3ph kwh

[PLR]

I need to figure out amperage from kwh ,the average use in 12 mouths was 49000 kwh 120/208 3PH

 

[augie47]

difficult to obtain from kwh as you don't know the load at any given time.

 

[raider1]

Welcome to the forum.

I don't understand what you are trying to get?

Kwh is the amount of power used and converting this to the amout of amperes used does not accomplish much.

Chris

 

[skeshesh]

For sizing conductors/OCPDs the average demand is not very usefull. What you need to find is the maximum demand.

max demand (kwh) / hours = kw / (208 x sq root(3)) = max current draw

But yea, as stated by Chris, this doesnt accomplish much. I also wonder, how did you get the 12-month average kWH for a single motor?

 

[Besoeker]

Kwh is the amount of power used and converting this to the amout of amperes used does not accomplish much.

Chris

It is the amount of energy, not power.
I'm not being pedantic - it's an important distinction.
To illustrate this, the 49,000 kWh over a year could be 49,000 kW for one hour or a constant 5.6 kW or 11.2kW for a 50:50 duty cycle or etc....
It doesn't really help in determining maximum currents.

 

[raider1]

It is the amount of energy, not power.
I'm not being pedantic - it's an important distinction.
To illustrate this, the 49,000 kWh over a year could be 49,000 kW for one hour or a constant 5.6 kW or 11.2kW for a 50:50 duty cycle or etc....
It doesn't really help in determining maximum currents.

Yes, you are correct it is the amout of energy used.

Chris

 

[charlie b]

The question is analygous to saying that you put 15,000 miles on your truck last year, and wanting to know what speed you were using. Sure, you could divide 15,000 by 365 days, then divide again by 24 hours, and come up with an average of 1.7 miles per hour. But that would not be representative of any real event. You weren't driving 24 hours a day, 365 days a year. Even when you were driving, your speed varied up and down. So an average speed cannot really be determined from only knowing the total miles driven in a year.

Power (KW) is the rate at which you are using energy. Since you know the voltage and the number of phases, and presuming you can make a reasonable estimate of power factor, knowing the KW is the same as knowing the amps. But KWHr is the total amount of energy used. So you can't tell, from that alone, whether you were using power at a high rate during the summer days, and at a slower rate on spring days, and at a still slower rate at night.

Welcome to the forum.