As a test case, consider a 100-hp 575V induction motor, running continuously, connected by 250m of cable installed in cable tray by trades who work for a rate of $100/ hr. Where electricity purchased at a rate of $0.10/kWh, a 3C#2/0 AWG cable (instead of a CE Code-recommended 3C#1/0 AWG) could pay for itself in as little as 5.3 years.
The U.S. Energy Policy Act (EPAct) of 1992 requires 1-hp to 200-hp general-purpose motors manufactured or imported for sale in the U.S. to meet minimum federally mandated efficiency levels. The 100-hp motor described above, running at 100 hp output consumes 89.7A at 575V, power factor 87.5% and is 95.5% efficient. A typical 1990s electric motor of the same size was about 89% efficient. Since 1997, energy-efficient motors have been mandated to be at least 93.6% efficient.
Thus, PF (power factor) = |cos φ|, where φ is the apparent power phase angle. The real power P in watts W is equal to the apparent power |S| in volt-amperes (VA) multiplied by PF.
|S(VA)| • PF = P(W)
(√3 • VL-L(V) • I(A)) x PF = P(W)
(√3 • 575 • 89.7 • 0 .875 ) = 78168 W
So for a year of continuous operation at $0.10/kWh, the 100-hp motor consumes:
(365.25 • 24 • 78.168 • 0.1) = $68,500
of electricity per year.
It becomes apparent that a motor costing $10,000 consumes almost 7x its purchase price in energy each year.
Cost of ownership has been carefully considered when it comes to large power transformers. IEEE C57.120 “Loss evaluation guide for power transformers and reactors” details methods for establishing the electric power used to supply the losses of a transformer or reactor. Users of this standard can determine the relative economic bene t of a high-first-cost, low-loss unit versus one with a lower-first-cost and higher losses.
This standard was first published more than 20 years ago, which supports the validity of examining the total cost of ownership (TCO). While transformer loss evaluation is at the discretion of the purchaser, motor loss prevention has been mandated, regardless of whether the user’s motor runs day in and day out, or rarely (e.g. fire pump motor).
Let’s examine the typical 3-5% voltage drop on supply wiring and evaluate opportunities to recover some of those losses.
Copper (and aluminum) conductors have resistance. Thicker conductors have lower resistance than thinner ones. Copper that is cool has lower resistance than when it is warm. An identical current passing through a thinner wire wastes more energy in heat than its thicker counterpart. To make matters worse, the additional dissipated energy causes the wire to increase in temperature resulting in a further increase in resistance, resulting in more heat until an equilibrium is reached. This is referred to as the steady-state operating temperature.
Traditional thinking guides the design engineer to choose wire sizes to limit the voltage drop (the difference in voltage from supply to load) to 3% for a specific device’s load cable, and an additional 2% for upstream equipment that lies between the utility connection and the break-out point for individual loads.
This article examines the TCO of wiring that supplies loads. To do that, let’s examine delivered power. Induction motors are constant power devices. Unlike a resistive load, an induction motor increases current draw when terminal voltage drops to maintain constant power output.
For this analysis—to ensure the motor delivers the same amount of useful work in each evaluated scenario—calculations have been normalized for losses and benefits.
Our test case compares copper conductors (#1/0 AWG to #4/0 AWG) supplying a 575V, 100-hp induction motor. The modelled test case uses 250m of 3-conductor Teck 90 aluminum-armoured copper cable installed in cable tray and operated at a 30°C ambient. At this operating current, 250m of cable exhibits slightly less than 3% voltage drop using standard 65°C voltage drop tables. In actual fact, equilibrium steady-state running temperatures are lower than 65°C, and have been calculated for each cable type and included in the evaluation.
The base case analysis is then subjected to adjustment of the assumptions to determine the sensitivity of outcome as a result of the change in assumption.
Sensitivities analyzed include electricity cost, cable cost, cable length, operating hours, termination hardware cost, cable tray fill, carbon tax and labour costs.
A summary of the anticipated advantages and disadvantages of installing larger-than-normal electrical cables is presented below.
To ensure that a sweet spot analysis was not presented, each cost was increased by a factor of two, then decreased by a factor of two, and analyzed to identify the sensitivity to each. Therefore, where electricity costs $0.20/kWh instead of $0.05/kWh, the user can determine how the difference will affect the outcome.
The evaluation is a 100-hp, 575V, 60Hz 3-phase induction motor connected with 250m of 3C#1/0 Teck cable in 24-in. aluminum ladder tray, using compression lugs and teck connectors sized appropriately at each end. The considered cable tray is 240- m in length, and has 4 horizontal elbows and 3 vertical elbows (approximating a typical installation).
