02-Parallel-Operation.pdf

02- Alternators

Alternators Parallel Operation of Alternators [3/21.1/p.471] Power plants whether in DC or AC station, will be generally found to have several smaller generators running in parallel rather than large single units capable of supplying the maximum peak load. These smaller units can be run singly or in various parallel combinations to suit actual load demand. Why are synchronous generators operated in parallel? There are several advantages to operate alternators in parallel: 1. Several generators can supply a bigger load than one machine by itself. 2. Having many generators increase the reliability of power system since the failure of any one of them does not cause a total power loss to the load. 3. Having many generators operating in parallel allows one or more of them to be removed for shutdown and preventive maintenance. 4. If only one generator is used and it is not operating near full load, then it will be relatively inefficient. But with several smaller machines it is possible to operate only a fraction of them. The ones that do operate are operating near full load and thus more efficiently. 5. Having many generators increase the flexibility of power system by selling to the industry or buying the power from the generation s tation of industry. 6. Connecting parallel the additional small unit future load demand can be fulfilled.

Requirements for Parallel Operation [3/21.2/p. 472] In order to operate properly, AC polyphase generators must first meet several requirements before they may be connected in parallel. 1. They must have the same voltage rating. 2. They must have the same frequency rating. The speeds need not necessarily be the same. 3. The waveform of the voltage should be the same, and therefore the generators should be of the same type, though their kVA ratings may be differ. 4. In order to divide load in proportion their ratings, the alternators should have prime movers whose speed-load speed-load characteristics are the same. The operation of connecting an alternator in parallel with another alternator or with common bus-bar is known as synchronizing. Generally, alternators are used in a power system where they are in parallel with many other alternators. This mode of operation takes place at power stations where many generators run in parallel. The generators are connected together via transformers and busbars and so a power network or grid is formed. It means that the alternator is connected to a live system of constant voltage and constant frequency. Often the electrical system, to which the alternator is connected, has already so many alternators and loads connected to it that no matter what power is delivered by the incoming alternator, the voltage and frequency of the system remain same. In that case, the alternator is said to be connected to infinite bus-bar or it is technically said that an alternator is synchronized with the infinite bus ( Infinite Infinite bus bars are those t hose whose frequency and the p.d.’s are not affected by changes in the conditions of any one machine connected in parallel to it. In other wards, they are constant-frequency, constant-frequency, constant-voltage constant-voltage bus bar.). It is never advisable to connect a stationary alternator to live bus-bar, because, stator induced emf being zero, a short-circuit will result.

Prepared by Dr. M. A. Mannan

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02- Alternators

Synchronizing Procedure [3/21.3/p. 472] One alternator is placed in parallel with one or more other alternators only when additional load requires it. Those alternators already carrying load are known as the running machines, while that which is to be placed in the system is known as the incoming machine. At the time of synchronizing, the following conditions must be met: 1. The effective voltage of the incoming generator must be exactly equal to that of the others or of the bus bars connecting them. 2. The individual phase voltages which are to be connected to each other must be in exact phase opposition. This is the same as saying that DC generators must be connected positive (+) to positive (+) and negative (–) to negative (–). 3. The phase rotation, or sequence of the running and incoming generators, must be the same. 4. The frequency should be the same, also it will be seen shortly that it is more desirable that the frequencies at the instant of paralleling be almost, but not quite, identical. The above conditions are all necessary in order to have no circulating currents between the generators at the time their terminals are connected together . In order to explain the above conditions, let us assume that if several generators are carrying load, they may be replaced by one equivalent running alternator. 1. If the effective voltage are unequal, it is obvious that there is then a resultant voltage between the two armatures, and hence a circulating current. Despite the fact that satisfying some of the other necessary conditions may allow the incoming generator to pick up some of the load, power may be lost within both generators because of this circulating current. 2. The armatures of the alternators are shown schematically in Fig. 21.1, and the triple pole single-throw switch is used to connect the like terminals together. If no current is to flow between the two armatures, then the resultant voltage in any closed path taken between the two generators should be zero. If path consisting of phases a and b of both generators is taken, and the starting point is point o running generator, the net voltage will be E R = E oa + E AO + E ob + E BO For proper paralleling, the phase rotation should be the same for both machines, or that shown in Fig. 21.2 . If the effective voltage values are same, it is seen that this figure that E oa and E AO add up to zero, and likewise E ob and E BO also add up to zero.


