How to Calculate the Power of Capacitors

How to Calculate the Power of Capacitors

ALPES-TECH-EP2 How to calculate the power of capacitors BASED ON ELectRICITY BILLS ON I T Calculation Example To calculate the capacitor banks to be installed, use the For the subscriber: RMA O following method: • Highest reactive energy bill: December F N • Select the month in which the bill is highest (kVArh • Number of kVArh to be billed: 70,000 to be billed) • Monthly operating times: • Assess the number of hours the installation operates high-load + peak times = 350 hours RAL I each month • Calculate the capacitor power Qc to be installed ENE G 70,000 Qc (bank to be installed) = = 200 kVAr kVArh to be billed (monthly) 350 Qc = No. of hours' operation (monthly) 10 ALPES-TECH-EP2 A Group brand BASED ON MEASUREMENTS TAKEN ON THE HV/LV TRANSFORMER SECONDARY: PkW-cos ON ON ø I I T T Example RMA RMA O O An establishment supplied from an 800 KVA HV/LV F F N N subscriber station wanting to change the power factor of its installation to: Qc (bank to be installed) = • Cos = 0.928 (tg = 0.4) at the primary RAL I RAL I ø ø PkW x (tg measured - tg to be obtained) • I.e. Cos ø = 0.955 (tg ø = 0.31) at the secondary, with ø ø the following readings: ENE Qc = 475 x (0.88 - 0.31) = 270 kVAr GENE G - Voltage: 400 V 3-phase 50 HZ - PkW = 475 - Cos (secondary) = 0.75 (i.e. tg ø = 0.88) CALCULATION FOR FUTURE INSTALLATIONS In the context of future installations, compensation is frequently required right from the engineering stage. In this case, it is not possible to calculate the capacitor bank using conventional methods (electricity bill). For this type of installation, it is advisable to install at least a capacitor bank equal to approximately 25% of the nominal power of the corresponding HV/LV transformer. Example 1000 kva transformer, Q capacitor = 250 kVAr Note: This type of ratio corresponds to the following operating conditions: • 1000 kVA transformer • Actual transformer load = 75% Qc = 1000 x 75% x 0.80 x 0.421 = 250 kVAr • Cos ø of the load = 0.80 } k = 0.421 • Cos ø to be obtained = 0.95 } (see table on page 12) 11.

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