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Power Factor Correction: a Guide for the Plant Engineer

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Power Factor Correction: a Guide for the Plant Engineer Effective August 2014 Capacitor banks and Technical Data SA02607001E Supersedes November 2010 passive harmonic filters Power factor correction: a guide for the plant engineer Contents Description Page Description Page Part one: power factor Part two: harmonics What is power factor? . 2 Introduction . 19 Should I be concerned about What are harmonics? . 19 low power factor? . 3 What are the consequences What can I do to improve power factor? . 4 of high harmonic distortion levels? . 20 How much can I save by installing IEEET 519 . 20 power capacitors? . 5 How are harmonics generated? . 21 How can I select the right capacitors What do power factor correction for my specific application needs? . 9 capacitors have to do with harmonics? . 22 How much kVAR do I need? . 9 How do I diagnose a potential Where should I install capacitors harmonics-related problem? . 22 in my plant distribution system? . 15 How can harmonics problems be eliminated? . 22 Can capacitors be used in nonlinear, What is a passive harmonic filter? . 22 nonsinusoidal environments? . 17 Do I need to perform a system analysis What about maintenance? . 17 to correctly apply harmonic filters? . 23 Code requirements for capacitors . 17 What is Eaton’s experience Useful capacitor formulas . 18 in harmonic filtering? . 23 Technical Data SA02607001E Power factor correction: Effective August 2014 a guide for the plant engineer Part One: power factor What is power factor? Special electrical requirement of inductive loads Fundamentals of power factor Most loads in modern electrical distribution systems are Power factor is the ratio of working power to apparent power . inductive . Examples include motors, transformers, gaseous tube It measures how effectively electrical power is being used . A high lighting ballasts, and induction furnaces . Inductive loads need power factor signals efficient utilization of electrical power, while a magnetic field to operate . a low power factor indicates poor utilization of electrical power . Inductive loads require two kinds of current: To determine power factor (PF), divide working power (kW) by apparent power (kVA) . In a linear or sinusoidal system, the result • Working power (kW) to perform the actual work of creating heat, is also referred to as the cosine light, motion, machine output, and so on . θ. • Reactive power (kVAR) to sustain the magnetic field kW PF = = cosine θ Working power consumes watts and can be read on a wattmeter . kVA It is measured in kilowatts (kW) . Reactive power doesn’t perform For example, if you had a boring mill that was operating at 100 kW useful “work,” but circulates between the generator and the load . and the apparent power consumed was 125 kVA, you would divide It places a heavier drain on the power source, as well as on the 100 by 125 and come up with a power factor of 0 .80 . power source’s distribution system . Reactive power is measured in kilovolt-amperes-reactive (kVAR) . (kW) 100 = (PF ) 0.80 Working power and reactive power together make up apparent (kVA) 125 power . Apparent power is measured in kilovolt-amperes (kVA) . NNote For a discussion on power factor in nonlinear, nonsinusoidal systems, turn to Page 17 . Heat Component = Work Done G Circulating Component = No Work Figure 3. kVA Power Resistive G Load Light kVA kW COSθ ==-----------PF kVAR Hot Plate kVA θ Figure 1. kW Power kW Motor Figure 4. Power Triangle M G Field NNote A right power triangle is often used to illustrate the relationship between kW, kVAR, and kVA . Figure 2. kVAR Power 2 EATON www.eaton.com Power factor correction: Technical Data SA02607001E a guide for the plant engineer Effective August 2014 Should I be concerned about low power factor? Low power factor means you’re not fully utilizing the electrical power you’re paying for . As the triangle relationships in Figure 5 demonstrate, kVA decreases as power factor increases . At 70% power factor, it requires 142 kVA to produce 100 kW . At 95% power factor, it requires only 105 kVA 142 to produce 100 kW . Another way to look at it is that at 70% power kVA factor, it takes 35% more current to do the same work . 100 kVAR θ 100 kW 100 PF ==-------- 70% 142 105 kVA 33 kVAR θ 100 kW 100 PF ==-------- 95% 105 Figure 5. Typical Power Triangles EATON www.eaton.com 3 Technical Data SA02607001E Power factor correction: Effective August 2014 a guide for the plant engineer What can I do to improve power factor? You can improve power factor by adding power factor correction capacitors to your plant distribution system. When apparent power (kVA) is greater than working power (kW), 18A M the utility must supply the excess reactive current plus the working current . Power capacitors act as reactive current generators . (See Figure 6 .) By providing the reactive current, they reduce the 10 hp, 480V Motor total amount of current your system must draw from the utility . at 84% Power Factor 95% power factor provides maximum benefit Theoretically, capacitors could provide 100% of needed reactive 16A M power . In practical usage, however, power factor correction to approximately 95% provides maximum benefit . The power triangle in Figure 7 shows apparent power demands 3.6A on a system before and after adding capacitors . By installing power capacitors and increasing power factor to 95%, apparent power is reduced from 142 kVA to 105 kVA—a reduction of 35% . 