Utility Analysis
Baseline Electricity AnalysisIntroductionAs a prologue to a doctor’s visit, a patient’s body weight, temperature, pulse and blood pressure are typically measured and recorded. The doctor uses these outward measures to begin to assess one’s inner health, and to help track changes throughout treatment. Similarly, collecting and analyzing energy billing data yields important information about a plant’s energy use. Understanding and documenting current energy use is called developing a baseline. Developing a baseline:Helps define potential energy savings Helps focuses efforts on the most important areasDetermines accurate avoided energy costs for calculating cost savingsHelps identify energy saving opportunitiesProvides a baseline from which to measure the effectiveness of energy management activities.This chapter discusses how utilities typically structure charges for electricity, how to calculate the avoided cost of electricity, and how to use utility billing analysis to help identify cost saving opportunities. Specific cost saving opportunities include:Rate StructureSwitch to electric rate structure with lower overall costsNegotiate electric rate structure with lower overall costsSwitch to electric rate structure with demand control discountBilling ErrorsReconcile billing error with utilityMeter ConsolidationConsolidate electrical metersPurchasing TransformerPurchase transformer and switch to primary servicePower Factor CorrectionCorrect power factor by downsizing over-sized motorsCorrect power factor by adding capacitorsDemand Saving PotentialReschedule operation of electrical equipment to reduce peak demand. Use control equipment to shed loads to manage peak demandElectricity IndustryBefore the 1990s, most U.S. electric utilities provided electricity within a defined service area and all customers within that service area purchased electricity from the utility. To ensure that the utilities did not abuse their monopoly power, most investor-owned utilities were regulated by a state public utility commission (PUC). The PUC typically approved rates that would protect customers while guaranteeing that the utility could cover the cost of providing electrical service and make a fair profit. Rate structures are public documents that describe how a utility charges for electricity. Most utilities have a different rate structure for each class of customer. For example, residential, small commercial, large commercial, and industrial users usually have different rate structures. Today, most utilities post their rate structures on the internet. As part of determining an energy baseline, it is important to find and understand the appropriate rate structure. In recent years, technology and market changes have created many new types of rate structures. For example, some utilities now charge customers using a rate that varies on an hour-by-hour basis to reflect the cost of electricity purchased by the utility. This type of rate structure is called ‘real-time pricing’. Sophisticated energy users can take advantage of real-time pricing by scheduling electricity using processes to run during times when electricity prices are lowest and curtailing electricity using processes when electricity prices are highest. In addition, some utilities are also offering incentives to reduce electrical demand during periods of high system-wide demand. Beginning in the 1990s, many states have begun deregulating electric utilities. The theory behind deregulation is that market competition between utilities will lower costs and increase reliability. For deregulation to work, multiple utilities must be able to compete for a customer’s business. To create competitive markets, most states have required utilities to separate the transmission and distribution (T&D) part of their business from the electrical generation part of their business. In these cases, a consumer’s cost for electricity is the sum of the T&D costs and the generation costs. In many cases, this creates complicated utility bills that show the breakdown of these costs. Unfortunately, this information is of little value to a consumer trying to manage their electricity use to reduce total cost. Instead, the most important information for managing electricity costs is the breakdown of costs according to electricity use patterns that a costumer has some control over: total electricity use, peak electrical demand and power factor.The average real cost of electricity in the U.S. has varied significantly over time; however, in recent years prices have been rising. From 2000 to 2008, real cost of electricity to the industrial sector rose at a rate of 2.7% per year and the nominal cost of electricity to the industrial sector rose at a rate of 5.3% per year. Source: U.S. Dept. of Energy, Annual Energy Review 2008, Report No. DOE/EIA-0384(2008)Electric Rate StructuresWith the exception of rate structures that employ real-time pricing, the total cost of electricity in most commercial and industrial rate structures is the sum of four components: A service chargeAn energy chargeA demand chargeA power factor chargeTo help companies manage electricity costs, it is extremely useful to reduce an electrical rate structure into these components. Service ChargeThe service charge is typically