Salient Pole Alternators: Solved Problems & Insights

Revision of Important Concepts

\[\begin{aligned} & \boxed{\tan\delta =\dfrac{I_{a}X_{q}\cos\Phi}{V\pm I_{a}X_{q}\sin\Phi}}~ \pm ~(\text{Gen/Motor})~ \Leftarrow \text{Neglecting} ~R_a \end{aligned}\]

Problem - 1

A 480 V, 60 Hz, \(\Delta\) connected four pole synchronous generator has a direct-axis reactance of \(0.1 \Omega\) and a quadrature-axis reactance of \(0.075 \Omega\). Its armature resistance may be neglected. At full load, this generator supplies \(1200 \mathrm{~A}\) at a power factor of \(0.8\) lagging.
1. Find the internal generated voltage \(\mathbf{E}_A\) of this generator at full load, assuming that it has a cylindrical rotor of reactance \(X_d\).
2. Find the internal generated voltage \(\mathbf{E}_A\) of this generator at full load, assuming it has a salient-pole rotor.

(a) Cylindrical rotor:

\[\begin{aligned} I_A & =\frac{1200 \mathrm{~A}}{\sqrt{3}}=693 \mathrm{~A} \qquad (\because \Delta) \\ \theta & =\cos ^{-1} 0.8=36.87^{\circ} \\ \mathbf{E}_A & =\mathbf{V}_\phi+j X_S I_A \\ & =480 \angle 0^{\circ} \mathrm{V}+j(0.1 \Omega)\left(693 \angle-36.87^{\circ} \mathrm{A}\right) \\ & =521.6+j 55.4=524.5 \angle 6.1^{\circ} \mathrm{V} \\ \delta & = 6.1^{\circ} \end{aligned}\]

(b) Salient-pole rotor:

To break down the current into direct and quadrature axis components, it is necessary to know the direction of \(\mathbf{E_A}\)

\[\begin{aligned} \mathbf{E}_A^{\prime \prime} & =\mathbf{V}_\phi+R_A \mathbf{I}_A+j X_q \mathbf{I}_A \\ & =480 \angle 0^{\circ} \mathrm{V}+0 \mathrm{~V}+j(0.075 \Omega)\left(693 \angle-36.87^{\circ} \mathrm{A}\right) \\ & =511.2+j 41.6=513 \angle 4.65^{\circ} \mathrm{V} \\ \text{direction of }~\mathbf{E_A}&\Rightarrow~\delta = 4.65^{\circ}\\ I_d & =I_A \sin (\theta+\delta) \\ & =(693 \mathrm{~A}) \sin (36.87+4.65) \\ & =459 \mathrm{~A} \\ I_q & =I_A \cos (\theta+\delta) \\ & =(693 \mathrm{~A}) \cos (36.87+4.65) \\ & =519 \mathrm{~A} \end{aligned}\]

\[\begin{aligned} \mathbf{I}_d & =459 \angle-85.35^{\circ} \mathrm{A} \\ \mathbf{I}_q & =519 \angle 4.65^{\circ} \mathrm{A} \\ \mathbf{E}_A & =\mathbf{V}_\phi+R_A \mathbf{I}_A+j X_d \mathbf{I}_d+j X_q \mathbf{I}_q \\ & =480 \angle 0^{\circ} \mathrm{V}+0 \mathrm{~V}+j(0.1 \Omega)\left(459 \angle-85.35^{\circ}\right)+ \\ & \qquad j(0.075 \Omega)\left(519 \angle 4.65^{\circ}\right) \\ & =522.6+j 42.49=524.3 \angle 4.65^{\circ} \mathrm{V} \end{aligned}\]

is not much affected but the angle is considerably different with salient poles as compared to that of non-salient machine. Note: The magnitude of

Problem - 2

A 3-phase alternator has a direct-axis synchronous reactance of 0.7 p.u. and a quadrature axis synchronous reactance of 0.4 p.u. For full-load 0.8 p.f. lagging determine (i) the load angle and (ii) the no-load per unit voltage.

