An alternator is connected to an infinite bus bar and delivers 1.0 p.u. current at 1.0 p.u. voltage and p.f. (i) 0.95 lag (ii) 0.95 lead. The reactance of the alternator $X_{d}$ = 1.0 p.u. Determine

(a) P, Q, E and δ

(b) If the excitation is reduced by 20%, calculate P, Q, E, δ, $\cos \phi_{1}$ and current I.

(c) If the system is as in (a) and the steam input is reduced by 20%, calculate Q, E, δ cos, I, P.

Question

An alternator is connected to an infinite bus bar and delivers 1.0 p.u. current at 1.0 p.u. voltage and p.f. (i) 0.95 lag (ii) 0.95 lead. The reactance of the alternator [latex]X_{d}[/latex] = 1.0 p.u. Determine

(a) P, Q, E and δ

(b) If the excitation is reduced by 20%, calculate P, Q, E, δ, [latex]\cos \phi_{1}[/latex] and current I.

(c) If the system is as in (a) and the steam input is reduced by 20%, calculate Q, E, δ cos, I, P.

Accepted Answer

(i) 0.95 lag

E = 1.0 + j 1.0 (0.95 – j 0.3122)

= 1.3122 + j 0.95

= 1.62 ∠35.9

[latex]\begin{aligned}P &=\frac{1.62 imes 1.0}{1.0} \sin 35.9 \&=0.95 \Q &=0.3122 \end{aligned}[/latex]

(b) [latex]\quad E_{1}[/latex] = 1.62 × 0.8 = 1.296

[latex]\begin{aligned}\sin \delta_{1} &=\frac{0.95 imes 1.0}{1.296 imes 1.0} \\delta_{1} &=47.14^{\circ} \Q_{1} &=\frac{1.296 imes 1.0}{1.0} \cos 47.14-1.0 \&=-0.118 \I &=\sqrt{\frac{0.95^{2}+0.118^{2}}{1.0^{2}}}=0.957 ext { p.u. } \\cos \phi &=\frac{0.95}{1.0 imes 0.957}=0.9926\end{aligned}[/latex]

(ii) 0.95 lead

We know from the previous example that for this condition E = 1.1728 ∠54°

[latex]E_{1}[/latex] = 1.1728 × 0.8 = 0.93824

[latex]\sin \delta_{1} =\frac{0.95 imes 1.0}{0 93824 imes 1.0}[/latex]

Here again a situation arises when 20% reduction in excitation below certain excitation level is not permissible.

Let us assume that the reduction is by 10% then

[latex]\begin{aligned}E_{1} &=1.1728 imes 0.9=1.055 ext { p.u. } \ herefore \quad \sin \delta_{1} &=\frac{0.95 imes 1.0}{1.055 imes 1.0}=0.9 \ ext {or} \quad \delta_{1} &=64.16 \Q_{1} &=\frac{1.055 imes 1.0}{1.0} \cos 64.16-1.0 \&=-0.54 ext { p.u. } \I &=\sqrt{\frac{0.95^{2}+0.54^{2}}{1.0^{2}}}=1.09 ext { p.u. } \ \cos \phi &=\frac{0.95}{1.09 imes 1.0}=0.9047\end{aligned}[/latex]

[latex] herefore \quad[/latex] P, V, f No change

E, cos Φ Decrease

I, δ, |Q| Increase

(c) Power decreased by 20%

[latex]P_{1}[/latex] = 0.95 × 0.8 = 0.76

[latex]E_{1}[/latex] = E = 1.62

(i) Power factor 0.95 lag

[latex]\begin{aligned} \sin \delta_{1} &=\frac{0.76 imes 1.0}{1.62 imes 1.0}, \delta_{1}=27.98 \simeq 28^{\circ} \ Q &=\frac{1.62 imes 1.0}{1.0} \cos 28-1.0 \ &=0.43 ext { p.u. } \ I &=\sqrt{\frac{0.76^{2}+0.43^{2}}{10^{2}}}=0.87 ext { p.u. } \ \cos \phi &=\frac{0.76}{0.87 imes 1.0}=0.87 \end{aligned}[/latex]

E, V, f No change

P, cos Φ, I, δ Decrease

Q Increase

(ii) p.f. 0.95 lead

[latex]E_{1}[/latex] = E = 1.1728 ∠54°

[latex]\begin{aligned} \sin \delta_{1} &=\frac{0.76 imes 1.0}{1.1728 imes 1.0}, \delta_{1}=40.4 \ Q &=\frac{1.1728 imes 1.0}{1.0} \cos 40.4-1.0 \ &=0.893-1.0 \ &=-0.107 ext { p.u. } \ I &=\sqrt{0.76^{2}+0.107^{2}}=0.768 ext { p.u. } \ \cos \phi & \simeq 1.0 \end{aligned}[/latex]

E, V, f No change

I, P, δ, |Q| Decrease

cos Φ Increase.

Question C.2: An alternator is connected to an infinite bus bar and delive...

Related Answered Questions

Verified Answer:

Verified Answer: