Saving Electicity by Upsizing the Wiring?

Here's my take (I hope I got all my arithmetic right!):

Consider a 20 Amp circuit with a 100 ft run of cable and that's under a constant 10 Amp load (that's high). You'd normally use #12 Romex cable for this run, but could upsize to #10 Romex.

Some specs we'll need:

#12-2 AWG Romex cable:
Resistivity: 1.558 Ω/1000 ft = 1.588 mΩ / ft.
source: http://en.wikipedia.org/wiki/American_wire_gauge
Cost (250+ ft): $0.82/ft
source: http://www.azpartsmaster.com/shopazp/Wire+%2D+Electrical.html

#10-2 AWG Romex cable:
Resistivity: 0.999 Ω/1000 ft = 0.999 mΩ/ft.
Cost (250+ ft): $1.33/ft

Losses
At 10Amps load, our losses are:

#12 Wire: 10² * 0.001588 Ω/ft * 100 ft = 15.9 Watts
#10 Wire: 10² * 0.000998 Ω/ft * 100 ft = 9.99 Watts

So we'd save: 15.9 - 9.99 = 5.91 Watts
Over an 8760 hour year, that's 51.8 kWH

At 10 cents/kWh, that energy costs: $5.18

Cost difference:
($1.33 - $0.82) * 100 = $51 to upsize the wire.

So your payback time is pretty much exactly 10 years, which doesn't seem like a great investment to me. In reality, the loading on the circuit is probably much lower so the losses will be lower and the payback time will be even longer.

Thanks so much Steve! I will forward your thorough calculation over to Mike, who was inquirying about the subject on greenbuildingtalk.com. Please visit the site (not selling anything) if you ever would like some advice on any kind of green construction project.

Thanks again,

Jamie

Steve,

"Here's my take (I hope I got all my arithmetic right!):"

You'll kick yourself for this one ... but you made the thing we all fear: the "stupid" obvious mistake (at least that's what I call it when I make one).

The 100' run of cable has a circuit length of 200', so the losses are twice as big, and the payback is in half the time using your 10 Amp constant load, which I agree is high for most situations.

In my area, your wire prices are about 30% high based on the web site of Lowe's Home Improvement Centers, for my zip code, and my electric costs are double your example ($.20 per kWh).

However, the 10 Amp average load is really high for the average case, and there are code issues and installation issues in how many wires of which gauge can be stuffed in the standard boxes for switches and receptacles. 10/2 with ground would be difficult if there is more than one cable going into a standard box for a receptacle, and I'm not sure it would pass the Underwriters inspection, depending on the particular box. In any case it could be a challenge to work with #10 for standard wiring in a house. Having said that, it could make make sense for some circuits in my area to upsize the wire, as expensive as copper has become lately, because of my electric costs and the trend of it only getting higher each year.

Greg

All of those changes are left as an exercise for the student.

You're right on the 100' thing, I missed that one. Thinking about it though, 100' is a long one-way run for a residentail building, so 100' round-trip may be just about right. To be honest, I used 100', 10 Amps, and 10 cents/kWh just to make the math easier (it was late, I was in a hurry). I wrote out all the calculations so that people could plug their own numbers in where appropriate.

SO why don't we use 220 volt in North America as they pretty much everywhere else in the world?

Wouldn't the current requirements drop

thinkin out loud

The 110V system came first, and at this point changing over would be hugely expensive.

Hi Steve,

It's a pity that the North America and parts of South America are stuck with the 110 V system because it makes designing equipment so much harder. I have worked for American companies for most of my career and converting things from 110 V to the 240 V we use in Australia can be a real nightmare. As I have shown earlier doubling the line voltage means you can attain the same cabling power loss with one quarter as much conductor in the cables. As a result trying to get the connectors we use onto those monstrous 110 V cables is usually an impossibility. I usually means pulling everything to bits and replacing the leads as well and on much of the equipment this can be nigh on impossible. Another hassle can be line filters as these are usually installed before any voltage selection circuitry. If the filters were not specifically designed to cope with the higher voltage the transients we see during normal operation can cause them to overload.

I know that North America also uses a split phase 220 V system but I don't know how prevalent this is. Is this split phase system used a lot or is if only used where the 110 V system is impractical?

