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Ok guys I need some help trying to figure out how much load this customer is using out of his 200amp service. He has provided me his kwh from the last 4 months.
5141 kwh
5620 kwh
5270 kwh
4191 kwh
Any help would be appreciated. His service is a single phase 200 amp.
 

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It's really simple..

5141kwh is 5,141,000Wh so divide by the number of days in the month and the number of hours in a day (24) and you'll get the watts used in an hour. Then divide by 240V and you'll get the amps used over a given hour.
 

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It's really simple..

5141kwh is 5,141,000Wh so divide by the number of days in the month and the number of hours in a day (24) and you'll get the watts used in an hour. Then divide by 240V and you'll get the amps used over a given hour.
Really? That comes out to around 29A per hour.

What if it was a 50MW motor that ran for 6 minutes a month? Still going to be less than 5141kWh at the end of the month, but isn't that going to be more than 29A peak?

You cannot determine anything useful about peak current from only having kWh measurements, it has zero reference to any instantaneous value.
 

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Really? That comes out to around 29A per hour.
29 amps per hour? I've been an electrician for 25 years and that's the first time I have ever heard of an amps per hour calculation. Is that even possible?

ETA, it's like saying my car goes 60 MPH per hour.
 

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JRaef is one of the smartest people on this forum.. we all have a lot to learn from him.

I think he was implying Ah and not A/h. :whistling2:
 

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Ok guys I need some help trying to figure out how much load this customer is using out of his 200amp service. He has provided me his kwh from the last 4 months.
5141 kwh
5620 kwh
5270 kwh
4191 kwh
Any help would be appreciated. His service is a single phase 200 amp.
That's vague. What prompted the customer's enquiry? Does he think he needs a service upgrade? Is his power bill too high? Background needed.
 

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29 amps per hour? I've been an electrician for 25 years and that's the first time I have ever heard of an amps per hour calculation. Is that even possible?

ETA, it's like saying my car goes 60 MPH per hour.
I was responding to FrunkSlammer, of course there is no value in knowing "A/hr".

FrunnkSlammer had said:
5141kwh is 5,141,000Wh so divide by the number of days in the month and the number of hours in a day (24) and you'll get the watts used in an hour. Then divide by 240V and you'll get the amps used over a given hour.
5141kWh per month / 30 days = 171kWh per day
171kWh per day divided by 24 hours = 7.14 kWh/Hour (no such thing, but roll with it here)
7.14 kWh = 7140Wh
7140Wh / 240V = 29.75 A/h

A completely bogus and meaningless value, but it was just the extension of (what turned out to be) FrunkSlammer's joking way of telling the OP the same thing... You can't get there from here.
 

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Ok guys I need some help trying to figure out how much load this customer is using out of his 200amp service.
He is using 100% sometimes and 0% other times.

That answer is as in-accurate as any other answer you will get. There is simply not enough information to answer the question.

It's like saying my customers car used 65 gallons of gas last month. How many horsepower did he use out of the 200 HP his car is rated for.

He may have drove like a granny and only peaked out a 40 HP, but drove a long time. Or he may have put the pedal to the metal, red-lined it often, used every one of those 200 HP at times, but didn't go very far.
 

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Ok guys I need some help trying to figure out how much load this customer is using out of his 200amp service. He has provided me his kwh from the last 4 months.
5141 kwh
5620 kwh
5270 kwh
4191 kwh
Any help would be appreciated. His service is a single phase 200 amp.
month 1: 5141 kwh
month 2: 5620 kwh
month 3: 5270 kwh
month 4: 4191 kwh
 

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watts per hour, kilowatts per hour or megawatts per hour do not tell you anything about service size or capacity. You need a demand meter which measures the highest load that service provided at a single point in time.

If my car get 20 miles per gallon how big does my gas tank need to be? Not enough information to answer the question.
 

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I think JRaef was correct at 50Mw. He said 50 Mw for 6 minuites /mo. That would be 5000KWh.

60 minutes at 5Mw (or 5,000kW) would be 5,000Kwh
6 minutes at 50 Mw (or 50,000kW) would be 5,000Kwh
Thank you, yes.

And I know it is an absurd example. THAT was my point!

Sheesh... :rolleyes:
 
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