How much a clothes dryer cost to use? Again ......

Page 4 of 5  

snipped-for-privacy@optonline.net writes:

To tell you the truth, I thought it would be obvious that you multiply by 120v rather than 240v since each leg is obviously only 120v. My apologies for not mentioning that distinction if that confused people.
Also, indeed there *is* a need to clamp around both conductors since the 120v leg feeding the motor only goes through one of the conductors. So if you measure only one conductor and multiply by 240v then you either are missing the 120v component completely or you are mistakenly doubling its power contribution.

You could do it in two steps. But the OP (or someone in the thread) mentioned that the power draw varied over time. So if you wanted to measure the power over time it might be easiest to clamp both wires in the way I suggested.
Anyway -- I think we all understand the point and I certainly have no desire to jump into the middle of a flame war here... But thanks to all who contributed positively to this thread.
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Yes, I agree 100%
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writes:

Yeah, but then you get a reading approx'ly double the true value!
Remember, in most cases, one hot is carrying only the heating element current, while the other is carrying the heating element plus the 120 V loads. Altho it is possible to have the 120 V loads distributed over both hots!
To check this, run the unit with air only (no heat), see what the power draw is on each hot. Usually one will be zero, but it could be that there is a little bit of current in each hot, even without the heating element, if they distribued the 120 V load..
But, with the heat off, and some draw in both hots, this *could* also indicate that the controls/motor are also 220V!! To rule that out, with the heat off, disconnect one hot at a time, see if everything goes off if *either* hot is removed. Usually, with one hot removed, everything except the heat will work, but if both hots are required for the motor/controls, then other stuff is operating at 220 in addition to the heating element.
Ultimately it's best -- and ultimately easiest -- to clamp each wire separately. You get more complete info this way. Clamp-ons are real cheap now, anyway -- used to be very expensive. $11 from HF -- I must have 6 of'em! Three are used to monitor the 3 phases on a rotary phase converter, the other three are in case I lose two.
The way to be able to use a clamp-on "at will", at least for 120 V circuits, is to fashion an adapter from a 2-prong plug with zipcord, going to a receptacle, and "split" the zip cord. Basically a mini-extension cord, with the cord split. Now, you can put the clamp-on easily around a single conductor.
You can do the same with a "regular" 220V plug/receptacle, but this starts getting a little expensive for 30 and 50 A plugs/receptacles.

Indeed. Hopefully Trader4 learned about Biot-Savart. :) :)
Oh, and to his comment about "we're not solving Maxwell's equations here....." Well, in a sense we did, cuz Maxwell's equations subsumes all those disparate experimental observations on E&M.
This whole thing really was a very neat, very elegant demonstration of physical principles. "Simple" for Trader, but neat for me.
--
EA



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Which is why I would consider them as 120v legs (which I guess would implicitly be dividing by 2 for the 240v circuit portion).

May be possible but I am not aware of any units that do this -- do you know of any that are wired this way?

Personally, I haven't seen this on US units -- but I am no expert. Are you aware of any US brands that have 220v controls?

And you are less likely to make a mistake. I was just trying to suggest a "clever" trick. In practice, I too would have done each leg separately. In part, because it would give me the added information of being able to separate the 120v component from the 240v component (by taking the difference of the current in the 2 legs, assuming that the 120v current is not "distributed").

Yup - I have a HF version -- but issue for me isn't price but room to store all the "wonderful stuff"/"junk" I buy at HF.

Actually, HF used to sell a cheap ($4.99) line splitter (#92072) that did effectively the same thing - they seem to have discontinued it. It also has two loops - one for 1x and one for 10x so that you can get better resolution of low currents. I found a pic of it on http://www.kindlachristmas.com/HTMultimeter.asp .
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wrote in message

No. I'm just a little bored, is all. :) Just an inneresting scenario for possible sleuthing.

No. I'm just a little bored, is all. :) :) More possible sleuthing.... :)

If you shop right, a cheap plug will be 50c, HD outlets are 50c, and the zip cord should be 12c. :)
It also

Now THIS is ingenious!!!! In Trader4-like hindsight, "super obvious", and I'm kicking myself for not having thought of it myself! Deeee-licious!! But more Biot-Savart!!! A potentially very useful little trick, and super-easy to do yourself -- just count loops for your multiplier!! Zip cord will loop just fine, too. Excellent!
--
EA

> I found a pic of it on http://www.kindlachristmas.com/HTMultimeter.asp .



