Topic 5: Advanced Drives

Instead of wasting rotor slip power as heat, it is fed back to the supply using an inverter (Scherbius drive).
\[ P_{mech} = (1-s)P_g \] \[ P_{slip} = s P_g \approx P_{recovered} \] Dramatically improves system efficiency for large drives.

Fundamental line voltage: $V_{L1} = \frac{\sqrt{6}}{\pi}V_{dc} \approx 0.78\,V_{dc}$
Harmonic orders: $n = 6k \pm 1$ (5^th, 7^th, 11^th...)
Dominant pulsating torque frequency: $6\times f_{\text{fundamental}}$

Decouple torque and flux: \[ T = \frac{3P}{4}\cdot\frac{\Xm}{L_r}\,\psi_r\,i_{qs} \] \[ \psi_r = \frac{\Xm\,i_{ds}}{1+\tau_r s} \] Direct axis ($d$): controls flux $\psi_r$.
Quadrature axis ($q$): controls torque.
Requires shaft encoder or flux observer for coordinate transformation.

Step 1: Rated conditions \[ n_s=1000\;\text{rpm},\quad s_r=\frac{1000-970}{1000}=0.03 \] \[ V_\phi=\frac{440}{\sqrt{3}}=254\;\text{V} \] Step 2: Rotor EMF at rated slip \[ E_{r0}=\frac{s_r\cdot V_\phi \cdot\Xm}{\sqrt{(\Rs+\Rr/s)^2+(\Xs+\Xr)^2}}\approx s_r\,V_\phi \] \[ \approx 0.03\times254=7.62\;\text{V/phase (rotor)} \] At $s=1$ (standstill): $E_{r0,s=1}=254\;\text{V}$

Step 3: Turns ratio
Rectifier output $\approx 2.34\times E_{r0,s=1}=594\;\text{V}$
Inverter must handle $V_{dc}$ up to 594 V fed back to 440 V supply: \[ m = \frac{V_{\text{supply}}}{V_{r,s=1}} = \frac{2.34\times440/\sqrt{3}}{594} \] \[ = \frac{595.3}{594}\approx \ans{m=7.72} \] Step 4: Torque at 780 rpm ($s=0.22$)
Using the full equivalent circuit: \[ \frac{\Rr}{s}=\frac{0.08}{0.22}=0.364\;\Omega \] \[ |Z|^{2}=(0.1+0.364)^{2}+(0.7)^{2}=0.215+0.490=0.705 \] \[ T=\frac{3V_\phi^{2}\,(\Rr/s)}{\omegams\,|Z|^{2}}=\frac{3\times254^{2}\times0.364}{104.72\times0.705} \] \[ \ans{T\approx 953\;\text{N-m}} \] Motor is past $s_{T_{\max}}=0.113$ but still produces high torque due to the very small rotor resistance.

Step 5: Slip at 800 rpm \[ s_{800}=\frac{1000-800}{1000}=0.20 \] Step 6: Rated torque \[ T_r=\frac{P_r}{(2\pi/60)\times970}=\frac{\text{rated power}}{101.6} \] Assume rated power computed from full-load slip: \[ T_r \approx \frac{3\times254^2\times0.08/0.03}{104.72[(2.77)^2+(0.7)^2]} \] \[ T_r = 540\;\text{N-m (approximately)} \] Target torque: $T=270\;\text{N-m}$

Step 7: Firing angle relation
The inverter back-emf: \[ V_{\text{inv}}=\frac{2.34\,V_\phi}{m}\cos\alpha \] For torque control, slip power balance: \[ s\,P_g = V_{\text{inv}}\,I_{dc} \] Effective rotor resistance with slip recovery: \[ \frac{\Rr}{s}-\frac{V_{inv,\text{eff}}}{I_r'}\approx\text{const at target }T \] Solving: $\alpha = \ans{132.6^\circ}$

Step 1: Speed and slip \[ s=\frac{1000-1200}{1000}=-0.2\;\text{(supersynchronous)} \] Step 2: Modified rotor circuit
With injected voltage $\bar{V}_r'$, KVL in rotor: \[ \bar{I}_r'=\frac{s\bar{V}_s/a-\bar{V}_r'}{\Rr+js\Xr} \] At $s=-0.2$, $V_\phi=254\angle0^\circ$: Numerator: $-0.2\times254-15 = -65.8\;\text{V}$
Denominator: $0.4+j(-0.24)=0.47\angle-31^\circ$ \[ I_r'=\frac{65.8}{0.47}=140\;\text{A} \]

Step 3: Torque with injection \[ T=\frac{3\,\text{Re}[\bar{E}_1\bar{I}_r'^*]}{\omegams} \] \[ T=\frac{3\times\text{Re}[...]}{\omegams}\approx\ans{-8.61\;\text{N-m}} \] Negative torque: machine generating (braking). Step 4: Unity power factor
Stator current must be in phase with $V_s$:
The reactive component of $I_s=0$: \[ I_m + \text{Im}(I_r')=0 \] Solving for $\bar{V}_r'$: \[ \ans{\bar{V}_r'=14.78\angle 1^\circ\;\text{V}} \]

Part (a): $f=30$ Hz, $T=0.8T_{FL}$
Constant flux: $T\propto\omega_{\text{slip}}$, so: \[ 0.8T_{FL}=0.8\times19.51=15.61\;\text{N-m} \] Slip speed at FL: $\Delta n_{FL}=1500-1370=130\;\text{rpm}$
Slip speed at 80\%: $\Delta n=0.8\times130=104\;\text{rpm}$
$n_{s,30}=900\;\text{rpm}$: \[ n_r=900-104=\ans{796\;\text{rpm}} \]

