ALTERNATORS II

ALTERNATORS PART 2 PREPARED BY: JCM

Alternator On Load • As the load on an alternator is varied, its terminal voltage is also found to vary as in dc generators. This variation in terminal voltage is due to the following reasons: – Voltage drop due to armature resistance, Ra – Voltage drop due to armature leakage reactance, XL – Voltage drop due to armature reaction

Alternator On Load Armature Resistance – The armature resistance/phase causes a voltage drop/phase of IRa which is in phase with the armature current. However, this voltage drop is practically negligible. Armature Leakage Reactance – When current flows through the armature conductors, fluxes are set up which do not cross the air-gap but take different paths. Such fluxes are known as leakage fluxes.

Alternator On Load Armature Leakage Reactance (cont...) – The leakage flux is practically independent of saturation but is dependent on I and its phase angle with terminal voltage V. – This leakage flux sets up an emf of self inductance which is known as reactance emf which is ahead of I by 900. – Hence, armature winding is assumed to possess leakage reactance, XL (also known as Potier reactance) such that the voltage drop due to this equal IXL.

Alternator On Load Armature Reaction – As in dc generators, armature reaction is the effect of armature flux on the main field flux. – In the case of alternators, the power factor of the load has a considerable effect on the armature reaction. – In a 3-phase machine the combined mmf wave is sinusoidal which moves synchronously. This mmf wave is fixed relative to the poles, its amplitude is proportional to the load current, but its position depends on the pf of the load.

Alternator On Load Armature Reaction (cont...) – Unity Power Factor • The armature flux is cross magnetizing. It is 900 space degrees with respect to the poles. • The result is that the flux at the leading tips of the poles is reduced while it is increased at the trailing tips. However, these two effects nearly offset each other leaving the average field strength constant. • Armature reaction for unity power factor is distortional.

Alternator On Load

Alternator On Load Armature Reaction (cont...) – Zero Power factor Lagging • The armature flux whose wave has moved backward by 900 is in direct opposition to the main flux. Hence, the main flux is decreased. • The armature reaction in this case is wholly demagnetizing, with the result that due to the weakening of the main flux, less emf is generated. • To keep the value of the generated emf the same, field excitation will have to be increased to compensate for this weakening.

Alternator On Load

Alternator On Load Armature Reaction (cont...) – Zero Power factor Leading • Armature flux wave has moved forward by 900 so that it is in phase with the main flux wave. This results in added main flux. • In this case, armature reaction is wholly magnetising, which results in greater emf. • To keep the value of generated emf the same, field excitation will have to be reduced somewhat.

Alternator On Load

Alternator On Load Armature Reaction (cont...) – Intermediate Power Factor • For lagging power factor, the effect is partly distortional and partly demagnetizing. • For leading power factor, the effect is partly distortional and partly magnetizing.

Synchronous Reactance • From the above discussion, it is clear that for the same field excitation, terminal voltage is decreased or increased from its no-load value Eg to V. This is because of: – Drop due to armature resistance, IRa – Drop due to leakage reactance, IXL – Drop due armature reaction

Synchronous Reactance • The drop in voltage due to armature reaction may be accounted for by assuming the presence of a fictitious reactance Xar in the armature winding. The value of Xar is such that I Xar represents the voltage drop due to armature reaction. • The leakage reactance XL and the armature reactance may be combined to give synchronous reactance XS.

Synchronous Reactance

X S = X L + X ar • Therefore, the total voltage drop in an alternator under load is: Vdrop = IR a + jIX S = I (Ra + jX S ) = IZ S where : Z S is called the synchronous impedance

Vector Diagrams of Loaded Alternator

Vector Diagrams of Loaded Alternator Example: A 3-phase, star-connected alternator supplies a load of 10 MW at 0.85 pf lagging at 11 kV. Its resistance is 0.1 ohm per phase and synchronous reactance of 0.66 ohm. Calculate the line value of emf generated.