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02- Alternators

There is thus no resultant voltage in the path that has been chosen, and there is no circulating current. If any other path is chosen, the result is that same. If, however, the incoming machine had a phase rotation of ACB, its phasor diagram would appear as that in Fig. 21.3. It is obvious that E oa+ E AO would again add up to zero. Now, however, we see that when we add E OB and E bo, the phasor addition does not give us zero. Instead, as seen from Fig. 21.4, the resultant E R has a relatively large magnitude, and at this instant is equal to 1.73 times the phase voltage. This is actually equal to the line voltage, and there will be a circulating current. Again, if any other path is chosen, it will be seen that there is always a resultant voltage when the phase sequence is different, and hence there is circulating current between the two machines. 3. When the phase sequences are correct and the voltages are in exact phase opposition, there is no circulation current. Suppose, however, that the incoming machine voltages lag the running machine voltages by an angle α, as shown in Fig. 21.5. Now each of the corresponding phases will have a resultant voltage, as seen when E ao and EOA are adding in Fig. 21.6 . The same magnitude of resultant voltage will be present for each of the other phase, and so circulating current will exist.


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02- Alternators

4. It can be assumed that the running machine is operating at rated frequency. If, however, the incoming machine is rotating slightly faster than its rated speed, its frequency will be higher than that of the running machine. Effectively we may say that the voltage phasors of the running machine are stationary, while of the incoming one rotate around them at frequency which is equal to the difference between the two machines. Thus, at some instant, the voltages may be in exact phase opposition, but at all other times there will be a phase difference such that a resultant voltage will exact o as shown in Fig. 21.7 . At some other instant, two phase voltages will be 180 apart, so that the potential difference will be their arithmetic sum. If both frequencies are exactly same, there is a very small chance that exact phase o opposition will exist. This is so, since it is only at one instant of the entire 360 phasor rotation (one cycle) that the phase opposition can occur, and therefore it is more likely that a resultant voltage will exist when the frequencies are identical. For this reason, why synchronizing two alternators, it is more convenient to allow a very slightly difference in frequencies to exist, and then choose the proper moment to place the two machines in parallel.

The General Procedure for Paralleling Generators Step 1: The primemover of the incoming alternator is started, and the alternator is brought up to near its rated speed. Step 2: Using voltmeters, the field current of the incoming generator should be adjusted until its terminal voltage is equal to the line voltage of the running system. Step 3: The phase sequence of the incoming generator must be compared to the phase sequence of the running system. If all three lamps get bright and dark together, then the systems have the same phase sequence. If this is not the case, then it means that the phase sequences are not correct, and it is simply necessary to interchange two leads of the incoming machine. Step 4: Further adjustment of the incoming primemover is now necessary, until the lamps flicker at a very low rate, usually less than one dark period per second . Step 5: Final adjustment of the incoming voltage again made and synchronizing switch is thrown in the middle of a dark period .


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02- Alternators

Synchronizing Lamps [3/21.4/p. 476] Dark Lamp Method: A voltmeter can be used to satisfy the first condition ( The effective voltage of the incoming generator must be exactly equal to that of the others or of the bus bars connecting them) of paralleling. It is best that one meter measure the voltages of the two machines. This is done with a voltmeter switch used for this purpose, which actually is nothing more than the double-pole double-throw switch shown in Fig, 21.8 . Satisfaction of the other conditions of phase sequence, voltage opposition and frequency may be determined by the use of incandescent lamps connected between the two machines as shown in Fig. 21.8. At any instant it is seen that the voltage across the lamps is the sum of the individual phase voltages.

