On Thu, 28 Jan 2010 09:12:29 -0800, terry wrote:

On Jan 28, 1:22Â*pm, JP wrote:
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.

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.

Our bills do end up all over the place - I just got the latest one a few
minutes ago, and it's currently 4.5 cents / kWh for the stuff we have on
off-peak (heaters, dryer and water heater), then 9.4c/kWh for the first
500 kWh used of 'normal' electric and 7.4c after that. We also get $9.50
off for a "water heater credit" and $4 off for a "dryer credit".

That off-peak rate's certainly less than it was the previous month.

We've just ordered a couple of hundred gallons of propane today, so it'll
be interesting to see what the prices there are like when that gets
delivered.

cheers

Jules

On Jan 28, 4:09*pm, "Existential Angst"
wrote:
"Jim Elbrecht" wrote in message

...





wrote:

On Jan 28, 11:44 am, "Existential Angst"
wrote:

-snip-
A much much bigger factor:
How corrupt is a state's legislature?
Given the corruption factor in NY, I spose 30c/kWhr (what it winds up
really
being, not the bull**** 9c) is still a bargain.
--
EA

Please post us a link to the utility in NY where electricity costs 30C
a kwh for any typical amount of residential usage. * I've seen many
tables showing electricity costs, highs, lows, etc and have never seen
anywhere that it costs 30c.

30cents wouldn't surprise me for ConEd, LIPA or Central Hudson. * The
average is supposedly 19cents;
http://www.eia.doe.gov/cneaf/electri...able5_6_a.html

I'm served by National Grid- one of the bigger upstate suppliers- and
usually on the low end. * They have on the bill that they are selling
me electricity for 6.8cents. * *But when I divide the bill by the KWH,
I find I'm paying over 15cents. * *LIPA and ConEd both advertise over
20 cents & they both have higher taxes than upstate- so 30cents is
real easy to believe.

But I think California and Hawaii are higher.

They are.
But I don't think Trader is really curious, as much as he just wants to
catch me in a lie, like SaltyAss, Ricodjour, ****ty Two, and their ilk.

Again, for the slow, my near-30c rate is not an explicit, published rate -- *
the published rate is 9c.
But the *real* rate I wind up paying, when I divide the $$ that I mail in by
the kWhr that I got, is near 30c.

It's good to hear that not everyone is f'd ita like this. *But CA, NY, and a
bunch of others are.
I suspect everyone's turn will come, tho. *After all, what corp. concern can
resist the easy fleecing of millions of sheeple? * What CongressShill has
the balls to fight it? *If balls is even an issue -- corruption is the
issue.

Free Money, Free Money!!!
--
EA


It's OK. Help is on the way. Last night I heard Obama say he wants
more nuclear power plants built. Oh, wait a minute. His actual
words were "a new generation of safe, clean nuclear power plants."
Doh! Looks like you're still f**d, because we know there will never
be one that's clean enough or safe enough for him or the environmental
extremists. That's change you can believe in.

wrote:
....

It's OK. Help is on the way. Last night I heard Obama say he wants
more nuclear power plants built. Oh, wait a minute. His actual
words were "a new generation of safe, clean nuclear power plants."
Doh! Looks like you're still f**d, because we know there will never
be one that's clean enough or safe enough for him or the environmental
extremists. That's change you can believe in.

Well, we're going to know pretty soon. There are 20-something [1]
applications for new licenses on file w/ NRC currently. When the formal
hearings begin we'll find out where the C sequestration people are in
terms of whether want to accomplish something or simply be obstructionists.

[1] www.nrc.gov has links to actual filings/numbers; I haven't looked at
precise numbers for a while

--

E Z Peaces 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.

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.

Tony wrote:

E Z Peaces 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.

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.

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.

Jules
wrote:

You can drastically shorten your drying times/costs with a front loader wash
machine.

Agreed

Why they cost so g-d much up front is another story.....

Also agreed. ****es me right off. A typical machine at a typical 'big

Agree on BOTH accounts!

I bought front loader in 2000..... Sears Kenmore

Lasted only 7 years before it fell apart!!

I sure didn't save any money as the washer was $600

I'm now back to a GE top loader.

writes:

On Jan 28, 10:15Â*am, blueman wrote:
Jules writes:
... ours is on off-peak so gets 6c/kWh, but I can't remember the wattage
on the heater for ours either (and it normally runs for about an hour for
a full load)

It's amazing (to me) that our cost is almost 3x as high.
The total cost per kwh (including tax, generation, transmission, fees)
is a whopping 17.7 cent/kwh without any possibility of off-peak.

It's hard to believe that the "free market" price (in the absense of
governmental regulation) would be 3x as large particularly given that
electricity is:
Â*- An almost pure commodity (a volt is a volt is a volt)

But what it takes for fuel to generate the power is a huge factor and
varies widely. The areas with the lowest electric prices are
usually the ones driven off hydro-electric. Unfortunately, because of
geography, most areas of the country don't have that available.

