Volt Amps = ??
Phil Allison
"Ross Irvine"
> Hi All,
>
> Ok, I know this is probably a silly question but what are "Volt Amps" as
> used in America.
** Same as anywhere else.
>
> Eg. I have an American made device that wants 24 Volts AC, 50or60 Hz,
> 40VA (Volt Amps)
>
> What does a Volt Amp translate to in what we'd normally use which is
> just AMPS (Eg 1.5 AMPS or something) or is it something different?
> Is there a formula to convert VA into AMPs?
** Volt Amps ( VA) = Voltage x Current ( with both given in rms
values.)
Usually the voltage is known - like 240 or 120 volts.
So, VA divided by the supply voltage = current draw in amps rms.
The current may or may not be a sine wave - ie it can be distorted
as with most electronic loads.
......... Phil
Wild Wizard
> Hi All,
>
> Ok, I know this is probably a silly question but what are "Volt Amps" as
> used in America.
exactly what they sound like Volt Amps
> Eg. I have an American made device that wants 24 Volts AC, 50or60 Hz,
> 40VA (Volt Amps)
VA = V * A
A = VA/V
V = 24
VA = 40
A = 40/24
A = 1.7
> What does a Volt Amp translate to in what we'd normally use which is
> just AMPS (Eg 1.5 AMPS or something) or is it something different?
> Is there a formula to convert VA into AMPs?
> Thanks.
Think i've just answered it :)
--
The Kennedy Constant:
Don't get mad -- get even.
C3
allows you to see the voltage and current components that it comprises.
C3
Bristan
"Ross Irvine" <rirvineDOT@netspaceDOT..DOTnet.DOT.au> wrote in message
news:3FF24343.E798DB6C@netspaceDOT..DOTnet.DOT.au...
> Hi All,
>
> Ok, I know this is probably a silly question but what are "Volt Amps" as
> used in America.
VoltAmps are watts (power = voltage times current)
VA are used instead of watts where there is a reactive component to the load
in ac circuits. eg inductance or capacitance and the watts used in the load
are not the total power used.
Transformers typically are rated in kVA not kw as a certain quantity of
power is taken up by inductance of the transformer itself. (power factor not
equal to one and the current lags the voltage).
> Eg. I have an American made device that wants 24 Volts AC, 50or60 Hz,
> 40VA (Volt Amps)
>
Phil Allison
"C3" <gn...@telsMONOPOLYtra.com>
> As far as I can tell, it just a way to express power .....
** VA is not power - a 55 uF capacitor across a 240 volt AC supply
dissipates no power but the effective load is nearly 1000 VA. VA is a
convenient way to describe the rms current draw of an item. This is
important to know so generator, cable and fuse ratings are not exceeded.
A sparkie just adds up all the VA numbers and compares that to the circuit
capacity in VA.
........... Phil
Phil Allison
"Bristan"
> VoltAmps are watts (power = voltage times current)
> VA are used instead of watts where there is a reactive component to the
load
> in ac circuits. eg inductance or capacitance and the watts used in the
load
> are not the total power used.
* The watts used in the load are always the total power - but not alway the
same as the VA.
There need be no reactive load either - electronic loads ( DC from AC
supplies) are rated in VA.
> Transformers typically are rated in kVA not kw as a certain quantity of
> power is taken up by inductance of the transformer itself. (power factor
not
> equal to one and the current lags the voltage).
** The VA rating of a transformer ( along with the specified operating
voltages ) tells you the amount of **rms** current the secondary can supply
long term - without overheating.
........ Phil
DJ
<rirvineDOT@netspaceDOT..DOTnet.DOT.au> wrote:
>Hi All,
>
>Ok, I know this is probably a silly question but what are "Volt Amps" as
>used in America.
>
>Eg. I have an American made device that wants 24 Volts AC, 50or60 Hz,
>40VA (Volt Amps)
>
>What does a Volt Amp translate to in what we'd normally use which is
>just AMPS (Eg 1.5 AMPS or something) or is it something different?
>Is there a formula to convert VA into AMPs?
>Thanks.
There is a comprehensive discussion of this and related topics at
http://www.splatco.com/tips/pwrfact/pfarticl.htm
dj
David
A sparkie may add up all the VA numbers, but an engineer (or really clever
sparkie) would know that as VA's are vector quantities, you can't just add
them up. Power factor correction devices are often added to commercial
installations to reduce the total VA, and bring the power factor closer to
unity.
David
Phil Allison
"David" <no_...@hotmail.com> wrote in message
news:pan.2004.01.01....@hotmail.com...
> On Wed, 31 Dec 2003 15:33:45 +1100, Phil Allison wrote:
>
> >
> > "C3" <gn...@telsMONOPOLYtra.com>
> >
> >> As far as I can tell, it just a way to express power .....
> >
> >
> > ** VA is not power - a 55 uF capacitor across a 240 volt AC supply
> > dissipates no power but the effective load is nearly 1000 VA. VA is a
> > convenient way to describe the rms current draw of an item. This is
> > important to know so generator, cable and fuse ratings are not exceeded.
> >
> > A sparkie just adds up all the VA numbers and compares that to the
circuit
> > capacity in VA.
> >
> >
> >
> >
> > ........... Phil
>
> A sparkie may add up all the VA numbers, but an engineer (or really clever
> sparkie) would know that as VA's are vector quantities, you can't just add
> them up.
** VAs may be quite resistive, have a known or unknown phase angle and
or/or have distorted (peaky) current waveforms. If you simply add them then
the answer is the same as the worst case value.
>Power factor correction devices are often added to commercial
> installations to reduce the total VA, and bring the power factor closer to
> unity.
** That is only true where there is a phase angle to be corrected - what
if the load is 1000s of electronic devices with AC to DC power supplies in
them that only draw current at the crest of the AC voltage ??
......... Phil
David
>
> "David" <no_...@hotmail.com> wrote in message
> news:pan.2004.01.01....@hotmail.com...
>> On Wed, 31 Dec 2003 15:33:45 +1100, Phil Allison wrote:
>>
>> >
>> > "C3" <gn...@telsMONOPOLYtra.com>
>> >
>> >> As far as I can tell, it just a way to express power .....
>> >
>> >
>> > ** VA is not power - a 55 uF capacitor across a 240 volt AC supply
>> > dissipates no power but the effective load is nearly 1000 VA. VA is a
>> > convenient way to describe the rms current draw of an item. This is
>> > important to know so generator, cable and fuse ratings are not exceeded.
>> >
>> > A sparkie just adds up all the VA numbers and compares that to the
> circuit
>> > capacity in VA.
>> >
>> >
>> >
>> >
>> > ........... Phil
>
>>
>> A sparkie may add up all the VA numbers, but an engineer (or really clever
>> sparkie) would know that as VA's are vector quantities, you can't just add
>> them up.
>
>
> ** VAs may be quite resistive, have a known or unknown phase angle and
> or/or have distorted (peaky) current waveforms. If you simply add them then
> the answer is the same as the worst case value.
>
If you simply add them for the worst case, then you will be over
specifying the power distribution system, requiring larger cables and
transformers etc, which will result in increased costs to the end user.
For a large industrial user this could well be significant.
>>Power factor correction devices are often added to commercial
>> installations to reduce the total VA, and bring the power factor closer to
>> unity.
>
>
> ** That is only true where there is a phase angle to be corrected - what
> if the load is 1000s of electronic devices with AC to DC power supplies in
> them that only draw current at the crest of the AC voltage ??
>
>
> ......... Phil
Non linear loads cause harmonic currents to appear in the electrical
distrubtion system. Harmonic filters can reduce the harmonic currents
generated by non linear loads in the phase and neutral conductors of the
three phase distrubution system. This will reduce the peak phase current,
and average phase current, which will reduce the system losses, and reduce
costs.
David
Phil Allison
"David" <no_...@hotmail.com>
Phil A wrote:
> > ** VAs may be quite resistive, have a known or unknown phase angle and
> > or/or have distorted (peaky) current waveforms. If you simply add them
then
> > the answer is the same as the worst case value.
> >
>
> If you simply add them for the worst case, then you will be over
> specifying the power distribution system,
** How so ?
You are assuming a MOST unlikely scenario where different loads onm the
SAME circuit have cancelling phase angles !!!!!
>
> >>Power factor correction devices are often added to commercial
> >> installations to reduce the total VA, and bring the power factor closer
to
> >> unity.
> >
> >
> > ** That is only true where there is a phase angle to be corrected -
what
> > if the load is 1000s of electronic devices with AC to DC power supplies
in
> > them that only draw current at the crest of the AC voltage ??
> >
> > ......... Phil
> Non linear loads cause harmonic currents to appear in the electrical
> distrubtion system. Harmonic filters can reduce the harmonic currents
> generated by non linear loads in the phase and neutral conductors of the
> three phase distrubution system. This will reduce the peak phase current,
> and average phase current, which will reduce the system losses, and reduce
> costs.
** Where does the "harmonic filter" go ??
Please describe such a filter .
BTW Are you quoting from some web site ??
It sure as hells looks to me like you are.
....... Phil
Bristan
I am not a power engineer and don't profess to be an expert however I am
prompted to recall an explanation from my university days, as you seem to
have a rather garbled idea of what is going on.
The average power in an ac circuit is given by
P =VIcos phi
where phi is the phase angle between the current and the voltage.
and the product VI is called the VOLT-AMPERES
cosphi is the power factor
When the current is in phase with the voltage then phi =0, power factor =1
and P =VI
In this case of a non reactive or balanced load the VA *ARE* equal to the
watts.
In most ac inductive or capacitive circuits this will not be the case and
the VA will be greater than the average power in watts. This leads to
larger currents and hence cables etc than are necessary for the power
provided by the device. This leads to power factor correction devices
(capacitors with inductive loads) As the current leads the voltage in a
capacitor this will bring the average current more in phase with the voltage
and make thew PF closer to one.