Vector Diagrams of Loaded Alternator Solution:

10 MW 0.85 p. f . lagging 11 kV (line - line)

Vector Diagrams of Loaded Alternator 11 ∠00 = 6.3509 ∠00 kV 3  10 MW 0 31 . 788 ∠ −   * S  0.85  IL = IA = = 3VL*LL 3 11∠00 kV

VLφ =

( )(

)

= 617.487∠ − 31.7880 A.

(

)

EGφ = 617.487∠ − 31.7880 A (0.1 + j 0.66Ω ) + 6.3509∠00 = 6.6255∠2.71540 kV . EGLL =

( 3 )(6.6255kV ) = 11.4757 kV

kV

Vector Diagrams of Loaded Alternator Alternative Solution:

I L RA = (617.487 )(0.1) = 61.7487 V

I L X S = (617.487 )(0.66 ) = 407.5414 V EGφ =

(6.3509 kV + 61.7487 cos 31.788

= 6.6255 kV EGLL =

( 3 )(6.6255kV ) = 11.4757 kV

0

)

+ 407.5414 cos 58.212 0 + (− 61.7487 sin 31.788 + 407.5414 sin 58.212 ) 2

2

Voltage Regulation • It is clear that with change in load, there is a change in terminal voltage of an alternator. The magnitude of this change depends not only on the load but also on the load power factor. • The voltage regulation of an alternator is defined as “the rise in voltage when full-load is removed (field excitation and speed remaining the same) divided by the rated terminal voltage.”

Voltage Regulation V'L − VFL %VR = × 100% VFL Note: • VNL – VFL is the arithmetical differenece and not the vectorial one. • In the case of leading load pf terminal voltage will fall on removing the full-load. Hence, regulation is negative in that case. • The rise in voltage when full-load is thrown off is not the same as the fall in voltage when full-load is applied.

Determination of Voltage Regulation • In the case of small machines, the regulation may be found by direct loading. The procedure is as follows: – The alternator is driven at synchronous speed and the terminal voltage is adjusted to its rated value. – The load is varied until the wattmeter and ammeter (connected for the purpose) indicate the rated values at desired pf.

Determination of Voltage Regulation – Then the entire load is thrown off while the speed and excitation are kept constant. – The open-circuit or no-load voltage is read. – The regulation can be found from:

V'L − VFL %VR = ×100% VFL

Determination of Voltage Regulation • In the case of large machines, the cost of finding the regulation by direct loading becomes prohibitive. Hence, other indirect methods are used. • It will be found that all these methods differ chiefly in the way the no-load voltage is found in each case.

Determination of Voltage Regulation • Indirect Methods – Synchronous Impedance or EMF Method. It is due to Behn Eschenberg. – The Ampere-Turn or MMF Method. This method is due to Rothert. – Zero Power Factor or Potier Method • All these methods require : – Armature Resistance, RA – Open Circuit/No-Load Characteristics – Short-Circuit Characteristics (but zero power factor lagging characteristic for Potier Method

Determination of Voltage Regulation • Value of RA – Armature resistance per phase can be measured directly by voltmeter-ammeter method or by using Wheatstone bridge. However, under working conditions, the effective value of RA is increased due to “skin effect”. The value of RA so obtained is increased by 60% or so to allow for this effect. – Generally, a value 1.6 times the dc value is taken.

Determination of Voltage Regulation • Open-Circuit (OC) Characteristic – As in dc machines, this is plotted by running the machine on no-load and by noting the values of the induced voltage and field excitation current.

• Short-Circuit (SC) Characteristic – It is obtained by short-circuiting the armature windings through a low resistance ammeter. The excitation is adjusted as to give 1.5 to 2 times the value of the full-load current. During this test, the speed which is not necessarily synchronous is kept constant.

Determination of Voltage Regulation Example: The effective resistance of a 2200 V, 50 Hz, 440 kV, 1-phase alternator is 0.5 ohm. On short circuit, a field current of 40 A gives the full-load current of 200 A. The emf on open circuits with the same field excitation is 1160 V. Calculate the synchronous impedance and reactance.

