Open circuit emf and mechanical power of 75 kW motor

Given: $P_{out} = 75$ kW, $V_L = 400$ V (star), $R_a = 0.04\ \Omega$, $X_s = 0.4\ \Omega$, pf = 0.8 leading, $\eta = 92.5\%$.

Input power and current

$$\begin{aligned} P_{in} &= \frac{P_{out}}{\eta} = \frac{75}{0.925} = 81.081\ \text{kW} \\ I_a &= \frac{P_{in}}{\sqrt3\,V_L\cos\phi} = \frac{81081}{\sqrt3 \times 400 \times 0.8} = 146.29\ \text{A} \\ V &= \frac{400}{\sqrt3} = 230.94\ \text{V per phase} \end{aligned}$$

Open-circuit (excitation) emf

For a motor $\vec V = \vec E_f + \vec I_a Z_s$, so $\vec E_f = \vec V - \vec I_a Z_s$. Take $V$ as reference; leading current is at $+36.87^\circ$.

$$\begin{aligned} Z_s &= 0.04 + j0.4 = 0.402\angle 84.29^\circ\ \Omega \\ \vec I_a &= 146.29\angle 36.87^\circ = 117.03 + j87.77\ \text{A} \\ \vec I_a Z_s &= 58.81\angle 121.16^\circ = -30.43 + j50.32\ \text{V} \\ \vec E_f &= 230.94 - (-30.43 + j50.32) \\ &= 261.37 - j50.32 \\ &= 266.17\angle -10.90^\circ\ \text{V} \end{aligned}$$

Line value: $\sqrt3 \times 266.17 = 461.0$ V. As expected for leading pf, $E_f > V$ (over-excited), and the load angle is $\delta = 10.90^\circ$.

Mechanical power developed

$$\begin{aligned} P_{cu} &= 3I_a^2R_a = 3 \times 146.29^2 \times 0.04 = 2568\ \text{W} \\ P_{mech} &= P_{in} - P_{cu} = 81081 - 2568 = 78513\ \text{W} \end{aligned}$$

Check: $3\,\text{Re}(\vec E_f \vec I_a^{*}) = 78.51$ kW, the same value.

(The difference between 78.51 kW and the 75 kW output is the friction, windage, core and stray loss.)

Answer: $E_f = 266.2$ V per phase (461.0 V line), load angle $10.9^\circ$; mechanical power developed $= 78.51$ kW.