And also factor in labor rates, materials costs, land costs, etc.
What it costs to build a sub-station or run a new transmission line
near Niagra falls is going to be a whole lot different than one in
northern NJ or San Francisco.

At best all of the above perhaps explains why we pay more for the
generation portion of our electric bill (12.4 cents/kWh). And I
understand why burning natural gas or oil which is supposedly the main
fuel here in New England would be more expensive than coal burnt in the
Midwest and certainly more expensive than Hydro from the Pacific
NorthWest.

However, why do we pay 7.6 cents/kWh for transmission which is *more*
than the total cost of 6-7 cents that other users here claim to be
paying.

If anything in a regulated monopoly, our transmission costs should be
*lower* than other parts of the country since our lines were built a
long time ago and presumably have already recouped their cost of
capital. Also, with low population growth here, there is not a
requirement for huge new investments in expansion.

I think the real problem here is government regulation and corruption
which accomplishes the threefold evil of keeping prices artificially
high and discouraging competition, and preventing investment in new
technologies or cheaper sources of power...

"Existential Angst" writes:

"blueman" wrote in message
...
"Existential Angst" writes:
"Bob F" wrote in message
...
Existential Angst wrote:
"E Z Peaces" wrote in message
...
terry wrote:
Our very conventional old style tumble dryer timer runs for some 45
minutes per load.

With heater cutting in and out (estimating it's on say 80%?).

Heater elements are either 3000 watts or maybe 4500, haven't had
this one apart yet, since I got it in exchange for a dozen beer!

Our domestic electricity costs a little over 10 cents per k.watt hr.

So one load of clothes 45/60 x 0.8 a cost of electricity = 0.75 x
0.8 x 0.12 = 7 cents per load.

Occasionally it is necessary to run a 'heavy' load, towels and
blankets etc. part of a second run.

So maybe that could be say 12 to 15 cents per load. In summer we
hang bedclothes and towels on outside lines.

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.

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.

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 *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 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...

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.

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

On Jan 29, 2:19*am, blueman wrote:
"Existential Angst" writes:
"blueman" wrote in message
...
"Existential Angst" writes:
"Bob F" wrote in message
...
Existential Angst wrote:
"E Z Peaces" wrote in message
...
terry wrote:
Our very conventional old style tumble dryer timer runs for some 45
minutes per load.

With *heater cutting in and out (estimating it's on say 80%?).

Heater elements are either 3000 watts or maybe 4500, haven't had
this one apart yet, since I got it in exchange for a dozen beer!

Our domestic electricity costs a little over 10 cents per k.watt hr.

So one load of clothes 45/60 x 0.8 a cost of electricity = 0.75 x
0.8 x 0.12 = 7 cents per load.

Occasionally it is necessary to run a 'heavy' load, towels and
blankets etc. part of a second run.

So maybe that could be say 12 to 15 cents per load. In summer we
hang bedclothes and towels on outside lines.

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.

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.

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 *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 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...



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












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.

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

- Show quoted text -

On Thu, 28 Jan 2010 18:04:18 -0500, Tony wrote:

E Z Peaces 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.

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.

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

blueman wrote:

writes:

On Jan 28, 10:15Â am, blueman wrote:
Jules writes:
... ours is on off-peak so gets 6c/kWh, but I can't remember the wattage
on the heater for ours either (and it normally runs for about an hour for
a full load)

It's amazing (to me) that our cost is almost 3x as high.
The total cost per kwh (including tax, generation, transmission, fees)
is a whopping 17.7 cent/kwh without any possibility of off-peak.

It's hard to believe that the "free market" price (in the absense of
governmental regulation) would be 3x as large particularly given that
electricity is:
 - An almost pure commodity (a volt is a volt is a volt)

But what it takes for fuel to generate the power is a huge factor and
varies widely. The areas with the lowest electric prices are
usually the ones driven off hydro-electric. Unfortunately, because of
geography, most areas of the country don't have that available.

And also factor in labor rates, materials costs, land costs, etc.
What it costs to build a sub-station or run a new transmission line
near Niagra falls is going to be a whole lot different than one in
northern NJ or San Francisco.

At best all of the above perhaps explains why we pay more for the
generation portion of our electric bill (12.4 cents/kWh). And I
understand why burning natural gas or oil which is supposedly the main
fuel here in New England would be more expensive than coal burnt in the
Midwest and certainly more expensive than Hydro from the Pacific
NorthWest.

However, why do we pay 7.6 cents/kWh for transmission which is *more*
than the total cost of 6-7 cents that other users here claim to be
paying.

