Electric Bill
bab...@my-dejanews.com
A couple of recent posts about electric bills brought something to my mind.
Anyways, I know homes are 220 service and the meter measures the load.
My thought was that say you have a 200 amp service. But you are unfortunate
enough to only load one leg say at 100 amps and 20 on the other. Now do you
pay for 60 amps of power you are actually using or 100 amps because the meter
only measures the higher loaded leg?
Just a thought...
-----------== Posted via Deja News, The Discussion Network ==----------
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BILL WATSON
>
> A couple of recent posts about electric bills brought something to my mind.
>
> Anyways, I know homes are 220 service and the meter measures the load.
>
> My thought was that say you have a 200 amp service. But you are unfortunate
> enough to only load one leg say at 100 amps and 20 on the other. Now do you
> pay for 60 amps of power you are actually using or 100 amps because the meter
> only measures the higher loaded leg?
>
> Just a thought...
>
>
Not sure where you get the 60 amp figure since you say you are using a
100 amp on one leg and 20 on the other that is a 120 amp load. Makes no
difference if all the load is on one leg you still pay the same as the
wattage is the same, amps x volts.
example: If you are running a 100 amp load on one leg at 120V the
wattage would be 100 x 120 = 12000 watts. If the load was balanced at 50
amps per leg it would be 50 x 120=6000 watts on each leg for a total of
12000 watts.
--
Bill Watson
Sam Goldwasser
on.
Figure it this way: The two wires run through the meter in such a way that
the current adds.
--- sam | Sci.Electronics.Repair FAQ Home Page: http://www.repairfaq.org/
Repair | Main Table of Contents: http://www.repairfaq.org/REPAIR/
+Lasers | Mirror Info: http://www.repairfaq.org/REPAIR/F_mirror.html
| Sam's Laser FAQ: http://www.misty.com/~don/lasersam.html
In article <75ltal$b4l$1...@nnrp1.dejanews.com> bab...@my-dejanews.com writes:
A couple of recent posts about electric bills brought something to my mind.
Anyways, I know homes are 220 service and the meter measures the load.
My thought was that say you have a 200 amp service. But you are unfortunate
enough to only load one leg say at 100 amps and 20 on the other. Now do you
pay for 60 amps of power you are actually using or 100 amps because the meter
only measures the higher loaded leg?
Just a thought...
-----------== Posted via Deja News, The Discussion Network ==----------
Doug
>A couple of recent posts about electric bills brought something to my mind.
>Anyways, I know homes are 220 service and the meter measures the load.
>My thought was that say you have a 200 amp service. But you are unfortunate
>enough to only load one leg say at 100 amps and 20 on the other. Now do you
>pay for 60 amps of power you are actually using or 100 amps because the meter
>only measures the higher loaded leg?
Because the coil inside the meter that makes the wheel go around is
220 volt powered.
It makes no differanc which leg has the higher amp draw because it's
all proportionate
bab...@my-dejanews.com
it can only measure only one side. I guessed this since I knew the two legs
can never touch. If I remeber what the backside of a meter looked like it
merely looked like current ran through it (2 in, 2 out) I figured it would
just measure the current in the wires inductivly and hence the meter would
spin. I thought it would be too complicated to measure both and divide by 2
multiplied by 240. So they would chose to measure just one leg.
It never occured to me that if you measure both and don't divide you get the
same result if you multiply by 120... And I thought I passed math.... Oh
well..... I guess I was making the problem bigger by thinking again......
Anyways, thanks.....
In article <SAM.98De...@colossus.stdavids.picker.com>,
s...@stdavids.picker.com (Sam Goldwasser) wrote:
> The meter mesaures the actual power you consume no matter which side it is
> on.
>
> Figure it this way: The two wires run through the meter in such a way that
> the current adds.
>
> --- sam | Sci.Electronics.Repair FAQ Home Page: http://www.repairfaq.org/
> Repair | Main Table of Contents: http://www.repairfaq.org/REPAIR/
> +Lasers | Mirror Info: http://www.repairfaq.org/REPAIR/F_mirror.html
> | Sam's Laser FAQ: http://www.misty.com/~don/lasersam.html
-----------== Posted via Deja News, The Discussion Network ==----------
Jean St-Pierre
you'll be billed for 120 amps! Power consumption adds up, it doesn't get
averaged.
bab...@my-dejanews.com wrote:<snip>
John Coggins
you can allegedly attach to your watt-hour meter and monitor it from your
computer.
http://www.measure.com/sensors/sensor-energy.html
bab...@my-dejanews.com wrote in message
<75ltal$b4l$1...@nnrp1.dejanews.com>...
