin× 100%

Or η =Pout

Pout+ PLosses=

Pin−PLosses

Pin× 100%

Where: Pin = Pout + Pcu + (Pi + Pmech) 3.5.1 Condition for Maximum Efficiency: The condition for maximum efficiency of DC generator is given by,

Variable loss = Constant loss.

Ia2 . Ra = W𝑐

Generator output = V. I

Generator input = P output + P losses

Pi = V. I + Ia2 . Ra + Wc = V. I + (I + Ish)2. Ra + Wc (∵ Ia = I + Ish)

However, if Ish is negligible as compared to load current, then Ia = I

∴ η =Pout

Pin × 100% =

V.I

V.I+(I )2.Ra+Wc × 100% (∵ Ia = I)

Mechanical

Input Power

Pin

Iron & fiction loss

Pi & P mech.

Armature

Power

Pg = Eg. Ia

Cupper loss

Pcu

Output

Power Pout

= It.Vt

A B C


53


η =1

1 + [I . Ra

V +

WcV. I

]× 100%

The load current corresponding to maximum efficiency is given by:

IL = √W𝑐

Ra

Ex.: A DC shunt generator delivers 195 A at terminal p.d. of 250 V. The armature resistance

and shunt field resistance are 0.02 Ω and 50 Ω respectively. The iron and friction losses (Stray)

equal 950 W. Find

(a) E.M.F. generated (b) Pcu (c) output of the prime mover (d) mechanical and electrical

efficiencies (e) total efficiency.

Sol.: (a) E. m. f = V + Ia. Ra

Ia = IL + Ish

Ish = Vt

Rsh=

250

50= 5 A.

∴ Ia = 195 + 5 = 200A.

∴ E. m. f = 250 + (200 × 0.02) = 254 V.

(b) PcuTotal loss = Pcu Arm. + Pcu Shu.

ArmatureCu loss = Ia2. Ra = (200)2 × 0.02 = 800 W

ShuntCu loss = Ish2 . Rsh = 52 × 50 = 1250 W

∴PcuTotal loss = 800 + 1250 = 2050 W.

(c) Stray losses = 950 W.

Pin G. = Pout P.M = Pout + PTotal losses

PTotal losses = PcuTotal loss + Pstray losses

PTotal losses = 2050 + 950 = 3000 W.

Pout = Vt . IL = 250 × 195 = 48750 W.

∴ Pout P.M = 48750 + 3000 = 51750 W.

(d) ηmech. = Pg

Pin × 100% =

Eg.Ia

Pin100% =

254×200

51750=

50800

51750× 100%


54


∴ ηmech. = 98.2%

∴ ηelec. = Pout

Pg × 100% =

48750

50800100% = 95.9%

(e) η =Pout

Pin × 100% =

48750

51750100% = 94.2%

Ex.: A shunt DC generator has a F.L. current of 196 A at 220 V. The stray losses are 720 W

and the shunt field resistance is 55 Ω. If it has a F.L. efficiency of 88%, find the armature

resistance. Also, find the load current corresponding to maximum efficiency.

Sol.: Pout = V. I = 220 × 196 = 43.12 kW

Where: Total efficiency is; η = 88%

η =Pout

Pin × 100%

∴ Pin = Pout

η=

43.12

0.88= 49 kW

PTotal losses = Pin − Pout = 49 kW – 43.12 kW = 588 W

Ish = Vt

Rsh=

220

55= 4 𝐴.

∴ Ia = IL + Ish = 196 + 4 = 200 A

∴ ShuntCu loss = V . Ish = 220 × 4 = 880 W

Stray losses = 720 W

Constant losses = Shunt Cu loss + Stray losses = 880 + 720 = 1.6 kW

∴ ArmatureCu loss = Total losses − Constant losses

= 588 kW − 1.6 kW = 4.28 kW

∴ Ia2. Ra = 4.28 kW

2002 .Ra = 4.28 kW

∴ Ra =4.28 Kw

2002= 0.107 Ω

For maximum efficiency, (Constant losses = Variable losses)

∴ I2. Ra = Constant losses = 1.6 kW

∴ IL = √Constant losses

Ra= √

1600

0.107= 122.28 A

ch. 3- d.c. generatorselearning.uokerbala.edu.iq/pluginfile.php/20807/mod...university of karbala dc...

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