--- title: "Slip power recovery for slip ring induction motor" subject: "Utilization of Electrical Energy" chapter: 3 key: "semester_7/utilization_of_electrical_energy:slip-power-recovery-slip-ring-motor" source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/slip-power-recovery-slip-ring-motor site: BCT Engineering --- # Slip power recovery for slip ring induction motor Utilization of Electrical Energy · Chapter 3: Control of Electric Drive - Asked in: 2071 Shrawan; 2069 Chaitra (asked 2 times) - Video: [Slip Power Recovery System | What is SPRS | Induction Motor in Hindi](https://www.youtube.com/watch?v=wZP_zyhzFZM) — Electrical With Abhishek Joshi - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/slip-power-recovery-slip-ring-motor - Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-3 ## Question Discuss the slip power recovery system for slip ring induction motor. ## Answer In a slip-ring induction motor, part of the air-gap power, the **slip power** $sP_g$, appears in the rotor circuit. With rotor-resistance speed control this power is wasted as heat in external resistors, making low-speed operation inefficient. In a **slip power recovery system**, the slip power is taken out of the rotor through power converters and **fed back to the supply** (or to the shaft), so speed can be controlled efficiently. #### Static Kramer drive (sub-synchronous) ```text 3-ph supply ---------+-----------------------+ | | [Stator] [Transformer] Slip-ring IM | [Rotor] | | slip rings | [Diode bridge]--[L]--[Line-commutated (rectifier) DC thyristor inverter] link ``` **Working** 1. Rotor voltage $sE_2$ at slip frequency is rectified by an uncontrolled diode bridge: $$ V_{d} = 1.35\,sE_2 \quad (E_2 = \text{standstill rotor line voltage}) $$ 2. A line-commutated thyristor inverter, operating with firing angle $90^\circ < \alpha < 180^\circ$, converts the DC power back to AC at supply frequency and returns it to the mains through a transformer (turns ratio adapts the voltage). Its DC voltage is $$ V_{I} = -1.35\,V_L'\cos\alpha $$ where $V_L'$ is the inverter-side AC line voltage. 3. Neglecting the drop in the DC-link inductor, $V_d = V_I$: $$ 1.35\,sE_2 = -1.35\,V_L'\cos\alpha \;\Rightarrow\; s = -\frac{V_L'}{E_2}\cos\alpha $$ 4. Changing $\alpha$ changes the counter-voltage seen by the rotor, so the slip, and hence speed, changes: at $\alpha = 90^\circ$, $s \approx 0$ (near synchronous speed); as $\alpha \to 180^\circ$, slip increases (lower speed). 5. Torque is proportional to DC-link current: $T \propto I_d$; the drive gives a shunt-like speed characteristic. **Features** - Speed control only **below synchronous speed** (power flows only out of the rotor through the diode bridge). - High efficiency, since slip power is recovered; converter rating is proportional to the speed range (for pumps and fans with limited range, a small converter suffices). - Drawbacks: poor power factor (inverter draws reactive power), harmonics; starting needs rotor resistors because the converter is rated for limited slip. #### Static Scherbius drive - The diode bridge and inverter are replaced by a **bidirectional converter** (cycloconverter, or back-to-back PWM converters, as in doubly-fed induction generators). - Slip power can flow both ways, so speed control is possible **below and above synchronous speed**, with regenerative braking and power-factor control. #### Applications Large pumps, fans, compressors, cement-mill drives, and wind turbines (doubly-fed induction generator), where a limited speed range is needed with high efficiency. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does slip power recovery do? A. Wastes slip power in resistors B. Feeds rotor slip power back to supply for efficient speed control C. Increases rotor resistance D. Supplies DC to stator Answer: B. Slip power sPg is taken from rotor via converters and returned. (Key point: Purpose) 2. Static Kramer drive controls speed: A. Only below synchronous speed B. Above and below C. Only above synchronous D. Only at synchronous Answer: A. Diode bridge allows power flow only out of the rotor. (Key point: Kramer drive range) 3. In Kramer drive, increasing alpha from 90° towards 180°: A. Slip increases (lower speed) B. Speed increases C. Slip stays zero D. Torque reverses Answer: A. s = -(VL'/E2) cos alpha, so slip rises. (Key point: Firing angle effect) 4. Static Scherbius drive allows: A. Only sub-synchronous B. Bidirectional slip power, speed control below and above synchronous C. No regeneration D. Only starting Answer: B. A bidirectional converter replaces diode bridge and inverter. 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