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In principle that will work, but if the waveform is complicated, soOn Jun 3, 6:38 pm, "Phil Allison" <philalli...@tpg.com.au> wrote:
"panfilero"
And then just take the product Pavg = Vrms*Irms
** The formula for Volt Amperes or VA.
.... I don't see why I would need the phase angles to find the power.
** Fraid you do and there will not be a simple number either.
...... Phil
ok phil, so you are telling me that I can not put a shunt resistor in
line with one of my fan windings and simply measure the voltage
waveform across that shunt, and convert that waveform to an RMS
voltage and then divide that by the shunt resistance in order to get
my RMS current? And once I have both the RMS voltage and current,
that I can't multiply those values together in order to see how much
power the winding is consuming?
I'm not just directing this at Phil, if anyone thinks this would not
work please let me know, I don't see what is wrong with this approach.
much thanks.
J
ok... after reading some of the documents you guys have pointed me to,On Jun 4, 8:50 am, panfilero <panfil...@gmail.com> wrote:
On Jun 3, 6:38 pm, "Phil Allison" <philalli...@tpg.com.au> wrote:
"panfilero"
And then just take the product Pavg = Vrms*Irms
** The formula for Volt Amperes or VA.
.... I don't see why I would need the phase angles to find the power.
** Fraid you do and there will not be a simple number either.
...... Phil
ok phil, so you are telling me that I can not put a shunt resistor in
line with one of my fan windings and simply measure the voltage
waveform across that shunt, and convert that waveform to an RMS
voltage and then divide that by the shunt resistance in order to get
my RMS current? And once I have both the RMS voltage and current,
that I can't multiply those values together in order to see how much
power the winding is consuming?
I'm not just directing this at Phil, if anyone thinks this would not
work please let me know, I don't see what is wrong with this approach.
much thanks.
J
In principle that will work, but if the waveform is complicated, so
will be the calculations.
There are cheap in-line meters intended for use by radio-controlledOn Jun 4, 11:01 am, Richard Henry <pomer...@hotmail.com> wrote:
On Jun 4, 8:50 am, panfilero <panfil...@gmail.com> wrote:
On Jun 3, 6:38 pm, "Phil Allison" <philalli...@tpg.com.au> wrote:
"panfilero"
And then just take the product Pavg = Vrms*Irms
** The formula for Volt Amperes or VA.
.... I don't see why I would need the phase angles to find the power..
** Fraid you do and there will not be a simple number either.
...... Phil
ok phil, so you are telling me that I can not put a shunt resistor in
line with one of my fan windings and simply measure the voltage
waveform across that shunt, and convert that waveform to an RMS
voltage and then divide that by the shunt resistance in order to get
my RMS current? And once I have both the RMS voltage and current,
that I can't multiply those values together in order to see how much
power the winding is consuming?
I'm not just directing this at Phil, if anyone thinks this would not
work please let me know, I don't see what is wrong with this approach.
much thanks.
J
In principle that will work, but if the waveform is complicated, so
will be the calculations.
ok... after reading some of the documents you guys have pointed me to,
what I'm gathering is that if I do Vrms*Irms I will just get the
apperant power, and in order to convert that to the real power I need
to multiply by the cosine of my phase shift, but that is only true for
sinusoids, since I have a messy pulse I think my situation has become
a lot more complicated.... I'm dealing with a pretty small fan and
feeding the windings from a microcontroller sending out pulses to each
winding, like a pwm to control the speed.... these watt meters look
like they are for big AC type circuits, are there watt meters for low
voltage non-sinusoidal applications?
appreciate the help
What they're getting at here is that you could do a Fourier analysis of"Phantom"
I gave you the answer in an earlier post.
Go read this page:
http://en.wikipedia.org/wiki/Power_factor
Then read this page for some more of the same explanation:
http://www.yokogawa.com/tm/tr/tm-tr0605_01.htm
followed by this page to explain how to actually measure the power
delivered to your 3 windings, except that you will probably need to use a
scope with trace math, rather than a wattmeter. The scope will have to be
able to multiply the instantaneous voltage and current and integrate
(average) that product. You can use the two-wattmeter method to get the
3-phase power delivered to the motor windings, with the scope taking the
place of the wattmeters.:
http://www.yokogawa.com/tm/tr/tm-tr0605_02.htm
Those links were really helpful, thanks for sending them... from what
I'm gathering according to the yokogawa links what I'm trying to do,
find the power going to the windings, can be done by "Active power is
calculated by averaging the products of the instantaneous voltages and
currents" and "Ideal active power is expressed as the product of the
instantaneous voltages and currents averaged over one period of
voltage or current. "
but then it also says "We know that the active power from the voltage
and current of the distorted wave is the sum of the active powers
obtained from the products of the voltages, currents, and power
factors of the same harmonic component (frequency)"
One way to look at the calculation of the RMS value of a voltage waveformso would you say I can just multipy the instantaneous values of V and
I to get the real power? I don't even need RMS anything? Weird... I
thought the point of RMS is cause you can't do that.......
thanks
Those links were really helpful, thanks for sending them... from whatI gave you the answer in an earlier post.
