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Join Date: Jul 2008
Posts: 13

Electrical Engineering - Calculate kwh/year

06/26/2010 1:04 PM

Dear friends,

If my Kw=2300@0.70 PF what will be my kwh/year,if my cost is 8$/unit

If i use capacitor bank and improve my PF to0.95 what will be my Kw and savings with the same cost 8$/unit.kindly show the formula /calculation.

With thanks and regards.

suresh.

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Guru

Join Date: Oct 2008
Posts: 42355
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#1

Re: electrical engineering

06/26/2010 1:11 PM

You'll need to provide more information about usage.

What is "8$/unit"?

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Commentator

Join Date: Jul 2008
Posts: 93
#2

Re: Electrical Engineering - Calculate kwh/year

06/27/2010 2:10 PM

You will pay the same thing-meter reads KWATT not KVA.

PF value does not make any difference.

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Guru
Technical Fields - Technical Writing - New Member Engineering Fields - Piping Design Engineering - New Member

Join Date: May 2009
Location: Richland, WA, USA
Posts: 21017
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#3

Re: Electrical Engineering - Calculate kwh/year

06/28/2010 4:38 AM

2300kw x 365 days x 24 hours = 20,148,000 kwh.

If the unit cost is US$8, then 20,148,000 x 8 = $161,184,000....

Somehow I doubt this. For one thing, the more likely unit cost is 8¢. If so, your bill would be $1,611,840. Not only that, but you might not be using the 2300 kw all of the time.

You need to find out whether and/or how much your utility charges for low PF. Then you can calculate how much you can save by improving the PF. The data so far are inconclusive.

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