terry wrote:

Weigh the load going in and coming out. Each pound lost takes 0.285kWh. Some cotton garments are very heavy going into the dryer, and they probably cost a lot to dry.

You would have to add the cost of turning the drum and blowing the air. I could get the wattage by timing my power meter after switching off all my other circuits and starting a load in with no heat. After the load dried, I'd run it without heat again and check the wattage again. I'd take the average and multiply it by the time a load ran.

The exit air is warmer than the entrance air. Without knowing the volume of air my dryer blows, I can't tell if that adds much to the cost.

I think the Kill-a-Watt EZ will do all that. I believe it measures instantaneous wattage AND accumulates kWhrs.... about $25 at Costco. I have one, but haven't used it yet.

--

EA

EA

Existential Angst wrote:

You have a 110 V dryer?

You have a 110 V dryer?

Heh.... good point.... I wonder if you could use two Kill-a-Watts, on different 120 V legs.... :) I might try that, cuz I bought one for my BIL.

--

EA

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Existential Angst wrote:

Mine is 15A max also, so that could limit you too.

Mine is 15A max also, so that could limit you too.

FOUR killawatt thingies, then?? googawd, nuthin is simple....

--

EA

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*> *

EA

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Existential Angst wrote:

I just checked mine. My meter is stamped 7.2Kh. Before I turned on the dryer, a meter revolution took 62 seconds, for 418 watts. With the empty dryer turning without heat, it was 42 seconds, for 617 watts. So the motor uses about 200 watts empty. Where I live, that's less than 2 cents an hour.

Next time I dry a load, I can see what the motor uses with a wet load. With the heat on, I can use an IR thermometer to check the temperature of the galvanized vent pipe, which would probably be near that of the air flowing inside. Then if I can estimate the volume, I can estimate how much that adds to the cost.

Suppose it's 5000 liters of air per hour (1.4 liters per second) at 122F. If room temperature is 68F, That would be 192 watts to warm the air. That's about what the motor uses. I wonder how much of the power that goes into the motor ends up heating the air.

If I put in a wet load at 20 pounds and take it out at 10 pounds half an hour later, that's 2.85kWh to evaporate the water and perhaps 0.25kWh for the motor and warm exhaust. In this case, 92% of the cost would come from the weight of water, which could be determined by weighing the laundry.

I just checked mine. My meter is stamped 7.2Kh. Before I turned on the dryer, a meter revolution took 62 seconds, for 418 watts. With the empty dryer turning without heat, it was 42 seconds, for 617 watts. So the motor uses about 200 watts empty. Where I live, that's less than 2 cents an hour.

Next time I dry a load, I can see what the motor uses with a wet load. With the heat on, I can use an IR thermometer to check the temperature of the galvanized vent pipe, which would probably be near that of the air flowing inside. Then if I can estimate the volume, I can estimate how much that adds to the cost.

Suppose it's 5000 liters of air per hour (1.4 liters per second) at 122F. If room temperature is 68F, That would be 192 watts to warm the air. That's about what the motor uses. I wonder how much of the power that goes into the motor ends up heating the air.

If I put in a wet load at 20 pounds and take it out at 10 pounds half an hour later, that's 2.85kWh to evaporate the water and perhaps 0.25kWh for the motor and warm exhaust. In this case, 92% of the cost would come from the weight of water, which could be determined by weighing the laundry.

Yeah but isn't the 220 draw from hot to hot with the only current returning through the neutral being the 110v leg that typically runs the light and the motor. So, I'm not sure that hooking up two of them would work (plus I'm pretty sure they are not rated at the 35A or so amperage of a dryer).

I think the easiest thing would be to put a clamp-on ammeter on it -- which shouldn't be too hard since one typically has easy access to the dryer end of the cord where the wires terminate. Don't forget to measure current in both legs of course.

Actually, with a clamp-on meter, would you get an accurate rating of 220v current if you clamped around both hot legs but with the direction of the wire in one of the legs reversed 180 degrees -- my thinking is that reversing the wire direction would make both currents appear in phase and hence be additive...

No, you still can't clamp around both wires. This is 220/240

But the clamp-on is a good idea, because it is esp. accurate on a purely resistive load -- no power factor to worry about. You would, however, have to subtract out the the motor current of one leg, tho. If one hot were reading 20 A, and the other leg were reading 16 A, the total wattage would be: 16 x 240 + 4 x 120 x .8 , where .8 would be a typical (inductive) power factor for motors. Presumably the neutral coming from the dryer would also read that same 4 A, but who knows what's going with grounds, etc.

--

EA

EA

wrote:

Not two PHASES. The TWO ends of the same single phase supply. Which also has a centre tap to provide a zero point for two 120 volt 'legs'. Have lived in a house that had a single neutral and three separate phases each 120 degrees apart! There was a 3 phase main breaker and then three separate sections to the breaker panel.

Not two PHASES. The TWO ends of the same single phase supply. Which also has a centre tap to provide a zero point for two 120 volt 'legs'. Have lived in a house that had a single neutral and three separate phases each 120 degrees apart! There was a 3 phase main breaker and then three separate sections to the breaker panel.

On Thu, 28 Jan 2010 11:27:12 -0500, "Existential Angst"

Since it is single phase, the current in both legs will be the same. And you don't add them to get the total.

Since it is single phase, the current in both legs will be the same. And you don't add them to get the total.

Metspitzer wrote:

The current will be slightly asymmetric since a dryer typically has 120V control loads in addition to 240V heating elements. This is where the neutral current on the ground issue arises and why the current code has 4 wire receptacles for dryers with separate neutral and ground conductors.

The current will be slightly asymmetric since a dryer typically has 120V control loads in addition to 240V heating elements. This is where the neutral current on the ground issue arises and why the current code has 4 wire receptacles for dryers with separate neutral and ground conductors.

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.

This seems to me to be basic physics... if I am wrong please explain where I am misunderstanding...

Good point about the power factor when trying to use current & voltage to calculate power used (and charged).

As the other folks have tried to point out, to calculate power you need to know the current, power factor, and VOLTAGE. It seems you understand the power factor part of the problem, but not the voltage issue. Let's assume for the moment that the dryer has no 120V loads. In that case, all the current is flowing from one hot lead to the other at 240V. To calculate the power you need only one amp meter on EITHER hot wire because the same current is coming in on one hot and going back out on the other. The power is then:

P = 240V X amps in either hot X power factor

If the dryer has a 120V load component, then the 120V current component is flowing in via one of the hots and back via the neutral. In that case you measure the current on BOTH hots, The power is then:

P = 240V X amps in lesser of the two hots X power factor + 120V X (higher amperage hot - lower amperage hot) X power factor

You could also measure the current in the neutral, but you don't have to, since the difference between the two must be on the neutral. But if you did, the following formula would apply:

P = 240V X amps in lesser of the 2 hots X power factor + 120V X amps in neutral X power factor

As the other folks have tried to point out, to calculate power you need to know the current, power factor, and VOLTAGE. It seems you understand the power factor part of the problem, but not the voltage issue. Let's assume for the moment that the dryer has no 120V loads. In that case, all the current is flowing from one hot lead to the other at 240V. To calculate the power you need only one amp meter on EITHER hot wire because the same current is coming in on one hot and going back out on the other. The power is then:

P = 240V X amps in either hot X power factor

If the dryer has a 120V load component, then the 120V current component is flowing in via one of the hots and back via the neutral. In that case you measure the current on BOTH hots, The power is then:

P = 240V X amps in lesser of the two hots X power factor + 120V X (higher amperage hot - lower amperage hot) X power factor

You could also measure the current in the neutral, but you don't have to, since the difference between the two must be on the neutral. But if you did, the following formula would apply:

P = 240V X amps in lesser of the 2 hots X power factor + 120V X amps in neutral X power factor

=================================================

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.

First, he's right in that something is cancelling, but it is not the phases, cuz, well, there are no phases to cancel.

The clamp-on meter reading cancels when measuring both hots because of a cancelled B field (magnetic field -- Biot's law, or sumpn), ie, the alternating B field is in opposite directions in each wire because the current is physically travelling in opposite directions in each wire, so there is no net field for the meter to read .

You would see this cancelling effect even on DC current (which traditional clamp-ons can't measure, anyway.).

