Electric kwh meter question

Andy asks:

I have a question regarding the electric meter that is used to monitor the power consumption of a house, as used by the power companies.

My question regards the meter reading if the power is reactive.

For example, If I have a 1000 watt incandescent load, the meter will register 1 kwh at the end of one hour..

If, however, I am driving a capacitive or inductive load of 1000 watts with a power factor of 0.707 , at the end of an hour will the meter have recorded 1 kwh or will it have recorded 0.707 kwh ?

Thanks for your informed answer ...

Andy W4OAH

Reply to
AndyS
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The electric power meter will only record and totalize actual power
consumed, so if your load is 1000 volt-amperes with a power factor of
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Reply to
John Fields

Not true in the uk, fwir. The British Standard spec for domestic meters requires them to read accurate kwh with leading or lagging power factor. Can't remember the actual figures, but must be within 5% or so over quite a wide range of power factor.

Industrial meters are different and companies get charged against maximum demand and extra for bad power factor...

Regards,

Chris

Reply to
ChrisQ

Which is what John said. Electricity meters read kWh (real power * time), not kVAh (apparent power * time). So, with a power factor of 0.707 (i.e. leading or lagging by 45 degrees), 1000VA is 707W; after 1 hour, the meter will read 0.707kWh, not 1kVAh.

Reply to
Nobody

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How is that different from what I stated?

There's no such thing as a leading or lagging power factor, since
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Reply to
John Fields

OK, here's the "real" problem, there is no such thing as a capacitive or inductive load of 1000 Watts. A pure reactive load draws NO real power, and the meter will sense zero power, as it is specifically designed to only read real power.

Now, the semantics are fuzzy. if you read 1000 Watts to mean Amps x Volts = 1000, that is NOT Watts, but 1 KVA. It is not related at all to real power drawn.

One could also mean a PF of .707 and a real power draw of 1000 W, in which case the KVA will be 1414 W. The meter would show 1 KW draw.

Jon

Reply to
Jon Elson

It is actually an energy meter and not a power meter, there is no record or display of power any more than the odometer on your car gives an indication of the speed used to total it. The original form of this meter was a squirrel cage ac-induction motor driving a geared totalizer register. The motor is powered by two windings: one winding directly in shunt with the line voltage and a the second winding in series with the load. The basic physics of magnetism and the arrangement of the windings result in a rotational torque on the motor rotor that is proportional to the product of the line voltage and the in-phase component of the current, i.e. the real power provided to the load. The torque causes the disk to rotate which drives the register gears to totalize the rotational displacement- this is directly analgous to an odometer of an automobile. Each unit of the register total represents a unit of angular displacement of the rotor, and this unit of angular displacement represents a torque x angular displacement product, or energy. A high power load will spin thru the displacement quickly and a flea power load will take a long time, no matter the case, the register records and totalizes the energy as it is consumed by the load. The reactive component of load power does not influence the motor rotation and therefore is not totalized with the energy consumed.

Reply to
Fred Bloggs

It isn't and I should have read what you *actually* said.

Second time i've done that in the past week, not a good sign...

Regards,

Chris

Reply to
ChrisQ

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Being able to admit to errors is always a good sign. :-)
Reply to
John Fields

Sorry, but actually it is, with a factor of the cosine of the phase angle.

Hope This Helps! Rich

Reply to
Rich Grise

He who never makes a mistakes never learns anything, even if it's just to remember to think before jumping in :-)...

Regards,

Chris

Reply to
ChrisQ

IFF all signals are sine waves.

Reply to
krw

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Normally we only use (VA/apparent power) for ac circuits. That is a combination of resistance and induction to form a Z..

In DC land, it would be True power. The Power factor (PF) is a ratio between the two, on motors for example.

Jamie

Reply to
Jamie

OK, what I was TRYING to say was that a current measurement and a voltage measurement tell you only KVA, but nothing about wattage. You can't determine Wattage or power factor with only a current and a voltage measurement. If you know the power factor, then you can make the conversion. or, if you can measure instantaneous current and instantaneous voltage, multiply, and integrate over the whole cycle, then you can compute power directly. This is what the all-electronic watthour meters do.

Jon

Reply to
Jon Elson

Hmmm, this OUGHT to mean that I'm REALLY smart, by now, shouldn't it?

Jon

Reply to
Jon Elson

Now that would be good. Just because we learn from mistakes doesn't mean at some time we are free from making more, or even repeating the same mistake again. Such is the human condition, being not machines and all that :-)...

Regards,

Chris

Reply to
ChrisQ

It does give an upper bound.

You can if you have a lot of measurements. ;-)

The "alot of measurements" method. ;-)

Reply to
krw

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Strange then that our kilowatt-hour meters aren't kilovolt-ampere-hour
meters.
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Reply to
John Fields

John, you do realize you are wasting your time on that dimwit, right?

John

Reply to
John - KD5YI

"John Fields"

** Both terms are in the language and have been use by engineers for many decades as a shorthand to describe the nature of the phase angle involved - when there IS one involved.

Google gives 191,000 hits for one and 141,000 hits for the other.

** The general definition involves only the ratio of VA and true power.

.... Phil

Reply to
Phil Allison

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