Basic AC wattage question: am I doing my math right?

H

HC

Guest
Hey, all, I'm trying to check something and I've come up with a number
and I'm not sure I've done my math right. What I want to do is figure
out how much a device is costing me to operate in electricity
charges. I'm not sure I'm doing the wattage calculations correctly
(I'm not well-versed with AC calculations and have only light hobbyist
experience with DC), and I'm not sure I'm going from the wattage to
the kW hours correctly.

I have a device that runs on 120v AC. I put a DMM in line with one of
the two conductors in the power cord and I measure 16.56 mA (my DMM
has a setting for measuring AC amps and that's what I used). I then
used the same DMM to measure AC voltage at the receptacle and I
measured 124.8 VAC. I'm not sure if this DMM does true RMS (as I see
that some manufacturers advertise that their meters measure "true
RMS") and the manual does not say yes or no.

Now, I take the amps times voltage to get watts so: 0.01656 x 124.8
and I get 2.066688 watts. I get kWh by 2.066688 x 24 hours / 1000
(the device is on constantly) and I get roughly 0.0496 kWh per day.
Say the billing cycle is for 31 days so that's 0.0496 x 31 = 1.5376
kWh each month and all I have to do is multiply that by my cost per
kWh and I should be good (for simplicity, say it costs me 10 cents per
kWh it's 15 cents a month).

Did I do all that right or am I mucking up the math or the wattage
calculation for AC?

Thanks in advance.

--HC
 
On Fri, 28 Dec 2007 09:34:24 -0800, HC wrote:

Hey, all, I'm trying to check something and I've come up with a number
and I'm not sure I've done my math right. What I want to do is figure
out how much a device is costing me to operate in electricity charges.
I'm not sure I'm doing the wattage calculations correctly (I'm not
well-versed with AC calculations and have only light hobbyist experience
with DC), and I'm not sure I'm going from the wattage to the kW hours
correctly.

I have a device that runs on 120v AC. I put a DMM in line with one of
the two conductors in the power cord and I measure 16.56 mA (my DMM has
a setting for measuring AC amps and that's what I used). I then used
the same DMM to measure AC voltage at the receptacle and I measured
124.8 VAC. I'm not sure if this DMM does true RMS (as I see that some
manufacturers advertise that their meters measure "true RMS") and the
manual does not say yes or no.

Now, I take the amps times voltage to get watts so: 0.01656 x 124.8 and
I get 2.066688 watts. I get kWh by 2.066688 x 24 hours / 1000 (the
device is on constantly) and I get roughly 0.0496 kWh per day. Say the
billing cycle is for 31 days so that's 0.0496 x 31 = 1.5376 kWh each
month and all I have to do is multiply that by my cost per kWh and I
should be good (for simplicity, say it costs me 10 cents per kWh it's 15
cents a month).

Did I do all that right or am I mucking up the math or the wattage
calculation for AC?

Thanks in advance.

--HC
Click to expand...
That is right as far as it goes, and assuming that your meter reads
correctly in RMS it will give you an upper bound.

The "real" number depends on the power factor of the device, which is
probably less than 1. AC power can be "reactive", where the current is
out of phase with the voltage, and the energy, instead of flowing into
your device, is just swishing in and out at 120Hz (or 100, depending on
where you live).

Unless the answer is critical, though, your number is probably more than
good enough.

--
Tim Wescott
Control systems and communications consulting
http://www.wescottdesign.com

Need to learn how to apply control theory in your embedded system?
"Applied Control Theory for Embedded Systems" by Tim Wescott
Elsevier/Newnes, http://www.wescottdesign.com/actfes/actfes.html
 
On Dec 28, 11:54 am, Tim Wescott <t...@seemywebsite.com> wrote:
On Fri, 28 Dec 2007 09:34:24 -0800, HC wrote:
Hey, all, I'm trying to check something and I've come up with a number
and I'm not sure I've done my math right. What I want to do is figure
out how much a device is costing me to operate in electricity charges.
I'm not sure I'm doing the wattage calculations correctly (I'm not
well-versed with AC calculations and have only light hobbyist experience
with DC), and I'm not sure I'm going from the wattage to the kW hours
correctly.

I have a device that runs on 120v AC. I put a DMM in line with one of
the two conductors in the power cord and I measure 16.56 mA (my DMM has
a setting for measuring AC amps and that's what I used). I then used
the same DMM to measure AC voltage at the receptacle and I measured
124.8 VAC. I'm not sure if this DMM does true RMS (as I see that some
manufacturers advertise that their meters measure "true RMS") and the
manual does not say yes or no.

