THE PHYSICS OF... POWER

[FionaZuppa]

no one has refuted
va = volt x amp = joule/coloumb x coulomb/sec = joule/sec = watt
NO ONE

absolutely true, except for one thing, "power" in physics is energy into work, but Q = I²√-1X = imaginary, meaning no power, you only have the real term of amps to account for.

if it were lossless pure jX then Q(t) = ±watts(sin), an oscillation of energy as joules move back and forth w/o loss = no work.

it really depends on where you look. the system in a box, Q is very real, needs to be accounted for in real terms. but draw a box just around jX of the load then Q suddenly looks imaginary because the VA input to load != to the kW output of the load, this VA however is very real to the generator.

 

[mbrooke]

There are none so blind as those who will not see.
VA is volt amperes. You can have VA and no power. Can't you see that??

No such thing. For a half cycle energy is transferred to the load, and then transferred back in the other half (if my timing is correct)


1+(-1)=0

 

[FionaZuppa]

From the perspective of instructing someone who is going to use or employ your information, for example loading on a UPS rated 100 kVA 80 kW where the VA rating is bolded on the equipment labels and marketing promotions but the kW rating is in fine print on the nameplate or specs ...

Would you say VA = watts ?

Is it true or is it false?

True or false?

if V_rms=1000

80kW is associated with 80amps = 80k(VA)_in-phase
the UPS then has 20kVA worth of headroom (capacity, or 20amps) to deal with load Q derived from load jX component.
pure resistive load the UPS can handle a 100kW load

if the UPS was spec'd that way then the UPS was designed around a PF_min
on the source side we only need to know VA rating and V_rms, is that good enough for the load? the load kW & PF ratings will tell us, and vice-versa.

 

[__dan]

pure resistive load the UPS can handle a 100kW load

No, I don't believe so but I am not the one who makes that finding. You would have to call the manufacturer, Liebert Emerson or whomever you prefer, and ask them. If the UPS is rated 100 kVA and 80 kW, which is a typical rating the kW is .8 the kVA, may you load it to 100 kW resistive?, as modern IT loads are pf corrected.

If it was a 100 kW rated machine, I believe the nameplate would say so. For me clearly, it is a code violation to load more than 80 Kw on the unit per nameplate, even as it says 100 kVA, 80 kW.

The UPS is three devices in series, rectifier, DC bus, inverter. Only the output section would see any pf. My guess is the manufacturer will say it's an 80 kW machine with an allowance for pf down to .8 at the output side.

 

[winnie]

Prove this to be false
va = volt x amp = joule/coulomb x coulomb/sec = joule/sec = watt

It cannot be proven. The statement is false by the definition of the terms as used.

The below statement is true, for DC and for instantaneous values:
1V * 1A = 1W

If you take the integral of volts * amps you will get the energy delivered to the load over the integration period. Divide by time to get average power.

However the term 'Volt Ampere' is commonly used to describe the product of V(rms) * A(rms).

In general, V(rms) * A(rms) != (the integral of V(t) * A(t))/(integration time)

In other words, V(rms) * A(rms) is in general not equal to the rate that energy is being delivered to the load. V(rms) * A(rms) is not equal to power.

Since 'Volt Ampere' is used to describe something that is not in general equal to power, 'Volt Ampere' is _not_ power. But as I said, that is the common usage of the term, not a proof.

Volt Ampere is used to describe something related to power, in particular it is used to describe things like circuit capacity used, or losses in the circuit wires.

On a given circuit, 1000VA will have roughly the same losses in the wires, no matter what the actual power delivered to the load. For purpose of sizing wires, transformers, generators, etc. VA is equivalent to watts. But for the purpose of sizing batteries, generator prime movers, and other actual sources of energy, VA is different.

-Jon

 

[FionaZuppa]

No, I don't believe so but I am not the one who makes that finding. You would have to call the manufacturer, Liebert Emerson or whomever you prefer, and ask them. If the UPS is rated 100 kVA and 80 kW, which is a typical rating the kW is .8 the kVA, may you load it to 100 kW resistive?, as modern IT loads are pf corrected.

