Confusion on CI Video - Power= VI (KVA or Watts)

[Owism]

Hello.

Regarding the CI video below: the example he finds P and gives it the Unit watts. Whereas the previous video ( https://www.youtube.com/watch?v=vxePVlQp4Mw ) he finds P and gives it the unit KVA. Is it because this example is finding line power and the previous is finding phase power? That doesnt make sense to me though. I thought KVA is units for S = VI which has a power factor cmoponent to it? S=Pcos(feta) 

 

[TNPE]

18 minutes ago, Owism said:

Hello.

Regarding the CI video below: the example he finds P and gives it the Unit watts. Whereas the previous video ( https://www.youtube.com/watch?v=vxePVlQp4Mw ) he finds P and gives it the unit KVA. Is it because this example is finding line power and the previous is finding phase power? That doesnt make sense to me though. I thought KVA is units for S = VI which has a power factor cmoponent to it? S=Pcos(feta) 

P=VIcos(theta)  (real power...i.e. usable power delivered to a load)  *Note: theta is the power factor angle, also inherently known as the angle between voltage and current, in which the current leads or lags the voltage, or even the impedance angle

S=VI (apparent power)

S=P+jQ (complex power...also, S=VI*)

Q=VIsin(theta)  (reactive power)

If you understand these equations and the relationships between them all, coupled with a thorough understanding of how these are applied and used for 3-phase and single phase applications, you should be good.  Not to sound blunt, but this should be the gimme stuff on the exam, if you're asked about it at all.

 

[TNSparky]

P=VIcos(theta)  (real power...i.e. usable power delivered to a load)  *Note: theta is the power factor angle, also inherently known as the angle between voltage and current, in which the current leads or lags the voltage, or even the impedance angle

S=VI (apparent power)

S=P+jQ (complex power...also, S=VI*)

Q=VIsin(theta)  (reactive power)

If you understand these equations and the relationships between them all, coupled with a thorough understanding of how these are applied and used for 3-phase and single phase applications, you should be good.  Not to sound blunt, but this should be the gimme stuff on the exam, if you're asked about it at all.

Yeah. This is basic stuff for power. If you're not familiar with it, pick up the Chapman machines book and the Blackburn protective relating book. They go over that stuff. It's not just for 3 phase applications. Single phase has complex (apparent) power too. 

 

[TNSparky]

Sorry, that should read "Chapman Protective Relaying"

 

[TNSparky]

Ugghhh....senior moment. "BLACKBRUN Protective RELAYING" 

Fat fingers....Lol.

 

[TNSparky]

Blackburn*

I give up.    :12:

 

[Owism]

Thanks,

I get it this is fundamental stuff, thats why I question why he uses different units for the same equation in the videos. I fully understand the P,Q,S triangle but also know that some engineers use P and S interchangably. P has units of watts, S has units of KVA , they should really keep it separate... So to clarify, why does he use P=IV and call it watts in the first video then P=IV and call it KVA in the second video... 

 

[KatyLied P.E.]

Actually, after taking a quick look at it, I also think he used the wrong unit.  If you look at some of the listed comments on YouTube at least a couple of folks also think that.  Apparent Power (S) is the vector representation and should be in KVA.  He states that he is solving for power (P) which will always be in W.  That's just taking a quick look at just his units.  I'm about to head out but figured I'd give a stab.  S= Square root of (Psq + Ssq).  It is a vector representation of power and would include some mention of the angle and/or magnitude of real and reactive power.  I welcome debate on it my brain is frazzled.  I'm in the middle of work related power outages believe it or not.  Needed something to take my mind off that. 

 

[Owism]

Thanks. Yes, just a correction on your statement:  S = sqrt ( P^2 + Q^2 ) 

 

[KatyLied P.E.]

My bad!  You are welcome and you are correct. 

 

[TNPE]

Units would be wrong; however, with that said, the only time P=S=VI is at unity power factor.  Much like Q=S=VI at 90 degrees, which is a purely reactive load.  Though these quantities are the same at these given conditions, the units are not the same.  100W is not the same as 100VA, just as 100VARs is not the same as 100VA.  From a quantifiable standpoint, they are the same, but from a specific, theoretical view, they aren't.  

I had a much more detailed response written earlier today, but my computer had other ideas and wet the bed.  If you have any other questions, let 'em rip!  That's what this board is for, other than the occasional troll-fest, cause I had to crawl my *** under the bridge a time or four to troll during this grueling wait!!  Best of luck going forward and kick this test's *** in the spring!

 

[Owism]

Thanks TNPE.

One of the folks on the board just sold me his books including the fundamental ones everyone recommends: Grainger, Chapman, Wildi, NESC, and Complex imaginary sets. 

Time to get cranking. Yes I'll keep the questions coming.