ECONOMICS OF DISTRIBUTED RESORUCES BFT 112 INTODUCTION TO

ECONOMICS OF DISTRIBUTED RESORUCES BFT 112 INTODUCTION TO ENGINEERING NOOR SYAFAWATI AHMAD SCHOOL OF ELECTRICAL SYSTEM ENGINEERING

DISTRIBUTED RESOURCES (DR) � Traditional focus in electric power planning has been on generation resources : forecasting demand, extrapolating existing trends and most cost-effective power plant. � In the final decades of twentieth century, important shift has focus on real need was for energy services : illumination, for example not raw kilowatt-hours and energy used efficiently. � Integrated resource planning (IRP) emerged both supply-side and demand-side resources were evaluated, including environmental and social cost.

DISTRIBUTED RESOURCES (DR) �Recently, with increased attention to the electricity grid and the emergence of efficient, cost-effective cogeneration, IRP now recognizes three kinds of electricity resources: generation resources, grid resources and demand-side resources. Figure 1: Example of distributed resources

ELECTRIC UTILITY RATE STRUCTURE � An essential step in any economic calculation for a DR project is an analysis of the cost of electricity and/or fuel that will be displaces by proposed system. � Electric rates vary considerably, depending not only on the utility itself, but also on the electrical characteristic of the specific customer purchasing the power. � The rate structure for a residential customer is typically include basic fee to cover cost of billing, meters and other equipment, plus an energy charge based on the number of kilowatt-hours of energy used.

ELECTRIC UTILITY RATE STRUCTURE �Commercial and industrial customers are usually billed not only for energy (kilowatt-hours) but also for peak amount of power that used (kilowatts). �Demand charge for power (RM/mo per k. W) is the most important difference between the rate structures designed for small customers versus large one. �Large industrial customers mat also pay additional fees if their power factor is outside of certain bounds.

STANDARD RESIDENTIAL RATES Table 1: Standard Residential Tariff Rates

Example 1 �Calculating a simple residential utility bill. Suppose that a customer subject to the rate structure in Table 1 uses 1300 k. Wh a month. a. What would be the total cost of electricity (RM/month)? b. What would be the value (sen/k. Wh) of an efficiency project that cuts the demand to 1000 k. Wh/month?

Solution 1 a. The total monthly bill include 200 k. Wh = 21. 80 sen/k. Wh, 100 k. Wh = 33. 40 sen/k. Wh, 300 k. Wh = 51. 60 sen/k. Wh, 300 k. Wh = 54. 60 sen/k. Wh, 400 k. Wh = 57. 10 sen/k. Wh, for a total of: (200 x RM 0. 218) + (100 x RM 0. 334) + (300 x RM 0. 516) + (300 x RM 0. 546) + (400 x RM 0. 571) = RM 624. 00 b. would be (200 x RM 0. 218) + (100 x RM 0. 334) + (300 x RM 0. 516) + (300 x RM 0. 546) + (100 x RM 0. 571) = RM 452. 70 The saving per k. Wh is (RM 624. 00 – RM 452. 70)/300 k. Wh = RM 0. 571/k. Wh

DEMAND CHARGES � The rate structure that apply to commercial and industrial customers usually include a monthly demand charge based on the highest amount of power drawn by the facility. � Demand charge may be especially severe of the customer’s peak corresponds to the time during which the utility has its maximum demand since at those times the utility is running its most expensive peaking power plants. � The demand charge is based on the peak demand in a given month, usually average over a 15 minute period, no matter what time of day it occurs.

DEMAND CHARGES Table 2: Standard Commercial Tariff

Example 2 �Impact of Demand Charges A small commercial building that uses 20, 000 k. Wh per month has a peak demand of 100 k. W. a. Compute the monthly bill (ignoring fixed customer charges). b. How much does the electricity cost for a 100 W computer that is used 6 hour a day for 22 days in the month? The computer is turned on during the period when the peak demand is reached for the building. How much is that in sen/k. Wh?

Solution 2 a. The monthly bill is made up of energy and demand charge: Energy charge = 20, 000 k. Wh x RM 0. 365/k. Wh = RM 7, 300. 00/month Demand charge = 100 k. W x RM 30. 30/k. W = RM 3, 030. 00/month For a total of RM 7, 300 + RM 3, 030 = RM 10, 330. 00/month (29% of which is demand) b. The computer uses 0. 10 k. W x 6 hr/day x 22 day/month = 13. 2 k. Wh/month Energy charge = 13. 2 k. Wh/month x RM 0. 365/k. Wh = RM 4. 82/month Demand charge = 0. 10 k. W x RM 30. 30/k. W = RM 3. 03/month Total cost = RM 4. 82 + RM 3. 03 = RM 7. 85/month On a per kilowatt-hour basis, the computer costs Electricity = (RM 7. 85/month) / (13. 2 k. Wh/month) = RM 0. 597/k. Wh Notice how the demand charge makes the apparent cost of electricity for the computer (RM 0. 597/k. Wh) nearly double the RM 0. 365/k. Wh price of the electric energy.

DEMAND CHARGES WITH A RATCHET ADJUSTMENT � The demand charge in the rate schedule shown in Table 2 applies to the peak demand for each particular month in the year. � The revenue derived form demand charges that may only be monetarily significant for just one month of the year, however, may not be sufficient for the utility to pay for the peaking power plant they had to build to supply that load. � For example, the monthly demand charges may be ratchet to a level of perhaps 80% of the annual peak demand. That is, if a customer reaches a highest annual peak demand of 1, 000 k. W, then for every month of the year the demand charge will be based on consumption of at least 0. 80 x 1, 000 k. W = 800 k. W. � This can lead to some rather extraordinary penalties for customers who add a few kilowatts to their load right at the time of their annual peak; conversely, it provides considerable incentive to reduce their highest peak demand.