For most general use receptacles, 10 A. is a good estimate. But, these receptacles represent the minority of the electrial use in homes, thus the return in investment time is reduced even more. Your RMS wattage loss calcs are right on, but your unit cost of electricity is low. This was a viable option in the past, but with the present price of copper it is less appealing to do this.

From the electricians viewpoint. There are usually an additional charge for the bigger wire, not because of material cost, but labor cost (around $0.07/lf).

Of course the thicker cables will have less loss but wiring Specs are also to be complied with. Switches, sockets & other fittings accept a max size, Wiring Ducts etc restrict the max size of cables. More over you have to compromise between cost & losses; & a limit reaches where increase in size gives a fintismal gain.

Most of the houses in N.America have 220V supply directly or 110+110V & hi consumption items may be connected .but here again is if you have to buy new ones; can you afford it

This is not a practical answer but increasing the supply voltage will have a much greater effect. By doubling the voltage you halve the current but the losses are proportional to the square of the current therefore reduce the losses by a factor of 4. By doubling the supply voltage you can halve the size of the conductor and still reduce the line losses by half.

So is it worth the handful of countries that still use 110 V upgrading to the 220 V distribution systems most of the world uses? Probably not but it's an interesting point though.

By doubling the voltage you halve the current

i think ive missed some thing here

V=I*R

the resistance of the wire is constant so when we double the voltage the current the current goes high too

please answer me

If we double the line voltage for a given output power then the current is halved. Since the power being lost in the cable is the current squared divided by the resistance then the power lost in the cable is only a quarter of the original.

If we wish to keep the line losses at the same level we can increase the resistance of the cable by a factor of 4 and that means the use of a quarter of the material. Therefore by doubling the voltage and halving the size of the cable we still end up halving the power lost in the line.

This is precisely why power transmission lines use incredibly high voltages. Step the voltage up from 110 V to 110,000 V and you drop the amount of power lost in the transmission by a factor of 10⁶.

These equations were accidentally left out of the previous post

V = I x R

&

P = V x I

.˙.

P = I² x R

Yes you have missed something here.

You are talking about the load restaiance - P = VI, if the resistance of the load stays the same - only the current can change which is then dependent upon the voltage.

Therefore if you increase the voltage to the load then its current consumption will halve.

This thread is talking about power losses in the conductor supplying the load - this is measured by

I squared R (I²R) where R is the resistance of the conductor. So if you increase the size of the conductor the resistance will be less and the power loss will be reduced.

The proposal is correct in theory but not worth while in a domestic practical application as the capital cost would never be recovered. It makes a little more sense in the USA where the voltage is 110v but not here in Europe where the voltage is 230v

Here in the US, we typically use 110 volts for most application. We only go to 220 volts for high power applications like clothes dryers, heaters, ovens,... Things like that.

I wonder Masu, how is the job market in Australia? I would dearly love to find a job in which I would spend April through September in the US, then October through March in Australia. In other words... PERPETUAL SUMMER!!!

Sincerely

Bill

Hi Bill,

I wonder Masu, how is the job market in Australia? I would dearly love to find a job in which I would spend April through September in the US, then October through March in Australia. In other words... PERPETUAL SUMMER!!!

Over the past few decades we have been subjected to way too many accountants with short sighted gains in mind. As a result nobody bothered to employee and cadets, trainees and apprentices and we are now suffering from a catastrophic shortage of skilled labor. There is a huge skilled workers migration program in place and if you can speak English and have a skill then you can pretty much get a job in Australia. Unfortunately the bean counters are still in charge and have learnt diddley squat by their mistakes so there is still precious little going on in the training department.

As for the perpetual summer, I had a had a job that based me in Adelaide but had about half of my work in Darwin. With a little creative scheduling I managed to avoid the bad weather in both places for about 4 years. It was a great job but having to commute the 3,000 Km from Adelaide to Darwin eventually got to be a bid of a drag.

Hi too..

Yes it's true because when there's more small sizes of wiring in the house, the more resistance of wires they have, and the more the resistance, the more the electric current cosumed.

I am involved in the electrcal / building field.

Trust me: Residential wiring is generally so oversized that, while there is a small
theoretical saving involved in upsizing your wiring, the costs & complcations,
would far outweigh the benefits. - An upgrade would be a major undertaking.
Some of the comments above address material costs without consideration for labor,
(a BIG mistake).

You get a far better cost/benefit and energy saving by converting to fluorescent lighting,
(those screw in modules). - A cheap & easy fix.