a nominal charge for each electrical service and meter. Energy ChargeUtilities measure the total electricity used during each billing period in units of kWhs, and base part of the total charge for electricity on the amount of energy consumed. The energy charge typically comprises about 50% of the total electricity bill. Block Structures: Sometimes, the cost of energy depends on how much is purchased. This type of pricing is called a block structure. In a block structure, the first "block" of energy usually costs more than energy in successively higher blocks. For example, the energy charge may be:$0.05 /kWh for the first 10,000 kWh$0.04 /kWh for the next 100,000 kWh$0.03 /kW for all remaining kWhSometimes, the size of each “block” of energy depends on the peak demand. For example, the energy charge may be:$0.05 /kWh for first 250 kWh/kW$0.04 /kWh for next 150 kWh/kW$0.03 /kWh for all additional kWhIf the peak demand during the billing period was 500 kW, the energy charge would be:$0.05 /kWh for first 250 kWh/kW x 500 kW = 125,000 kWh$0.04 /kWh for next 150 kWh/kW x 500 kW = 75,000 kWh$0.03 /kWh for all additional kWhFuel Cost Adjustments and Taxes: Because the cost of fuel for a utility may vary over time, utilities sometimes modify the energy costs in the rate schedule with a “fuel cost adjustment”. The fuel cost adjustment may be a credit or charge depending on the cost of fuel. In addition, there may be other charges or taxes based on energy use added to the cost of energy.Demand ChargeBecause electricity cannot be stored like natural gas, electric utilities continually vary the amount of electricity they generate in order to meet the system wide demand for electricity. Typically the demand for electricity varies over the day and year. Thus, a utility must purchase and maintain enough generating, transmission and distribution equipment to meet the peak demand (plus a safety allowance) even though peak demand typically occurs for only a few hours per year. Purchasing and maintaining this equipment are real costs to the utility, and these costs would be reduced if the peak demand were reduced. Hence, utilities seek to send a price signal that will encourage customers to reduce peak demand. They do this through a demand charge. To calculate peak demand, utilities record how much electricity is consumed during every 15-minute or 30-minute interval throughout a billing period. Peak electrical demand is calculated by identifying the single interval in which the most electrical energy (kWh) was consumed, and calculating the average power consumption during this interval. For example, if 100 kWh were consumed during a 15-minute demand period, the electrical demand during this interval would be:100 kWh / 15 minutes x 60 minutes/hour = 400 kWThus, near-instantaneous power spikes, such as when motors startup, have little effect on the peak demand since the duration of a short power spike is small compared to the demand interval. Moreover, longer demand periods generally result in lower peak demand.Block Structures: Sometimes, the cost of demand depends on how much power is used. This type of pricing is called a block structure. In a block structure, the first "block" of demand usually costs more than the demand in successively higher blocks. For example, the cost of demand may be:$10.00 /kW-month for the first 500 kW$8.00 /kW-month for all demand in excess of 500 kWFor example, if the peak demand was 700 kW during a monthly billing period, the demand charge would be:($10 /kW-month x 500 kW) + ($8 /kW-month x 200 kW) = $6,600 /month On Peak-Off Peak Rates: To encourage users to reduce electrical demand during periods of peak system usage, some utilities offer different rates or methods of calculating billing demand for on-peak and off-peak periods. For example, a utility may offer a rate in which the ‘billing demand’ is calculated as the greater of:the actual on-peak demand, or 50% of the actual off-peak demand. Thus, if a user were able to schedule production such that actual on-peak demand was 500 kW and the actual off-peak demand was 1,500 kW, the billing demand would only be 750 kW rather than 1,500 kW.Seasonal Demand Charge: Some utility rate structures also have a seasonal demand charge, especially if the utility’s peak load is much higher during some parts of the year than during other parts of the year. For example, a utility may offer a rate in which the ‘billing demand’ is calculated as the greater of:the actual demand, or75% of the peak monthly demand during the previous 12 months.Thus, if the actual demand during July was 160 kW and the actual demand during October was 100 kW, the “billing” demand during October would be:0.75 x 160 kW = 120 kW instead of the 100 kW “actual” demand. In this case, a seasonal demand penalty of 20 kW would be assessed in October because of the high summer demand in July.Power Factor ChargeIn general terms, electrical power P is calculated as the product of current I and voltage V. In direct current (DC) systems, this is always strictly true:P = I V (for DC systems)In alternating current (AC) systems, current and voltage vary sinusoidally. In resistive loads, such as those from electric resistance heating elements, the current and voltage are in phase with each other and power is