Now, I take the amps times voltage to get watts so: 0.01656 x 124.8 and
I get 2.066688 watts. I get kWh by 2.066688 x 24 hours / 1000 (the
device is on constantly) and I get roughly 0.0496 kWh per day. Say the
billing cycle is for 31 days so that's 0.0496 x 31 = 1.5376 kWh each
month and all I have to do is multiply that by my cost per kWh and I
should be good (for simplicity, say it costs me 10 cents per kWh it's 15
cents a month).

Did I do all that right or am I mucking up the math or the wattage
calculation for AC?

Thanks in advance.

--HC

That is right as far as it goes, and assuming that your meter reads
correctly in RMS it will give you an upper bound.

The "real" number depends on the power factor of the device, which is
probably less than 1. AC power can be "reactive", where the current is
out of phase with the voltage, and the energy, instead of flowing into
your device, is just swishing in and out at 120Hz (or 100, depending on
where you live).

Unless the answer is critical, though, your number is probably more than
good enough.

--
Tim Wescott
Control systems and communications consultinghttp://www.wescottdesign.com

Need to learn how to apply control theory in your embedded system?
"Applied Control Theory for Embedded Systems" by Tim Wescott
Elsevier/Newnes,http://www.wescottdesign.com/actfes/actfes.html
Click to expand...
Hey, Tim, thank you for your response. It's not critical, I just
wanted to know. The device is a television that is plugged in but not
turned on. Somewhere I once read that the power consumption of
televisions that are plugged in but not turned on is still significant
(as far as the whole United States is concerned). I have always
wanted to verify that by testing my own TV and finally have gotten
around to testing it. Then I thought it would be fun to check how
much it costs me in terms of money, hence what I've done here. FWIW,
the Hz is 60.

If the number I've come up with is 0.0496 kWh per day and if working
the number in a more precise way as you have alluded to (considering
power factor and whether or not the device is subject to the
reactivity of AC) would only get me maybe 10% more precision then my
number is fine for my purposes. I only need it to be close. If my
method gets me 0.0496 and the best way gets 0.0495 then I don't need
to do it the best way. If my way gets 0.0496 kWh/day and the best way
gets me a number like 2.6 kWh/day then I would want to do it the
"best" way because the inaccuracy is so great that it voids the
purpose of me doing the testing at all.

Thanks again for your reply.

--HC
 
"HC"
I have a device that runs on 120v AC.
Click to expand...

** Everything depends on what that "device" is - you fool.

Pathetic of you to hide what it is.



........ Phil
 
"HC"

Hey, Tim, thank you for your response.
Click to expand...

** Don't thank someone for supplying bad advice.


The device is a television that is plugged in but not
turned on.
Click to expand...

** Then your test method is completely worthless.

Somewhere I once read that the power consumption of
televisions that are plugged in but not turned on is still significant
Click to expand...

** So you prefer vague and false information.


If the number I've come up with is 0.0496 kWh per day and if working
the number in a more precise way as you have alluded to (considering
power factor and whether or not the device is subject to the
reactivity of AC) would only get me maybe 10% more precision then my
number is fine for my purposes.
Click to expand...

** The likely error is hundreds of percent.

Not using a true rms meter means your amp reading may well be out by factor
of 2 or 3.

The power factor of the load is also unknown and can alter the result by
similar factor.

A special wattmeter is needed measure the standby energy consumption of such
an appliance.

See the one in the pic here:

http://sound.westhost.com/articles/external-psu.htm#test



...... Phil
 
On Dec 28, 3:48 pm, "Phil Allison" <philalli...@tpg.com.au> wrote:
"HC"



I have a device that runs on 120v AC.

** Everything depends on what that "device" is - you fool.

Pathetic of you to hide what it is.

....... Phil
Click to expand...
Wow, so you're this group's neighborhood troll. Nice to meet you.
Hey, when you grow up your dicky might get bigger than it is now. Be
patient. In the meantime, find another way to compensate for it.

--HC
 
On 12/28/07 1:48 PM, in article 5tl98mF1dfjliU1@mid.individual.net, "Phil
Allison" <philallison@tpg.com.au> wrote:

"HC"

I have a device that runs on 120v AC.


** Everything depends on what that "device" is - you fool.

Pathetic of you to hide what it is.