They also lower your wiring losses due to the decrease in load, even allowing for
the harmonics in the neutral.

Pragmatist: Good point. Converting lighting to compact flourescent lamps (CFL) and turning them off when you're not using them is much more cost effective.

In my analysis above I was implicitly assuming that this was for a new wiring installation. There's absolutely no possible way that re-wiring a house would be cost effective!

All of you have made excellent points, however I do not know of an electrical component manufacture that is making switches or duplex outlet that will accept AWG#10 wire let alone the ape that would be strong enough to bend the end around the screw terminal.

As to the skilled labor issue, it is not just "Down Under", it is world wide. All of the trade magazines (ie Plant Engineering, Maintenance Technology, Reliable Maintenance etc.) are chanting the same mantra "NO SKILLED LABOR FORCE IN FIVE TO TEN YEARS" and the trade schools are shutting down due to lack of interest. We will all be in a wage and benefits biding war for the few that remain by 2012 - 2017.

I bid you peace, Deepfreezssailor.

The responses so far have focused on the costs and payback time for upsizing home wiring. While these are always important considerations, what impact on power use does upsizing have? In orther words, does upsizing reduce power consumption, and thereby pollution created by power sources? Of course, one should also include the cost of producing the additional copper, etc. in an attempt to "calculate" the total environmental impact - i.e. contribution to global warming. And let's not forget the common pollution problems created by power production. Acid rain has caused considerable damage in the NE United States.

Also, ten years is not a long payback period. One often lives in a house for many more years than that. This raises the question of how to calcuate the return on investment when constructing more environmentaly kinder and gentler homes.

Using only payback period to ignores market forces invovled in the valuation of real property. A better model is the calculation one makes when renovating. FOr example, in the NE US one considers using granite counters when renovating a kitchen. It is very expensive, but one gets to enjoy them while living in the home, and it increases resale value - to the extent that there is usually a possative return on the initial investment.

One should be able to invest in energy efficiency consturuction, recover a portion of the up-front costs while living in the home (lower energy costs) and pass on uncaptured costs to the buyer. This works, of course, only when buyers place a value on energy efficiency.

While these are always important considerations, what impact on power use does upsizing have?

Lets look at this in a little and with purely resistive loads for the moment. Lets assume that we have a 120 V supply voltage driving a 118 Ω load with a 2Ω resistance in the cabling in the house. The current would therefore be

I = V_Supply ÷ (R_Load + R_Cabling) = 120 ÷ (118 + 2) = 1 Amp

Power = V x I = 120 Watts

Now lets look what happens if we double the size of the cables and reduce the resistance by 50%. Remember we cant change the supply voltage so

I = V_Supply ÷ (R_Load + R_Cabling) = 120 ÷ (118 + 1) = 1.008 Amp

Power = V x I = 121 Watts

Now this is a simplified situation and only looks at the load in a purely resistive manner but by reducing the resistance in the cables by 50% we have ended up increasing the amount of power consumed by 1% and that is not what we set out to do. So unless you can re-tap the last transformer and reduce the mains supply voltage to accommodate the reduced resistance you could end up using more power.

A typical home owner in the US doesn't stay in his home for 10 years. Usually, they buy, then sell and move up or move on. So, they are looking to minimize the initial cost, which must be paid for upfront, and the interest cost, which is amortized over the mortgage period. So, unless it's required by local codes, most will never consider adding to the cost for something they'll never see as savings.

When calculating 'payback` it is a reality that 'costs` are financed.
This means that approx. 7% of the cost is continuing interest that must
be overcome before any 'payback` is realized.
Financially, increasing the wire size will, therefore, never show a benefit.

As to enviornmental gain only consideration, we must also consider the
enviornmental cost of the extra copper itself, (sulfur emission, energy, etc.,
not large but real).

WAG.: Some small enviornmental gain might be realized by the increase of wiring
for 'large, constant load` loads such as refigerator or window Air Conditioner circuits,
but not for general lighting & receptacle circuits which are mostly in the
'they also serve who only stand and wait` category.

Even heavy load appliances like refrigerators and air conditioners aren't as "heavy" as you would think. According to the Department of Energy, a typical 16 cu. ft. refrigerator draws 750 Watts, or about 6.25 Amps from a 120 volt supply. That refrigerator is also only "on" about 50% of the time (source: EnergyIdeas.org). So its net average current is about 3 amps. You can plug that into the analysis at the top of this thread to see what savings you'll get - it's insignificant. Spend your money on compact flourescent lamps, better insulation, and more efficient appliances instead.

