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From: michael@m-net.UUCP (Michael McClary)
Newsgroups: sci.electronics
Subject: Re: Domestic KiloWatt Hour Meters
Message-ID: <1216@m-net.UUCP>
Date: Sat, 25-Apr-87 03:36:39 EDT
Article-I.D.: m-net.1216
Posted: Sat Apr 25 03:36:39 1987
Date-Received: Sat, 2-May-87 09:21:32 EDT
References: <923@mhuxh.UUCP>
Reply-To: michael@node.UUCP (Michael McClary)
Organization: McClary Associates, Ann Arbor, MI
Lines: 71
Keywords: meters,power,watthour
Summary: Here's how they work (in excruciating detail).

In article <923@mhuxh.UUCP> oso@mhuxh.UUCP) (Edward J. Osoliniec) asks:

>I need some help in understanding the operation of the
>domestic KW-Hr. meters the electric utility companies
>use to calculate my electric bill every month.

You can find descriptions of watthour meters in electrical engineering
texts.  Check at your local university's library.

First, you referred to "two phases".  Well, yes, you could think of
a single-phase service as two phases 180-degrees apart, but we'd better
stick to power company nomenclature and think of it as a single-phase
supply, with two hot conductors carrying current in opposite directions.
(Otherwise we'd have somewhere between 9 and 12 phases running around
by the time we'd dealt with a 3-phase meter.)

I will Keep It Simple and Stupid by discussing only an ideal meter,
neglecting the dozen or so minor ways they deviate from perfection.

A (single-phase) watthour meter is a two-phase electric motor:
 - The spinning disk is the rotor.
 - A forward force proportional to the power consumed is applied to it
   by a pair of electromagnets driven by currents 90 degrees out-of-phase.
 - A retarding force proportional to the speed of the rotor is applied
   by a permanent magnet.

Thus the rotor rapidly reaches an equilibrium speed proportional to the
power consumed, and its rotation (which drives a gear-train to an analog
counter) integrates the power.  It will also run backward if you feed
power to the grid.

(A polyphase watthour meter has two or three separate disks on a common
 shaft, one to receive a forward force proportional to the power delivered
 by each phase of the service.  Under some conditions you can get away
 with two disks in a 3-phase service.)

The two electromagnets that apply the force are a "voltage" magnet and a
"current" magnet.  Assuming they are driven by sine waves, the force they
apply is proportional to the product of their field strengths times the
sine of the phase angle between them.  Thus if the "voltage" magnet has
a field strength proportional to the voltage, but delayed by 90 degrees,
and the "current" magnet has a field strength proportional to, and in
phase with, the current, the total force will be proportional to the power
through the meter.

The "voltage" magnet is wound with many turns of thin wire, placing a large
resistance in series with its inductance.  This causes it to carry a small
current, retarded about 90 degrees from the voltage's actual phase.  It is
connected between the two "hot" wires on the input side of the meter (thus
making the assumption that the "neutral" wire is at a voltage exactly
midway between the two "hot" wires.)  The field strength is proportional
to the current times the number of turns, which is what we wanted.  (The
current is small, but there are a >lot< of turns, so the field is usable.)

The "current" magnet has two windings, one for each of the "hot" wires.
There are very few (3-ish) turns of very heavy wire, so there is negligible
voltage drop through the meter.  The windings are connected so that if you
pull your current from hot wire to hot wire, the fields will add.  Thus if
you pull one ampere from hot to hot (220 volts), you'll cause twice as much
of a field as if you pull one ampere from one hot wire to neutral (110 volts).

Does it all make sense now?

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Above opinions are the official position of McClary Associates.  Customers
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