For an example if the current is lagging the voltage by say 60 degrees then
the PF will be 0.5 and the VA will be twice the average power in the
circuit.
Therefore as I said before, electrical machinery for ac is usually rated in
kva to take account of this occurrence.
"Phil Allison" <phila...@optusnet.com.au> wrote in message
news:3ff25348$0$18747$afc3...@news.optusnet.com.au...
Phil Allison
"Bristan" <daed>
> To make it clear
> I am not a power engineer and don't profess to be an expert however I am
> prompted to recall an explanation from my university days, as you seem to
> have a rather garbled idea of what is going on.
** Sorry - you have the garbled idea, and a very common form of garble
it is.
>> The average power in an ac circuit is given by
> P =VIcos phi
> where phi is the phase angle between the current and the voltage.
** Who says there is **any phase angle** or the current is a sine wave
?????????????????????????????????????
What you have forgotten from your Uni days is the crucial qualification: "
.... in an all sine wave system.... "
> Therefore as I said before, electrical machinery for ac is usually rated
in
> kva to take account of this occurrence.
** So what ?????????
Since when is this NG or was the OP speaking of "electrical machinery "
?????
Cripes - even my floor fan blowing on me as I type has a PF of 0.97.
......... Phil
David
>
> "David" <no_...@hotmail.com>
>
> Phil A wrote:
>
>
>> > ** VAs may be quite resistive, have a known or unknown phase angle and
>> > or/or have distorted (peaky) current waveforms. If you simply add them
> then
>> > the answer is the same as the worst case value.
>> >
>>
>> If you simply add them for the worst case, then you will be over
>> specifying the power distribution system,
>
>
> ** How so ?
>
> You are assuming a MOST unlikely scenario where different loads onm the
> SAME circuit have cancelling phase angles !!!!!
>
Example case. Assume pure resitive load of 1000VA, and a pure
inductive load of 1000VA. Simply adding the VA's would result in 2000VA,
where the real VA would only be 1414VA. This would result in approx 70%
less current requirement than your method, which represents a real
saving.
>
>>
>> >>Power factor correction devices are often added to commercial
>> >> installations to reduce the total VA, and bring the power factor closer
> to
>> >> unity.
>> >
>> >
>> > ** That is only true where there is a phase angle to be corrected -
> what
>> > if the load is 1000s of electronic devices with AC to DC power supplies
> in
>> > them that only draw current at the crest of the AC voltage ??
>
>> >
>> > ......... Phil
>
>
>
>> Non linear loads cause harmonic currents to appear in the electrical
>> distrubtion system. Harmonic filters can reduce the harmonic currents
>> generated by non linear loads in the phase and neutral conductors of the
>> three phase distrubution system. This will reduce the peak phase current,
>> and average phase current, which will reduce the system losses, and reduce
>> costs.
>
>
>
> ** Where does the "harmonic filter" go ??
>
> Please describe such a filter .
>
The harmonic filter is on the three phase system. Depending on the
harmonics, the filter may be a delta transformer, which will eliminate
zero sequence harmonics (3rd, 6th, 9th etc). Other techniques can be used
to eliminate positive and negative sequence harmonics if needed. These can
range from simple LC shunt and series filters, to sophisticated active
power filters which are DSP based, and measure the harmonic currents and
inject currents into the supply to cancel out the harmonics.
A google search on Harmonic Mitigating Transformers and Harmonic
Filters will yeild plenty of examples.
David
Phil Allison
"David" <no_...@hotmail.com
> > You are assuming a MOST unlikely scenario where different loads onm
the
> > SAME circuit have cancelling phase angles !!!!!
> >
>
> Example case. Assume pure resitive load of 1000VA, and a pure
> inductive load of 1000VA. Simply adding the VA's would result in 2000VA,
> where the real VA would only be 1414VA.
** Pure inductive loads are just a tad rare - so your example is not
typical.
If you use a 1000 VA load with a reactive PF of 0.7 then add a resistive
load of 1000 VA the sum is 1800VA.
If you have electronic loads with whatever PF then the VAs just add.
> >> Non linear loads cause harmonic currents to appear in the electrical
> >> distrubtion system. Harmonic filters can reduce the harmonic currents
> >> generated by non linear loads in the phase and neutral conductors of
the
> >> three phase distribution system. This will reduce the peak phase
current,
> >> and average phase current, which will reduce the system losses, and
reduce
> >> costs.
> >
> >
> >
> > ** Where does the "harmonic filter" go ??
> >
> > Please describe such a filter .
> >
> The harmonic filter is on the three phase system.......
** Bad luck for single phase loads.
> A google search on Harmonic Mitigating Transformers and Harmonic
Filters will yeild plenty of examples.
BTW Are you quoting from some web site ??
It still sure as hells looks to me like you are.
....... Phil
David
>
> "David" <no_...@hotmail.com
>
>
>> > You are assuming a MOST unlikely scenario where different loads onm
> the
>> > SAME circuit have cancelling phase angles !!!!!
>> >
>>
>> Example case. Assume pure resitive load of 1000VA, and a pure
>> inductive load of 1000VA. Simply adding the VA's would result in 2000VA,
>> where the real VA would only be 1414VA.
>
>
> ** Pure inductive loads are just a tad rare - so your example is not
> typical.
>
> If you use a 1000 VA load with a reactive PF of 0.7 then add a resistive
> load of 1000 VA the sum is 1800VA.
>
Still a saving of 10%. Well worthwhile saving.
> If you have electronic loads with whatever PF then the VAs just add.
>
Only if the the power factors are the same. Transformers supplied loads
tend to be lagging, and switch mode supplies tend to be leading, so
they will cancel to some degree. Properly designed high power factor
electronic loads will have a power factor very close to unity.
>
>
>
>> >> Non linear loads cause harmonic currents to appear in the electrical
>> >> distrubtion system. Harmonic filters can reduce the harmonic currents
>> >> generated by non linear loads in the phase and neutral conductors of
> the
>> >> three phase distribution system. This will reduce the peak phase
> current,
>> >> and average phase current, which will reduce the system losses, and
> reduce
>> >> costs.
>> >
>> >
>> >
>> > ** Where does the "harmonic filter" go ??
>> >
>> > Please describe such a filter .
>
>> >
>> The harmonic filter is on the three phase system.......
>
>
> ** Bad luck for single phase loads.
>
Most single phase loads are supplied from three phase switchboards, where
the filter would be installed. Single phase active filters are available,
but cheaper to put in main swithboard, which for large loads are three
phase.
David
Phil Allison
"David" <no_...@hotmail.com>
Phil A wrote:
> > ** Pure inductive loads are just a tad rare - so your example is not
> > typical.
> >
> > If you use a 1000 VA load with a reactive PF of 0.7 then add a
resistive
> > load of 1000 VA the sum is 1800VA.
> >
>
> Still a saving of 10%. Well worthwhile saving.
** Save 10% of what exactly ?
If the sparkie does nothing but add VAs he may slightly under-utilise a
circuit.
If he uses his trusty clamp meter - then guess what ?
> > If you have electronic loads with whatever PF then the VAs just add.
> >
>
> Only if the the power factors are the same.
** Same current wave shape ???
>Transformers supplied loads tend to be lagging, and switch mode supplies
tend to be leading, so
> they will cancel to some degree. Properly designed high power factor
> electronic loads will have a power factor very close to unity.
** There is no PFC in the vast majority of electronic items - including
some that draw up to 4000 VA from a single phase circuit. Also there is
virtually no phase angle, peak current draw co-insides exactly with peak
voltage. The PF of such loads is about 0.5.
........ Phil
David
>
> "David" <no_...@hotmail.com>
>
> Phil A wrote:
>
>
>> > ** Pure inductive loads are just a tad rare - so your example is not
>> > typical.
>> >
>> > If you use a 1000 VA load with a reactive PF of 0.7 then add a
> resistive
>> > load of 1000 VA the sum is 1800VA.
>> >
>>
>> Still a saving of 10%. Well worthwhile saving.
>
>
> ** Save 10% of what exactly ?
As the current will 10% less, then the conductor size can be smaller.
As coppper cables typically come in standard size, ie 1mm^2,
1.5mm^2, 2.5mm^2 and so on, the savings can be large.
Assuming the above example, 2000VA at 240V is 8.3A, 1800VA is 7.5A.
Depending on the installation method, and voltage drop requirements,
1.5mm^2 cable may only have a rating of say 8 Amps, calculated using
AS3008.1 If you assume 2000VA, you would have to use 2.5mm cable, which
costs more than 1.5mm cable.
Also in larger installations, a distribution transformer or switchboard
may have to be upgraded. If the incorrect calculations are used,
then a larger, and more expensive upgraded may be required. A correct
calculation may indicate existing equipment is adequate.
> If the sparkie does nothing but add VAs he may slightly under-utilise a
> circuit.
>
Or cost the customer extra installation costs, or over price a quote and
do himself out of a job. His calculation may indicate that the existing
cable can not cope with the current, and needs replacing. Correct
calculation may indicate that the existing cable is still within its
rating, and does not need replacement.
> If he uses his trusty clamp meter - then guess what ?
>
If it's not a true RMS clamp meter then he is stuffed.
Also, a bit late to use a clamp meter after the installation has been
completed, and he has discovered that he has installed cable heavier than
required, and has cost himself profit by using oversized cables.
>
>
>
>> > If you have electronic loads with whatever PF then the VAs just add.
>> >
>>
>> Only if the the power factors are the same.
>
>
> ** Same current wave shape ???
>
>
>>Transformers supplied loads tend to be lagging, and switch mode supplies
> tend to be leading, so
>> they will cancel to some degree. Properly designed high power factor
>> electronic loads will have a power factor very close to unity.
>
>
> ** There is no PFC in the vast majority of electronic items - including
> some that draw up to 4000 VA from a single phase circuit. Also there is
> virtually no phase angle, peak current draw co-insides exactly with peak
> voltage. The PF of such loads is about 0.5.
