Electrical Business

The economics of upsizing your conductors

April 4, 2018
April 4, 2018
By Duane Grzyb


April 4, 2018 — Upsizing the cross-sectional area of current-carrying conductors in power electrical cables can reduce the I2R loss which, in certain cases, can offset the higher material and installation costs.

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:

Solving for Y0-00 (which represents recovery time in years):

{gallery}Upsizing cables March 2018 article calculation 2

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