--- title: "Capacitor bank for 340 kW induction motor" subject: "Utilization of Electrical Energy" chapter: 6 key: "semester_7/utilization_of_electrical_energy:capacitor-bank-340kw-induction-motor" source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/capacitor-bank-340kw-induction-motor site: BCT Engineering --- # Capacitor bank for 340 kW induction motor Utilization of Electrical Energy · Chapter 6: Demand Side Management - Asked in: 2080 Baisakh; 2072 Kartik (asked 2 times) - 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/capacitor-bank-340kw-induction-motor - Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-6 ## Question A 340 kW, 50 Hz, 3-phase star connected induction motor has full load efficiency of 85% and p.f of 0.8 lagging. It is desired to improve the power factor to 0.96 lagging by using bank of three capacitors. Calculate: i) The kVAR rating of the capacitor bank. ii) The capacitance of each limbs of the condenser bank connected in delta. iii) The capacitance of each capacitor, if each one of the limb of the delta-connected condenser bank is formed by using 6 similar 3300 V capacitors. ## Short answer The capacitor bank rating is 183.33 kVAR. Each delta-connected limb has a capacitance of 17.86 μF, and each individual capacitor is 2.98 μF. ## Answer The supply voltage is not stated; since the capacitors are rated 3300 V, the motor is taken as a **3300 V, 3-phase** motor and the 6 capacitors in each delta limb are taken as connected **in parallel** (each sees the full 3300 V line voltage). #### Motor input $$ P = \frac{\text{output}}{\eta} = \frac{340}{0.85} = 400\ \text{kW} $$ #### i) kVAR rating of the capacitor bank $$ \begin{aligned} \tan\phi_1 &= \tan(\cos^{-1}0.8) = 0.75, \quad \tan\phi_2 = \tan(\cos^{-1}0.96) = 0.2917 \\ Q_C &= P(\tan\phi_1 - \tan\phi_2) = 400(0.75 - 0.2917) = 183.33\ \text{kVAR} \end{aligned} $$ #### ii) Capacitance of each limb (delta connected) kVAR per limb $= 183.33/3 = 61.11$ kVAR; voltage across each limb $= V_L = 3300$ V. $$ \begin{aligned} C_{limb} &= \frac{Q_{limb}}{\omega V^2} = \frac{61\,111}{2\pi \times 50 \times 3300^2} \\ &= 17.86 \times 10^{-6}\ \text{F} = 17.86\ \mu\text{F} \end{aligned} $$ #### iii) Capacitance of each capacitor Each limb has 6 equal 3300 V capacitors in parallel: $$ C = \frac{17.86}{6} = 2.98\ \mu\text{F} $$ ```text L1 /\ C=17.86 uF each limb = 6 x 2.98 uF / \ in parallel, 3300 V L2 ------ L3 ``` **Answer:** (i) 183.33 kVAR, (ii) 17.86 μF per limb, (iii) 2.98 μF per capacitor. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the motor input power found for the capacitor calculation? A. Input = output / efficiency = 340/0.85 = 400 kW B. 340 x 0.85 = 289 kW C. 340 / 0.8 = 425 kW D. 340 kW, since output is used Answer: A. The pf correction acts on input power, so output is divided by efficiency. (Key point: Motor input power) 2. Which gives the capacitor bank rating? A. Q_C = 340(0.75 - 0.2917) = 155.8 kVAR B. Q_C = 400(0.75 - 0.2917) = 183.33 kVAR C. Q_C = 400(0.96 - 0.8) = 64 kVAR D. Q_C = 400 x 0.2917 = 116.7 kVAR Answer: B. Q_C = P(tan phi1 - tan phi2) with tan(cos^-1 0.8) = 0.75 and tan(cos^-1 0.96) = 0.2917. (Key point: Capacitor kVAR) 3. In delta, what voltage and kVAR does each limb have? A. Phase voltage 3300/sqrt3 V and the full 183.33 kVAR B. 3300 V and 183.33 kVAR C. Line voltage 3300 V and one third of the kVAR (61.11 kVAR), giving C = Q/(omega V^2) = 17.86 uF D. Line voltage divided by 3 Answer: C. In delta each limb sees line voltage and carries Q/3. (Key point: Delta limb capacitance) 4. Why is each capacitor 17.86/6 = 2.98 uF? A. They are in series, so the reciprocals add B. Each carries 6 times the limb voltage C. The limb capacitance is multiplied by 6 D. The 6 equal capacitors in each limb are in parallel, so capacitances add Answer: D. Parallel capacitors add, so each is one-sixth of the limb value. (Key point: Individual capacitor) ## Related in Utilization of Electrical Energy - [Line pf of 35 kW induction motor with capacitors](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Ainduction-motor-35kw-capacitor-line-pf) — Utilization of Electrical Energy, chapter 6, asked 3× - [Induction motor with square-law load: voltage at 1300 rpm](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Ainduction-motor-fan-load-1300rpm-voltage) — Utilization of Electrical Energy, chapter 3 - [Induction motor plugging: braking current and torque](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Ainduction-motor-plugging-braking-current) — Utilization of Electrical Energy, chapter 3 - [Annual saving by capacitors for 200 kW factory](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Afactory-200kw-capacitor-annual-saving) — Utilization of Electrical Energy, chapter 6 - [Nichrome strip width for 40 kW 3-phase oven](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Anichrome-strip-40kw-oven-width) — Utilization of Electrical Energy, chapter 5, asked 2× - [Low frequency induction furnace with half-full hearth](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Ainduction-furnace-half-full-hearth) — Utilization of Electrical Energy, chapter 5, asked 2× ## Related in other subjects - [Delta capacitor per phase for 50 kW motor](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Acapacitor-delta-415v-50kw-0-72) — Basic Electrical Engineering, chapter 8 - [Capacitor bank for 3-phase motor pf](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Acapacitor-bank-60kw-induction-motor) — Basic Electrical Engineering, chapter 8 - [2.2 kW 440 V cage motor (star-delta) design](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-2-2kw-440v-star-delta) — Electric Machine Design, chapter 4, asked 3× - [2.2 kW cage motor designed for best power factor](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-2-2kw-best-power-factor) — Electric Machine Design, chapter 4, asked 2×