again the simple product of the two.P = I V (for AC systems with resistive loads)However, many electrical devices work by utilizing the magnetic field produced when current flows through a conductor. For example, motors induce a large magnetic field by wrapping current conducting wire around an iron core. In AC systems, the magnetic field is pushed into and out of the iron core as the polarity of the voltage varies from positive to negative. However, the iron core offers some resistance to the changing magnetic field, which causes the current to lag behind the voltage. Thus, current and voltage are slightly out of phase in inductive loads. As before, power is the product of voltage and current. However, when voltage and current are out of phase, some of the power is unusable.The concept of actual and reactive power can be visualized using current and voltage wave forms. In resistive loads, the current and voltage are in phase. Hence, the sign of the power waveform, which is product of voltage and current, remains positive. All power with a positive sign can be used by electrical equipment and is measured as kW. Inductive loads, such as motors, cause the current waveform to lag behind the voltage waveform. This causes the sign of a portion of the power waveform to be negative. Power with a negative sign is unusable, and is called reactive power (kVAr). The quantity of unusable reactive power generated by an inductive load is determined by the phase angle between the voltage and current. The total power must be greater than the reactive and usable power and is measured in units kVA. Power factor = 1.0: all power is + and usable Power factor = 0.7: some power – and unusableThe mathematical relationships between total power (kVA), reactive power (kVAr), and useful power (kW) can be modeled using a right triangle with phase angle ?. The quantity of each type of power can be calculated using the trigonometric relations defined by the power triangle.Power factor, PF, is ratio of the useful power, kW, and total power, kVA. The power factor can be calculated using the following relationships: PF = kW / kVA = cos (?) = cos (tan-1 (kVAr/kW)) (1)Reactive power, kVAr, can be calculated using the following relationships:kVAr = kW x tan (?) = kW x tan (cos-1 (PF))(2)The total power, kVA, can be calculated from the reactive power and power factor from the following relationships: kVA = kVAr / sin (?) = kVAr / sin (cos-1(PF))(3)Most utilities charge for low power because they must supply total power even though only a portion of the power is consumed by the user. Three common methods of charging for low power factor are:Adding a charge for reactive power (kVAr)Adding a demand penalty when the power factor is below a set amountBasing the demand charge on supplied power (kVA) rather than actual power (kW) In most cases, the power factor, kVAr and kVA used to determine power factor are the average values over the billing period rather than the peak values used to determine billing demand. The examples below demonstrate how to calculate the power factor charge.Some utilities charge for each kVAr of reactive power generated. ExampleIf the actual demand is 1,000 kW, the power factor is 80%, and the power factor charge is $0.30 per kVAr each month, calculate the power factor charge:kVAr = kW x Tan[Cos-1(PF)] = 1,000 (kW) x Tan[Cos-1(0.8)] = 750 kVArPF charge: 750 kVAr x $0.30 /kVAr-month = $225 /monthSome utilities add a demand penalty when the power factor is less than a set amount, usually 90%. For example, a utility may specify a demand penalty of:Demand penalty = kW (0.90 – PF) / PF when PF < 0.90ExampleIf actual demand is 1,000 kW, the power factor is 80%, and the cost of demand is $15 /kW-month, calculate the power factor charge:Demand penalty = 1,000 kW (0.90 – 0.80) / 0.80 = 125 kWPF charge = 125 kW x $15 /kW-month = $1,875 /monthSome utilities base the demand charge on the total supplied power (kVA) rather that actual power (kW). Basing the demand charge on kVA implicitly includes a power factor charge since kVA = kW / PF. Example:If actual demand is 1,000 kW, the power factor is 80%, and the cost of demand is $16 /kVA-month, calculate the implicit power factor charge:PF charge = (kVA – kW) x $16 /kVA-month [(1,000 kW / 0.8 kW/kVA) – 1,000 kW) x $16 /kVA-month = $4,000 /monthOther AdjustmentsUtilities typically transmit electricity at high voltages to reduce line losses. Most industrial plants use electricity at either 240 V or 480 V. Thus, the voltage of electricity must typically be reduced before it enters the plant. Transformers reduce the voltage of electricity. Transformers are typically rated in terms of supplied power kVA, and must be sized to meet peak demand with a reasonable safety factor. Utilities generally have different rates depending on who owns and maintains the transformer, and the placement of the meter with respect to the transformer.Primary and Secondary Service: When a customer owns and maintains the transformer, it is called “primary service”. When the utility owns and maintains the transformer, it is called “secondary service”. Many rate structures offer lower electricity rates for primary service, since the customer must purchase and maintain the transformer. In many electric rate structures, it