Way OT: Calculating watts from voltage + amps on motor nameplate

Sun Oct 14 15:57:05 MST 2007

--- KevinO <kevin at kevino.org> wrote:

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> Kurt Granroth wrote:
> > Okay, this is quite a bit off-topic but I imagine
> there are some people
> > here with a decent bit of electrical knowledge so
> I'll try anyway.
> > 
> > I'm trying to calculate how many watts my pool
> pump and a/c systems use
> > based on the info on their nameplates.  I
> understand the theory, but I'm
> > having a hard time translating what the nameplates
> are telling me to the
> > theory that I understand!
> > 
> > Starting with my pool pump.  It claims that the
> voltage is 208-230/115.
> >  I am guessing that that that means it will
> typically use between 208
> > and 230 volts, depending on load.  Alternatively,
> it can run on 115
> > volts?  Am I right so far?
> > 
> > Then there are the amps.  It says "SF
> 8.5-7.8/15.6".  Hmm... SF probably
> > means "Service Factor".  So maybe it's using
> between 8.5 and 7.8 amps
> > (load based) when connected to the 220v system and
> 15.6 amps if
> > connected to a 120v?
> > 
> > Okay, I'll use Watt = Volt * Amp.
> > 
> > 7.8 * 208 = 1.62kW
> > 8.5 * 230 = 1.96kW
> > 15.6 * 115 = 1.79kW
> > 
> > Is that all right?  If my pump is connected to a
> 220v circuit, can I
> > accurately say that it draws between 1.62kW and
> 1.96kW?  If so, short of
> > a watt meter hooked up to the pump, is there any
> way to calculate the
> > kWh?  Do you typically just take the lower number?
>  Or some average?
> > 
> > The AC systems are a bit more complicated.  Both
> say that the Voltage is
> > 197-253.  Neither out-right say the amperage,
> though.  They both say the
> > "minimum amperage circuit" and the maximum.  Do
> those values really
> > correspond to the minimum and maximum amp range? 
> If not, then how can I
> > tell what it would be within that range?
> > 
> > So the 2-Ton unit has an amp circuit breaker range
> of 17.9-30 and the
> > 5-Ton unit ranges between 33.1-50.  Naively, I
> could calculate:
> > 
> > 2-Ton Min: 17.9 * 197 =  3.53kW
> > 2-Ton Max: 30.0 * 253 =  7.59kW
> > 5-Ton Min: 33.1 * 197 =  6.52kW
> > 5-Ton Max: 50.0 * 253 = 12.65kW
> > 
> > But is that even remotely right for the 5-Ton? 
> I'm assuming it uses a
> > 3-phase motor at that size so would I need to
> multiply that by 1.73?
> > And in any case, what values should I use to
> calculate average kWh for them?
> > 
> > Any help (including links to really good sites
> that discuss this type of
> > stuff) will be greatly appreciated!
> > 
> You are pretty much on the right track, but there
> are a couple caveats.
> 
> When you are dealing with an inductive load running
> on AC, and your motors fit
> this description, the voltage and current are not in
> phase. This means that Watts
> does not equal Volts x Amps, but will be something
> less. This is why UPS's are
> often rated in volt-amps.
> 
> In order for any of your motors to be three-phase,
> you would have to have
> three-phase power at the house, and I seriously
> doubt it.
> 
> You cannot go by the circuit breaker rating. The
> rating of the breaker is set by
> the electrical codes and the current limitation of
> the wiring in your unit, and
> the wiring to your unit.
> 
> You can use a clamp-on ammeter to measure the actual
> current draw of your loads,
> but you will need access to each of the individual
> power leads.
> 
> To calculate true RMS (root-mean-square) power in
> watts, you will need an
> oscilloscope, voltage probes, at least one clamp on
> current probe, be able to
> measure the phase angle between the voltage and
> current using the scope, and then
> use some math to calculate the true figure. Better
> is to just use the 'scope to
> figure the phase angle, and use true RMS meters to
> measure voltage and current.
> 
> There are dedicated meters to measure power all in
> one shot, but you will
> probably have to rent one, or hire a SERIOUS
> electrician to do the measurement
> for you.

Breaking in here. To measure just ONE item using the
below method, you MUST turn off or disconnect
EVERYTHING from the meter except whatever you want to
check the power usage on. Even refrig's, a/c on the
roof (kill the circuit breaker) etc. Then take and use
the method below as the only load turning the power
company meter then will be that device. Just don't
forget to plug in the freezer's and refrig's..hate to
waste food!!

The pool stuff should be on its own circuit breaker,
so just switch off all others AFTER shutting down your
computer(s).

Ed Culbertson.

> If you go look at your utility power meter, you will
> see that is has a
> multiplying factor printed on the unit, and you can
> actually use the power meter
> itself to figure out the power drain. Calculate a
> reading with the AC off, then
> do it again with the AC on. (Count the revolutions
> of the disk per minute, then
> scale it up to get kilowatt hours per hour by
> multiplying by 60 minutes/hour,
> after scaling it for the multiplication factor of
> the meter itself.)
> 
> Wikipedia has some entries relating to electrical
> power, phase angle, etc..
> - --
> KevinO
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Ed/ke7feg
Ah, dripping weather gone! Was nice and cool this morning  

 .. ...   --. --- -. .   ..-. --- .-.   - .... .   -.-- . .- .-.  
-.-. .--   .-. ..- .-.. . ... 
--. --- --- -..   -.. .- -.-- --..--   .-- .. ... ....   -.-- --- ..-   - .... .   -... . ... - 

PS use the following if curious about the cw above and don't know it.

http://www.onlineconversion.com/morse_code.htm

Did I mention, 2/23/07 the FCC droped all testing for any license as a ham?
Just pass the written and "Yo's a ham"!!

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