CR4 Admin: removed broken link

I agree that the gains would be very small, but I remind you that I was comparing
them 'enviornmentally only` to the one time cost of the extra copper.

The question I was answering was as to enviornmental concern above & beyond
the financial practicalities which were addressed in the first paragraph..

I understand. My point was that if you want to do something to help your environmental footprint, there are much more effective ways to spend your money.

Well put. Modern 500 litre (>18cft) fridges draw as little as 200 watts. Forget the wire change the fridge and other old inefficient appliances at their use by date with better items.

For those obsessed with wire (nor me after 34 years) get your hands on a cable selection book from one of the cable manufacturers such as Pirelli. In this book you will find the figures for each type of cable configuration (Single core double insulated ((SDI)), flat twin, three core circular etc, etc.) in each size. The most useful figure in this case is the voltage drop per ampere metre (foot). To get the accuracy you need for decision making some equations accounting for the length to each outlet or appliance on the run. Now for the current to choose, an arbitary 10 amps or socket maximum capacity is a waste of time. You will need to either do a maximum demand calculation or somehow actually measure consumption.

For long cable runs we always calculate the current requirement and look up the size that complies with voltage drop requirements of either legislation or the sensitivity of the power consuming device to voltage drop. A good example is running a cable to someone's shed. Always assume they will one day want to weld there (without frustration). The primary current on 240v for a smallish welder can be assumed at 28 amps at full noise. In some instances 2.5 sqmm is rated at 28 amps before derating factors, but voltage drop is the killer, so run a minimum of 4 sqmm.

Where cable size really matters is for solar and other extra low voltage applications. A voltage drop of 5% is allowed at 240v ie. 12v, try running your 12v or 24v system with 12v voltage drop.

Where cable size really matters is for solar and other extra low voltage applications. A voltage drop of 5% is allowed at 240v ie. 12v, try running your 12v or 24v system with 12v voltage drop.

A simple rule for checking cabling size is multiply the diameter of the cable by the factor you lower the voltage. In other words half the voltage means twice the diameter one twentieth means twenty times the diameter.

The rule works the other ways too and gives you a feeling for why it is so desirable to have transmission line voltages ahs high as possible. I you increase the voltage by a factor of 3,000 you only need a conductor 1/3,0000^th the diameter to maintain the same losses in the cable.

Now before everybody jumps in at the deep end this is only for guesstimation purposes and not meant to replace the proper voltage drop and power dissipation calculations.

Thanks for the rule of thumb. Doubling diameter equals quadrupling the copper content. Even 240 has huge advantages in environmental efficiency over 110 as you pointed out earlier. Luckily I only use 110 for control (then it's centre tapped to comply with ELV touch potentials, in fact now 55v to ground ((ELV)), earth leakage protection, IP2x and door interlocks are all getting integrated into the package especially underground).

I remember some of the old European stuff that was floating around in the 70s with single strand conductors around 4 sqmm, about the only advantage is it stayed where it was put. Don't know if skin effect was the total culprit but the insulation used to just fall off. I never did any temperature checks just rewired it with decent cable.

I have spent most of my working life employed by US based companies and some of the stuff that they try and run of 110 V is ridiculous. There was a computer system I worked on a few years back that pulled 15 A at 240V on three phases. Because of the way the power supplies was configured it could not be wired to use phase to phase voltages so that meant in North America they couldn't use their 220 V split phase or 190 V phase to phase systems. As a result the power cable to the thing was about as thick as my wrist so you can imagine the hassle trying to get a Clipsal 20 A three phase 5 pin plug on the end of it. Talk about trying to thread a camel through the eye of a needle.

We ended up replacing all the power cables then sold the original cables for scrap and put the proceeds in the Christmas party fund. We had one mother of a Christmas party that year!

How is the water going up in Queensland, have they drowned all the people that are complaining about the new dam yet or is Brisbane going to run out of water like Sydney? There is a plan to build an additional dam and reservoir for Sydney in the Hunter Valley that has been on the drawing board since the 1950s. All the people that purchased property in the associated area were told clearly and had to agree that when the dam was required it would be necessary for them to sell their properties back to the government. Now they are winging because the government is talking about building a dam there.