To ensure that the same amount of useful work is delivered by the motor in each calculated scenario—despite the argument that the motor is already a constant-power device—the losses and benefits were based on constant motor terminal voltage.The constant terminal voltage approach is conservative for the purpose of determining time to break even, because decreased motor terminal voltage results in increased supply cable current, which results in increased I2R losses in the cable and motor. This is further compounded by higher amperage, causing increased copper conductor temperatures, thereby increasing resistance and increasing losses. So the results obtained are conservative!
Conductor operating temperatures in this examination have been calculated using 30°C free air operation. A cable in free air will run cooler than it would in a random-filled tray. An increase in cable temperature using random- filled installation further enhances quicker payback. Thus, the free air assumption is conservative for the purposes of this evaluation.
Motor data was supplied courtesy of Teco Westinghouse and cable temperature rise calculations have been performed courtesy of Nexans. Other factors that have considered include:
• It costs more to install larger copper cables because they are bigger/heavier.
• It takes longer to terminate them.
• The termination hardware costs more.
• Fewer cables fit into a similar-sized tray.
The total cost to install 3C#1/0 cable ($TIC0) is the sum:
• cost of 250m of 3C#1/0 Teck 90 cable,
• labour cost to install 250m of 3C#1/0 cable,
• cost of 2 teck connectors to t 3C#1/0 cable,
• labour cost to install 2 size 3C#1/0 teck connectors,
• cost of 6 compression lugs for #1/0 conductor,
• labour cost to install 6 compression lugs on #1/0 conductors,
• 1/56th* the cost of 240m of a 24-in. cable tray system, and
• 1/56th* the cost of labour to install the 240m tray system
* Accounting for differing cable tray capacity has been accomplished by determining the quantity of each cable type that could be installed in a 24-in. cable tray, and using the appropriate proportional cost assigned to each cable’s total installed cost (TIC).
Figure 1 is a graphical representation illustrating various cable/tray relationships.
$TIC00 is the total installed cost of a 3C#2/0 cable, terminations and 1/55th the cost of the tray
$TIC000 is the total installed cost of a 3C#3/0 cable, terminations and 1/54th the cost of the tray
$TIC0000 is the total installed cost of a 3C#4/0 cable, terminations and 1/48th of the cost of the tray
For interest, some consultants propose minimizing cable size to cut capital costs. Using smaller cables allow more to be put in each tray. So just how long do they save us money?
$TIC1 is the total installed cost of a 3C#1 cable, terminations and 1/75th of the cost of the tray
$TIC2 is the total installed cost of a 3C#2 cable, terminations and 1/85th of the cost of the tray
The upfront cost ($UFC) that must be recovered to offset the TIC differences between 3C#1/0 and 3C#2/0 cables are:
$UFC0-00 = $TIC00 - $TIC0
And, similarly, to upgrade from #1/0 to #3/0
$UFC0-000 = $TIC000 - $TIC0
and from 3C#1/0 to 3C#4/0
$UFC0-0000 = $TIC0000 - $TIC0
In the case of downsizing cables to save capital costs (Installed Cost Savings), does running more, smaller cables in a cable tray result in permanent savings, or does the choice have repercussions after a calculable period?
$ICS0-#1 = $TIC0 - $TIC#1
$ICS0-#2 = $TIC0 - $TIC#2
The calculation of the payback period to recover capital costs:
View the embedded image gallery online at:
Solving for Y0-00 (which represents recovery time in years):
View the embedded image gallery online at:
D is the per unit value of 1 year or 8766 hours. (Running 50% of the time D would be 0.5)
$/kWh is the cost of electricity energy
WL0 is the number of watts lost in I2R losses of a 3C#1/0 cable 250m long, operating at 87.73A.
WL00 is the number of watts lost in I2R losses of a 3C#2/0 cable 250m long, operating at 87.73A.
(WL0 - WL00) / 1000 is the difference in kW (I2R losses) between a 3C#1/0 and a 3C#2/0 cable,
each 250m long operating at 87.73A.
GL is the average percentage of electricity that is lost in the AIS transmission system.
1 + GL is the ratio of electricity produced to the amount received by the end user.
TonC / kWh is the average amount of CO2 emissions in tons/kWh generated based on the heat ratio
of the source generator.
$ / TonC is an empirical value of a tax that could be levied against end users of fossil fuel energy
(i.e. carbon tax). Rates of $15 per metric ton of CO2 produced are typical values.
Results: upsizing from the base case
Using the base case, 100-hp motor running continuously—with no carbon tax—using 250m of cable in 24-in. cable tray, installed by trades at the rate of $100/hr, the cost equalization of upsizing from #1/0 AWG conductors to #2/0 AWG conductors is 5.36 years.
