one of three categories; resistive or capacitive. In your own plant, the most common is likely to be inductive. Typical examples of this include transformers, fluorescent lighting and AC induction motors. Most inductive loads use a conductive coil winding to produce an electromagnetic field, allowing the motor to function. Power factor is a measure of how efficiently you are using the electricity you are purchasing from your electric utility, Equipment such as motors, HVAC blowers, pumps, and fans affects your power factor. Depending on your electric utility, a poor power factor can be costly. If you are large enough for your electric utility to have a demand meter at your facility that is used for billing purposes, there is a chance you are being penalized for a poor power factor and not even know it. Reactive Power (kvarr) is of no benefit to the user, but does cost the consumer money. The less kvarr used the greater the savings, with no reduced consumption, real Power and Apparent Power are said to be at unity, at 1.0 PF, if they share the same plane. All electrical energy is then going towards real work. If there is an opportunity for savings on your electric bill by correcting a poor power factor, we will notify you. And we will present you with options as to how you might wish to pay for needed equipment. Whether it is a one-time situation or an ongoing maintenance agreement, we will work with you to determine what services best fit your needs. All inductive loads require two kinds of power to operate: Active Power (kwatts) - to produce the motive force Reactive Power (kvar) - to energise the magnetic field The operating power from the distribution system is composed of both active (working) and reactive (non-working) elements. The active power does useful working in driving the motor whereas the reactive power only provides the magnetic field. As the power factor drops the system becomes less efficient. A drop from 1.0 to 0.9 results in 15% more current being required for the same load. A power factor of 0.7 requires approximately 43% more current; and a power factor of 0.5 requires approximately 100% (twice as much) to handle the same load. The objective, therefore, should be to reduce the reactive power drawn from the supply by improving the power factor. If an AC motor were 100% efficient it would consume only active power but, since most motors are only 75% to 80% efficient, they operate at a low power factor. This means poor energy and cost efficiency because the Regional Electricity Companies charge you at penalty rates for a poor power factor. By installing capacitors to improve your power factor you could SAVE MONEY on your electricity bill( 2- 3%). Additional potential benefits include: Reduction of heating losses in transformers and distribution equipment Longer plant life Stabilised voltage levels Increase in capacity of your existing system and equipment Improved profitability The bad news is that you are charged for both ! Why Improve Your Power Factor ? Some of the benefits of improving your power factor are as follows: Your utility bill will be smaller. Low power factor requires an increase in the electric utilities generation and transmission capacity to handle the reactive power component caused by inductive loads. Utilities usually charge a penalty fee to customers with power factors less than 0.95. You can avoid this additional fee by increasing your power factor. Your electrical system's branch capability will increase. Uncorrected power factor will cause power loss in your distribution system. You may experience voltage drops as power losses increase. Excessive voltage drops can cause overheating and premature failure of motors and other inductive equipment. What do I have to do to save money ? Simply contact our office and we will arrange for an engineer to call who will be able to advise you of the savings you can achieve which could be as much as 8 -20% of your current electricity bill up to load condition.. |
| G E E T E C H |
| Power Factor & Calculation |
and the displacement power factor, calculate the KVAR as above and you have the required correction. To calculate the correction from a known pf, first calculate the KVAR in the load at the known power factor, then calculate the KVAR in the load for the target power factor and the required correction is the difference between the two. i.e Measured load condition: KVA = 100 pf = 0.65 Target pf = 0.95 (a). KVAR = KVA x sqart ( 1 - pf x pf ) 560 x sqrt (1 - 0.55 x 0.55) = 560 x 0.835 = 467 KVAR (b). KVAR = KVA x sqrt ( 1-pf x pf ) = 560 x sqrt (1 - 0.95 x 0.95) = 560 x 0.3122 = 174.86 VAR (c). Correction required to correct from 0.55 to 0.95 is (a) -(b) = 292.8 KVAR ( =300 VAR ) To calculate the reduction in line current or kVA by the addition of power factor correction for a known initial kVA and power factor and a target power factor, we first calculate the KW from the known KVA (or line current ), i.e Initial KVA = 560 Initial pf = 0.55 Target pf = 0.95 (a). KW = KVA x pf = 560 x 0.55 = 308 kVA (b). KVA = KW / pf = 308 / 0.95 = 324 kVA KVA reduction from 560 KVA to 324 kVA Current reduction to 57 % |
| Power Factor Calculation Sheet |
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