The prime mover of the incoming machine is started, and the alternator is brought out to near its rated speed. By adjusting field current, its terminal voltage is made the same as that of the running alternator. The lamps in the circuit will now flicker at a rate equal to the difference in the frequency of the alternator. If the phases are properly connected, all the lamps will be bright and dark at the same time. If this is not the case, then it means that the phase sequences are not correct, and it is simply necessary to interchange two leads of the incoming machine. Further adjustment of the incoming prime mover is now necessary, until the lamps flicker at a very low rate, usually less than one dark period per second . Final adjustment of the incoming voltage is again made and the synchronizing switch is thrown in the middle of a dark period . Advantages of this lamp method : 1. The equipment is inexpensive, and 2. The proper phase sequence is readily determined. Disadvantage of this lamp method :


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02- Alternators

1. Lamps go dark at some what less than half their rated voltage, and so the synchronizing switch may be closed when there is a considerable phase difference between the machines, with a high circulating current resulting in possible damage to the machines. 2. The lamp filaments might burn out. For this reason it is desirable to have the two frequencies nearly equal, so that synchronizing is not dependent on a constant dark period. 3. The flicker of the lamps does not indicate which machine has the higher frequency. If the rate of lamp flicker is increased due to the slightly increased the speed of the incoming generator, then the incoming alternator is above rated frequency. On the other hand, if the rate of lamp flicker is decreased due to the slightly increased the speed of the incoming generator, then the incoming alternator is above rated frequency. The above method of synchronization is known as the dark-lamp method .

Three-Bright Method (Bright Lamp Method): If the lamps are connected across the phases, that is a to B, b to C , and c to A, they will again flicker in unison. This time however, the proper synchronizing moment occurs when all three are in the middle of a bright period. This has the advantage of avoiding synchronizing when the lams may have burned out, but it is more difficult to estimate the middle of a bright period than the middle of a dark one. This method is known as the threebright method.

Two-Bright One-Dark method (Rotating lamp method): In this method only two of the lamps are cross connected. The connections are phase a to A, b to C , and c to B. this is known as two-bright one-dark method. It avoids the disadvantages of both previous lamp methods, but actually be more confusing to the eye.

Synchroscope [3/21.5/p. 478] An instrument which not only indicates the exact moment for synchronizing, but also tells whether the incoming machine is fast or slow, is the synchronscope as shown in Fig. 21.9. When the pointer moves, it indicates that the two frequencies are different. When it is stationary, the frequencies are the same. However, it is only when the pointer is in the vertically upright position that the voltages are in exact phase opposition, and this is the proper synchronizing moment. The synchroscope is a single-phase instrument having two coils, both on the stator, one connected across two lines of the running machine or bus bar and one connected across two lines of the incoming alternator. The synchroscope is thus actually a single-phase motor, setting up a rotating field revolving at a speed that is the difference between the frequencies of the two stator coils. Therefore, before it is first used, the proper phase sequence must be established by some


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02- Alternators

other method. Hence, one of the lamp methods is commonly used together with the synchroscope.

Synchronizing Current, I sy [1/37.32/p. 1459] When in exact synchronism, the two alternators have equal p.d.’s (potential differences) and are in exact phase opposition, so far as the local circuit is concerned. Hence, there is no current circulating round the local circuit. As shown in Fig. 35.82 (b) emf E 1 of machine No. 1 (running machine) is in exact phase opposition to the emf of machine No. 2 (incoming machine) i.e. E 2.

It should be clearly understood that the two emf’s are in opposition, so far as their local circuit is concerned but in the same direction with respect to the external circuit. Hence, there is no resultant voltage (assuming E 1= E 2 in magnitude) round the local circuit. But now suppose that due to change in speed of the governor of second (incoming) machine, E 2 falls back by a phase angle of α electrical degrees as shown in Fig. 35.82 (c) (though still E 1= E 2). Now, they have a resultant voltage E r which when acting on the local circuit, circulates a current known as synchronizing current. The value of this current is given by E I sy = r where, Z s is the synchronous impedance of the phase windings of both Z s machines (or of one machine only if it is connected to infinite bus-bar). The current I sy lags behind E r by an angle θ given by tan θ= X s/ Ra where X s is the combined synchronous reactance of the two machines and Ra is the combined resistance of o o their armature. Since Ra is negligibly small, θ is almost 90 . So I sy lags E r by 90 . It is seen that I sy is generating current with respect to machine No. 1 (running machine) and motoring current with respect to machine No. 2 (incoming machine) (remember when the current flows in the same direction as emf, then the alternator acts as a generator, and when it flows in the opposite direction, the machine acts as a motor ). Thus current I sy sets up a synchronizing torque, which tends to retard the generating machine and accelerate the motoring machine. Similarly, if E 2 tends to advance in phase [ Fig. 35.82(d)], then I sy, being generating current for machine No. 2, tends to retard it and being motoring current for machine No. 1 tend to accelerate it. Hence, any departure from synchronism results in the production of a synchronizing current I sy which set up synchronizing torque. This re-establishes synchronism between the two machines by retarding the leading machine and by accelerating the lagging machine.