>
>
>
>
>
>
>A couple of recent posts about electric bills brought something to my mind.
>
>Anyways, I know homes are 220 service and the meter measures the load.
>
>My thought was that say you have a 200 amp service. But you are
unfortunate
>enough to only load one leg say at 100 amps and 20 on the other. Now do
you
>pay for 60 amps of power you are actually using or 100 amps because the
meter
>only measures the higher loaded leg?
>
>Just a thought...
John Coggins
sites might help. Both are from government agencies so they're your tax
dollar at work (on a subject other than sexual promiscuity). The second one
is more scientific. Enjoy.
Watt-hour meter calibration standards
http://www.eeel.nist.gov/
Watt-hour meter testing and adjustments: braking, speed adjustments,
resistors and shunts, etc.
http://www.usbr.gov/power/data/fist/fist3~10/3~10_con.htm
danh...@infonet.isl.net
>A couple of recent posts about electric bills brought something to my mind.
>
>Anyways, I know homes are 220 service and the meter measures the load.
>
>My thought was that say you have a 200 amp service. But you are unfortunate
>enough to only load one leg say at 100 amps and 20 on the other. Now do you
>pay for 60 amps of power you are actually using or 100 amps because the meter
>only measures the higher loaded leg?
You pay for what you actually use.
Dan Hicks
Hey!! My advice is free -- take it for what it's worth!
http://www.millcomm.com/~danhicks
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Sam Goldwasser
figured out that they wouldn't get billed for any power if they just used
the phase that wasn't metered?!!!
The utility meters are actually quite sophisticated and by law must be
guaranteed accurate to within a small percentage regardless of the imbalance
in the phase load as well as for a wide range of power factors (for the
advanced course - motors and many other types of equipment result in the
voltage and current not being in phase which means that the real (useful)
power is less than what would be computed by multiplying V x A. However,
residential customers are only billed for the real power). It's all quite
clever actually.
--- sam | Sci.Electronics.Repair FAQ Home Page: http://www.repairfaq.org/
Repair | Main Table of Contents: http://www.repairfaq.org/REPAIR/
+Lasers | Mirror Info: http://www.repairfaq.org/REPAIR/F_mirror.html
| Sam's Laser FAQ: http://www.misty.com/~don/lasersam.html
Anyways, thanks.....
-----------== Posted via Deja News, The Discussion Network ==----------
Dan Lanciani
| In <75ltal$b4l$1...@nnrp1.dejanews.com>, bab...@my-dejanews.com writes:
| >A couple of recent posts about electric bills brought something to my mind.
| >
| >Anyways, I know homes are 220 service and the meter measures the load.
| >
| >My thought was that say you have a 200 amp service. But you are unfortunate
| >enough to only load one leg say at 100 amps and 20 on the other. Now do you
| >pay for 60 amps of power you are actually using or 100 amps because the meter
| >only measures the higher loaded leg?
|
| You pay for what you actually use.
At the risk of speaking without full knowledge of the operation of the meter's
clever magnetic circuit, I have to wonder if the original poster might have a
valid concern, though not for the reason he states (and not of the magnitude
he supposes).
Meters that I have seen have no neutral connection and thus no knowledge of
the actual voltages delivered on the L1 and L2 legs. Only the leg-to-leg
(i.e., nominal 240V) measurement is available to them. The meter is presumably
calibrated to accumulate the instantaneous product of the sum of L1 and L2
currents multiplied by one half of the leg-to-leg voltage.
Now, let's suppose that you have a fairly long run from your meter to the
transformer such that IR losses are non-trivial. Let's further suppose,
just to pick some numbers, that you want to draw a total of 100A at 120V,
that the nominal L1 & L2 voltages are 120V, and that the resistance of each
run of wire from meter to transformer is 0.01 ohm. For the sake of simplicity,
assume your load's current draw is largely independent of small voltage
changes. (You can run the numbers with a purely resistive load; the math
just looks more complicated.) Also, ignore parallel-to-neutral ground
paths.
