Way OT: Calculating watts from voltage + amps on motor namep…

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Author: Kurt Granroth
Date:  
To: Main PLUG discussion list
Subject: Way OT: Calculating watts from voltage + amps on motor nameplate
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!

Kurt

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