Go read this page:
http://en.wikipedia.org/wiki/Power_factor
Then read this page for some more of the same explanation:
http://www.yokogawa.com/tm/tr/tm-tr0605_01.htm
followed by this page to explain how to actually measure the power
delivered to your 3 windings, except that you will probably need to use a
scope with trace math, rather than a wattmeter. The scope will have to be
able to multiply the instantaneous voltage and current and integrate
(average) that product. You can use the two-wattmeter method to get the
3-phase power delivered to the motor windings, with the scope taking the
place of the wattmeters.:
http://www.yokogawa.com/tm/tr/tm-tr0605_02.htm
so would you say I can just multipy the instantaneous values of V and
I to get the real power?
Thanks Mark,so would you say I can just multipy the instantaneous values of V and
I to get the real power?
Yes... but they have to be the V and I taken _at the same instnat in
time__, you multiply them and that gives you the instantaneous power
at that time....then repeat for many periods.. take all the power
results and average them and you will have the true average power...
but why do you want to measure the true average power of a small fan
motor?
lets get to the meat of the real problem you are trying to solve..
Mark
I didn't say I bought it off ebay. I bought it for work, new."Phil Hobbs"
Phil Allison wrote:
** No - because HP 400As are simply not true rms meters.
Post a link to your example.
I'm travelling at the moment, but I'll see when I get back to the lab. I
bought it in 1991-ish.
** Not off eBay you didn't.
You know, Phil, I can be wrong sometimes. Perhaps I'm wrong about the"Phil Hobbs"
Phil Allison wrote:
** No - because HP 400As are simply not true rms meters.
Post a link to your example.
I'm travelling at the moment, but I'll see when I get back to the lab. I
bought it in 1991-ish.
** Not off eBay you didn't.
I didn't say I bought it off ebay.
* Quote from YOU here, in this thread on May 3.
" I dunno--I have an HP 400A with true-RMS, bandwidth ~ 10 MHz, cost on
Ebay ~ $75. "
** ????????????????????????????????
I bought it for work, new.
** If and when you ever arrive back on planet earth ........
Do try to get your FUCKING facts straight !!
IDIOT !!!!!!!!!!
Apology? In what way have I injured you, Phil?"Phil Hobbs"
snip
However, the fact that I bought my work meter new doesn't contradict the
statement that such things (analogue meters with no digital interface) are
going cheap on eBay at the moment.
** Yep a ** HP 3400A ** might go for such a low figure.
Rusty, bent needle and all.
Assuming that this discussion stays vaguely civil, I'll check the model
number, look up the specs, and post the results when I get back to the
lab.
** I won't hold my breath expecting an apology out of you.
So, can't you measure _that_?... the controller
gets power from somewhere else...
On Jun 3, 6:38 pm, "Phil Allison" <philalli...@tpg.com.au> wrote:
And then just take the product Pavg = Vrms*Irms
A whirling magnet/coil gizmo can be a motor or generator..... I don't see why I would need the phase angles to find the power.
** Fraid you do and there will not be a simple number either.
If you have access to the voltage and current waveforms, simplyOn Jun 4, 11:01 am, Richard Henry <pomer...@hotmail.com> wrote:
On Jun 4, 8:50 am, panfilero <panfil...@gmail.com> wrote:
On Jun 3, 6:38 pm, "Phil Allison" <philalli...@tpg.com.au> wrote:
"panfilero"
And then just take the product Pavg = Vrms*Irms
** The formula for Volt Amperes or VA.
.... I don't see why I would need the phase angles to find the power.
** Fraid you do and there will not be a simple number either.
...... Phil
ok phil, so you are telling me that I can not put a shunt resistor in
line with one of my fan windings and simply measure the voltage
waveform across that shunt, and convert that waveform to an RMS
voltage and then divide that by the shunt resistance in order to get
my RMS current? And once I have both the RMS voltage and current,
that I can't multiply those values together in order to see how much
power the winding is consuming?
