Okay so you haven't actually determined the true power consumption from the AC side (wall plug side side) of the power supply. This is important to determine actual energy usage due to losses in the power supply.. Back to the beginning.

Thanks again for your help. As soon as I get the figure I'll post it.

SMPS architecture has efficiencies around 90%. Not exact, but neither are your computer and monitor estimates.

I have my SD3 plugged into a KILLAWATT meter.  For the last 37.5 hrs it had used 2.08kw.About 10 hrs of print time , the rest idle.

Was the idle with heaters on or off?

Ecarots wrote:

Thanks,

now to do the math to determine Kwh so I can calculate actual cost.

hmm the basic DC formula is A x V=P So 7.25 X 120 = 900 / 1000 = .9kw

E(kWh) = P(kW) × t(hr)= .9 X 1= .9 Kwh

So adding in .4Kwh for the PC and .1Kwh for the monitor the total consumption is 1.4Kwh

si given Consumer's Energy elctric costs of 12.5 cents per KWH the cost is 17.5 cents per hour.

not exact but close enough to calculate for me.

Nope.

The 7.25A is at 12 volts.

So, your calcs are off by a factor of 10, roughly (also power supply efficiency to account for)

Okay so you haven't actually determined the true power consumption from the AC side (wall plug side side) of the power supply. This is important to determine actual energy usage due to losses in the power supply.. Back to the beginning.

Thanks again for your help. As soon as I get the figure I'll post it.

Thanks,

now to do the math to determine Kwh so I can calculate actual cost.

hmm the basic DC formula is A x V=P So 7.25 X 120 = 900 / 1000 = .9kw

E(kWh) = P(kW) × t(hr)= .9 X 1= .9 Kwh

So adding in .4Kwh for the PC and .1Kwh for the monitor the total consumption is 1.4Kwh

si given Consumer's Energy elctric costs of 12.5 cents per KWH the cost is 17.5 cents per hour.

not exact but close enough to calculate for me.

Nope.

The 7.25A is at 12 volts.

So, your calcs are off by a factor of 10, roughly (also power supply efficiency to account for)

now to do the math to determine Kwh so I can calculate actual cost.

hmm the basic DC formula is A x V=P So 7.25 X 120 = 900 / 1000 = .9kw

E(kWh) = P(kW) × t(hr)= .9 X 1= .9 Kwh

So adding in .4Kwh for the PC and .1Kwh for the monitor the total consumption is 1.4Kwh

si given Consumer's Energy elctric costs of 12.5 cents per KWH the cost is 17.5 cents per hour.

not exact but close enough to calculate for me.

http://www.soliforum.com/post/4053/#p4053