Cheated by the Electric Company?

tootie; don't you think that after over one hundred year, they would know how to build a motor with dials? if your driving a load with a electric motor and the voltage drops to 170 the motor would slow down, now to keep the customer from cheating the power company their going to all electronic meters that will record over 63 different items, as time time of day useage,power factor, lo/hi voltage, lo/hi current & when! perry

The meters that you describe (with the spinning disc) work correctly across a variety of voltages. The speed of the disc is proportional to the product of the actual voltage and actual curent going through the meter.

Thus if you have a lower voltage, the meter will still read correctly.

Those meters however in some special instances actually cheat the electricity supply companies and that is one reason that Australia has moved away from them.

but what about reactive power? all those pcs on standby, I measured one; it only consumes 5 W but its apparent power is 33.6 VA. That current if charged for by what is supplied over what is actually used would be quite expensive.

and plug-top power supplies with outrageous power factors, each producing a dc voltage, why not replace them with dc-dc supplied by 12V battery charged by a solar panel? or at least persuade manufacturers to produce switched-mode versions!

The speed of the rotation of the meter disk depends on voltage, so no cheating.

See the immediate previous posting (http://cr4.globalspec.com/thread/43599) and there is the steps of a easy way of the meter testing procedure.

- MS

well, I see all comments are quite similar and that's as what I thought before...

that's true the meter works correctly across a variety of voltages if that run in a capacitive load and even slowing down by E2/R ,but how if the load that require certain amount of current to operate, such as a welding machine or particular inductive loads ??...

let me put it this way,

with pressure of 220 ,water flow in amount of 5 through 1" of pipe we get 1100 drums of water in an hour.

if then the pressure drop to 170 ,by same dimension & amount of flows the result will be 850 drums of water in an hour...until here that's correct what you consume is what you pay.

but on the contrary if when you need 1100 drums in an hour, but the pressure stay put at 170.....wouldn't it be made the water sucked out to fulfill that?

because what happen is, the more current flows due to lower voltage through series of inductive coils in the meter increase the magnetic field and speed up the disc.

thank you.

dear,

infact, it is just the reverse of what you have felt to assume!

as you said, by lowering the voltage, power demand is lowered by the supply co. true,but the conclusion is electricity co. gets less billing ! less revenue, then if they would have provided the correct higher voltage.

In India, one of the private electricity supplier has played a simple trick - they just increaed the voltage by 10 volts, to 230 volts exact. Result, consumer's bills shot upiu by 30 - 40 % ! Now, can you call it as a cheating ? or a smart business strategy ?

hi harry potter,

what I meant by power demand is at the customer's side, because of that condition certainly customer can't use more loads due to amperage limitation.

in my opinion, the electricity company can play smart by simply just lowering down the voltage at the customer's grid but overall total power capacity they can sell stay the same as long as less power loss in the distribution lines.

refer to the simple trick of private electricity supplier in your country, I think it would only impact to customers who use much capacitive loads.

Hi msamad, I think everybody here knows the formula you mentioned and most of them got to be agree with that ,however I don't think that kWh (actually kVAh) meter that we are talking about has capable of either measuring the rms value and reactive power…on top of that how do you explain P = I x E ? Power is the rate of doing work or the rate of expending energy, doesn't it mean for let's say 1100W/220V/50Hz/1 ph load has capable of expending energy of 1100W/hour? Because of its ability to expend that amount of energy for doing work per hour then you might not say when the pressure (voltage) drops, simply lowering its capability especially to that the inductive loads such as transformer in electronics, ballasts or current transformer used by welding machine.….I agree to what Perry said back there that if your driving a load with a electric motor and the voltage drops to 170 the motor would slow down ..It's true because electric motor would only rotate based on the rotating field that created by electrons flow in the stator and rotor windings while its quantity depends on how much pressure (V) you can give. So what I'm trying to say in that case is when something down another must up …that is current (I) or in other words what you cannot push you absorb.