frequently becomes advantageous for a customer to purchase and maintain a transformer, and consequently qualify for the cheaper “primary” rate, when the average monthly demand exceeds about 1,000 kVA.Primary and Secondary Metering: Transformers are typically about 99% efficient; 1% of the electrical energy is lost as heat when the voltage is reduced. Thus, a utility meter placed on the utility side of a transformer would record more electricity use than a meter placed on the customer side of a transformer. When the utility meter is placed on the utility side of a transformer it is called “primary metering”; when the meter is placed on the customer side of the transformer it is called “secondary metering”. To account for this difference, some rate structures adjust the actual energy use and demand depending on the location of the meter. For example, in some primary service rate structures with “secondary metering”, the “billing” energy use and demand are calculated as:Billing kWh = 1.01 x Actual kWhBilling kW = 1.01 x Actual kWSimilarly, in some secondary service rate structures with “primary metering”, the “billing” energy use and demand are calculated as:Billing kWh = 0.99 x Actual kWhBilling kW = 0.99 x Actual kWExample Electric Rate StructuresIt is useful to summarize the information in published rate structures in terms of the primary costs: service, energy, demand, power factor and other adjustments. Examples of summarized rate structures are shown below. The general service primary rate shown below applies to customers who own the transformer. It has fixed service and power factor charges. The energy charge varies by a small amount on a month to month basis due to the variable fuel cost adjustment. Peak demand is measured over 30-minute intervals during on-peak and off-peak periods. The billing demand is the greatest of 100% of the on-peak demand or 75% of the off-peak demand, which creates an incentive to move demand to the off-peak times. In addition, the billing demand includes a seasonal charge and would be adjusted upward if measured demand is less than 75% of the billing demand during the previous 11 months. General Service Primary RateService: $95 /monthEnergy:$0.008 /kWh (base)$0.012 /kWh (approximate fuel adjustment)$0.001 /kWh (taxes)Total: $0.021 /kWhDemand: $13.86 /kW-monthGreatest average power during any 30-minute periodGreatest of:100% of on-peak (weekdays: 8 am to 8 pm)75% of off-peak (all other times)75% of max Jun, Jul, Aug, Dec, Jan, Feb in last 11 months Power Factor: $0.30 /kVAr-month Secondary Metering Adjustment:Billing kWh = 1.01 x Actual kWhBilling kW = 1.01 x Actual kWThe general service secondary rate shown below applies to customers who do not own the transformer. It has no service charge, and the power factor charge is implicit in the demand charge, since the demand charge is based on supplied power measured in kVA. The energy charge is based on a block structure, and the size of the blocks depends on the peak demand. Peak demand is measured over 15-minute intervals, and the rate is based on a fixed block structure. General Service Secondary RateService:No chargeEnergy:$0.0256 /kWh for first 250 kWh/kVA$0.0092 /kWh for all additional kWhDemand: $18.36 /kVA-mo for first 4,000 kVA: $14.45 /kVA-mo for all additional kVAGreatest average power during any 15-minute period Power factor:ImplicitPrimary Metering Adjustment:Billing kWh = 0.99 x Actual kWhBilling kW = 0.99 x Actual kWAvoided Cost of ElectricityTo calculate cost savings from reducing electricity usage, it is common practice to multiply the average cost of electricity times the electricity savings. Unfortunately, estimating cost savings using the average cost of electricity usually inflates the estimated cost savings because the average cost of electricity includes fixed costs such as service charge and because many rate structures employ block structures in which the first block of electricity costs more than subsequent blocks. For example, if a plant used 150,000 kWh and the rate structure specified energy charge of $0.04 per kWh for the first 100,000 kWh and $0.03 per kWh for usage in excess of 100,000 kWh, the average cost of electricity would be:[($0.04 /kWh x 100,000 kWh) + ($0.03 /kWh x 50,000 kWh)] / 150,000 kWh = $0.0367 /kWhHowever, if the electricity usage were reduced by 1 kWh, the savings would only be $0.03 /kWh. Thus, use of the average cost of electricity to estimate savings from energy conservation retrofit would inflate the estimate of cost savings by 22%.In addition, use of the average cost of electricity to estimate cost savings from energy conservation retrofits may lead to even greater errors if the energy conservation retrofit does not affect peak demand. For example, use of occupancy sensors to turn lights off at night would reduce electricity usage, but would probably not reduce electrical demand. In this case, use of the average cost of electricity to estimate savings would greatly inflate the estimate of savings, since the plant would probably see no reduction in the demand charge.Thus, the most accurate way to estimate cost savings from reducing electricity usage is to calculate the reduction in demand and energy costs separately based on the average demand and energy use and the rate structure. Avoided