BUT, since there is extra current in one wire, due to the 120 loads included by one wire, when you measure both wires together, you in fact get the DIFFERENCE of the current in the two wires, which approx. equals the neutral current -- why it's not exactly the same as the neutral current beats me.

NOW, here's the REAL inneresting part:

Well, if the B fields are cancelling because of current travelling in opposite directions, then if I could pull out ONE of the 220V wires and twist it once so that when I place on the clamp-on meter, the meter 'sees" two currents travelling in the

And they DO!! With about 21 A in one wire, and 23 in the other, I measured 44 with a twist in one wire.

To better help visualize this, or duplicate this, you'll need to be able to pull out a bit of individual #12 or #10 wire, proly in the breaker panel. Then, draw arrows on each wire, in opposite directions, with a Sharpie. Then, manipulate one wire so that the arrows point in the same direction. This will demonstrate "the physics".

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.

I was able to do this cuz I have a breaker panel right by the dryer, which happened to be opened, for another outlet being added. You couldn't verify this with a regular dryer "cord", unless you go to the breaker box, or go to the dryer terminal block, and splice in individual wires.

All the above applies to 120V hots and neutrals, as well.

--

EA

*>*

*> > But the clamp-on is a good idea, because it is esp. accurate on a purely*

EA

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Exactly - I am well aware of voltage and power factor.

Correct - I shouldn't have said phase so much as direction of current flow (which is still technically phase in the mathematically trivial sense of being 180 degree different)

That is why I suggested reversing the direction of one of the wires.

Probably due to how the "integration" is done in a clamp-on meter so.

That was the ENTIRE point of my suggestion. Reverse the direction of one of the wires so that the currents flow in the same direction and add rather than cancel. I suggested that as a "clever" response to those suggesting the need for multiple clamp-ons.

So, then my hypothesis was right. Thanks for checking it!

Thanks for the doing the experiment to verify my intuition based on my long-ago learned university E&M courses.

Too bad it took so many posts and tangents to verify this very simple point.

Get used to it! Also get used to the dick-waving, altho there is less of that here than it many places. I guess when one's house is falling apart, there's less active testosterone....

Ackshooly, I hadn't realized that this is what you were driving at in your original post, but mebbe it registered subliminally, and unknowingly was the root of my li'l sleuthing.

But it was a very good insight! And neat experiment. Altho, by reversing one of the wires you are approx. doubling the current reading. Your point proly didn't register because current directionality hadn't registered, cuz you normallly don't think of such in AC circuits. But indeed there is physical directionality in two wires, even if that directionality is at 60 Hz!

But, iirc, wasn't it multiple Kill-a-watt ditties we were talking about, not multiple clampons??

--

EA

EA

wrote:

I'd say the more fundamental problem in how blueman is looking at things is that he's trying to add the current in the two hots as if they were totally different currents, or out of phase currents, etc. In fact, with a pure 240V load or a 120V balanced load, all he's doing is measuring the same exact current as it flows through the circuit. It's like counting the current twice in a flashlight, once as it enters the bulb, once at it leaves.

If the load has a 120V unbalanced component, then the current will be higher in one hot than the other, with the difference being in the neutral.

I agree, you could do the power calcualtion by using the sum of the current measurements in the two hot legs and then multiplying by 120Volts. Which gets back to what I said before, that to correctly calculate the power, you need to correctly factor in both the voltage and the current.

To recap, let's say there is a 20 amp 240V load and a 3 amp 120V load on the circuit. You'd mesure 23 amps in one hot, 20 amps in the second hot, 3 amps in the neutral. So, per the formulas I gave above, by measuring the two hots you'd get P = 240VX20 + 120VX3. Or you could do it by adding the current measurements in both hots and then USING 120V, P= 120VX43

I don't know why you find this so interesting, nothing new here, it's just basic physics.

After trying to impress us by presenting basic physics as if it were something surprising, are you sure you want to start hurling insults?

I'd say the more fundamental problem in how blueman is looking at things is that he's trying to add the current in the two hots as if they were totally different currents, or out of phase currents, etc. In fact, with a pure 240V load or a 120V balanced load, all he's doing is measuring the same exact current as it flows through the circuit. It's like counting the current twice in a flashlight, once as it enters the bulb, once at it leaves.