>
>
Florescent lamps are common devices which have power factor correction. In
the older types capacitors were fitted. Electronic ballasts can achieve
high power factors, very close to unity. Electrical Authorities do
require minimum power factors for loads.
Computer power supplies are also available which have power factor
correction to comply with harmonic current emission standards (they just
cost more than your el cheapo chinese power supply).
David
Bristan
>
>
> ** Who says there is **any phase angle** or the current is a sine wave
> ?????????????????????????????????????
What are you talking about ???????????????????????
If the PF is less than one then there is a phase angle.
You are on a different planet.....
> What you have forgotten from your Uni days is the crucial qualification:
"
> .... in an all sine wave system.... "
>
The OP was talking about an electric globe operating from the 240v mains.
last time I checked it was a sine wave........
In any case I was endeavouring to give a not too technical explanation to
the OP and not get involved in another one of your wanking sessions where
you try to boost your low self esteem by dropping lots of jargon and
revealing your lack of real knowledge.
if you don't know what the phase angle between current and voltage in a
(single)phase reactive load is all about then you should read up on it or
shut up.
to sum and give a simple explanation of volt amperes.
in a dc circuit the VA is identical to the watts and IS the power
in an ac circuit with PF =1 the VA is identical to the watts and IS the
power
in an ac circuit with PF less than 1 the VA is the watts multiplied by the
reciprocal of the power factor and IS the power with an allowance for the
increased current due to the reactive load.
The VA is equal to the max power which the device will use IF the power
factor is one.
So to the layman the VA IS the power.
and the va marking on the OP light globe IS to most people ,the power.
Not of course to pedantic chip on shoulder wankers like you Allison who
immediatley jump on to it and perform another one of you pathetic,chest
beating, moronic ego trip attempts.
To which you will no doubt reply with a childish and illogical denial and
attempt to discredit the above.
If you ever grow up and overcome your inferiority complex you may be able to
make a positive contribution, then would you care to do that anyway?.....I
doubt not.
>
Phil Allison
> >
> > ** Save 10% of what exactly ?
>
> As the current will 10% less, then the conductor size can be smaller.
** That is sheer pedantry.
> Assuming the above example, 2000VA at 240V is 8.3A, 1800VA is 7.5A.
> Depending on the installation method, and voltage drop requirements,
> 1.5mm^2 cable may only have a rating of say 8 Amps, calculated using
> AS3008.1 If you assume 2000VA, you would have to use 2.5mm cable, which
> costs more than 1.5mm cable.
** Yawn. You are inventing fanciful examples again.
> > If the sparkie does nothing but add VAs he may slightly under-utilise
a
> > circuit.
> >
>
> Or cost the customer extra installation costs, or over price a quote and
> do himself out of a job. His calculation may indicate that the existing
> cable can not cope with the current, and needs replacing. Correct
> calculation may indicate that the existing cable is still within its
> rating, and does not need replacement.
** More fanciful stuff.
>
> > If he uses his trusty clamp meter - then guess what ?
> >
>
> If it's not a true RMS clamp meter then he is stuffed.
* Standard rms calibrated meters are fine for any non distorting load.
> Also, a bit late to use a clamp meter after the installation has been
> completed, and he has discovered that he has installed cable heavier than
> required, and has cost himself profit by using oversized cables.
** You seem to be assuming something without ever saying it. Are you
assuming some industrial situation with the permanently wired loads ?? I
was not.
> >> > If you have electronic loads with whatever PF then the VAs just add.
> >> >
> >> Only if the the power factors are the same.
> >
> > ** Same current wave shape ???
** No answer noted.
There was nothing about it on the web site.
> >
> >>Transformers supplied loads tend to be lagging, and switch mode supplies
> > tend to be leading, so
> >> they will cancel to some degree. Properly designed high power factor
> >> electronic loads will have a power factor very close to unity.
> >
> >
> > ** There is no PFC in the vast majority of electronic items -
including
> > some that draw up to 4000 VA from a single phase circuit. Also there is
> > virtually no phase angle, peak current draw co-insides exactly with
peak
> > voltage. The PF of such loads is about 0.5.
> >
> >
** No answer noted - terrible how web site cannot answer
questions.
> Florescent lamps are common devices which have power factor correction. In
> the older types capacitors were fitted. Electronic ballasts can achieve
> high power factors, very close to unity. Electrical Authorities do
> require minimum power factors for loads.
** This David fuckwit is now quoting whole chunks from some web site -
parrot fashion.
> Computer power supplies are also available which have power factor
> correction to comply with harmonic current emission standards (they just
> cost more than your el cheapo chinese power supply).
** Don't ya just hate talking to a parrot.
.......... Phil
Phil Allison
Bristan the wanker:
> "Phil Allison" <phila...@optusnet.com.au> wrote in message
> news:3ff3c321$0$18690$afc3...@news.optusnet.com.au...
> >
> >
> > ** Who says there is **any phase angle** or the current is a sine wave
> > ?????????????????????????????????????
>
> What are you talking about ???????????????????????
If the PF is less than one then there is a phase angle.
** No fucking way.
There is NO mention of phase angle in the general definition of power
factor.
> You are on a different planet.....
>
> > What you have forgotten from your Uni days is the crucial
qualification:
> "
> > .... in an all sine wave system.... "
> >
>
> The OP was talking about an electric globe operating from the 240v mains.
> last time I checked it was a sine wave........
** Bullshit - the word used by the OP was "device".
BTW I aint seen to many bulbs with VA figures on them !!!!!!!!
> In any case I was endeavouring to give a not too technical explanation to
> the OP and not get involved in another one of your wanking sessions where
> you try to boost your low self esteem by dropping lots of jargon and
> revealing your lack of real knowledge.
** Well fuck you Mr Arrogant Prick - the one who lacks basic knowledge is
YOU.
> if you don't know what the phase angle between current and voltage in a
> (single)phase reactive load is all about then you should read up on it or
> shut up.
** You are the one who needs to read up - on the definition of VA for a
start.
See any textbook - look for two formulae.
One which is general - the other for "all sine wave" systems.
> in an ac circuit with PF less than 1 the VA is the watts multiplied by
the
> reciprocal of the power factor and IS the power with an allowance for the
> increased current due to the reactive load.
** The load does NOT have to be reactive in the LEAST.
> The VA is equal to the max power which the device will use IF the power
> factor is one.
> So to the layman the VA IS the power.
> and the va marking on the OP light globe IS to most people ,the power.
** VA markings on a light globe ????
> Not of course to pedantic chip on shoulder wankers like you Allison who
> immediately jump on to it and perform another one of you pathetic,chest
> beating, moronic ego trip attempts.
** You are a cretin, Bristan .
> To which you will no doubt reply with a childish and illogical denial and
> attempt to discredit the above.
** Sorry if basic electrical theory goes over the head of a posturing
software wanker.
( snip rest of Bristan's juvenile gratuitous insults)
.......... Phil
Kevin Ettery
I AM an Electrical Engineer and do know a bit about this subject.
Real Power (what provides the apparent power output) in an ac circuit is
given by:
P =VI cos theta
where theta is the angle between the current and the voltage.
and the product VI is called the VOLT-AMPS (written as VA).
cos theta is the power factor.
When the current is in phase with the voltage then theta =0, power factor =1
(sometimes referred to as unity power factor), and P =VI
In this case, a non-reactive load the VA *ARE* equal to the watts.
In most ac circuits there is an inductive or capacitive component to the
load and so watts will not be the same as VA. Thus the VA will be greater
than the watts. Assuming the volts remain the same (usually valid) this
means that there will be larger currents flowing than the watts measured
would indicate, and thus the cables used MUST be rated for this increased
current. Failure to do so can result in the cable being overstressed and
failing - not a pretty sight!
Power factor correction devices are capacitor banks, as it is very very rare
for a ac load to be capacitive (causes a host of other problems when it is
too!!). They work on the fact that, in an ac circuit, capacitance can
cancel out inductance. By switching in enough capacitance (which is what
the power factor correction control circuitry does) it is possible to cancel
out the inductance and return the power factor to unity. The reason it is
done is that the closer to unity you can get the power factor the lower the
VA for the same power output requirements, and so you can reduce the rating
of the power conductors. Also by reducing the current demand you reduce the
voltage drop in the cable - every cable has a certain resistance per metre
and if you can reduce the current you can thus reduce the voltage drop in
the cable.
Harmonics are totally unrelated to power factor, and are in fact a measure
for the distortion of the ac waveform which should be a pure 50 Hz
sinusoid - harmonics are multiples of the fundamental frequency (50 Hz) with
usually only odd harmonics (150, 350, 450, etc) being present (even
harmonics are bad news in a power system). Removing harmonics is done for
many of the same reasons power factor correction in done, and also because
the harmonic frequencies can play merry hell with many instruments and even
power factor correction devices can have trouble handling harmonics (the
apparent capacitance or inductance - called reactive impedance - is a
function of frequency, so at each harmonic these values appear different).
A few minor additions.
There is such a term as Reactive Power, which is given by VI sin theta.
Real Power is measured in Watts, Reactive Power is measured in VARs (Volt
Amps Reactive).
Reactive power is usually written as P with a subscript 'r'.
In vector terms both the current and voltage are vectors rotating in space.
Leading or lagging Power Factor indicates whether the current vector is in
front of (leading) or behind (lagging) the voltage vector. The following is
a good memory aid.
CIVIL
C (Capacitive)
I (Current)
V (Voltage)
I (Current)
L (Inductive)
So if load Capacitive then Current is before (leads) Voltage.
If load Inductive then Current is after (lags) Voltage.
So there ends a potted version of the fundamentals of AC power terminology
and calculations.
Hope it cleared a few things up.
Regards
Kevin
"Phil Allison" <phila...@optusnet.com.au> wrote in message
news:3ff3c321$0$18690$afc3...@news.optusnet.com.au...
Phil Allison
"Kevin Ettery" <kpet...@dcsi.net.au>
> Guys,
>
> I AM an Electrical Engineer and do know a bit about this subject.
