There are two types of power - real power and reactive power. Real power is power that is used, such as the power used to light a light bulb. Reacitve power is power that is held and released by a reactive element (capacitor or inductor), thus is not actually used to do any work. The reactive elements cause a phase shift between voltage and current, which manifests itself as a change in power factor.
Power companies must supply both reactive and active power. Total power is equivalent to sqrt( Reactive^2 + active^2). Not only do they need to supply the reactive power, their equipment must be sized to handle a larger total power. Reactive power is generated by installing extra equipment - capacitor banks or inductor banks - or by running generation in such a way that more reactive power is created (this will lower the power plants' real power output).
Depending on who the customer is, they may not be billed for the reactive power, thus the oversizing of equipment, and the supplying of reactive power does not generate any revenue (but costs them). If power factor is bad enough, the power company will lose money, and thus require power factor correction, or will bill on both real and reactive power usage.
It's simple economics (money)!
I assume that you are asking how to calculate the 'value' of a capacitor? Well, it depends what it is used for. If, for example, it is used to improve the power factor of a load, then it is first necessary to determine what the load's existing reactive power is; then, you need to know what reactive power is necessary with the power factor at its desired value; finally you need to difference between the actual and the desired values of reactive power -and this will be the necessary value for the capacitor. Power factor correction capacitors are rated in reactive volt amperes, not farads.
When you need to calculate the estimated load or load current during initial design phase or so, you need to assume a value for power factor which is realistic. 0.95 is a realistic value.
Power factor improvement means the power factor has been made closer to 1. If a power station delivers energy to places where the load is significantly reactive (contains capacitative or inductive components) it is expensive for them because they deliver current which actually doesn't do any work. The cosine of the phase angle between voltage and current is the power factor.
The capacitor size for a 5 kVA generator typically depends on the power factor and the specific application. A general rule of thumb is to use about 0.5 to 1 microfarad (μF) of capacitance per kVA for power factor correction. Therefore, for a 5 kVA generator, you might consider a capacitor size ranging from 2.5 μF to 5 μF. However, it's essential to consult with a professional or refer to the generator's specifications for precise requirements.
It can supply 2.5 kW to a load with a power factor of 1. Otherwise the power available must be multiplied by the power factor which is always less than 1. If uncertain assume 0.8 which makes it 2 kW.
To calculate the correction factor for a freezer, you need to compare the setpoint temperature of the freezer to the actual temperature inside. The correction factor is the difference between the setpoint and actual temperature. Adjust the setpoint temperature by this correction factor to ensure the freezer maintains the desired temperature consistently.
I assume that you are asking how to calculate the 'value' of a capacitor? Well, it depends what it is used for. If, for example, it is used to improve the power factor of a load, then it is first necessary to determine what the load's existing reactive power is; then, you need to know what reactive power is necessary with the power factor at its desired value; finally you need to difference between the actual and the desired values of reactive power -and this will be the necessary value for the capacitor. Power factor correction capacitors are rated in reactive volt amperes, not farads.
Most of the commercial and Industrial installations have large electrical loads which are severely inductive in nature, such as motors, large machines, air conditioners, drivers etc. Which results in a severely lagging power factor. This means loss and wastageof energy and heavy penalties by electricity boards. In case of fixed loads this can be taken care by manual switching of capacitors.However in case of rapidly varying and scattered loads it becomes difficult to maintain a high power factor by manually switching on/off the capacitors in proportion to variation of load within an installation. This drawback is overcome by using an APFC panel (Automatic Power Factor Correction Panel) which not only maintains a high power factor but also eliminates the eliminates the need for constant manual intervention.The ratio of Active Power to Apparent Power is called Power Factor (PC) :Power Factor = Active Power/ Apparent Power = KW/KVAIf you want to know more about PFC panels please visit www.powerfactorcorrector.com or email me on info@urjaghar.com.
Can somerone please explain how to do a factor fireworks problem? I am completely lost on how they work. I need two different factor fireworks of 64.
A poor power factor that occurs when a load draws a lagging current is corrected by the addition of a parallel capacitor to draw a leading current. If the load draws a leading current, an inductor in parallel would be used. In each case the added component is chosen to draw the same number of VARs as the original load. Some loads like TV sets have a poor power factor that arises from harmonic currents in the load, and this method is not amenable to correction by an added component. Power-factor correction is not normally carried out in domestic supplies.
To measure power, you need the Voltage, and Current. In the case of AC you need the number of phases, and the power factor. Once you have these, you can find the proper formula.
When you need to calculate the estimated load or load current during initial design phase or so, you need to assume a value for power factor which is realistic. 0.95 is a realistic value.
In order to answer this question, you need to know (a) its efficiency, and (b) its power factor at full load. 11 kW (not 'Kw') is its output power, so you need to know its efficiency in order to determine its input power. Then, because, for a three-phase system, power is equal to 1.732 times the product of the line voltage, line current, and power factor, you also need to know its power factor.
ratio between true power and apparent power is called the power factor for a circuit Power factor =true power/apparent power also we conclude PF=power dissipated / actual power in pure resistive circuit if total resistance is made zero power factor will be zero
Power factor improvement means the power factor has been made closer to 1. If a power station delivers energy to places where the load is significantly reactive (contains capacitative or inductive components) it is expensive for them because they deliver current which actually doesn't do any work. The cosine of the phase angle between voltage and current is the power factor.
Watts = Volts x Amps x Power Factor To make your calculation you need to plug-in the Voltage and Power factor. The Power Factor is a value from zero to one with one being a pure resistive load.
Watts = Voltage x Current x Power Factor 1000 Watts = 1 Kilowatt Therefore, you need to know current and Power Factor to answer your question.