What is a watt is a watt?

[davidej]

If the load is not purely resistive, then watts in not an accurate measure. VA (volt amperes) is a better measure.

Despite a physics degree (some years ago) this is new to me.

Can you explain the difference please.

 

[lw395]

Despite a physics degree (some years ago) this is new to me.

Can you explain the difference please.

In AC, the current and voltage can be out of phase to a greater or lesser degree.
A pure resistive load will have them in phase, a highly reactive e.g. inductive load will have them moving towards 90degrees out of phase.
A current 90 degrees out of phase transfers no power to the load, but still heats up the wiring etc in the generator. Hence the total current must be limited to what the generator can handle, not just the useful, in-phase current. Hence generators are often rated in kVA not kW.

 

[kunyang]

In AC, the current and voltage can be out of phase to a greater or lesser degree.
A pure resistive load will have them in phase, a highly reactive e.g. inductive load will have them moving towards 90degrees out of phase.
A current 90 degrees out of phase transfers no power to the load, but still heats up the wiring etc in the generator. Hence the total current must be limited to what the generator can handle, not just the useful, in-phase current. Hence generators are often rated in kVA not kW.

In an inductive circuit component, voltage leads current and in a capacitive circuit component current leads voltage, therefor,as an aid, think of CIVIL, if we add a capacitive component to an inductive circuit (coil) we can obtain more or less unity power factor and therefor KVA, reactive power is KVAr (reactive). This thread may never stop?

But basically, both loads are too big, one is a bit bigger than the other and the friction component of belts, motors etc in a hoover makes the difference.

 

[kunyang]

Oh, and for the answer to the question, one Watt is a Joule per second

 

[BlueSkyNick]

In AC, the current and voltage can be out of phase to a greater or lesser degree.
A pure resistive load will have them in phase, a highly reactive e.g. inductive load will have them moving towards 90degrees out of phase.
A current 90 degrees out of phase transfers no power to the load, but still heats up the wiring etc in the generator. Hence the total current must be limited to what the generator can handle, not just the useful, in-phase current. Hence generators are often rated in kVA not kW.

To keep it simple, this is referred to as Power Factor. Because of the phasing as described, some motors are more efficient than others, ie the closer PF to 1.0 or 100%, the better.

(I have replied to lw395's post, but intended the answer for davidej)

 

[davidej]

Thanks for the excplanations - I understand this so far but how is VA measured.

For AC and a resistive load, I seem to remember that watts is on an RMS basis - effectively the area under a sine wave graph. But if V and A are out of sinc, I assume to measure power theoretically you multiplying them at each point of the cycle and average out the product. How do you actually do this in practice and is the answer watts or VA?

Sorry for fred drift - perhaps I should ask Prof Hawkins -he is on the radio at the moment

 

[ianj99]

Thanks for the excplanations - I understand this so far but how is VA measured.

For AC and a resistive load, I seem to remember that watts is on an RMS basis - effectively the area under a sine wave graph. But if V and A are out of sinc, I assume to measure power theoretically you multiplying them at each point of the cycle and average out the product. How do you actually do this in practice and is the answer watts or VA?

Sorry for fred drift - perhaps I should ask Prof Hawkins -he is on the radio at the moment

Its the product of the rms values of current and voltage, therefore any phase shifts are irrelevant.

 

[CreakyDecks]

Thanks for the excplanations - I understand this so far but how is VA measured.

For AC and a resistive load, I seem to remember that watts is on an RMS basis - effectively the area under a sine wave graph. But if V and A are out of sinc, I assume to measure power theoretically you multiplying them at each point of the cycle and average out the product. How do you actually do this in practice and is the answer watts or VA?

Sorry for fred drift - perhaps I should ask Prof Hawkins -he is on the radio at the moment

To calculate the real power you multiply the voltage x current x cosine of the difference in phase between them(usually given greek letter phi))

The cos(phi) term is the power factor and varies between one for a resistive load (phi = 0) and zero for a reactive load (phi = 90deg).

 

[mucklestone]

I think you need to take the rated power ratings with a pinch of salt. They do not mean much!

 

[Bilgediver]

I pose this question as I am again without power-just isolate my supply and plug in the 800watt generator and I am back in business.
However the 1500 watt hoover will not work(oh what joy to me but annoyance to my wife) because of course its 1500 watts-but then again my big freezer should not work as it needs about 1000 watts to start up.
My big freezer does work all be it with a total power drop when it starts so what about the 1500 watt hoover-well I know its slow start and possibly brushless and yes it works!
I am sure we all are attracted to both power tools and domestic appliances which tend to be priced according to wattage.
So whats going on-my Hoover does note that its also 200 air watts which sounds like its true wattage so what is the 1500 watts?

Now that you are totally confused you have to remember also thet the Genny maker likes to exagerate too.... The 800 watts is no doubt the rating of the genny at a power factor of 1!!!! You could be dragging the power factor down to less than 0.75 which means you may only be able to deliver 600 Watts and even a demand of 800 could cause overload.

 

[kunyang]

Thanks for the excplanations - I understand this so far but how is VA measured.

For AC and a resistive load, I seem to remember that watts is on an RMS basis - effectively the area under a sine wave graph. But if V and A are out of sinc, I assume to measure power theoretically you multiplying them at each point of the cycle and average out the product. How do you actually do this in practice and is the answer watts or VA?

Sorry for fred drift - perhaps I should ask Prof Hawkins -he is on the radio at the moment

The average value of a sign wave can be found by using a multiple of 0.637 peak and the rms value approximates to 0.707 peak. Get the scope out and have a go. (pure sine wave only, impure sine waves get into fights, normally bad drivers)

Watts are purely resistive (no reactance) VA are Watts (no reactance) VAr uses the reactive component.

 

[Steve Clayton]

So, go on then, who's going to pipe up about Hole and Electron Flow, Co-valent bonding and Crystal Lattice structures????....Arghhhhhhhhhhhh

I will hole flow is faster than electron flow

 

[Amulet]

Its the product of the rms values of current and voltage, therefore any phase shifts are irrelevant.

'snot true. If you are so out of phase so that the peak current is at zero voltage due to the delay imposed by the inductive load, then the integral of AV is going to be clearly less than the RMS product would have you think. You are chucking energy into the opposition of the inductive load which is stopping the current rising in synch with the voltage.

I'm not being a smart ass. I argued blue in the face that it was simply about RMS products until some physicist mates bound and gagged me and made me listen to the explanation. They reminded me about the cute hysteresis loops that we used to display on the oscilliscope as kids.