If the load has a 120V unbalanced component, then the current will be higher in one hot than the other, with the difference being in the neutral.

I agree, you could do the power calcualtion by using the sum of the current measurements in the two hot legs and then multiplying by 120Volts. Which gets back to what I said before, that to correctly calculate the power, you need to correctly factor in both the voltage and the current.

To recap, let's say there is a 20 amp 240V load and a 3 amp 120V load on the circuit. You'd mesure 23 amps in one hot, 20 amps in the second hot, 3 amps in the neutral. So, per the formulas I gave above, by measuring the two hots you'd get P = 240VX20 + 120VX3. Or you could do it by adding the current measurements in both hots and then USING 120V, P= 120VX43

I don't know why you find this so interesting, nothing new here, it's just basic physics.

After trying to impress us by presenting basic physics as if it were something surprising, are you sure you want to start hurling insults?

wrote:

I'd say the more fundamental problem in how blueman is looking at things is that he's trying to add the current in the two hots as if they were totally different currents, or out of phase currents, etc. In fact, with a pure 240V load or a 120V balanced load, all he's doing is measuring the same exact current as it flows through the circuit. It's like counting the current twice in a flashlight, once as it enters the bulb, once at it leaves.

If the load has a 120V unbalanced component, then the current will be higher in one hot than the other, with the difference being in the neutral.

I agree, you could do the power calcualtion by using the sum of the current measurements in the two hot legs and then multiplying by 120Volts. Which gets back to what I said before, that to correctly calculate the power, you need to correctly factor in both the voltage and the current.

To recap, let's say there is a 20 amp 240V load and a 3 amp 120V load on the circuit. You'd mesure 23 amps in one hot, 20 amps in the second hot, 3 amps in the neutral. So, per the formulas I gave above, by measuring the two hots you'd get P = 240VX20 + 120VX3. Or you could do it by adding the current measurements in both hots and then USING 120V, P= 120VX43

I don't know why you find this so interesting, nothing new here, it's just basic physics. ================================================= Mebbe, but I didn't see***you*** mention the Biot-Savart law.......
In fact, you're ***still*** multiplying volts x amps in various forms, when that
wasn't even the Q.
The Q was: why can't you put a clamp-on around ***both*** wires at the same
time?

After trying to impress us by presenting basic physics as if it were something surprising, are you sure you want to start hurling insults? ===================================================== The surprise here, which has apparently gone clear over your head (much the same way climb/conventional cutting went over the heads of the above Assaholic Triumvirate), is how "basic" physics can be so neatly demonstrated by, well, an***appliance***.

If you aren't positively tickled by this clamp-on thing, then, well, you really don't appreciate "basic" physics, and anyone who doesn't appreciate "basic" physics proly doesn't understand "basic" physics.

In fact, imo, the very term "basic physics" is a kind of high-handed and pretentious dismissal of what the greatest minds in history labored for decades to develop. Newton himself did not properly grok "energy". Yet, you and your ilk will call energy/momentum "basic" or "simple".

Newton's Cradle http://scientificsonline.com/product.asp_Q_pn_E_3081502 is "simple", a "simple toy" (a so-called "executive toy", as executives are, well, pretty simple....), yet it demonstrates the fundamental (mathematical) laws of our universe. Well, just how "simple" is that??

Btw, Wiki has a nice article on Newton's Cradle, which was in fact not invented by Newton, but was a wonderful homage to Newton by the inventor. http://en.wikipedia.org/wiki/Newton's_cradle . For anyone interested, there are on-line interactive demo's and explanations of the cradle -- just google around, you'll find them. Some are certainly better than others.

Imo, it is the philistine who confuses "simple" and "basic" with "profound" and "fundamental". And Einstein himself observed, If you think something is simple, you probably don't understand it.

Ultimately, "basic" physics is VERY surprising.

I'd say the more fundamental problem in how blueman is looking at things is that he's trying to add the current in the two hots as if they were totally different currents, or out of phase currents, etc. In fact, with a pure 240V load or a 120V balanced load, all he's doing is measuring the same exact current as it flows through the circuit. It's like counting the current twice in a flashlight, once as it enters the bulb, once at it leaves.