Cost of Electrical Energy UseThe avoided cost of electrical energy is the cost savings from reducing electrical energy usage by one kWh. The avoided cost of electrical energy should be calculated based on the appropriate block structure if applicable, and should include average fuel cost adjustments and taxes. For example, consider the electric rate structure summarized in the following figure. If the electrical demand were 100 kW, the first 30,000 kWh would be charged at a rate of $0.032 per kWh and all additional kWh would be charged at $0.021 per kWh. If the average monthly electrical energy use were 20,000 kWh, the avoided cost of energy would be $0.032 per kWh. If the annual average monthly electrical energy use were 40,000 kWh, the avoided cost of energy would be $0.021 per kWh. Avoided Cost of Electrical DemandThe avoided-cost of demand is the cost savings from reducing demand by one kW. The avoided cost of demand should include provisions for power factor charge and block structures if applicable. Provisions for Power Factor Charge: Consider the rate structure shown below.Demand: $13.86 /kW-moPower Factor: $0.30 /kVArIn this rate structure, the avoided cost of demand should include a provision for power factor charges because reducing the demand by one kW would also reduce the reactive power, kVAr. Thus, if the average power factor is 90%, the average kVAr reduction associated with a reduction in one kW of demand is: kVAr = kW x Tan[Cos-1(PF)] kVAr/kW = Tan[Cos-1(PF)] = Tan[Cos-1(0.90)] = 0.4843 kVAr/kWThe associated savings from reducing the reactive power by this amount would be:kVAr/kW x $0.30 /kVAr = 0.4843 kVAr/kW x $0.30 /kVAr = $0.15 /kWThus, the total avoided cost of demand would be:$13.86 /kW-mo + $0.15 /kW-mo = $14.01 /kW-moProvision for Demand-Dependent Energy Block Structures: If an electrical rate structure bases the kWh charge on the demand, then reducing the demand will also reduce the overall cost of energy consumption. This affect should be included in the avoided cost of demand, as long as the electrical energy use (kWh) is large enough to benefit from the smaller block. For example, consider the following rate schedule in which the block structure for energy is: $0.032 /kWh for first 300 kWh/kW of demand$0.021 /kWh for all remaining kWhIn this rate structure, reducing electrical demand by one kW, moves 300 kWh into the lower cost block. Thus, the avoided cost of demand, ACD, should include this adjustment to the base cost:ACD = 300 kWh/kW x ($0.032 - $0.021) kWh = $3.30 /kWThis adjustment should be applied only if some electrical energy use (kWh) moves into the less expensive block. For example, if the demand is 100 kW and energy use is 20,000 kWh, then the size of the energy use block is 100 kW x 300 kWh/kW = 30,000 kWh. In this case, the energy use is too small to benefit from the change in size of the base block, and the adjustment should not be applied. In general, the avoided cost of demand for a demand-dependent block structure is:ACD = B1 (kWh/kW) x (E1 $/kWh - E2 $/kWh)Sometimes the sizes of multiple blocks of energy depends on the demand. For example, consider the following rate structure for energy. $0.037 /kWh for first 250 kWh/kW$0.034 /kWh for next 150 kWh/kW$0.029 /kWh for all additional kWhIn this case, the avoided cost of demand to account for moving more energy into a cheaper pricing blocks is:ACD = B1 (kWh/kW) x (E1 $/kWh - E2 $/kWh) + B2 (kWh/kW) x (E2 $/kWh – E3 $/kWh)ACD = 250 kWh/kW x ($0.037/kWh - $0.034/kWh) + 150 kWh/kW x ($0.034/kWh – $0.029/kWh)ACD = $1.50 /kWLoad FactorLoad factor is the average fraction of the peak electrical demand used by a facility. The load factor, LF, is defined as the ratio of average power consumption to maximum power consumption.LF = (kWh/period) / (peak kW x hours/period) Load factor can be used to predict of how many shifts per day a plant is running or to gauge the occupancy of a building. High load factors indicate multiple operating shifts. For example, consider a plant with power use of 100 kW for one 8-hour shift per day, 5 days per week, and no power use on off-shifts or weekends. The plant’s load factor would be:LF = (100 kW x 8 hours/day x 5 days/week) / (100 kW x 24 hours/day x 7 days/week) = 24% If the plant went to a two shift per day operation, the load factor would increase to:LF = (100 kW x 16 hours/day x 5 days/week) / (100 kW x 24 hours/day x 7 days/week) = 48% Based on this analysis, typical relationships between load factor and number of production shifts are shown in the table below. In general, increasing the number of production shifts per day and subsequently increasing load factor, enables a plant to purchase more kWh for the same demand charge, and lowers the average cost per kWh. Thus, high load factors correlate with lower average electricity costs.Load FactorProbable Operating Schedule20 %1 shift per day30 %1+ shifts per day or 1 shift + equipment left on at night40 %2 shifts per day45 %2+ shifts per day60 %3 shifts per dayLoad factor is also an important indicator of the potential for load shifting to achieve peak demand reduction. It is generally easier to shift the operation of large pieces of electrical equipment to the off-peak shift in two-shift per day or three-shift per day operations. Peak