Why is VA used instead of Watts?

E

Eric R Snow

Guest
Volts times amps = watts. So it would seem to me that the spec. on a
transformer would be watts. But I see VA specified. So I'm thinking
that VA does not have to be equal to watts. Is this correct? If so,
why?
Thank You,
Eric R Snow
 
"Eric R Snow" <etpm@whidbey.com> wrote in message
news:vo3tp01sd92kjole6mc6etmb6nau7hea8m@4ax.com...
Volts times amps = watts. So it would seem to me that the spec. on a
transformer would be watts. But I see VA specified. So I'm thinking
that VA does not have to be equal to watts. Is this correct? If so,
why?
Click to expand...
That IS correct, and the reason it is is that loads don't have
to be purely resistive. As noted in another thread in this
group, reactive loads (i.e., capacitances and inductances)
make for a condition in which the current waveform and
voltage waveform are out of phase. Only the resistive
part of the load actually "consumes power", and that power
is in watts. But if you multiply together voltage and current
that are not in phase, you get a quantity that has both
magnitude and phase, and therefore both a "real" and
an "imaginary" component. The "real" part of this value
is the "resistive" power, the power actually being consumed;
the "imaginary" part is the "reactive" power, and actually
represents energy that is being taken out of the circuit by
the reactive elements only to be returned later.

The point of all this is that volt-amps can have a peak value
that's greater than the resistive part alone, even though it's
the only part of power that is actually being "consumed."
Hence, transformers and other elements in an AC power
system have to be rated in VA and not just watts.

For more information, look up "power factor" and the
reasons for "power factor correction."

Bob M.
 
Eric R Snow wrote:
Volts times amps = watts. So it would seem to me that the spec. on a
transformer would be watts. But I see VA specified. So I'm thinking
that VA does not have to be equal to watts. Is this correct? If so,
why?
Click to expand...
If you say something is rated for X watts, you generally mean that the
thing is going to dissipate that amount of power.

However, the transformer isn't doing that. It's simply transporting the
power, and the circuit connected to the secondary is doing the
dissipation. Thus, they use VA, which has the units of power. The
transformer is not going to generate the heat indicated by its VA rating.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.
 
On Fri, 19 Nov 2004 16:29:43 -0800, Eric R Snow <etpm@whidbey.com>
wrote:

Volts times amps = watts. So it would seem to me that the spec. on a
transformer would be watts. But I see VA specified. So I'm thinking
that VA does not have to be equal to watts. Is this correct? If so,
why?
Click to expand...
---
The transformer's spec is based on how hot it's allowed to get, and
that spec spells out how much voltage and how much current the
transformer can supply to a load and stay within its (the
transformer's) temperature rise limitations. Since loads are
frequently reactive, the transformer will be required to supply curent
into the load out of phase with its output voltage, and since



P = EI cos(phi)


if the phase angle between current and voltage is 90°, P (power
dissipated by the load, in watts) will be equal to zero no matter what
E (the voltage across the load) and I (the current in the load) happen
to be. However, current will still be flowing into and out of the
transformer's winding resistances, so even if the load is dissipating
no power at all, the transformer will be getting hot and, if the
current is more than it's specified to be able to handle, its lifetime
will be shortened.

There is one case, though, where the transformer's rating, in
VoltAmperes will be equal to watts, and that's when the load is
resistive and cos(phi) will be equal to 1 because the voltage across
it and the current through it will be in phase.

--
John Fields
 
VA vs Watts
Simply stated, it refers to whether or not the current wave form and the
voltage
wave form are running concurrently (or in phase) with each other. If they
are, we
speak of power as Watts, if not, we refer to power as Volt Amperes. It all
depends on the characteristics of the load that the power is looking into.

If the load is a resistor, the current and voltage wave forms run
together so
we have Watts. As soon as inductance or capacitance is added to the load,
then
we are dealing with Volt Amperes because the current will either lag or lead
the
voltage. This is where power factor becomes an issue. As the power factor
deteriorates, so does the efficiency of the power
supply.



"Eric R Snow" <etpm@whidbey.com> wrote in message
news:vo3tp01sd92kjole6mc6etmb6nau7hea8m@4ax.com...
Volts times amps = watts. So it would seem to me that the spec. on a
transformer would be watts. But I see VA specified. So I'm thinking
that VA does not have to be equal to watts. Is this correct? If so,
why?
Thank You,
Eric R Snow
Click to expand...
 
the difference between w and va is the phase
for dc w=va
for ac w=va x (cos of the phase)

"Eric R Snow" <etpm@whidbey.com> wrote in message
news:vo3tp01sd92kjole6mc6etmb6nau7hea8m@4ax.com...
Volts times amps = watts. So it would seem to me that the spec. on a
transformer would be watts. But I see VA specified. So I'm thinking
that VA does not have to be equal to watts. Is this correct? If so,
why?
Thank You,
Eric R Snow
Click to expand...
 
"hotkey" <adennel@jeack.com.au> wrote in message
news:41b9a840$1_1@news.iprimus.com.au...
the difference between w and va is the phase
for dc w=va
for ac w=va x (cos of the phase)
Click to expand...
I've seen this answer posted more than once in this thread and I don't
understand why these posters assume that there has to be a phase angle
involved. Electricians talk about volt-amps reactive or VARs when they
have a power factor of less than 1. But VAs don't have to have any
power factor, which can be assumed as 1.