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02- Alternators

This current I sy is superimposed on the load currents in the case the machines are loaded.

Synchronizing Power, P sy [1/37.33/p. 1460] Consider Fig. 35.82 (c), where machine No. 1 is generating and supplying the synchronizing power, Psy = E 1 I sy cos φ1 which is approximately equal to Psy = E 1 I sy since φ1 is small. Since φ1=(90o-θ), synchronizing power Psy = E 1 I sy cos φ1 = E 1 I sy cos(90° − θ) = E 1 I sy sin θ = E 1 I sy because

θ ≅ 90°

This power output from machine No. 1 goes to supply (a) power input to machine No. 2 (which is motoring), and (b) the Cu losses in the local armature circuit of the two machines. Power input to machine No. 2 is P2in = E 2 I sy cos φ 2 which is approximately equal to P2in = E 2 I sy since φ 2 is small. Psy

= P2in + Cu

Losses;

E 1 I sy

= E 2 I sy + Cu

Losses

Now, let E 1= E 2= E (say), then E r = 2 E cos[(180° − α ) / 2] = 2 E cos[90° − α / 2] = 2 E sin α / 2 = 2 E × α / 2 = α E Considering α is very small. E E α E Now, I sy = r = r = if Ra of both machines is negligible. 2 X s 2 X s Z s Synchronizing power (supplied by machine No. 1) is α E α E 2 sin θ = sin θ Psy = EI sy sin θ = E 2 X s 2 X s Psy

= EI sy = E

α E 2 X s

=

α E 2 2 X s

(Approximately)

Total synchronizing power of three-phases is α E 3α E 2 sin θ = sin θ P3 sy = 3Psy = 3 EI sy sin θ = 3 E 2 X s 2 X s P3 sy

≅

3α E 2 2 X s

(Approximately)

This is the value of synchronizing power when two alternators are connected in parallel and are on no-load.

Alternators Connected to Infinite Bus-bars [1/37.34/p. 1461] Now, consider the case of an alternator which is connected to infinite bus-bars. The expression for Psy given above is still applicable but with one important difference i.e. impedance (or reactance) of only that one alternator is considered (and not of two as done above). Hence, expression for synchronizing power in this case becomes E r = α E Now, I sy

=

E r Z s

=

E r X s

=

α E X s

if Ra of incoming machine is negligible.

α E α E 2 Synchronizing power is Psy = EI sy = E = (Approximately) X s X s


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02- Alternators

Now, S. C. (Short Circuit) current I SC =

∴ Psy = ∴ Psy =

α E 2 X s

α E 2 X s

= α E

E

Z s

=

E X s

= α EI SC (Approximately)

X s

= α E

E

E

X s

= α EI SC sin θ (More accurately)

Total synchronous power for three phases is:

∴ P3sy = 3Psy = 3α EI SC sin θ

∴ P3sy = 3α EI SC (Approximately)

Synchronizing Torque, T sy [1/37.35/p. 1461] Let T sy be the synchronizing torque per phase in Newton-meter (N-m) and N s is synchronous speed in rpm (120 f /P)

(a) When there are two alternators in parallel Psy

= T sy ×

2π N s 60

;

T sy

=

Psy

2π N s / 60

=

α E 2 / 2 X s 2π N s / 60

= 3T sy =

Total torque due to three phases: T 3 sy

N − m

3Psy 2π N s / 60

=

3α E 2 / 2 X s 2π N s / 60

N − m

(b) Alternator connected to infinite bus-bars Psy

= T sy ×

2π N s 60

;

T sy

=

Psy

=

2π N s / 60

Total torque due to three phases 3Psy 3α E 2 / X s = T 3 sy = 3T sy = 2π N s / 60 2π N s / 60

