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From Wikipedia

Electric power

Electric power is the rate at which electrical energy is transferred by an electric circuit. The SI unit of power is the watt.

When electric current flows in a circuit, it can transfer energy to do mechanical or thermodynamic work. Devices convert electrical energy into many useful forms, such as heat (electric heaters), light (light bulbs), motion (electric motors), sound (loudspeaker), information technological processes (computers), or even chemical changes. Electricity can be produced mechanically by generation, or chemically, or by direct conversion from light in photovoltaic cells, also it can be stored chemically in batteries.

Mathematics of electric power

Circuits

Electric power, like mechanical power, is represented by the letter P in electrical equations. The term wattage is used colloquially to mean "electric power in watts."

Direct current

In direct current resistive circuits, electrical power is calculated using Joule's law:

P = VI \,

where P is the electric power, V the potential difference, and I the electric current.

In the case of resistive (Ohmic, or linear) loads, Joule's law can be combined with Ohm's law (I = V/R) to produce alternative expressions for the dissipated power:

P = I^2 R = \frac{V^2}{R},

where R is the electrical resistance.

Alternating current

In alternating current circuits, energy storage elements such as inductance and capacitance may result in periodic reversals of the direction of energy flow. The portion of power flow that, averaged over a complete cycle of the AC waveform, results in net transfer of energy in one direction is known as real power (also referred to as active power). That portion of power flow due to stored energy, that returns to the source in each cycle, is known as reactive power.

The relationship between real power, reactive power and apparent power can be expressed by representing the quantities as vectors. Real power is represented as a horizontal vector and reactive power is represented as a vertical vector. The apparent power vector is the hypotenuse of a right triangle formed by connecting the real and reactive power vectors. This representation is often called the power triangle. Using the Pythagorean Theorem, the relationship among real, reactive and apparent power is:

\mbox{(apparent power)}^2 = \mbox{(real power)}^2 + \mbox{(reactive power)}^2

Real and reactive powers can also be calculated directly from the apparent power, when the current and voltage are both sinusoids with a known phase angle between them:

\mbox{(real power)} = \mbox {(apparent power)}\cos(\theta)
\mbox{(reactive power)} = \mbox {(apparent power)}\sin(\theta)

The ratio of real power to apparent power is called power factor and is a number always between 0 and 1. Where the currents and voltages have non-sinusoidal forms, power factor is generalized to include the effects of distortion.

In space

Electrical power flows wherever electric and magnetic fields exist together and fluctuate in the same place. The simplest example of this is in electrical circuits, as the preceding section showed. In the general case, however, the simple equation P = IV must be replaced by a more complex calculation, the integral of the cross-product of the electrical and magnetic field vectors over a specified area, thus:

P = \int_S (\mathbf{E} \times \mathbf{H}) \cdot \mathbf{dA}. \,

The result is a scalar since it is the surface integralof thePoynting vector.



From Yahoo Answers

Question:here in our place we have two types of power voltage supplying our community. one is three phase 220v / 60 hz.the line to ground or neutral voltage is 110v and phase to phase is 220v. the other supply is 3 phase 380v. the voltage from line to neutral is 220v and line to line is 380v. Now according to my electrical instructor that the formula of computing total load in ampere for three phase supply is.. I = total load in watts times power factor divided by square root of 3 times 220V.. question is where did 220v came from.. what about the 3 phase 380v or the 3 phase 220v power supply.. please help me

Answers:For 3-phase calculations: P = sqrt(3) V I pf where P = 3-phase power load, watts V = Line-to-line voltage, volts I = Line current, amperes pf = power factor The 380/220 is indeed a 3-phase system because 380/1.732 = 220 But the 220/110 is not a 3-phase system, 220/1.732 is not equal to 110. The secondary winding (220 volts) of the single-phase distribution transformer is center-tapped to produce 220/110 volts supply.

Question:We have to improve power factor of our industrial installation thats why question is been asked.

Answers:Power factor is a calculation of actual wattage and designed wattage....if the motor is a 30 hp and the wattage is P=i x e x1.732 then the designed wattage is 22,680...if you put an amp meter on it and you get a reading that when substituted into your equation the wattage is 27,999 the pf is .81... to correct this you need to stabilize the voltage with a PF-CC.. here is an example Let's take an example. A 3/4 HP electric motor has a power factor of .85. The nameplate current is 10 Amps at 115 Volts, or 1150 Volt Amps. Apparent power = 1150 Volt Amps Active power (P) = .85 * 1150 = 977.5 Watts Reactive Power (Q) = sq rt(1150^2 - 977.5^2) = 605 VAR this is for single phase so don't forget the sq rt of 3... this example uses a .6 kVar cap some severe examples can use up to 20kvar...Wesco is a leading supplier and will send you information on their equipment... there ya go from the from the E..

Question:question is calculate from the following given information the maximum number of lamps that may be installed according to regulations -10kva/220 v single phase transformer with a power factor of 0.95 -Floor area 399.75 m2 under cover - Two electric water heaters of 1Kw each -One electrical oil heater of 500 w -One cooking unit at 3kW -One electric single phase motor at 0.75 KW floor area is 399 5 kw for the 1 st 100 m2 and 1kw of every 100m2 = 8 KW two electrical heaters 1KW x 2 = 2 KW one oil heater 500 watt = 500 W one cooking unit 3Kw = 3KW one motor at 0.75 KW = 750 W + ------------------ 14 250 WATT What is the answer and how and why. I have 3 different questions based like this one but the other 2 questions total wattage is 9400 watt which is close to the 9500 watt considered with the power factor of 0.95 and a lamp is 60 watt each according to regulations can anyone explain and give me the answer so that i can confirm it with mine and explain. cause the electrical code or book SANS 10142 does not explain it.

Answers:Your home is about 3600 square feet (I use this for the following example). The National Electrical Code here in the USA, uses an example for a single family homes based on 1500 square feet. The NEC calculates the 1500 square feet home will require 18.6 KVA. That compares with the 10 KVA in your text above. Based on this, either the NEC is generous with their estimate, or you are very conservative with your 10 KVA transformer. This is just food for thought. I don't feel qualified to comment on your comments and sizing parameters. Regards.

Question:The answer should be to the point.For Eg: For a connected load of 268KW,3phase supply,length of the cable is 221mtrs 1.what will be the size of the cable? 2.What will be the voltage drop for 1KMs of the same cable? 3.Please send me the formula and factors to be considered also?

Answers:size (cable thickness, gauge) and voltage drop are interrelated. Increase the size and decrease the voltage drop, as anyone with more than a passing interest in electricity knows. So you have to pick a size, then calculate the voltage drop, or pick a voltage drop and calculate a size. In your example, the number of phases are irrelevant. The load is also irrelevant. What is needed is the load and the voltage, so that the current can be calculated. You tell us the load (268kW) but not the voltage, so we cannot calculate the current, and thus can't calculate voltage drop, and thus cannot calculate wire size. Bottom line, not enough data. need line voltage and allowed voltage drop. ______________________ I'm not going to write any formula, as all I'll use is Ohm's law. So I'll make up an example. You have a 1000 volt line, with a load of 10 amps, 1000 meters long.. (one phase or 3 phase, doesn't matter) and want to keep the voltage loss to 10%. 10% of 1000 volts is 100 volts max drop. with 10 amps of current, max resistance is 100 volts divided by 10 amps or 10 ohms. With 1000 meters of line, the max resistance is 0.01 ohm per meter. Just take that number and plug it into a wire table in a reference book, or use an online calculator to look up the wire gauge. In this case #15 wire Copper is 0.0104 ohms/meter.

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