** Shame you are not an electronics engineer - then you might just get it
right.
>
> Real Power (what provides the apparent power output) in an ac circuit is
> given by:
> P =VI cos theta
> where theta is the angle between the current and the voltage.
> and the product VI is called the VOLT-AMPS (written as VA).
> cos theta is the power factor.
** Tell me Kevin - in the EE game, is the phrase "ac circuit" code speak
for something ? Is is code speak for those systems where voltages and
current waveforms are all sinusoidal ???????????
If so - get outta here.
......... Phil
David
>
> "David" <no_...@hotmail.com> wrote in message
> news:pan.2004.01.01....@hotmail.com...
>> On Fri, 02 Jan 2004 00:11:07 +1100, Phil Allison wrote:
>
>> >
>> > ** Save 10% of what exactly ?
>>
>> As the current will 10% less, then the conductor size can be smaller.
>
>
> ** That is sheer pedantry.
>
>
>
>> Assuming the above example, 2000VA at 240V is 8.3A, 1800VA is 7.5A.
>> Depending on the installation method, and voltage drop requirements,
>> 1.5mm^2 cable may only have a rating of say 8 Amps, calculated using
>> AS3008.1 If you assume 2000VA, you would have to use 2.5mm cable, which
>> costs more than 1.5mm cable.
>
>
> ** Yawn. You are inventing fanciful examples again.
>
This is an example from your figures. Real examples from real situations
can have similar outcomes. Some of us do have practical experience.
>
>> > If the sparkie does nothing but add VAs he may slightly under-utilise
> a
>> > circuit.
>> >
>>
>> Or cost the customer extra installation costs, or over price a quote and
>> do himself out of a job. His calculation may indicate that the existing
>> cable can not cope with the current, and needs replacing. Correct
>> calculation may indicate that the existing cable is still within its
>> rating, and does not need replacement.
>
>
> ** More fanciful stuff.
>
>
>>
>> > If he uses his trusty clamp meter - then guess what ?
>> >
>>
>> If it's not a true RMS clamp meter then he is stuffed.
>
>
> * Standard rms calibrated meters are fine for any non distorting load.
>
Opps - You slipped up there Phil, assuming a sine wave are we?
>
>
>> Also, a bit late to use a clamp meter after the installation has been
>> completed, and he has discovered that he has installed cable heavier than
>> required, and has cost himself profit by using oversized cables.
>
>
> ** You seem to be assuming something without ever saying it. Are you
> assuming some industrial situation with the permanently wired loads ?? I
> was not.
I have assumed nothing. Domestic installations have permanently wired
loads.
>
>
>> >> > If you have electronic loads with whatever PF then the VAs just add.
>> >> >
>> >> Only if the the power factors are the same.
>> >
>> > ** Same current wave shape ???
>
>
> ** No answer noted.
>
> There was nothing about it on the web site.
>
>
>
>
>> >
>> >>Transformers supplied loads tend to be lagging, and switch mode supplies
>> > tend to be leading, so
>> >> they will cancel to some degree. Properly designed high power factor
>> >> electronic loads will have a power factor very close to unity.
>> >
>> >
>> > ** There is no PFC in the vast majority of electronic items -
> including
>> > some that draw up to 4000 VA from a single phase circuit. Also there is
>> > virtually no phase angle, peak current draw co-insides exactly with
> peak
>> > voltage. The PF of such loads is about 0.5.
>> >
>> >
>
> ** No answer noted - terrible how web site cannot answer
> questions.
>
You may need a web site to answer your questions. I don't.
>
>> Florescent lamps are common devices which have power factor correction. In
>> the older types capacitors were fitted. Electronic ballasts can achieve
>> high power factors, very close to unity. Electrical Authorities do
>> require minimum power factors for loads.
>
>
> ** This David fuckwit is now quoting whole chunks from some web site -
> parrot fashion.
>
I don't need to quote a web site. If I did I would post a URL. If you are
so smart and think it is from a web site then prove it and quote a URL,
instead of using foul language.
>
>
>> Computer power supplies are also available which have power factor
>> correction to comply with harmonic current emission standards (they just
>> cost more than your el cheapo chinese power supply).
>
>
> ** Don't ya just hate talking to a parrot.
I certainly do! Do you want a cracker?
David
Phil Allison
> > "David"
> >> On Fri, 02 Jan 2004 00:11:07 +1100, Phil Allison wrote:
> >
> >> >
> >> > ** Save 10% of what exactly ?
> >>
> >> As the current will 10% less, then the conductor size can be smaller.
> >
> >
> > ** That is sheer pedantry.
> >
> >> Assuming the above example, 2000VA at 240V is 8.3A, 1800VA is 7.5A.
> >> Depending on the installation method, and voltage drop requirements,
> >> 1.5mm^2 cable may only have a rating of say 8 Amps, calculated using
> >> AS3008.1 If you assume 2000VA, you would have to use 2.5mm cable,
which
> >> costs more than 1.5mm cable.
> >
> >
> > ** Yawn. You are inventing fanciful examples again.
> >
>
> This is an example from your figures.
** No it dam well is not
Real examples from real situations can have similar outcomes. Some of us do
have practical experience.
** But at what ??
> >> > If he uses his trusty clamp meter - then guess what ?
> >> >
> >>
> >> If it's not a true RMS clamp meter then he is stuffed.
> >
> >
> > * Standard rms calibrated meters are fine for any non distorting load.
> >
>
> Opps - You slipped up there Phil, assuming a sine wave are we?
** The slip was yours arsehole - you assumed non sine wave without any
mention.
Sparkies have been successfully using standard clamp meters for many
decades.
> >> Also, a bit late to use a clamp meter after the installation has been
> >> completed, and he has discovered that he has installed cable heavier
than
> >> required, and has cost himself profit by using oversized cables.
> >
> >
> > ** You seem to be assuming something without ever saying it. Are you
> > assuming some industrial situation with the permanently wired loads ??
I
> > was not.
>
> I have assumed nothing.
** You have assumed lots of things - and mentioned none of them.
>Domestic installations have permanently wired loads.
** Yes - nice, resistive water heaters and stoves .
No need for fancy phasor maths there.
You just shot both feet off.
> You may need a web site to answer your questions. I don't.
** Liar.
> >> Florescent lamps are common devices which have power factor correction.
In
> >> the older types capacitors were fitted. Electronic ballasts can achieve
> >> high power factors, very close to unity. Electrical Authorities do
> >> require minimum power factors for loads.
> >
> >
> > ** This David fuckwit is now quoting whole chunks from some web
site -
> > parrot fashion.
>
> I don't need to quote a web site.
** No need ? Then why do it ?
> >> Computer power supplies are also available which have power factor
> >> correction to comply with harmonic current emission standards (they
just
> >> cost more than your el cheapo chinese power supply).
> >
> >
> > ** Don't ya just hate talking to a parrot.
>
>
> I certainly do! Do you want a cracker?
** Piss off polly.
....... Phil
David
>
>
>> >> > If he uses his trusty clamp meter - then guess what ?
>> >> >
>> >>
>> >> If it's not a true RMS clamp meter then he is stuffed.
>> >
>> >
>> > * Standard rms calibrated meters are fine for any non distorting load.
>> >
>>
>> Opps - You slipped up there Phil, assuming a sine wave are we?
>
>
> ** The slip was yours arsehole - you assumed non sine wave without any
> mention.
>
> Sparkies have been successfully using standard clamp meters for many
> decades.
>
Pity switch mode power supplies havent. Sparkies need to keep up to
date. Be carefull playing with that cotton covered rubber insulated cable
in split conduit stuff you use, you may hurt yourself.
>
>> >> Also, a bit late to use a clamp meter after the installation has been
>> >> completed, and he has discovered that he has installed cable heavier
> than
>> >> required, and has cost himself profit by using oversized cables.
>> >
>> >
>> > ** You seem to be assuming something without ever saying it. Are you
>> > assuming some industrial situation with the permanently wired loads ??
> I
>> > was not.
>>
>> I have assumed nothing.
>
>
> ** You have assumed lots of things - and mentioned none of them.
>
>
>>Domestic installations have permanently wired loads.
>
>
> ** Yes - nice, resistive water heaters and stoves .
>
> No need for fancy phasor maths there.
>
> You just shot both feet off.
Forgotten about airconditioners, swimming pool pumps, fridges, Florescent
lamps etc ?
>
>> You may need a web site to answer your questions. I don't.
>
>
> ** Liar.
>
>
>
>> >> Florescent lamps are common devices which have power factor correction.
> In
>> >> the older types capacitors were fitted. Electronic ballasts can achieve
>> >> high power factors, very close to unity. Electrical Authorities do
>> >> require minimum power factors for loads.
>> >
>> >
>> > ** This David fuckwit is now quoting whole chunks from some web
> site -
>> > parrot fashion.
>
>>
>> I don't need to quote a web site.
>
>
> ** No need ? Then why do it ?
>
Please answer the question instead of ignoring it.
I don't need to quote a web site. If I did I would post a URL. If you are
so smart and think it is from a web site then prove it and quote a URL,
instead of using foul language.
David
Phil Allison
"David" <no_...@hotmail.com
> Pity switch mode power supplies havent.
** Pity you are posturing jerk.
> Forgotten about airconditioners,
** They plug in.
> fridges,
** Plug in.
>Florescent lamps etc ?
** Like homes have thousands of VA worth - get real.
> >> >> Florescent lamps are common devices which have power factor
correction.
> > In the older types capacitors were fitted. Electronic ballasts can
achieve
> >> >> high power factors, very close to unity. Electrical Authorities do
> >> >> require minimum power factors for loads.
> >> >
> >> > ** This David fuckwit is now quoting whole chunks from some web
> > site - parrot fashion.
> >
> >>
> >> I don't need to quote a web site.
> >
> > ** No need ? Then why do it ?
>
> Please answer the question instead of ignoring it.