I'm not just directing this at Phil, if anyone thinks this would not
work please let me know, I don't see what is wrong with this approach.
much thanks.
J
In principle that will work, but if the waveform is complicated, so
will be the calculations.
ok... after reading some of the documents you guys have pointed me to,
what I'm gathering is that if I do Vrms*Irms I will just get the
apperant power, and in order to convert that to the real power I need
to multiply by the cosine of my phase shift, but that is only true for
sinusoids, since I have a messy pulse I think my situation has become
a lot more complicated.... I'm dealing with a pretty small fan and
feeding the windings from a microcontroller sending out pulses to each
winding, like a pwm to control the speed.... these watt meters look
like they are for big AC type circuits, are there watt meters for low
voltage non-sinusoidal applications?
--appreciate the help
You don't need to do all that work. Sample the currend and voltageOn Jun 4, 11:01 am, Richard Henry <pomer...@hotmail.com> wrote:
On Jun 4, 8:50 am, panfilero <panfil...@gmail.com> wrote:
On Jun 3, 6:38 pm, "Phil Allison" <philalli...@tpg.com.au> wrote:
"panfilero"
And then just take the product Pavg = Vrms*Irms
** The formula for Volt Amperes or VA.
.... I don't see why I would need the phase angles to find the power.
** Fraid you do and there will not be a simple number either.
...... Phil
ok phil, so you are telling me that I can not put a shunt resistor in
line with one of my fan windings and simply measure the voltage
waveform across that shunt, and convert that waveform to an RMS
voltage and then divide that by the shunt resistance in order to get
my RMS current? And once I have both the RMS voltage and current,
that I can't multiply those values together in order to see how much
power the winding is consuming?
I'm not just directing this at Phil, if anyone thinks this would not
work please let me know, I don't see what is wrong with this approach.
much thanks.
J
In principle that will work, but if the waveform is complicated, so
will be the calculations.
ok... after reading some of the documents you guys have pointed me to,
what I'm gathering is that if I do Vrms*Irms I will just get the
apperant power, and in order to convert that to the real power I need
to multiply by the cosine of my phase shift, but that is only true for
sinusoids, since I have a messy pulse I think my situation has become
a lot more complicated.... I'm dealing with a pretty small fan and
feeding the windings from a microcontroller sending out pulses to each
winding, like a pwm to control the speed.... these watt meters look
like they are for big AC type circuits, are there watt meters for low
voltage non-sinusoidal applications?
--appreciate the help
the Fluke 87, which could cover the audio bandwidth, using an analogpanfilero wrote:
I have a very noisy square pulse wave that I'm trying to measure the
RMS current and RMS votlage of... my pulses can have up to 24 volt
peaks. and a frequency of 1kHz.... I was wondering if anyone could
reccomend a Multi-Meter for doing these measurements, or do I need to
use a Scope?
From what I've gathered in trying to find a meter is that meters are
made for Sinusoids, and True RMS isn't always very true... this
article was interesting http://www.enginova.com/true_rms_volts.htm .
And since there is noise on my signal, I'd have different spikes and
things happening that would be beyond my fundamental frequency of
1kHz, and I don't know what a multi-meter does with that.
I've heard some meters assume that your signal is centerd around the
zero-axis and therefore return bad results, I've heard other meters do
internal calculations assuming a sinusoid, and give erroneous
results.... anyone have any suggestions?
The most popular hand-held meter that gave true rms voltage used to be
You really do have to worry about the phase shift. When the voltage"You can't just multiply an RMS current and RMS voltage measurement to
get power." - RL
So.... I've read that Vrms*Irms = Pavg..... but..... you're saying I
need to include the phase shift?! Isn't the phase shift implied in the
waveform? I have two meters hooked up, and I'm reading a Vrms and an
Irms, if I multiply these together don't I get the Apparent Power?
Aren't RMS values scalars? No phaeses involved.... and isn't
Vrms*Irms = P apparent?
On Mon, 02 Jun 2008 18:24:25 -0400, Jamie
jamie_ka1lpa_not_valid_after_ka1l...@charter.net> wrote:
panfilero wrote:
I have a very noisy square pulse wave that I'm trying to measure the
RMS current and RMS votlage of... my pulses can have up to 24 volt
peaks. and a frequency of 1kHz.... I was wondering if anyone could
reccomend a Multi-Meter for doing these measurements, or do I need to
use a Scope?