And don't you forget, since the load current is through coils in series inside the kWh meter so the magnetic field produced by these coils is proportional to the load current, while on the other hand winding coil to the armature which is connected directly across the line produces smaller magnetic field due to smaller voltage resulting unbalancing between both stator & rotor. That's why the greater the current in the stator coils the bigger the magnetic flux in it, and the greater the magnetic flux in the stator than rotor's then it potentially slippery jump and faster the rotation will be.

Refer to your answer to Harry's, I agree that power consumed by resistor is dramatically affected by the voltage applied. As the formula implied, the power is affected by the square of the voltage & directly affected by the resistance….so if the voltage is doubled, the power will increase four times and if the voltage is decreased by 50%, the power consumed will decrease by 75%. In his case voltage is increased by 10Volt equal to 4.55% so increase of the bill with assumption of full resistive loads is only 9.3% the most

hi Mitsurati,

It's not true that I'm confusing power with current, so please refer to my argument on the previous post.

and talking about power company, I'm fully agree that they would keep good voltage to avoid power loss a long the transmission line....the higher the voltage the lower the current the smaller the resistance the lesser the power loss and the cheaper the transmission cost will be. but it can make the huge different if they drop it at the distribution line.

cheers,

Hello again tootje.

The spinning disk in the meter is part of a motor. One coil of that motor has your load current passing through it and the other coil has the supplied voltage across it. The interaction of the two magnetic fields (effectively the product of current and voltage) determines the speed that the disc spins.

Lower supply voltage means SLOWER spinning for the same current.

The meters with spinning discs (generally) only have difficulty when the load is not at proper power factor and in that instance it's the power company that is cheated. That's one reason why they place significant penalties on bad power factor.

The meter spins based on the ACTUAL voltage and ACTUAL current passing through it instantaneously and the number of revolutions turns the dials on the meter effectively integrating the total power used.

I've seen 240 volt meters used in a 120 volt system with no deterioration in accuracy. (They did have trouble when we tried them at 32 volt, and I suspect they would have issues beyond 300V in high current situation as the armature of the meter would reach magnetic saturation.)

If you are truly only getting 170V, I'd be focussed on other practical aspects of the business, like stall protection on motors to cut power before they burn out, protection of refrigeration compressors to make sure they don't try to start on such undervoltage conditions and so on.

Hello again to you too Just an engineer,

You are absolutely right with your explanation but…

There are 2 points I should pin from it,

• Lower supply voltage means SLOWER spinning for the same current
• The meter spins based on the ACTUAL voltage and ACTUAL current through it instantaneously

So my question is, do you agree when the Power stay, voltage drop and the result current will increase? If not, why & how? We are not talking about the resistive load here remember? If you agree, then in the case of lower supply power but current is higher & the actual voltage is low but the actual current through it increases, what is your opinion?

Thank you.

Tooje:

Hundreds of thousands of physicists and engineers have understood the relationship for centuries; are you trying to say that they were all wrong and that when voltage goes down power goes up? The equation is simple, P=VI. Making V smaller doesn't make P larger. Basic arithmetic.

The reactance of capacitive loads depends on frequency, not voltage. If the frequency is kept constant while voltage drops, the current will also drop.

Do you think it's the hundreds of thousands of physicists and engineers (and for that matter young children learning arithmetic) that have it wrong, or is it perhaps you?

Please read your own signature line.

mitsurati,

I don't think you really get the point here....please read me carefully.

which of my arguments said voltage goes down power goes up ?? check my posts body before you judge me wrong and embarrass yourself..........

thank you

"which of my arguments said voltage goes down power goes up ?? check my posts body before you judge me wrong and embarrass yourself.........."

This argument: "dropping down the voltage at customer's side from standard 220 V to let's say 170 V as it happened most of the time here in my country,wouldn't it give result to the faster spinning of the meter"

You are wrong. Everyone on this forum says you are wrong. Every competent electrical engineer would say you are wrong. Every physicist would say you are wrong. Are you just waiting for someone to tell you you are right?

