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

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

Reply to
Jules
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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.

Reply to
trader4

snipped-for-privacy@optonline.net wrote: ...

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]
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has links to actual filings/numbers; I haven't looked at precise numbers for a while

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Reply to
dpb

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.

Reply to
Tony

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.

Reply to
Pete C.

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...

Reply to
blueman

But clamping measures current, not voltage. So clamping both wires (with the direction reversed in one wire so that the opposite phases cancel as I mentioned before) should give you a measurement of the total current. Then you can multiply by the voltage (or maybe rms voltage) to get the total watts consumed.

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

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

Reply to
blueman

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 =3D 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 =3D 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 =3D 240V X amps in lesser of the 2 hots X power factor + 120V X amps in neutral X power factor

Reply to
trader4

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

Reply to
Jules

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.

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.

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.

Reply to
Pete C.

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.

Reply to
Pete C.

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.

Reply to
Tony

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.
Reply to
Pete C.

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

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

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

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

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

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

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

Blueman asked a very good Q: WHY can't you measure both together? And, afaict, it wasn't the notion of voltage that he was missing, as you stated. And since you just repeated my answer, his Q wasn't really answered.

But he also mis-stated his own premise, in a couple of ways.

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

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

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

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

NOW, here's the REAL inneresting part:

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

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

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

Now, Ricodjour, SaltyAss, and ShittyTwo are proly getting blisters from their bunched-up panties, screaming, Liar, Where's the Citation????? Proly Trader would too, if he didn't already agree -- altho he don't know the physics, apparently.

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

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

Reply to
Existential Angst

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

Enron. Except I don't think the retail end WAS regulated -- effing consumers were clobbered, iiuc, and NOW they are STILL paying 30c+ , in a tier system that very quickly reaches that 30c rate, or more.

Heh, and you heard that the supremes struck down limitations on corporate campaign contributions? You think it was bad then, or now.... you ain't seen nuthin yet. Orwell Lives.

Reply to
Existential Angst

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

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

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

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

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

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

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

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

Reply to
blueman

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

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one can assume that the inductive load is only about 6.4% of the power consumption. So, assuming the motor has a power factor of 0.8, the weighted average power factor would be: 0.99 which is pretty close to 1 in my book...

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

Reply to
blueman

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.

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

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.

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.

Reply to
blueman

No, the CA scam was indeed to deregulate the wholesale end while still keeping the retail end regulated and rate capped. When the wholesale power market prices exceeded the capped retail price, CAs mock "deregulation" scam unraveled since the utilities were not about to sell the power at retail for less than they had to pay for it wholesale. The exact reasons for the wholesale price increase aren't really relevant since it was only a matter of time until CAs house of cards collapsed.

CA is in a similar situation with their welfare state. The high CA taxes, combined with overall economy issues has hastened the exodus of productive working taxpayers from CA, leaving fewer and fewer productive workers to pay the high taxes to support the welfare state for all the non productive workers. The same economy issues have also led to the loss of a lot of low end jobs, meaning more on CAs welfare rolls. Some of those low end workers are leaving CA, but most don't have that mobility, so they remain a drain on CAs remaining taxpayers.

Reply to
Pete C.

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.

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.

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.

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.

Reply to
Pete C.

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