wrv@ihlpm.ATT.COM (Vogel) (04/12/89)
Thanks to everyone who responded to my pump question. I've
learned alot just from all the responses. I tried to send email to
everyone who responded but about half of it bounced back.
Here's a quick summary:
I got a bunch of responses saying that the main difference comes
from the lower wire resistance loss at the higher voltage, specifically
1/4 the loss. Most everyone said that the motor would run more
efficiently too. I decided to convert it. Then rsd@sei.cmu.edu
did a detailed analysis (which actually I could have done, but just
never thought of it). This analysis showed an $11 savings over the
entire season. Then I decided not to convert it. Then gary@bellcore.UUCP
responded saying that my electric bill would be one half (@220v) of what
it is at 110v, due to the way the meter measures the power used. So I
decided to convert it again.
Anyway, I got to thinking about it again (dangerous for someone with
only 1 year of EE (switched to CS!)), and I wonder just how an electric
meter does measure the power you use. I realize it is measuring current
over time, but current from where to where? It seems to me that there
are three possible current paths:
line from power co.
110v neutral 110v'
| | |
| | |
| | |
| | |
| | |
v v v
^-------^ ^-------^
Path #1 Path #2
^-----------------^
Path #3
So how does the power company measure current? If its just on the
ground wire, then it would miss all current on Path #3. If its the
sum of 110v and 110v', then current on Path #3 would be metered twice
(wouldn't it?), once goin' out and once commin' back! And since the
voltage there is 220, wouldn't the power be 4X what it really was?
Oh I'm so confused.......... Sorry if these questions are boring
everybody.
Bill [just trying to understand reality] Vogel, ihlpm!wrv