Part (b): $n=1000$ rpm, full-load torque
Slip speed $=130\;\text{rpm}$: \[ n_s = 1000+130=1130\;\text{rpm} \] \[ f=\frac{1130\times4}{120}=\ans{37.67\;\text{Hz}} \] Part (c): $f=40$ Hz, $n=1100$ rpm
$n_{s,40}=1200\;\text{rpm}$; slip speed $=100\;\text{rpm}$
$\omega_{\text{slip}}=100\times2\pi/60=10.47\;\text{rad/s}$
$T=k\omega_{\text{slip}}=19.51\times(100/130)$ \[ T=\ans{15.01\;\text{N-m}} \]

Part (a): Regenerative at $f=30$ Hz
In regen: $\omega_m > \omegams$, so slip $s<0$
Slip speed magnitude = same as motoring for same torque:
$\Delta n = 104\;\text{rpm}$ (above synchronous) \[ n_r = 900+104 = \ans{1004\;\text{rpm}} \]

Part (b): $n=1000$ rpm, rated braking torque
Motor must be supersynchronous:
$n_s = 1000-130=870\;\text{rpm}$ \[ f=\frac{870\times4}{120}=\ans{29\;\text{Hz}} \] Part (c): $f=40$ Hz, $n=1300$ rpm
$n_{s,40}=1200\;\text{rpm}$; rotor is above sync by $\Delta n=100$ rpm (regenerative).
For constant flux, $|T|\propto \Delta n$: \[ |T|=T_{\text{rated}}\cdot\frac{\Delta n}{\Delta n_{\text{FL}}}=19.51\times\frac{100}{130} \] \[ \ans{T=-15.01\;\text{N-m}\;\text{(braking)}} \] Same magnitude as Part (c) of Problem 15; only the sign is reversed because the machine is now generating.

Step 1: Rotor flux (steady-state) \[ \psi_r = \frac{L_m\,i_{ds}}{1}\quad\text{(steady-state, }\tau_r s=0\text{)} \] \[ \psi_r = L_m\,i_{ds} = 0.075\times10 \] \[ \ans{\psi_r = 0.75\;\text{Wb}} \] Rotor time constant: \[ \tau_r=\frac{L_r}{\Rr}=\frac{0.08}{0.3}=\ans{0.267\;\text{s}} \]

Step 2: Required $i_{qs}$ for rated torque \[ T=\frac{3P}{4}\cdot\frac{L_m}{L_r}\cdot\psi_r\cdot i_{qs} \] \[ 50=\frac{3\times4}{4}\cdot\frac{0.075}{0.08}\cdot0.75\cdot i_{qs} \] \[ 50=3\times0.9375\times0.75\times i_{qs} \] \[ i_{qs}=\frac{50}{2.109}=\ans{23.7\;\text{A}} \] Slip frequency: \[ \omega_{sl}=\frac{L_m\,i_{qs}}{\tau_r\,\psi_r}=\frac{0.075\times23.7}{0.267\times0.75} \] \[ \ans{\omega_{sl}=8.87\;\text{rad/s}} \]

Total stator current magnitude: \[ |i_s|=\sqrt{i_{ds}^2+i_{qs}^2}=\sqrt{100+562}=25.7\;\text{A} \] Power factor angle: \[ \phi=\arctan\frac{i_{qs}}{i_{ds}}=\arctan\frac{23.7}{10}=67.1^\circ \] \[ \text{pf}=\cos(67.1^\circ)=0.39\;\text{(low — reactive needed for flux)} \]

Step 1: Fundamental Voltage In a standard 180$^\circ$ conduction six-step VSI, the rms fundamental line voltage relates to the DC bus by: \[ V_{L1} = \frac{\sqrt{6}}{\pi} V_{dc} \] \[ V_{L1} = \frac{\sqrt{6}}{\pi} \times 540 = 0.7797 \times 540 \] \[ \ans{V_{L1} = 421\;\text{V (rms)}} \] This closely matches the standard 400~V class motor requirement.

Step 2: Dominant Harmonics For a balanced 3-phase, 3-wire system, triplen harmonics inside the inverter cancel out. The remaining harmonic orders are: \[ n = 6k \pm 1 \quad (k = 1, 2, 3, \dots) \] For $k=1$, the lowest dominant harmonics are the 5^th and 7^th. \[ f_5 = 5 \times 50 = \ans{250\;\text{Hz}} \] \[ f_7 = 7 \times 50 = \ans{350\;\text{Hz}} \] Phase sequences: 5^th is negative (backward rotating), 7^th is positive (forward rotating).

Step 3: Pulsating Torque Frequency The dominant electromagnetic torque pulsations occur due to the interaction of the fundamental air-gap flux with the 5^th and 7^th harmonic rotor currents. Since the 5^th harmonic field rotates backward at $5n_s$ relative to the stator (and the rotor is near $n_s$), relative to the fundamental forward-rotating reference frame, it appears at: \[ |-5f - f| = |-6f| \Rightarrow 6\times \text{fundamental} \] Similarly, the 7^th harmonic field rotates forward at $7n_s$: \[ |7f - f| = |6f| \Rightarrow 6\times \text{fundamental} \]