Energy management: 2011/2012

Transcription

1 Energy management: 2011/2012 Class# 4/10/2011 & 6/10/2011 Energy prices (electricity) João Rodrigues

2 Problem 1: Refrigerator selection Silva family intend to buy a refrigerator, selecting the following two in a supermarket: Brand Bosch (model KDV29V13): Volume of refrigeration and freezer: 199 l and 68 l. Price 399, Energy classification: A+ Annual consumption 250 kwh/y Brand Samsung (model RT25): Volume of refrigeration and freezer: 161 l and 56 l Price 319 Energy classification: B Annual consumption 423 kwh/y Assuming that the daily power consumption of a refrigerator is 25% off-peak (vazio) and 75% in peak hours (não vazio), give aid to the Silva family in respect of electrical consumption: a)best solution for tarifa simples b)best solution for tarifa bi-horária (ciclo diário) Energy Management: Energy prices Slide 2 of 53

3 Solution 1: Refrigerator selection a) In the simple fare always costs /kwh. Thus, future electric bills associated with refrigerators is: -Bosch: 250 kwh/y x 0,1285 /kwh = 32 /y -Samsung: 423 kwh/y x 0,1285 /kwh = 54 /y The savings will be 22 /y We conclude that the added cost of a refrigerator Bosch is 80 and is recovered in less than four years of operation. b) In bi-hourly rate electricity costs: -Off peak hours (10h/day) = 0,0742 /kwh -Peak hours (24h/day) = 0,1382 /kwh Consumption has a weighting with 25% for peak and 75% for off-peak. The average price will be / kwh = 25% x % x ). For this value the difference in favor of Bosch refrigerator is 21 / y (0.122 x ( )), against 22 / y for the simple fare. Thus, the savings are not significantly influenced by the contract options. Energy Management: Energy prices Slide 3 of 53

4 Problem 2: Electrical heating Identify the best heating solution for an house with the following heat needs: 0 kw between 22h00-8h00 4 kw between 8h00-10h30 2 kw between 10h30-18h00 4 kw between 18h00-22h00 Solution A is based on heating by heat accumulation. The consumption is done during off-peak, releasing heat during the day. Solution B is based on heat pump heating with a COP of 2. Consider that the house has a contract of kva and measurement is trihorária, with daily cycle. Use winter schedule. a)what is the electrical power required in each solution, can it be supplied? b)what is the most economical solution? c)and what if the measurement was simples instead of tri-horária? Energy Management: Energy prices Slide 4 of 53

5 Solution 2: Electrical heating Assuming no heat losses during off-peak hours, the accumulation will be equal to the needs of heat during one day: 2,5 h x 4 kw + 7,5 h x 2 kw + 4h x 4 kw = 41 kwh. Since the off-peak period lasts 10h, the electric power in A is 4.1 kw. The COP is 2, so in solution B the electric power is either 2 or 1 kw. b) Accumulation: the price is kwh, so the daily cost is 3.0 / day Heat pump: requires the assessment of consumption at peak hours (ponta) and day hours (cheias) and COP = 2: Peak hours (ponta): (1.5 x 4 kw kwh x 4 kw) / 2 = 8 kwh / day Day hours (cheia): 41 / 2-8 = 12.5 kwh / day The daily consumption is 20.5 kwh which is exactly half the consumption in the solution of accumulation, due to the COP be 2. So the cost will be: 8 kwh/day x /kwh kwh/day x /kwh = 2.9 /day The result is very similar in both situations. However, if the heat pump had a better COP then 2 it would be the most economical solution. c) In the simple fare the kwh price is always equal throughout the day. In this case as the heat pump has a COP of 2, the cost will be half the cost of the accumulator. Energy Management: Energy prices Slide 5 of 53

6 Problem 3: Lighting António moved to a new apartment and decided to change the lighting of the living room. He is contemplating one of the following options: 1 Incandescent light bulb of 100 W: unit cost of 0.60 and life time of 1000h. 5 Halogen light bulbs of 50 W: unit cost of 1.45 and life time of 4000h. 1 Fluorescent light bulb of 25 W: unit cost of 6.45 and life time of 8000h. Assume simple tariff and 4h of daily use. a) If António rents the house for ½ year, what lighting should he choose? b) And if he buys the house? c) If the contract is for bi-hourly counting and all consumption occurs in the vazio period, would each of the previous decisions change? Tarifa simples: 0,1285 Tarifa bi-horária: Vazio (22h-8h) = 0,0742 /kwh Fora de vazio (8h-22h) = 0,1382 /kwh Energy Management: Energy prices Slide 6 of 53

7 Solution: Lighting Use of 4h/day, after half an year 4h/day x 365/2 day/year = 730 h/year, it's not necessary to change any light bulb. Cost Incandescent = 1 x h x kw/h x 0.1 kw = 9.98 Cost Halogen = 5 x h x kw/h x 0.25 kw = 30.7 Cost Fluorescent = 1 x h x kw/h x kw = 8.8 Fluorescent is the most economic solution. b) Consumption after 8000 h of use: Cost Incandescent = 8 x h x kw/h x 0.1 kw = Cost Halogen = 2 x 5 x h x kw/h x 0.25 kw = Cost Fluorescent = 1 x h x kw/h x kw = Fluorescent is the most economic solution. c) In the long run there is no change because the determining factor is energy consumption. After half an year the answer changes: Cost Incandescent = 1 x h x kw/h x 0.1 kw = 6.04 Cost lower than the cost of the halogen or fluorescent light bulbs. Energy Management: Energy prices Slide 7 of 53