You pay more for transmission in the frozen northeast because:

1. The heavily forested states cost a lot more for tree trimming around
the power lines.

2. The ice storms and falling branches cost a lot more to repair the
power lines.


If anything in a regulated monopoly, our transmission costs should be
*lower* than other parts of the country since our lines were built a
long time ago and presumably have already recouped their cost of
capital. Also, with low population growth here, there is not a
requirement for huge new investments in expansion.

Old decrepit infrastructure costs more to repair and maintain, as well
as costing more for rebuilding as sections become overloaded or
otherwise unserviceable. Look at how often you see crews replacing mile
of poles and stringing new lines, often with taller poles and double
circuits to meet the increasing power demands.


I think the real problem here is government regulation and corruption
which accomplishes the threefold evil of keeping prices artificially
high and discouraging competition, and preventing investment in new
technologies or cheaper sources of power...

Government regulation is certainly part of the problem, but it is
typically keeping prices artificially low and discouraging system
upgrades. Look at how CA got into their power mess with their mock
"deregulation", where they deregulated the wholesale end but kept the
retail end regulated and capped and caused companies pull back rather
than absorb losses so the politicians could buy votes.

Jules wrote:

On Thu, 28 Jan 2010 18:04:18 -0500, Tony wrote:

E Z Peaces 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.

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.

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

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.

Pete C. wrote:
Jules wrote:
On Thu, 28 Jan 2010 18:04:18 -0500, Tony wrote:

E Z Peaces 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.
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.
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

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.

Your last paragraph doesn't hold true with my electric company. If I
had 2 200 amp meters, the second service would have a surcharge of
something like $10/month even if no power was used, or if a million
KWH's were used they would still charge that extra $10/month. So I paid
to have the current transformer installed and have no monthy fee except
for how many total KWH's I use. The one thing I dislike is the current
transformer divides the power by 20, so the "digital spinning wheel"
only runs at 1/20th the speed which is not easy to read to get that
"instant look" at how much power I'm using at the moment.

Tony wrote:

Pete C. wrote:
Jules wrote:
On Thu, 28 Jan 2010 18:04:18 -0500, Tony wrote:

E Z Peaces 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.
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.
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

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.

Your last paragraph doesn't hold true with my electric company. If I
had 2 200 amp meters, the second service would have a surcharge of
something like $10/month even if no power was used, or if a million
KWH's were used they would still charge that extra $10/month. So I paid
to have the current transformer installed and have no monthy fee except
for how many total KWH's I use. The one thing I dislike is the current
transformer divides the power by 20, so the "digital spinning wheel"
only runs at 1/20th the speed which is not easy to read to get that
"instant look" at how much power I'm using at the moment.

You didn't understand what I was saying in that last paragraph.

A 400A panelboard cost a *lot* more than two 200A panels, so even if you
have a single CT metered 400A service, in most cases it feeds two 200A
panels as this is less expensive.

Whether the two 200A panels are fed from one 400A service or separate
200A services is just a matter of whether you prefer two monthly service
charges, or one upfront CT meter installation charge.

writes:

On Jan 29, 2:19Â*am, blueman wrote:
"Existential Angst" writes:
"blueman" wrote in message
...
"Existential Angst" writes:
"Bob F" wrote in message
...
Existential Angst wrote:
"E Z Peaces" wrote in message
...
terry wrote:
Our very conventional old style tumble dryer timer runs for some 45
minutes per load.

With Â*heater cutting in and out (estimating it's on say 80%?).

Heater elements are either 3000 watts or maybe 4500, haven't had
this one apart yet, since I got it in exchange for a dozen beer!

Our domestic electricity costs a little over 10 cents per k.watt hr.

So one load of clothes 45/60 x 0.8 a cost of electricity = 0.75 x
0.8 x 0.12 = 7 cents per load.

Occasionally it is necessary to run a 'heavy' load, towels and
blankets etc. part of a second run.

So maybe that could be say 12 to 15 cents per load. In summer we
hang bedclothes and towels on outside lines.

Weigh the load going in and coming out. Â*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. Â*After the load dried, I'd run it without heat again and check
the wattage again. Â*I'd take the average and multiply it by the time
a load ran. The exit air is warmer than the entrance air. Â*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. Â*I believe it measures
instantaneous wattage AND accumulates kWhrs.... Â*about $25 at Costco.
I have one, but haven't used it yet.

You have a 110 V dryer?

Heh.... Â*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.

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. Â*This is 220/240 *single
phase* -- true, you are using two legs of opposite phase, but the net
voltage is still single phase. Â*Iow, there is single phase and three phase,
but no two phase -- heh, funny how that works.

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...

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

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...

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


All very true but it has nothing to do with my question or my
understanding of the physics.