If it was a 100 kW rated machine, I believe the nameplate would say so. For me clearly, it is a code violation to load more than 80 Kw on the unit per nameplate, even as it says 100 kVA, 80 kW.

The UPS is three devices in series, rectifier, DC bus, inverter. Only the output section would see any pf. My guess is the manufacturer will say it's an 80 kW machine with an allowance for pf down to .8 at the output side.

its 80kW with expected PF >= 0.8. if the PF is say 0.9 then so be it, the UPS would provide max 80kW and would just do less work (0.9PF - 0.8PF) to deliver that 80kW. simple math on this one.

70kW .75PF yields max VA
72kW .75PF exceeds max VA

On a given circuit, 1000VA will have roughly the same losses in the wires, no matter what the actual power delivered to the load. For purpose of sizing wires, transformers, generators, etc. VA is equivalent to watts. But for the purpose of sizing batteries, generator prime movers, and other actual sources of energy, VA is different.

-Jon

what?? that doesnt make sense. the losses are highly dependent on real power that the load uses. how bad the impedance mismatch is depends on PF which dictates kVAr

my 1kVA xfrmer delivers 10watt to a load with PF 0.9
my 1kVA xfrmer delivers 100watt to a load with PF 0.9

condition 2 will waste more power than condition 1, my 1kVA xfrmer stayed the same

 

[mivey]

Using the definition of power, the time averaged case is in the definition and the instantaneous case is not. 6c, http://www.merriam-webster.com/dictionary/power

Looked like it was there to me. Energy is transferred back and forth in the instantaneous/reactive case.

To me, winnie has had the best posts on the topic and covered it quite well.

Distance and altitude have the same unit dimension, but would you say travelling 10 meters in altitude is the same as travelling 10 meters in distance? The effect and meaning is completely different.

Not quite as it would be back and forth rather than up and down. Let distance be energy (quantity) and mph be power (rate).

For active power we have a net distance traveled and the person walks to a destination, say 4 miles in 2 hours or an average speed of 2 mph.

For reactive power the person paces back and forth at 2 mph with no net distance traveled. There is definitely a rate involved so we have power, just not active power.

VA would be like the person's top speed or something they could run and is a possible power but may not be something they actually did.

Active and reactive power are definitely happening as energy is moving at some rate. One is average and the other instantaneous but both are power. Both are not active power.

 

[mivey]

One is power. The other isn't. No matter what you call it.

One is active power, the other isn't.

 

[mivey]

VA is not a measure of power.

But it can be a quantity representing power that could be (potential power if you will).

 

[FionaZuppa]

One is active power, the other isn't.

true, but only if you are looking at the load.
the kVAr does no work at the load and therefore is not power by the real definition of power, yet it can in fact still hold unit of watts. we discussed this 32 pages back.

to push amps to supply kVAr and kW you need to do work on the gen side for kVAr+kW. kVAr "power" does not traverse the gen & transmission w/o loss, its wasted power. we also saw back 35pages or so that in a lossless pure jX look at gen & load the kVAr would cost poco nothing, input to gen would = load kW, but the work profile on gen would be in-phase with amps and would fluctuate as function of sin, the avg work thus would = load kW. why? because input work needs a tad more than avg to push, but a tad less than avg as jX does kickback. exact same model for a perfect ideal lossless physical spring.

But it can be a quantity representing power that could be (potential power if you will).

any energy stored by jX is exactly potential energy, which is in fact wasted at some point. potential energy cannot be power by true definition of power until such time the potential energy is converted to work (aka power).

 

[FionaZuppa]

someone mentioned you can go gool searching to check my statements, this should be easy to find

In real-world environments where resistors, inductors and capacitors are all combined within complex machines, the power that is wasted and not used to do work on the load is represented by the Reactive Power

might as well just re-label Q as "Wasted Power" :thumbsup:
get it, not used to do work on the load, but wasted. or, imaginary at the load, real on the poco side.

this thread title should say "Impedance Mismatch, Practical Example"

 

[GoldDigger]

No such thing. For a half cycle energy is transferred to the load, and then transferred back in the other half (if my timing is correct)


1+(-1)=0

For a total of half a cycle energy goes one way, but not for a continuous half cycle.
What actually happens is that energy goes one way for 1/4 cycle, then the other way for 1/4 cycle, continuing on in that way.