First, consider the case where you divide your load evenly between L1 &
L2 such that 50A are drawn from each and there is no neutral current. There
will be a 0.5V drop in each of the L1 & L2 runs, so you will get 2*119.5V*50A=
11.95KW of real power delivered to your load. The meter will see a leg-to-leg
voltage of 239V and thus charge you at a rate of 119.5V*(50A+50A)=11.95KW as
you might expect.
Next, consider the case of drawing the whole 100A from L1. There will be
a 1V drop in each of the L1 & neutral runs, so you will get 118V*100A=11.8KW
of real power delivered to your load. The meter will again see a leg-to-leg
voltage of 239V and thus charge you at a rate of 119.5V*100A=11.95KW. Oops.
Where did the 150W go? Well, clearly the utility does not eat the 100W lost
in the neutral; you do. But you are also paying for 50W of the 100W lost
in L1 whereas in the example above the utility absorbed the cost of all power
lost in L1 & L2 (a total of 50W). In effect, you are penalized for not
minimizing the losses in the utility's lines, and thus it does pay to balance
your L1 and L2 loads.
Ok, the above discussion was theoretical. For all I know there is some
clever mechanism that compensates for this effect. If so, I'd love to
hear the details. Alternately, I may be making some obvious mistake...
Dan Lanciani
ddl@danlan.*com
Terry Kennedy
> One more site. Home automation geeks will love his one. They have a device
> you can allegedly attach to your watt-hour meter and monitor it from your
> computer.
> http://www.measure.com/sensors/sensor-energy.html
This probably won't help the original poster - it's a photocell that
keeps track of meter revolutions, and if the meter is reading incorrectly,
this will be off by the same amount, since it uses the meter.
I found a site that offers an outboard box that duplicates the meter's
function along with computing the bill amount and other stuff. The page
is at http://www.energymonitor.com. Has anyone used this unit? I'm think-
ing of getting one as I have been having an ongoing battle with PSE&G
regarding a fast meter here.
Terry Kennedy Operations Manager, Academic Computing
te...@spcvxa.spc.edu St. Peter's College, Jersey City, NJ USA
+1 201 915 9381 (voice) +1 201 435-3662 (FAX)
John Coggins
>Meters that I have seen have no neutral connection and thus no knowledge of
>the actual voltages delivered on the L1 and L2 legs. Only the leg-to-leg
>(i.e., nominal 240V) measurement is available to them. The meter is
presumably
>calibrated to accumulate the instantaneous product of the sum of L1 and L2
>currents multiplied by one half of the leg-to-leg voltage.
Think of these meters as small induction motors, the rate of spin being
proportion to RMS power (as opposed to instantaneous). In general, they
respond to in-phase components of voltage and current at the line frequency.
Guess that's why they're watt-meters.
>Now, let's suppose that you have a fairly long run from your meter to the
>transformer ....that you want to draw a total of 100A at 120V, that the
nominal L1 & L2 >voltages are 120V, and that the resistance of each run of
wire from meter to >transformer is 0.01 ohm.
>
Cable length isn't a significant factor if the wire is correctly sized. If
you draw 100 amps across 0.01 ohms you get a 1 volt drop -- one way. But
there's another equal voltage drop on the return wire for a total of 2
volts. This makes sense: total resistance is 0.01 + 0.01. Power loss is
(100^2)*0.02 = 200 watts.
>First, consider the case where you divide your load evenly between L1 &
>L2 such that 50A are drawn from each and there is no neutral current.
Energy is conserved. If you drew 100A @ 240V (24 KVA coming in), your
balanced load is 12 KVA on each bus. At 120V, this is 100 amps per leg --
not 50. Since impedances never exactly balance, there is also some neutral
current.
>The meter will see a leg-to-leg voltage of 239V and thus charge you at a
rate of 119.5V*(50A+50A)=11.95KW as you might expect.
Depending on how it's geared, the meter will turn at a rate proportional to
your use of ~24 KVA (minus wire losses of 0.2 KVA and some negligible power
used to turn the meter itself).