α E 2 / X s 2π N s / 60

N − m

N − m

Effect of Load on Synchronizing Power [1/37.36/p. 1462] 2

In this case, instead of Psy=α E /X s, the approximate value of synchronizing power would be Psy=α EV/X s, where V is bus-bar voltage, is α load angle or torque angle and E is the alternator induced emf per phase. The value of E =V + IZ s. As seen from Fig. 35.83 , for a lagging p.f., E = [V cos φ + IRa ]

2

+ [V sin φ + IX s ]2


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02- Alternators

Example 37.43 . Find the power angle when 1500 kVA, 6.6 kV, 3-phase, Y-connected alternator having a resistance of 0.4 ohm and a reactance of 6 ohm per phase delivers full-load current at normal rated voltage and 0.8 PF lag. Draw the phasor diagram. Solution: Given, VA= 1500×103; V L=66×103; Ra=0.4 ohm; X s=6 ohm; cos φ = 0.8

The angle α between V and E 0 is known as power angle, load angle or torque angle. 1500 × 10 3 = 131 A Full load current is, I = 3 × 66 × 10 3 Voltage/phase, V =

66 × 10 3 3

= 3810

V

φ = cos − (0.8) = 36.86° = 36°50' ∴ sin φ = 0.6 From the vector diagram, we obtain AB V sin φ + IX S 3810 × 0.6 + 131 × 6 = = tan(φ + α ) = OA V cos φ + IRa 3810 × 0.8 + 131 × 0.4 1

= 0.991

φ + α = tan −1 (0.991) = 44.741 α = tan − (0.991) = 44.741 − φ = 44.741 − 36.86 = 7.85 1

Parallel Operation of Two Alternators [1/37.38/p. 1463] Consider two alternators with identical speed/load characteristics connected in parallel as shown in Fig. 35.86 . The common terminal voltage V is given by V = E1 − I 1 Z1 = E 2 − I 2 Z 2

∴ E1 − E 2 = I 1 Z1 − I 2 Z 2 Also, I = I 1 + I 2 and V = IZ ∴ E1 = I 1 Z1 + IZ = I 1 Z1 + ( I 1 + I 2 ) Z = I 1 ( Z1 + Z ) + I 2 Z ∴ E 2 = I 2 Z 2 + IZ = I 2 Z 2 + ( I 1 + I 2 ) Z = I 2 ( Z 2 + Z ) + I 1 Z ( E1 − E 2 ) Z + E1 Z 2 ( E 2 − E1 ) Z + E 2 Z 1 ; I 2 = I 1 = Z 1 Z 2 + Z ( Z 1 + Z 2 ) Z 1 Z 2 + Z ( Z 1 + Z 2 ) E1 Z 2 + E 2 Z 1 E1 Z 2 + E 2 Z 1 ; I = V = IZ = ( Z 1 Z 2 / Z ) + Z1 + Z 2 Z 1 Z 2 + Z ( Z 1 + Z 2 ) E − V E − V ; I 1 = 1 I 2 = 2 Z 1

Z 2

The circulating current under no-load condition is, I C =

E1 − E 2 Z 1 + Z 2

Using Admittances The terminal voltage may also be expressed in terms of admittances as shown below: ∴ I = I 1 + I 2 = V / Z = VY V = IZ = ( I 1 + I 2 ) Z; I 1

=

E1

− V

Z 1

I = I 1 + I 2

= ( E1 − V )Y 1 ;