"Pete C." writes:
blueman wrote:

writes:

On Jan 28, 10:15Â am, blueman wrote:
Jules writes:
... ours is on off-peak so gets 6c/kWh, but I can't remember the wattage
on the heater for ours either (and it normally runs for about an hour for
a full load)

It's amazing (to me) that our cost is almost 3x as high.
The total cost per kwh (including tax, generation, transmission, fees)
is a whopping 17.7 cent/kwh without any possibility of off-peak.

It's hard to believe that the "free market" price (in the absense of
governmental regulation) would be 3x as large particularly given that
electricity is:
 - An almost pure commodity (a volt is a volt is a volt)

But what it takes for fuel to generate the power is a huge factor and
varies widely. The areas with the lowest electric prices are
usually the ones driven off hydro-electric. Unfortunately, because of
geography, most areas of the country don't have that available.

And also factor in labor rates, materials costs, land costs, etc.
What it costs to build a sub-station or run a new transmission line
near Niagra falls is going to be a whole lot different than one in
northern NJ or San Francisco.

At best all of the above perhaps explains why we pay more for the
generation portion of our electric bill (12.4 cents/kWh). And I
understand why burning natural gas or oil which is supposedly the main
fuel here in New England would be more expensive than coal burnt in the
Midwest and certainly more expensive than Hydro from the Pacific
NorthWest.

However, why do we pay 7.6 cents/kWh for transmission which is *more*
than the total cost of 6-7 cents that other users here claim to be
paying.

You pay more for transmission in the frozen northeast because:

1. The heavily forested states cost a lot more for tree trimming around
the power lines.

I live in Eastern MA - I wouldn't say we were heavily
forrested. Certainly much less forrested than Pacific Northwest where
the total power cost is less than 7 cents/kwh.


2. The ice storms and falling branches cost a lot more to repair the
power lines.

Much less ice than Quebec where total cost is also less than our
transmission costs.



If anything in a regulated monopoly, our transmission costs should be
*lower* than other parts of the country since our lines were built a
long time ago and presumably have already recouped their cost of
capital. Also, with low population growth here, there is not a
requirement for huge new investments in expansion.

Old decrepit infrastructure costs more to repair and maintain, as well
as costing more for rebuilding as sections become overloaded or
otherwise unserviceable. Look at how often you see crews replacing mile
of poles and stringing new lines, often with taller poles and double
circuits to meet the increasing power demands.

The big cost is in installing new high tension lines. Replacing the
occassional pole or transformer pales in comparison. And we are not
doing much of either around here. You rarely see power crews out doing
much of anything.


I think the real problem here is government regulation and corruption
which accomplishes the threefold evil of keeping prices artificially
high and discouraging competition, and preventing investment in new
technologies or cheaper sources of power...

Government regulation is certainly part of the problem, but it is
typically keeping prices artificially low and discouraging system
upgrades. Look at how CA got into their power mess with their mock
"deregulation", where they deregulated the wholesale end but kept the
retail end regulated and capped and caused companies pull back rather
than absorb losses so the politicians could buy votes.

Not true here in Northeast. Environmental regulation and political hacks
keep costs high. For example, we have dumb laws requiring at least one
official policeman directing traffic at every single roadside job site.

blueman wrote:

"Pete C." writes:
blueman wrote:

writes:

On Jan 28, 10:15Â am, blueman wrote:
Jules writes:
... ours is on off-peak so gets 6c/kWh, but I can't remember the wattage
on the heater for ours either (and it normally runs for about an hour for
a full load)

It's amazing (to me) that our cost is almost 3x as high.
The total cost per kwh (including tax, generation, transmission, fees)
is a whopping 17.7 cent/kwh without any possibility of off-peak.

It's hard to believe that the "free market" price (in the absense of
governmental regulation) would be 3x as large particularly given that
electricity is:
 - An almost pure commodity (a volt is a volt is a volt)

But what it takes for fuel to generate the power is a huge factor and
varies widely. The areas with the lowest electric prices are
usually the ones driven off hydro-electric. Unfortunately, because of
geography, most areas of the country don't have that available.

And also factor in labor rates, materials costs, land costs, etc.
What it costs to build a sub-station or run a new transmission line
near Niagra falls is going to be a whole lot different than one in
northern NJ or San Francisco.

At best all of the above perhaps explains why we pay more for the
generation portion of our electric bill (12.4 cents/kWh). And I
understand why burning natural gas or oil which is supposedly the main
fuel here in New England would be more expensive than coal burnt in the
Midwest and certainly more expensive than Hydro from the Pacific
NorthWest.

However, why do we pay 7.6 cents/kWh for transmission which is *more*
than the total cost of 6-7 cents that other users here claim to be
paying.

You pay more for transmission in the frozen northeast because:

1. The heavily forested states cost a lot more for tree trimming around
the power lines.

I live in Eastern MA - I wouldn't say we were heavily
forrested. Certainly much less forrested than Pacific Northwest where
the total power cost is less than 7 cents/kwh.