From what I've gathered in trying to find a meter is that meters are
made for Sinusoids, and True RMS isn't always very true... this
article was interestinghttp://www.enginova.com/true_rms_volts.htm.
And since there is noise on my signal, I'd have different spikes and
things happening that would be beyond my fundamental frequency of
1kHz, and I don't know what a multi-meter does with that.
I've heard some meters assume that your signal is centerd around the
zero-axis and therefore return bad results, I've heard other meters do
internal calculations assuming a sinusoid, and give erroneous
results.... anyone have any suggestions?
The most popular hand-held meter that gave true rms voltage used to be
the Fluke 87, which could cover the audio bandwidth, using an analog
rms converter (100KHz limit). It produced rms current measurements bt
measuring a shunt resistor voltage, which is the most common current
measurement method.Other meters are available that use the same kind
of hardware.
The old Fluke 8922a bench powered meter used to be usefull with higher
voltage/power circuits. It used a thermal converter, followed by
conventional voltage metering and is highly sensitive to calibration.
This has a 1MHz or 11MHz bandwidth, depending on input voltage range.
There are other similar meters.
More recently, given the availability of digital scopes with good
bandwidth and math capability, mathematical derivation is more common.
You can't just multiply an RMS current and RMS voltage measurement to
get power. Their phase relationship is important. As an extreme
example, an amplifier driving an inductor with 1Vrms will dissipate
internal losses equivalent to driving a short circuit, while the load
power is minimal.
A digital scope math calculation uses time-coherent multiplication to
produce a fairly reliable power indication for related waveforms in
its display/memory.
DC-offset will also affect measurements, the ability of the meter to
produce meaningful readings and the operator's ability to understand
what the reading means.
RL
Hmmmmm..... ok, well my voltages will definitely be out of phase sincepanfilero wrote:
"You can't just multiply an RMS current and RMS voltage measurement to
get power." - RL
So.... I've read that Vrms*Irms = Pavg..... but..... you're saying I
need to include the phase shift?! Isn't the phase shift implied in the
waveform? I have two meters hooked up, and I'm reading a Vrms and an
Irms, if I multiply these together don't I get the Apparent Power?
Aren't RMS values scalars? No phaeses involved.... and isn't
Vrms*Irms = P apparent?
You really do have to worry about the phase shift. When the voltage
and current are in phase, the power is going in the same direction as
the current. When they're out of phase, the power is going in the
opposite direction. From the wall socket's perspective, that's the
difference between a motor and a generator.
If you imagine a load that drew some enormous current right when the
voltage was near 0, and much less current near the peaks, you could be
off by a factor of 100 or more if you just multiplied RMS voltage times
RMS current. The same is true of a purely reactive load, in which the
voltage is 1/4 cycle out of phase with the current. For instance, this
is what it would look like with a capacitor across the line.
I___ ___
/ X \ V
/ / \ \
/---/---\---\---/----------------------
/ \ \ /
/ \ X
-- --- --
| | | | |
| | | | |
1 2 3 4
In quarter-cycle 1, I >0 and V < 0, so the VI < 0 : power is going from
the capacitor into the wall socket.
In QC 2, I>0 and V>0, so VI > 0: the power is going from the wall socket
into the capacitor.
In QC 3, I<0 and V>0, so VI < 0 again, and
In QC 4, I<0 and V<0, so VI > 0.
Since the waveforms are symmetric, the net power transferred over a
cycle is zero.
Cheers,
Phil Hobbs
over one cycle, it's zero, period. Sorry about that.On Jun 18, 3:08 pm, Phil Hobbs <pcdhSpamMeSensel...@pergamos.net
wrote:
panfilero wrote:
"You can't just multiply an RMS current and RMS voltage measurement to
get power." - RL
So.... I've read that Vrms*Irms = Pavg..... but..... you're saying I
need to include the phase shift?! Isn't the phase shift implied in the
waveform? I have two meters hooked up, and I'm reading a Vrms and an
Irms, if I multiply these together don't I get the Apparent Power?
Aren't RMS values scalars? No phaeses involved.... and isn't
Vrms*Irms = P apparent?
You really do have to worry about the phase shift. When the voltage
and current are in phase, the power is going in the same direction as
the current. When they're out of phase, the power is going in the
opposite direction. From the wall socket's perspective, that's the
difference between a motor and a generator.
