How do you calculate 3 phase kva to kw?

Answer 1

First determine what voltage is needed for phase to phase. Where you work on or where the gears are coming from would help to answer this question. In Southeast Asia, where power is 220~240 volts, a 3 phase system would be 415v phase to phase. In North America, a 3 phase system would be 208v phase to phase. So now you know the voltage that you are working with on a 3 phase system. Next use this formula, kVA x 1000 / voltage required / 1.73 = amp 3 phase Example, you have a 60kVA generator. And you working in North America, which is 208v phase to phase. 60 kVA x 1000 / 208 v / 1.73 = 166amp 3 phase Written by: Rony Xu Hope this helps!

Answer 2

In a three phase system, power P = U * I * sqrt(3) * cos(phi), where U is the voltage (rms), I the amperage (rms), sqrt(3) is the square root of three, and cos(phi) is cosine of phi, where phi is the angle between U and I. For pure resistive load, phi is zero (i.e. cos(phi) = 1), making the calculation pretty easy in this case. For inductive or capacitive loads such as motors or flourescent lights, phi moves to different values and must be estimated, measured, or calculated.

Find the Amperage of an 7200w inductive (i.e. motors, pumps, fluorescents etc.) load on a 415 Volt,

3 phase

branch circuit.

I = 7200 / (415 x 1.732) = 10.01 Amps per phase.

In practical we have to consider P.F (i.e cos(phi) = 0.95) for inductive loads.

I = 7200 / (415 x 1.732*0.95) = 13.83 Amps per phase.

Answer 3

Use this formula; KW= (volts(avg) x amps(avg) x power factor x 1.732) divided by 1,000. If you don't have a power factor use 0.9. Also make sure you use the larger voltage of the system voltages, ex) if it's 480/277 use 480 in the formula...208/120 use 208...240/120 use 240. If possible use a volt meter to read the actual voltage between each phase. Use the average of the three voltage readings (V1+V2+V3 / 3). Take the amperage reading from each phase and average them (A1+A2+A3 / 3). To find KWh, just multiply KW by the amount of hours your system will be serving the load.

Example:

An operating three phase motor has voltages measured with a voltmeter on each phase of 453, 458, and 461 volts, amperage measured on each phase with an ammeter are 14.1, 13.9, and 13.8 amps, power factor was measured as 0.82.

(453 + 458 + 461) / 3 = 457V

(14.1 + 13.9 + 13.8) / 3 = 13.9A

(457V x 13.9A x .82pf x 1.732) / 1000 = 9.02 Kwatts

To calculate the kWh for one days use:

9.02 Kw x 24hrs = 216.48 kWh

If you are serving unbalanced loads and the amps vary between phases, you can calculate each phase separately. In this case, you are calculating three separate single-phase values, so the 1.73 multiplier is not used, and you use the phase to neutral voltage.

Example:

Phases measured with a voltmeter at 272, 264.4, and 266.2 volts (A-N, B-N, C-N), amperage measured on each phase with an ammeter are 10, 25, and 16 amps, power factor is assumed as 0.9:

(272V x 10A x .9pf) / 1000 = 2.45kw

(264.4V x 25A x .9pf) / 1000 = 5.95kw

(266.2V x 16A x .9pf) / 1000 = 3.83kw

2.45 + 5.95 + 3.83 = 12.23kw total

To calculate the kWh for one days use:

12.23 Kw x 24hrs = 293.52kWh (IF the load remains constant!)

This method also works for balanced loads, it just takes longer and is more work, but you get the same answer (actually slightly more accurate, since you are not averaging) as the first method!

Modification of the example above for an accurate total kw:

If you are serving unbalanced loads and the amps vary between phases, you can calculate each phase separately, however you must still average the phases, and multiply by the square root of 3.