I lived in CT for 34 years and still have property there. I spent plenty
of time with my chain saw cleaning up after the various ice storms over
the years. Eastern MA may be less forested than western MA, but it still
gets hit with ice storms which tear down lines.



2. The ice storms and falling branches cost a lot more to repair the
power lines.

Much less ice than Quebec where total cost is also less than our
transmission costs.

Their power is damned near free from Hydro-Quebec. Comparing rate
structures in other countries also tends to be deceptive based on how
things are structured.




If anything in a regulated monopoly, our transmission costs should be
*lower* than other parts of the country since our lines were built a
long time ago and presumably have already recouped their cost of
capital. Also, with low population growth here, there is not a
requirement for huge new investments in expansion.

Old decrepit infrastructure costs more to repair and maintain, as well
as costing more for rebuilding as sections become overloaded or
otherwise unserviceable. Look at how often you see crews replacing mile
of poles and stringing new lines, often with taller poles and double
circuits to meet the increasing power demands.

The big cost is in installing new high tension lines. Replacing the
occassional pole or transformer pales in comparison. And we are not
doing much of either around here. You rarely see power crews out doing
much of anything.

Building new transmission lines to support growing demand is a huge
expense as is building generation facilities to provide the power.

I think you may not be paying attention to what is being done, since in
my 34 years in CT I saw hundreds of miles of complete street level
distribution replacement. In most cases it was old single circuit runs
being replaced with new taller poles and double circuits in heavier
gauge was well, probably ~4X the previous capacity on the primaries. I
also so the secondaries replaced with much heavier ones and the
transformer count roughly doubled.



I think the real problem here is government regulation and corruption
which accomplishes the threefold evil of keeping prices artificially
high and discouraging competition, and preventing investment in new
technologies or cheaper sources of power...

Government regulation is certainly part of the problem, but it is
typically keeping prices artificially low and discouraging system
upgrades. Look at how CA got into their power mess with their mock
"deregulation", where they deregulated the wholesale end but kept the
retail end regulated and capped and caused companies pull back rather
than absorb losses so the politicians could buy votes.

Not true here in Northeast. Environmental regulation and political hacks
keep costs high. For example, we have dumb laws requiring at least one
official policeman directing traffic at every single roadside job site.

Yep, those are some of the biggest costs. Here in TX we have far less
headaches with that nonsense and better rates as a result. We also have
a hefty percentage of the countries wind generation here.

On Fri, 29 Jan 2010 16:37:15 -0500, blueman wrote:

writes:

On Jan 29, 2:19*am, blueman wrote:
"Existential Angst" writes:
"blueman" wrote in message
...
"Existential Angst" writes:
"Bob F" wrote in message
...
Existential Angst wrote:
"E Z Peaces" wrote in message
...
terry wrote:
Our very conventional old style tumble dryer timer runs for some 45
minutes per load.

With *heater cutting in and out (estimating it's on say 80%?).

Heater elements are either 3000 watts or maybe 4500, haven't had
this one apart yet, since I got it in exchange for a dozen beer!

Our domestic electricity costs a little over 10 cents per k.watt hr.

So one load of clothes 45/60 x 0.8 a cost of electricity = 0.75 x
0.8 x 0.12 = 7 cents per load.

Occasionally it is necessary to run a 'heavy' load, towels and
blankets etc. part of a second run.

So maybe that could be say 12 to 15 cents per load. In summer we
hang bedclothes and towels on outside lines.

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.

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.

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 *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 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...

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

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.

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.

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...

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


All very true but it has nothing to do with my question or my
understanding of the physics.

On Jan 29, 2:19*pm, "Existential Angst"
wrote:
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.


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





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 *same* direction, the currents should ADD,
right??

And they DO!!

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



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 ****tyTwo 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.

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 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

Strange that nobody considered the COST of the machine. Suppose you
buy a dryer for $500 which includes installation costs and it lasts 15
years. During its lifetime, it dried 2000 loads of clothes, making
the average cost of the load (based on the washer's initial cost)
500/2000 = 25 cents a load. You have to add the depreciated cost to
the energy consumed, unless you got a free dryer.

Pete C. wrote:
Tony wrote:
Pete C. wrote:
Jules wrote:
On Thu, 28 Jan 2010 18:04:18 -0500, Tony wrote:

E Z Peaces 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.
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.
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
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.
Your last paragraph doesn't hold true with my electric company. If I
had 2 200 amp meters, the second service would have a surcharge of
something like $10/month even if no power was used, or if a million
KWH's were used they would still charge that extra $10/month. So I paid
to have the current transformer installed and have no monthy fee except
for how many total KWH's I use. The one thing I dislike is the current
transformer divides the power by 20, so the "digital spinning wheel"
only runs at 1/20th the speed which is not easy to read to get that
"instant look" at how much power I'm using at the moment.