>Next, consider the case of drawing the whole 100A from L1.
Dan, this is not really a practical case. The largest loads in your home are
220V (water heater, range, dryer). These being line-to-line, example #1
dominates. You can unbalance 120V loads but they're such a small percentage
of your total energy usage the losses would be insignificant to you. Large
collective imbalances would have more meaning to the utility since they tend
to overload one transformer winding.
Everything you wanted to know about watt-hour meters:
http://www.usbr.gov/power/data/fist/fist3~10/3~10_con.htm
>
> Dan Lanciani
> ddl@danlan.*com
Dan Lanciani
| Dan Lanciani wrote in message <343...@news.IPSWITCH.COMM>...
| >Meters that I have seen have no neutral connection and thus no knowledge of
| >the actual voltages delivered on the L1 and L2 legs. Only the leg-to-leg
| >(i.e., nominal 240V) measurement is available to them. The meter is
| presumably
| >calibrated to accumulate the instantaneous product of the sum of L1 and L2
| >currents multiplied by one half of the leg-to-leg voltage.
|
|
| Think of these meters as small induction motors, the rate of spin being
| proportion to RMS power (as opposed to instantaneous). In general, they
| respond to in-phase components of voltage and current at the line frequency.
| Guess that's why they're watt-meters.
The problem with that is that it is very painful to define RMS power when
the waveforms aren't nice sinusoids. However, the net result is the same.
I was just proposing an easy way to look at the operation. For this analysis
you can assume a perfect power factor; it has no impact on the problem.
| >Now, let's suppose that you have a fairly long run from your meter to the
| >transformer ....that you want to draw a total of 100A at 120V, that the
| nominal L1 & L2 >voltages are 120V, and that the resistance of each run of
| wire from meter to >transformer is 0.01 ohm.
| >
|
| Cable length isn't a significant factor if the wire is correctly sized.
However, the utilities are well known for using drop wire that is far
smaller than what we are required to use for service entrance cable
by the NEC. If the utility is paying for all of the losses caused by
the smaller wire it might be better, but I believe I have shown that
they aren't in all cases.
| If
| you draw 100 amps across 0.01 ohms you get a 1 volt drop -- one way. But
| there's another equal voltage drop on the return wire for a total of 2
| volts. This makes sense: total resistance is 0.01 + 0.01. Power loss is
| (100^2)*0.02 = 200 watts.
This statement is true and reflects the second case with neutral as one
of the runs. The problem is that you get to pay for some of that power
if you unbalance the loads.
| >First, consider the case where you divide your load evenly between L1 &
| >L2 such that 50A are drawn from each and there is no neutral current.
|
| Energy is conserved. If you drew 100A @ 240V (24 KVA coming in), your
| balanced load is 12 KVA on each bus. At 120V, this is 100 amps per leg --
| not 50. Since impedances never exactly balance, there is also some neutral
| current.
This is also true, but the example I gave involves only about half this power,
so I'm not sure how these numbers are relevant.
| >The meter will see a leg-to-leg voltage of 239V and thus charge you at a
| rate of 119.5V*(50A+50A)=11.95KW as you might expect.
|
|
| Depending on how it's geared, the meter will turn at a rate proportional to
| your use of ~24 KVA (minus wire losses of 0.2 KVA and some negligible power
| used to turn the meter itself).
I'm not sure if you are now talking about your example or mine. Certainly
my example did not reflect a usage of 24KW.
| >Next, consider the case of drawing the whole 100A from L1.
|
| Dan, this is not really a practical case.
But that was the whole point of the original question. The poster thought
something bad would happen wrt billing if the loads were severely unbalanced.
His reasons were wrong and there is no such large-scale problem. However,
I claim that there is indeed a difference and you do pay more in the unbalanced
case. The difference may be small, but I have yet to see anything to convince
me that it doesn't exist.
|The largest loads in your home are
| 220V (water heater, range, dryer). These being line-to-line, example #1
| dominates. You can unbalance 120V loads but they're such a small percentage
| of your total energy usage the losses would be insignificant to you.
That may well be true for any actual setup, but I believe the question
was hypothetical.
Dan
John Coggins
KVA. However, you probably intended this as another hypothetical, since
power is usually delivered and metered at a nominal of 240V.
