Economics of Cable Upsizing

Transcription

1 Economics of Cable Upsizing February 2, 2014 Duane Grzyb, P.Eng, Senior Standards Engineer Magna IV Engineering

2 2 Table of Contents 1. EXECUTIVE SUMMARY 3 2. PROJECT BACKGROUND 3 3. INTRODUCTION 5 4. CALCULATION METHOD 6 5. CALCULATIONS 7 6. RESULTS UPSIZING FROM THE BASE CASE RESULTS DOWNSIZING FROM THE BASE CASE SENSITIVITIES CONCLUSIONS AND RECOMMENDATIONS AREAS THAT DESERVE FURTHER INVESTIGATION 13

3 3 1. EXECUTIVE SUMMARY A rigorous examination of the new construction economics of installing increased gauge Teck 90 power cables supplying induction motors in a typical industrial application. This paper explores the hypothesis of obtaining an acceptable return on investment as well as a perceivable reduction in CO 2 equivalent emissions by upsizing cable conductors in new designs. The analysis method is described, identifying the various factors that need to be considered in this analysis, both positive and negative, to determine the realistic benefits. A test case is examined in detail, comparing several gauges of copper conductors (#1/0 - #4/0 AWG) supplying a 575V, 100HP induction motor. The modelled scenario uses 250 meters of Teck 90 aluminum armoured copper cable installed in cable tray at a 30 C ambient. Steady state running temperatures are calculated for each cable type and included in the evaluation. Analysis is done based on present day economic factors. 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. The variables examined include electricity cost, hours of operation per year, material costs (tray, cable, fittings, terminations), installation labour (tray, cable, fittings, terminations) as well as the impact of conceivable carbon taxes. The evidence suggests that upsizing cables, hence reducing their I 2 R losses does offset increased installation costs in five to seven years. Upgrade Duration to Pay for Upgrade* Yearly $ Recovered Upgrade #1/0 to #2/0 AWG 5.3 years $399 Upgrade #1/0 to #3/0 AWG 5.7 years $711 Upgrade #1/0 to #4/0 AWG 6.7 years $953 *assuming base case described in section 4. Sensitivities are explored in section 5. It is evident from the analysis that the amount of time that a load is operated is critical to achieving timely payback. Some guidelines are provided which may help owners and engineers identify applications that are candidates for further consideration. 2. PROJECT BACKGROUND 2.1. An example where choosing premium cost, energy efficient equipment has become mainstream is the application of NEMA Premium Efficiency induction motors. In directives from the EPA and CEMEP, minimum efficiency has been specified for electrical motors A 100 HP motor operating at nameplate output consumes (89.7 amps at 575 Volts 3Ø Power Factor of 87.5%) For sinusoidal current, the power factor PF is equal to the absolute value of the cosine of the

4 4 apparent power phase angle φ (which is also is impedance phase angle): PF = cos φ PF is the power factor. φ is the apparent power phase angle. The real power P in watts (W) is equal to the apparent power S in volt-ampere (VA) times the power factor PF: S(VA) PF = P(W) ( 3 VL-L(V) I(A)) x PF = P(W) ( x ) = 78,168 Watts For a year of continuous operation at 10 per kwh ( x 24 x /.01) = $68,500 per year of continuous operation It becomes apparent that a motor costing $10,000 consumes almost 7 times it s initial cost in energy each year. In 1992 the U.S. Congress, as part of the Energy Policy Act (EPAct) set minimum efficiency levels for electric motors. In 1998 the European Committee of Manufacturers of Electrical Machines and Power systems (CEMEP) issued a voluntary agreement of motor manufacturers on efficiency classification. Several statements have been made regarding motor use and the advantages of using premium-efficiency or higher efficiency motors. These include: Based on U.S. Department of Energy data, it is estimated that in the U.S. the NEMA Premium motor program will save 5,800 gigawatt hours ($580M) of electricity and prevent the release of nearly 80 million metric tons of carbon into the atmosphere over the next ten years. This is equivalent to keeping 16 million cars off the road. Pages/NEMA-Premium-Motors.aspx Roughly 30 million new electric motors are sold globally each year for industrial purposes. Some 300 million motors are in use in industry, infrastructure and large buildings. These electric motors are responsible for 40% of global electricity used. By using best practice energy efficiency of electrical motors the incremental cost typically has a payback time of 1 to 3 years. In addition there is reduction of global greenhouse gas emissions IEEE C IEEE Loss Evaluation Guide for Power Transformers and Reactors. Abstract: A method for establishing the dollar value of the electric power needed to supply the losses of a transformer or reactor is provided. Users can use this loss evaluation to determine the relative economic benefit of a high-first-cost, low-loss unit versus one with a lower first cost and higher losses, and to compare the

