Topic 1: Equivalent Circuit & Steady-State

\[ n_s = \frac{120\,f}{P}, \quad \omegams = \frac{2\pi n_s}{60} \] \[ s = \frac{n_s - n_r}{n_s}, \quad \omegam = \omegams(1-s) \] \[ f_r = s\cdot f \quad \text{(rotor freq.)} \]

Stator impedance: $Z_s = \Rs + j\Xs$
Rotor (referred): $Z_r = \dfrac{\Rr}{s} + j\Xr$
\[ V_s = I_s Z_s + E_1, \quad E_1 = I_m\,(j\Xm) \]

\[ P_g = \frac{3|\Rr|^2}{\Rr/s}=3|I_r'|^2\frac{\Rr}{s} \] \[ P_{\text{rotor copper}} = s\,P_g \] \[ P_{\text{mech}} = (1-s)\,P_g \]

$P_{\text{in}} \xrightarrow{-P_{\text{stator Cu}}} P_g \xrightarrow{-P_{\text{rotor Cu}}} P_{\text{mech}} \xrightarrow{-P_{\text{friction}}} P_{\text{output}}$
$\eta = P_{\text{output}}/P_{\text{in}}$; Rotor Cu loss $= s\,P_g$; $P_{\text{mech}} = (1-s)\,P_g = T\,\omegams(1-s)$

\[ T = \frac{P_g}{\omegams} = \frac{3}{\omegams}\cdot\frac{|V_s|^2\,\Rr/s} {(\Rs+\Rr/s)^2+(\Xs+\Xr)^2} \]

\[ s_{T_{\max}} = \frac{\Rr}{\sqrt{\Rs^2+(\Xs+\Xr)^2}} \] \[ \Tmax = \frac{3}{2\omegams}\cdot\frac{|V_s|^2}{\Rs+\sqrt{\Rs^2+(\Xs+\Xr)^2}} \]

\[ T \approx \frac{3}{\omegams}\cdot\frac{|V_s|^2\,\Rr/s}{(\Xs+\Xr)^2},\quad s\ll s_{T_{\max}} \] \[ \Tmax \approx \frac{3}{2\omegams}\cdot\frac{|V_s|^2}{\Xs+\Xr} \]

\[ \frac{T}{\Tmax} = \frac{2}{\dfrac{s}{s_{T_{\max}}}+\dfrac{s_{T_{\max}}}{s}} \]

\[ \Tst = \frac{3}{\omegams}\cdot\frac{|V_s|^2\,\Rr}{(\Rs+\Rr)^2+(\Xs+\Xr)^2} \] Starting current: $I_{st}=\dfrac{|V_s|}{\sqrt{(\Rs+\Rr)^2+(\Xs+\Xr)^2}}$

Step 1: Synchronous speed & slip \[ n_s = \frac{120\times50}{6}=1000\,\text{rpm} \] \[ \omegams = \frac{2\pi\times1000}{60}=104.72\,\text{rad/s} \] \[ s = \frac{1000-950}{1000}=\mathbf{0.05} \]

Step 2: Phase voltage \[ V_s = \frac{440}{\sqrt{3}}=254.0\,\text{V (phase)} \] Step 3: Rotor current \[ |I_r'|=\frac{254}{\sqrt{(0.5+8)^2+(2.4)^2}} \] \[ \ans{|I_r'| = 28.76\,\text{A}} \]

Step 4: Rotor current phasor
With $V_s = 254\angle 0^\circ$: \[ Z_{\text{total}} = (0.5+8)+j(1.2+1.2) \] \[ =8.5+j2.4\,\Omega \Rightarrow |Z|=8.83\,\Omega \] \[ \bar{I}_r' = 28.76\angle -15.77^\circ\,\text{A} \]

Step 5: Magnetising current \[ \bar{I}_m = \frac{254\angle 0^\circ}{j50}=5.08\angle -90^\circ\,\text{A} \]

Step 6: Stator current \[ \bar{I}_s = \bar{I}_r'+\bar{I}_m \] \[ = (27.68-j7.82)+(0-j5.08) \] \[ = 27.68 - j12.90\,\text{A} \] \[ \ans{|I_s| = 30.5\,\text{A},\;\angle-25^\circ} \]

Step 1: Slip \[ n_s=\frac{120\times50}{4}=1500\,\text{rpm} \] \[ s = \frac{1500-1440}{1500} = \mathbf{0.04} \] Step 2: Air-gap power \[ P_g = 3\,|I_r'|^2\frac{\Rr}{s} \] \[ \approx 3\times(20)^2\times\frac{0.8}{0.04} \] \[ P_g = 24{,}000\,\text{W} = \mathbf{24\,\text{kW}} \]

Step 3: Rotor copper loss \[ P_{\text{Cu,r}} = s\,P_g = 0.04\times24000 \] \[ \ans{P_{\text{Cu,r}} = 960\,\text{W}} \] Step 4: Mechanical power \[ P_{\text{mech}} = (1-0.04)\times24000 \] \[ \ans{P_{\text{mech}} = 23{,}040\,\text{W}} \] Step 5: Shaft (output) power \[ P_{\text{out}} = 23040-80 = \ans{22{,}960\,\text{W}} \]

Step 6: Stator copper loss \[ P_{\text{Cu,s}} = 3\times(20)^2\times1 = 1200\,\text{W} \] Step 7: Total input power \[ P_{\text{in}} = P_g + P_{\text{Cu,s}} + P_{\text{core}} \] \[ = 24000+1200+150 \] \[ P_{\text{in}} = \mathbf{25{,}350\,\text{W}} \]

Step 8: Efficiency \[ \eta = \frac{P_{\text{out}}}{P_{\text{in}}} = \frac{22960}{25350} \] \[ \ans{\eta = 90.6\%} \] Torque at shaft: \[ T_{\text{shaft}}=\frac{22960}{2\pi\times1440/60}=152.3\,\text{N-m} \]