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wrote:

So, here we have it again folks. EA goes around hurling insults at a bunch of us here and he once again shows us that he is totally clueless and doesn't understand basic electricity on a 240V circuit.
One more time: Blueman is 100% correct. You could reverse the direction of one of the dryer hot leads, put both wires in a clamp on amp meter and measure the current. But then you treat the total current as a 120V load when calculating the power. Both blueman and I have stated that at least 6 times. It's clearly explained by blueman right above. Do we have to draw cartoons for you?
Say you have 23 amps flowing in one hot, 20 in the other, 3 in the neutral. Using the above method you would measure 43 amps using the clamp-on. The power is 120V X 43 amps. Capiche?

You figure that out all by yourself? Geez, the clueless guy who told us to use a Kill-a-Watt meter on a dryer is now trying to explain to us how a 240V circuit works.

No need to run experiments, the rest of us here know this. Again, last time I checked, in the USA the standard we use is 240V. How the hell do you get 120V off one leg and 220V between the two? Where do you live?

You really think so? Instead of reversing wires to clamp them both at the same time or playing with your panel wiring, which you seem to find so fascinating? LOL

More likely they'll get smoked like that Kill-a-Watt when you stick them and your fingers someplace they don't belong.
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writes:

So, here we have it again folks. EA goes around hurling insults at a bunch of us here and he once again shows us that he is totally clueless and doesn't understand basic electricity on a 240V circuit.
One more time: Blueman is 100% correct. You could reverse the direction of one of the dryer hot leads, put both wires in a clamp on amp meter and measure the current. But then you treat the total current as a 120V load when calculating the power. Both blueman and I have stated that at least 6 times. It's clearly explained by blueman right above. Do we have to draw cartoons for you?
Say you have 23 amps flowing in one hot, 20 in the other, 3 in the neutral. Using the above method you would measure 43 amps using the clamp-on. The power is 120V X 43 amps. Capiche?
======================================================= Except for one thing: iirc, Blueman explicitly stated that calculating power was not his agenda -- all he was simply speculating on was the viability of measuring current of two wires simultaneously with a clamp-on. And I never disputed the above arithmetic.
Btw, what was YOUR response to his speculation???? iirc, he stated your response was irrelevant, ie, you missed the point altogether. Altho he said this a little more diplomatically, wihch, in your case, was wasted diplomacy.
Your example above is arithmetically correct, but still near pointless, becuz, well.....in your example, you don't *have* 43 amps!!!! So yeah, the power would be the same, but who was asking?? Dats why people run shit at 240 V, to avoid large currents, which results in better efficiency, ito wire size economy, IR drop, I^2R power losses, etc.

You figure that out all by yourself? Geez, the clueless guy who told us to use a Kill-a-Watt meter on a dryer is now trying to explain to us how a 240V circuit works. ================================================ Just because you didn't figger out the Biot-Savart stuff is no reason to become a hater....

No need to run experiments, the rest of us here know this. ============================================================ How could anyone possibly know this without doint the experiment? If you have 20 A in one dryer leg, and 23 in the other, we are ASSUMING that difference of 3 amps is the 120 V load in one leg, and that 20 A is running at 240 V.
It COULD be 16 A of heater, 4 amps of 120 V load in one leg, and 7 amps in the other. OR, it could be a *mixture* of 120 V loads on one or both legs, 240V heater load, and add'l 240 V control loads.
Not saying this is likely, or that it makes design sense, but be very clear that you *could not* know for sure unless you did the sleuthing.... Well, not you....
Again, our assumptions are reasonable assumptions, and likely to be accurate, but not absolutely know-able without the experiment.
Again, last time I checked, in the USA the standard we use is 240V. How the hell do you get 120V off one leg and 220V between the two? Where do you live?
================================================ Many electricians use 220/230/240 V interchangeably, and 110/115/120 V interchangeably, as the basic point is that single hot to ground is 1/2 the voltage between two hots. Don't be a nitpicker, don't be a hater....
And if the Assaholic Triumvirate want citations, just look at many motor nameplates, which will say "220-240", with no change in wiring. Some, I believe, can even tolerate 208-240. Others will have specific taps.
But you get the point..... well, mebbe you don't.... hating does that....

