Calculating Wattage on a 208v 2 Pole Circuit

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fdirla

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Does anyone one know how to calculate the wattage on a 2 pole 208volt circuit.

I have seen 2 formulas:

i = current
v = line to line volts
pf = power factor
w= watts
COS 30 deg = .86025404

1. the most common i * v * pf = w

2. a not so common that calculates approximatly 15% higher wattage is

((i x v) / COS 30deg) * pf = w

Can anyone shed which one is the correct answer?
 
What type of load are you trying to calculate? In any event you would multiply current times voltage. I don't know why you would need to be concerned with power factor in an every day field calculation.

When I hear "2 pole 208 volt" circuit the first thing that comes to mind is ballasts. In that case you would use the current rating of the ballast to determine circuit size. Power factor is irrelevant.
 
It is a Data Center load for Blade Servers. Very clean power and very critical thus the reason the PF is used...The wattage is needed to see the rate growth for watts per square foot.

Hope that clearifies a little..

FD
 
Here we go again!

Here we go again!

fdirla said:
Does anyone one know how to calculate the wattage on a 2 pole 208volt circuit.

I have seen 2 formulas:

i = current
v = line to line volts
pf = power factor
w= watts
COS 30 deg = .86025404

1. the most common i * v * pf = w

2. a not so common that calculates approximatly 15% higher wattage is

((i x v) / COS 30deg) * pf = w

Can anyone shed which one is the correct answer?
Click to expand...

The first formula computes real power delivered to a single phase load:

Preal = Irms x 208Vrms x PF in watts

The second formula computes apparent power provided by the two transformers which provide the 208V.

Pa = Irms x 120 x 2 = Irms x 240 in VA, NO PF

instead of,

Irms x 208V

The ratio, 240/208 is 1.154, therefore,

Pa = Irms x 208 x 1.154

This question has been discussed at length in a thread about the Oregon Fudge Factor.
 
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