\[\begin{aligned} V & =1 \text { p.u.}~~ X_d=0.7 \text { p.u.}~~ X_q=0.4 \text { p.u. } \\ \cos \phi & =0.8 ~~ \sin \phi=0.6 ~~\phi=\cos ^{-1} 0.8=36.9^{\circ} \\ I_a&=1 \text { p.u. } \\ \tan \delta & =\frac{I_a X_q \cos \phi}{V+I_q \sin \phi}=\frac{1 \times 0.4 \times 0.8}{1+0.4 \times 0.6}=0.258 \\ \Rightarrow ~\delta&=16.5^{\circ} \\ I_d & =I_a \sin (\phi+\delta)=1 \sin \left(36.9^{\circ}+14.9^{\circ}\right)=0.78 \mathrm{~A} \\ E_0 & =V \cos \delta+I_d X_d=1 \times 0.966+0.78 \times 0.75=1.553 \end{aligned}\]

Problem - 3

A 3-phase, star-connected, 50-Hz alternator has direct-axis synchronous reactance of 0.6 p.u. and quadrature-axis synchronous reactance of 0.45 p.u. The generator delivers rated kVA at rated voltage. At full-load 0.8 p.f. lagging calculate the open-circuit voltage and voltage regulation. Resistive drop at full-load is 0.015 p.u

\[\begin{aligned} I_a & =1 \text { p.u. } \quad V=1 \text { p.u. } \\ X_d&=0.6 \text { p.u. } \quad X_q=0.45 \text { p.u. } \quad R_a=0.015 \text { p.u. } \\ \tan \psi & =\frac{V \sin \phi+I_a X_q}{V \cos \phi+I_a R_a}=\frac{1 \times 0.6+1 \times 0.45}{1 \times 0.8+1 \times 0.015}=1.288 \\ \Rightarrow~ \psi & =52.2^{\circ} \\ \delta & =\psi-\phi=52.2^{\circ}-36.9^{\circ}=15.3^{\circ} \\ I_d & =I_a \sin \psi=1 \times 0.79=0.79 \mathrm{~A} \\ I_q&=I_a \cos \psi=1 \times 0.61=0.61 \mathrm{~A} \end{aligned}\]

\[\begin{aligned} E_0 & =V \cos \delta+I_q R_a+I_d X_d \\ & =1 \times 0.965+0.61 \times 0.015+0.79 \times 0.6=1.448 \\ \% \text { regn. } & =\frac{1.448-1}{1} \times 100=44.8 \% \end{aligned}\]

Problem - 4

A 3-phase, Y-connected syn. generator supplies current of 10 A having phase angle of \(20^{\circ}\) lagging at 400 V. Find the load angle and the components of armature current \(I_d\) and \(I_q\) if \(X_d\) = 10 ohm and \(X_q\) = 6.5 ohm. Assume arm. resistance to be negligible

\[\begin{aligned} \cos \phi & =\cos 20^{\circ}=0.94 \quad \sin \phi=0.342 \\ I_a & =10 \mathrm{~A} \\ \tan \delta & =\frac{I_a X_q \cos \phi}{V+I_a X_q \sin \phi}=\frac{10 \times 6.5 \times 0.94}{400+10 \times 6.5 \times 0.342}=0.1447 \\ \delta & =8.23^{\circ} \\ I_d & =I_a \sin (\phi+\delta)=10 \sin \left(20^{\circ}+8.23^{\circ}\right)=4.73 \mathrm{~A} \\ I_q & =I_a \cos (\phi+\delta)=10 \cos \left(20^{\circ}+8.23^{\circ}\right)=8.81 \mathrm{~A} \\ I_d \cdot X_d & =4.73 \times 10=47.3 \mathrm{~V} \end{aligned}\]

\[\begin{aligned} E_0 & =V \cos \delta+I_d X_d=400 \cos 8.23^{\circ}+47.3=443 \mathrm{~V} \\ \% \text { regn. } & =\frac{E_0-V}{V} \times 100 \\ & =\frac{443-400}{400} \times 100=\mathbf{1 0 . 7 5 \%} \end{aligned}\]