Topic 4: Speed Control Methods

$T \propto V_s^2$ at a given slip.
Used for fan/pump loads; poor efficiency at reduced speeds.

\[ \frac{V_s}{f} = \text{const} \Rightarrow \phi_m \approx \text{const} \] \[ \Tmax \approx \text{const} \quad \forall\; f \] Boost required at low $f$ to overcome $I\Rs$ drop.

At constant flux: $T = k\,\omega_{\text{slip}}$
Rated torque at any speed if slip speed = rated slip speed.

Introduces external resistance $R_e$ into the rotor circuit (slip-ring motors).
Increases starting torque and reduces starting current.
Poor efficiency due to $I^2R$ loss in the external resistor.

Part (a): $n=1200$ rpm, find $V$ \[ n_s=1500\;\text{rpm},\quad s=\frac{300}{1500}=0.2 \] At rated torque: $T_{\text{rated}}=\frac{2800}{2\pi\times1370/60}=19.51\;\text{N-m}$ Setting up torque equation with $V$ as unknown: \[ T=\frac{3V_\phi^2\,\Rr/s}{\omegams\left[(\Rs+\Rr/s)^2+(\Xs+\Xr)^2\right]} \] \[ 19.51=\frac{3V_\phi^2\times25}{157.1\times\left[27^2+100\right]} \] \[ V_\phi=\ans{253.2\;\text{V (phase)}} \] Line: $V_L=253.2\;\text{V}$ ($\Delta$: $V_\phi=V_L$)

Stator current at (a): \[ I_r'=\frac{253.2}{\sqrt{27^2+100}}=9.28\;\text{A} \] \[ I_m=\frac{253.2}{80}=3.165\;\text{A} \] \[ I_s\approx\sqrt{9.28^2+3.165^2}=9.81\;\text{A} \] Line: $I_L=\sqrt{3}\times9.81=\ans{17\;\text{A}}$
Part (b): $V=300\;\text{V}$, find speed
At rated torque ($s=x$): \[ 19.51=\frac{3\times300^2\times(5/s)}{157.1[(2+5/s)^2+100]} \] Solving iteratively: $s=0.147$ \[ n=1500(1-0.147)=\ans{1279\;\text{rpm}} \] $I_s\approx\ans{16.9\;\text{A}}$

Step 1: Rated slip \[ n_s=1000\;\text{rpm},\quad s_1\approx 0.03\text{ (assumed)} \] Step 2: Required slip at 850 rpm \[ s_2=\frac{1000-850}{1000}=0.15 \] Step 3: For constant torque, slip proportional to rotor resistance \[ \frac{\Rr+R_e}{\Rr}=\frac{s_2}{s_1} \] \[ \frac{0.2+R_e}{0.2}=\frac{0.15}{0.03}=5 \] \[ R_e=0.8\;\Omega\;\text{(referred to stator)} \]

Step 4: Actual rotor resistance
If turns ratio $a=1$ (delta/delta): \[ \ans{R_e=0.8\;\Omega\;\text{per phase (rotor)}} \] Efficiency penalty: \[ \eta_{\text{mech}}=\frac{P_{\text{mech}}}{P_g}=1-s_2=0.85 \] \[ \text{vs. rated: }1-s_1=0.97 \] Power wasted in $R_e$: \[ P_{R_e}=(s_2-s_1)P_g \]

Step 1: Rated conditions \[ n_s=1500\;\text{rpm},\quad n_{FL}=1370\;\text{rpm} \] \[ \omega_{\text{slip,rated}}=\omegams\cdot s_{\text{rated}}=157.1\times0.0867 \] \[ \omega_{\text{slip,rated}}=13.62\;\text{rad/s} \] Step 2: At $f=30\;\text{Hz}$
$V/f$ constant: $V_{30}=400\times30/50=240\;\text{V}$
Synchronous speed: $n_{s,30}=\frac{120\times30}{4}=900\;\text{rpm}$ \[ \omegams^{30}=94.25\;\text{rad/s} \]

Step 3: Motor speed at $f=30$ Hz
Keep same slip speed: $\omega_{\text{slip}}=13.62\;\text{rad/s}$ \[ n_r=900-\frac{13.62\times60}{2\pi}=900-130=\ans{770\;\text{rpm}} \] Step 4: Torque at 30 Hz
Reactances scale: $X'=X\times(30/50)=0.6X$ \[ T=\frac{3\times(138.6)^2\times(3\times0.0867)}{94.25[(2+3/0.0867)^2+(0.6\times7)^2]} \] Evaluating: $\ans{T\approx 22.7\;\text{N-m}}$
Step 5: Frequency for 1200 rpm \[ n_s-n_r=130\;\text{rpm (slip speed)} \] \[ f=\frac{(1200+130)\times P}{120}=\frac{1330\times4}{120}=\ans{44.3\;\text{Hz}} \]

$f$ (Hz)	$V$ (V)	$n_s$ (rpm)	$T$ (N-m)
50	400	1500	19.5 (rated)
40	320	1200	$\approx$19.5
30	240	900	$\approx$19.5
20	160	600	$\approx$19.5
10	80	300	reduces (boost needed)

Stator current remains approximately constant: \[ I_s \approx \frac{V_s/f}{L_s+L_m}\cdot\frac{1}{\omegams}\approx \text{const} \] At low frequencies: Voltage boost required to overcome $I\Rs$ drop and maintain constant flux: \[ V_{\text{boost}} = \frac{V_s}{f_{\text{rated}}}\cdot f + I_s\Rs \]