Formula to convert Amps to Kw/hr?

[lombardo]

I want to run a test in order to verify energy savings related with a high performance oil in a recip compressor, I have an equiptment to meassure amperage but I want to know how to transform the amps to kilowatts/hour in order to know how much money I`ll save.
Thanks

 

[Skogsgurra]

You need to measure A and V and you need to do that in a way that gives you the AxVxcos(phi) where phi is phase angle between voltage and current. To have a correct result, you also need to measure all three phases and on top of that accumulate over time to get kWh (not kW/hr),

I have used "you need to" a lot of times here - and now one last: You probably need to hire a technician and the right instrumentation to do the job.

Gunnar Englund

http://www.gke.org

 

[lombardo]

Thanks skogsgurra.
I can collect the amperage per phase the thime I select with an equipment called pace data logger.
I found in a lubricant magazine article this info.
documenting energy savings (nergy savings 2.2 amp)
all figures are based on 24 hour/day at 365 days/year volts x 3 x service factor x hour/1000= kwh/day x 365days/year= $KWH/ year at $0.04 kwh

460 x 1.5 x 1.73 x 1.15 x 24/1000=32.946 KWH/day x 0.4=1.32/day x 365 day/year= $481 usd per year (savings)

I only want to estimate the energy savings I do not need a perfect result, if you know a formula that I can use please let me know.
Thanks again

 

[DanDel]

To truly measure kWh, you need to do what skogsgurra and anthony2005 have already mentioned.
If you want to see the difference in the same machine on two different lubricants, the change in current draw will give you an indication of the kWh savings, since they both are directly proportional and the savings afforded by the decreased machanical resistance should be direct kW and should not affect the Power Factor(the angle between voltage and current) very much.

 

[ccjersey]

Anthony's formulas-
"Power = voltage x currrent x power factor x 1.73 for a 3phase machine

Power = voltage x current x powerfactor for a single phase machine"

The voltage as used in these formulas is the voltage line-line on the 3 phase machine and line-line or line-neutral as the case may be on the single phase machine.

I cannot see how the original formula posted by lombardo will give any meaningful result since it doesn't use actual amperage, only service factor amps, maybe it got garbled along the way.

Jim

 

[lombardo]

thanks for the answers,
The formula posted requiere the current but my equipment collect potency (amperage), so I want to know if I can convert to kwh this savings, Thanks a lot for your responses , are very helpfull for me

 

[ccjersey]

"
The formula posted requiere the current but my equipment collect potency (amperage),
"
"
Current = I = amperage = amperes
Potential = E = voltage = volts

Jim

 

[lombardo]

Thanks a lot ccjercey, I really apreciate your support, as you can see electrical is not my area and your support is very very important, one more question is
where can I get the power factor of this equipment? or that power factor mean in order to use your posted formula.

 

[itsmoked]

It's hard to get the power factor given to you... As it depends on lots of things including the load your particular machine runs at. Any values anyone here can give you would only be guesses that would in all probability have larger errors then the savings you are trying to equate.

You want dah power factor? You gots to measure it!

One thing not mentioned here yet is buy a watt-hr meter and install it on the machine! Run the machine for a week or two keeping very close tabs on the total time of use. Change oil and repeat the measure. You will have precisely the energy you have saved.... or not.

You can always use the watt-hr meter to check other things in the plant too.

 

[lombardo]

Thanks itsmoked,
So I can´t meassure the savings with my ampers collector?

 

[jraef]

As has already been said, yes you can, but you can't directly convert it to a "kWH" number without other measurements.

But what DanDel said above is just as valid. What I see you wanting is just a comparison. Since the power supply is going to be the same in both tests, the watts value is really unimportant. What you really want to know is the delta (difference) in the two tests, and current is a perfectly valid measurement tool all by itself in this instance.

For example, if it takes 2.2A when using oil #1, and 2.1A using oil #2, then the delta is 2.1/2.2, or .955 (95.5%), so oil #2 saves 4.45% energy compared to oil #1. If you then learn the kWH of either one at a later date, you can extrapolate the actual monitary value of the other or the savings by applying that ratio.

"Our virtues and our failings are inseparable, like force and matter. When they separate, man is no more."
Nikola Tesla

 

[advidana]

Just rent the proper equipment and be done with it.

 

[JBinCA]

Gettinig a power meter is the right answer (and someone to show you how to take the readings), but if you expect the difference to be on the order of a few percent, the Amp/Amp comparison that Jraef describes will probably work out fine.

The question here is whether we electrical guys expect a significant change in pf as a result of a minor load change in a motor.

JMB