Fine. Reducing voltage increases power consumption. James Joule had it all wrong, and you cleverly figured it out. Happy? Now stop belaboring your idiotic point and go finish designing your perpetual motion machine. Here's a hint: try applying zero volts to a motor to make it spin at infinite speed.

hahahahahaha...I've told you got me wrong Mit,...and c'mon man, they're 2 question marks right??

Well I think you've misunderstood bro....here is my very basic point as the matter of fact, let's start it over Ok:

"dropping down the voltage at customer's side from standard 220 V to let's say 170 V as it happened most of the time here in my country, wouldn't it give result to the faster spinning of the meter & so that customers would pay more than as it supposed to be ??....."

and my argument was not that by speeding up of the kWh meter the power goes up bro but instead the current increases.., that's the point !!! Isn't this right? P=V x I, I= P/V..

P (Power) is the rate of doing work or the rate of expending energy

V (Voltage) is Electron pressure

I (Current) is Volume of electrons

thus if V has not enough pressure, some loads (non resistive) would absorb certain amount of electrons to fulfill its rate of expending energy, so V drops x Current ups (=) equal Power rates of doing work.

As the way of the induction kWh meter works has explained clearly by deept ^{http://cr4.globalspec.com/comment/454001/Re-Cheated-by-the-Electric-Company} and added by me ^{http://cr4.globalspec.com/comment/454087/Re-Cheated-by-the-Electric-Company} confirmed that the induction kWh meter which consists of a potential coil & the two current coils connected in series with the load take responsible to the rotation of the motor consisting of an aluminum disk of this energy meter.

So now let's take a closer look at those coils that acting as the moving magnetic field behavior.

• Potential coil connected across the load is composed of many turns of relatively small wire; it is wound on one leg of the laminated magnetic circuit. Because it's many turns, this coil has high impedance and high inductance and therefore the current through it lags the applied voltage by nearly 90º.
• The two Current coils connected in series with the load are composed of few turns of heavy wire and they wound on two legs of laminated magnetic circuit...few turns mean low inductance & impedance

Since rotation of the aluminum disk is accomplished by eddy current that are established by the current coils & the potential coil thus what happen if the applied voltage to the potential coil is under its rating? would it results to drop of its magnetic field, inductance & impedance (lower inductance, lighten its inertia) plus isn't that more current flows through the current coils resulting to strengthen or give greater intensity of its magnetic fields may probably at the end give impact to its spinning speed ?

That was actually my point mit, & frankly speaking before starting this thread I've done some experiments with my own correctly registered 230 V, 5(20)A, 50 Hz,900rev/kWh meter that similar with electricity Co's, by using a single same load 4x36W fluorescent lamps and use a small good gen-set 220V and with the Electricity co as another power source which voltage wasn't steady around 175 ~ 190V, I found that the spinning of the meter was really a bit different between both sources.....the result was 213 rev by electricity co and 196 by gen-set so 17 revolutions leads the good voltage in an hour.....

so does anyone can tell of what phenomenon is this ????!!

I hope by this explanation will clarify the 'untold story'...tq

Hi Joe,

As already described by Just an Engineeer . Operating mechanism of ElectroMechanical type energy meter available on Wikipedia are as under.

The electromechanical induction meter operates by counting the revolutions of an aluminium disc which is made to rotate at a speed proportional to the power. The number of revolutions is thus proportional to the energy usage. It consumes a small amount of power, typically around 2 watts.The metallic disc is acted upon by two coils. One coil is connected in such a way that it produces a magnetic flux in proportion to the voltage and the other produces a magnetic flux in proportion to the current. The field of the voltage coil is delayed by 90 degrees using a lag coil. This produces eddy currents in the disc and the effect is such that a force is exerted on the disc in proportion to the product of the instantaneous current and voltage. A permanent magnet exerts an opposing force proportional to the speed of rotation of the disc. The equilibrium between these two opposing forces results in the disc rotating at a speed proportional to the power being used. The disc drives a register mechanism which integrates the speed of the disc over time by counting revolutions, much like the odometer in a car, in order to render a measurement of the total energy used over a period of time.

