-
Controlling the Machine is no longer being updated. Don't worry, though, you can still follow Nate Holt at his new blog, AutoCAD Electrical Etcetera. You'll find it at http://nateholt.wordpress.com. Or you can subscribe to his feed to get latest words of wisdom automatically: http://nateholt.wordpress.com/feed/. You also can continue to view the Controlling Machine archives for Nate's AutoCAD Electrical tips and tricks.
-
Voltage drop and Green - AutoCAD Electrical
March 20, 2008, 10:55 PM Nate HoltThis writer is a product of parents whose childhood was totally shaped by the Great Depression. Their mindset was passed on to me. "Waste not, want not...". If you didn't clean your plate, odds are you'd see it again. And you'd always stoop to pick up a penny.
And this one specifically from my mom... "You're burning a hole in the day!" - translation: "There's some daylight coming in through the windows, you don't need that electric light!".
My first real job out of school, electrical engineer for a huge, brand-new flat glass manufacturing plant in the middle of the Carolinas. When things were running smoothly, I would look for stuff to do. There were some large, rarely visited utility rooms along one side of the factory. Every time I walked into one of them, I could just hear my mother's voice. Each room had long banks of overhead fluorescent lights... and they were always on! There was no light switch at the door - the lights were wired directly to the circuit breaker panel (cheaper to install this way), and it was very inconvenient to find the CB panel to turn the lights on and off . The result was they were always on, burning a huge hole in the day.
Ran the numbers... 40 fixtures, two bulbs each, 40 watt cool white. 40 x 2 x 0.04kW = 3.2kW.
Always on, so 24 x 365 x 3.2kW = 28000 kWh per year
Back then we were probably paying 3 cents per kWh = $840/year.
PLUS... the facility had to pay a "peak kW demand" every month, peaks that would happen where there was a major disruption in the glass flow through the furnace and annealing area. The 3.2kW in the utility room just added on top of this peak value charge each month.
So, I got my electrician buddy to help and we ran a half dozen sticks of conduit, pulled a little wire, and mounted a wall switch at the door. That was 30 years ago and I believe the plant is still going strong. So, $840 x 30 = $25,200 and counting... all for a couple hours' work.
Going green with Voltage Drop calculations
Here's a less visible place where electrical losses can pile up... conductor heating losses. Let's say we have a 100 horsepower, 480 volt, 3 phase motor that pretty much runs all the time at 285 amps full load at 0.85 power factor. The power wiring to the motor is about 400 feet. Based upon various factors, the National Electrical Code might show that 500KCMIL conductors should be able to handle the current.
But what about the voltage drop in this length of wire run? And what is the "cost" of this voltage drop loss in the wiring, dissipated as heat in the wiring instead of reaching the motor to do useful work? Would it make "green" sense to go to the next larger wire size to reduce the voltage drop and thereby reduce the annual IR losses in the wiring?
Here's a little AutoLISP utility that could help give a clue. I tried to make this follow what I found in the NEC. Hopefully it is accurate. The tricky part is including the power factor into the equation.
The UI "user interface" is pretty crude. It just displays in the command window.
To use, just APPLOAD this file (revised, see below)
Then type PF3[Enter] at the "Command:" prompt.
Let's run it on the example above...
Command: pf3
Wire size [? for list] =500
Select 1=PVC, 2=Alum conduit, 3=Steel conduit:3
Power factor [default=0.85]:0.85
Amp current:285
Wire length (feet):400
voltage drop=9.84, kW loss=4.9Hmm... according to this little utility, will only have about 10 volt drop at the motor, well within spec. But this voltage at this 285 amp current, 3 phase, calculates out to almost a continuous 5 kW loss in the wiring.
Let's say today's electrical rate is 7 cents per kWh. Our wiring loss (assuming motor is always running) will be 5 x 24 x 365 x 0.07 = $3066 per year. Seems like a lot.
Let's run two 250KCMIL conductors per phase (142.5 amps in each wire) and try again.
Command: pf3
Wire size [? for list] =250
Select 1=PVC, 2=Alum conduit, 3=Steel conduit:3
Power factor [default=0.85]:0.85
Amp current:142.5
Wire length (feet):400
voltage drop=7.24, kW loss=1.8Okay, voltage drop is better now, went from 9.84 volts down to 7.24. The kW loss is now 3.6kW (1.8 in each of the doubled up conductors). Our wiring loss cost will be 3.6 x 24 x 365 x 0.07 = $2208.
So, according to this little calculation, we might save $800 per year by running double 250KCMIL wires instead of single 500KCMIL wires to our 100HP motor. Same amount of copper, just different conductor characteristics at this power factor value.
Mom would be proud.
PS. If anyone sees a flaw in my logic, please don't hesitate to raise an issue. Thanks! Nate.
Update: revised the Lisp utility to deal with either Copper or Aluminum wiring (first version only dealt with copper conductors)
files/21101_21200/21190/file_21190.lsp (rename to pf3.lsp and APPLOAD it)
You must be logged in to post a comment.