Putting it another way, out of each cycle, 1/2 has energy going one way and 1/2 has it going the other way, but energy does not keep going one way for a half cycle and then reverse direction.

 

[Besoeker]

ay be.

y
no one has refuted
va = volt x amp = joule/coloumb x coulomb/sec = joule/sec = watt
NO ONE

I have. More than once. Others have too.
It's just plain wrong. You ought to know that. Ingenieur you are not.
Disingenuous maybe.

 

[romex jockey]

does this fit here?

does this fit here?

Perhaps i can slip a Q in here , one that was forwarded to me by an apprentice that had me stumped?

'Why does Hot to Neutral trip a breaker w/ a load , as opposed to a simple load" ? (think light fixture)

As a career Master , i'm embarrassed to say i was at a loss to respond...

~R(egg on face)J~

 

[Sahib]

Phil. take heart. The opinions range from denying reactive/ apparent power as power to rejection of possibility of conversion of reactive power to active power.

 

[Besoeker]

Phil. take heart. The opinions range from denying reactive/ apparent power as power to rejection of possibility of conversion of reactive power to active power.

How would you do that?

 

[FionaZuppa]

Phil. take heart. The opinions range from denying reactive/ apparent power as power to rejection of possibility of conversion of reactive power to active power.

not opinions, just facts based on physics

post #1 boils down to post #471

 

[Ingenieur]

ay be.
I have. More than once. Others have too.
It's just plain wrong. You ought to know that. Ingenieur you are not.
Disingenuous maybe.

simmer down with the insults fancy pants lol
no you haven't

va = volt x amp = joule/coloumb x coulomb/sec = joule/sec = watt

show me using unit analysis what is not correct
very simple, except you can't because it is valid

 

[winnie]

Responding to my statement that 1000VA and 1000W are similar in terms of losses in the wires feeding the load

what?? that doesnt make sense. the losses are highly dependent on real power that the load uses. how bad the impedance mismatch is depends on PF which dictates kVAr

my 1kVA xfrmer delivers 10watt to a load with PF 0.9
my 1kVA xfrmer delivers 100watt to a load with PF 0.9

condition 2 will waste more power than condition 1, my 1kVA xfrmer stayed the same

I agree with your example above, with different load currents being served, you will see different losses.

What I was trying to say was that for the _same_ load current but different power factor, the losses in the wires will be roughly the same.

This is different from saying that kVAr directy means power 'lost'. In principal, the kVArs mean energy that is flowing back and forth without being consumed. kVArs are associated with _increased_ losses, but are not losses themselves.

I am uncomfortable with your use of 'impedance mismatch', but don't have a good enough understanding of the implications to say that your use is wrong.

In RF work, the goal is to match impedance because impedance mismatch results either in reduced power output or standing waves on transmission lines. But in 60 Hz power distribution you have different goals. You want a 'stiff' low impedance supply, you don't worry about the 'characteristic impedance' of the wires, and you don't match the load to the supply, and you don't match the impedance of the wires to the load. The closest thing you do is to match capacitive and inductive loading, without matching the resistive part of the impedance.

-Jon

 

[winnie]

Phil. take heart. The opinions range from denying reactive/ apparent power as power to rejection of possibility of conversion of reactive power to active power.

How would you do that?

If you consider any energy stored in a magnetic field as 'reactive' then clearly that energy can be delivered as real power.

If you consider only the energy being shuttled back and forth and not actually being consumed by the load as 'reactive', then _by definition_ it cannot be converted to active power, because as soon as you do something to _use_ that energy it is no longer reactive power.

Speaking to Sahib:
The energy coupled between primary and secondary of a transformer is very clearly being moved magnetically.
But if you place a resistive load on the secondary of a transformer, the primary current will be very nearly in phase with the primary voltage, with only a small reactive component for transformer magnetization.

-Jon