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Joined: Mar 2005
Posts: 2
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Greetings,
I have been stumped at what calculations I should be performing to determine power consumption from an electrical main. I have 3 CT's giving me a 0-5V feedback and I want: 1. Watts 2. Peak 3. Peak Demand 4. Watt-hours
I think I know how to get #1 W = V x I x 1.73 for 3 phase power - is this going to give me an accurate relationship?
#3 seems simple enough - storing the highest #2 value for a given period - ie. peak for a day, week, month and year - but how do i Calculate #1,2 and 4?
Any help would be appreciated. Even if it just points me to some other refrence material.
Thank you, Drew
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Joined: May 2004
Posts: 50
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1&4 are the same, 2&3 will be the same!!!!
You already know how to calculate the power supplied..you stated the formula..
kVa = kW = W. ie: 2kVa = 2kW = 2000W = 2kW/hr = 2000W/hr
The peak demand of the installation will be the point in time when the installation will be subjected to the maxmimum possible loading, usually whn all electrical equipment attached to it is turned on initially. This will also be your Peak Usage.
The only way to calculate the Peak Demand is to calculate the loading of the installtion and then take into account start-up times for given equipment, when you find that you have identified a time in the day or week that has the most equipment start-ups, that will be the peak demand time, simply add it all together and convert into kW. This should also give you the peak usuage at the same time, although onlgoing electronic measurement is the best way to determine if the two do actually coincide or whether some other factor is causing a slight discrepancy..unlikely, but can happen for very brief periods.
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Joined: Mar 2005
Posts: 2
OP
Junior Member
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I actually do have a controller that is continuously recording the 0-5V feedback.
For best results should I also factor in power factor? Thus: W = I x V x 1.732 x pf ?
Also, lets say that I have a 500 Amp circuit that I am stepping down to a 50:5 current ratio. So then do I simply interpolate what my primary side current is based on my secondary side output? Thus if I get 2.5 VDC feedback I figure 250 Amps on the primary and my calculation would be (assume pf = 0.95): W = 250 Amp x 460 Volt x 1.732 x 0.95 and I would have 3 of these calculations, one for each phase - then do I add the 3 up and divide by 3 to get the average?
How would I measure power factor and is it dynamic? Assuming with lighting load it would be fairly static, but how about with motors, especially one's with VFD's. I would think that it would be dynamic - so how could I measure it in real time to keep my #1 Watts calculation as accuarate as possible?
#2 I also meant to put: Calculate Demand Present. What is the calculation to get Demand Present? (not Peak) I have an electrical power book coming that I ordered - but after ordering it I found this great forum. If I don't hear back I will try and figure it out when I get the book and I will post the answer. Thanks for all the great help!
Drew
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Joined: Feb 2002
Posts: 182
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You posted "I have 3 CT's giving me a 0-5V feedback". If you have CT's you are getting current not voltage. If you have a 200/5 CT the multipler is 40.(200/5 = 40) Multiply the current reading by 40 for the total amperage. There are recorders that monitor voltage with PT's and Current with CT's and multiply the total to get the total watthrs and peak power in kw.
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Posts: 1,803
Joined: March 2005
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