5 5 offerings of two or more manufacturers to aid in making the best purchase choice. This standard was first published in 1991, more than 20 years ago, which supports the concept of total ownership cost considerations. It is interesting to note that transformer loss evaluation is at the discretion of the purchaser, and motor loss prevention has been mandated Data centers which operate hundreds if not thousands of computers mounted in high density racking systems have evolved with a sensitivity to energy loss reduction. It is becoming accepted practice to purchase three phase transformers with atypical secondary voltages (380 Volt Line to Line) to correspond with single phase 220 Volt input settings on commonly available commercial dual voltage switching power supplies. This seemingly innocuous change reduces current draw in infrastructure wiring, resulting in reduced losses thereby justifying the additional effort to utilize non-standard equipment With these things in mind, it is time to rigorously evaluate the benefits of low-loss power cabling. 3. INTRODUCTION 3.1. Copper conductors have resistance. Thicker copper conductors have lower resistance than thinner ones. Copper that is cool has lower resistance than that which is warm. An identical current passing through a thinner wire wastes more energy than its thicker counterpart. To make matters worse, the additional dissipated energy causes the wire to increase its temperature. Higher temperature causes 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 Not unlike motors, transformers and data centers, this paper examines the total ownership cost of the supply to load portion of cabling and its infrastructure, focusing on reducing the target value of 3% to less Delivered Power getting more for your money Induction motors are constant power devices. Typically and by its nature an induction motor automatically increases current draw when terminal voltage drops to maintain constant power output. A decrease in supply voltage is compensated by increased current For this paper, to ensure that the same amount of useful work is delivered by the motor in each evaluated scenario, the approach is to calculate losses and benefits based on constant motor terminal voltage. This approach removes any ambiguities surrounding the nature of an induction motor as a constant power device Heating loads such as electric heat tracing or space heating is dependent on delivered watthours. A decrease in device terminal voltage results in a decrease in output heat, ultimately resulting in an increased duty cycle for thermostatically controlled applications. In essence,

6 6 becoming a constant power device Different types of lighting would require additional analysis, resistive types may not reflect the potential savings, due to an increased brightness level not have a perceived higher value. HID lighting would likely be dependent on ballast type for their perceived output differences Non Linear it is likely that the electrical power in is equivalent to mechanical power out, however detailed analysis is recommended Saving Opportunities / Value Enhancers Long term electricity savings that never end Decreased cable operating temperature contributes to further increased savings as a result of lower conductor resistance Lower operating temperature = longer cable life Lower heat loss = reduced AC load in summer (for indoor portions of cables) Potential Carbon Tax savings Perceived Public Image Marketing opportunities 3.6. Cost Increases / Value Detractors More expensive cable (Copper, Armour, Insulation, Freight) Increase in required cable tray infrastructure (or conduit) Increased labour to install larger cables, connectors, terminations etc Increased cost of larger cable and conductor connectors Increased pipe rack / building space requirements for cabling. 4. CALCULATION METHOD 4.1. To evaluate real world economics of this hypothesis, a typical installation was modelled to evaluate factual motor data, cable data, material costs, electricity rates and labour rates to determine a meaningful calculation The typical installation examined is a 100 HP, 575 volt, 60Hz, three phase induction motor, connected with 250 meters of 3c#1/0 Teck cable in 24 aluminum ladder tray, using compression lugs, and teck connectors sized appropriately at each end. The cable tray is 240 meters in length, and has 4 horizontal elbows and 3 vertical elbows to approximate a typical installation.

7 To ensure that the same amount of useful work is delivered by the motor in each scenario, the approach is to calculate losses and benefits based on constant motor terminal voltage. This approach removes any ambiguities surrounding the nature of an induction motor as a constant power device. Typically and by its nature an induction motor automatically increases current draw when terminal voltage drops to maintain constant power output Using 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 I 2 R losses. This is further compounded by higher amperage causing increased cable conductor temperature thus increasing resistance, thus increasing voltage drop, thus increasing losses Conductor operating temperatures in these calculations are calculated using 30 C free air operation. In typical installations these cables are random filled. Random filled tray is extremely difficult to calculate conductor operating temperatures due to the variable nature of cable loading, thermal conductivity and cable placement. However a cable in free air will likely run cooler than it would in a random filled tray. The electrical code mandates reductions in current per conductor as much as 40% for multi-cable and multi-conductor installations. It is evident from the calculations performed that equal increases in conductor temperatures of base and upsized cables results in small additional net energy savings The net energy saving of larger conductors was further supported by comparing calculations using tabulated voltage drop tables for C conductor temperatures rather than actual calculated conductor temperatures. Therefore the calculated break even times presented in this analysis are shorter for the more typical random fill scenario Motor data sheets are attached (courtesy of Teco Westinghouse) Cable temperature rise calculations are attached (courtesy of Nexans). 5. CALCULATIONS 5.1. Calculation of Up-Front Cost to Install Larger Cable. Factors included are: More copper, insulation, armour in the cable Fewer cables can be run in the same size tray Teck connectors are larger Terminations are larger Labour to install larger materials takes longer