When Warragamba Dam was completed in the late 1950s they said that Sydney's water supply was assured till the turn of the century and that there was plenty of time to develop strategies. Well guess what the turn of the century came and went and nothing was done. As per usual everybody sat on their posteriors for the best part of half a century and Sydney is now running out of water. Didn't they shoot people for stuff ups less than this not that long ago?

You've summed it up well on all fronts there mate.

How's this for a high current low voltage doozy. A system I have had to bring under control uses a VVVF powering a 30kw 3 phase squirrel cage motor to power the cooling fans. The Motor started life as 460vac somewhere in GE's parts bin. The alternator at idle is putting out about 444vac at about 27hz and at full noise it runs over 1000vac and 63 hz. Obviously enough the 444 won't give enough DC link voltage to successfully run the 460vac motor at engine idle without something like a buck converter. So what do they do? Do they source a 440 or 415 vac motor? Nope, they delta connect the motor for 270 vac and set the DC link at 390 then 410 when that doesn't work. Alright for a one off? No again this is a production item.

Of course the Thyristors and IGBTs have to be rated for the full voltage, so by not selecting the correct motor they also have to rate for current.

If the US ever wakes up on voltage levels among other anachronisms the copper price will plummet.

Guys you are all forgetting an important fact. The current that you are being billed for is limited by the sum of the load impedance and cabling impedance. If you reduce the cable impedance then the overall impedance will reduce ant the current will increase. Put mathematically

Power = V² / R

So since you have no control over the voltage that you are supplied with a reduction of the total impedance you will increase the amount of power you consume and pay for. To be able to reduce the energy loss by increasing the cable thickness you also need to reduce the mains voltage to maintain the same current through the load.

This is a completely bogus idea. By increasing the cable size you will not only have the added cost of the materials but you will also consume more electrical energy and increase operating costs. You loose both ways.

In the case of a purely resistive load such as lighting (not compact flourescent lamps), you have a point. However many of the largest household loads aren't resistive, they're constant power loads, or nearly so. As such they look like a negative impedance. As the voltage at the terminals of the load drops due to IR losses in the wiring, the current drawn by the load increases.

In the case of a motor running at constant power (e.g., a constant speed blower, like in your heat/AC system), a drop in the motor terminal voltage causes the current drawn to increase. Your refrigerator compressor is similar. It uses a low slip induction motor (for high efficiency) that runs at very nearly constant speed, no matter what the input voltage - it'll just draw more current if it needs to. That's why motor overheating is a potential problem during power system brownouts.

Likewise, the computer I'm typing this on (like many electronic devices) uses a switching power supply that draws constant power from the line. As the input voltage drops due to cable IR losses, the supply just increases its duty cycle to keep the logic bus voltages constant, so it draws more current, not less.

So, I don't buy the "consume more electrical energy" argument, although I will concede that it depends on the balance of types of load you have in your house.

Hi Masu,

"This is a completely bogus idea. By increasing the cable size you will not only have the added cost of the materials but you will also consume more electrical energy and increase operating costs. You loose both ways."

I have to disagree to a somewhat.

Reducing voltage drop (increasing the voltage to the load) makes most appliances more efficient. Motors run cooler, air conditioners cool faster (than on a too low voltage) toasters will toast your bread at a lower setting (less time on), heaters of all kinds will be on less (with the possible exception of a few things like hairdryers). Microwave ovens are also particularly sensitive to voltage changes.

A part of it has to do with your incoming voltage also. I am at the end of the run from a utility transformer and my voltage tends to be on the low side. The guy whose house sits adjacent to the transformer runs on the high side. Lessening voltage drop in his lines would put him even higher which isn't good either for many devices. Incandescent light bulbs, electric range elements etc, will have considerably reduced lifetimes if run at full line voltage.

I think there is a middle ground here.

If you are running a line to a high current draw load(s) a larger cable could well make good sense. If you have a kitchen and an adjacent main bathroom (common for plumbing reasons) it might well make sense to install a sub-panel nearby fed from 240 VAC and split off the 120 V legs there. The AC common running between the sub-panel and main panel will typically only carry 1/2 or less of the combined current from both legs (If the loads are actually balanced, it carries no current). We have a "problem" here in the states with 120 VAC when it comes to kitchens. During meal prep time, you need a number of circuits if you don't want to have to watch what you have on at the same time or where its plugged in (toaster, microwave, electric coffee maker, toaster oven, refrigerator etc). The rest of the time, except for the refrigerator they are hardly used and that has to added in to the equation.