Further, in response to your example of filling the water drums , it is true that more current will be needed & extra time also , at low voltage( Pressure) but energy consumed(KWH) will be equal in both cases.

cheers,

Hi deept,

Thank you so much for your plain explanation about the kWh meter that you quoted from wikipedia if I'm not wrong. All the process and how it works is definitely correct as what it is in your post except several matters that I want to argue…

actually one coil is connected in such a way that it produces a magnetic flux in proportion to the voltage is what so called the potential coil that is wound on one leg of the laminated magnetic circuit and because of its many turns, it has high impedance and high inductance and therefore, the current through it lags the applied voltage by nearly 90 deg.

and the other produces a magnetic flux in proportion to the current is actually the two current coils connected in series with the load that composed of a few turns of heavy wire and because of the few turns, this current coils have low inductance and low impedance.

So if you agree to the very basic ohm's law P=I x E, isn't that if E drops than I increases?? And as you mentioned "it is true that more current will be needed & extra time also" but tell me, which comes first…??? More current or extra time??

As you know that the strength or intensity of coil's field depends on a number of factors and one of them is the amount of current flow through the coil.

And do you know Rowland's law for the magnetic circuit states that the number of lines of magnetic flux in Maxwell (φ) is directly proportional to the magneto motive force in gilberts (F) ??

Well, I think we can further discuss this if you guys first have accepted that by lowering down the voltage makes the current increase, otherwise as we do not have the same perception it's useless to describe furthermore.

Cheers,

Hi Tootje,

The discussion is very interesting , thank you for raising this topic.

First , no one in this forum is denying Ohm's law . Cureent will certainly increase ,when voltage drops, But only to some extent .

Second , about your experiment ,

I've done some experiments with my own correctly registered 230 V, 5(20)A, 50 Hz,900rev/kWh meter that similar with electricity Co's, by using a single same load 4x36W fluorescent lamps and use a small good gen-set 220V and with the Electricity co as another power source which voltage wasn't steady around 175 ~ 190V, I found that the spinning of the meter was really a bit different between both sources.....the result was 213 rev by electricity co and 196 by gen-set so 17 revolutions leads the good voltage in an hour.....

so does anyone can tell of what phenomenon is this ????!!

Please read carefully the specs. of KWH meter , a small factor 50Hz is there.

I hope that you kept the same in both cases in your experiment . If not ,please do & let us know the results. ( You have to use same Power source & Votage variac)

Thank You

Firstly, congratulations on using some relatively scientific method and sharing the details with us.

My "thought" on your experiment is that the "purity of waveform" is the major reason for the observed difference, compounded then by the variability that would be present in your test piece (The fluorescent lamp set), the wiring used and so on.

Hi,

Just an Engineer,

This was the experiment done by Tootje , on which discussion is based . This in not about purity of waveform , output load, wiring used etc. I am advising him to keep frequency same in both cases. This could only done when single source is being used. Frequency at 220 V DG source & 170 V of Electric company could not be same in any case.

I advise him to use any one source at 220 V for keeping same frequency , install KWH meter before load & take readings. Now drop voltage by Variac , Install KWH meter after Variac ( to avoid measuring losses) , use same load & take reading.

Then let us know the results.

Thank you,

Hi just,
as the matter of fact, most of your previous inputs and comments are quite in line with my mind in some sorts but for your thought about the purity of wave form become the major reason for the difference of kWh meters spin on that very small load was not appropriate in my opinion …besides that I actually use the same load with the same wiring during that experiment too.

And somehow I also suspect too many losses between those testing, that's why I decided to do the test again with some methods, including that as suggested by deept shortly.

Thank you.

Hi Tootje,

Now, Let us consider all factors affecting ratating speed of KWh meter. First current then Voltage ( as we are considering KWh Meter not Ampere Hour meter), Frequency, Power Factor etc.

In India, it is a General problem of low grid frequency due to overloading ,mainly at peak summer/ winter. The voltage drops upto 180 V at aprox. 45 Hz. insted of 220V, 50 hz. ( Any power Generator will do the same , overload cause less rotating speed of alternator & hence drop in output voltage and in frequency).