8 Therefore: $TIC 0 = $ c0 + $ ic0 + $ t0 + $ it0 + $ l0 + $ il0 + $ a0 + $ ia0 Where: $TIC 0 is the total installed cost of the cable, tray, and hardware for the 3c#1/0 base case. Where: $ c0 is the cost of 250 meters of 3c#1/0 Teck 90 cable. $ ic0 is the labour cost to install 250 meters of 3C#1/0 cable. $ t0 is the cost of 2 teck connectors to fit 3c#1/0 cable $ it0 is the labour cost to install 2 size 3c#1/0 teck connectors $ l0 is the cost of 6 compression lugs for #1/0 conductor $ il0 is the labour cost to install 6 compression lugs on #1/0 conductors $ a0 is 1/56th of the cost of 240 meters of a 24 cable tray system $ ia0 is 1/56th of the cost of labour to install the 240 meter tray system Similarly: $TIC 00 is the total installed cost of a 3C#2/0 cable, terminations and 1/55th of the cost of the tray system. And $TIC 000 is the total installed cost of a 3C#3/0 cable, terminations and 1/54th of the cost of the tray system. $TIC 0000 is the total installed cost of a 3C#4/0 cable, terminations and 1/48th of the cost of the tray system. $TIC 1 is the total installed cost of a 3C#1 cable, terminations and 1/75th of the cost of the tray system. $TIC 2 is the total installed cost of a 3C#2 cable, terminations and 1/85th of the cost of the tray system. Differing cable tray capacity has been accounted for. This has been accomplished by determining the quantity of each cable type that could be installed in a 24 cable tray, and using the appropriate proportional cost assigned to each cable s total installed cost (TIC)

9 9 A graphical representation is shown below which illustrates the following relationships: 5.2. It is now required to determine the cost to be recovered to offset up-front cost differences between 3C#1/0 and 3C#2/0 cables and their required proportional amount of extra or less cable tray system. $UFC 0-00 = $TIC 0 -$TIC 00 And similarly to upgrade from #1/0 to #3/0 $UFC = $TIC 0 -$TIC 000 And #1/0 to #4/0 $UFC = $TIC 0 - $TIC Which brings up a further question, are up-front capital cost savings, resulting from choosing smaller cables, and being able to put more cables in each tray permanent savings or does the choice have repercussions after a calculable period of time? $ICS 0-#1 = $TIC 0 - $TIC #1 And $ICS 0-#2 = $TIC 0 - $TIC # Calculation of Payback Period to Recover Capital Costs:

10 10 Solving for Y 0-00 (which represents recovery time in years): Where: D is the per unit value of 1 year or 8,766 hours. (Running 50% of the time D = 0.5) $/kwh W L0 W L00 is the cost of electricity energy the motor is used. is the number of watts lost in I 2 R losses of a 3c#1/0 cable 250 meters long, operating at amps. is the number of watts lost in I 2 R losses of a 3c#2/0 cable 250 meters long, operating at amps. (W L0 - W L00 )/1000 is the difference in kilowatts ( I 2 R losses ) between a 3c#1/0 and a 3c#2/0 cable each 250 meters long, each operating at amps. G L is the average percentage of electricity that is lost in the AIS transmission system. 1+G L is the ratio of electricity produced to the amount received by the end user. Ton c /kwh $/Ton c is the average amount of CO 2 emissions in tons per kilowatt hour generated based on the heat ratio of the source generator. is an empirical value of a tax that could be levied against end users of fossil fuel energy. Commonly known as Carbon Tax. Rates of $15 per metric ton of CO 2 produced are typical values. Currently Alberta does not track or collect this tax. 6. RESULTS UPSIZING FROM THE BASE CASE 6.1. Using the base case, 100 HP, running continuously, with no carbon tax, using 250 meters of cable in cable tray, installed by trades at the rate of $85 per hour, the cost equalization of upsizing from #1/0 AWG conductors to #2/0 AWG conductors is 5.3 years. Y 0-00 = 5.3 years After the additional capital cost has been recovered each subsequent years operating costs are reduced by $399 per year Similarly;

