Calculations for Three-Phase Power

"will some one tell me, which formula use with which concept..."

It is exactly the "concept" that you are missing. Simply memorizing which "formula" to use for a specific situation does absolutely nothing for you and in fact is a detriment to both you and whoever hired you as an electrical engineer. Having said that, I will try to assist you.

The first step is to understand that average power is the product of RMS phase voltage and RMS phase current and power factor ALWAYS, ALWAYS, ALWAYS. It doesn't matter if the system is three phase or single phase, this is always true. The "3" is used in three phase calculations because there are three phases.

Now look at the relations between:

• phase voltage and line voltage in a wye system
• phase current and line current in a wye system
• phase voltage and line voltage in a delta system
• phase current and line current in a delta system.

Take a good hard look at those. From where the root 3 comes should become clear to you keeping in mind that in the "3VI", the V and I are phase values. I hope in doing a little work on this yourself, you will come to understand the concept.

RMS is .707 and average is .637 of peak to peak or peak sine wave. I am confused.

You say that RMS of voltage combined with RMS of current will give average power and then adjusted by the power factor?

.707=RMS=Effective values of peak or peak to peak, .637=average=volume of peak or peak to peak.

I just had this discussion about calculating the cost of power in a three phase system and started to research it.

I guess I will continue!

I have read the wikipedia and agree that a little understanding goes along way. However, familiarity and confidence goes a long way for those of us who have not worked with 3 Phase interactions in our day to day activities.

That being said, how does wye connection (Vph-ph only) voltage imbalance such as 205 Vac, 208Vbc, and 206Vca coupled with the unbalanced current loading such as 100A ph-A, 90 ph-B, and 105A ph-C. All values are RMS actuals.

What is the total power?

What is the stranded power by phase A and B?

I have attempted to work out the old fashion way but trip myself up several times.

A pspice calculation seem to be consistent with a power meter. Can you shed some light on the calculation method?

Thanks, Olddog

additional idea from me, may be can helpfull

note : this reply for #1.

Total Power = Power A + Power B + Power C

elect ok,

apologies but I can not read the text. could you enlarge and please post again?

thanks

The power that a three phase source can deliver to a load is 1.732 times more efficient then a single phase source of the same voltage.

For example...

Given a 50kW resistive load (a big heater) at 208 volts...

Single phase...

50,000 / 208 = 240.3 amps.

Three phase...

50,000 / (208 * 1.732) = 138.8 amps

So... using the square root of 3 ,or 1.732, is proper for your calculations.

I found it and understand your answer better. Thanks

but its also depends on specification of that heater. because if that is 3 phase heater, we should give 3 phase voltage. (it means that heater has 3 terminal (R,S,T). and if that is 1 phase heater, we should give 1 phase voltage.

for example : heater with specification 50KW, 3ph, 50Hz. 380 Volt. must be connected with 3 phase voltage. CANNOT be connected with 1 phase voltage (220 volt).

so we cannot compare same heater by using 2 different kind of voltage.unless if we had modified internal wiring of that heater by ourself.

Yes, a 3-phase load requires 3 phases; and yes, a single phase load requires a single phase. Beyond that, I really don't know what you are trying to convey.

Regarding the comment by North of 60, his example of a 50KW, 208V, 1_Φ load and a 50KW, 208V, 3_Φ load is perfectly valid. I have seen numerous examples of each type of load. Particularly in commercial HVAC, you will find single phase 208V. There is nothing in the rule book that says you can't take two legs from a 208V, 3_Φ service and connect them to a single phase load.

Yes of course it not an "efficiency" factor. I presented it in a way that was easier to understand.

I agree with everything you've shown here. This is basic stuff! You are right that the original guy should understand the basic concept. Then he would be able to apply that to any situation without having to rely on a specific formula for a specific situation.

Apply Pythagoras' Theorem to the triangles on a phase diagram, and all will become clear.

First of all total power is always the sum of individual power drains regardless of how the loads are connected. Power has a real component (watts) and a reactive component (vars) in the case of AC. In a three phase system total power is the sum of power usage on each phase, hence if power on each phase is VI then 3phase power is 3*VI. If the load is balanced we may calculate the total power using line voltages and current which is more accessible then phase voltages and currents. Since there is a sq. root of 3 factor difference between phase and line voltage and current depending on whether de load is delta or star it follows that total 3phase power equals sq. root of 3 V*I (line values). In other words, we use the sq. root formula for balance loads only. There is also a power factor associated with power due to the type of load which needs to be considered if loads are different.

For my part--THANK YOU, Mark and North--good stuff and I hope to put it to good use.

Hi there,

The complete formula for the active power calculation is:

P = sqrt3 * V * I * cos phi

The apparent power is:

S = 3 * V * I

and the reactive power:

Q = sqrt3 * V * I * sin phi

Wow, very expanded answers, a wide range of detail, and some incorrect terminology and theory for a very simple question.

The reality that both Line voltage and line current are not readily measurable in any system is the foundation for a simple answer to the questioner:

Any time a phase value for either current or voltage is used in the power formula (with an exception to follow) then the relationship of the line to phase value for E or I varies by the value of the tangent of 60 degrees (1.732), or √3.

Further detail and the exception:

In a delta circuit, line voltage = phase voltage and is measured directly across 2 lines, however the current that is measured in either line is shared by 2 phases and the net result varies by 1.732. (√3*V_LI_L)

In a wye circuit, the line current = phase current and is directly measurable, however the line voltage is shared by 2 phases and varies from the phase voltages by 1.732. (√3*V_LI_L)

The exception mentioned to the above statement is that when both values are stated in phase values (wye circuits can be measured this way as can the stinger leg of delta, or if the values are given in class for learning purposes) then the power formula for three phases would be (E1ph * I1ph)+(E2ph * I2ph)+(E3ph * I3ph) for unbalanced systems, and Ephase * Iphase *3 for a balanced system. (3*V_pI_p)

All of the above will also be modified by the cosine of the angle of lag or lead of the current to the voltage (also known as the power factor).

Simple answer to "which formula use with which concept..." is:

Delta, P=(√3)EI(cos Θ), using Line values, √3*240*8.66=3600

Wye, P=(√3)EI(cos Θ), using Line values, √3*208*10=3600, or

Wye, P=(3)EI(cos Θ), using phase values, 3*120*10=3600

Values that will highlight themselves in your studies and be useful to remember are:

cosine 0f 30deg = (.866).

2*.866 = 1.732 = √3 = Tan 60⁰ = Inverse Tan 120⁰

Trigonometry of right triangles and pythagorem's theory applied to phasor diagrams is the language of electricity measurement.

3*VI when Voltage & current are measured from line to neutral (VI of 3 individual phases are summed up)

Sqrt(3)*VI when Voltage & current are measured from line to line.