Consider a three-phase alternator with a cylindrical rotor and is operating at a lagging power factor.

Let,

• E = per phase induced EMF

• V = per phase terminal voltage

• I_a = per phase armature current

• cos$phi$ = power factor (lagging) of load

• $delta$ = power angle (angle between E and V)

Therefore, the output power of a three-phase alternator is given by,

$$mathrm{mathit{P_{0}}:=:3mathit{VI_{a}cosphi }cdot cdot cdot (1)}$$

Approximate Output Power of 3-Phase Alternator

In a three-phase alternator, the resistance $R_{a}$ of the armature circuit is very small as compared to the synchronous reactance $X_{s}$ of the machine. Thus, we can neglect the armature resistance ($R_{a}$), after that we get approximate equivalent circuit of the alternator as shown in Figure-1. The phasor diagram of the circuit is also shown in Figure-1.

From the phasor diagram, we get,

$$mathrm{mathit{AB}:=:mathit{I_{a}X_{s}cosphi }:=:mathit{E}:mathrm{sindelta }}$$

$$mathrm{Rightarrow mathit{I_{a}cosphi }:=:frac{mathit{E:mathrm{sindelta }}}{mathit{X_{s}}}cdot cdot cdot (2)}$$

Now, from equation (1) & (2), we get,

$$mathrm{mathit{P_{0}}:=:frac{3mathit{EV:mathrm{sindelta }}}{mathit{X_{s}}}cdot cdot cdot (3)}$$

The expression in Equation-3 gives the approximate output power of a three-phase alternator.

When the alternator is operating at a constant speed with a constant field current, then X_s and E both are constant and therefore the terminal voltage V is also constant. Thus, from Equation-3, we can observe,

$$mathrm{mathit{P_{0}}propto :mathrm{sindelta }}$$

We know, when $delta$ = 90°, then

$$mathrm{mathrm{sin:90^{circ}}:=:1}$$

Thus, the alternator supplies maximum power at $$ =90°, and it is given by,

$$mathrm{mathit{P_{max}}:=:frac{3mathit{EV}}{mathit{X_{s}}}}cdot cdot cdot (4)$$

The maximum power given by Equation-4 is called the static stability limit of the alternator.

Numerical Example

A 3-phase, 11 kV, 3 MVA star-connected alternator has a per phase synchronous reactance of 10 $Omega$. Its excitation is such that the generated line EMF is 15 kV. When the alternator is connected to infinite busbars. Calculate the maximum output power of the alternator at the given excitation when armature resistance is neglected.

Solution

Given data,

$$mathrm{mathrm{Line:voltage,}mathit{V_{L}}:=:11:kV:=:11000:V}$$

$$mathrm{therefore:mathrm{Terminal:voltage:per:phase,}mathit{V}:=:frac{11000}{sqrt{3}}:=:6350.85:V}$$

$$mathrm{mathrm{Generated:line:EMF}:=:15:kV:=:15000:V}$$

$$mathrm{therefore:mathrm{Generated:EMF:per:phase,}mathit{E}:=:frac{15000}{sqrt{3}}:=:8660.25:V}$$

$$mathrm{mathrm{Synchronous: reactance: per: phase,}:mathit{X_{s}}:=:10:Omega }$$

Therefore, the maximum power output of the alternator will be,

$$mathrm{mathit{P_{max}}:=:frac{3mathit{EV}}{mathit{X_{s}}}:=:frac{3times 8660.25times 6350.85}{10}}$$

$$mathrm{therefore mathit{P_{max}}:=:16499times 10^{3}W:=:16499:mathrm{kW}}$$