How do you calculate reactive power QC?

As shown in the figure, tracing a line segment from the value of the initial cosφ to the value to be obtained, the intersection of the line with the middle graduated scale, gives the value of K which, multiplied by the active power P of the load, defines the necessary reactive power Qc.

How do you calculate capacitive power?

The k factor is read from a table 1 – Multipliers to determine capacitor kilovars required for power factor correction (see below) and multiplied by the effective power. The result is the required capacitive power. For an increase in the power factor from cosφ = 0.75 to cosφ = 0.95, from the table 1 we find a factor k = 0.55:

What is the maximum reactive power of a shunt capacitor bank?

This discharge may cause a rupture of the failed unit with possible damage to the rest of the bank. To prevent it, the maximum reactive power of one series section should not be higher than 4,650 kvar at a rated voltage and 60 Hz frequency. Refer to IEEE Std. C37.99-1990 “IEEE Guide for Protection of Shunt Capacitor Banks 1.

How to calculate capacitance?

Enter the power in kW, Current in Amps, Voltage in Volts either line or phase, choose the phase, and frequency (required for capacitance calculator). Press the calculate button. Also, enter the value kW value that near to the multiplication of current and voltage.

How do you calculate C/K in a power factor relay?

The so-called C/k value is calculated by the step size C divided by the ratio k of the current transformer. It is clear that a capacitor with, for instance, 50 kvar may not be switched in if the power factor relay measures a deviation of just 10 kvar reactive power with regard to the preadjusted power factor target.

How to calculate reactive power in kvar / capacitor bank?

A three-phase motor has 100kW real power load at operating at 0.7pf, we need to improve the power factor to 0.96. Let we calculate the required reactive power in kVAR or capacitor bank to be connected across the motor? Here, PF 1 = 0.7 PF 2 = 0.96 Required capacitor bank = 100 x tan (cos -1 (0.7)- cos -1 (0.96)) = 72.85 kVAR.