In most cases it doesn't make sense to increase wire size and as so many have said, focus on energy saving bulbs, appliances etc.

Like so many things, it comes down to particulars and common sense.

Regards, Greg

Hi Steve & Greg G,

So, I don't buy the "consume more electrical energy" argument, although I will concede that it depends on the balance of types of load you have in your house.

I agree, and the calculations I did in an earlier post were as I stated for a purely resistive load. I restated it in less qualified terms because it was apparent that few had realized the implications of my earlier post.

I just did a quick check and we have 8 240 V 50 Hz single phase circuits as follows

32 A Oven and range top with purely resistive load

20 A Hot water heater with purely resistive load

16 A General purpose outlets max load item 10 A for heating

16 A General purpose outlets max load item 10 A for heating

16 A General purpose outlets max load item 10 A for heating

16 A General purpose outlets

10 A Lighting about 50% of draw is incandescent

10 A Lighting about 50% of draw is incandescent

So the biggest current gain items are resistive heating elements and I would guess that roughly 66% of the energy consumed would go to these loads. Even if it were as low as 50% any gains in the power going to regulated items would pretty much be cancelled out by the increase in the power going to the resistive loads.

In reality if you weren't using something like a air conditioning system or large freezers etc. I doubt if you would see a reduction in your bill and you may end up actually paying more.

Masu,

That P=V>2/R is for resistance loads.

Most larger of the larger constant household loads are inductive,

i.e. air conditioners, well pumps, refrigerators.

(I do assume that electrical heating & cooking, a financial disaster in

most areas, is not a factor).

In our house is the stove/oven followed closely by the water heater are the heaviest loads and these are both purely resistive loads. The next biggest single culprits are the heaters which are also resistive items.

Now admittedly this will vary depending what you use as a primary heat source and whether of not gas is available for cooking and to heat water but even a single electric oven can draw a huge amount of power compared to the compressor in a refrigerator. In the USA where you are using 120 V supply compared to the 240 V supply the losses due to the cabling is going to be four times the problem that it is in Australia.

Ultimately though we can break the load up into three groups

PURELY RESISTIVE these are thing like stoves, ovens, water heaters, electric space heaters etc.

REGULATED LOADS This group include things like TVs, radios, computers and other anything that has a regulated power supply.

INDUCTIVE LOADS This group contains all the things that have a primarily inductive load like compressors motors, refrigerators, fans, pumps etc.

Now the purely resistive load will end up drawing more current with the upgraded cabling as the reduced voltage drop in the cable will result in a higher voltage at the appliance, hence greater current through the resistive load. Now this group contains the items that if used would draw the greatest currents.

The regulated load group will can be divided into two sub groups. The will either have series pass or switch mode power supplies. The switch mode power supplies will draw less power while the series pass will draw slightly more. The thing is though that the items with regulated power supplies usually draw the least power of the three groups and ultimately any increase in the size of cabling would most likely have little to no net effect.

Finally we have the inductive load group that contains items like electric motors that will actually draw more current at lower voltages. This group is where the gains are made as the increased cable size will increase the voltage to the motors and that results in a decrease in the current. This group however usually doesn't draw the high currents that you see with the purely resistive loads and for most items the reduced current draw would be less than the increase to the items in the first group.

All up it depends on the balance between the purely resistive and inductive loads, if you have more purely resistive loads than inductive then you will end up using more power. If however the opposite is true and the inductive items are the greatest load then you will use less power..

Ultimately when you add it all up increasing the size of the cabling in a domestic environment is more than likely going to have little appreciable effect on the total amount of energy consumed. The gains in one area would likely be cancelled by losses in another and even then since most of the world uses 220-240 V any effect would be ¼ that of those in North America where you use a 110-120 V supply. This would tend to indicate that increasing the size of the cabling would most likely be a waste of time, money, effort and energy.

Hi Masu,

You are conveniently leaving out of your equation the fact that all the resistive loads you mention, by consuming more power, will simply be required to be on less time. (Ex: The kWh required to heat a given volume of water to a set temperature remains constant.) The total power consumption of the devices themselves would therefore be essentially equal, whether they consumed less power for a longer period of time, or more power for a shorter period, and only the cabling losses would actually come into play. I know full well you know this so I suspect you are having some fun here.