You didn't understand what I was saying in that last paragraph.

A 400A panelboard cost a *lot* more than two 200A panels, so even if you
have a single CT metered 400A service, in most cases it feeds two 200A
panels as this is less expensive.

Whether the two 200A panels are fed from one 400A service or separate
200A services is just a matter of whether you prefer two monthly service
charges, or one upfront CT meter installation charge.

OK, I get it now. I think I'm set for life with a 200 amp service/panel
in the garage.

On Jan 30, 12:06*pm, "Existential Angst"
wrote:
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.

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.




The Q was: *why can't you put a clamp-on around *both* wires at the same
time?

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.



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.

writes:
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.

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.

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.




The Q was: Â*why can't you put a clamp-on around *both* wires at the same
time?

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.

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.

Metspitzer writes:
On Fri, 29 Jan 2010 16:37:15 -0500, blueman wrote:

writes:

On Jan 29, 2:19Â*am, blueman wrote:
"Existential Angst" writes:
"blueman" wrote in message
...
"Existential Angst" writes:
"Bob F" wrote in message
...
Existential Angst wrote:
"E Z Peaces" wrote in message
...
terry wrote:
Our very conventional old style tumble dryer timer runs for some 45
minutes per load.

With Â*heater cutting in and out (estimating it's on say 80%?).

Heater elements are either 3000 watts or maybe 4500, haven't had
this one apart yet, since I got it in exchange for a dozen beer!

Our domestic electricity costs a little over 10 cents per k.watt hr.

So one load of clothes 45/60 x 0.8 a cost of electricity = 0.75 x
0.8 x 0.12 = 7 cents per load.

Occasionally it is necessary to run a 'heavy' load, towels and
blankets etc. part of a second run.

So maybe that could be say 12 to 15 cents per load. In summer we
hang bedclothes and towels on outside lines.

Weigh the load going in and coming out. Â*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. Â*After the load dried, I'd run it without heat again and check
the wattage again. Â*I'd take the average and multiply it by the time
a load ran. The exit air is warmer than the entrance air. Â*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. Â*I believe it measures
instantaneous wattage AND accumulates kWhrs.... Â*about $25 at Costco.
I have one, but haven't used it yet.

You have a 110 V dryer?

Heh.... Â*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.

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. Â*This is 220/240 *single
phase* -- true, you are using two legs of opposite phase, but the net
voltage is still single phase. Â*Iow, there is single phase and three phase,
but no two phase -- heh, funny how that works.

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...

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

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.

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.

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.

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...

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


All very true but it has nothing to do with my question or my
understanding of the physics.

Phisherman wrote:

Strange that nobody considered the COST of the machine. Suppose you
buy a dryer for $500 which includes installation costs and it lasts 15
years. During its lifetime, it dried 2000 loads of clothes, making
the average cost of the load (based on the washer's initial cost)
500/2000 = 25 cents a load. You have to add the depreciated cost to
the energy consumed, unless you got a free dryer.

I got a free dryer...

On Jan 30, 6:36*pm, blueman wrote:
writes:
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.

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.





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.

The Q was: *why can't you put a clamp-on around *both* wires at the same
time?

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.

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.- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

Yes, I agree 100%

"blueman" wrote in message
...
writes:
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.

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.

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.




The Q was: why can't you put a clamp-on around *both* wires at the same
time?

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.

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.

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.


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.

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

Pete C. wrote:
Phisherman wrote:
Strange that nobody considered the COST of the machine. Suppose you
buy a dryer for $500 which includes installation costs and it lasts 15
years. During its lifetime, it dried 2000 loads of clothes, making
the average cost of the load (based on the washer's initial cost)
500/2000 = 25 cents a load. You have to add the depreciated cost to
the energy consumed, unless you got a free dryer.

I got a free dryer...

I was gonna say that. A lot of people get a new matching set of washer
and dryer even though the dryer still works fine. There are always free
or cheap dryers available.

Tony wrote:

Pete C. wrote:
Phisherman wrote:
Strange that nobody considered the COST of the machine. Suppose you
buy a dryer for $500 which includes installation costs and it lasts 15
years. During its lifetime, it dried 2000 loads of clothes, making
the average cost of the load (based on the washer's initial cost)
500/2000 = 25 cents a load. You have to add the depreciated cost to
the energy consumed, unless you got a free dryer.

I got a free dryer...

I was gonna say that. A lot of people get a new matching set of washer
and dryer even though the dryer still works fine. There are always free
or cheap dryers available.

They are also fairly common I think as housewarming gifts from parents.

"Existential Angst" writes:
"blueman" wrote in message
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.

Yeah, but then you get a reading approx'ly double the true value!

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


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

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!!
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?