demand reduction is generally harder to achieve in one shift per day operations.Interpreting Electricity Billing DataGraphical AnalysisThe importance of graphing energy use data cannot be overstated. Our eyes are much better at identifying patterns and trends from graphical information than from tables of numbers. For example, the demand spike in the figure below was discovered only after graphing monthly electrical demand data. In this case, electrical demand spiked in the middle of the winter in a production facility located in a warm climate. The cause of the demand spike was subsequently discovered to be a short scheduled shutdown of steam service, which caused electrical resistance heaters throughout the facility to operate at full load. Simple changes to the HVAC system during subsequent scheduled shutdowns resulted in large demand cost savings.It is generally advantageous to plot at least one year of monthly electrical demand and monthly average energy per day on the same graph. Plotting average monthly energy use per day instead of monthly energy use removes the effect of unequal billing periods from the trend. Plotting data versus the meter reading date, instead of the billing date, improves interpretation. In many cases, electrical demand is less volatile than electrical energy use since the same major electrical equipment is typically operated simultaneously at least once during each month. In contrast, electrical energy use often varies with operating hours and levels of production. This pattern is evident in the figure shown below; electrical demand remains relatively stable while electrical energy use varies with production.A patterned increase in electrical demand and/or energy use during the summer months is often associated with air conditioning. This pattern is evident in the figure shown below. The patterned increase in actual electrical demand and energy use during the summer months is associated with air conditioning. The seasonal demand charge of 75% of the previous 11 month peak electrical demand is also evident in the flat pattern exhibited by the billed demand. Quick Electrical Demand and Energy Use BreakdownsA patterned increase in electrical demand and/or energy use during the summer months is often associated with air conditioning. The approximate quantities of air conditioning and production related electricity demand and use can be quickly estimated by drawing lines through average winter demand and energy use. For example, in the figure below, electrical demand below the line is associated with production and demand above the line is associated with air conditioning. Thus, air conditioning demand is about:4,700 – 3,900 = 800 kWThe size of the peak air conditioning load can be estimated by multiplying air conditioning demand by air conditioning energy intensity. For example, assuming the air conditioning equipment power use is 1.2 kW per ton of cooling, the peak air conditioning load is about800 kW x 1.2 kW/ton = 670 tons coolingSimilarly, electrical energy use below the line is associated with production and electrical energy use above the line is associated with air conditioning. In the figure below, winter energy use is about 78,000 kWh/day and annual average energy use is about 83,000 kWh/day. Thus, the fractions of electricity associated with production and air conditionings are about:Fraction Production = 78,000 kWh/day / 83,000 kWh/day = 94% Fraction Air Conditioning = 6%Disaggregating electrical demand and energy use into production and air conditioning components can identify the magnitude of possible savings opportunities in each area and improve the accuracy of disaggregating energy use by equipment and end use. Interpreting Interval DataSome utilities also provide electrical data measured over short time intervals. Short-time interval data is called interval data. Interpretation of interval data can lead to many valuable insights. Consider, for example, the following graph of 10 days of electricity interval data from a plastic blow-molding plant. The top graph shows plant electrical energy use (kWh) over every 30 minute demand period for 10 days. The pattern shows that the plant operates three shifts per day during weekdays and shuts down on weekends. The sharp spikes on Monday mornings are caused when the barrels and heads of the blow molding equipment are brought up to operating temperature using electrical resistance heating bands. The second graph shows actual electrical demand (kW) and total electrical demand (kVA) over the same period. The high total demand (kVA) relative to actual demand (kW) is caused by the low power factor associated with under-loaded hydraulic and grinder motors. The third graph shows power factor (kW/kVA) over the same period. The spikes in power factor during Monday morning startups are caused when plant demand is dominated by the electrical resistance heating bands with a power factor of unity. The power factor falls as the hydraulic and grinder motors are brought online when production begins. Electrical System Cost Saving OpportunitiesSeveral cost saving opportunities can be identified through billing analysis.Billing ErrorsBilling errors can be identified by calculating energy costs using the electrical rate structure and comparing calculated