Otherwise, since the issue relates to losses, it would be appropriate to
consider the utility's loss when a user's power factor (and/or sinsusoidal
crest factor) exceeds the limits for which the meter is calibrated. This
loss exceeds any customer loss due to small unbalanced 120V loads.
Dan Lanciani wrote in message <344...@news.IPSWITCH.COMM>...
>In article <75tts8$51l$1...@ash.prod.itd.earthlink.net>, cog...@earthlink.net
(John Coggins) writes:
>| Dan Lanciani wrote in message <343...@news.IPSWITCH.COMM>...
>I'm not sure if you are now talking about your example or mine. Certainly
>my example did not reflect a usage of 24KW.
>
> Dan
Dan Lanciani
| Dan, you are correct. You specifically said 100 amps at 120V which is 12
| KVA. However, you probably intended this as another hypothetical, since
| power is usually delivered and metered at a nominal of 240V.
Well, the whole thing was hypothetical... But I had to specify 120V
loads or else there would be no choice involved in how to allocate
them and thus the entire issue would be moot. :) On the other hand,
don't imagine that such loading is *that* unusual. Looking at the
200A sub-panel in my workshop you will see only a single 240V breaker
(25A for the heat pump). The rest are 120V: 30A circuits for the
bigger computer racks, 20A for the general-purpose outlets, and a few
15A lighting circuits. It would be pretty easy to create a severely
unbalanced load depending on what I turned on. Fortunately, my service
is underground and thus my responsibility, so the wire is not undersized...
| Otherwise, since the issue relates to losses, it would be appropriate to
| consider the utility's loss when a user's power factor (and/or sinsusoidal
| crest factor) exceeds the limits for which the meter is calibrated.
Note, though, that the losses we are discussing have nothing to do with
power factor or unfortunate current waveforms. The analysis remains unchanged
with perfectly in-phase sinusoidal voltage and current waveforms.
| This
| loss exceeds any customer loss due to small unbalanced 120V loads.
But it isn't clear that this difference in scale entitles the utility
to turn the customer's loss into its gain. :)
Again keeping in mind that this is just an exercise and that the actual
percentage losses are relatively small, there are still at least two points
of interest (IMHO):
1. Most residential electric service is tariffed to be charged based on
real power delivered to the user's loads. It is true that the utility
can charge you extra for such things as a bad power factor or peak demand
(though these elements are usually confined to commercial service), but
extra charges generally have to be explicit. I'll bet that most tariffs
do not allow the utility to charge you for line losses on their side of
the meter, and that is effectively what happens when the loads are not
balanced. If nothing else, I could imagine using this issue to push
for a bigger drop cable, something a number of people would like.
2. Regardless of the tariff, you do pay more real dollars for your
real power if you don't balance the loads. It may not be much, but
it could add up. Of course, it's always been good practice to balance
the loads--here is a financial reason if you need one...
Dan Lanciani
ddl@danlan.*com
John Coggins
>there are still at least two points of interest (IMHO):
>
>1. Most residential electric service is tariffed to be charged based on
>real power delivered to the user's loads. It is true that the utility
>can charge you extra for such things as a bad power factor or peak demand
>(though these elements are usually confined to commercial service), but
>extra charges generally have to be explicit. I'll bet that most tariffs
>do not allow the utility to charge you for line losses on their side of
>the meter, and that is effectively what happens when the loads are not
>balanced.
I'll bet the utility may legally charge you for delivered power, regardless
of how loads are balanced on your side of the meter. Moreover, delivered
power measured at the meter does not include line losses in the service
cable: it is a function of the current and the [remaining] line-to-line
voltage at the meter, not back at the transformer terminals.
I just don't see where a utility has anything to gain by undersizing service
cable.
>2. Regardless of the tariff, you do pay more real dollars for your
>real power if you don't balance the loads. It may not be much, but
>it could add up. Of course, it's always been good practice to balance
>the loads--here is a financial reason if you need one...
This is perhaps the most hypothetical aspect since it's not possible to
perfectly balance 220V loads, much less intermittent dynamic ones at 120V.
Maybe someone could optimize the imbalance by installing their own meters
across each service leg and then running around adjusting loads, but cost
would far exceed value.
