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

[KevinO]

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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.

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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