and comparing calculated and billed costs.Meter ConsolidationPlants served by multiple meters typically have higher electricity costs than plants served by a single meter. The cost savings from consolidating meters typically come from: eliminating service charges for multiple meters, pushing more energy use and/or demand into blocks with lower unit costs, and from non-coincident demand. Non-coincident demand savings occur because in many cases, the peak monthly demand from multiple meters is not set during the same demand periods. Thus, the sum of the peak demands when measured by separate meters is usually higher than the peak demand measured by a single meter. Consider, for example the peak hourly demands from two meters shown in the first figure below. If these peaks were recorded by separate meters, the billable demand would be 80 kW for the first meter and 50 kW for the second meter for a total of 130 kW of billable demand. However, if the meters were consolidated, the total billable demand would be the maximum of the sum of the meters, which would be 100 kW, as shown in the second figure below. Thus, meter consolidation would reduce peak demand by 30 kW.Purchasing TransformerUtilities generally have different rates depending on whether the utility of the customer owns and maintains the transformer. When the customer owns and maintains the transformer, it is called “primary service”. When the utility owns and maintains the transformer, it is called “secondary service”. Many rate structures offer lower electricity rates for secondary service, since the customer must purchase and maintain the transformer. In many electric rate structures, it frequently becomes advantageous for a customer to purchase and maintain a transformer, and consequently qualify for the cheaper “primary” rate, when the average monthly demand exceeds about 1,000 kVA.Power Factor CorrectionDevices which generate large amounts of reactive power in relation to actual power consumed have low power factors. Such devices include under-loaded motors and devices which convert AC power to DC power such as DC drives, welding machines and induction furnaces.Many utilities have explicit or implicit charges for low power factor. In addition, low power factor increases the line current, and hence losses, in transformers and the electrical distribution system. Correcting power factor by adding capacitors to the electrical systems can reduce or eliminate these costs.Line Losses:The quantity of line losses, LL, associated with low power factor correction can be calculated as follows:LL = I2 R = (kVA/V)2 R = [(kW/PF) / V]2 R = [kW2 R / V2] / PF2Thus, the fraction of line loss before and after power factor correction is:% Line Loss Savings = (LL1 – LL2) / LL1 = [1 / PF12 – 1 / PF22] / 1 / PF12 = 1 – (PF1/ PF2)2For example, if the power factor were improved from 80% to 90%, the percent line loss savings would be: % Line Loss Savings = 1 – (PF1/ PF2)2 = 1 – (80%/ 90%)2 = 21%Although percent line loss savings are relatively high, total energy savings are typically small since line losses are small. For example, if line losses are 2% of the total power draw, the total power savings from correcting the power factor would be:2% x 21% = 0.42%Capacitors:Capacitors are typically parallel plates that temporarily store electrons before the electrons jump the gap between the plates. This action delays voltage behind current, producing the opposite effect of inductance. Hence capacitors effectively cancel reactive power and current on the primary or upstream side of the capacitor. For example, if a motor operates at 70% power factor, installing a capacitor in the power supply line to the motor would reduce reactive power and line current on the primary side of the capacitor, but would not change the line current on the secondary (motor) side of the capacitor. Thus, installing capacitors directly upstream from low-power-factor loads reduces line current throughout the plant’s electrical distribution system; whereas installing capacitors directly down stream of the utility meter at the electrical service entrance to the plant, results in power factor correction for utility billing purposes, but will not reduce line losses and overheating throughout the plant.Capacitors are sized by the amount of reactive power (kVAr) they cancel. Simple capacitors are sized to compensate for a fixed amount of power. “Stepped” capacitors have internal controls that adjust the amount of reactive power compensation. Adding too much capacitance pushes the system from ‘current lagging voltage’ to ‘current leading voltage’. In either case, the voltage and current are out of phase and reactive power is produced. The resulting power factor is less than 1.0. Thus, installing purchasing excess capacitance does not harm equipment; however, it is expensive and serves no useful purpose.The method for determine the fixed amount of capacitor kVAr required is described below:Find the reactive power , Pr (kVAr), for each month: Pr (kVAr) = Pa (kW) x Tan[Cos-1(PF)]To increase PF as close to 1.0 as possible, recommend additional capacitance equal to minimum monthly kVAr during the past 12 months. This approach minimizes the possibility of adding too much capacitance.Subtract the recommended capacitance (kVAr) from recorded (kVAr) for each month. This difference represents the reactive power (kVAr) if the recommended capacitance were added.Recalculate PF, kVA or kVAr and electricity costs for each month, using the reactive power calculated in the previous step. These costs represent the costs if the recommended capacitance had been added.Calculate savings as the difference between the actual costs and the costs calculated in the previous step.If the quantity of reactive power varies significantly from month to month, then stepped capacitors that vary the capacitance with load would provide additional savings.ExampleConsider a case where the power factor charge is $0.30 per kVAr each month. If the actual demand is 1,000 kW, the power factor is 80%, and the, calculate the size of capacitance to correct the power factor to 1.0, and the savings from reducing the power factor charge:The reactive power is:kVAr = kW x Tan[Cos-1(PF)] = 1,000 (kW) x Tan[Cos-1(0.8)] = 750 kVArHence, the capictor should be sized at 750 kVAr. The reduction in power factor charge would be:750 kVAr x $0.30 /kVAr-month = $230 /monthExampleConsider a case where the power factor charge is determined by the following demand penalty.Demand penalty = kW (0.90 – PF) / PF when PF < 0.90If the actual demand is 1,000 kW, the power factor is 80%, and the, calculate the size of capacitance to correct the power factor to 0.90, and the savings from reducing the power factor charge:The current demand penalty and power factor charge are:Demand penalty = 1,000 kW (0.90 – 0.80) / 0.90 = 125 kWPF charge = 125 kW x $15 /kW-month = $1,875 /monthThe demand penalty and power factor charge if the power factor were increased to 0.90 would be:Demand penalty = 1,000 kW (0.90 – 0.90) / 0.90 = 0 kWPF charge = 0 kW x $15 /kW-month = $0 /monthHence the demand penalty savings would be $1,875 /month.The current reactive power is:kVAr1 = kW x Tan[Cos-1(PF)] = 1,000 (kW) x Tan[Cos-1(0.8)] = 750 kVArThe reactive power if the power factor increased to 0.90 would be :kVAr2 = kW x Tan[Cos-1(PF)] = 1,000 (kW) x Tan[Cos-1(0.9)] = 484 kVArThus, the quantity of capacitance needed is:750 kVAr1 – 484 kVAr2 = 266 kVArIn practice, capacitors are available in discrete sizes. Hence, for this application the closest size capacitor that does not overcorrect may be 200 kVAr. If a 200 kVAr capacitor were installed, the resulting reactive power would be 550 kVAr. The resulting power factor would be:PF = Cos (Tan-1 (kVAr/kW)) = Cos (Tan-1 (550/1000)) = 0.876The demand penalty and power factor charge if the power factor were increased to 0.876 would be:Demand penalty = 1,000 kW (0.90 – 0.876) / 0.90 = 27 kWPF charge = 27 kW x $15 /kW-month = $405 /monthHence the demand penalty savings would be: $1,875 /month - $405 /month = $1,470/monthDemand Saving PotentialDemand costs typically make up about 50% of the total electricity costs. Thus, reductions in peak demand can significantly reduce electricity costs even when total electricity use remains the same. Three general methods of reducing peak demand are:Rescheduling operation of electrical equipment in order to reduce peak demand. Sometimes this can be done by moving operations from first shift, when the peak demand typically occurs, to second or third shift. Using simple control equipment that ensures that equipment does not run simultaneously. Using energy management and control systems (EMCS) programmed to strategically shed loads to keep the peak demand within a specified range.In general, look for opportunities to reschedule operations to reduce peak demand in the following circumstances:Plant operates more than one shift per day.Electrical demand is higher during one shift than the others.Plant has electricity-using operation(s) that currently run during the shift with peak demand, but not during off-peak shifts.Consider the following cases. When a plant only operates for one shift or with three evenly loaded shifts, no demand reduction through shifting loads is possible. When the plant operates uneven shifts, demand shifting is possible. For example in the case shown below, plant demand is reduced by the full 20 kW of shiftable load.When the plant operates uneven shifts, demand shifting is possible, but may be limited by the demand during other shifts. For example in the case shown below, plant demand is reduced by only 10 kW of the 20 kW of shiftable load, because first shift now sets the peak load.The savings from rescheduling operations to reduce peak demand may be even larger when the utility offers an “off-peak” demand rate that is lower than “on-peak” demand rate. Consider, for example, the following graph of demand during first, second and third shifts, with on-peak and off-peak demand rates. When moving demand from first to second shift, maximum “peak demand” savings are 500 kW. When moving demand from first to third shift, maximum “peak demand” savings are 1,000 kW.In addition, the savings from rescheduling operations to reduce peak demand are larger in the following circumstances:Relatively high peak demand chargesThe rate schedule includes a minimum monthly demand charge that increases the demand charges for several months based on a peak during a single month. ................
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