Since these Electromechanical meters are designed & calliberated for 220V , 50 Hz ,supply . V/F ratio . 10 % Less frequency should slower down the speed of meter by 10 %, but on the other hand due to increased current , overall effect is almost nil.

Now, they are replacing these meters with solid state meters. Simple prodect of voltage , current & time .They are not cheater & facing more problems due to under voltage & under frequency.

Thank You,

Deept

Hi deept,

You are right that no one denies ohm's law here, and the main purpose of this topic is certainly to share and discuss something we dealt about.

And for my experiment, I did not find any problem with the frequency factor of the kWh meter and as far as I know the frequency of electricity co is also relatively similar to the generator's 50Hz....

I totally agree with your suggestion of using the variac Voltage for more reliable test ... but unfortunately I do not have one at this moment....

Anyhow, thanks for your comment and support.

Cheers,

Regards

Energy meter runs not only on current through it but also depends on the Voltage at every instant.

However the problem is that most of the consumer are not aware of POWER_FACTOR

In old days ; some 30 years back we in Pakistan we were billed the Meter-Readings x PF [that is always less than unity] but now we are being billed on only

Meter-Readings

It is where the looters are doing their job in addition to fixing non-standards meters

[ faster ones].

No because watt= V X I so if increase V, I will drop.

most country use 220 to save copper.

In AC another factor ie PF power factor

VxA is called VA not Watt

V x A x PF = Watt

From my limited knowledge of watt-hour meters, it seems to me that a W-Hr meter measures a combination of voltage and amperes (aka Watts). So if the voltage dropped, the current would increase and the wattage would remain the same. If no power is being used, the meter disc will be stationary. No matter what the combination of volts and amps is, the watts will always be the same. (within the limits of the power grid) Sounds pretty straight forward to me.

hi welcome aboard ronseto,

that's it, but remember this only occur to the non resistive load because for the resistive Power implies the formula E²/R.

thank you.

Hello Mr rosento

The idea you say is for motors where the speed is decreased with fall of line voltage resulting in more current [due to low back-emf ] & a condition results of overheating of motor which may result burnt-out motor.

In resistive load like a room heater with the decrease / increase in line voltage the power consumed will decrease / increase as factor 'V' will has 2 effects with its decrease / increase the 2nd factor 'I' the current will also decrease / increase

Regards

Have a fine day

OK, now that we know how our service drop meters work, it's time to jump off-topic for a minute, although really IS about the power companies cheating us. In Florida, the power commissioners plan on charging the customers, meaning me (and a few others) NOW for two nuke plants they are considering building in the future. If they decide not to build them, they will keep the money. In my little mind, this is really cheating, OK a frigging ripoff. Don't get me wrong, I am definitely for building these plants; them being our only current efficient technology for the way out of our petroleum/coal fired plant realm. I just want my money back when (1) the NEC won't approve them, or (b) they figure out they are too expensive to build.

Since I raised this issue, there are so many unfriendly responses that go like pursing on what I reveal, even some of them as though shouted angrily for not being able to accept my argumentations.

so I think it's need to be emphasized again that the main point of the topic of this thread as titled is just deliver something that is very possible that for pursuing large profit or for covering losses due to load inefficiency for instance etc. the electricity company that sells energy to customers have great potential of cheating the unconscious customers.

And because in general the energy value is paid from the calculation of kWh meter revolutions then it is a possibility the cheat begins from here. Because in addition, the majority of ordinary people do not understand electricity, thus this kWh meter can be deliberately set to spin improperly.... just look around, how many people out there who just really submit to their fate to electricity co. and how confident you are with a proportional rotation of the meter is really in accordance with the energy you consume and last but not least, how many of you guys witnessing the meter installed at your house being registered before you?