Determination of Voltage Regulation Solution : VOC 1160 ZS = = = 5.8Ω I SC 200 X S = Z − R = 5.8 − 0.5 = 5.78Ω 2 S

2

2

2

Determination of Voltage Regulation Example: A 100-kVA, 3000-V, 50-Hz, 3-phase, star-connected alternator has effective armature resistance of 0.2 ohm. The fieldcurrent of 40 A produces a short-circuit current of 200 A and an open-circuit emf of 1040 V (line value). Calculate the generated line emf at rated load and 0.8 pf lagging.

Determination of Voltage Regulation Solution : Z Sφ =

VOCφ I SC

1040 = 3 = 3.00Ω 200 2

X Sφ = Z S2φ − Rφ = 32 − 0.2 2 = 2.99Ω  3000  0 EG φ = 19.245∠ − 36.87 0 (0.2 + j 2.99) +   ∠0  3  = 1770.196∠1.4150 V EGLL =

( 3 )(1770.196) = 3066.068 V

Synchronous Impedance Method • Following procedural steps are involved in this method: – OCC is plotted from the given data. – SCC is plotted from the given data by the shortcircuit test. Both these curves are drawn on a common field current base. – XS is obtained from

XS = Z − R 2 S

2

Synchronous Impedance Method • Knowing RA and XS, a vector diagram can drawn for any load and any power factor.

Synchronous Impedance Method • This method is not accurate because the value of ZS so found is always more than its value under normal voltage conditions and saturation. Hence, the value of regulation so obtained is always more than that found from an actual test. That is why it is called a pessimistic method. • The value of ZS usually taken is that obtained from full-load in the short-circuit test.

Synchronous Impedance Method • Here, the XAR has not been treated separately but along with leakage reactance XL. Example: The following test results are obtained from a 3-phase, 6000-kVA, 6.6kV, star connected, 2-pole, 50 Hz turbo-alternator: With a field current of 125 A, the open-circuit voltage is 8kV at the rated speed; with the same field current and rated speed, the shortcircuit current is 800 A. At rated full-load, the resistance drop is 3%. Find the regulation of the alternator on full-load and at a pf of 0.8 lagging.

Synchronous Impedance Method Solution : 8000 Z Sφ =

800

3 = 5.774Ω

 6.6kV  I L RAφ = (0.03)  = 114.315V  3  114.315 RAφ = = 0.218Ω 524.864 X Sφ = 5.774 2 − 0.2182 = 5.770Ω

Synchronous Impedance Method Solution (cont...) :

(

)

EGφ = 524.864∠ − 36.8690 (0.218 + j 5.77 ) +

6.6kV ∠00 3

= 6184.66∠22.3730 V 6184.66 − 3464.102 %VR = ×100% = 78.536% 3464.102

Synchronous Impedance Method Example: A 3-phase 50-Hz star-connected 2000 kVA, 2300 V alternator gives a short-circuit current of 600 A for a certain field excitation. With the same excitation, the open circuit voltage was 900 V. The resistance between a pair of terminals was 0.12 Ω. Find full-load regulation at 0.8 pf leading.

Synchronous Impedance Method Solution : 900 3 = 0.866Ω Z Sφ = 600  0.12  RAφ = 1.6  = 0.096Ω  2  X Sφ = 0.866 − 0.096 = 0.861Ω 2

2

Synchronous Impedance Method Solution (cont...) :

(

)

EGφ = 502.044∠36.8690 (0.096 + j 0.861) +

2.3kV ∠00 3

= 1168.811∠18.7 0 V 1168.811 − 1327.906 %VR = × 100% = −11.981% 1327.906

Synchronous Impedance Method Example: A 3-phase, star-connected alternator is rated at 1600 kVA, 13500 kV. The armature resistance and synchronous reactance are 1.5 Ω and 30 Ω respectively per phase. Calculate the % regulation for a load of 1280 kW at 0.8 leading power factor.