I 2

=

E 2

− V

Z 2

= ( E 2 − V )Y 2

= ( E1 − V )Y 1 + ( E 2 − V )Y 2


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02- Alternators

VY = I ; VY = ( E1 − V )Y 1 + ( E 2 − V )Y 2 VY = E1Y 1 − VY 1 + E 2Y 2 − VY 2 VY + VY 1 + VY 2 V =

= V (Y + Y 1 + Y 2 ) = E1Y 1 − VY 1 + E 2Y 2 − VY 2

E1Y 1 + E 2Y 2 Y + Y 1

+ Y 2

Using Parallel Generator Theorem

⎡ E − V E 2 − V ⎤ ⎡ E − V ⎤ ⎡ E − V ⎤ = ( I 1 + I 2 ) Z = ⎢ 1 + Z ; V = ⎢ 1 Z + ⎢ 2 ⎥ ⎥ ⎥ Z Z Z Z Z 2 ⎣ 1 ⎦ ⎣ 1 ⎦ ⎣ 2 ⎦ ⎡ E V ⎤ ⎡ E ⎡ E E ⎤ ⎡1 V ⎤ 1 ⎤ + ⎥ Z V = ⎢ 1 − Z + ⎢ 2 − Z ; V = ⎢ 1 + 2 ⎥ Z − V ⎢ ⎥ ⎥ ⎣ Z 1 Z1 ⎦ ⎣ Z 2 Z 2 ⎦ ⎣ Z1 Z 2 ⎦ ⎣ Z 2 Z 2 ⎦ ⎡1 1 1 ⎤ ⎡ E E ⎤ + ⎥ Z = ⎢ 1 + 2 ⎥ = I SC 1 + I SC 2 = I SC V ⎢ + ⎣ Z Z 2 Z 2 ⎦ ⎣ Z1 Z 2 ⎦ V = IZ

I SC 1

=

E1

;

I SC 2

=

E 2

Z1 Z 2 where, I SC 1 and I SC 2 are the short-circuit currents of the two alternators. 1 1 1 1 = + + If then V = Z 0 I SC Z 0 Z Z 2 Z 2 Example 37.44. A 3000 kVA, 6 pole alternator runs at 1000 rpm in parallel with other machine on 3300 V busbars. The synchronous reactance of both machines is 25% of the voltage per phase. Calculate the synchronizing power for one mechanical degree of displacement and the corresponding synchronous torque. Solution: It may please be noted that here the alternator is working in parallel with many alternators. Hence, it may be considered to be connected to infinite bus-bars. 3300 = 1905 V Voltage/phase, E = 3

Full load current, I =

kVA

3V L

=

3000 × 10 3 3 × 3300

= 525 A

Now, synchronous reactance drop is 25% of phase voltage. Thus IX s

T sy

=

0.25 × 1905

= 0.9075 Ω 525 Here, α = 1° (mech.); α = 1 × (6 / 2) = 3° (elec.) 3 × π π α = rad = 180 60 2 2 3α E 3 × π × 1905 Also, Psy = = = 628.4 kW 60 × 0.9075 × 1000 X s

X s

=

60 Psy 2π N s

= 9.55

Psy

N s

= 9.55

628.4 × 10 3


1000

= 6000 N − m

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= 0.25 ×1905

02- Alternators

Example 37.50. Two single-phase alternators operating in parallel have induced EMFs an o o open circuit of 230 ∠0 and 230∠10 volts and respective reactances of j2 oham and j3 ohm. Calculate (i) terminal voltage, (ii) currents, and (iii) power delivered by each of alternators to a load of impedance 6 ohm (resistive). o Solution: Given, Z 1= j2; Z 2= j3; Z =6; E 1=230∠0 =230+ j0 o and E 2=230∠10 230(0.985+ j0.174)=(226.5+ j39.9) ( E1 − E 2 ) Z + E1 Z 2 [230 + j 0 − (226.5 + j 39.9)]6 + (230 + j 0) j 3 (ii) = I 1 = Z 1 Z 2 + Z ( Z 1 + Z 2 ) j 2 j 3 + 6( j 2 + j 3)

= 14.3 − j 3.56 = 14.73∠ − 14° ( E 2 − E1 ) Z + E 2 Z1 [(226.5 + j 39.9) - (230 + j 0)]6 + (226.5 + j 39.9) j 2 = I 2 = j 2 j 3 + 6( j 2 + j 3) Z 1 Z 2 + Z ( Z 1 + Z 2 ) I 2 = 22.6 − j1.15 = 22.63∠ − 3.4° I = I 1 + I 2 = (14.3 − j 3.56) + ( 22.6 − j1.15) = 36.9 − j 4.71 = 37.2∠ − 7.3° V = IZ = (36.9 − j 4.71)6 = 221.4 − j 28.3 = 223.237.2∠ − 7.3° P1 = VI 1 cos φ 1 = 223.2 × 134.73 × cos14° = 3190 W P2 = VI 2 cos φ 2 = 223.2 × 22.63 × cos 3.4° = 5040 W I 1

(i) (iii)

Distribution of Load [1/37.40/1468] The amount of load taken up by an alternator running, in parallel with other machines, is solely determined by its driving torque i.e. by the power input to its prime mover (by giving it more or less steam, in the case of s team drive). Any alternation in its excitation merely changes its kVA output, but not its kW output. In other words, it merely changes the power factor at which the load is delivered.