Ultimately it's best -- and ultimately easiest -- to clamp each wire
separately. You get more complete info this way.
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").
Clamp-ons are real cheap now, anyway -- used to be very expensive. $11 from
HF -- I must have 6 of'em!
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.

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.

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.

"blueman" wrote in message
...
"Existential Angst" writes:
"blueman" wrote in message
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.

Yeah, but then you get a reading approx'ly double the true value!

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


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

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


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!!
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?

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


Ultimately it's best -- and ultimately easiest -- to clamp each wire
separately. You get more complete info this way.
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").
Clamp-ons are real cheap now, anyway -- used to be very expensive. $11
from
HF -- I must have 6 of'em!
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.

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.

Actually, HF used to sell a cheap ($4.99) line splitter (#92072) that did
effectively the same thing - they seem to have discontinued it.

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

It also
has two loops - one for 1x and one for 10x so that you can get better
resolution of low currents.

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.

On Jan 31, 11:05*am, "Existential Angst"
wrote:
"blueman" wrote in message

...





writes:
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.

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.

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.

The Q was: why can't you put a clamp-on around *both* wires at the same
time?

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.

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.

Yeah, but then you get a reading approx'ly double the true value!


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?




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!

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.



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!!

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?


*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 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





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.

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

On Sun, 31 Jan 2010 11:23:35 -0500, Tony
wrote:

Pete C. wrote:
Phisherman wrote:
Strange that nobody considered the COST of the machine. Suppose you
buy a dryer for $500 which includes installation costs and it lasts 15
years. During its lifetime, it dried 2000 loads of clothes, making
the average cost of the load (based on the washer's initial cost)
500/2000 = 25 cents a load. You have to add the depreciated cost to
the energy consumed, unless you got a free dryer.

I got a free dryer...

I was gonna say that. A lot of people get a new matching set of washer
and dryer even though the dryer still works fine. There are always free
or cheap dryers available.


I don't understand the need to have matching washer and dryer. These
appliances are typically hidden behind doors, garage, or
utility/laundry room. A "matching set" does nothing else than
increase appliance dealer profit. I can't recall ever having a
"matching set," let alone the same brand.

Phisherman wrote:

On Sun, 31 Jan 2010 11:23:35 -0500, Tony
wrote:

Pete C. wrote:
Phisherman wrote:
Strange that nobody considered the COST of the machine. Suppose you
buy a dryer for $500 which includes installation costs and it lasts 15
years. During its lifetime, it dried 2000 loads of clothes, making
the average cost of the load (based on the washer's initial cost)
500/2000 = 25 cents a load. You have to add the depreciated cost to
the energy consumed, unless you got a free dryer.

I got a free dryer...

I was gonna say that. A lot of people get a new matching set of washer
and dryer even though the dryer still works fine. There are always free
or cheap dryers available.

I don't understand the need to have matching washer and dryer. These
appliances are typically hidden behind doors, garage, or
utility/laundry room. A "matching set" does nothing else than
increase appliance dealer profit. I can't recall ever having a
"matching set," let alone the same brand.

I have a very nice "stack" unit, so it inherently matches. I find the
stack design is much more space efficient than the separate units.

Phisherman wrote:
On Sun, 31 Jan 2010 11:23:35 -0500, Tony
wrote:

Pete C. wrote:
Phisherman wrote:
Strange that nobody considered the COST of the machine. Suppose you
buy a dryer for $500 which includes installation costs and it lasts 15
years. During its lifetime, it dried 2000 loads of clothes, making
the average cost of the load (based on the washer's initial cost)
500/2000 = 25 cents a load. You have to add the depreciated cost to
the energy consumed, unless you got a free dryer.
I got a free dryer...
I was gonna say that. A lot of people get a new matching set of washer
and dryer even though the dryer still works fine. There are always free
or cheap dryers available.


I don't understand the need to have matching washer and dryer. These
appliances are typically hidden behind doors, garage, or
utility/laundry room. A "matching set" does nothing else than
increase appliance dealer profit. I can't recall ever having a
"matching set," let alone the same brand.

It's a strange world out there!

On Feb 2, 11:40*am, Phisherman wrote:
On Sun, 31 Jan 2010 11:23:35 -0500, Tony
wrote:

Pete C. wrote:
Phisherman wrote:
Strange that nobody considered the COST of the machine. *Suppose you
buy a dryer for $500 which includes installation costs and it lasts 15
years. *During its lifetime, it dried 2000 loads of clothes, making
the average cost of the load (based on the washer's initial cost)
500/2000 = 25 cents a load. *You have to add the depreciated cost to
the energy consumed, unless you got a free dryer.

I got a free dryer...

I was gonna say that. *A lot of people get a new matching set of washer
and dryer even though the dryer still works fine. *There are always free
or cheap dryers available.

I don't understand the need to have matching washer and dryer. *These
appliances are typically hidden behind doors, garage, or
utility/laundry room. * A "matching set" does nothing else than
increase appliance dealer profit. * I can't recall ever having a
"matching set," *let alone the same brand.