Synchronous Impedance Method Solution : 1280kW

0.8 ∠36.87 0 = 68.427∠36.87 0 A 3 (13500) 13500 0 0 EGφ = 68.427∠36.87 (1.5 + j 30 ) + ∠0 3 = 6859.624∠14.3820 V 13500 6859.624 − 3 ×100% = −11.99% %VR = 13500 3 IL =

( )

(

)

Synchronous Impedance Method Example: The effective resistance of a 1200-kVA, 3.3 kV, 50-Hz, 3-phase, Y-connected alternator is 0.25 Ω per phase. A field current of 35 A produces a current of 200 A on short-circuit and 1.1 kV on open circuit. Calculate the power angle and p.u change in magnitude of the terminal voltage when the full load of 1200 kVA at 0.8 pf lagging is thrown off. Draw the corresponding phasor diagram.

Synchronous Impedance Method Solution : 1100 3 = 3.175Ω Z Sφ = 200 X Sφ = 3.1752 − 0.252 = 3.165Ω 1200kVA IL = ∠ − 36.87 0 = 209.946∠ − 36.87 0 A. 3 (3300 )

( )

Synchronous Impedance Method Solution (cont...) :

(

)

EGφ = 209.946∠ − 36.87 0 (0.25 + j 3.165) + = 2398.644∠12.034 0 V 2398.644 − 1905.256 = 0.259 1905.256 δ = 12.0340

p.u =

3.3kV ∠00 3

Synchronous Impedance Method

MMF or Ampere-Turns Method (Rothert’s Method) • This method also utilizes OC and SC data, but is the converse of the Synchronous Impedance (EMF) method in the sense that the armature leakage reactance is treated as an additional armature reaction. • Therefore, it is assumed that the change in terminal voltage on load is due entirely to armature reaction (and due to the ohmic resistance drop which, in most cases, is negligible).

MMF or Ampere-Turns Method (Rothert’s Method) • In the MMF method, use is made of a vector diagram of magneto motive forces. The theory upon which this method is developed is based on the assumption that for every voltage vector of the alternator diagram there is a corresponding magneto motive force.

MMF or Ampere-Turns Method (Rothert’s Method)

MMF or Ampere-Turns Method (Rothert’s Method) • Calculation procedure for MMF method – Terminal voltage per phase (V) is used as the reference phasor. – Lay off the armature IARA drop in phase with the current, at the pf angle for which the regulation is desired. Determine the field current If ’ required to produce the voltage E1 using the OC curve. – It is assumed that on short-circuit all the excitation is opposed by the mmf of armature reaction and armature reactance.

MMF or Ampere-Turns Method (Rothert’s Method) • Calculation procedure for MMF method (cont...) – Hence A represents the mmf (or field current) required to produce rated current on short circuit. It is the field current required to overcome the IAXS drop and it is constructed opposite to the current IA . – The excitation If required to produce terminal voltage at no load is then the vector sum of If ’ and A. EO is obtained from the OC curve.

MMF or Ampere-Turns Method (Rothert’s Method) • Calculation procedure for MMF method (cont...)

MMF or Ampere-Turns Method (Rothert’s Method) Example: A 3.5-MVA, Y-connected alternator rated at 4160 V at 50-Hz has the OC characteristic given by the following data: Field Current, A

50

100

150

200

250

300

EMF, V

1620

3150

4160

4750 5130 5370

350

400

450

5550

5650

5750

A field current of 200 A is found necessary to circulate full-load on short-circuit of the alternator. Calculate the full-load regulation at 0.8 pf lagging.

MMF or Ampere-Turns Method (Rothert’s Method) Solution: • Neglect RA • It is seen from the given data that for normal voltage of 4160 V, the field current needed is 150 A. • The field current necessary to circulate FL current is 200 A.

MMF or Ampere-Turns Method (Rothert’s Method) Solution (cont...):

MMF or Ampere-Turns Method (Rothert’s Method) Solution (cont...): θ = cos − 0.8 = 36.870 If =

(200) + (150) 2

2

− 2(200 )(150 )cos(90 + 36.87 )

= 313.85 A.

The generated emf EO corresponding to this excitation as found from OCC if drawn is 5440 V.