You really think so? Instead of reversing wires to clamp them both at the same time or playing with your panel wiring, which you seem to find so fascinating? LOL ======================================================== Heh, it was news to you, wadnit? Mr. Everything is obvious/oh-so simple in hindsight.....

More likely they'll get smoked like that Kill-a-Watt when you stick them and your fingers someplace they don't belong. ================================================= Well, what I would love to do is pluck out a wire from my 3 ph control panel (which switches in multiple staged 3-ph motors, and a switchable bank of capacitors for voltage/load balancing), bring you over to fix it, and watch you spend the next two months tryna figger out WTF's going on.
--
EA




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wrote:

Now you're simply lying. Here again is the exact question that Blueman asked:
"But clamping measures current, not voltage. So clamping both wires (with the direction reversed in one wire so that the opposite phases cancel as I mentioned before) should give you a measurement of the total current. Then you can multiply by the voltage (or maybe rms voltage) to get the total watts consumed. "
So, it's very clear Blueman was interested in calculating power. And he also never stated that he wasn't. Liar.

Yes you did. Blueman had outlined how to measure the power using his method. And he was correct. You could reverse the direction of one hot, put them together in a clamp on amp meter and determine the power. In fact we had been through the math, how you need to then calculate the power using 120V and I thought the matter was settled.
But then you chimed in with this:
"Yeah, but then you get a reading approx'ly double the true value! "

I seem to recall your being a bit confused about the premise of the question, etc.
"Blueman asked a very good Q: WHY can't you measure both together? And, afaict, it wasn't the notion of voltage that he was missing, as you stated. And since you just repeated my answer, his Q wasn't really answered. But he also mis-stated his own premise, in a couple of ways. "
I'd also point out that voltage is an important part of the whole thing and if you take the sum of the currents in both hots, YOU THEN MUST USE 120V TO CALCULATE POWER, NOT 240V. But 20 posts later, you come back and start telling him all over again that if he uses his method he's going to be counting the current twice. Which by the way is also factually inaccurate, because he's only counting the 240V current twice, not the 120V component. Capiche?

Blueman was asking

More tellling us what we already know. From the genius that first told us to use a Kill-a-Watt meter on a dryer. And then suggested using 2 of them. But maybe there is hope for you yet. I see you are now finally starting to use the term 240V instead of using 220V. You do much looking at any of those meters you have?


Anyone who cares to look can see it was you who started hurling insults:
"Now, Ricodjour, SaltyAss, and ShittyTwo are proly getting blisters from their bunched-up panties, screaming, Liar, Where's the Citation????? Proly Trader would too, if he didn't already agree -- altho he don't know the physics, apparently. "
Just for the record the first 3 guys you insulted were not even part of this discussion. But I can assure you that all 3 of them know:
A - You can't use a Kill-a-Watt to measure power in a dryer
B - You can't hook up two of them, one on each leg either
C - Voltage pairing at a dryer is 240V/120V, not 220V/120V

Because we understand basic electricity. Do you need to drive a car off a cliff to know it will fall? BTW, one more time, you can't have 220V and 120V on a dryer circuit. There's an experiment you can try.

Sure, in the case of a theoretical load, that is true. However, it matter not a wit in terms of measuring the power of the dryer. As Blueman pointed out, if you measure the current on the two hot legs, you can then calculate the power. In your own example above, all we need to know is (20+23amps)*120V.

Maybe the dryer has a little built-in nuke to run it. Better go check that too.

You're the only guy I've seen using the pairing 220V/120V.

And you call me a hater? LOL

It was never news to me that the easiest thing to do was use a clamp on meter on one dryer wire at a time.