** Not worth answering - the likes of you are not worth feeding.
> I don't need to quote a web site.
** What you do need is very painful.
......... Phil
David
>
>
>> Forgotten about airconditioners,
>
> ** They plug in.
Most dont.
>>
>
> ** What you do need is very painful.
>
>
Speaking from experience Phil ?
David
Phil Allison
> On Fri, 02 Jan 2004 15:04:37 +1100, Phil Allison wrote:
>
> >
> >
> >> Forgotten about airconditioners,
> >
> > ** They plug in.
>
> Most dont.
** At least 90 % do - you really are just another Hotmailer idiot.
......... Phil
David
Please provide a reference to back up this figure. Apart from very small
systems, most draw too much current for a 10A socket outlet.
David
Phil Allison
"David" <no_...@hotmail.com> wrote in message
news:pan.2004.01.02....@hotmail.com...
> On Fri, 02 Jan 2004 16:15:52 +1100, Phil Allison wrote:
>
> >
> > "David" <no_...@hotmail.com> wrote in message
> > news:pan.2004.01.02....@hotmail.com...
> >> On Fri, 02 Jan 2004 15:04:37 +1100, Phil Allison wrote:
> >>
> >> >
> >> >
> >> >> Forgotten about airconditioners,
> >> >
> >> > ** They plug in.
> >>
> >> Most dont.
> >
> >
> > ** At least 90 % do - you really are just another Hotmailer
idiot.
> >
>
> Please provide a reference to back up this figure.
** Open your eyes wanker - or have you already completed the job.
......... Phil
David
>
> "David" <no_...@hotmail.com> wrote in message
> news:pan.2004.01.02....@hotmail.com...
>> On Fri, 02 Jan 2004 16:15:52 +1100, Phil Allison wrote:
>>
>> >
>> > "David" <no_...@hotmail.com> wrote in message
>> > news:pan.2004.01.02....@hotmail.com...
>> >> On Fri, 02 Jan 2004 15:04:37 +1100, Phil Allison wrote:
>> >>
>> >> >
>> >> >
>> >> >> Forgotten about airconditioners,
>> >> >
>> >> > ** They plug in.
>> >>
>> >> Most dont.
>> >
>> >
>> > ** At least 90 % do - you really are just another Hotmailer
> idiot.
>> >
>>
>> Please provide a reference to back up this figure.
>
>
>
> ** Open your eyes wanker - or have you already completed the job.
Can't provide any reference to back up you fiction can you? Caught you out
have I ?
David
Phil Allison
"David" <no_...@hotmail.com>
** Open your eyes wanker - and stop farkking about with your little
sister too.
....... Phil
Tony Pearce
>
> ** Open your eyes wanker - or have you already completed the job.
This is Phil's way of saying you are right David, but he has never owned an
air conditioner, let alone know anything about them!
Better not discuss the inverter types and reason for soft start circuits
etc.
TonyP.
Phil Allison
"Tony Pearce"
aka 'Trevor"
aka " roverT"
aka biggest fake and cretin on Australian usenet.
>
> This is Phil's way of saying you are right David,
** Fuck you Pearceoid
You are one stinking turd.
.......... Phil
James
> If so - get outta here.
>
>
>
>
> ......... Phil
G'day Mates, Happy new year to everyone first up :)
Phil - please be nice to people, personal insults are unprofessional
to say the least and you are un doing some of the good advice you
give.
In smaller installations you do just add up all the VA's because you
know it will be a safe assumption. Especially if the installation will
use less than 100 amps per phase because thats the size you would
normally install anyway if the load is 100 amps or less so you are
just checking. With larger installations its different because the
costs are much higher. Also with long cable runs, they can get very
expensive very quickly so its a good idea check properly, the savings
are significant.
Sometimes when adding load to the installation it can be a very fine
line whether to upgrade the power supply or not. This is when careful
calculations are critical as you can save a lot of money. Installing
power factor correction equipment can also be done to get out of an
upgrade.
In most office installations I have seen, the load drawn by computers
is not really significant compared to the other loads in the building.
Having said that, they certainly cant be forgotten about, but compared
to the hundred of amps per phase the combined aircon load is, its
pretty small.
Anyway, that is based on what I have seen.
Cheers
James
Phil Allison
"James" <j...@octa4.net.au>
> snip snip snip
> > If so - get outta here.
> >
>
> > ......... Phil
>
> G'day Mates, Happy new year to everyone first up :)
>
> Phil - please be nice to people,
** There are no "people" here - just egos with keyboards.
> personal insults are unprofessional to say the least ...
** I agree - so I react strongly against them.
> and you are un doing some of the good advice you give.
** No point in me giving good advice when there are countless fuckwits
here undoing it all.
........... Phil
David
I know! Thanks Tony.
David
Phil Allison
"David" <no_...@hotmail.com>
> I know! Thanks Tony.
** Tony is a fake name - it belongs to a fake and a monstrous liar.
You and Tony make a fine pair - of arseholes.
.......... Phil
Kevin Ettery
>
> "Kevin Ettery" <kpet...@dcsi.net.au>
> > Guys,
> >
> > I AM an Electrical Engineer and do know a bit about this subject.
>
>
> ** Shame you are not an electronics engineer - then you might just get
it
> right.
Phil,
My degree is a Bachelor of Engineering (Electrical/Electronic) - what's your
qualification? Also I have practical working experience with the systems I
described in my email.
> > Real Power (what provides the apparent power output) in an ac circuit is
> > given by:
> > P =VI cos theta
> > where theta is the angle between the current and the voltage.
> > and the product VI is called the VOLT-AMPS (written as VA).
> > cos theta is the power factor.
>
> ** Tell me Kevin - in the EE game, is the phrase "ac circuit" code
speak
> for something ? Is is code speak for those systems where voltages and
> current waveforms are all sinusoidal ???????????
>
> If so - get outta here.
Phil, I used the term "ac circuit" so my email would be comprensible to the
majority of people reading it - "ac systems" is the more correct term. And
it is not a game, its a profession.
Also, in ac systems the waveforms are very rarely pure 50 Hz sinusoids,
however almost any waveform can be very closely approximated by the summing
a fundamental and various harmonics of suitable magnitudes, using Fourier
analysis (even a square wave can be approximated so). If you are not aware
of this then you are not really in a position
I mentioned power factors as I deal with these every day in my work, and
because they're fairly simple to explain.
I only described the steady-state conditions (normal running conditions), as
dealing with non-steady-state conditions (ie switching transients, fault
behaviour) is fairly specialised and difficult to describe in comprehensible
layman terms. I have experience with such non-steady-state conditions as
one of my jobs, for a few years, was in power system protection which, among
other things, required analysis of the fault response of three phase
systems - 3phase faults, phase to phase faults and phase to ground faults.
Regards
Kevin
Phil Allison
"Kevin Ettery" <
>
> "Phil Allison" <...
> > > Guys,
> > >
> > > I AM an Electrical Engineer and do know a bit about this subject.
> > ** Shame you are not an electronics engineer - then you might just
get
> it right.
>
> Phil,
>
> My degree is a Bachelor of Engineering (Electrical/Electronic) - what's
your
> qualification? Also I have practical working experience with the systems
I
> described in my email.
** But still you are unaware that Power Factor is simply Watts divided by
VA ???
And that VA is simply rms Volts multiplied by rms Amps ????
> > ** Tell me Kevin - in the EE game, is the phrase "ac circuit" code
> speak for something ? Is is code speak for those systems where voltages
and
> > current waveforms are all sinusoidal ???????????
> >
> > If so - get outta here.
>
> Phil, I used the term "ac circuit" so my email would be comprensible to
the
> majority of people reading it - "ac systems" is the more correct term.
** A rose by any name .......
But you did not answer my question.
> And it is not a game, its a profession.
** Yawn.
> Also, in ac systems the waveforms are very rarely pure 50 Hz sinusoids,
** Still wondering if "ac systems" includes ones with non sinewave
currents ??
Hmmmmmm, have patience Phil.
> however almost any waveform can be very closely approximated by the
summing
> a fundamental and various harmonics of suitable magnitudes, using Fourier
> analysis (even a square wave can be approximated so). If you are not
aware
> of this then you are not really in a position
** Cripes - now he drags old Fourier out of his grave !!!!
A good mathematician never dies ........
> I mentioned power factors as I deal with these every day in my work, and
> because they're fairly simple to explain.
** But you got the explanation wrong - considering this is an
***electronics*** NG.
> I only described the steady-state conditions (normal running conditions),
as
> dealing with non-steady-state conditions (ie switching transients, fault
> behaviour) is fairly specialised and difficult to describe in
comprehensible
> layman terms.
** What about non sine loads ??????
I have experience with such non-steady-state conditions as
> one of my jobs, for a few years, was in power system protection which,
among
> other things, required analysis of the fault response of three phase
> systems - 3phase faults, phase to phase faults and phase to ground faults.
** Snore ......
......... Phil
Tony Pearce
> You and Tony make a fine pair - of arseholes.
This is Phil's way of admitting he cannot refute the argument so must attack
the poster, where he has much more experience than any other Usenet poster.
*Words of wisdom. : Never argue with Phil A, he will drag the argument down
to personal abuse, then beat you with his greater experience.
*With apologies to Dilbert :-)
TonyP.
Phil Allison
>
> "Phil Allison" <phila...@optusnet.com.au> wrote in message
> news:3ff56a26$0$18752$afc3...@news.optusnet.com.au...
> > You and Tony make a fine pair - of arseholes.
>
> This is Phil's way of admitting he cannot refute the argument ....
** More lies, all lies, nothing but lies......
This guy is store window dresser - hes like to fondle the mannequins.
......... Phil
Kevin Ettery
>
> "Kevin Ettery" <
> >
> > "Phil Allison" <...
>
> > > > Guys,
> > > >
> > > > I AM an Electrical Engineer and do know a bit about this subject.
>
> > > ** Shame you are not an electronics engineer - then you might just
> get
> > it right.