3 kVAR Capacitor Power Factor Improved to 95% Line Current Reduced 11% Note: Current into motor does not change. Figure 6. Capacitors as kVAR Generators COSθ ==--------100--70% PF 1 142 COSθ ==--------100--95% PF 2 105 67 kVAR Capacitor Added 70% PF 100 kVAR Before Before 142 kVA Before 105 kVA After 33 kVAR θ1 95% PF After θ 2 After Figure 7. Required Apparent Power Before and After Adding Capacitors 4 EATON www.eaton.com Power factor correction: Technical Data SA02607001E a guide for the plant engineer Effective August 2014 How much can I save by installing power capacitors? Case 2 Power capacitors provide many benefits: Assume the same conditions except that: 400 kW @ 87% = 460 kVA • Reduced electric utility bills 400 kW @ 97% = 412 kVA corrected billing • Increased system capacity • Improved voltage kVA demand charge: • Reduced losses $1 .91 / kVA / month (112,400 kWh / month energy consumed) Reduced utility bills Energy charge: Your electric utility provides working (kW) and reactive power (kVAR) $0 .0286 / kWh (first 200 kWh / kVA of demand) to your plant in the form of apparent power (kVA) . While reactive $0 .0243 / kWh (next 300 kWh / kVA of demand) power (kVAR) doesn’t register on kW demand or kW hour meters, $0 .021 / kWh (all over 500 kWh / kVA of demand) the utility’s transmission and distribution system must be large enough to provide the total power . Utilities have various ways Uncorrected: of passing the expense of larger generators, transformers, cables, 460 kVA × $1.91 = $878.60 switches, and the like, along to you . –$786.92 $ 91.68 savings in demand charge As shown in the following case histories, capacitors can save you money no matter how your utility bills you for power . Corrected: kVA billing 412 kVA × $1 .91 = $786 .92 The utility measures and bills every ampere of current, including Uncorrected energy: reactive current . kWh = 112,400 460 × 200 = 92,000 kWh Case 1 @ 0.0286 = $2631.20 Assume an uncorrected 460 kVA demand, 480V, three-phase 460 × 300 = 138,000 at 0 .87 power factor (normally good) . but balance only = 20,400 Billing: @ $0.0243 = $495.72 $4 .75/kVA demand $2631.20 Correct to 0 .97 power factor +$ 495.72 $3126.92 uncorrected energy charge Solution: Corrected energy: kVA × power factor = kW 460 × 0 .87 = 400 kW actual demand kWh = 112,400 460 × 200 = 82,400 kWh kW = kVA @ 0.0286 = $2356.64 PF 460 × 300 = 123,600 but balance only = 30,000 400 = 412 corrected billing demand 0.97 @ $0.0243 = $729.00 From Table 6 kW multipliers, to raise the power factor from $2356.64 0 .87 to 0 .97 requires capacitor: +$ 729.00 $3085.64 corrected energy charge Multiplier of 0 .316 x kW 0 .316 x 400 = 126 kVAR (use 140 kVAR) $3126.92 –$3085.64 Uncorrected original billing: $ 41.28 savings in energy charge due to rate charge 460 kVA × $4.75 = $2185 / month (9600 kWh in first step reduced by $0 .0043) –$1957 $ 228 / month savings × 12 This is not a reduction in energy consumed, but in billing only . $2736 annual savings $ 41.28 energy –$ 91.68 demand Corrected new billing: $ 132.96 monthly total savings × 12 412 kVA × $4 .75 = $1957/month $1595.52 140 kVAR, 480V capacitor cost: $1600 (installation extra) . This capacitor pays for itself in less than eight months . A 130 kVAR capacitor can be paid for in less than 14 months . EATON www.eaton.com 5 Technical Data SA02607001E Power factor correction: Effective August 2014 a guide for the plant engineer kW demand billing with power factor adjustment The utility charges according to the kW demand and adds a Case 2 surcharge or adjustment for power factor . The adjustment may be a multiplier applied to kW demand . The following formula shows With the same 400 kW load, the power factor is only 81% . a billing based on 90% power factor: In this example, the customer will pay an adjustment on: 400 × 0.90 kW demand × 0.90 = 444 billing kW demand 0.81 actual power factor (From Case 1: When the power factor = 96%, the billing demand If power factor was 0 .84, the utility would require 7% increase is 375 kW = $3495 .00 per month .) in billing, as shown in this formula: kW × 0.90 = 107 (multiplier) First 40 kW @ $10.00 = $ 400.00 0.84 Next 160 kW @ $ 9.50 = $1520.00 Next 244 kW @ $ 9.00 = $2196.00 Some utilities charge for low power factor but give a credit or bonus Total 444 kW $4116.00 – $3495.00 = $621.00 x 12 for power above a certain level .
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