After the additional capital cost has been recovered, each subsequent year’s operating costs is reduced by $399 per year.
Payback period Upsize After payback recovery
Y0-00 = 5.36 years 3c#1/0 AWG to 3c#2/0 AWG $399/year
Y0-000 = 5.85 years 3c#1/0 AWG to 3c#3/0 AWG $711/year
Y0-0000 = 6.74 years 3c#1/0 AWG to 3c#4/0 AWG $954/year
Results: downsizing from the base case
In an effort to reduce capital costs, many variances have been written on mega-projects to reduce cable size by eliminating de-rating factors and stretching allowable voltage drops. Using the same methodology, changing the cable from the base case 3C#1/0 AWG to 3C#1 AWG results in savings for only the first 3.75 years. After which, increased energy cost is continuously paid over the life of the installation.
3.75 years is a conservative analysis, as the mean temperature of a tray of reduced sized conductors operates warmer than a tray of increased-size conductors. As well, no value has been deduced for reduced cable life, decreased cable longevity of surrounding cables or carbon tax costs. Decreasing cable size from #1/0 to #1 costs an additional $517 per year in electricity.
Downsizing even further to 3C#2 AWG results in energy cost overtaking savings in 3.72 years. Decreasing to #2 costs $1192 extra in electricity per year.
Not every situation is the same. Motors run less, cable and labour costs vary, electricity costs change. The following variables have been adjusted and recalculated to determine their effect: motor operating time 25-100%; electricity price $0.05/kWh to $0.20/kWh; cable and termination hardware prices 50% to 200%; installation labour $50/ hr to $200/hr; carbon tax $15 to $30 per ton; allowable cable tray fill 50% to 100%.
The variable having the most significant effect on the payback period is energy cost: the higher it is, the sooner a larger cable pays for itself.
Motor operation time is also a key factor in obtaining payback. Motors that are used all the time save energy all the time. High usage contributes to quicker payback.
Cable cost is a major factor. Larger conductors have more copper in them, more insulation and more armour. When cable price doubles, the payback period nearly doubles.
As for cable termination material cost, increased connection costs at each cable end become more relevant as the overall cable installed length is reduced. Allowable cable tray fill has a small effect on the payback period.
Labour cost changes had a small effect on the payback period. The increased labour costs for installing larger, heavier cables, bigger terminations, and additional cable tray was small in comparison to cable and energy cost.
As expected, a carbon tax shortens the payback period, and continues to pay dividends in the long term. However, the contribution is comparatively small.
Amperage of loads with proportionally sized cables does not appear to affect the outcome of the study. There is a curious relationship between the ratios of cable cost (primarily affected by copper commodity price) and the amount of cable losses that can be recovered for any particular load. Motors from 10-hp to 1000-hp have an eerie similarity of payback periods when upsizing one, two or three sizes from the typically accepted cable size.
Some less-tangible benefits
Upsized cable runs cooler and, therefore, has extended life. Cables in proximity to upsized cables operate cooler and have extended life. Energy efficiency and reduction in greenhouse gases may have a goodwill factor with the public. Random-filled cables run hotter and increase the benefits gained by upsizing. Motor terminal voltages will typically be higher in upsized cable installations, which has a positive effect on payback periods and further reduced cable heating.
Conclusions and recommendations
Upsizing power cables can have meaningful cost savings in just a few years. Look for motors that run most of the time, as the payback for cable upsizing is heavily reliant on long hours of operation. Increases in electricity rates shorten payback times.The relationship is nearly linear. Double the price of electricity results in break-even in half the time. Plus, the yearly after break-even savings increases, too.
Double the price of the cables and you add about 80% more time to recover the extra cost.
Look for motors (and other loads) that have cable lengths of 50m or more.The additional work and material cost of terminating larger cables can only be o set by savings that grow with the number of metres of cable.
Lastly, caution should be exercised when selecting smaller cables in an effort to save money; in certain cases, cost savings can be very short-lived.
View the embedded image gallery online at:
Duane Grzyb is an electrical engineer (Alberta) with 25 years of consulting engineering experience, and has worked on energy projects all over Western Canada. His specialties include area classification, failure analysis, renewables, VFD applications, power quality, harmonics mitigation and project management. He is also the current president of the Electrical Technical Exchange Group (ETEG) for Northern Alberta. This article is adapted from a paper Duane presented at the IEEE ESTMP Workshop in Calgary, 2014.
This article originally appeared in the March 2018 issue of Electrical Business Magazine.