If you imagine a load that drew some enormous current right when the
voltage was near 0, and much less current near the peaks, you could be
off by a factor of 100 or more if you just multiplied RMS voltage times
RMS current. The same is true of a purely reactive load, in which the
voltage is 1/4 cycle out of phase with the current. For instance, this
is what it would look like with a capacitor across the line.
I___ ___
/ X \ V
/ / \ \
/---/---\---\---/----------------------
/ \ \ /
/ \ X
-- --- --
| | | | |
| | | | |
1 2 3 4
In quarter-cycle 1, I >0 and V < 0, so the VI < 0 : power is going from
the capacitor into the wall socket.
In QC 2, I>0 and V>0, so VI > 0: the power is going from the wall socket
into the capacitor.
In QC 3, I<0 and V>0, so VI < 0 again, and
In QC 4, I<0 and V<0, so VI > 0.
Since the waveforms are symmetric, the net power transferred over a
cycle is zero.
Cheers,
Phil Hobbs
Hmmmmm..... ok, well my voltages will definitely be out of phase since
there's some reactance involved.... and I agree with everything you
are saying here, but this is looking at just one cycle of each.... I'm
assuming a meter is constantly sampling the signal and so the RMS
value is some sort of average that is constantly being updated, and
the average is maybe taken over a hundred cycles or somethign like
that not just one cycle, so do you think that since the meter is
taking an rms average of many cycles that somehow the phase shift
would then be included in there when I multiply the values together?
I'm fine with the fact that I'm getting the reacitve power (the
negative power going back to the source) in my readings, I'm fine with
the fact that I'm getting apparent power instead of true/active
power... but I only have meters and am trying to avoid the phase
issue.... I'm using the two-wattmeter method to do this (thanks
Phantom)
J.
No, since every cycle is just like every other, if the total is zero
yes, completely agree, that a purely reactive circuit has no real/truepanfilero wrote:
On Jun 18, 3:08 pm, Phil Hobbs <pcdhSpamMeSensel...@pergamos.net
wrote:
panfilero wrote:
"You can't just multiply an RMS current and RMS voltage measurement to
get power." - RL
So.... I've read that Vrms*Irms = Pavg..... but..... you're saying I
need to include the phase shift?! Isn't the phase shift implied in the
waveform? I have two meters hooked up, and I'm reading a Vrms and an
Irms, if I multiply these together don't I get the Apparent Power?
Aren't RMS values scalars? No phaeses involved.... and isn't
Vrms*Irms = P apparent?
You really do have to worry about the phase shift. When the voltage
and current are in phase, the power is going in the same direction as
the current. When they're out of phase, the power is going in the
opposite direction. From the wall socket's perspective, that's the
difference between a motor and a generator.
If you imagine a load that drew some enormous current right when the
voltage was near 0, and much less current near the peaks, you could be
off by a factor of 100 or more if you just multiplied RMS voltage times
RMS current. The same is true of a purely reactive load, in which the
voltage is 1/4 cycle out of phase with the current. For instance, this
is what it would look like with a capacitor across the line.
I___ ___
/ X \ V
/ / \ \
/---/---\---\---/----------------------
/ \ \ /
/ \ X
-- --- --
| | | | |
| | | | |
1 2 3 4
In quarter-cycle 1, I >0 and V < 0, so the VI < 0 : power is going from
the capacitor into the wall socket.
In QC 2, I>0 and V>0, so VI > 0: the power is going from the wall socket
into the capacitor.
In QC 3, I<0 and V>0, so VI < 0 again, and
In QC 4, I<0 and V<0, so VI > 0.
Since the waveforms are symmetric, the net power transferred over a
cycle is zero.
Cheers,
Phil Hobbs
Hmmmmm..... ok, well my voltages will definitely be out of phase since
there's some reactance involved.... and I agree with everything you
are saying here, but this is looking at just one cycle of each.... I'm
assuming a meter is constantly sampling the signal and so the RMS
value is some sort of average that is constantly being updated, and
the average is maybe taken over a hundred cycles or somethign like
that not just one cycle, so do you think that since the meter is
taking an rms average of many cycles that somehow the phase shift
would then be included in there when I multiply the values together?
I'm fine with the fact that I'm getting the reacitve power (the
negative power going back to the source) in my readings, I'm fine with
the fact that I'm getting apparent power instead of true/active
power... but I only have meters and am trying to avoid the phase
issue.... I'm using the two-wattmeter method to do this (thanks
Phantom)
J.
No, since every cycle is just like every other, if the total is zero
over one cycle, it's zero, period. Sorry about that.
Cheers,
Phil Hobbs