Two Reaction Theory of Salient Pole Synchronous Machine (Alternator)

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Two Reaction Theory of Salient Pole Synchronous Machine (Alternator)

In a salient-pole rotor synchronous machine, the air-gap is highly non-uniform. Consider a synchronous machine having a 2-pole salient-pole rotor rotating in the anti-clockwise direction within a 2-pole stator, as shown in Figure-1.

In Figure-1, the axis shown along the axis of the rotor is known as direct axis or d-axis and the axis perpendicular to the d-axis is called quadrature axis or q-axis. It can be seen that the two small air-gaps are involved in the path of d-axis flux (φ_d), thus the reluctance of the path is minimum. The q-axis flux (φ_q) path has two large air-gaps and it is the path of maximum reluctance.

The rotor magnetic field (B_r) is shown directed vertically upward in Figure-2. This rotor magnetic field induces an EMF in the armature (or stator winding).

If a lagging power factor load is connected to the alternator, an armature current (I_a) will flow. The armature current (I_a) lags behind the excitation voltage (E_f) by an angle ψ (see Figure-2). The armature current (I_a) produces the stator MMF (F_s) which lags behind I_a by 90°. The stator MMF (F_s) produces the stator magnetic field (B_s) along the direction of F_s.

According to Blondel's Two Reaction Theory, the stator MMF (F_s) can be resolved into two components viz. the direct-axis component (F_d) and the quadrature axis component (F_q).

If

• φ_d = Direct axis flux
• φ_q = Quadrature axis flux
• S_d = Reluctance of direct axis flux path
• S_q = Reluctance of quadrature axis flux path

Then,

$$\mathrm{\text{Direct Axis Flux, } \: \phi_{d} \:=\: \frac{F_{d}}{S_{d}} \:\:\:\dotso\: (1)}$$

And,

$$\mathrm{\text{Qaudrature Axis Flux, }\:\phi_{q} \:=\: \frac{F_{q}}{S_{q}} \:\:\:\dotso\: (2)}$$

Since S_d < S_q , the direct axis component (F_dd) of stator MMF produces more flux than the quadrature axis component (F_q) of the stator MMF. The direct axis and quadrature axis components of the stator fluxes produce voltages in the stator winding by armature reaction.

Let,

• E_ad = Direct axis component of armature reaction voltage
• E_aq = Quadrature axis component of armature reaction voltage

Since each armature reaction voltage is directly proportional to respective armature current and lags behind the armature current by 90°, the armature reaction voltages can be written as,

$$\mathrm{E_{ad} \:=\: - jI_{d}X_{ad} \:\:\:\dotso\: (3)}$$

$$\mathrm{E_{aq} \:=\: - jI_{q}X_{aq} \:\:\:\dotso\: (4)}$$

Where,

• X_ad is the armature reaction reactance in the direct axis per phase.
• X_aq is the armature reaction reactance in the quadrature axis per phase.

Here, X_aq < X_ad because the EMF induced by a given MMF acting on the direct axis is smaller than the EMF on the quadrature axis due to its higher reluctance.

Now, the resultant EMF induced in the machine is,

$$\mathrm{E_{R} \:=\: E_{f} \:+\: E_{ad } \:+\: E_{aq}}$$

$$\mathrm{\Rightarrow\:E_{R}\:=\:E_{f}\:-\:jI_{d}X_{ad}\:-\:jI_{q}X_{aq} \:\:\:\dotso\: (5)}$$

Also, the resultant voltage (E_R) is equal to the phasor sum of terminal voltage and the voltage drops in the resistance and leakage reactance of the armature, thus,

$$\mathrm{E_{R}\:=\:V \:+\: I_{a}R_{a} \:+\: jI_{a}X_{l} \:\:\:\dotso\: (6)}$$

The armature current (I_a) is split into two components, one in phase with the excitation voltage (E_f) and the other in phase quadrature to it.

If

• I_q = quadrature axis component of I_a in phase with E_f
• I_d = direct axis component of I_a lagging E_f by 90°

Then, the total armature current is the phasor sum of I_q and I_d, i.e.,

$$\mathrm{I_{a} \:=\: I_{q} \:+\: I_{d} \:\:\:\dotso\: (7)}$$

Now, from eqns. (5) and (6), we get,

$$\mathrm{E_{f} \:=\: V \:+\: I_{a}R_{a} \:+\: jI_{a}X_{l} \:+\: jI_{d}X_{ad} \:+\: jI_{q}X_{aq} \:\:\:\dotso\: (8)}$$

And, from Eqns. (7) and (8), we get,

$$\mathrm{E_{f} \:=\: V \:+\: (I_{q} \:+\: I_{d})R_{a} \:+\: j(I_{q} \:+\: I_{d})X_{l} \:+\: jI_{d}X_{ad} \:+\: jI_{q}X_{aq}}$$

$$\mathrm{\Rightarrow \: E_{f } \:=\: V \:+\: (I_{q } \:+\: I_{d})R_{a} \:+\: jI_{d}(X_{l} \:+\: X_{ad}) \:+\: jI_{q} (X_{l} \:+\: X_{aq} )}$$

$$\mathrm{\Rightarrow\: E_{f} \:=\: V \:+\: (I_{q} \:+\: I_{d})R_{a} \:+\: jI_{d}X_{d } \:+\: jI_{q}X_{q } \:\:\:\dotso\: (9)}$$

Where,

$$\mathrm{X_{d} \:=\: X_{l} \:+\: X_{ad} \:\:\: and \:\:\: X_{q} \:=\: X_{l} \:+\: X_{aq}}$$

The reactance X_d is known as the direct-axis synchronous reactance and the reactance X_q is called the quadrature-axis synchronous reactance.

$$\mathrm{\therefore \: E_{f} \:=\: V \:+\: I_{a}R_{a} \:+\: jI_{d}X_{d} \:+\: jI_{q}X_{q} \:\:\:\dotso\: (10)}$$

Equation (10) is the final form of the voltage equation for a salient-pole synchronous generator.

Phasor Diagram

Figure-3 shows the complete phasor diagram of a salient-pole synchronous generator based on the Blondel's Two Reaction Theory.