A synchronous motor is connected to an infinite bus and draws 1.0 p.u. current at 1.0 p.u. Voltage (i) 0.9 p.f. lag and (ii) 0.9 p.f. lead. The synchronous reactance of the motor is 1.0 p.u. (a) Determine E, Q, I, δ, P (b) If the excitation is increased by 10%, determine E, Q, I, δ P, cos Φ (c) If

Question

A synchronous motor is connected to an infinite bus and draws 1.0 p.u. current at 1.0 p.u. Voltage (i) 0.9 p.f. lag and (ii) 0.9 p.f. lead. The synchronous reactance of the motor is 1.0 p.u. (a) Determine E, Q, I, δ, P (b) If the excitation is increased by 10%, determine E, Q, I, δ P, cos Φ (c) If the system is as in (a) and the loading on the motor is increased by 10%, determine E, Q, I, δ, P, cos Φ

Accepted Answer

(a) (i) p.f. 0.9 lag

E = V – j I [latex]X_{d}[/latex]

= 1.0 – j (0.9 – j 0.436) × 1.0

= 0.564 – j 0.9

= 1.062 ∠– 57.9°

δ = – 57.9°, E = 1.062

[latex]P=\frac{1.063 imes 1.0}{1.0} \sin 57.9=0.9[/latex] p.u.

Here we are taking δ as positive, even though P is –ve as it is motoring action

[latex]Q=\frac{1.063 imes 1.0}{1.0} \cos 57.9-1.0[/latex]

= – 0.4356 p.u. (under excitation)

(b) Excitation increased by 10%

[latex]E_{1}[/latex] = 1.1 E = 1.1682

[latex]\begin{aligned} \sin \delta_{1} &=\frac{0.9 imes 1.0}{1.1682 imes 1.0}, \delta_{1}=50.39 \ Q_{1} &=\frac{1.1682 imes 10}{1.0} \cos 50.39-1.0 \ &=-0.255 ext { p.u. } \ I_{1} &=\sqrt{\frac{0.9^{2}+0.255^{2}}{1.0^{2}}}=0.935 \ \cos \phi_{1} &=\frac{0.9}{0.935}=0.96 \end{aligned}[/latex]

Summarising

[latex]P_{1}[/latex], f, V No change

[latex]|Q|, \delta_{1}, I_{1}[/latex] Decrease

E, [latex] \cos \phi_{1}[/latex] Increase

Similarly, if p.f. is leading 0.9, the back e.m.f. originally is

[latex]\begin{aligned}E &=V-j I X_{d} \&=1.0-j(0.9+j 0.436) imes 1.0 \&=1.436-j 0.9 \&=1.69 \angle-32^{\circ} \E &=1.69, \delta=-32^{\circ}, Q=0.436 ext { p.u. }\end{aligned}[/latex]

Now if [latex]E_{1}[/latex] =1.1 × 1.69=1.859

[latex]\begin{aligned}\sin \delta_{1} &=\frac{0.9 imes 1.0}{1.859 imes 1.0}, \delta_{1}=-28.9 \Q &=\frac{1.859 imes 1.0}{1.0} \cos 28.9-1.0 \&=0.627 ext { p.u. }\ I &=\sqrt{\frac{0.9^{2}+0.627^{2}}{1.0^{2}}}=1.076 ext { p.u. } \ \cos \phi_{1} &=\frac{0.9}{1.0 imes 1.096}=0.82\end{aligned}[/latex]

Summarising

P, V, f No change

δ, cos Φ Decrease

E, I, Q Increase

(c) [latex]P_{1}[/latex] = 1.1 × 0.9 = 0.99

(i) p.f. lag [latex]E_{1}[/latex] = E = 1.062

[latex]\begin{aligned} \sin \delta_{1} &=\frac{0.99 imes 1.0}{1.062 imes 1.0}, \delta_{1}=-68.78 \ Q &=\frac{1.062 imes 1.0}{1.0} \cos 68.78-1.0 \ &=-0.615 ext { p.u. } \ I &=\sqrt{\frac{0.99^{2}+0.615^{2}}{1.0^{2}}}=1.165 ext { p.u. } \ \cos \phi &=\frac{0.99}{1.165}=0.849 \end{aligned}[/latex]

Summarising

E, V, f No change

cos Φ Decrease

P, δ, |Q|, I Increase

(ii) p.f. 0.9 lead

[latex]\begin{aligned} E_{1} &=E=1.69 \ \sin \delta_{1} &=\frac{0.99 imes 1.0}{1.69 imes 1.0}, \delta_{1}=-35.86 \ Q &=\frac{1.69 imes 1.0}{1.0} \cos 35.86-1.0 \ &=0.37 \mathrm{p} . \mathrm{u} . \ I &=\sqrt{\frac{0.99^{2}+0.37^{2}}{1.0^{2}}}=1.057 \mathrm{p} . \mathrm{u} . \ \cos \phi &=\frac{0.99}{1.057}=0.936 \end{aligned}[/latex]

Summarising

E, V, f No change

Q Decrease

P, δ, I, cos Φ Increase

Similarly, the effect of decreasing excitation and/or load on the motor performance can be studied.

Question C.3: A synchronous motor is connected to an infinite bus and draw...

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