....... Phil
Click to expand...
Hey Phil.... Merry Christmas and Happy New Year.

Now, do your best insult. Have at it Big Guy.
 
On 12/28/07 2:01 PM, in article
8f8d8a71-362e-4212-ad7d-cd89cce3ac4e@s19g2000prg.googlegroups.com, "HC"
<hboothe@gte.net> wrote:

On Dec 28, 3:48 pm, "Phil Allison" <philalli...@tpg.com.au> wrote:
"HC"



I have a device that runs on 120v AC.

** Everything depends on what that "device" is - you fool.

Pathetic of you to hide what it is.

....... Phil

Wow, so you're this group's neighborhood troll. Nice to meet you.
Hey, when you grow up your dicky might get bigger than it is now. Be
patient. In the meantime, find another way to compensate for it.

--HC
Click to expand...
Ignore what he says at your peril.
 
"HC" = Google Groping moron.


** Wow, so you're ANOTHER of this group's FUCKWIT trolls.

Peeeeeuuukkkeeeee ....



...... Phil
 
"HC" = Google Groping TROLL

Ignore what he says at your peril.

Don, I appreciate your comment and I will look for more information to
try to get a better understanding of my question and the answers I've
received, including Phil's. I just didn't appreciate being spoken to
harshly for asking a simple question and being maligned for leaving
out details that, in my ignorance, are perhaps important.
Click to expand...


** You are one, pig arrogant, posturing bloody idiot.

PISS OFF




........ Phil
 
"HC" = another Google Groping SHITHEAD



** Wot a pathetic wanker.

Yaaaaawnnnnnnnnnnnnnnnnnnnn ...





........ Phil
 
On Dec 28, 6:28 pm, Don Bowey <dbo...@comcast.net> wrote:
On 12/28/07 2:01 PM, in article
8f8d8a71-362e-4212-ad7d-cd89cce3a...@s19g2000prg.googlegroups.com, "HC"



hboo...@gte.net> wrote:
On Dec 28, 3:48 pm, "Phil Allison" <philalli...@tpg.com.au> wrote:
"HC"

I have a device that runs on 120v AC.

** Everything depends on what that "device" is - you fool.

Pathetic of you to hide what it is.

....... Phil

Wow, so you're this group's neighborhood troll. Nice to meet you.
Hey, when you grow up your dicky might get bigger than it is now. Be
patient. In the meantime, find another way to compensate for it.

--HC

Ignore what he says at your peril.
Click to expand...
Don, I appreciate your comment and I will look for more information to
try to get a better understanding of my question and the answers I've
received, including Phil's. I just didn't appreciate being spoken to
harshly for asking a simple question and being maligned for leaving
out details that, in my ignorance, are perhaps important.

--HC
 
On Dec 28, 7:10 pm, "Phil Allison" <philalli...@tpg.com.au> wrote:
"HC" = Google Groping moron.

** Wow, so you're ANOTHER of this group's FUCKWIT trolls.

Peeeeeuuukkkeeeee ....

..... Phil
Click to expand...
Does mommy know you're using her computer? Parental filters are so
underutilized.

--HC
 
In article <5tlopsF1e8ke6U1@mid.individual.net>, "Phil Allison"
<philallison@tpg.com.au> wrote:

<A lot of his usual crap>

Howdy, Phil. Did ya get some more of those majuhkul stereo resistors for
Christmas? You know, the ones you love to use to parallel all kinds of
stuff?

I bet you asked Santa for a few rolls of that thar oxygen-free hookup wire, too.

Didja get that, too?

Be careful now, you don't want to cause a fire at the home.

Merry Christmas, YOF.
 
"Joke" &lt;none@given.now&gt;



** This congenital dope is drowning in the shallow end of the gene pool.





....... Phil
 
On Fri, 28 Dec 2007 14:01:20 -0800 (PST), HC &lt;hboothe@gte.net&gt; wrote:

On Dec 28, 3:48 pm, "Phil Allison" &lt;philalli...@tpg.com.au&gt; wrote:
"HC"



I have a device that runs on 120v AC.

** Everything depends on what that "device" is - you fool.

Pathetic of you to hide what it is.

....... Phil

Wow, so you're this group's neighborhood troll. Nice to meet you.
Hey, when you grow up your dicky might get bigger than it is now. Be
patient. In the meantime, find another way to compensate for it.

--HC
Click to expand...
Phil has a problem; always has, always will.