If you're really screwing around with that kind of stuff, at least we won't have to put up with you much longer.
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snipped-for-privacy@optonline.net writes:

NO I am NOT missing the point. I know very well that the real power is equal to voltage times amps times power factor. I didn't talk about voltage since the voltage is a known quantity. Also, the power factor is an assumption (and a relatively small overall correction -- see below) and in any case is not easily measurable with simple electrical tools. So, I focused on current since that is the unknown variable that drives power.
However, all of that is in any case irrelevant since my one and only point is that I believe that you can measure the current in both "phases" with a single clamp-on lead by *reversing* the direction of one of the leads.
Then the real power would be: 120V * (total current through both leads) * (weighted average power factor).
Note for simplicity, I would probably plug in a power factor of 1 since the resistive heating load will dominate the much smaller inductive load due to the motor. So, any errors would be second order. In fact, based on http://michaelbluejay.com/electricity/dryers.html , one can assume that the inductive load is only about 6.4% of the power consumption. So, assuming the motor has a power factor of 0.8, the weighted average power factor would be: 0.99 which is pretty close to 1 in my book...

All very true but it has nothing to do with my question or my understanding of the physics.
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Knowing the current is both legs doesn't tell you anything about power. The current is the same in both legs. If you put both phases in an amp probe, you will get double, but it is a useless value.

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The whole point is you *don't* get exactly double because the 120v leg feeding the motor only goes through one of the legs.
And calculating the power is a trivial calculation that follows from knowing the voltage in each leg (120v) and assuming a power factor (which I proved in the previous post can be assumed to be just 1 and not lose more than 1% accuracy). To tell the truth I'm puzzled why people seem to be hung up by the voltage which is the one known factor while the clear unknown is the current.

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IMO I think you would be measuring it twice. the current flows from one end of the 220 volt supply, thorugh the the machine, mainly the heaters, and back through the other 220 volt as a return. Some dryers do have some 110 volt that would unbalnce that a bit but not much.
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wrote:

I'd recommend trying them both out soon and make sure you save the receipt until you do. I bought one at Costco last summer. Right out of the box, one of the keys didn't work, so it was impossible to use. Took it back and they gave me another one. On that one a different key didn't work. So, I decided to wait a few months in the hope that maybe they just had a bad batch. Bought one a couple months ago and that one is working fine.
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Cost to run dryer
I live on the Central Oregon Coast & my electric company says its about27.6 cents per HR our rates are 6.14 cents per kilowatt hr.
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That's relatively cheap; but does that include ALL the charges. To get ours I divide the total bill (which includes a per account and sales tax charges) by total k.w.hrs. Since we are on an even monthly charge plan, the same every month, adjusted annually, it's accurate.
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terry wrote:

Rather than run a cable to the new outbuilding at his farm, my brother considered having the power company install a service drop. If he'd used no electricity in the building, that second account would have cost him $9 a month.
Electricity is 8 cents per kWh. If in a given month his only use in the building was a little lighting, his he might have been billed $9.08 for 1kWh.
Suppose his tractor was parked in the building, and the battery ran down. He figures it will take 2kWh to charge it. That would add 16 cents to his bill.
Dividing the total bill by the total kWh would lead him to estimate that charging his battery would add $18.12 to his bill.
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E Z Peaces wrote:

I recently had the option of a second service with a separate meter, and that would have had a monthly charge somewhere around $10 even if no power was used. The other choice was for them to install what they call a "current transformer". The secondary on the current transformer powers the meter and i get 1 bill. I think I can have as many lines tapped off that as I need. The only catch is they have a charge to install the current transformer, but the savings pays it off in I believe less than 1 year. I forget the actual figures.
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Tony wrote:

Yep, that's the change when your service size transitions from the size where unit meters are available, 200A and under I think, to where meters with separate current transformers are required.
You can have as many "lines" off either (various code provisions apply), but it's uncommon on small services. I believe it is fairly common for McMansions to have a 400A service and metering, feeding two 200A service panels since the two 200A panels are cheaper than one 400A panel.
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On Thu, 28 Jan 2010 18:04:18 -0500, Tony wrote:

That kind of sounds like our off-peak setup; we have two meters - one primary, which measures everything, and one secondary which slaves off the primary and measures the off-peak use. For billing they just subtract the off-peak reading from the primary to get "non off-peak" usage.
No charge for having the off-peak metering/equipment installed. Given the stories on here, I'm starting to think we have an exceptionally good power company :-)
cheers
Jules
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Jules wrote:

Quite different really and nothing to do with peak / off peak rates.
If you have loads that require a 400A service, you can have a 400A service installed which requires a meter with separate current transformers, or you can have two 200A services which use separate unit meters. In either case you end up feeding two separate 200A service panels as the most cost effective option.
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