Totally agree. We are are 'on' our second dryer (which 'cost' a dozen
beer) in 40+ years.
Our second, second hand washer (repaired with the drum/tub from
another some five/six years ago) must be 20+ years old now. The
cooking stove; can't remember but IIRC it's the third 'free' one we
have had.
The used dishwasher was a simple fix, one component! Hated to throw
out the old one, which still worked but this one does wash better and
IS quieter.
Neighbour is planning to change out her microwave; I'll be tempted,
although we have a couple of spares around now. Thre is a plenitude of
free and cheap appliances and other items! Sometimes needing the
occasional repair but most of them are pretty simple devices. And
stick to white.
PS. Haven't had to 'accept' yet any of those stoves with digital
display oven timers/thermostats .............. so keeping a couple of
the older style thermostats around just in case! haven't 'bought' an
appliance for the last 20 years or so.
One of thesed days probably be faced with multiple replacements!

terry wrote:

PS. Haven't had to 'accept' yet any of those stoves with digital
display oven timers/thermostats .......

OK, here's a good mystery. My stove/oven has a digital clock with a
digital timer. The timer justs beeps, it can't be set to turn off the
oven. Anyway here is the really weird part. The digital clock TICKS!
Sixty ticks/minute! One of these days I'm going to have to open it up
just to see it. It is LED display and I certainly don't think it has a
mechanical timer running a digital display, so I'm guessing they put a
tiny pizo or speaker hooked up to a "tick" circuit. Maybe some old
timers didn't like it if they couldn't hear it tick? I've shown it to a
few people and they started looking around for where I hid the wind up
ticking timer. :-) No. No wind up ticker, a digital ticker.

On Feb 3, 8:46*am, "Existential Angst"
wrote:

ConManTrader:

You are indeed an artful liar, and utterly disingenuous. *You quoted stuff
that proves yourself wrong, to make ME look wrong. *Very good!!!!!
You almost had ME thinking I was fulla****.


You obviously are. You made this totally bogus claim:

" 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 that is a lie for two reasons. First, his question clearly shows
the point of his idea was a way to measure power. Notice the last
sentence where he talks about calculating WATTS. Perhaps you are not
aware of it, but watts are units of power.

Here is his question again:

"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. "


And second, I'd like you to show us where he ever said that the
objective of his question and suggestion was not to measure power.
You can do that with a reference, not by hurling insults and changing
post topics to vulgarity.



You quoted (correctly) me saying this:

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.

And THEN proceeded to try to make me look like an idiot, when in fact you
*deliberately omitted* Blueman's immediate response (1/29), which was:

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



There you go again further proving yourself a liar, without even
realising it. Why would Blueman be talking about power factor if as
you claim, he were not interested in calculating power? Hmm?



Blueman:
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.

Talking about clever, how about YOUR clever idea of using a Kill-a-
Watt meter to measure power on a dryer? When that was quickly
exposed as idiotic, you then suggested using TWO of them, which is
probably even more stupid, because for starters, the dryer far exceeds
the current carrying rating of the Kill-a-Watt.

And why is it that you felt the need to hurl insults at 4 of us here
in the group? I don't think Ricodjour, Smitty, or SaltyDog were even
in this discussion. But I know they all know more about electricity
than you do.

BTW, I see you still haven't learned how to trim posts either.

On Feb 2, 3:27*pm, Tony wrote:
terry wrote:
PS. Haven't had to 'accept' yet any of those stoves with digital
display oven timers/thermostats .......

OK, here's a good mystery. *My stove/oven has a digital clock with a
digital timer. *The timer justs beeps, it can't be set to turn off the
oven. *Anyway here is the really weird part. *The digital clock TICKS!
Sixty ticks/minute! *One of these days I'm going to have to open it up
just to see it. *It is LED display and I certainly don't think it has a
mechanical timer running a digital display, so I'm guessing they put a
tiny pizo or speaker hooked up to a "tick" circuit. *Maybe some old
timers didn't like it if they couldn't hear it tick? *I've shown it to a
few people and they started looking around for where I hid the wind up
ticking timer. :-) *No. *No wind up ticker, a digital ticker.

"The digital clock TICKS!"

I often run the clock/scoreboard controller for the basketball games
at our schools.

It too has a LCD display yet you can hear it ticking inside the unit.

On top of that, when it drops below one minute and the 10ths of the
seconds show up, the ticking increasing 10 fold!

In article , wrote:

Careful Doug, you're about to be put on his enemies list.

I'm trembling.

Let's see,
it now includes SaltyDog, Smitty, RicodJour, the HVAC group, and me.
At least those are the ones we know about. Seems he can't get along
with anyone. You think this is how Nixon got started?

LOL