(a) Effect of Change in Excitation Suppose the initial operating conditions of the two parallel alternators are identical i.e. each alternator supplies one half of the active load (kW) and one half of the reactive load (kVAR), the operating PF thus being equal to the load PF. In other words, both active and reactive powers are divided equally thereby giving equal apparent power (kVA) triangles for the two machines as shown in Fig. 37.92(b). As shown in Fig. 37.92(a), each alternator supplies a load current I so that total output current is 2 I . Now, let excitation of alternator No. 1 be increased, so that E 1 becomes greater than E 2. The difference between two emf’s set up a circulating current I C= I sy=( E 1- E 2)/2 Z s which is the original current distribution. As seen, I C is vectorially added to the load current of alternator No. 1 and subtracted from that of alternator No. 2.


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02- Alternators

The two machines now deliver load currents I 1 and I 2 at respective PF”ss of cos φ1 and cosφ2. These changes in the load currents lead to changes in PF’s, such that cos φ1 is reduced, whereas cosφ2 is increased. However, effect on the kW loading of the two alternators is negligible, but kVAR1 supplied by alternator No. 1 is increased, w hereas kVAR2 supplied by alternator No. 2 is correspondingly decreased as shown by triangles of F ig. 37.92(c). Now, suppose that excitations of the two alternators are kept the same but steam supply to alternator No. 1 is increased i.e. power input to its prime mover is increased. Since the speeds of the two machines are tied together by their synchronous band, machine No. 1 can not overrun machine No. 2. Alternatively, it utilizes its increased power input for carrying more load than No. 2.

Effect of Change in Steam Supply This can be made possible only when rotor No. 1 advances its angular position w.r.t. No. 2 as shown in Fig. 37.93(b) where E 1 is shown advanced ahead of E 2 by an angle α. Consequently, resultant voltage E r (or E sy) is produced which, acting on the local circuit, sets up a current I sy which lags by almost 90 o behind E r but is almost in phase with E 1 (so long as angle is α small). Hence, power phase of No. 1 is increased by an amount of E 1 I sy whereas that of No. 2 is decreased by the same amount (assuming total power demand to remain unchanged). Since I sy has no appreciable reactive (or quadrature) component, the increase in steam supply does not disturb the division of reactive powers, but it increases the active power of alternator No. 1 and decreases that of No. 2. Load division, when steam supply to alternator No. 1 is increased, is shown in Fig. 37.93(c). So, it is found that by increasing the input of prime mover, an alternator can be made to take a greater share of the load, though at a different PF.


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02- Alternators

The points worth remembering are: 1. The load taken up by an alternator directly depends on its deriving torque or in other words, upon the angular advanced of its rotor. 2. The excitation merely changes the PF at which the load is delivered without affecting the load so long as steam supply remains unchanged. 3. If input of the prime mover of an alternator is kept constant, but its excitation is changed the load so component of its output is changed, not kW. Example 37.51. Two identical 3-phase alternators work in parallel and supply a total load of 1500 kW at 11 kV at a PF of 0.867 lagging. Each machine supplies half the total power. The synchronous reactance of each is 50 ohm per phase and the resistance is 4 ohm per phase. The field excitation of the first machine is so adjusted that its armature current is 50 A lagging. Determine the armature current of the second alternator and the generated voltage of the first machine. 1500 × 10 3 = 90.8 A Solution: Load current at 0.867 PF lagging is I = 3 × × × × 3 11 10 0.867

cos φ = 0.867;

sin φ = 1 − cos 2 φ = 1 − (0.867) 2

= 0.498 Wattful component of the current I cos φ = 90.8 × 0.867 = 78.72 A Wattless component of the current I sin φ = 90.8 × 0.498 = 45.218 A Each alternator supplies half of each of the above two component when conditions are identical (Fig. 37.94). 90.8 Current supplied by each machine, I 1 = I 2 = = 45.4 A 2 Since steam supply of first machine is not changed, the working components of both machines would remain the same at I 1 cos φ = I 2 cos φ = 78.72 / 2 = 39.36 A . But the wattles or reactive components would be redivided due to change in excitation. The armature current of the first machine is changed from 45.2 A to 50 A.

