active power filter
  • Equipment to solve Power Quality problems:Power factor.

    Equipment to solve Power Quality problems:Power factor.

    Reducing power factor requires producing reactive energy as close as possible to connected loads. Installing capacitors on the network, such as ZDDQ  Low voltage capacitor banks, Medium voltage capacitor banks,low voltage static var generator, medium voltage STATCOM are the easiest and most common way to achieve this goal.
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  • Equipment to solve Power Quality problems:Harmonics.

    Equipment to solve Power Quality problems:Harmonics.

    Active filters are the recommended solution for harmonic mitigation, thanks to their flexibility and high correction performance. Alternative approaches could involve passive filters, multi-pulse arrangement transformers or harmonic correction at the equipment level (for example, by integrating harmonic filtering into variable speed drives). ZDDQ Active Power Filter can solve this probelm perfectly
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  • Equipment to solve Power Quality problems: Transients.

    Equipment to solve Power Quality problems: Transients.

    Transient voltage surge suppressors are the best option for protecting against transients in a power system.
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  • why focus on power quality improvement

    why focus on power quality improvement

    A large interest has been focused on power quality domain due to: disturbances caused by the non-linear loads, Increase in the number of electronic devices and growth of renewable energy sources. Power quality measures the efficiency of electric power transmitted from generation to the industrial, domestic and commercial consumers. At least 50% of power quality problems are of voltage quality type.
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  • Static Var Generator (SVG) Benefits

    Static Var Generator (SVG) Benefits

    Static Var Generator (SVG) 1. New approach to power factor correction and load balancing 2. Fast response time < 5 ms, with dynamic reaction time less than 10 µs 3. Precise compensation after compensating the target power factor can reach a value of unity 4. Capable of both inductive and capacitive compensation and will avoid under and over compensation issues 5. Minimal loss, better energy efficiency, long-term safe and reliable operation 6. Modular design, compact structure, small footprint, simple operation, easy maintenance
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  • How much kVAR do I need?

    How much kVAR do I need?

    The unit for rating power factor capacitors is a kVAR, equal to 1000 volt-amperes of reactive power. The kVAR rating signifies how much reactive power the capacitor will provide. Sizing capacitors for individual motor loads To size capacitors for individual motor loads, use Table 3 on the following page. Simply look up the type of motor frame, RPM, and horsepower. The charts indicate the kVAR rating you need to bring power factor to 95%. The charts also indicate how much current is reduced when capacitors are installed. Sizing capacitors for entire plant loads If you know the total kW consumption of your plant, its present power factor, and the power factor you’re aiming for, you can contact with us.
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  • Consider the particular needs of your plant

    Consider the particular needs of your plant

    When deciding which type of capacitor installation best meets your needs, you’ll have to weigh the advantages and disadvantages of each and consider several plant variables, including load type, load size, load constancy, load capacity, motor starting methods, and manner of utility billing. Load type If your plant has many large motors, 50 hp and above, it is usually economical to install one capacitor per motor and switch the capacitor and motor together. If your plant consists of many small motors, 1/2 to 25 hp, you can group the motors and install one capacitor at a central point in the distribution system. Often, the best solution for plants with large and small motors is to use both types of capacitor installations. Load size Facilities with large loads benefit from a combination of individual load, group load, and banks of fixed and automatically-switched capacitor units. A small facility, on the other hand, may require only one capacitor at the control board. Sometimes, only an isolated trouble spot requires power factor correction. This may be the case if your plant has welding machines, induction heaters, or DC drives. If a particular feeder serving a low power factor load is corrected, it may raise overall plant power factor enough that additional capacitors are unnecessary. Load constancy If your facility operates around the clock and has a constant load demand, fixed capacitors offer the greatest economy. If load is determined by eight-hour shifts five days a week, you’ll want more switched units to decrease capacitance during times of reduced load. Load capacity If your feeders or transformers are overloaded, or if you wish to add additional load to already loaded lines, correction must be applied at the load. If your facility has surplus amperage, you can install capacitor banks at main feeders. If load varies a great deal, automatic switching is probably the answer. Utility billing The severity of the local electric utility tariff for power factor will affect your payback and ROI. In many areas, an optimally designed power factor correction system will pay for itself in less than two years.
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  • How can I select the right capacitors for my specific application needs?

    How can I select the right capacitors for my specific application needs?

    Advantages of individual capacitors at the load: • Complete control; capacitors cannot cause problems on the line during light load conditions • No need for separate switching; motor always operates with capacitor • Improved motor performance due to more efficient power use and reduced voltage drops • Motors and capacitors can be easily relocated together • Easier to select the right capacitor for the load • Reduced line losses • Increased system capacity Advantages of bank installations at the feeder or substation: • Lower cost per kVAR • Total plant power factor improved—reduces or eliminates all forms of kVAR charges • Automatic switching ensures exact amount of power factor correction, eliminates over-capacitance and resulting overvoltages
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  • How much can I save by installing power capacitors?

    How much can I save by installing power capacitors?

    Power capacitors provide many benefits: • Reduced electric utility bills • Increased system capacity • Improved voltage • Reduced losses
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  • What can I do to improve power factor?

    What can I do to improve power factor?

    You can improve power factor by adding power factor correction(PFC) capacitors to your plant distribution system. When apparent power (kVA) is greater than working power (kW), the utility must supply the excess reactive current plus the working current. Power capacitors act as reactive current generators. Solutions of power factor correction:SVC,TSC,SVG,STATCOM
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  • Should I be concerned about low power factor?

    Should I be concerned about low power factor?

    Low power factor means you’re not fully utilizing the electrical power you’re paying for. Take an exapmle from the triangle relationships 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 to produce 100 kW. Another way to look at it is that at 70% power factor, it takes 35% more current to do the same work. ZDDQ Solutions: Low voltage capacitor banks;Medium voltage capacitor banks;low voltage static var generator;medium voltage STATCOM. 
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  • Practical calculation of reactive power

    Practical calculation of reactive power

    Please refer to the following calculation examples Type of circuit Apparent power S (kVA) Active power P (kW) Reactive power Q (kVAr) Single phase (Ph + N) S = V x I P = V x I x cos φ Q = V x I x sin φ Single phase (Ph + Ph) S = U x I P = U x I x cos φ Q = U x I x sin φ Example:                                   5 kw load Cos φ= 0.5 10 kVA 5 Kw 8.7 kVAr Three-phase (3Ph or 3Ph+N) S = √3 x U x I P = √3 x U x I x cos φ Q = √3 x U x I x sin φ Example of Motor with               Pn = 51kW cos φ = 0.86                         efficiency = 0.91 65 kVA 56 kW 33 kVAr Calculations in the three-phase example were as follows: Pn = power supplied to the rotary axis = 51 kW P = active consumed power = Pn/ρ = 56 kW S = apparent power = P/cos φ = P/0.86 = 65 kVA Hence: Q = √(S² - P²) = √(65² - 56²) = 33 kVAr The average power factor values for various loads are given below.
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