"Eric R Snow" <etpm@whidbey.com> wrote in message
news:vo3tp01sd92kjole6mc6etmb6nau7hea8m@4ax.com...
Volts times amps = watts. So it would seem to me that the spec. on a
transformer would be watts. But I see VA specified. So I'm thinking
that VA does not have to be equal to watts. Is this correct? If so,
why?
Thank You,
Eric R Snow
Click to expand...
 
My 2cents
Power factor. Power factor of 1 means all the power being supplied (by
generator or transformer...) is being dissipated by the load. This is the
most desired effect. Power factor less than 1, not all of the energy being
supplied is being used, wasted energy or "stored" energy. Power factor
greater than one = Bill Gates will be cleaning your home... it won't
happen.

Algebraically on a cartesian plane
VA=Watts+-j VAR

If none of these posts mean anything or are just confusing, just think of
it as watts in AC.


--
-Ray
 
"Ray" <ray.straub@gmail.com> wrote in message
news:Xns9605A73EF59D1raystraubgmailcom@199.45.49.11...
My 2cents
Power factor. Power factor of 1 means all the power being supplied (by
generator or transformer...) is being dissipated by the load. This is the
most desired effect. Power factor less than 1, not all of the energy
being
supplied is being used, wasted energy or "stored" energy. Power factor
greater than one = Bill Gates will be cleaning your home... it won't
happen.
Click to expand...
Mathematically, it CAN'T happen; the maximum value of
cos(x) is exactly 1.000...

Bob M.
 
"Eric R Snow" <etpm@whidbey.com> wrote in message
news:vo3tp01sd92kjole6mc6etmb6nau7hea8m@4ax.com...
Volts times amps = watts. So it would seem to me that the spec. on a
transformer would be watts. But I see VA specified. So I'm thinking
that VA does not have to be equal to watts. Is this correct? If so,
why?
Thank You,
Eric R Snow
Click to expand...
It's actually more technically correct to mark a "VA" rating on a
transformer instead of an identical "Watts" rating.
It's common practice though just to mark the transformer power rating as
"Watts" (just look through any electronic parts catalogue). Everyone seems
more familiar with a "Watt" rather than the cumbersome, pedantic "Volt-amp".
The VA-Watts distinction is only noticed when a transformer is connected to
none-resistor loads.
Example ... A transformer (say 50Hz) with a 10V secondary winding capable of
2amps, could equally as well be marked as "20Watts" or "20VA".

Hang a 5ohm resistor across the secondary and 2amp flows from the 10V
winding. the resistor gets hot as it is dissipating 20Watts of heat. The
transformer is now at its spec' limit of 10V at 2amps.
Change the load to a 2.5ohm resistor and the tranformer would overload and
cook as the windings now try to supply 4amps at 10V. (It's the amps that's
the killer)

Now just hang a 650u capacitor on the secondary. 10V still feeds the
capacitor and 2amps of current will again flow and yet again the transformer
is its spec' limit.
Trouble is, no Watts power is being used. The cap' just borrows current on
one half cycle and generously returns it the next. The transformer is
technically supplying a 0.0 Watt load but sure as hell notices its full
rated load of 2amps.

Hang a 1000u cap on the secondary and the transformer is well overloaded and
will start cooking, as way too much current is being taken out of the
windings. The capacitor though is still consuming 0.0 Watts. The
transformer still supplying 0.0 Watts.

In these cases (all cases!) it's maybe better to rate the transformer as
capable of supplying a max of "20Volt-amps". The Volt-amps rating can thus
technically apply to any kind of load, whereas a Watt rating implies just
resistor loads.

The peversity of tradition has therefore decreed that "Watts" will be marked
on a transformer but we must remember that it isn't really Watts but a
Volt-amp rereading of the same number.

(Also ... by marking in "Watts", transformer manufacturers can take benefit
from a cop-out against customers 'misusing' their transformers in real world
applications. )

regards
john
 
"Watson A.Name - \"Watt Sun, the Dark Remover\"" wrote:

I've seen this answer posted more than once in this thread and I don't
understand why these posters assume that there has to be a phase angle
involved. Electricians talk about volt-amps reactive or VARs when they
have a power factor of less than 1. But VAs don't have to have any
power factor, which can be assumed as 1.
Click to expand...
I don't understand what you mean by, "which can be assumed as 1". A
VA measurement is the product of an RMS volt measurement and an
independent RMS current measurement. It contains no information about
the phase angle between voltage and current or the power factor.
Watts are constrained by a VA reading to be anywhere between the VA
and the negative of VA, including zero.

--
John Popelish
 
"John Popelish" <jpopelish@rica.net> wrote in message
news:41BB26CC.E1156F7C@rica.net...
"Watson A.Name - \"Watt Sun, the Dark Remover\"" wrote:

I've seen this answer posted more than once in this thread and I
don't
understand why these posters assume that there has to be a phase
angle
involved. Electricians talk about volt-amps reactive or VARs when
they
have a power factor of less than 1. But VAs don't have to have any
power factor, which can be assumed as 1.

I don't understand what you mean by, "which can be assumed as 1".
Click to expand...
In other words, the PF is unity, the phase angles between V and I is
zero, and the load is resistive.

A VA measurement is the product of an RMS volt measurement and an
independent RMS current measurement. It contains no information about
the phase angle between voltage and current or the power factor.
Watts are constrained by a VA reading to be anywhere between the VA
and the negative of VA, including zero.
Click to expand...
That's what I meant, but seems to have got snipped. Other people keep
answering the VA question with this additional cosine of the phase
angle, which isn't applicable.

--
John Popelish
Click to expand...
 
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