Chapter Three DC Generators PDF

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LECTURE NOTE (3) DC GENERATORS 3. Introduction Although a far greater percentage of the electrical machines in service are a.c.machines, the d.c. machines are of considerable industrial importance. The principal advantage of the d.c. machine, particularly the d.c. motor, is that it provides a fine control of speed. Such an advantage is not claimed by any a.c. motor. However, d.c. generators are not as common as they used to be, because direct current, when required, is mainly obtained from an a.c. supply by the use of rectifiers. Nevertheless, an understanding of d.c. generator is important because it represents a logical introduction to the behaviour of d.c. motors. 3.1. Types of D.C. Generator and their Characteristics (a) Separately-excited generator A typical separately-excited generator circuit is shown in Fig. 3.0 When a load is connected across the armature terminals, a load current

I a will flow. The terminal voltage V will fall from its open-circuit e.m.f. E due to a

volt drop caused by current flowing through the armature resistance, shown as Ra .

Figure 3.0 Separately Excited Generator

I.e

terminal voltage, V = E − I a Ra Or generated e.m.f., E = V + I a Ra

3.0

Example 1 Determine the terminal voltage of a D.C generator which develops an e.m.f. of 200V and has an armature current of 30A on load. Assume the armature resistance is 0.30Ω. Solution V=E−

I a Ra

= 200 – (30)(0.3) = 200 – 9 = 191 volts. Example 2 A separately-excited generator develops a no-load e.m.f. of 150V at an armature speed of 20 rev/s and a flux per pole of 0.10Wb. Determine the generated e.m.f. when (a) the speed increases to 25 rev/s and the pole flux remains 1

unchanged, (b) the speed remains at 20 rev/s and the pole flux is decreased to 0.08 Wb, and (c) the speed increases to 24 rev/s and the pole flux is decreased to 0.07Wb. Solution (a) Generated e.m.f E α ɸn From which,

From which

E1 E2

=

ɸ1 N 1 ɸ2 N 2

=

150 E2

=

(0.10)(20) (0.1)(25)

(150 )(0.10 )( 25 ) (0.10)(20)

E2 =

= 187.5 volts

(b)

150 E3

=

From which, emf.,

(0.10 )( 20 ) (0.08 )( 20 ) E3 =

(150)(0.08 )(20 ) (0.10)(20)

= 120 volts

(c)

150 E4

=

From which, emf.,

(0.10)(20) (0.07)(24) E4 =

(150 )(0.07 )( 24 ) (0.10 )(20 )

= 126 volts Characteristics The two principal generator characteristics are the generated voltage/field current characteristics, called the opencircuit characteristic and the terminal voltage/load current characteristic, called the load characteristic. A typical separately-excited generator open-circuit characteristic is shown in Fig. 3.1(a) and a typical load characteristic is shown in Fig. 3.1(b)

Figure 3.1 Separately Excited Generator Characteristics A separately-excited generator is used only in special cases, such as when a wide variation in terminal p.d. is required, or when exact control of the field current is necessary. Its disadvantage lies in requiring a separate source of direct current. 2

(b) Shunt wound generator In a shunt wound generator the field winding is connected in parallel with the armature as shown in Fig. 3.2 The field winding has a relatively high resistance and therefore the current carried is only a fraction of the armature current.

Figure 3.1 Shunt Wound Generator. For the circuit shown in Fig. 3.1, terminal voltage, V = E -

I a Ra

or generated e.m.f., E = V + I a Ra

Ia =

If

+ I from Kirchhoff’s current law, where

I a = armature current,

I f = field current (= V / Rf ) and I = load current.

Example 3 A shunt generator supplies a 20kW load at 200V through cables of resistance, R = 100mΩ. If the field winding resistance,

Rf

= 50Ω and the armature resistance,

Ra

D 40mΩ, determine (a) the terminal voltage, and (b)

the e.m.f. generated in the armature.

Solution (a)

Load current, I =

20000 watts 200 volts

= 100A

Volt drop in the cables to the load = IR = (100) (100 x

−3

10

) = 10V.

