Explanation of Definitions on Power

[1fastlook]

Would someone that is smarter than me please give me an elementary explanation of how to understand the following:

Power Factor= Real Power / Apparent Power, but the definition of
Real Power = V*I*1.732*PF

You cannot calculate the power factor without knowing the real power, but you cannot calculate the real power without knowing the power factor.

This seems to me much like the old question "which came first the chicken or the egg."

Thanks in advance for your response.

 

[winnie]

The equations that you've given connect real power and power factor, but neither equation says what you are measuring. It is as if you use Ohm's Law: Voltage = Current * Resistance and then write Resistance = Current / Voltage

You need to somehow separately measure real power and apparent power, or apparent power and power factor, or any of the other sets of related measurements, and then you can use the equations to derive the other related terms.

For example, a standard electromechanical KWH meter measures _real_ energy consumption. By timing the rotation of the spinning disk you get a measurement of _real_ power consumption. If you then separately measure volts and amps, and multiply, then you will have a measurement of _apparent_ power consumption. Using your equations, you could then figure out power factor.

In the DC world, power is the product of voltage and current. In the AC world, we generally measure both RMS voltage and RMS current; these are smoothed readings that represent continuously changing values as single numbers. The problem is that if you multiply RMS voltage times RMS current, you still don't know the _real_ power. This is because the _timing_ of the current flow does not necessarily match up with the timing of the voltage. If the two are not perfectly synchronized, then you will find that for a portion of the AC cycle power is flowing from the source to the load, but for a portion of the cycle the power is actually flowing in the reverse direction.

Because of the power flow from load back to source, the net power delivered by a given rms current (at a given voltage) is reduced.

-Jon

 

[hockeyoligist2]

Would someone that is smarter than me please give me an elementary explanation of how to understand the following:

Power Factor= Real Power / Apparent Power, but the definition of
Real Power = V*I*1.732*PF

You cannot calculate the power factor without knowing the real power, but you cannot calculate the real power without knowing the power factor.

This seems to me much like the old question "which came first the chicken or the egg."

Thanks in advance for your response.

I had that same question on my exam. Never figured it out.

 

[Mayimbe]

Would someone that is smarter than me please give me an elementary explanation of how to understand the following:

Power Factor= Real Power / Apparent Power, but the definition of
Real Power = V*I*1.732*PF

You cannot calculate the power factor without knowing the real power, but you cannot calculate the real power without knowing the power factor.

if we know that "t" is the angle between the voltage phasor and the current phasor, then

pf = cos (t)

Apparent Power = S
Real power = P
Reactive Power = Q

S=V*I(conjugate)

S=S*cos(t) + j*S*sin(t)

S=P+j*Q

From the real part we have

S*cos(t)=P

P/S=cos(t)=pf

when you say

Real Power = V*I*1.732*PF

thats P=(V*I*1.732)*pf and we know that

S = V*I*1.732 = V*I*sqrt(3) and

pf=cos(t)

then P=S*cos(t) like we said before.

you are telling the same in the 2 expresions.

do the power triangule, you will see why the power factor is inherently known in this equations

 

[drbond24]

Power Factor= Real Power / Apparent Power, but the definition of
Real Power = V*I*1.732*PF

You just stated the same equation twice because:

Apparent power = V*I*1.732

As with any equation, you can only solve for one varible. You will have to measure either real power or power factor to be able to calculate the other, and that is assuming you know either the apparent power or both the voltage and current.

I had that same question on my exam.

What exam? If you gave the exact question in more detail we could probably be more helpful.

 

[__dan]

Power factor = kW/kVA

Power factor = kW/kVA

Power factor = kW/kVA

kW, or real power, is the rate of doing work, the power or work consumed by the load.

kVA is the component of real power + a component of reactive power. They are a vector sum on a right triangle. For x, y coordinates, if you imagine real power (kW) as an x component, reactive power is the y component, and the vector sum, kVA, is the hypotenuse of the triangle.

Reactive power flow is cause by an imbalance between the capacitive and inductive elements in the circuit, the reactive flow tries to equalize these. If you have a large capacitive element, like a large UPS with capacitor filters, you may see a large (huge) reactive current flow due to imbalance. For loading the UPS use the kW rating and not the kVA. You are allowed a real power load plus an allowance for reactive current flow due to capacitance - inductance imbalance between elements in the circuit.

Your formula "Real Power = V*I*1.732*PF" is not the one I would use.

Understand that kW is flowing to the load and the reactive power component is flowing probably elsewhere, due to imbalance between the capacitive and inductive circuit elements, not necessarily at the load.

Reactive current flow can be large enough in amps to occupy conductor ampacity and OCPD trip settings.

Basically, if power factor, kW/kVA, is less than .9, the equipment is out of spec and calling for service or corrective action.

If you have a big difference between kW and KVA ratings, say 750 kVA 600kW rating on a UPS and want to know if you can load to the kVA rating, you can not. Max load rating is at the kW rating.

 

[hockeyoligist2]

[QUOTE
What exam? If you gave the exact question in more detail we could probably be more helpful.[/QUOTE]

My Journeyman test a few years ago. Detail was the problem, the question had no other details. I copied it word for word on a piece of scrap paper and asked several people to look at it. They all said there is no answer because there is not enough info.

 

[K8MHZ]

[QUOTE
What exam? If you gave the exact question in more detail we could probably be more helpful.

My Journeyman test a few years ago. Detail was the problem, the question had no other details. I copied it word for word on a piece of scrap paper and asked several people to look at it. They all said there is no answer because there is not enough info.[/QUOTE]

What exactly, was the question?

I saw not so much as a single question mark in either your or the OP's post.

 

[K8MHZ]

Power Factor= Real Power / Apparent Power, but the definition of
Real Power = V*I*1.732*PF

Yes, for three phase.

For single phase, leave out the 1.732

Example: 120 volt, 16 amps, PF of .78 will give you 1497.6 watts of real power.

 

[zog]

It is like asking to solve for Y (or X) in Y=X-4

You need more info to solve it.

 

[zog]

This seems to me much like the old question "which came first the chicken or the egg."

.

The rooster of course.

 

[1fastlook]

Thank you all for your replies!

 

[drbond24]

It is like asking to solve for Y (or X) in Y=X-4

You need more info to solve it.

If they are wanting an actual value for Y then you would indeed need a numerical value for X. However, if the purpose is just to check your algebra skills then Y=X-4 is the answer. I suspect that might be what happened here. Perhaps the question was just to solve for the equation that would give you the numerical value of the power factor assuming you already had all of the other variables.

 

[1fastlook]

My question is simply this: How can you calculate the real power when a portion of the formula requires power factor, when you cannot calculate power factor without knowing real power???

Thanks again.

Jay