∴ Wattless component of the first machine= 50 2 − 39.36 2 = 30.83 ∴ Wattless component of the second machine=45.2-30.83=14.37 A.

A

The new current diagram is swown in Fig. 37.95 (a) (i)

Armature current of the second machine, I 2

=

39.36

2

+ 14.37 2 = 41.9 A

(ii) Terminal voltage/phase=11,000/ √3=6350 V Considering the first machine, IR drop= 4×50=200 V; IX drop=50×50=2,500 V;

(a) Fig. 37.94


(b) Fig.37.95

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02- Alternators

cos φ 1 = 39.36 / 50 = 0.7872; sin φ 1 Then, as seen from Fig. 37.95(b) E =

(V cos φ 1 + IR) 2

E =

(6350 × 0.7872 + 200) 2

=

1 − 0.7872 2

= 0.616;

+ (V sin φ 1 + IX ) 2 + (6350 × 0.616 + 2500) 2 = 8254.41

V

Line voltage=√3×8254.41=14297.06 V

AT: ampere-turn; OC: open-circuit; SC: short-circuit; w.r.t.: with respect to;

FL: full-load; OCC: open-circuit current; SCC: short-circuit current; PD: potential differece

NL: no-load; PF: power factor; m/c: Machine;

References [1] B. L. Theraja, A. K. Theraja, “ A Textbook of ELECTRICAL TECHNOLOGY in SI Units , AC & DC Machines”, S. Chand & Company Ltd., ( Multicolour illustrative Volume II Edition). [2] A. F. Puchstein, T. C. Lloyd, A.G. Conrad, “ Alternating Current Machines”, © 1942, Asia Publishing House, Third Edition (Fully revised and corrected Edition 2006-07). [3] Jack Rosenblatt, M. Harold Friedman, “ Direct and Alternating Current Machinery”, nd Indian Edition ( 2 Edition), CBS Publishers & Distributors. th [4] A. E. Fitzgerald, Charles Kingsley, Jr. Stephen D. Umans, Electric Machinery, 5 Edition in SI units, ©1992 Metric Edition, McGraw Hill Book Company. [5] Irving L. Kosow, Electrical Machinery and Transformers, Second Edition, Prentice –Hall India Pvt. Limited.

Synchronizing Current: After a suitable synchronism, the current due to the any departure from synchronism is called synchronizing current I sy. Synchronizing Torque: The torque which is set up by synchronizing current is called synchronizing torque.

This re-establishes synchronism between the two machines by retarding the leading machine and by accelerating the lagging machine.

Dark Lamp Method To check the first condition ( The effective voltage of the incoming generator must be exactly equal to that of the others or of the bus bars connecting them) a voltmeter can be used as shown in Fig. 21.8. By using the voltmeter, we can measure the effective voltages of incoming machine and running machine. If the measured voltages are not same, by adjusting field current, its terminal voltage is made the same as that of the running alternator. Satisfaction of the other conditions of phase sequence, voltage opposition and frequency may be determined by the use of incandescent lamps connected between the two machines as shown in Fig. 21.8 .


Page 15 of 16

02- Alternators

If the second condition (the individual phase voltages which are to be connected to each other must be in exact phase opposition) is not met, the lamps are shown always bright. By changing the connection, the individual phase voltages can be made exact phase opposition. The lamps in the circuit will now flicker at a rate equal to the difference in the frequency of the alternator. If the phases are properly connected, all the lamps will be bright and dark at the same time. If this is not the case, then it means that the phase sequences are not correct, and it is simply necessary to interchange two leads of the incoming machine. Further adjustment of the incoming prime mover is now necessary, until the lamps flicker at a very low rate, usually less than one dark period per second . Final adjustment of the incoming voltage is again made and the synchronizing switch is thrown in the middle of a dark period .


Page 16 of 16

02-Parallel-Operation.pdf

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