Electrical Machines · Synchronous Machines

Understanding Salient Pole AlternatorsTwo-Reaction Theory

Dr. Mithun Mondal BITS Pilani, Hyderabad Campus Electrical Machines

Demonstrative Video

SECTION 01

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)

salient rotor
Salient Rotor

  • d-axis \(\Rightarrow\) Both field and armature mmfs

  • q-axis \(\Rightarrow\) Only armature mmf

SECTION 02

Two-Reaction Theory (proposed by Blondel)

two reaction
Two Reaction

  • \(I_a \Rightarrow\) \(\left[I_d \perp E_0\right]\) 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 \(I_q\)

  • \[\begin{array}{ccc} X_{d}=X_{ad}+X_{L} & \mbox{and} & X_{q}=X_{aq}+X_{L}\end{array}\]
    \(X_L\)

  • q-axis reluctance higher owing to larger air-gap, \(X_d>X_q\)

Salient Phasor
\[\begin{aligned} \mbox{internal power-factor angle,}~\Psi & =\mbox{between}~E_{0}~\mbox{and}~I_{a}\\ \mbox{power angle,}~\delta & =\mbox{between}~E_{0}~\mbox{and}~V\\ E_{0} & =V+I_{a}R_{a}+jI_{d}X_{d}+jI_{q}X_{q}\\ I_{a} & =I_{d}+I_{q} \end{aligned}\]
salient phasor
Salient Phasor
Key Concepts
\[\begin{aligned} E_{0} & =V\cos\delta+I_{q}R_{a}+I_{d}X_{d}~\mbox{generating}\\ & =V\cos\delta-I_{q}R_{a}-I_{d}X_{d}~\mbox{motoring} \end{aligned}\]
for motor for generator and If we neglect
SECTION 03

Power Developed by Syn Generator

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

  • The total power = \(3 \times P_d\)

consists of power due to (first term) + (second term)
\[P_{d}=\dfrac{E_{0}V}{X_{d}}\sin\delta+\dfrac{V^{2}\left(X_{d}-X_{q}\right)}{2X_{d}X_{q}}\sin2\delta\]

  • \(P_d\) 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