> >
> > Phil,
> >
> > My degree is a Bachelor of Engineering (Electrical/Electronic) - what's
> your
> > qualification? Also I have practical working experience with the
systems
> I
> > described in my email.
>
>
> ** But still you are unaware that Power Factor is simply Watts divided by
> VA ???
>
> And that VA is simply rms Volts multiplied by rms Amps ????
Duh!! Hardly surprising that PF = watts/VA, cosidering that the calculation
for Watts is VA multiplied by Cos theta (which is the power factor). Any
Year 7 maths student could do that formula juggling.
What do you think harmonic analysers use when calculating harmonic
components - they use Fourier analysis, not black magic.
> > I mentioned power factors as I deal with these every day in my work, and
> > because they're fairly simple to explain.
>
> ** But you got the explanation wrong - considering this is an
> ***electronics*** NG.
Pray tell what did I get wrong with my explanation?
> > I only described the steady-state conditions (normal running
conditions),
> as
> > dealing with non-steady-state conditions (ie switching transients, fault
> > behaviour) is fairly specialised and difficult to describe in
> comprehensible
> > layman terms.
>
> ** What about non sine loads ??????
Non-sine ! What do you mean by 'Non-sine'. I suspect what you're referring
to are non-linear loads (inductive and capacitive loads are actually linear
loads). They're a whole level of complexity higher - I'd have to brush up
on the descriptions of these systems since its been a few years since I've
had to deal with them.
> I have experience with such non-steady-state conditions as
> > one of my jobs, for a few years, was in power system protection which,
> among
> > other things, required analysis of the fault response of three phase
> > systems - 3phase faults, phase to phase faults and phase to ground
faults.
>
> ** Snore ......
At least I understand fault calculations Phil. Do you.
Phil Allison
"Kevin Ettery" <
> "Phil Allison" <
> > ** But still you are unaware that Power Factor is simply Watts divided
by
> > VA ???
> >
> > And that VA is simply rms Volts multiplied by rms Amps ????
>
> Duh!! Hardly surprising that PF = watts/VA, considering that the
calculation
> for Watts is VA multiplied by Cos theta (which is the power factor).
** You are still wrong: Watts = VA times PF.
No angles involved - see.
> > > I mentioned power factors as I deal with these every day in my work,
and
> > > because they're fairly simple to explain.
> >
> > ** But you got the explanation wrong - considering this is an
> > ***electronics*** NG.
>
> Pray tell what did I get wrong with my explanation?
** I have told you already.
> >
> > ** What about non sine loads ??????
>
> Non-sine ! What do you mean by 'Non-sine'. I suspect what you're
referring
> to are non-linear loads...
** Yes.
>
> > I have experience with such non-steady-state conditions as
> > > one of my jobs, for a few years, was in power system protection which,
> > among other things, required analysis of the fault response of three
phase
> > > systems - 3phase faults, phase to phase faults and phase to ground
> faults.
> >
> > ** Snore ......
>
> At least I understand fault calculations Phil. Do you.
** Go annoy the folk on "aus.electrical.pedants" - not me you wanker.
......... Phil
David
Kevin, This is Phil's way of saying he has no idea what a fault
calculation is and why it is important, yet alone how to do them.
David
Phil Allison
"David" <no_...@hotmail.com>
>
> Kevin, This is Phil's way of saying he has no idea what a fault
> calculation is and why it is important, yet alone how to do them.
** The issue is not relevant to the topic - it is a complete red
herring.
Funny how many red-herrings are produced by bloated toad fish.
........ Phil
Tony Pearce
> This guy is store window dresser - hes like to fondle the
mannequins.
** More lies, all lies, nothing but lies......
It's Phil who likes his blow up dolls :-)
TonyP.
David
>
> "Kevin Ettery" <
>
>> "Phil Allison" <
>
>
>> > ** But still you are unaware that Power Factor is simply Watts divided
> by
>> > VA ???
>> >
>> > And that VA is simply rms Volts multiplied by rms Amps ????
>
>>
>> Duh!! Hardly surprising that PF = watts/VA, considering that the
> calculation
>> for Watts is VA multiplied by Cos theta (which is the power factor).
>
>
> ** You are still wrong: Watts = VA times PF.
>
> No angles involved - see.
>
The angle in the power factor, as Kevin has stated. If the voltage and
current waveforms are in phase, then regardless of the waveform, the power
factor will be unity. There HAS to be a phase difference between the
voltage and current waveform for their to be reactive power, and thus a
power factor of less than one.
David
Phil Allison
"David" <no_...@hotmail.com
> The angle in the power factor, as Kevin has stated. If the voltage and
> current waveforms are in phase, then regardless of the waveform, the power
> factor will be unity.
** Wrong - you are totally wrong.
There is no reference to phase angle in the general definition of PF.
There HAS to be a phase difference between the
> voltage and current waveform for their to be reactive power,
** There is no reference to "'reactive power" in the general definition of
PF.
A non unity power factor exists whenever the current waveform does not
follow the voltage waveform - though they may be exactly in phase. This is
the case with typical electronic devices where current is drawn only at AC
voltage peaks. Such loads have PFs of around 0.5.
How many times do I have to say it.
....... Phil
Vermin
Please don't feed the troll Tony.
I've got Pill killfiltered but unfortunately I still get to see his
antisocial ramblings when others get sucked into his "little mind"
games.
Phil Allison
"Vermin" <Ver...@nowhere.com>
>
> Please don't feed the troll Tony.
>
> I've got Pill killfiltered but unfortunately I still get to see his
> antisocial ramblings when others get sucked into his "little mind"
> games.
** Look what is talking and what it calls itself.
Look at what it is talking to as well.
.......... Phil
Kevin Ettery
Phil,
Repetition does not make it so.
Your description "a non unity power factor exists whenever the current
waveform does not follow the voltage waveform - though they may be exactly
in phase" sounds more like the dscription of a factor called a Form Factor
(FF). This is usually relevant to thyristor (SCRs, Triacs, etc.) circuits,
but could be applied to circuits with similar functions.
For a description and definition of power factor I refer the learned
gentleman to the following book:
"Schaum's Outline Series - Theory and Problems of Electric Circuits" (the
edition I have is written by Joseph A. Edminster). It is a fairly simple
and straight-forward text and devotes an entire chapter to power and power
factor (it was one of my second year Degree texts).
Also can you please advise what "typical electronic devices" only draw
current at AC voltage peaks (unloaded full wave rectifiers are the only
things I'm aware of that do this).
BTW loads with a PF of 0.5 are seriously inductive wouldn't you agree? Most
transformers are between 0.8 and 0.95 (some of the toroidals push to about
0.98).
Kevin
Phil Allison
"Kevin Ettery" <kpet...@dcsi.net.au
> Repetition does not make it so.
** You are ignoring facts and the formulae.
( snip irrelevant stuff )
> Also can you please advise what "typical electronic devices" only draw
> current at AC voltage peaks (unloaded full wave rectifiers are the only
> things I'm aware of that do this).
** All of them - excepting only ones that have "active power factor
correction". The vast majority draw current in pulses of about 2 mS
duration 100 times per second. Pulses have a higher rms value than a sine
wave which inflates the VA figure over the watts consumed - hence the poor
PF figure.
> BTW loads with a PF of 0.5 are seriously inductive wouldn't you agree?
** There are none so blind as those who will not see - and pompous damn
fools like you.
........... Phil
Kevin Ettery
>
> "Kevin Ettery" <kpet...@dcsi.net.au
>
>
>
> > Repetition does not make it so.
>
>
> ** You are ignoring facts and the formulae.
>
I was just curious about your source of facts and formulae.
>
> ( snip irrelevant stuff )
Of course you're going to ship a reference that contradicts what you say
>
> > Also can you please advise what "typical electronic devices" only draw
> > current at AC voltage peaks (unloaded full wave rectifiers are the only
> > things I'm aware of that do this).
>
>
> ** All of them - excepting only ones that have "active power factor
> correction". The vast majority draw current in pulses of about 2 mS
> duration 100 times per second. Pulses have a higher rms value than a sine
> wave which inflates the VA figure over the watts consumed - hence the
poor
> PF figure.
>
100 times a second (which sounds like twice the 50Hz ac supply frequency)
and drawing current for 2mS. Sounds like we are talking about a full wave
rectifier with moderately large smoothing capacitors - those parameters you
describe sound about right for this type of rectifier. It appears the
values you are using to calculate what you call PF are ac input volts, ac
input current and dc output power.
And for this sort of circuit phase angles, leading or lagging, are not
relevant as they apply only to ac circuits, not this which is essentially a
dc circuit with a little 100Hz ripple.
You should have indicated at the start of this thread that you were
discussing rectifiers and dc circuits. Now that I have it in the correct
context I can see what your comments referred to. What misled me is that
you referred to Power Factor - what you refer to as Power Factor is a power
in Vs power out ratio akin to efficiency. I think my texts call it a Power
Efficiency Ratio [PER], but I'd have to do a bit of looking as its been a
while since I had to refer to this field, and particularly these
parameters - its pertinent if you're designing rectifiers, but most of my
work has just been using them.
> > BTW loads with a PF of 0.5 are seriously inductive wouldn't you agree?
>
> ** There are none so blind as those who will not see - and pompous damn
> fools like you.
In an ac circuit a Power Factor of 0.5 is seriously inductive. In an
essentially dc circuit what you are calling Power Factor is not related to
inductance or capacitance at all (once the initial transients settle down).
And Phil, based on some of the comments directed to others on this group,
you're obviously speaking from first-hand knowledge of pompous fools.
Phil Allison
"Kevin Ettery" <
> "Phil Allison"
> >
> > > Repetition does not make it so.
> >
> >
> > ** You are ignoring facts and the formulae.
> >
>
> I was just curious about your source of facts and formulae.
