How to Correct a Power Factor With Capacitors


The power factor, or pf, is the condition associated with the load powered by a power supply. It causes the actual power received by the load to be less than the power supplied by the power supply. The pf is a number between zero and one. The higher the power factor, the closer actual power is to supplied power. For this reason, the ideal power factor is one, where actual equals supplied. Low power factors waste power, use up resources and cost money. To correct for low power factors, you need to add capacitance to the power system. The amount of capacitance will vary depending on the correction factor you choose.

  • Find the power needed by the load, the voltage needed and the power factor of the load. Refer to the manufacturer's specifications for the load. The power is measured in units of kilowatts, or KW. The voltage is measured in units of volts, and the power factor is a constant between zero and one. As an example, let's assume the load needs 1200 KW, 480 volts and has a power factor of 0.6.

  • Calculate the power supply size needed, in kilovolt-amperes or "KVA (old)," to deliver the needed power at the current power factor. Use the formula: KVA (old) = KW/pf (old). Using the example numbers:

    KVA (old) = 1200/0.6 = 2000 KVA

  • Calculate the reactive power, or KVAR (old), for the current power factor: KVAR = sqrt (KVA (old)^2 - KW^2). Continuing with the example:

    KVAR = sqrt (2000^2 - 1200^2) = sqrt (4,000,000 - 1,440,000) = sqrt (2,560,000) = 1600 KVAR

  • Choose the corrected power factor. This is the power factor you want to move to by adding pf correction capacitors. The goal is to get the power factor closer to one as possible. As an example, let's increase the pf from 0.6 to 0.9.

  • Calculate the power supply size needed with the new power factor, or KVA (new):

    KVA (new) = KW/pf (new) = 1200/0.9 = 1333 KVA. As you can see, with a higher power factor, less power is needed to deliver 1200 KW to the load, which is more efficient and saves money.

  • Calculate the new reactive power, or KVAR (new), for pf at 0.9:

    KVAR (new) = sqrt (KVA (new)^2 - KW^2) = sqrt (1333^2 - 1200^2) = sqrt (1,776,889 - 1,440,000) = sqrt (336,889) = 580.42 KVAR

  • Calculate the difference between the old KVAR and the new KVAR using the formula: KVAR (diff) = KVAR (old) - KVAR (new). In this example:

    KVAR (diff) = 1600 - 580.42 = 1019.58.

  • Calculate the capacitance needed to produce KVAR (diff): C = [(KVAR (diff) x 10^3)]/[(2pi x f x kV^2] where pi is 3.1415, f is frequency and KV is voltage/1000 (this is the voltage from Step 1). Assuming f is 60 hz and continuing with our examples:

    C = [(1019.58 x 10^3)]/[(2)(3.1415)(60)(0.480^2)] = [(1019.58 x 10^3)]/[(2)(3.1415)(60)(0.23)] = [(1019.58 x 10^3)]/86.85 = 11739.55 microfarads or 11739.55 x 10^-6 farads or 0.011739 farads


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