Example:

Phases measured with a voltmeter at 272, 264.4, and 266.2 volts (A-N, B-N, C-N). We will assume 480v phase to phase. Phase to neutral voltages will not be used in the calculation. Amperage measured on each phase with an ammeter are 10, 25, and 16 amps, power factor is assumed as 0.9:

(10 + 25 + 16) / 3 x 480 x 1.732 x 0.9 / 1000 = 12.71 kw total

To calculate the kWh for one days use:

12.71 Kw x 24hrs = 305.26 kWh (IF the load remains constant!)Answer

ANSWER">ANSWERThe above answers are confusing because they use incorrect terminology. The three energised conductors supplying a three-phase load are called LINE CONDUCTORS, which is why the voltages across them are called LINE VOLTAGES and the currents through them are called LINE CURRENTS. The PHASES are represented by the three loads, and the voltages across individual loads, and the currents through those individual loads are, therefore, called PHASE VOLTAGES and PHASE CURRENTS. If there is a neutral conductor, then the voltage measured between any LINE CONDUCTOR and the NEUTRAL CONDUCTOR are called PHASE VOLTAGES.

So if, in your question, you are referring to the currents flowing in the supply conductors, then your question should read, 'LINE CURRENTS', not 'phase currents'! Incidentally, from the practical point of view, it's quite difficult to access the phases of a machine in order to measure phase current, whereas it's very easy to measure a line current.

Having got the terminology out of the way, let's answer the question. For a BALANCED LOAD (i.e. each load being identical in all respects), the total power is given by the following equation:

P = 1.732 UL IL xpower factor

...where the suscript, L, represents LINE quantities.

If the loads are unbalanced, you cannot use this equation. Instead, you have to measure the phase currents (IP) -i.e. the currents in the individual loads or PHASES, and the voltages across them (UP) -i.e. the phase voltages. To find the total power, you must apply the following equation to each phase, and add them together:

P = UP IP x power factor.

Finally, to find the energy consumed, in kilowatt hours, you must multiply the total power (expressed in kilowatts) by the time for which the load is operating (expressed in hours).

Answer 4

Instantaneous power dissipationIn a one phase circuit the instantaneous power dissipated by a resistor is the Voltage, V [V], over the resistor times the current, I [A], through the resistor. Power, P, is measured in Watt [W]. So in symbols U x I = P with units [V]x[A]=[W].

Instantaneous power dissipation with an alternating voltage

For an alternating voltage the voltage can be written as V(t)=Vp x cos(t)

Where Vp is a constant value. When the Voltage V(t) is applied over a fixed value resistor the current through the resistor will also alternated. By ohm's law the current will be I(t)=V(t)/R = (Vp/R) x cos(t).

the instantaneous power dissipated will be P(t)=V(t) x I(t) = V(t) x (V(t)/R) [W]

So also the power will alternate.

Metric system

In the metric system the k stands for kilo which means 1000.

So 1 kW means 1000. So far calculating W into kW you can simply divided by 1000.

Three phase system

For a three phase system Ohm's law is still valid but is a little bit harder to calculate. The instantaneous power will be the sum of the dissipation in all three phases.

The all line voltage are sinusoidal but have a phase difference of 120 degrees = (360 degrees/3). When working with radial the phase shift is (2 x pi/3).

The line voltage are named A,B and C and can be written as:

Vla(t) = Vp x cos(t)

Vlb(t) = Vp x cos(t + 2 x pi)

Vlc(t) = Vp x cos(t - 2 x pi)

Star connection

When all lines are connected in a star connection and all load (resistors) are the same. The dissipation can be written as:

Pa(t) = Va(t) x (Va(t)/R)

Pb(t) = Vb(t) x (Vb(t)/R)

Pc(t) = Vc(t) x (Vc(t)/R)

The total power dissipated is then the sum:

P(t) = Pa(t) + Pb(t) + Pc(t)

when using some mathematics this equation can be simplified to:

P=(3/2) x Vp x (Vp/R)

N.B. that the power is now constant and independent of the time and therefore also independent of the frequency of the cosine.

Answer 5

current = 1000 x kilowatts divided by ( 1.73 x voltage x power factor )

Answer 6

You don't.

Volts and amps are separate entities. There are no direct translations.