Hence terminal voltage, V = 200 + 10 = 210 volts.

(b) Armature current

Ia =

If

+I

3

Field current, I f Hence I a

=

If

V Rf

=

210 50

=

= 4.2 A

+ I = 4.2 + 100 = 104.2A

Generated e.m.f E = V +

I a Ra

= 210 + (104.2)(40 x

10−3 )

= 210 + 4.168 = 214.17 volts Characteristics The generated e.m.f., E, is proportional to proportional to field current

If

ɸω , hence at constant speed, since ω = 2πn, E α Φ. Also the flux Φ is

until magnetic saturation of the iron circuit of the generator occurs. Hence the

open circuit characteristic is as shown in Fig. 22.9(a).

Figure 3.2 Shunt Wound Generator Characteristics. As the load current on a generator having constant field current and running at constant speed increases, the value of armature current increases, hence the armature volt drop,

I a Ra increases. The generated voltage E is larger

than the terminal voltage V and the voltage equation for the armature circuit is V = E -

I a Ra . Since E is

constant, V decreases with increasing load. The load characteristic is as shown in Fig. 3.2 (b). In practice, the fall in voltage is about 10 per cent between no-load and full-load for many d.c. shunt-wound generators. The shunt-wound generator is the type most used in practice, but the load current must be limited to a value that is well below the maximum value. This then avoids excessive variation of the terminal voltage. Typical applications are with battery charging and motor car generators. (c) Series-wound generator In the series-wound generator the field winding is connected in series with the armature as shown in Fig. 3.3.

Figure 3.3 Series Wound Generator. Characteristic 4

The load characteristic is the terminal voltage/ current characteristic. The generated e.m.f. E, is proportional to

ɸω

and at constant speed ω = 2πn is a constant. Thus E is proportional to Φ. For values of current below

magnetic saturation of the yoke, poles, air gaps and armature core, the flux Φ is proportional to the current, hence E α I. For values of current above those required for magnetic saturation, the generated e.m.f. is approximately constant. The values of field resistance and armature resistance in a series wound machine are small, hence the terminal voltage V is very nearly equal to E. A typical load characteristic for a series generator is shown in Fig. 3.4.

Figure 3.4 Series Wound Generator Characteristics. In a series-wound generator, the field winding is in series with the armature and it is not possible to have a value of field current when the terminals are open circuited, thus it is not possible to obtain an open-circuit characteristic. Series-wound generators are rarely used in practice, but can be used as a ‘booster’ on D.C. transmission lines. (d) Compound-wound generator In the compound-wound generator two methods of connection are used, both having a mixture of shunt and series windings, designed to combine the advantages of each. Fig. 3.5 (a) shows what is termed a long-shunt compound generator, and Fig. 3.5 (b) shows a short-shunt compound generator. The latter is the most generally used form of d.c. generator.

Figure 3.5 Compound Wound Generator. Example 4 A short-shunt compound generator supplies 80A at 200 V. If the field resistance,

Rf

= 40Ω, the series

resistance,

Rse = 0.02Ω and the armature resistance, Ra = 0.04Ω, determine the e.m.f. generated.

5

Solution Volt drop in series winding =

IR se = (80)(0.02) = 1.6V.

P.d. across the field winding = p.d. across armature = V 1 = 200 + 1.6 = 201.6V Field current

If

=

Armature current, I a

V1

=

Rf

=I+

Generated e.m.f., E =

201.6 40

= 5.04 A

I f = 80 + 5.04 = 85.04A V1 +

I a Ra

= 201.6 + (85.04)(0.04) = 201.6 + 3.4016 = 205 volts Characteristics In cumulative-compound machines the magnetic flux produced by the series and shunt fields are additive. Included in this group are over-compounded, level-compounded and under compounded machines – the degree of compounding obtained depending on the number of turns of wire on the series winding. A large number of series winding turns results in an over-compounded characteristic, as shown in Fig. 3.6, in which the full-load terminal voltage exceeds the no-load voltage. A level-compound machine gives a full-load terminal voltage which is equal to the no-load voltage, as shown in Fig. 3.6

