[jtn275]

What is the total power consumed (in watts) of a 220vac well pump motor if each leg has 7 amps running through it?

770watts
Or
1540watts

[parttime]

Hi jtn and welcome, I'm curious as to how you arrived at 770 Watts? There should be an expert along shortly, so hang tight.

[jtn275]

7 amps (x) 110vac = 770 watts

[parttime]

Why would you use 110 volts on what you called a 220 volt motor? Are you a student?

[jtn275]

why would you use 110 volts on what you called a 220 volt motor? are you a student?

A 220vac supply consists of a pair (2) of wires each at 110vac.
Voltage measured between the pair will be 220vac.
Voltage measured between either one of the pair and ground/neutral will be 110vac

[hfcarson]

P = E x I = 220V x 7A = 1540W
of course you could look at each phase leg and say 770W was being drawn by each so the total was 770W + 770W = 1540W...

[jtn275]

P = E x I = 220V x 7A = 1540W
of course you could look at each phase leg and say 770W was being drawn by each so the total was 770W + 770W = 1540W....

Thanks that's kind of what I was thinking but just wanted to double check.
One other thing if you don't mind.
Are the two legs 180 degrees out of phase with each other? (assuming standard 220vac American residential supply)
Thanks again

[hfcarson]

.. I hope this isn't a test question..
If you're referring to common residential supply from a single transformer, (common pole mounted), then, yes...
If you see (3) transformers grouped together on a pole then you will "most likely" have three-phase in which each phase is 120 degrees from the others.

So did I give away a homework question? (be honest!)

[jtn275]

No test questions promise, just running a well pump nearly nonstop and wanted to know exactly how much its costing me.
Since we are discussing AC electricity led to the phase question.
Just wondering, since the leg A and B sine waves are 180 out of phase if at some point they would cancel each other resulting on a lower total amperage/wattage draw between the two. No I don't believe perpetual motion/free energy :)
The motor is 1hp which in a perfect loss free world would draw 746 watts if working at rated hp.
So I am still a little puzzled at the 1540 watt figure.
Anyway thanks again!
JTN

[hfcarson]

looking at A phase and B phase at the same time on a scope... A starts at zero and climbs to approx +170 volts... B starts at zero at the same time and falls to approx -170 volts... now the definition for the difference of potential is Va - Vb = 170 - (-170) = 340 Volts...
Now these values are the "peak" values, if you want the RMS values multiply 340 times the [sqrt (2) / (2)] which is about 0.7071... so 340 x .7071 = 240Vrms...

and you are correct, in a perfect world a one hp motor would generate 746 Watts of "Output" power... the input will always be subject to the losses of the wires, the slip rings and altered by the power factor... horsepower refers to a measure of the output so the input is always more...

perpetual motion? Free energy? First show me the math...

[jtn275]

"Show me the math..."
Exactly!
That's how this whole thing came up.
I figured this would be simple problem to solve, A x V = Watt.
Got wrapped around the axle over two live legs 180 out of phase. DC seems much simpler!
I would have thought a water cooled AC motor would be more efficient than 50%.
But then again I am not using a true RMS meter and my Amp meter is a $20.00 cheapo model so my readings could be completely slewed.

[hfcarson]

Even the lower priced meters are surprisingly accurate...
Although I like my Fluke 92 and scope...

Keep asking questions, there are a lifetime of answers...