If anyone is really concerned about this <g>, they should install a
relay/transfer switch at each 120V load and wire up both hots everywhere.
Have a computer monitor each bus in your panel and, whenever a device is
switched on, select the source that gives the best balance.
John Coggins
well below accuracy of 0.5% to 0.25% that should be inherent in the utility's
meter.
Terry Kennedy
> FYI, they advertise "KWh Accuracy: +/- 2%, based on unity power factor." This is
> well below accuracy of 0.5% to 0.25% that should be inherent in the utility's
> meter.
I'd be happy if I had a utility meter with a 2% error rate! I own a 2-family
house (I'm using both halves) and one bill is ~$200/mo and the other is ~$50/
mo. Before they changed the meter on the $50/mo floor 5 years ago, it was up
to about $800/mo. The $200/mo floor was up to $1400/mo before they changed the
meter a year ago. Unfortunately they forgot to write down the initial reading
on that meter when they changed it, so they changed it again 2 months ago.
It's to the point where I simply don't trust the utility meters and I want
some sort of independent confirmation of the meter readings. Stick-on sensors
that watch the meter disc won't do that, as it's just reading the same wrong
reading as the meter.
I probably wouldn't protest to the utility unless the difference between
my readings and their readings were on the order of 5-10%.
If you have any suggestions for more accurate devices with the features of
this gadget in the same price range (< $200), I'm all ears.
Terry Kennedy
> I just don't see where a utility has anything to gain by undersizing service
> cable.
I guess in your area they're not just plain lazy like they are up here in
NY/NJ.
In Queens, NY (Con Ed), a friend upgraded from 60A service (nominally 120/
240, but actually 2 legs of a 3-phase feeder, so 120/208) to 200A service.
Con Ed declined to replace the feeders or meter at the same time, saying that
they would replace them if demand increased. What do you think would happen
if this house starts pulling 200A through a 60A meter base and a set of 60A
aluminum feeders? BTW, the upgrade was to install a computer room, year-round
air conditioning for the computer room, and NYNEX fiber optic equipment, so
the load is definitely there.
In Jersey City, NJ (PSE&G), a house is being upgraded from 2 100A services
(inddor meters, Federal Pacific panels) to 2 200A services (outdoor meters,
since PSE&G is so primitive they don't have meter remotes, though just about
every other utility does, and Square D panels). In order to get PSE&G to re-
place the existing drop cable (rated 60A total, currently feeding the 2 100A
services), the owner had to tell PSE&G that they were converting to all-
electric (2 stoves, 2 ovens, hot water heater, and electric heat).
I expect that in both these cases, the utility would wait until the drop
or meter base failed/burned before they'd replace them. I visited an apart-
ment house with 80-odd 30A meters (utility meters, not submeters) in the
basement and a 125A main feed. The conduit for the main feeders was hot to
the touch (not hot enough to raise blisters, but too hot to comfortably
hold). There was also a smell of overheated insulation. I advised the super
to call PSE&G. When they came out, they said there was no problem and if
it failed, to call them back.
Dan Lanciani
| Dan Lanciani wrote in message <345...@news.IPSWITCH.COMM>...
| >there are still at least two points of interest (IMHO):
| >
| >1. Most residential electric service is tariffed to be charged based on
| >real power delivered to the user's loads. It is true that the utility
| >can charge you extra for such things as a bad power factor or peak demand
| >(though these elements are usually confined to commercial service), but
| >extra charges generally have to be explicit. I'll bet that most tariffs
| >do not allow the utility to charge you for line losses on their side of
| >the meter, and that is effectively what happens when the loads are not
| >balanced.
|
| I'll bet the utility may legally charge you for delivered power, regardless
| of how loads are balanced on your side of the meter.
I think we agree on this.
| Moreover, delivered
| power measured at the meter does not include line losses in the service
| cable: it is a function of the current and the [remaining] line-to-line
| voltage at the meter, not back at the transformer terminals.