If the load has a 120V unbalanced component, then the current will be higher in one hot than the other, with the difference being in the neutral.

I agree, you could do the power calcualtion by using the sum of the current measurements in the two hot legs and then multiplying by 120Volts. Which gets back to what I said before, that to correctly calculate the power, you need to correctly factor in both the voltage and the current.

To recap, let's say there is a 20 amp 240V load and a 3 amp 120V load on the circuit. You'd mesure 23 amps in one hot, 20 amps in the second hot, 3 amps in the neutral. So, per the formulas I gave above, by measuring the two hots you'd get P = 240VX20 + 120VX3. Or you could do it by adding the current measurements in both hots and then USING 120V, P= 120VX43

I don't know why you find this so interesting, nothing new here, it's just basic physics. ================================================= Mebbe, but I didn't see

After trying to impress us by presenting basic physics as if it were something surprising, are you sure you want to start hurling insults? ===================================================== The surprise here, which has apparently gone clear over your head (much the same way climb/conventional cutting went over the heads of the above Assaholic Triumvirate), is how "basic" physics can be so neatly demonstrated by, well, an

If you aren't positively tickled by this clamp-on thing, then, well, you really don't appreciate "basic" physics, and anyone who doesn't appreciate "basic" physics proly doesn't understand "basic" physics.

In fact, imo, the very term "basic physics" is a kind of high-handed and pretentious dismissal of what the greatest minds in history labored for decades to develop. Newton himself did not properly grok "energy". Yet, you and your ilk will call energy/momentum "basic" or "simple".

Newton's Cradle http://scientificsonline.com/product.asp_Q_pn_E_3081502 is "simple", a "simple toy" (a so-called "executive toy", as executives are, well, pretty simple....), yet it demonstrates the fundamental (mathematical) laws of our universe. Well, just how "simple" is that??

Btw, Wiki has a nice article on Newton's Cradle, which was in fact not invented by Newton, but was a wonderful homage to Newton by the inventor. http://en.wikipedia.org/wiki/Newton's_cradle . For anyone interested, there are on-line interactive demo's and explanations of the cradle -- just google around, you'll find them. Some are certainly better than others.

Imo, it is the philistine who confuses "simple" and "basic" with "profound" and "fundamental". And Einstein himself observed, If you think something is simple, you probably don't understand it.

Ultimately, "basic" physics is VERY surprising.

--

EA

*>*

*> I was able to do this cuz I have a breaker panel right by the dryer, which*

EA

Click to see the full signature.

wrote:

Excuse me, but the question under discussion was how to measure the POWER going to a dryer. At which point you posted this gem:

"I think the Kill-a-Watt EZ will do all that. I believe it measures instantaneous wattage AND accumulates kWhrs.... about $25 at Costco. I have one, but haven't used it yet. "

Which obviously won't work for a whole host of very simple reasons:

1 - The Kill-a-Watt is not a 240V device

2 - Dryers far exceed the current limitations of your Kill-a-Watt

3 - The plug from a dryer won't fit into the Kill-a-Watt

4 - The Kill-a-Watt will not plug into a dryer outlet

Need, I go on?

Then blueman posted this:

"I think the easiest thing would be to put a clamp-on ammeter on it -- which shouldn't be too hard since one typically has easy access to the dryer end of the cord where the wires terminate. Don't forget to measure current in both legs of course.

Actually, with a clamp-on meter, would you get an accurate rating of 220v current if you clamped around both hot legs but with the direction of the wire in one of the legs reversed 180 degrees -- my thinking is that reversing the wire direction would make both currents appear in phase and hence be additive... "

To which you replied:

"No, you still can't clamp around both wires. This is 220/240***single
phase*** -- true, you are using two legs of opposite phase, but the net
voltage is still single phase. Iow, there is single phase and three
phase,
but no two phase -- heh, funny how that works.

But the clamp-on is a good idea, because it is esp. accurate on a purely resistive load -- no power factor to worry about. You would, however, have to subtract out the the motor current of one leg, tho. If one hot were reading 20 A, and the other leg were reading 16 A, the total wattage would be: 16 x 240 + 4 x 120 x .8 , where .8 would be a typical (inductive) power factor for motors. "

In fact, both of you have parts of it wrong. Blueman is wrong because if it's 220v current he's measuring, there is no need to clamp around BOTH conductors. What is coming in on one is going out on the other. But if he did what he described, by reversing the direction of one hot wire, he'd be measuring 2X the actual current. And yes, he could calculate the power that way if he treated it as all 120V load instead of 240V load.

