Potential electrical power saving by increasing the size of conductors NEC REQUIRED

[Rjryan]

I have found a few articles about the electrical use savings by increasing the size of the conduct past NEC requirements.
It makes sense to me larger wire has less resistance, will not heat up as much under the same load, because heat causes resistance.
I questioned an Electrical Inspector from Fort Worth, Texas several years ago about this. Everything has gone up, but at that time he said a rough calculation
is for every size you increase the wire it will pay for itself in 5 years.
Besides the resistance of the conductors, I have not found a formula or method to accurately calculate electrical use savings by increasing conductor
size beyond the NEC.
Does anyone have a formula or method for these calculations?

 

[LarryFine]

There is no one formula because there is no one answer. Whether lessening voltage drop is financially beneficial depends largely on the nature of the load.

For example, you won't keep a light on several minutes longer because it's dimmer, but you would keep the electric heat running longer if the voltage sags.

 

[petersonra]

You have to make so many assumptions that any real numbers are hard to come by.

Keep in mind that most load calculation overstate the actual loads the electrical system will usually see by a factor of two or three.

So suppose you had a 100 amp service and you had a calculated load of 70 amps. Most of the time your load will be more like 25 amps or maybe 30.

How much heat is generated in a #2 wire running 20 or 30 amps? How much less heat if it is a #1 wire.

 

[wwhitney]

How much heat is generated in a #2 wire running 20 or 30 amps? How much less heat if it is a #1 wire.

The last question is easy: #1 wire has an area that is very close to the cube root of 2 times the area of #2 wire. [AWG is geometric, and 3 sizes up is slightly over double the area, 2.0073 times to be more precise.] So with an area 1.26 times larger, the resistance of #1 is 79.3% of the #2, and the heat generated will be 20.7% less for a given current.

Cheers, Wayne

 

[hillbilly1]

I attended a continuing education class many years ago where the instructor stated that running 3/0 to a stove would pay off in two years……….LOL!

 

[LarryFine]

I attended a continuing education class many years ago where the instructor stated that running 3/0 to a stove would pay off in two years……….LOL!

And if you ran 250mcm, your meter would run backwards.

 

[drcampbell]

I attended a continuing education class many years ago where the instructor stated that running 3/0 to a stove would pay off in two years……….LOL!

I would have insisted that he show his work.
How many actual watts? How many actual hours?
Commercial kitchen running nearly continuously, or residential?

I'd also ask him to demonstrate connecting 3/0 cable to a stove receptacle.

 

[roger]

And if you ran 250mcm, your meter would run backwards.

Mine did

 

[don_resqcapt19]

About the one time this works is for circuits that run pretty close to full load 24/7.
You have to run all of the numbers based on the cost of power, the load and how many hours a year the load will run, then look at the increased labor and material costs.
For industrial motor loads that run pretty much 24/7 I could get a 18 to 24 month payback by going up one wire size as long as that wire size increase did not trigger an increase in the conduit size. But even on industrial projects there is a large percentage of the loads that are not even close to full load for 24/7.

 

[hillbilly1]

I would have insisted that he show his work.
How many actual watts? How many actual hours?
Commercial kitchen running nearly continuously, or residential?

I'd also ask him to demonstrate connecting 3/0 cable to a stove receptacle.

He was a residential guy! LOL! The company I worked for started paying for a group instructor that was one of the CMP members. It cost them a lot of money, but he was good! I always liked taking his classes.

 

[Rjryan]

This is not going to be an easy answer, because a number of parameters have to be included:
1. Estimated load
2. Continuous or noncontinuous load
3. Total load time
4. Length the conductors
5. Number of conductors in conduit
6. Ambient temperature
7. Voltage
8. Estimated voltage drop
9. The difference in costs of wire size
10. The cost of electricity at the time of comparison

These parameters should be known or be estimated fairly accurately.
Two calculations would be necessary. You would have to determine the efficiency of both circuits to be able to have a known quantitive
difference.

It’s not going to be easy answer, but with all the variable known the results should be calculable.
In a formula you would plug the known value in an equation.

Have to take into consideration if a larger conduit is needed to accommodate larger wire.
Only considering conduit circuits calculations at this time.

My hope is someone with a knowledge in estimating and is good with mathematics could derive a formula that you could
just plug the known variables in to get an answer.
Maybe someone in estimating that has access to AI.

 

[Rjryan]

I was always fascinated how researchers trying to develop a super conductor, dropped the temperature of conductors to extremely low
temperature to achieve a conductor with no resistance. Heat that was a limiting factor on conductivity.
Lowering the temperature by increasing the wire size and decreasing resistance probably reduces electrical usage.
I worked for a large school district, if like the Electrical Inspector for Fort Worth, Texas speculated a single jump in wire size payed
for itself in 5 years and the average of a school is 50+ years, that would be a huge saving in electrical use, which translate into greener
schools. I was never able to catch the attention of the school district, because I had no definitive mathematic calculations to back me
up.
It probably would not be cost effective for all circuits, but probably very effective for services, branch feeders and circuits to large usage
items like like AC units.
Something to think about. It seems like if you have all the variables, a formula or formulas are possible.

 

[jaggedben]

I don't think anyone has mentioned the most important thing, which is the characteristics of the load. For old fashioned resistive loads larger conductors will actually increase the voltage at the load which will cause greater current to flow which will increase the overall watts drawn and thus *increase* the cost. Granted most loads these days aren't old fashioned resistive loads, but I suspect a lot of LED lamps still behave like them, while others like computer power supplies are more or less constant wattage and with others like motors the efficiency of the work they perform is a factor. I don't know how anyone could develop a rule of thumb.

 

[wwhitney]

I don't think anyone has mentioned the most important thing, which is the characteristics of the load. For old fashioned resistive loads larger conductors will actually increase the voltage at the load which will cause greater current to flow which will increase the overall watts drawn and thus *increase* the cost.

Not when they are thermostatically controlled, like an oven, a water heater, etc. As Larry observed in post #2.

Cheers, Wayne

 

[Rjryan]

I don’t understand your inability to calculate loads. You have to determine loads to size the conductor for the NEC.
I also don’t get the fact that lowering the resistance of a circuit, by size of conductors or temperature achieved by the circuit
will not make the circuit more efficient.

 

[jim dungar]

Some 30 years ago the DOE introduced rules requiring dry type transformers to be most energy efficient at 35% loading as that is what researched showed was their typical usage in schools and other public buildings. This would imply that most NEC conductors, especially feeders and services, are already oversized compared to their actual loading.

If voltage drop was increasing heat loss through conductors why isn't it typically considered when performing heat loss/gain calculations or mentioned as an energy savings practice?

 

[jaggedben]

Not when they are thermostatically controlled, like an oven, a water heater, etc. As Larry observed in post #2.

Cheers, Wayne

Another good case of a detail that makes a difference. I had more in mind lighting loads.

 

[wwhitney]

If voltage drop was increasing heat loss through conductors why isn't it typically considered when performing heat loss/gain calculations

It is sometimes--maybe not for code minimum houses, where the effect is comparatively small, but in high performance, well insulated, well air-sealed houses. And surely for things like office towers, where the extra heat gain can increase A/C demand.

Cheers, Wayne

 

[Birken Vogt]

you would keep the electric heat running longer if the voltage sags.

Another factor to consider: Where is the heat going?

If the heat loss for the electric heater is still being dissipated into the building that it heats, then the loss is zero.

 

[gene6]

If they want bigger wire fine as long as they make a stranded romex cable too. I am tired of working 12 & 10 solid what a pain that stuff is compared to stranded MC cable we use down in the city.