Givens:
Voltage 424V
Current 354A
Wiring used 500MCM copper
Length of wires 500'
Number of conductors 3
price per KWH $.15
Hours of operation 24Hrs/ day 30days/month

Question: How much will I save in line losses a month if I increase the voltage to 480V?

Non-resistive loads....

Any takers?

Dnk....

The "new truck" post is easier than this one.

Dnk - Are we to presume 424V 3Ø 3W with 354 line amps for the initial condition?

And entirely motor loads so we can say total connected VA = 424V x 354A x 1.73, which would also = 480V x 313A x 1.73 for the new condition?

Radar

How much (% or $ wise) do we get for answering this?

I humbly withdraw my reply.

[This message has been edited by Redsy (edited 03-03-2006).]

Dave T, that was toooo funny

Celtic, you get nothing but a "atta boy"

Redsy, your way off..or I'm way off...

Radar, just assume 354A per wire, nothing more than that yet. Assume all electrons are oscillating clockwise if you want...

Dnk...

[This message has been edited by Dnkldorf (edited 03-03-2006).]

That would be incorrect, I believe they'd go ccw. eheheh

Seriously, my guess is approximately $117.55.

Radar

Atta-boy Radar...

I got $118.05....

For my next one, I'll come up with something like:
If I have twice as many sp 20a breakers than sp 30's, but 1/2 as many 2 pole 20A than 50's?

Give me the load on each breaker?

You game?


Dnk..

I'm game - maybe we shouldda let this thread develope a little more before giving away the asnwer tho . . .

Radar

As the computer on 'Star Trek' says: "Insufficient Data for Meaningful Response".

The problem is that we don't know how the load will respond to the change in voltage. If these are a bunch of lightly loaded motors, or resistance heaters, then increasing the voltage might result in _increased_ current flow to the load. On the other hand, if these are heavily loaded motors, or thermostat controlled resistance heaters, or other _constant power_ loads, then increasing the voltage means an automatic reduction in current flow. If these are motors, then increasing the supply voltage will mean reducing the 'load current' but increasing the 'magnetizing current' (reducing the power factor).

For the purpose of calculation, we have to assume something, so I am going to assume a simple constant power load; increasing the voltage means reducing the current and the same total VA supplied to the load.

The resistance of 500 MCM conductor at 25C is 0.0129 ohm per 500 feet. The voltage drop in each conductor is 354A*0.0129 Ohm= 4.57V, and the voltage delivered to the load is 424-(root3 * 4.57) = 416V. The VA delivered to the load is 416*354*root3=255kVA. The power dissipated in the wires is 4.57*354*3 = 4850W

Now we change things so that we have 476V delivered to the load (I am guessing here, rather than solving the equation) so we get 255kVA/root3/476=309A. Check the voltage drop and we get 309 * 0.0129 = 3.98V The power dissipated in the wires is now 3700W. So you end up saving about 1150W. 8766 hours in a year, so you end up saving about 10000 kWh per year. So I guess a savings of about $1500 per year.

-Jon

It doesn't matter Radar, the thread usually ends when -jon comes in with another brilliant explanation like that anyway...

Atta boy winnie


Dnk..

Well at least our answers agree....I didn't read the question closely enough, and got a savings of $1500 per year...which works out to about $125 per month

-Jon