GATE 2023 EE Q.47: solution and answer

GATE 2023 · EEQ.47 · 2 marks · NAT

A three-phase synchronous motor with synchronous impedance of $0.1 + j 0.3$ per unit per phase has a static stability limit of $2.5$ per unit. The corresponding excitation voltage in per unit is ___________ (Round off to 2 decimal places).

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Answer and solution

ANSWER

The excitation voltage is $E = \frac{10}{2} \approx 1.58 p.u.$

Official answer: 1.58 to 1.59Source: Official answer key, GATE 2023 EE (IIT Kanpur)

Take the terminal voltage as the per-unit base: $V = 1∠ 0^{\circ} p.u.$ The synchronous impedance is $Z_{s} = R + j X_{s} = 0.1 + j 0.3 p.u.$ , so $∣ Z_{s} ∣ = 0. 1^{2} + 0. 3^{2} = 0.1 = 0.3162 p.u.$ and the impedance angle is $θ = tan^{- 1} \frac{0.3}{0.1} = 71.5 7^{\circ}$ , with $cos θ = \frac{R}{∣ Z _{s} ∣}$ .
The armature resistance is not negligible, so the simple formula $P_{ma x} = \frac{E V}{∣ Z _{s} ∣}$ does not apply. For a motor, $V = E + I Z_{s}$ . With $E = E ∠ - δ$ , the armature current is $I = \frac{V ∠ 0 ^{\circ} - E ∠ - δ}{∣ Z _{s} ∣∠ θ}$ .
The mechanical power developed is the power delivered to the excitation emf, $P = Re (E I^{*})$ : $E I^{*} P = \frac{E ∠ - δ ( V - E ∠ δ )}{∣ Z _{s} ∣∠ - θ} = \frac{E V}{∣ Z _{s} ∣} ∠ (θ - δ) - \frac{E ^{2}}{∣ Z _{s} ∣} ∠ θ = \frac{E V}{∣ Z _{s} ∣} cos (θ - δ) - \frac{E ^{2}}{∣ Z _{s} ∣} cos θ = \frac{E V}{∣ Z _{s} ∣} cos (θ - δ) - \frac{E ^{2} R}{∣ Z _{s} ∣ ^{2}}$
The static stability limit is the largest value of $P$ , which occurs at $δ = θ$ (where $cos (θ - δ) = 1$ ): $P_{ma x} = \frac{E V}{∣ Z _{s} ∣} - \frac{E ^{2} R}{∣ Z _{s} ∣ ^{2}}$
Substitute $V = 1$ , $∣ Z_{s} ∣ = 0.1$ , $\frac{R}{∣ Z _{s} ∣ ^{2}} = \frac{0.1}{0.1} = 1$ and $P_{ma x} = 2.5$ : $2.5 = \frac{E}{0.1} - E^{2} \Rightarrow E^{2} - 10 E + 2.5 = 0$
The discriminant is $(10)^{2} - 4 (2.5) = 10 - 10 = 0$ , so there is exactly one root: $E = \frac{10}{2} = 1.5811 p.u.$
Check: $\frac{1.5811}{0.3162} - 1.581 1^{2} = 5.0 - 2.5 = 2.5 p.u.$ This agrees with the given limit. Because the discriminant is zero, $2.5 p.u.$ is the largest power this motor can develop at any excitation, $\frac{V ^{2}}{4 R}$ , and it is reached only at $E = \frac{V ∣ Z _{s} ∣}{2 R}$ .
Excitation voltage $E \approx 1.58 p.u.$

KEY IDEA

When armature resistance is included, the maximum mechanical power a synchronous motor develops is $P_{ma x} = \frac{E V}{∣ Z _{s} ∣} - \frac{E ^{2} R}{∣ Z _{s} ∣ ^{2}}$ , and it occurs at a load angle equal to the impedance angle, $δ = θ$ . Setting this equal to the static stability limit gives a quadratic in $E$ .

WATCH OUT FOR

A common mistake is to use the lossless formula $P_{ma x} = \frac{E V}{∣ Z _{s} ∣}$ (or $\frac{E V}{X _{s}}$ ) and drop the $- \frac{E ^{2} R}{∣ Z _{s} ∣ ^{2}}$ term. That gives $0.79$ , or $0.75$ with $X_{s}$ . Here $R = 0.1 p.u.$ is a third of $X_{s}$ and cannot be ignored. Another mistake is to maximise the input power, $\frac{V ^{2} R}{∣ Z _{s} ∣ ^{2}} + \frac{E V}{∣ Z _{s} ∣}$ , which gives $0.47$ . The static stability limit refers to the power developed at the shaft side of $E$ .

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GATE 2023 EE Q.47: solution and answer

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