Figure 3.6 Compound Wound Generator. An under-compounded machine gives a full-load terminal voltage which is less than the no-load voltage, as shown in Fig. 3.6. However even this latter characteristic is a little better than that for a shunt generator alone. Compoundwound generators are used in electric arc welding, with lighting sets and with marine equipment. 3.2 Voltage Regulation

6

The change in terminal voltage of a generator between full and no load (at constant speed) is called the voltage regulation, usually expressed as a percentage of the voltage at full-load. % Voltage Regulation = Where

V NL−V FL V FL

X 100

V NL = Terminal voltage of generator at no load.

V FL

= Terminal voltage of generator at full load.

Note that voltage regulation of a generator is determined with field circuit and speed held constant. If the voltage regulation of a generator is 10%, it means that terminal voltage increases 10% as the load is changed from full load to no load.

3.3 Parallel Operation of D.C. Generators In a d.c. power plant, power is usually supplied from several generators of small ratings connected in parallel instead of from one large generator. This is due to the following reasons: (i) Continuity of service If a single large generator is used in the power plant, then in case of its breakdown, the whole plant will be shut down. However, if power is supplied from a number of small units operating in parallel, then in case of failure of one unit, the continuity of supply can be maintained by other healthy units. (ii) Efficiency Generators run most efficiently when loaded to their rated capacity. Electric power costs less per kWh when the generator producing it is efficiently loaded. Therefore, when load demand on power plant decreases, one or more generators can be shut down and the remaining units can be efficiently loaded. (iii) Maintenance and repair Generators generally require routine-maintenance and repair. Therefore, if generators are operated in parallel, the routine or emergency operations can be performed by isolating the affected generator while load is being supplied by other units. This leads to both safety and economy. (iv) Increasing plant capacity In the modern world of increasing population, the use of electricity is continuously increasing. When added capacity is required, the new unit can be simply paralleled with the old units. (v) Non-availability of single large unit In many situations, a single unit of desired large capacity may not be available. In that case a number of smaller units can be operated in parallel to meet the load requirement. Generally a single large unit is more expensive. 3.4 Connecting Shunt Generators in Parallel The generators in a power plant are connected in parallel through bus-bars. The bus-bars are heavy thick copper bars and they act as +ve and -ve terminals. The positive terminals of the generators are .connected to the +ve side of bus-bars and negative terminals to the negative side of bus-bars. Fig. 3.7 shows shunt generator 1 connected to the bus-bars and supplying load. When the load on the power plant increases beyond the capacity of this generator,

7

the second shunt generator 2 is connected in parallel wish the first to meet the increased load demand. The procedure for paralleling generator 2 with generator 1 is as under: (i)

The prime mover of generator 2 is brought up to the rated speed. Now switch

S 4 in the field circuit

of the generator 2 is closed. (ii) Next circuit breaker CB-2 is closed and the excitation of generator 2 is adjusted till it generates voltage equal to the bus-bars voltage. This is indicated by voltmeter

V2 .

(iii) Now the generator 2 is ready to be paralleled with generator 1. The main switch

S 3 , is closed, thus

putting generator 2 in parallel with generator 1. Note that generator 2 is not supplying any load because its generated e.m.f. is equal to bus-bars voltage. The generator is said to be “floating” (i.e., not supplying any load) on the bus-bars. (iv) If generator 2 is to deliver any current, then its generated voltage E should be greater than the bus-bars voltage V. In that case, current supplied by it is I = (E - V)/ Ra

where

Ra is the resistance of the armature circuit. By

increasing the field current (and hence induced e.m.f. E), the generator 2 can be made to supply proper amount of load. (v) The load may be shifted from one shunt generator to another merely by adjusting the field excitation. Thus if generator 1 is to be shut down, the whole load can be shifted onto generator 2 provided it has the capacity to supply that load. In that case, reduce the current supplied by generator 1 to zero (This will be indicated by ammeter A1) open C.B.-1 and then open the main switch

S1 .