** So PF = watts / VA is a mystery to you ???
> > ( snip irrelevant stuff )
>
> Of course you're going to ship a reference that contradicts what you say
** It was irrelevant - same as all your points.
> > > Also can you please advise what "typical electronic devices" only draw
> > > current at AC voltage peaks (unloaded full wave rectifiers are the
only
> > > things I'm aware of that do this).
> >
> >
> > ** All of them - excepting only ones that have "active power factor
> > correction". The vast majority draw current in pulses of about 2 mS
> > duration 100 times per second. Pulses have a higher rms value than a
sine
> > wave which inflates the VA figure over the watts consumed - hence the
> poor PF figure.
>
> 100 times a second (which sounds like twice the 50Hz ac supply frequency)
> and drawing current for 2mS. Sounds like we are talking about a full wave
> rectifier with moderately large smoothing capacitors - those parameters
you
> describe sound about right for this type of rectifier. It appears the
> values you are using to calculate what you call PF are ac input volts, ac
> input current and dc output power.
** Only one figure needs measurement.
The rms current draw from the AC supply.
One just needs a "true rms" amp meter.
>
> You should have indicated at the start of this thread that you were
> discussing rectifiers and dc circuits.
** What do you imagine the load presented by an electronic device is ??
What is the title of this newsgroup ????
I mentioned AC to DC conversion and current draw only during voltage
peaks several times.
> Now that I have it in the correct context I can see what your comments
referred to.
** You are getting further away - not closer.
> > > BTW loads with a PF of 0.5 are seriously inductive wouldn't you agree?
> >
> > ** There are none so blind as those who will not see - and pompous
damn
> > fools like you.
>
> In an ac circuit a Power Factor of 0.5 is seriously inductive.
** All you do is repeat endlessly your stupid one eyed view of PF.
PF = watts / VA ........ there is no phase angle to be seen
!!!!!!!!
>
> And Phil, based on some of the comments directed to others on this group,
> you're obviously speaking from first-hand knowledge of pompous fools.
** Usenet is full of them - you are one of thousands talking on a
topic you have NOT got the first clue about.
........... Phil
Franc Zabkar
<phila...@optusnet.com.au> put finger to keyboard and composed:
>
>"David" <no_...@hotmail.com
>
> > The angle in the power factor, as Kevin has stated. If the voltage and
>> current waveforms are in phase, then regardless of the waveform, the power
>> factor will be unity.
>
>
> ** Wrong - you are totally wrong.
>
> There is no reference to phase angle in the general definition of PF.
>
>
> There HAS to be a phase difference between the
>> voltage and current waveform for their to be reactive power,
>
>
> ** There is no reference to "'reactive power" in the general definition of
>PF.
>
> A non unity power factor exists whenever the current waveform does not
>follow the voltage waveform - though they may be exactly in phase. This is
>the case with typical electronic devices where current is drawn only at AC
>voltage peaks.
In typical electronic devices current pulses are drawn *before* the
voltage peaks. They are not *exactly* in phase.
>Such loads have PFs of around 0.5.
Computer PSUs are typically quoted as having a capacitive PF of 0.65
at full load, so your estimate of 0.5 may not be far off the mark.
Intuitively, however, I find it hard to correlate a 2ms current pulse
(as you state elsewhere) with an 0.5 PF. A duration of 2ms equates to
36 degrees at 50Hz, which would in the normal textbook "sinusoidal"
case result in a PF of 0.81. An 0.5 PF would normally suggest a phase
angle of 60 deg.
Having said that, here is an article that attempts to explain this
apparent anomaly:
http://www.st.com/stonline/books/pdf/docs/3711.pdf
It suggests that the actual PF is somewhat less than that which could
be expected by considering only the phase angle of the fundamental
component of the current pulse. Instead, the article shows that the
harmonic components of the current pulse tend to reduce the overall
PF.
> How many times do I have to say it.
Not surprisingly, people prefer an independent, authoritative
reference.
- Franc Zabkar
--
Please remove one 's' from my address when replying by email.
Phil Allison
"Phil Allison"
> >
> > ** There is no reference to "'reactive power" in the general definition
of
> >PF.
> >
> > A non unity power factor exists whenever the current waveform does not
> >follow the voltage waveform - though they may be exactly in phase. This
is
> >the case with typical electronic devices where current is drawn only at
AC
> >voltage peaks.
>
> In typical electronic devices current pulses are drawn *before* the
> voltage peaks. They are not *exactly* in phase.
** The current peak occurs *very close* to the voltage peak.
SMPS and transformer PSUs vary slightly.
>
> >Such loads have PFs of around 0.5.
>
> Computer PSUs are typically quoted as having a capacitive PF of 0.65
> at full load,
** The PF of a SMPS is not capacitive - it is due to waveform
distortion.
Also, the actual PF worsens at lower ie more typical loads as the charging
pulse gets narrower.
> so your estimate of 0.5 may not be far off the mark.
** It was not an "estimate".
> Intuitively, however, I find it hard to correlate a 2ms current pulse
> (as you state elsewhere) with an 0.5 PF.
** Simple. Imagine the current pulse is rectangular and centred on the AC
voltage peak.
Then the power consumed in watts is V peak x I peak x 0.2
For 240 AC and a 1 amp pulse this is 335 x 1 x 0.2 = 67 watts.
***Now for the VA:
The rms current for a rectangular pulse train is I peak x sq rt duty
cycle.
So in the example 1 x sq rt 0.2 = 0.447 amps rms.
The AC supply is 240 V rms so the load = 107.3 VA.
Such a load has a PF of 67 / 107.3 = 0.62
Change the pulse duration to 1 mS ( ie when the supply is lightly loaded)
and do the same calculation.
The new PF = 0.44.
>A duration of 2ms equates to
> 36 degrees at 50Hz, which would in the normal textbook "sinusoidal"
> case result in a PF of 0.81. An 0.5 PF would normally suggest a phase
> angle of 60 deg.
** The sub unity PF is not due to phase angle but wave shape and the
higher rms value of that wave shape.
>
> > How many times do I have to say it.
>
> Not surprisingly, people prefer an independent, authoritative
> reference.
** Shame those folk here have no brains to see the obvious nor can do a
simple Google on "active PFC".
............ Phil
Franc Zabkar
<phila...@optusnet.com.au> put finger to keyboard and composed:
>"Franc Zabkar" <
>"Phil Allison"
>
>> >
>> > ** There is no reference to "'reactive power" in the general definition
>of
>> >PF.
>> >
>> > A non unity power factor exists whenever the current waveform does not
>> >follow the voltage waveform - though they may be exactly in phase. This
>is
>> >the case with typical electronic devices where current is drawn only at
>AC
>> >voltage peaks.
>>
>> In typical electronic devices current pulses are drawn *before* the
>> voltage peaks. They are not *exactly* in phase.
>
>
> ** The current peak occurs *very close* to the voltage peak.
>
> SMPS and transformer PSUs vary slightly.
>
>
>>
>> >Such loads have PFs of around 0.5.
>>
>> Computer PSUs are typically quoted as having a capacitive PF of 0.65
>> at full load,
>
>
> ** The PF of a SMPS is not capacitive - it is due to waveform
>distortion.
OK, substitute "leading" for "capacitive".
> Also, the actual PF worsens at lower ie more typical loads as the charging
>pulse gets narrower.
>
>
>
>> so your estimate of 0.5 may not be far off the mark.
>
>
> ** It was not an "estimate".
>
>
>
>> Intuitively, however, I find it hard to correlate a 2ms current pulse
>> (as you state elsewhere) with an 0.5 PF.
>
>
> ** Simple. Imagine the current pulse is rectangular and centred on the AC
>voltage peak.
>
> Then the power consumed in watts is V peak x I peak x 0.2
>
> For 240 AC and a 1 amp pulse this is 335 x 1 x 0.2 = 67 watts.
>
>***Now for the VA:
>
> The rms current for a rectangular pulse train is I peak x sq rt duty
>cycle.
>
> So in the example 1 x sq rt 0.2 = 0.447 amps rms.
>
> The AC supply is 240 V rms so the load = 107.3 VA.
>
> Such a load has a PF of 67 / 107.3 = 0.62
>
> Change the pulse duration to 1 mS ( ie when the supply is lightly loaded)
>and do the same calculation.
>
> The new PF = 0.44.
I guess that one could argue, from an intuitive standpoint, that the
filter capacitor is now representing a greater component of the load,
and hence the PF falls as a consequence of this.
>>A duration of 2ms equates to
>> 36 degrees at 50Hz, which would in the normal textbook "sinusoidal"
>> case result in a PF of 0.81. An 0.5 PF would normally suggest a phase
>> angle of 60 deg.
>
>
> ** The sub unity PF is not due to phase angle but wave shape and the
>higher rms value of that wave shape.
>
>
>>
>> > How many times do I have to say it.
>>
>> Not surprisingly, people prefer an independent, authoritative
>> reference.
>
>
> ** Shame those folk here have no brains to see the obvious nor can do a
>simple Google on "active PFC".
Some may not be as conversant with this subject as you, and that
includes me, but that does not make us stupid. If you had illustrated
your case with your example early in this thread, there would have
been no need for a protracted slanging match. I wager that most people
would agree that the PF concept presented in your example is
counter-intuitive, and therefore not obvious, but the maths
nevertheless makes sense. The only thing that bothers me is that you
have used an artificial, unrepresentative pulse shape. If instead you
use a triangular pulse of 2A amplitude, then the wattage remains the
same, but the VA increases somewhat, ie by a factor of 1.16. This
would bring your 0.62 PF figure closer to 0.53, which is what you
stated from the outset.
Phil Allison
"Franc Zabkar"
"Phil Allison"
>
> > ** The PF of a SMPS is not capacitive - it is due to waveform
> >distortion.
>
> OK, substitute "leading" for "capacitive".
** It is not leading either - it is due to waveshape distortion.
>
> I guess that one could argue, from an intuitive standpoint, that the
> filter capacitor is now representing a greater component of the load,
> and hence the PF falls as a consequence of this.
