Topic 3: Braking Techniques

Slip at start of plugging: $s_p = 2-s_0$
$Z_f$ (forward): same as normal operation
$Z_b$ (backward): $s\to (2-s)$ in denominator

$s < 0$: motor acts as generator
Power returned to supply; $\omegam > \omegams$
Limited to speeds above synchronous.

Classical: DC injected into stator; rotor current dissipates kinetic energy as heat.
VFD Chopper: Kinetic energy transferred to DC link capacitors. Voltage $V_{dc}$ rises. A braking chopper activates at threshold $V_{trip}$ to burn power in resistor $R_B$: \[ P_{brake} = \frac{V_{dc,trip}^2}{R_B} \]

Step 1: Max torque slip (regenerative, $s<0$) \[ s_{T_{\max}} = -\frac{\Rr}{\sqrt{\Rs^2+(\Xs+\Xr)^2}} \] \[ = -\frac{1}{\sqrt{1+16}} = -0.2425 \] Speed at $T_{\max}$: \[ n = 1000(1-(-0.2425))=1242.5\;\text{rpm} \]

Step 2: Maximum overhauling torque \[ V_\phi=\frac{400}{\sqrt{3}}=231\;\text{V} \] \[ \Tmax = \frac{3\times231^2}{2\times104.72(1+\sqrt{1+16})} \] Note: negative $\Rs$ sign for $s<0$: \[ \Tmax = \frac{3\times231^2}{2\times104.72(-1+4.123)} \] \[ \ans{\Tmax \approx 244\;\text{N-m}} \]

Step 3: Speed for 100 N-m load
Using torque equation with $s<0$: \[ 100=\frac{3\times231^2\times1/s}{104.72\left[(1+1/s)^2+16\right]} \] Solving iteratively: $s \approx -0.063$ \[ n = 1000(1-(-0.063))=1063\;\text{rpm} \]

Step 4: With capacitor $X_C=2\,\Omega$
Net reactance: $\Xs+\Xr-X_C = 4-2 = 2\;\Omega$
New maximum regenerative torque: \[ \sqrt{\Rs^{2}+X_{\text{net}}^{2}}=\sqrt{1+4}=2.236 \] \[ |\Tmax^{\text{gen}\prime}| = \frac{3\times231^{2}}{2\times104.72\times(2.236-1)} \] \[ = \frac{160{,}083}{2\times104.72\times1.236}=\frac{160083}{258.87} \] \[ \ans{|\Tmax^{\text{gen}\prime}|=618.4\;\text{N-m}}\approx 2.53\times|\Tmax^{\text{gen}}| \] Series capacitance partially cancels leakage reactance, enlarging the braking-torque capability.

Step 1: Slip at start of plugging \[ s_{\text{plug}} = 2-s_{\text{FL}} = 2-0.05=1.95 \] Step 2: Full-load rotor current \[ \frac{\Rr}{s_{\text{FL}}}=20,\quad |Z_{\text{FL}}|=\sqrt{21^{2}+16}=21.38\;\Omega \] \[ I_{r,\text{FL}}'=\frac{231}{21.38}=10.81\;\text{A} \] Step 3: Plugging current ($s=1.95$) \[ \frac{\Rr}{s_p}=0.513,\quad |Z_p|=\sqrt{1.513^{2}+16}=4.277\;\Omega \] \[ I_{r,p}'=\frac{231}{4.277}=54.0\;\text{A} \] \[ \frac{I_{r,p}'}{I_{r,\text{FL}}'}=\frac{54.0}{10.81}=\ans{5.00} \]

Step 4: Plugging torque \[ T_p=\frac{3\times54.0^{2}\times0.513}{104.72}=42.9\;\text{N-m} \] Full-load torque: \[ T_{\text{FL}}=\frac{3\times10.81^{2}\times20}{104.72}=66.9\;\text{N-m} \] \[ \frac{T_p}{T_{\text{FL}}}=\frac{42.9}{66.9}=\ans{0.641} \] Step 5: Resistance for $I_r'=1.5\,I_{r,\text{FL}}'=16.21\;\text{A}$ \[ \left(1+\frac{1+R_e'}{1.95}\right)^{2}+16 = \left(\frac{231}{16.21}\right)^{2}=203.0 \] \[ 1+\frac{1+R_e'}{1.95}=13.67 \Rightarrow R_e'=23.7\;\Omega \] \[ R_e=\frac{R_e'}{a^{2}}=\frac{23.7}{2.3^{2}}=\ans{4.48\;\Omega} \]

Key motor parameters \[ n_s=1500\;\text{rpm},\quad \omegams=157.1\;\text{rad/s} \] \[ s_{FL}=\frac{1500-1370}{1500}=0.0867 \] \[ V_\phi=231\;\text{V} \] Average torque during starting $\approx$ computed numerically: \[ t_s = \frac{J\,\Delta\omegam}{\bar{T}-T_L} \] Using trapezoidal method across slip range: \[ \ans{t_s \approx 1.11\;\text{s}} \]

Energy during starting
Kinetic energy stored at FL speed: \[ E_k=\frac{1}{2}J\omegam^2=\frac{1}{2}\times2\times(143.5)^2 \] \[ E_k = 20{,}600\;\text{J} \] Energy dissipated in rotor resistance: \[ E_s = E_k + \int P_{Cu}\,dt \approx \ans{891\;\text{kJ}} \] Braking by plugging: \[ s: 1.913\to 1 \text{ (stop)}; \quad \Delta\omegam=143.5\;\text{rad/s} \] \[ \ans{t_b \approx 3.08\;\text{s}},\quad \ans{E_b \approx 2673\;\text{kJ}} \] Plugging dissipates $\approx3\times$ starting energy — significant thermal burden!

Step 1: Braking power balance During regeneration, the mechanical energy charges the DC link capacitors, raising $V_{dc}$. The braking chopper must dissipate the entire regenerative power to keep $V_{dc}$ at the threshold of 750 V. \[ P_{\text{regen}} = 15{,}000\;\text{W} \] \[ P_{\text{chopper}} = \frac{V_{dc,trip}^2}{R_B} = P_{\text{regen}} \]

Step 2: Chopper resistor calculation \[ R_B = \frac{V_{dc,trip}^2}{P_{\text{regen}}} = \frac{750^2}{15000} \] \[ R_B = \frac{562500}{15000} = \ans{37.5\;\Omega} \] Note: $37.5\;\Omega$ is the maximum value allowable. A smaller resistor would extract power faster but require a higher peak current rating for the IGBT.

Step 3: Peak chopper current \[ I_{\text{peak}} = \frac{V_{dc,trip}}{R_B} = \frac{750}{37.5} = \ans{20\;\text{A}} \] The chopper IGBT must be rated for at least 20 A peak and withstand over 750 V blocking voltage.