Your math is right for a simple (ie, resistive) load, but might be off
by, say, 2:1 in either direction, depending on the actual load
waveform and/or power factor.

John
 
In article &lt;5tm4bdF1e27e9U1@mid.individual.net&gt;, "Fill-up with Gas" wrote:

&lt;Something about "jean pull".&gt;

Oh goodness, playing near operating farm equipment, again?

And got yor'n britches caught by a moving tractor?
 
On Dec 28, 11:54 pm, John Larkin
&lt;jjlar...@highNOTlandTHIStechnologyPART.com&gt; wrote:
On Fri, 28 Dec 2007 14:01:20 -0800 (PST), HC &lt;hboo...@gte.net&gt; wrote:
On Dec 28, 3:48 pm, "Phil Allison" &lt;philalli...@tpg.com.au&gt; wrote:
"HC"

I have a device that runs on 120v AC.

** Everything depends on what that "device" is - you fool.

Pathetic of you to hide what it is.

....... Phil

Wow, so you're this group's neighborhood troll. Nice to meet you.
Hey, when you grow up your dicky might get bigger than it is now. Be
patient. In the meantime, find another way to compensate for it.

--HC

Phil has a problem; always has, always will.

Your math is right for a simple (ie, resistive) load, but might be off
by, say, 2:1 in either direction, depending on the actual load
waveform and/or power factor.

John
Click to expand...
Hey, John, the more I've been reading the more I think my math and
measurements have been badly wrong because they're rooted in DC
methodology which seems to be quite different from the AC world (or
can be for equipment/circuits that are not 100% resistive). So, as
you say, I could be quite wrong by just simply not knowing the
intricacies of AC power flow. I found a formula for calculating AC
current consumption on single-phase which is P = V x I x cosine Theta
which is great except that Theta is the "power factor angle" of the
equipment which I don't currently know. I approached this problem
thinking like I do about DC and I've found that it is wrong and the
problem is considerably more complex. I'm going to do a lot more
reading about this and "reboot" my whole test.

Thank you for your reply.

--HC
 
On Sat, 29 Dec 2007 00:08:08 -0800 (PST), HC &lt;hboothe@gte.net&gt; wrote:

On Dec 28, 11:54 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com&gt; wrote:
Your math is right for a simple (ie, resistive) load, but might be off
by, say, 2:1 in either direction, depending on the actual load
waveform and/or power factor.

John

Hey, John, the more I've been reading the more I think my math and
measurements have been badly wrong because they're rooted in DC
methodology which seems to be quite different from the AC world (or
can be for equipment/circuits that are not 100% resistive). So, as
you say, I could be quite wrong by just simply not knowing the
intricacies of AC power flow. I found a formula for calculating AC
current consumption on single-phase which is P = V x I x cosine Theta
which is great except that Theta is the "power factor angle" of the
equipment which I don't currently know. I approached this problem
thinking like I do about DC and I've found that it is wrong and the
problem is considerably more complex. I'm going to do a lot more
reading about this and "reboot" my whole test.
Click to expand...
Check out the "Kill-A-Watt", it does *exactly* what you want.
I bought mine on-line for about $20 (with free shipping!) for
uses like this. I tested 2 different TV models of the same size
(27") and similar age and found that one used nearly 16 watts
when off, while the other used less than 8.

The Kill-A-Watt shows RMS volts, amps, watts, KWH,
power factor, line frequency, and probably some others
I've forgotten. You plug the Kill-A-Watt into the wall, then
plug the thing you want to test into the Kill-A-Watt, and
push a button to select what you want to read.

I've wanted a watt-meter for *years* and always
figured I'd have to bite the bullet and build a crude
one. But the Kill-A-Watt does more than anything
I'd ever have built... and for only $20 !!! This is by
far my favorite toy of the year.

(PS: I have no affiliation with Kill-A-Watt, just
a delighted customer.)

Best regards,


Bob Masta

DAQARTA v3.50
Data AcQuisition And Real-Time Analysis
www.daqarta.com
Scope, Spectrum, Spectrogram, FREE Signal Generator
Science with your sound card!
 
On Fri, 28 Dec 2007 22:23:01 -0800, none@given.now (Joe) wrote:

In article &lt;5tm4bdF1e27e9U1@mid.individual.net&gt;, "Fill-up with Gas" wrote:

Something about "jean pull".

Oh goodness, playing near operating farm equipment, again?

And got yor'n britches caught by a moving tractor?
Click to expand...

No, even worse, a run in his panty hose.

John
 
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