** The issue is the distorted waveshape and rms current values thereof.
It has NOTHING to do with angles, reactances, leading, lagging etc.
> >>
> >> > How many times do I have to say it.
> >>
> >> Not surprisingly, people prefer an independent, authoritative
> >> reference.
> >
> >
> > ** Shame those folk here have no brains to see the obvious nor can do a
> >simple Google on "active PFC".
>
> Some may not be as conversant with this subject as you, and that
> includes me, but that does not make us stupid.
** When you stubbornly ignore the facts that are posted you are being
stupid.
When you stubbornly and mindlessly repeat a basic error in face of the
correct answer your are being stupid.
When you publicly insult the person presenting the truth - then you are
an utter arsehole.
> If you had illustrated your case with your example early in this thread,
there would have
> been no need for a protracted slanging match.
** You were the first here to even hint that you wanted to see one.
The other two morons were hell bent on ignoring my case and burying the
truth in order to protect their pathetic egos.
>I wager that most people would agree that the PF concept presented in your
example is
> counter-intuitive, and therefore not obvious, but the maths nevertheless
makes sense.
** Most people associate PF only with reactive phase angles, but that is an
error. People who have engineering backgrounds and qualifications should
not make that error, or at least when it is pointed out realise it and not
persecute the person saying the truth.
> The only thing that bothers me is that you
> have used an artificial, unrepresentative pulse shape.
** The pulse can be modelled as a simple rectangle - just square off the
top of the pulse and sharpen up the sides and you have it.
> If instead you use a triangular pulse of 2A amplitude,
** But it is NOT a triangle - the pulse shape resembles the last 1.5
cm of your little finger.
>then the wattage remains the same, but the VA increases somewhat, ie by a
factor of 1.16. This
> would bring your 0.62 PF figure closer to 0.53, which is what you stated
from the outset.
** The was no debate over the actual PF figure - you maliciously
misrepresent things.
The issue was the inability of two posturing fools to concede the existence
sub unity PF with NO PHASE ANGLE at all.
Something that nearly all mains powered electronic devices ever made present
as a load, something which is a recognised problem in electricity supply and
for which there is a modern solution - ie active PFC.
......... Phil
Peter_purple
Phil Allison wrote:> > If you have electronic loads with whatever PF then the
VAs just add.
>
> ** There is no PFC in the vast majority of electronic items - including
> some that draw up to 4000 VA from a single phase circuit. Also there is
> virtually no phase angle, peak current draw co-insides exactly with peak
> voltage. The PF of such loads is about 0.5.
>
> ........ Phil
Wrong, if there is no phase angle the pf = 1
thats its definition.
--
that is because,
peter purple proclaims
The infamous big dollar scam list of email addresses, growing,
appended here for those nice news bots to collect their email addresses...
benjamin...@yahoo.com, williams...@hotmail.com, frpie...@yahoo.fr,
"fprohanp" <fpro...@libero.it>, "mariam seseseko" <mase...@yahoo.ca>,
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<geral...@fsmail.net>,
<geral...@yahoo.co.uk>, <tariq...@gawab.com>, <rome...@freesurf.fr>,
"tariq ani" <tariqa...@fsmail.net>, "Dr. Isa Gambo"
<gamb...@zwallet.com>,
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From: "Dr. Ahmed Banigo" <ahm...@aol.com>, "Dr. Ahmed Banigo"
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"TEMBA BAKO NGANBA" <tawand...@netscape.net>, "mariam abacha"
<mariam_66...@fsmail.net>
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"ddlucian" <ddlu...@libero.it>, tony...@fsmail.net, tony...@maktoob.com,
jerry brown <jerry...@fsmail.net>, "DR HARRISON OHALE......."
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Peter_purple
>
> In vector terms both the current and voltage are vectors rotating in space.
No, they are not rotating in space, there are no coordinates, cartesian or
otherwise !full stop.
Its sheer imagination to say they are rotating in space, they are only
ever rotating with respect to each other, i have ben abstract maths
expert so I know this to be a mental construct.
> Hope it cleared a few things up.
with the exception of your appreciation of space,
Phil Allison
"Peter_purple" <pur...@203.0.178.192> wrote in message
news:3FFBAA77...@203.0.178.192...
>
>
> Phil Allison wrote:> > If you have electronic loads with whatever PF then
the
> VAs just add.
>
> >
> > ** There is no PFC in the vast majority of electronic items -
including
> > some that draw up to 4000 VA from a single phase circuit. Also there is
> > virtually no phase angle, peak current draw co-insides exactly with
peak
> > voltage. The PF of such loads is about 0.5.
> >
> > ........ Phil
>
> Wrong, if there is no phase angle the pf = 1
>
> thats its definition.
** Go drop dead Moron.
......... Phil
Franc Zabkar
<phila...@optusnet.com.au> put finger to keyboard and composed:
>
>"Franc Zabkar"
>
>"Phil Allison"
>
>
>>
>> > ** The PF of a SMPS is not capacitive - it is due to waveform
>> >distortion.
>>
>> OK, substitute "leading" for "capacitive".
>
>
> ** It is not leading either - it is due to waveshape distortion.
>
>
>>
>> I guess that one could argue, from an intuitive standpoint, that the
>> filter capacitor is now representing a greater component of the load,
>> and hence the PF falls as a consequence of this.
>
>
> ** The issue is the distorted waveshape and rms current values thereof.
>
> It has NOTHING to do with angles, reactances, leading, lagging etc.
OK, I see that now.
The idea that PF is related to phase angles is a consequence of the
way the concept is taught. If students were instead taught that
lagging and leading PFs were merely special cases of a more general
principle, then such ingrained misconceptions would not arise. You are
the first person to challenge my understanding (sic!) of PF, and to
demonstrate to my satisfaction that there need not be any phase angle
between current and voltage peaks for a sub-unity PF to exist.
I may not like you, but that does not prevent me from acknowledging
you when you are right. I have no ego to protect, and I in no way feel
diminished when I confess my ignorance, especially if I can learn
something in the process.
>> The only thing that bothers me is that you
>> have used an artificial, unrepresentative pulse shape.
>
>
> ** The pulse can be modelled as a simple rectangle - just square off the
>top of the pulse and sharpen up the sides and you have it.
>
>
>> If instead you use a triangular pulse of 2A amplitude,
>
>
> ** But it is NOT a triangle - the pulse shape resembles the last 1.5
>cm of your little finger.
Once again I am basing my statement on the ideas presented in basic
engineering texts. Apparently reality is different (no sarcasm
implied).
>>then the wattage remains the same, but the VA increases somewhat, ie by a
>factor of 1.16. This
>> would bring your 0.62 PF figure closer to 0.53, which is what you stated
>from the outset.
>
>
> ** The was no debate over the actual PF figure - you maliciously
>misrepresent things.
You are paranoid. There was no malice in any of my statements. In fact
I was conceding that your originally quoted figure of 0.5 sounded
reasonable after I had initially doubted it.
>The issue was the inability of two posturing fools to concede the existence
>sub unity PF with NO PHASE ANGLE at all.
The difficulty with this concept is that it bears no relation to
actual electronic loads. All common PSUs have a phase angle, however
small, between voltage and current peaks. The existence of such a
phase angle tends to reinforce the misconception that it alone is
responsible for the PF. That is why your statement appears
counterintuitive. In fact it is only now that I can understand why a
PSU with a PF of 0.65 need not necessarily have a massive ripple
voltage on the mains filter cap.
>Something that nearly all mains powered electronic devices ever made present
>as a load, something which is a recognised problem in electricity supply and
>for which there is a modern solution - ie active PFC.
- Franc Zabkar
Phil Allison
"Franc Zabkar" <fza...@optussnet.com.au>
> >The issue was the inability of two posturing fools to concede the
existence
> >sub unity PF with NO PHASE ANGLE at all.
>
> The difficulty with this concept is that it bears no relation to
> actual electronic loads. All common PSUs have a phase angle, however
> small, between voltage and current peaks.
** A small amount of asymmetry in the charging pulse shape about the AC
supply voltage peak has no effect on the wattage or PF. In a sine wave
system even a small phase angle immediately creates a situation were the
voltage and current are of opposing polarity for part of the cycle - this
is not so for the electronic load case.
>
>The existence of such a phase angle tends to reinforce the misconception
that it alone is
> responsible for the PF.
** There are cases where is is exact - does this make the PF drop to zero
?
The power factor of even a 10 degree phase angle is 0.99.
You are clutching at straws.
> That is why your statement appears counterintuitive. In fact it is only
now that I can understand why a
> PSU with a PF of 0.65 need not necessarily have a massive ripple
> voltage on the mains filter cap.
** Oh no !! Not the filter capacitor value affects the rms draw debate
again !!!!!
BTW There is a simple way to measure the true watts consumed of a
typical electronic load - armed with the knowledge that phase angle is not
relevant.
The duration of the current pulse matches the nearly flat top of the AC
supply wave - so you can take the supply voltage as being fixed at a value
like 330 volts. All you need is the average rectified value of the current
pulses - easily measured with a standard meter ( not true rms) and then
multiply the reading by 0.9 to remove the sine wave rms calibration factor.
The *average* current value times a fixed voltage gives true watts -
same as from a DC supply.
........ Phil
Brian Goldsmith
"Franc Zabkar" wrote about Phil Allison:-
The idea that PF is related to phase angles is a consequence of the
way the concept is taught. If students were instead taught that
lagging and leading PFs were merely special cases of a more general
principle, then such ingrained misconceptions would not arise. You are
the first person to challenge my understanding (sic!) of PF, and to
demonstrate to my satisfaction that there need not be any phase angle
between current and voltage peaks for a sub-unity PF to exist.
I may not like you, but that does not prevent me from acknowledging
you when you are right. I have no ego to protect, and I in no way feel
diminished when I confess my ignorance, especially if I can learn
something in the process.
***** Took a big man to write the above! More power to him!!!!!
Brian Goldsmith.