Excitation emf and maximum power of 4 kVA motor

Given: 4 kVA, 110 V, star, $X_d = 3\ \Omega$, $X_q = 2\ \Omega$, full load at 0.8 lagging, $R_a$ neglected.

$$\begin{aligned} V &= \frac{110}{\sqrt3} = 63.51\ \text{V},\quad I_a = \frac{4000}{\sqrt3 \times 110} = 20.99\ \text{A} \\ \vec I_a &= 20.99\angle -36.87^\circ = 16.80 - j12.60\ \text{A} \end{aligned}$$

Method (two-reaction theory, motor, $R_a = 0$): $\vec V = \vec E_f + jX_d\vec I_d + jX_q\vec I_q$.

1. $\vec E' = \vec V - jX_q\vec I_a$ gives the direction of $E_f$ (the q-axis), so its angle is $-\delta$.
2. Angle between $I_a$ and the q-axis: $\psi = \delta - \phi$ ($\phi$ positive for lagging, negative for leading); $I_d = I_a\sin\psi$, $I_q = I_a\cos\psi$.
3. $E_f = E' + (X_d - X_q)I_d$.

Load angle

$$\begin{aligned} jX_q\vec I_a &= j2(16.80 - j12.60) = 25.19 + j33.59 \\ \vec E' &= 63.51 - 25.19 - j33.59 = 38.32 - j33.59 = 50.96\angle -41.24^\circ\ \text{V} \end{aligned}$$

Load angle $\delta = 41.24^\circ$.

Excitation emf

$$\begin{aligned} \psi &= \delta - \phi = 41.24^\circ - 36.87^\circ = 4.37^\circ \\ I_d &= 20.99\sin4.37^\circ = 1.600\ \text{A},\quad I_q = 20.99\cos4.37^\circ = 20.93\ \text{A} \\ E_f &= 50.96 + (3 - 2)(1.600) = 52.56\ \text{V per phase} \end{aligned}$$

Line value $= \sqrt3 \times 52.56 = 91.03$ V.

Maximum power for this excitation

$$P = 3\left[\frac{E_fV}{X_d}\sin\delta + \frac{V^2}{2}\left(\frac{1}{X_q}-\frac{1}{X_d}\right)\sin2\delta\right] = 3\left[1112.57\sin\delta + 336.11\sin2\delta\right]$$

Check at $\delta = 41.24^\circ$: $3(733.5 + 334.2) = 3200$ W $= 4000 \times 0.8$. Correct.

Setting $dP/d\delta = 0$ with $a = 1112.57$, $b = V^2(1/X_q - 1/X_d) = 672.22$:

$$\begin{aligned} \cos\delta_m &= \frac{-a + \sqrt{a^2 + 8b^2}}{4b} = \frac{-1112.57 + \sqrt{1112.57^2 + 8(672.22)^2}}{4(672.22)} = 0.4055 \\ \delta_m &= 66.08^\circ \\ P_{max} &= 3[1112.57\sin66.08^\circ + 336.11\sin132.15^\circ] \\ &= 3[1017.0 + 249.2] = 3798\ \text{W} \end{aligned}$$

Answer: $E_f = 52.56$ V per phase (91.0 V line), load angle $= 41.24^\circ$, maximum power $\approx 3.80$ kW at $\delta = 66.1^\circ$.