What are the factors govern the terminal voltage of an alternator?
Answers
Answered by
2
The efficiency of an alternator always increases as its power increases. For example, if an alternator of 1 kW has an efficiency of 50%, then one of 10 MW will inevitably have an efficiency of about 90%. It is because of this improvement in efficiency with size that alternators of 1000 MW and above possess efficiencies of the order of 99%.
Another advantage of large machines is that power output per kilogram increases as the alternator power increases. If 1 kW alternator weighs 20 kg (i.e. 50 W/kg), then 10MW alternator weighing 20,000 kg yields 500 W/kg. In other words, larger alterna- tors weigh relatively less than smaller ones and are, consequently, cheaper.
However, as alternator size increases, cooling problem becomes more serious.

Since large machines inherently produce high power loss per unit surface area (W/m2), they tend to overheat.
To keep the temperature rise within acceptable limits, we have to design efficient cooling system which becomes ever more elaborate as the power increases. For cooling alternators of rating upto 50 MW, circulating cold-air system is adequate but for those of rating between 50 and 300 MW, we have to resort to hydrogen cooling. Very big machines in 1000 MW range have to be equipped with hollow water-cooled conductors. Ultimately, a point is reached where increased cost of cooling exceeds the saving made elsewhere and this fixes the upper limit of the alternator size.
So for as the speed is concerned, low-speed alternators are always bigger than high speed alternators of the same power. Bigness always simplifies the cooling

problem. For example, the large 200-rpm, 500-MVA alternators installed in a typical hydropower plant are air-cooled whereas much smaller 1800-r.p.m., 500-MVA alternators installed in a steam plant are hydrogen cooled.
Alternator on Load
As the load on an alternator is varied, its terminal voltage is also found to vary as in d.c. generators. This variation in terminal voltage V is due to the following reasons:
1. voltage drop due to armature resistanceRa
2. voltage drop due to armature leakage reactance XL
3. voltage drop due to armature reaction
(a) Armature Resistance
The armature resistance/phase Ra causes a voltage drop/phase of IRa which is in phase with the armature current I. However, this voltage drop is practically negligible.
(b) 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. Various types of leakage fluxes are shown in Fig. 37.22.

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 e.m.f. of self-inductance which is known as reactance e.m.f. and which is ahead of I by 90°. Hence, armature winding is assumed to possess leakage reactance XL (also known as Potier rectance XP) such that voltage drop due to this equals IXL. A part of the generated e.m.f. is used up in overcoming this reactance e.m.f.

This fact is illustrated in the vector diagram of Fig. 37.23.
(c) Armature Reaction
As in d.c. 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.We will consider three cases :
(i) when load of p.f. is unity
(ii) when p.f. is zero lagging and
(iii) when p.f. is zero leading.
Before discussing this, it should be noted that in a 3-phase machine the combined ampere-turn wave (or m.m.f. wave) is sinusoidal which moves synchronously. This amp-turn or m.m.f. wave is fixed relative to the poles, its amplitude is proportional to the load current, but its position depends on the p.f. of the load.
Consider a 3-phase, 2-pole alternator having a single-layer winding, as shown in Fig. 37.24 (a). For the sake of simplicity, assume that winding of each phase is concentrated (instead of being distributed) and that the number of turns per phase is N. Further suppose that the alternator is loaded with a resistive load of unity power factor, so that phase currents Ia, Ib and Ic are in phase with their respective phase voltages. Maximum current Iawill flow when the poles are in position shown in Fig.
(a) or at a time t1 in Fig. 37.24 (c). When Ia has a maximum value, Ib and Ic have one-half their maximum values (the arrows attached to Ia , Ib andIc are only polarity marks and are not meant to give the instantaneous directions of these currents at time t1). The instantaneous directions of currents are shown in Fig. 37.24 (a). At the instant t1, Iaflows
mark me as a brainlist
Another advantage of large machines is that power output per kilogram increases as the alternator power increases. If 1 kW alternator weighs 20 kg (i.e. 50 W/kg), then 10MW alternator weighing 20,000 kg yields 500 W/kg. In other words, larger alterna- tors weigh relatively less than smaller ones and are, consequently, cheaper.
However, as alternator size increases, cooling problem becomes more serious.

Since large machines inherently produce high power loss per unit surface area (W/m2), they tend to overheat.
To keep the temperature rise within acceptable limits, we have to design efficient cooling system which becomes ever more elaborate as the power increases. For cooling alternators of rating upto 50 MW, circulating cold-air system is adequate but for those of rating between 50 and 300 MW, we have to resort to hydrogen cooling. Very big machines in 1000 MW range have to be equipped with hollow water-cooled conductors. Ultimately, a point is reached where increased cost of cooling exceeds the saving made elsewhere and this fixes the upper limit of the alternator size.
So for as the speed is concerned, low-speed alternators are always bigger than high speed alternators of the same power. Bigness always simplifies the cooling

problem. For example, the large 200-rpm, 500-MVA alternators installed in a typical hydropower plant are air-cooled whereas much smaller 1800-r.p.m., 500-MVA alternators installed in a steam plant are hydrogen cooled.
Alternator on Load
As the load on an alternator is varied, its terminal voltage is also found to vary as in d.c. generators. This variation in terminal voltage V is due to the following reasons:
1. voltage drop due to armature resistanceRa
2. voltage drop due to armature leakage reactance XL
3. voltage drop due to armature reaction
(a) Armature Resistance
The armature resistance/phase Ra causes a voltage drop/phase of IRa which is in phase with the armature current I. However, this voltage drop is practically negligible.
(b) 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. Various types of leakage fluxes are shown in Fig. 37.22.

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 e.m.f. of self-inductance which is known as reactance e.m.f. and which is ahead of I by 90°. Hence, armature winding is assumed to possess leakage reactance XL (also known as Potier rectance XP) such that voltage drop due to this equals IXL. A part of the generated e.m.f. is used up in overcoming this reactance e.m.f.

This fact is illustrated in the vector diagram of Fig. 37.23.
(c) Armature Reaction
As in d.c. 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.We will consider three cases :
(i) when load of p.f. is unity
(ii) when p.f. is zero lagging and
(iii) when p.f. is zero leading.
Before discussing this, it should be noted that in a 3-phase machine the combined ampere-turn wave (or m.m.f. wave) is sinusoidal which moves synchronously. This amp-turn or m.m.f. wave is fixed relative to the poles, its amplitude is proportional to the load current, but its position depends on the p.f. of the load.
Consider a 3-phase, 2-pole alternator having a single-layer winding, as shown in Fig. 37.24 (a). For the sake of simplicity, assume that winding of each phase is concentrated (instead of being distributed) and that the number of turns per phase is N. Further suppose that the alternator is loaded with a resistive load of unity power factor, so that phase currents Ia, Ib and Ic are in phase with their respective phase voltages. Maximum current Iawill flow when the poles are in position shown in Fig.
(a) or at a time t1 in Fig. 37.24 (c). When Ia has a maximum value, Ib and Ic have one-half their maximum values (the arrows attached to Ia , Ib andIc are only polarity marks and are not meant to give the instantaneous directions of these currents at time t1). The instantaneous directions of currents are shown in Fig. 37.24 (a). At the instant t1, Iaflows
mark me as a brainlist
Answered by
0
Answer:
armature resistive drop
Similar questions