MMF or Ampere-Turns Method (Rothert’s Method) Solution (cont...):

5440 − 4160 %VR = × 100% 4160 = 30.78%

MMF or Ampere-Turns Method (Rothert’s Method) Example: The open-circuit and short-circuit test readings for a 3-phase, star-connected, 1000-kVA, 2000 V, 50-Hz, synchronous generator are: Field Current, A

10

20

25

30

40

50

OC terminal V

800

1500

1760

2000 2350 2600

SC current A

____

200

250

300

____ ____

The armature effective resistance is 0.2 ohm per phase. Estimate the full-load %VR at 0.8 pf leading.

MMF or Ampere-Turns Method (Rothert’s Method) Solution: The OCC and SCC curves are plotted:

MMF or Ampere-Turns Method (Rothert’s Method) Solution (cont...): 2000 = 1154.70V 3 1000kVA Full - load current = = 288.7 A 3 (2000 ) Full - load phase voltage =

( )

E=

(1154.7 + 288.7 × 0.2 × 0.8) + (288.7 × 0.2 × 0.6) 2

2

= 1201.4V

MMF or Ampere-Turns Method (Rothert’s Method) Solution (cont...): • From OCC curve, the field excitation necessary to produce E is 32 A. • From SCC curve, the field excitation necessary to produce full-load current is 29 A.

MMF or Ampere-Turns Method (Rothert’s Method) Solution (cont...): φ = 54.780 If =

(32) + (29) 2

2

− 2(32 )(29)cos(54.78)

= 28.19 A.

The generated emf EO corresponding to this excitation as found from OCC if drawn is 1080 V per phase or 1870.61 V line-to-line.

MMF or Ampere-Turns Method (Rothert’s Method) Solution (cont...):

1870.61 − 2000 %VR = × 100% 2000 = −6.47%

Zero Power Factor Method (Potier Method) • This method is based on the separation of armature-leakage reactance drop and the armature reaction effects. Hence, it gives more accurate results. • The experimental data required is: – No-load curve – Full-load zero pf curve (not the SCC) also called wattless load characteristic. It is the curve of terminal voltage against excitation when armature is delivering FL current at zero pf.

Zero Power Factor Method (Potier Method) • The zero pf curve can be obtained: – If a similar machine is available which may be driven at no-load as a synchronous motor at practically zero pf – By loading the alternator with pure reactors – By connecting the alternator to a 3-phase line with ammeters and wattmeters and so adjusting the field current that we get full-load current with zero wattmeter reading.

Zero Power Factor Method (Potier Method)

Zero Power Factor Method (Potier Method) • Procedural Steps for Potier Method – Suppose we are given V- the terminal voltage per phase. – We will be given or else we calculate armature leakage reactance XL and hence can calculate IXL. – Adding IXL (and IRA if given) vectorially to V, we get voltage E. – We will next find from OC curve, field excitation for voltage E. Let this be If1.

Zero Power Factor Method (Potier Method) • Procedural Steps for Potier Method (cont...) – Further, field current If2 necessary for balancing armature reaction is found from Potier triangle. – Combine If1 and If2 vectorially to get If. – Read from OC curve the emf corresponding to If. This gives us EO. Hence, regulation can be found.

Zero Power Factor Method (Potier Method) Example: A 3-phase , 6000-V alternator has the following OCC at normal speed: Field Current, A Terminal volts, V

14

18

23

4000

5000

6000

30

43

7000 8000

With armature short-circuited and FL current flowing the field current is 17 A and when the machine is supplying FL of 2000kVA at zero pf, the field current is 42.5 A and terminal voltage is 6000 V. Determine the field current required when the machine is supplying the full-load at 0.8 pf lagging.

Zero Power Factor Method (Potier Method) Solution:

Zero Power Factor Method (Potier Method) Solution (cont...): In the Potier Triangle BDH, line DE represents the leakage reactance drop (IXL) and is (by measurement) equal to 450 V.