I'm not going to beat this to death, so this will be my last posting on
the subject. If you look again at my original posting, you will see
that the power being measured at the meter *does* include line losses
in the neutral of the service cable. Moreover, when the loads are
unbalanced (which is the same as saying that there is neutral current)
the power being measured at the meter also includes a fraction of the
line losses in one of the hot conductors of the service cable. That was
the entire point of the calculation. It is an unavoidable consequence of
the fact that the meter has no way to measure the "remaining" line-to-neutral
voltage.
| I just don't see where a utility has anything to gain by undersizing service
| cable.
They save money on the cable, and they save money on the labor of (not)
replacing the cable when you ask them for a bigger service. I assume they
think that the occasional fire loss does not swamp these savings. :(
Dan Lanciani
ddl@danlan.*com
John Coggins
bills of $800 and $1400 per month. And, yes, the stick-on optical sensor is
a toy. But I'm not so sure of the other one either with the disclaimer in
there about power factor.
Terry Kennedy wrote in message ...
John Coggins
a meter. Unless you think it just runs fast some times and not others, this
may save you some money:
3.4. FRICTION. To compensate for friction, additional torque must be
introduced. This usually is accomplished by placing a movable
short-circuited turn of large cross section in part of the field of the
voltage (potential) coil. This also serves as a "light-load" adjustment.
3.12. BASE-LOAD SPEED. Another useful characteristic of the meter, if known,
is the base-load speed. Base-load speed is rpm at 115 volts and 5 (or 2-1/2)
amperes, unity power factor. On a steady load, by timing the meter speed
with a stopwatch, a very accurate value of kilowatt load may be obtained in
the absence of a wattmeter; or the same method may be used to check the
accuracy of an indicating wattmeter. (See Paragraph 4.1.4.) Base-load speeds
commonly used by various manufacturers follow, but should not be construed
as applying in all cases. For example, Westinghouse also uses a speed of
33-1/3 rpm in their Type CA-8 meter.
Base-load speeds - 500 (or 250) watts/element
General Electric ......................16-2/3 rpm
Westinghouse ............................... 25 rpm
Sangamo .................................16-2/3 rpm
Duncan ......................................... 25 rpm
As noted previously, see this site:
http://www.usbr.gov/power/data/fist/fist3~10/3~10_con.htm
John Coggins
Dan Lanciani wrote in message <348...@news.IPSWITCH.COMM>...
>In article <760hr0$nr4$1...@holly.prod.itd.earthlink.net>,
cog...@earthlink.net (John Coggins) writes:
>| Moreover, delivered
>| power measured at the meter does not include line losses in the service
>| cable: it is a function of the current and the [remaining] line-to-line
>| voltage at the meter, not back at the transformer terminals.
>I'm not going to beat this to death, so this will be my last posting on
>the subject. If you look again at my original posting, you will see
>that the power being measured at the meter *does* include line losses
>in the neutral of the service cable.
On voltage: V2, the voltage at your meter (on which your bill is based), is
less than V1, the voltage at the utility transformer because a portion of
the voltage IR was dropped across the supply lines. Therefore, the loss in
the utility service ([V1-V2]^2/R) is not power that you used or that you
will be directly billed for. It was wasted by your utility, which means you
will pay for it -- indirectly -- in higher rates.
On neutral currents. Even if your loads were in perfect balance there would
be some neutral current because the impedances in the utility's transformer
windings do not exactly balance. It's an imperfect world. However, ignoring
these neutral currents, any others are returns from your loads, which means
you used that power. Pay for it.
Terry Kennedy
> In your case, I can see the point: 2% sounds like a major improvement with
> bills of $800 and $1400 per month. And, yes, the stick-on optical sensor is
> a toy. But I'm not so sure of the other one either with the disclaimer in
> there about power factor.
Well, after the holidays I'll probably call and order a bunch of 'em. The
only other alternative I've come up with is to install additional meter sock-
ets on the interior and do my own sub-metering, which is going to cost me a
good deal more than these gadgets.
I assume that they use some sort of Amprobe-style clip-on sensors, which
is probably where they lose their accuracy. I'll ask them if they also offer
complete toroids to be installed when new wiring is installed.
Heron's Nest
several 100 watt lamps, through one meter for a measured amount of time (perhaps
an hour) and read the energy consumption. Then repeat the experiment through the
other meter. If the results are the same the odds are that the meters are ok.
If not have your power company correct the problem.
Phil