Quick Revision — Braking Techniques
Plugging (Reverse Phase Braking)
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
Regenerative Braking
$s < 0$: motor acts as generatorPower returned to supply; $\omegam > \omegams$
Limited to speeds above synchronous.
Dynamic Braking (Classical vs VFD)
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 $R_B$: \[ P_{brake} = \frac{V_{dc,trip}^2}{R_B} \]
Problems: Braking Techniques
Problem 6 — Regenerative Braking \& Overhauling Torque
Problem Statement
A 400 V, 6-pole, 50 Hz, Y-connected motor: $\Rs=\Rr=1\,\Omega$, $\Xs=\Xr=2\,\Omega$.
(a) Find maximum overhauling (regenerative braking) torque.
(b) Speed at which motor holds 100 N-m overhauling load.
(c) Effect of adding $X_C=2\,\Omega$ capacitive reactance in series.
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 $s<0$:
\[
\Tmax = \frac{3\times231^2}{2\times104.72(-1+4.123)}
\]
\[
\ans{\Tmax \approx 244\;\text{N-m}}
\]
Problem 6 — Solution (continued)
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: $s \approx -0.063$ \[ n = 1000(1-(-0.063))=1063\;\text{rpm} \]
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.
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.
Summary: Without capacitor $|\Tmax^{\text{gen}}|=244.7\;\text{N-m}$; with $X_C=2\,\Omega$ the maximum regenerative torque rises to $618.4\;\text{N-m}$ — a ratio of about 2.53. The speed at which the motor holds a 100 N-m overhauling load is approximately 1063 rpm. Caution: operating near resonance ($X_C\to\Xs+\Xr$) excites large currents.
Problem 7 — Plugging (Reverse Voltage Braking)
Problem Statement
The motor of Problem 6 is braked by plugging from full-load speed of 950 rpm.
Stator-to-rotor turns ratio $a=2.3$.
(a) Initial braking current and torque as multiples of full-load values.
(b) External resistance to limit initial braking current to 1.5$\times$ full-load.
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 $I_r'=1.5\,I_{r,\text{FL}}'=16.21\;\text{A}$
\[
\left(1+\frac{1+R_e'}{1.95}\right)^{2}+16 = \left(\frac{231}{16.21}\right)^{2}=203.0
\]
\[
1+\frac{1+R_e'}{1.95}=13.67 \Rightarrow R_e'=23.7\;\Omega
\]
\[
R_e=\frac{R_e'}{a^{2}}=\frac{23.7}{2.3^{2}}=\ans{4.48\;\Omega}
\]
Problem 8 — Braking Time \& Energy (Plugging)
Problem Statement
A 400 V, 50 Hz, 4-pole, 1370 rpm, Y-connected motor: $\Rs=2\,\Omega$, $\Rr=3\,\Omega$,
$\Xs=\Xr=3.5\,\Omega$. System inertia $J=2\;\text{kg-m}^2$. Find:
(a) Starting time and energy dissipated during starting.
(b) Time and energy during braking by plugging.
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!
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!
Problem 9 — VFD Dynamic Braking Chopper
Problem Statement
A 400 V (line), 50 Hz, 4-pole induction motor is driven by a VFD with a diode front-end. During braking, it returns 15 kW of regenerative power to the DC link. The nominal DC link voltage is $V_{dc,nom} = 1.35 \times 400 = 540$ V. The braking chopper switches ON when $V_{dc}$ rises to 750 V to prevent overvoltage trip.
Determine the maximum braking resistor $R_B$ required to dissipate this power safely, and the peak current through the chopper switch.
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.