Performance of 7.5 kW induction motor from equivalent circuit

Use the exact per-phase equivalent circuit (magnetizing reactance in parallel with the rotor branch; core loss is included in the 405 W constant loss).

 I1   R1=0.29  jX1=0.503       jX2=0.209
 o--[====]---[~~~~~]---+-----[~~~~~]----+
                       |                |
 V1=230.94 V       jXm=13.25      R2/s = 7.0
                       |                |
 o---------------------+----------------+

Given: $V_L = 400$ V (Y), $f = 50$ Hz, $P = 6$, $s = 0.02$.

Speed

$$\begin{aligned} N_s &= \frac{120 \times 50}{6} = 1000\ \text{rpm} \\ N &= (1 - s)N_s = 0.98 \times 1000 = 980\ \text{rpm} \end{aligned}$$

Impedances

$$\begin{aligned} V_1 &= \frac{400}{\sqrt{3}} = 230.94\ \text{V} \\ Z_2 &= \frac{R_2}{s} + jX_2 = \frac{0.14}{0.02} + j0.209 = 7 + j0.209\ \Omega \\ Z_p &= \frac{jX_m\, Z_2}{jX_m + Z_2} = \frac{j13.25 (7 + j0.209)}{7 + j13.459} = 5.340 + j2.983\ \Omega \\ Z_{in} &= (0.29 + j0.503) + Z_p = 5.630 + j3.486 = 6.622\angle 31.77^\circ\ \Omega \end{aligned}$$

Stator current and power factor

$$\begin{aligned} I_1 &= \frac{V_1}{Z_{in}} = \frac{230.94}{6.622} = 34.88\ \text{A}\ (\angle -31.77^\circ) \\ \text{pf} &= \cos 31.77^\circ = 0.850\ \text{lagging} \end{aligned}$$

Powers

Air-gap power (all power into $Z_p$ goes to $R_2/s$):

$$\begin{aligned} P_{ag} &= 3 I_1^2 R_p = 3 \times 34.876^2 \times 5.340 = 19\,485\ \text{W} \\ P_{m} &= (1 - s)P_{ag} = 0.98 \times 19\,485 = 19\,096\ \text{W} \\ P_{out} &= P_m - 405 = 18\,691\ \text{W} \approx 18.69\ \text{kW} \end{aligned}$$

Output torque

$$\begin{aligned} \omega &= \frac{2\pi \times 980}{60} = 102.63\ \text{rad/s} \\ T_{out} &= \frac{P_{out}}{\omega} = \frac{18\,691}{102.63} = 182.1\ \text{N·m} \end{aligned}$$

Efficiency

$$\begin{aligned} P_{in} &= 3V_1 I_1\cos\phi = 3 \times 230.94 \times 34.876 \times 0.8502 = 20\,543\ \text{W} \\ \eta &= \frac{18\,691}{20\,543} = 0.9098 = 90.98\% \end{aligned}$$

Check of losses: stator Cu $= 3 \times 34.876^2 \times 0.29 = 1058$ W; rotor Cu $= sP_{ag} = 390$ W; fixed $= 405$ W; total $= 1853$ W $= 20\,543 - 18\,691$. (checks)

Note: at 400 V the computed output at 2% slip exceeds the 7.5 kW rating; these are the values the given circuit data produce.

Answer: N = 980 rpm, $T_{out}$ = 182.1 N·m, $P_{out}$ = 18.69 kW, $I_1$ = 34.88 A, pf = 0.850 lagging, η = 90.98%.