Example 3 � Impact of Ratcheted Demand Charges on an Efficiency Project. A customer’s highest demand for power comes in August when it reaches 100 k. W. The peak in every other month is less than 70 k. W. A proposal to dim the lights for 3 hours during each of the 22 workdays in August will reduce the August peak by 10 k. W. The utility’s energy charge is RM 0. 365/k. Wh and it demand charge is RM 30. 3/k. W with an 80% ratchet on the demand charges. a. b. c. What is the current annual cost due to demand charges? What annual savings in demand energy charges will result from dimming the lights? What is the equivalent savings expressed in sen/k. Wh?

Solution 3 a. At RM 30. 30/k. W a month, the current demand charge in August will be August = 100 k. W x RM 30. 30/k. W-mo = RM 3, 030 For the other 11 months, the minimum demand charge will be based on 80 k. W, which is higher than the actual demand: Sept – July demand charge = 0. 8 x 100 k. W x RM 30. 30/k. W-mo x 11 mo = RM 26, 664 So the total annual demand charge will be: Annual = RM 3, 030 + RM 26, 664 = RM 29, 694

Solution 3 b. By reducing the August demand by 10 k. W, the annual demand charges will now be: August = 90 k. W x RM 30. 30/k. W-mo = RM 2, 727 Sept – July = 0. 8 x 90 k. W x RM 30. 30/k. W-mo x 11 mo = RM 23, 997. 60 Total annual demand charge = RM 2, 727 + RM 23, 997. 60 = RM 26, 724. 60 Annual demand savings = RM 29, 694 – RM 26, 724. 60 = RM 2, 969. 4 August energy savings = 3 hrs/d x 10 k. W x 22 days x RM 0. 365/k. Wh =RM 240. 90 Total savings = RM 2969. 40 + RM 240. 90 = RM 3210. 30 Notice that the demand savings is 92. 5 % of the total savings!

Solution 3 c. Dimming the lights saved 3 hr/d x 10 k. W x 22 d = 660 k. Wh and RM 3210. 30, which on a per k. Wh basis is Saving = RM 3210. 30 / 660 k. Wh = RM 4. 86/k. Wh In the other words, the business saves RM 4. 86 for each k. Wh that it saves, which is about 13 times more than would be expected if just the RM 0. 365/k. Wh cost of the energy is considered.

LOAD FACTOR �The ratio of a customer’s average power demand to its peak demand, called the load factor, is a useful way for utilities to characterize the cost of providing power to that customer: (eq 1)

LOAD FACTOR � For example, a customer with a peak demand of 100 k. W that uses 876, 000 k. Wh/yr (8760 h/yr x 100 k. W) would have an annual load factor of 100%. Another customer using the same 876, 000 k. Wh/yr with the peak demand of 200 k. W would have a load factor of 50%. � For this example, the utility would need twice as much generation capacity and twice as much transmission and distribution capacity to serve the customer with the lower load factor, which means that the rate structure must be designed to help recover those extra costs. � Since they use the same amount of energy, the RM/k. Wh charge would not differentiate between the two, but the demand charge RM/k. W-mo will. A stiff demand charge will encourage customers to shed some of their peak power, perhaps by shifting it to other times of day to even out their demand.

Example 4 �Impact of Load Factor on Electricity Costs. Two customers each use 100, 000 k. Wh/mo. One (customer A) has a load factor of 15% and the other (customer B) has a 60% load factor. Using a rate structure with energy charge of RM 0. 202/k. Wh and demand charges of RM 35. 50/k. W-mo, compare their monthly utility bills.

Solution 4 They both have the same energy costs : 100, 000 k. Wh/mo x RM 0. 202/k. Wh = RM 20, 200/mo Using (eq 1), the peak demand for A is which, at RM 35. 50/k. W-mo, will incur demand charge of RM 32, 869. 45/mo.

Solution 4 The peak demand for B is The demand charges of RM 8, 218. 25 The total monthly bill for A with poor load factor is nearly twice as high as for B (RM 53, 069. 45 for A and RM 28, 418. 25 for B)

REAL-TIME PRICING (RTP) �While time-of-use (TOU) rates attempt to capture the true cost of utility service, they are relatively crude since they only differentiate between relatively large blocks of time (peak, partial-peak, and off-peak, for example). �The ideal rate structure would be one based on realtime pricing (RTP) in which the true cost of energy is reflected in rates that change throughout the day, each and every day. With the price of electricity more accurately reflecting the real, almost instantaneous, cost of power, it is hoped that market forces will encourage the most efficient management of demand.

OFF-PEAK HOURS SCHEDULE OPERATIONS � During peak hours (8. 00 a. m. to 10. 00 p. m. ), TNB has to generate a higher amount of electricity to meet the demands of our usual work day. As the day progresses, more electricity is required, so more power plants are fired up to generate more supply. � During off-peak hours when most of us are asleep, TNB generates less electricity, but most of the time this supply is not used to its fullest. � By re-scheduling your operations to run heavy equipment during off-peak hours, you will reduce the amount of electricity required during the day, and making the most of electricity generated at night. Fewer power plants will need to be turned on, which is good for the environment. � With TNB's Off-Peak Tariff Rider (OPTR) scheme, you can also enjoy up to 20% savings for electricity usage during offpeak hours.