As in my opinion trough some observations, playing tricks on the type in this context is the type of induction kWh meter is most frequent. And because of the kWh meter working principle is based on the magnetic flux created and induced by the potential coil and the current coils that rotate the aluminum disk with the speed retarded by C shape magnet so it very easy to set up the speed of the meter…besides that I suspect the meter would run faster under the voltage drop that increasing the current value.
And for those who rebuttal, then let's see the facts below:

This table is the results of testing conducted by Stefan Fassbinder Deutsches Kupferinstitut and it clearly shows that for some lighting products that use particular ballast with 10% of voltage drop, current logged nearly equal or even exceed some portion of the normal voltage.

And then about the fact of the transformer, when a load is connected to secondary of transformer, current will flow through the secondary conductor winding. The current flow in the secondary creates an electromagnetic field that opposes the primary electro magnetic field. The secondary flux lines effectively reduce the strength of the primary flux lines. As a result, less CEMF is generated in the primary winding conductors. With less CEMF to oppose the primary applied voltage, the primary current automatically increases in direct proportion to the secondary current, that's why as I told in my previous post that there are certain burden that act to absorb current whenever the voltage drop.

Same thing happen to the motor because the output mechanical work of a motor is rated in horse power and the actual current drawn is dependent on the load on the motor.

Inductive load equipment such as motor and ballasts can overheat at reduced voltage because more current flow in it and thus it means part of the energy paid for useless VA measured by kWh. In the dc circuits the only property that opposes current flow is resistance. But in ac circuits because electrons move in different directions and intensities, the magnetic fields of individuals electrons combine to produce an expanding and collapsing magnetic field that opposes the current flow and this opposition to current flow results in eddy current that even though small, erratic, independent currents that oppose the current flow it consumes power.

The conclusion from the above explanation is that in my opinion, with voltage drop conditions that I believe seldom happen in your country, will greatly affect the work function of kWh meter induction, since the fall of the current voltage increases according to the law applicable ohms.

Alternating current produces a magnetic field through the conductor which produces an induced voltage that opposes the change in current thus if the smaller voltage applied to the potential coil then the back emf produces to oppose the change of current decrease and the result will increase the amount of current flow through it which in turn strengthening the intensity of coil's field.

At the other hands, as the current increases due to the voltage drops to some particular loads as described above and it flows through the current coils then as Rowland's law for the magnetic circuit states that the number of lines of magnetic flux in Maxwell (φ) is directly proportional to the magneto motive force will result to the increasing of magnetic fields induced and speed up the rotation of the aluminum disk.

Well, I think that is my simple explanation regarding to the topic I made on this forum ... and certainly if there are mistakes or lack conformity with the facts, of course I'll be happy to receive any arguments and inputs that should be very meaningful & enlightenment to all of us and on top of that I commit to my signature bro, do you ?.
Thanks.

Hi Toot, Yes,it is a fact some time our basic knowledge on the subject will not accept.Being in utility if we want to cheat the consumers it can be done many ways,very popular one to order meters with high positive errors-no one can easily know,if engineers like you want to challenge the same you have to shell out 4times of the meter cost to calibrate the same in a Regulator authorised lab and you have to spend at least 4days to get it done. As there is no takers for running around utility people,they pay.It can happen with operators looking for huge profit.The limits of error and repeatability of error in electro-mechanical meters are prone for drift as it ages.Few good quality meters are available,but utility normally go for the cheapest. There are lots of discussion going in developing countries to switch over KVAH billing to avoid administrative issues in billing and to discipline the consumers. Again the issue of whether to consider phasor apparent,arithematic apparent or average apparent for billing is yet to be decided. I would be happy if members able to start a new thread on this since use of electo-mechanical meters by utilites will be phased out in couple of years.Every one will have a static meter to record energy. cheers

Hey,

Are you in Florida? I've had Progress Energy out numerous times due to the electric going off for seconds then back. they keep saying it's not thier problem.

By state Florida's rates are above the national average and on top of it they want to increase rates...time for the government to check into their practices...federal trade commission...who ever will listen. I'm getting the ball rolling. I'm fed up.

No body knows who is fed up? (as you are guest! anonymous)

Nobody knows who knows who is fed up knows...