Voltage Regulation of an Alternator

Voltage Regulation of an Alternator AU : Oct.-97, April-01, 04, May-05, 07,16, Dec.-06

Under the load condition, the terminal voltage of alternator is less than the induced e.m.f. E_ph. So if load is disconnected, V_ph will change from V_ph to E_ph, if flux and speed is maintained constant. This is because when load is disconnected, I_a is zero hence there are no voltage drops and no armature flux to cause armature reaction. This change in the terminal voltage is significant in defining the voltage regulation.

Key Point : The voltage regulation of an alternator is defined as the change in its terminal voltage when full load is removed, keeping field excitation and speed constant, divided by the rated terminal voltage.

So if V_ph = Rated terminal voltage

E_ph = No load induced e.m.f.

then voltage regulation is defined as,

% Regulation = E_ph – V_ph /  V_ph × 100

The value of the regulation not only depends on the load current but also on the power factor of the load. For lagging and unity p.f. conditions there is always drop in the terminal voltage hence regulation values are always positive. While for leading capacitive load conditions, the terminal voltage increases as load current increases. Hence regulation is negative in such cases. The relationship between load current and the terminal voltage is called load characteristics of an alternator. Such load characteristics for various load power factor conditions are shown in Fig. 2.10.1.

1. kVA Rating of an Alternator

The alternators are designed to supply a specific voltage to the various loads. This voltage is called its rated terminal voltage denoted as V_L. The power drawn by the load depends on its power factor. Hence instead of specifying rating of an alternator in watts, it is specified in terms of the maximum apparent power which it can supply to the load. In three phase circuits, the apparent power is √3 V_L I_L, measured in VA (volt amperes). This is generally expressed in kilo volt amperes and is called kVA rating of an alternator where I_L is the rated full load current which alternator can supply. So for a given rated voltage and kVA rating of an alternator, its full load rated current can be decided.

Consider 60 kVA, 11 kV, three phase alternator. 

In this case  kVA rating = 60

but kVA = √3 V_L I_L × 10^-3

... 10^-3 to express the product in kilo volt amperes

60 = √3 × 11 × 10³ × I_L × 10^-3

I_L  = 3.15

This is the rated full load current of an alternator. But load current is same as the armature current. So from kVA rating, it is possible to determine full load armature current of an alternator which is important in predicting the full load regulation of an alternator for various power factor conditions. Similarly if load condition is different than the full load, the corresponding armature current can be determined from its full load value.

Key Point : I_a at half load = 1/2 × I_a at full load. It reduces in the same proportion in which load condition reduces.

Hence regulation at any p.f. and at any load condition can be determined.

Example 2.10.1 A 600 kVA, 125 V alternator connected in delta is reconnected in star. Calculate its new rating in i) volts iii) amperes and iii) kVA.

Solution :

Example 2.10.2 A 3 phase, 8 pole, 50 Hz, star connected alternator has 96 slots with 4 conductors per slot. The coil pitch is 10 slots. If the flux per pole is 60 mWb find

 i) The phase voltage

ii) The line voltage

iii) If each phase is capable of carrying 650 A, what is the kVA rating of the machine ?

Solution :

Example 2.10.3  A three phase star connected alternator is rated at 1500 kVA, 12000 V. The armature effective resistance and symchronous reactance are 2 Ω and 35 Ω respectively per phase. Calculate the percentage regulation for a load of 1200 kW at power factor of 0.8 lagging.

Solution :

kVA = 1500, V_line = 1200 volts

Example 2.10.4 A 3 phase star connected alternator has an open circuit line voltage of 6599 V. The armature resistance and synchronous reactance are 0.6 Ω and 6 Ω per phase respectively. Find terminal voltage regulation and if load current is 180 A at a power of i) 0.9 lagging ii) 0.8 leading

Solution :

Example 2.10.5 In a 50-kVA, Y-connected, 440-V, 3-phase, 50 Hz alternator, the effective armature resistance is 0.25 Ω/phase. The synchronous reactance is 3.22 Ω/phase. and leakage reactance is 0.52 Ω/phase. Determine at rated load at unity power factor : a) Internal e.m.f. Ea, b) No-load e.m.f., Eŋ, c) Percentage regulation on full load, d) Value of synchronous reactance which replaces armature reaction. AU May-16, Marks 8

Solution :

The phasor diagram for unity p.f. is shown in the Fig. 2.10.4.

Example 2.10.6 A 2000 kVA, 11 kV, 3 ϕ star connected alternator has a resistance of 0.3 Ω and reactance of 5 Ω/ phase. It delivers full load current at 0.8 lagging power factor at rated voltage. Compute the terminal voltage for the same excitation and load current at 0.8 power factor leading. AU : April-01, May-07, Marks 8

Solution :

Examples for Practice

Example 2.10.7 A 2000 kVA, 22 kV, 3-phase star connected alternator has a resistance of 0.3 ohm and reactance of 5 ohms/phase. It delivers full load current at p.f. of 0.8 lagging and normal rated voltage. Compute the terminal voltage for same excitation and load current at 0.8 pf. leading. Comment on the results obtained. UPTU : 2006-07

[Ans.: V = 12.086 kV]

Example 2.10.8 A three phase star connected alternator is rated at 1600 kVA, 13.5 kV. The per phase armature effective resistance and synchronous reactance are 1.5 Ω and 30 Ω respectively. Calculate voltage regulation for a load of 1.280 MW at power factors of,

i) 0.8 leading, ii) Unity p.f. and iii) 0.8 lagging UPTU : 2005-06

[Ans. : i) – 12 %, ii) + 3.227 %, iii) + 18.602 %)

Example 2.10.9 A 1500 kVA, 6600 V, three phase, star connected alternator with R = 0.5 Ω/ph and X_s = 5 Ω/ph delivers a full load current at 0.8 lagging p.f. load. Find the induced e.m.f. per phase. If this induced e.m.f. per phase is maintained constant but load is changed to full load unity p.f. estimate terminal voltage per phase.

[Ans. : +4168.1 V]

Example 2.10.10 A 3 phase, 11 kV, 1 MVA, Y-connected cylindrical rotor alternator has a resistance of 2.2 Ω/ph. The percentage regulation at 0.8 p.f. lagging is 24 %. Calculate the percentage regulation at 0.8 p.f. lead. VTU : August-02

[Ans. : - 13.89%]

Example 2.10.11 A 220 V, 100 kVA, star connected alternator has an effective resistance of 0.1 ohm, and leakage reactance of 0.5 ohm. Assume that when it is connected to a 0.4 power factor lagging load and delivers rated current , the armature reaction has twice the effect of armature reactance. Neglecting the effect of saturation. Calculate i) No load voltage when the load is suddenly thrown off with the field current and speed being the same. ii) No load voltage required to produce rated current assuming the alternator was short circuited. VTU : Feb.-06

[Ans. : 515.8458 V, 394.521 V]

Example 2.10.12 A 1200 kVA, 6600 V, 3 phase star connected alternator has its armature resistance as 0.25 Q per phase and its synchronous reactance as 5 Q per phase. Calculate its regulation if it delivers a full load at i) 0.8 lagging and ii) 0.8 leading p.f. 

[Ans.: + 9.33 %, - 7.00 %]

Review Question

1. What is voltage regulation of an alternator ?