3.5. Load Sharing The load sharing between shunt generators in parallel can be easily regulated because of their drooping characteristics. The load may be shifted from one generator to another merely by adjusting the field excitation. The analysis of load sharing of two generators which have unequal no-load voltages. Let

E1 ,

E2 = no-load voltages of the two generators

R1 ,

R2 = their armature resistances

V = common terminal voltage (Bus-bars voltage) Then

I1 =

E 1−V R1

and

I2

=

E 2−V R2

Thus current output of the generators depends upon the values of

E1 and

E3 . These values may be changed

by field rheostats. The common terminal voltage (or bus-bars voltage) will depend upon (i) the e.m.f.s of individual generators and (ii) the total load current supplied. It is generally desired to keep the busbars voltage constant. This can be achieved by adjusting the field excitations of the generators operating in parallel. 3.6. Compound Generators in Parallel Under-compounded generators also operate satisfactorily in parallel but over compounded generators will not operate satisfactorily unless their series fields are paralleled. This is achieved by connecting two negative brushes together as shown in Fig. (3.6) (i). The conductor used to connect these brushes is generally called equalizer bar. Suppose that an attempt is made to operate the two generators in Fig. (3.6) (ii) in parallel without an equalizer bar. 8

If, for any reason, the current supplied by generator 1 increases slightly, the current in its series field will increase and raise the generated voltage. This will cause generator 1 to take more load. Since total load supplied to the system is constant, the current in generator 2 must decrease and as a result its series field is weakened. Since this effect is cumulative, the generator 1 will take the entire load and drive generator 2 as a motor. Under such conditions, the current in the two machines will be in the direction shown in Fig. (3.6) (ii). After machine 2 changes from a generator to a motor, the current in the shunt field will remain in the same direction, but the current in the armature and series field will reverse. Thus the magnetizing action, of the series field opposes that of the shunt field. As the current taken by the machine 2 increases, the demagnetizing action of series field becomes greater and the resultant field becomes weaker. The resultant field will finally become zero and at that time machine 2 will short circuit machine 1, opening the breaker of either or both machines.

Figure 3.6 Generators in Parallel. 3.7 Efficiency of a d.c. generator The efficiency of an electrical machine is the ratio of the output power to the input power and is usually expressed as a percentage. The Greek letter, ‘η’ (eta) is used to signify efficiency and since the units are, power/power, then efficiency has no units. Thus,

Efficiency, η =

power ( output input power )

x 100%

If the total resistance of the armature circuit (including brush contact resistance) is Ra , then the total loss in the armature circuit is

I2a Ra .

If the terminal voltage is V and the current in the shunt circuit is I f , then the loss in the shunt circuit is

I f V.

If the sum of the iron, friction and windage losses is C then the total losses is given by: 2

I a Ra

+

I f V + C ( I2a Ra

+

I f V is, in fact, the ‘copper).

9

2

If the output current is I, then the output power is VI. Total input power = VI + I a Ra

+

I f V + C.

Hence

output input

Efficiency, η =

η=

(

, i.e.

VI VI + I R a+ I f V + C 2 a

)

x 100%

3.1

The efficiency of a generator is a maximum when the load is such that: 2

I a Ra

= V If

+C

i.e. when the variable loss = the constant loss.

Example 1 A 10 kW shunt generator having an armature circuit resistance of 0.75Ω and a field resistance of 125Ω, generates a terminal voltage of 250V at full load. Determine the efficiency of the generator at full load, assuming the iron, friction and windage losses amount to 600 W.

Solution Output power = 10 000W = VI from which, load current I = 10 000/V = 10 000/250 = 40 A. Field current, I f

= V/ Rf

Armature current, I a

(

Efficiency, η =

=

= 250/125 = 2A.

If

+ I = 2 + 40 = 42A

VI VI +I R a+ I f V + C 2 a

=

(

=

( 10000 12423 )

)

x 100%

10000 10000+( 42 ) ( 0.75) +( 2 ) ( 250) +600 2

)

x 100

x 100%

= 80.50%

10...