E = 3464.102 + 4502 − 2(450)(3464.10 )cos126.87 0 = 3751.41 V. From OCC curve, it is found that the field amperes required for this voltage = 26.5 A. Field amperes required for balancing armature reaction = BE = 14.5 A (by measurement from Potier triangle BDH)

Zero Power Factor Method (Potier Method) Solution (cont...):

φ = 132.380 If =

(26.5) + (14.5) 2

= 37.82 A.

2

− 2(26.5)(14.5)cos(132.38)

Zero Power Factor Method (Potier Method) Example: A 600-kVA , 3300-V, 8-pole, 3-phase alternator has the following characteristics: AmpTurns per pole

4000

5000

7000

1000

Terminal volts, V

2850

3400

3850

4400

There are 200 conductors in series per phase. Find the full-load voltage regulation at 0.8 pf lagging having given that the inductive voltage drop at full load is 7% and that the equivalent armature reaction in amp-turns per pole = 1.06 X ampere-conductors per phase per pole.

Zero Power Factor Method (Potier Method) Solution: - OC terminal voltages are first converted into phase voltages and plotted against field amp-turns.

Zero Power Factor Method (Potier Method) 600kVA Full - load current = = 104.97 A 3 × 3300 Demagnetizing Amp - turns per pole per phase for full - load at zero pf

( 1.06 )(104.97 )(200 ) = = 2781.71 AT

8 3300 'ormal phase voltage = = 1905.26 V 3 Leakage reactance drop = (0.07 )(1905.26) = 133.4 V

Zero Power Factor Method (Potier Method) E = 1905.26 + 133.4 − 2(133.4)(1905.26)cos126.87 2

2

0

= 1988.17 V. From OCC curve, we find that 1988.17 V corresponds to 5100 AT.

Zero Power Factor Method (Potier Method) φ = 129.950 mmfO =

(5100) + (2781.71) 2

2

− 2(5100 )(2781.71)cos(129.95)

= 7208.82 AT. From OCC curve, it is found that this corresponds to an OC voltage of 2242 V per phase.

2242 − 1905.26 %VR = × 100% 1905.26 = 17.67%

Zero Power Factor Method (Potier Method) Example: The following figures give the open-circuit and full-load zero pf saturation curves for a 15000 kVA, 11000 V, 3-phase, 50-Hz star-connected turbo-alternator: Field Ampturns in 103

10

18

24

30

40

45

50

OC terminal, kV

4.9

8.4

10.1

11.5

12.8

13.3

13.65

Zero pf fullload line kV

____

0

____

____

____

10.2

____

Find the armature reaction, the armature reactance and the synchronous reactance. Deduce the regulation for full-load at 0.8 pf lagging.

Zero Power Factor Method (Potier Method) Solution: - First OCC is drawn between phase voltages and field ampere-turns. - Full-load zero pf curve can be drawn because two points are known, i.e., A(18,0) and C(45, 5890). Other points on this curve by transferring the Potier triangle.

Zero Power Factor Method (Potier Method)

Zero Power Factor Method (Potier Method) Solution (cont...): – In the Potier Triangle CDE, line EF = GH represents the leakage reactance drop (IXL) and is (by measurement) equal to 640 V. – Field ampere-turns required for balancing armature reaction = CF = 15700 AT (by measurement from Potier triangle CDE). – Short-circuit ampere-turns required = OA = 18000 AT.

Zero Power Factor Method (Potier Method) 15000kVA Full - load current = = 787.3 A. 3 (11000) 640 XL = = 0.813 Ω 787.3 8400 3 = 6.16 Ω Z Sφ = 787.3 As RA is negligible X Sφ ≅ Z Sφ .

( )

Zero Power Factor Method (Potier Method E = 6350.85 + 640 − 2(640)(6350.85)cos126.87 2

2

0

= 6754.28 V. From OCC curve, we find that 6754.28 V corresponds to 30800 AT.

Zero Power Factor Method (Potier Method φ = 131.220 mmfO =

(30800) + (15700) 2

2

− 2(15700 )(30800)cos(131.22 )

= 42806.73 AT.

From OCC curve, it is found that this corresponds to an OC voltage of 7540 V per phase.

7540 − 6350.85 %VR = × 100% 6350.85 = 18.72%