Power angle characteristics of salient pole machine

The power angle characteristic gives the power of a salient pole machine as a function of the load angle $\delta$. Because the air gap is not uniform, the armature current is split into a direct-axis part $I_d$ (sees $X_d$) and a quadrature-axis part $I_q$ (sees $X_q$), with $X_d > X_q$ (two-reaction theory).

Derivation (generator, $R_a$ neglected)

             E (q-axis)
            /|
  jXq Iq  /  |
         /   |  jXd Id
        /    |
  V    /delta|
 -----+------+------> d-axis
   I lags V by phi; psi = delta + phi

Take $E$ along the q-axis and $V$ lagging $E$ by $\delta$. From the phasor diagram, resolving $V$ along the two axes:

$$\begin{aligned} V\cos\delta &= E - X_d I_d \ \Rightarrow\ I_d = \frac{E - V\cos\delta}{X_d} \\ V\sin\delta &= X_q I_q \ \Rightarrow\ I_q = \frac{V\sin\delta}{X_q} \end{aligned}$$

Power per phase is $P = VI\cos\phi$. Projecting $I_d$ and $I_q$ on $V$ (the angle between $V$ and the q-axis is $\delta$):

$$\begin{aligned} P &= V\left(I_q\cos\delta + I_d\sin\delta\right) \\ &= V\left[\frac{V\sin\delta}{X_q}\cos\delta + \frac{E - V\cos\delta}{X_d}\sin\delta\right] \\ &= \frac{EV}{X_d}\sin\delta + V^2\sin\delta\cos\delta\left(\frac{1}{X_q} - \frac{1}{X_d}\right) \end{aligned}$$

Using $\sin\delta\cos\delta = \tfrac12\sin2\delta$:

$$P = \frac{EV}{X_d}\sin\delta + \frac{V^2}{2}\left(\frac{1}{X_q}-\frac{1}{X_d}\right)\sin 2\delta \quad \text{(per phase)}$$

For three phases multiply by 3. The same expression holds for a motor (with $\delta$ the angle by which $E$ lags $V$).

• First term: excitation power, depends on field current.
• Second term: reluctance power, due to saliency; it exists even when $E = 0$ and vanishes for a cylindrical rotor ($X_d = X_q$).

 P |      .--.  resultant
   |    .'    '.
   |  .'  exc.  '.
   | /  .-.reluct.\
   |/.-'   '-.     \
   +-----------+----+---- delta
   0   ~60-75  90  180 deg

Condition for maximum power

$$\frac{dP}{d\delta} = \frac{EV}{X_d}\cos\delta + V^2\left(\frac{1}{X_q}-\frac{1}{X_d}\right)\cos2\delta = 0$$

Let $a = \dfrac{EV}{X_d}$ and $b = V^2\left(\dfrac{1}{X_q}-\dfrac{1}{X_d}\right)$. With $\cos2\delta = 2\cos^2\delta - 1$:

$$2b\cos^2\delta + a\cos\delta - b = 0 \ \Rightarrow\ \cos\delta_{m} = \frac{-a + \sqrt{a^2 + 8b^2}}{4b}$$

So maximum power occurs at $\delta_m < 90^\circ$ (unlike the cylindrical machine, where $P_{max}$ is at $90^\circ$). The salient pole machine is therefore "stiffer": it gives more synchronising power for a given $\delta$.