Understanding Salient Pole Alternators: Unveiling Two-Reaction Theory

Demonstrative Video

Operation of Salient-Pole Machine

$\Rightarrow$ uniform air-gap $\Rightarrow$ reactance remains the same irrespective of the spatial position of the rotor $\Rightarrow$ possess one axis of symmetry (pole or direct axis)
$\Rightarrow$ non-uniform air-gap $\Rightarrow$ reactance varies $\Rightarrow$ two axes
- field pole axis (direct or d-axis)
- axis passing through center of inter-polar space (quadrature or q-axis)

d-axis $\Rightarrow$ Both field and armature mmfs
q-axis $\Rightarrow$ Only armature mmf

Two-Reaction Theory (proposed by Blondel)

$I_a \Rightarrow$ $\left[I_d \perp E_0\right]$ and $\left[I_q ~ \mbox{along} ~ E_0\right]$
armature reactance $\Rightarrow$ q-axis ( $X_{ad}$ ) associated with $I_d$ and d-axis ( $X_{aq}$ ) associated with $I_q$
$X_L$ remains same for both axes, we get $\begin{array}{ccc} X_{d}=X_{ad}+X_{L} & \mbox{and} & X_{q}=X_{aq}+X_{L}\end{array}$
q-axis reluctance higher owing to larger air-gap, $X_d>X_q$

$\begin{aligned} \mbox{internal power-factor angle,}~\Psi & =\mbox{between}~E_{0}~\mbox{and}~I_{a}\\ \mbox{power angle,}~\delta & =\mbox{between}~E_{0}~\mbox{and}~V\\ E_{0} & =V+I_{a}R_{a}+jI_{d}X_{d}+jI_{q}X_{q}\\ I_{a} & =I_{d}+I_{q} \end{aligned}$

$\begin{aligned} I_{d} & =I_{a}\sin\Psi\\ I_{q} & =I_{a}\cos\Psi\\ \tan\Psi & =\dfrac{AD+AC}{OE+ED}\\ & =\dfrac{V\sin\Phi+I_{a}X_{q}}{V\cos\Phi+I_{a}R_{a}}~\mbox{generating}\\ & =\dfrac{V\sin\Phi-I_{a}X_{q}}{V\cos\Phi-I_{a}R_{a}}~\mbox{motoring}\\ \delta & =\Psi-\Phi~\mbox{generating}\\ & =\Phi-\Psi~\mbox{motoring} \end{aligned}$

$\begin{aligned} E_{0} & =V\cos\delta+I_{q}R_{a}+I_{d}X_{d}~\mbox{generating}\\ & =V\cos\delta-I_{q}R_{a}-I_{d}X_{d}~\mbox{motoring} \end{aligned}$ If we neglect $R_a$ $\begin{aligned} V\sin\delta & =I_{q}X_{q}=I_{a}X_{q}\cos\left(\Phi\pm\delta\right)\\ \Rightarrow V\sin\delta & =I_{a}X_{q}\left(\cos\Phi\cos\delta\pm\sin\Phi\sin\delta\right)\\ \Rightarrow V & =I_{a}X_{q}\left(\cos\Phi\cot\delta\pm\sin\Phi\right)\\ \Rightarrow V\pm I_{a}X_{q}\sin\Phi & =I_{a}X_{q}\cos\Phi\cot\delta\\ \tan\delta & =\dfrac{I_{a}X_{q}\cos\Phi}{V\pm I_{a}X_{q}\sin\Phi} \end{aligned}$ $+$ for generator and $-$ for motor

Power Developed by Syn Generator

Neglecting $R_a$ and hence Cu loss, per-phase power developed ( $P_d$ ): $\begin{aligned} P_{d} & =VI_{a}\cos\Phi\\ I_{q}X_{q} & =V\sin\delta; ~~~ I_{d}X_{d}=E_{0}-V\cos\delta\\ I_d & = I_a\sin \left(\Phi + \delta \right); ~~~ I_q = I_a\cos \left(\Phi + \delta\right) \end{aligned}$
Substituting Eq. (3) in (1) and solving $I_{a}\cos\phi=\dfrac{V}{X_{d}}\sin\delta+\dfrac{V}{2X_{q}}\sin2\delta-\dfrac{V}{2X_{d}}\sin2\delta$
Finally, Substituting in Eq. (1), we get $P_{d}=\dfrac{E_{0}V}{X_{d}}\sin\delta+\dfrac{V^{2}\left(X_{d}-X_{q}\right)}{2X_{d}X_{q}}\sin2\delta$
The total power = $3 \times P_d$

$P_{d}=\dfrac{E_{0}V}{X_{d}}\sin\delta+\dfrac{V^{2}\left(X_{d}-X_{q}\right)}{2X_{d}X_{q}}\sin2\delta$

$P_d$ consists of power due to (first term) + (second term)
If $X_d = X_q$ , for cylindrical rotor, second term = 0, and the power is due to first
If no field excitation, $E_0 =0$ , then first term = 0, and power is due to second