And you have it wrong, because despite what you said, he could do what he described. In fact, the measurement technique he described is exactly what you are now harping about as surprising physics.

It's also rather strange that you accuse me of talking about power, when you yourself answered about power and gave the correct power calculation.

Actually, it was why can't you put it around both wires at the same time after reversing one to measure the 220V current. The answer is still the same, if it's 220V current he's interested in, there is no need to clamp both. Just take the lower reading of the two. If it's all a 240v load, then they are the same. If not, the lower is the 220V current. And last time I checked, the standard is actually 240V, at least everywhere that I'm familiar with in the USA.

Glad you've found something to amuse and fascinate yourself with. That's still no excuse for hurling insults at a bunch of us who did nothing to you. Did any of us launch off at you after that whopper about using the Kill-a-Watt meter on a dryer?

There's your problem. You've been tickling yourself too much.

Look, we're not solving Maxwell's equations here. The concepts involved with correctly measuring dryer current are basic physics.

It would also help if you could learn how to trim posts.

Excuse me, but the question under discussion was how to measure the POWER going to a dryer. At which point you posted this gem:

"I think the Kill-a-Watt EZ will do all that. I believe it measures instantaneous wattage AND accumulates kWhrs.... about $25 at Costco. I have one, but haven't used it yet. "

Which obviously won't work for a whole host of very simple reasons:

1 - The Kill-a-Watt is not a 240V device

2 - Dryers far exceed the current limitations of your Kill-a-Watt

3 - The plug from a dryer won't fit into the Kill-a-Watt

4 - The Kill-a-Watt will not plug into a dryer outlet

Need, I go on?

Then blueman posted this:

"I think the easiest thing would be to put a clamp-on ammeter on it -- which shouldn't be too hard since one typically has easy access to the dryer end of the cord where the wires terminate. Don't forget to measure current in both legs of course.

Actually, with a clamp-on meter, would you get an accurate rating of 220v current if you clamped around both hot legs but with the direction of the wire in one of the legs reversed 180 degrees -- my thinking is that reversing the wire direction would make both currents appear in phase and hence be additive... "

To which you replied:

"No, you still can't clamp around both wires. This is 220/240

But the clamp-on is a good idea, because it is esp. accurate on a purely resistive load -- no power factor to worry about. You would, however, have to subtract out the the motor current of one leg, tho. If one hot were reading 20 A, and the other leg were reading 16 A, the total wattage would be: 16 x 240 + 4 x 120 x .8 , where .8 would be a typical (inductive) power factor for motors. "

In fact, both of you have parts of it wrong. Blueman is wrong because if it's 220v current he's measuring, there is no need to clamp around BOTH conductors. What is coming in on one is going out on the other. But if he did what he described, by reversing the direction of one hot wire, he'd be measuring 2X the actual current. And yes, he could calculate the power that way if he treated it as all 120V load instead of 240V load.

And you have it wrong, because despite what you said, he could do what he described. In fact, the measurement technique he described is exactly what you are now harping about as surprising physics.

It's also rather strange that you accuse me of talking about power, when you yourself answered about power and gave the correct power calculation.

Actually, it was why can't you put it around both wires at the same time after reversing one to measure the 220V current. The answer is still the same, if it's 220V current he's interested in, there is no need to clamp both. Just take the lower reading of the two. If it's all a 240v load, then they are the same. If not, the lower is the 220V current. And last time I checked, the standard is actually 240V, at least everywhere that I'm familiar with in the USA.

Glad you've found something to amuse and fascinate yourself with. That's still no excuse for hurling insults at a bunch of us who did nothing to you. Did any of us launch off at you after that whopper about using the Kill-a-Watt meter on a dryer?

There's your problem. You've been tickling yourself too much.

Look, we're not solving Maxwell's equations here. The concepts involved with correctly measuring dryer current are basic physics.

It would also help if you could learn how to trim posts.

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