Regards, Greg

Hi Greg,

I thought about the time base part of it shortly after I entered the last post. In effect items that are controlled by a thermostat are actually working as a switch mode regulated power supply and would therefore draw the higher currents for a shorter amount of time. However not all of them are thermostat regulated, for example stove top elements and which would generally be wasted the extra power.

Lets have a look at the actual figures for my house where we have a 240 V 50 Hz AC supply to a 17 room all electric, house occupied by 6 adults and 2 children in Sydney Australia.

First of we pay various rates for electricity depending on when the power is consumed and how much is consumed but over a period of 12 months we used 23,927 Kwh for which we paid AU$2,468.03 or 10.31 cents per Kwh

As I stated in post #31 we have 8 circuits that have various ratings. Lets look at the 16 A circuits and assume that the power is distributed on a circuit rating basis. That means that the 4 16 A circuits would each draw roughly 12% of the total power over a year. These 16 A circuits use a 4.5 mm² copper cable and are roughly 30 m in length

Cable_Resistance = 2 x ρ x Cable_Length ÷ Cable_Area

Cable_Resistance = 2 x 1.72 x 10^-8 x 30 ÷ 0.0045

Cable_Resistance = 2.933 x 10^-4

Cable_Resistance ≈ 300 μΩ

Now given the power over the year and dividing by the number of hours in the year will give us the average current draw which can then be split to look at a particular circuit as follows

Cable_Current = Power_Annual ÷ Hours_Consumed x Circuit_Ratio

Cable_Current = 23,927 Kwh ÷ 8,760 h x 12%

Cable_Current = 0.3277 A ≈ 330 ma

Now calculating the amount of energy lost in the cabling each hour gives us

Cable_Power = Cable_Current ² x Cable_Resistance

Cable_Power = 330 ma² x 300 μΩ = 0.00003267 W

Cable_Power ≈ 33 μw

So by multiplying by the number of hours in a year the cost of electricity and the circuit current ratio we will know how much it costs for the power lost in all the cabling each year

Cable_PowerCost = Cable_Power x Time_Used x Power_Cost ÷ Circuit_Ratio

Cable_PowerCost = 33 μw x 8,760 h x 10.31¢ ÷ 12% = 24.40317

Cable_PowerCost ≈ AU$0.25 per year

So if we doubled the size of the cable we would effectively halve this cost which would mean a phenomenal saving of 13 cents a year which isn't great and that's before we take any of the negative aspects into account. Now admittedly we use the highest mains voltage in the world and the savings in North America would be about five times those in Australia but it still isn't a great saving.

I don't know about you but even if I managed to reduce the cable losses by 100% it would take one hell of a long time to pay back the additional costs so I don't think I will be worrying about upgrading the cabling any time soon. Actually after seeing this I am seriously thinking looking for a way to downgrade the cabling and sell the excess copper as scrap.

Hi Masu,

"However not all of them are thermostat regulated, for example stove top elements and which would generally be wasted the extra power."

Not the way I cook, which is by pan/pot temperature. If boiling a pot of water on high, it would take less time, so in both cases it would be like the switching mode regulated power supply.

I am not saying that increasing cable size is practical generally, especially in a case like yours where you have 220-240 VAC per circuit and the savings are so very small.

However, you minimized them quite a bit by your method of calculation. By averaging out the power over the year, and dividing it on the basis of the circuit ampacities you probably under calculated by a factor of four at least. (But whether its 13 cents, or 52 cents, or $1.04, or even $5 a year matters not at all really, because it is still such a trivial amount).

Let's take the 20 Amp water heater cable for instance. If the heater is on 3 hrs a day, and draws 15 Amps, that would average out to:

45 amp * hrs / 24 hrs or, 1.875 Amps average.

Since cable losses are I²*R, and R is constant we can ignore R for now. If we calculate losses (15² / hr for 3 hrs) before averaging, we get:

225*3 hrs/24 hrs = 28.125 average I² losses,

but if we do it after it has been averaged, as you did, we get:

1.875² = 3.516 average I² losses, which is a factor of 8 times lower!

In any case, I think we both agree that under no circumstances would it pay for you to upsize your cabling, and that even here in the the U.S., its unlikely to pay except in certain relatively rare circumstances.

Regards, Greg

Please disregard the last post I was just re-reading it an realized it all crap so forget it and I will start over

We have a 17 room house occupied by 6 adults and 2 children and we consumed 23,927 Kwh at a cost of AU$2,468.03 which is in average 10.31¢ per Kwh.