11 11 Payback Period Upgrade After Payback Recovery Y 0-00 = 5.3 years Y = 5.8 years Y = 6.7 years Upsizing from 3c#1/0 AWG to 3c#2/0 AWG Upsizing from 3c#1/0 AWG to 3c#3/0 AWG Upsizing from 3c#1/0 AWG to 3c#4/0 AWG $399 / year $711 / year $953 / year 7. RESULTS DOWNSIZING FROM THE BASE CASE 7.1. In an effort to reduce capital costs it has been suggested that cables be reduced in size. 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 two and a half years. Y 0-1 = 2.52 years* *After 2.52 years, initial cost savings from reducing the size of the cable is consumed by increased energy years is a conservative analysis, as the mean temperature of a tray of reduced sized conductor would naturally operate warmer than a tray of increased size conductors. No value has been deducted for reduced cable life, decreased cable longevity of surrounding cables or potential carbon tax Downsizing even further to 3C#2 AWG Y 0-2 = 2.98 years* *the significant number of additional 3C#2 cables that can be installed into a 24 tray may contribute to the counterintuitive trend of the problem getting better not worse Based on a continued cost of electricity of $0.10 / kwh, 100% operating time and excluding possible carbon tax benefits and increased cable life, the yearly savings in net electricity are: Y$ 0-00 $399 per year of electricity saved Y$ $711 per year of electricity saved Y$ $953 per year of electricity saved 7.4. Or in the case of decreasing cable size: Y$ 0-#1 An additional $811 of electricity consumed per year Y$ 0-#2 An additional $1516 of electricity consumed per year

12 12 8. SENSITIVITIES 8.1. The variable having the most significant effect on the payback period is the motor duty cycle. The payback period is inversely proportional to the hours of use. The fastest payback occurs on motors run continuously 8.2. Cost of energy is significant sensitivity in this analysis. The more expensive electricity cost is the quicker the return on capital investment Material cost. Increased connection costs at each cable end become more significant as the cable length is shortened. In the extreme case, a short cable run, of a few meters would take much longer to recover the connection costs which are independent of cable length. Increased connection costs at each cable end become more relevant as the installed cable length is shortened. In the extreme case, a very short cable run of a few meters would take much longer to recover the connection costs which are independent of cable length. Increased cable cost is a relevant factor. Larger conductors have more copper in them, more insulation, and more armour. There is a careful balance to maintain. When cable price doubles, the payback period nearly doubles Cable tray fill has a small effect on the payback period. A 50% fill results in double the increased tray differential cost, however, the fractional tray cost is very small compared to other costs and as a result the payback period is affected very little Labour cost changes made small effects 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. The contribution is small Amperage of loads 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. Very similar relationships occur for motors and cables of varying sizes from 1/10 to 10 times in rated power Decreased operating voltage negatively affects efficiency. Voltage drop and therefore I 2 R losses are the same on a cable operating at 480 or 600 volts. It requires higher current deliver equal power to a load, therefore lower voltage installations benefit quicker from cable upsizing. (A general recommendation is to choose the highest reasonable voltage). 9. CONCLUSIONS AND RECOMMENDATIONS 9.1. Look for motors that run most of the time. Payback for cable upsizing is heavily reliant on long hours of operation.

13 Significant changes in electricity rates do affect payback. The relationship is nearly linear. Double the price of electricity results in breakeven in half the time. Plus double the yearly after breakeven savings Cable cost is close to linear as well; double the price of the cables, add about 80% more time to recover Look for motors that have cable length 50 meters or more. The additional work of terminating larger cables needs to be offset with savings that are dependent on the number of meters of cable Labour, cable tray materials, cable tray fill, connectors, terminations and Carbon Taxes are all secondary factors. If the case works after looking at duty cycle, cable and electricity costs these items make little difference in the outcome Unless the loads are short duty cycle (less than.25 pu) or capital is very scarce, do not use the smallest conductor sizes that code allows. Operations budgets will suffer after only a few years Other less tangible results: Upsized cable runs cooler and therefore has extended life Cables in proximity to upsized cable operate cooler and have extended life Reduction in greenhouse gases, may have goodwill factor to public Energy efficiency may have goodwill factor to public Random filling of trays result in average cable temperature increases, which favor quicker payback as well as quicker losses in the case of downsizing conductors. 10. AREAS THAT DESERVE FURTHER INVESTIGATION At motor power factor correction decreasing the I in I 2 R losses Increasing cable self-cooling decreasing the R in I 2 R losses One diameter cable spacing vs single layer tray fill (cables touching) versus random tray fill. Payback occurs in all cases, but which might be quickest Is there value in reducing the Impedance of MCC and Switchgear Buswork? Is there value in fan or refrigeration cooling of MCC and Switchgear Buswork? Are the conclusions the same for Aluminum? Life extension total cost of ownership Other load types.