How to Use Capacitors to Boost Efficiency in AC Motors
- 1). Determine the power (in kilowatts, or KW), the voltage (in volts) and power factor details associated with the load powered by the AC motor. Refer to the manufacturer's specifications. The power factor will be a constant between 0 and 1. As an example, assume the power requirement is 20 KW, the voltage is 120 volts and the power factor, or "pf (old)," is 0.75.
- 2). Calculate the power level the motor must produce in kilovolt-amperes, or "KVA (old)," to deliver the needed power to the load with the current pf. Use the formula "KVA (old) = KW/pf(old)." Using our example numbers: KVA (old) = 20/0.75= 26.6 KVA. This says the motor must produce 26.6 KVA to deliver 20 KW to the load, which reflects a poor efficiency caused by the load's power factor.
- 3). Determine the reactive power, or KVAR (old), associated with the current power factor using the formula "KVAR = sqrt (KVA (old)^2 - KW^2)." Continuing with our example: KVAR (old) = sqrt (26.6^2 - 20^2) = sqrt (707.56 - 400) = sqrt (307.56) = 17.53 KVAR.
- 4). Decide the new power factor. Your objective is to increase the power factor closer to 1. As an example, let's assume you want to increase it from 0.75 to 0.9, where the new power factor is 0.9.
- 5). Calculate the new power level, or KVA (new), with the new power factor using the formula "KVA (new) = KW/pf (new)." Continuing with our example: KVA (new) = 20/0.9 = 22.2 KVA. From this, you can see that a higher power factor increases the motor efficiency when compared to the other power factor.
- 6). Calculate the new reactive power, or KVAR (new), using the formula "KVAR (new) = sqrt (KVA (new)^2 - KW^2)." Continuing with our example: KVAR (new) = sqrt (22.2^2 - 20^2) = sqrt (492.84 - 400) = sqrt (92.84) = 9.63 KVAR.
- 7). Calculate the difference between the old and new KVAR using the formula "KVAR (differ) = KVAR (old) - KVAR (new)." Continuing with our example: KVAR (diff) = 17.53 - 9.63 = 7.9 KVAR.
- 8). Calculate the capacitance needed to improve the pf using the formula "C = [(KVAR (diff) x 10^3)]/[(2pi x f x kV^2]" where pi is 3.1415, f is the frequency and KV is V/1000. V is the voltage from Step 1. As an example, we chose it to be 120 volts. Assuming f is 60 hz and continuing with our example: C = [(7.9 x 10^3)]/[(2)(3.1415)(60)(0.120^2)] = [(7.9 x 10^3)]/[(2)(3.1415)(60)(0.0144)] = [(7.9 x 10^3)]/5.43 = 1454 microfarads. In summary, to boost the efficiency of the AC motor in this example, you need to add a 1454 microfarad capacitor across the motor output terminals.
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