The part about the 8 circuits as posted in #31 is correct so if we divide the power based on the circuit rating the 16 A circuits will each draw 12% of the total power. The 16 A circuits are roughly 30 m long and made of 4.5 mm² copper therefore the resistance of each circuit is

Resistance_Circuit = 2 x Conductivity_Copper x Cable_Length ÷ Cable_Area

Resistance_Circuit = 2 x 1.72 x 10^-8 x 30 ÷ 4.5 x 10^-6

Resistance_Cable = 22.93 x 10^-2

Resistance_Circuit ≈ 230 mΩ

The error in the previous calculation was in the cross sectional area as 4.5 mm² needs to be multiplied by 10^-6 and not 10^-3 as I did.

Now given that the circuit consumes 12% of the overall energy at 240 V over a year the average current flowing down the circuit is

Power_Circiut = Power_Total x Ratio_Circuit

Power_Circiut = 23,927 Kwh x 12%

Power_Circiut = 2,871.24 Kwh

Power_Circiut ≈ 2,900 Kwh

We can only have a guess at the mean current as the rate the power is consumed will directly effect this so we will need to take a guess and say that the duty cycle is roughly 50% so the mean current over a year would be

Current_Circuit = Power_Circuit ÷ Time_Consumed ÷ Voltage_Mains

Current_Circuit = 2,900 Kwh ÷ 4,380 hours ÷ 240 V

Current_Circuit = 2.75 A

Therefore we can now calculate the power that would be lost in each circuit as followes

PowerLoss_Circuit = Current_Circuit² x Resistance_Circuit

PowerLoss_Circuit = 2.75² A x 230 mΩ

PowerLoss_Circuit = 1.739357

PowerLoss_Circuit = 1.74 watts

We can now calculate the cost of the energy that is lost in each circuit as

Cost_Circuit = PowerLoss_Circuit x Time_Consumed x Price_Electricity

Cost_Circuit = 1.74 w x 4,380 h x 10.31¢

Cost_Circuit = 78.57¢

And if we now scale this with the factor we calculated earlier for the power going to each circuit we get

Cost_Total = Cost_Circuit ÷ Ratio_Circuit

Cost_Total = 78.57¢ ÷ 12%

Cost_Total ≈ $6.55

Now while this is around 20 times my original calculation and because of the considerably higher voltage will be around a fifth what you would get for the same consumption profile in North America it is still not a great expense compared to the total outlay for electricity and cabling. If you doubled the diameter of the cable you would only be able to save AU$3.28 a year I doubt it would be worth installing thicker cables in a new house let alone replacing existing cabling.

All this is before allowing for the negative aspects I have mentioned in other posts but the figures are taken from real live number, for a real house, for 2006 in Sydney Australia. While Australia has the highest mains voltage of any country and this reduces the losses considerably the savings that could be achieved are too small compared with the overall cost of electricity and of upgrading the cabling to make the concept viable.

So sorry for the incorrect calculations in the previous post, I don't know what I was thinking about at the time, it must have been the screaming kids running around that distracted me.

Hello

I have answered your Post/comments,

Generally best to use the Cable Maker of your particular cable, but try:

http://relemaccables.com/

If you are still needing help, reply with

Kind Regards....

I can add some "practical" aspects to the discussion since I had to rewire my house last year. I live on the NE corner of Indy in a private community.

Place was built in 1957-58 with a later add-on that used Aluminum wire. Horror story, but the botom line was about 7500$ total. Work was done by a licenced electrician and a helper.

No way you amortize that!! And it's cheap compared to say NY City or LI.

What I did invest in was a Power Line EMI filter on the 220V Mains with an active (light) indicator and Lighting arrestor. Now I wish I'd had a Transfer swith put in, but who knew those little generators would grow-up & run on gas? Wonder if some one would complain about a Wind generator? That would be cool!

As far as know you have the benefit of selling the excess generator power back to the utility. Thats immensly cool!!!!!! I'm retiring some time, so I buy a farm, rent the fields out and put up a 25KW windmill. Electric is free to me and I get an income from the excess! Plus Field rental and SSI.

Question, How large would a 25KW Windmill be? Who make them? What would it cost?

Don't you guys think we've beat this thing pretty much to death by now

Time to move on.