Energy Management :: 2007/2008

Transcription

1 :: 2007/2008 Class # 3P Energy Prices João Parente joao.parente@dem.ist.utl.pt

2 Energy transformation PROBLEM 01 Consider a Combined Heat and Power Plant (CHP), based on a micro gas turbine recuperated cycle. The main components and thermodynamic processes are described below: Main components in micro gas turbine: Compressor; Combustion chamber; Turbine; Recuperator (heat exchanger gas-gas) Economizer (heat exchanger gas-liquid) Alternator Summary description of the process: Atmospheric air is compressed in the compressor and heated by the exhausts gas at the recuperator, then the compressed air passes through the combustor where natural gas is burned. Combustion products are then expanded in the turbine that feeds the compressor and the alternator. Before being sent to the stack, expanded exhausts are cooled in the recuperator and in the economizer. Finally, in the economizer, using the sensible heat of the exhausts water is heated for heat production. Class # 3P :: Energy Prices Slide 2 of 53

3 Energy systems representation Gas Turbine Recuperated cycle in CHP Functional scheme Fuel C T ~ Electricity Air REC Cold water ECO Ht Hot water Eh Exhausts Class # 3P :: Energy Prices Slide 3 of 53

4 Energy transformation PROBLEM 01 Depending on the outputs that are considered, the efficiency of the CHP system will be: a) Electricity production only ~ 30% b) Combined heat and power production ~ 85% Considering all of this information, estimate the Primary Energy saves obtained by the full use of the CHP system when compared to an alternative system that uses an independent gas boiler (85% efficiency) and electricity supplied by the grid. Assume the following values: LHV NG = 39,5 MJ/m3 Primary energy conversion coefficients: Electricity 0,29 kgoe/kwhe Natural gas 0,82 kgoe/m 3 Class # 3P :: Energy Prices Slide 4 of 53

5 Energy transformation SOLUTION P01 Assuming100 kwh NG in the combustion chamber (CHP): Electricity: 100 kwh NG * η e = 30 kwh e Heat: 100 kwh NG * (η CG - η e ) = 55 kwh th Pi Primary energy: 1) CHP: Natural gas: 100 kwh NG * 3600 kj/kwh / LHV NG * 0,82 kgoe/m 3 = = 100 kwh NG * 3600 kj/kwh / (39500 kj/m 3 ) * 0,82 kgoe/m 3 = 7,5 kgoe 1) Boiler + grid: Natural gas: (55 kwh th / 0,85) * 3600 kj/kwh / LHV NG * 0,82 kgoe/m 3 = 4,8 kgoe Electricity: 30 kwhe * 0,29 kgoe/kwhe = 8,7 kgoe Nt Natural gas + electricity: it 48 4,8 kgoe ,7 kgoe = 13,5 kgoe (almostl t 2 twice the primary energy) ) Class # 3P :: Energy Prices Slide 5 of 53

6 Tariff in the SEP Normal Low voltage BTN (< 41,4 kva) Active energy (Variable term - VT) Contracted Power (Fixed Term - FT) BTN = E (kwh) * VT ( /kwh) + FT ( /month) Low, average, high and very high voltage BTE (> 41,4 kva), MT, AT and MAT Energy Active energy (Variable term - VTae) Reactive energy (Variable term - VTre) Contracted Power (Variable term - TVcp) Peak hours power (Variable term - TVpp) Fixed term (FT) BTE, MT, AT, MAT = Ea (kwh) * TVea ( /kwh) + Er (kvarh) * TVer ( /kvarh) + Pc (kw) * VTcp ( /kw.month) + Pp (kw) * VTpp ( /kw.month) + FT ( /month) Class # 3P :: Energy Prices Slide 6 of 53

7 Tariffs in SEP - BTN Normal Low Voltage Tariff (BTN): Power scale TARRIFFS CONTRACTED POWER (kva) Social Tariff 115 1, ,3 Simple Tariff Dual Tariff 1,15 2,3 3,45 4,6 5,75 6,9 10,35 13,8 17,25 20,7 3,45 4,6 5,75 6,9 10,35 13,8 17,25 20,7 Simple Tariff 27,6 34,5 41,4 Non Intensive Use Tariff 27,6 34,5 41,4 Intensive Use Tariff 27,6 34,5 41,4 Sazonal Triple Tariff 27,6 34,5 41,4 Sazonal Simple Tariff Sazonal Dual Tariff Sazonal Triple Tariff 3,45 4,6 5,75 6,9 10,35 13,8 17,25 20,7 3,45 4,6 5,75 6,9 10,35 13,8 17,25 20,7 3,45 4,6 5,75 6,9 10,35 13,8 17,25 20,7 Seasonal tariffs: these tariffs can only be applied to entities with seasonal consumptions - characterized by a period of at least 5 consecutive months per year without t any consumption. In this sense, these tariffs do not apply to the residential sector. Weekly cycle Winter legal time Summer legal time Monday Friday Monday Friday Peak 09.30/12.00 h 18.30/21.00 h Peak 09.15/12.15 h Full 07.00/09.30 h 07.00/09.15 h 12.00/18.30 h Full 12.15/24.00 h 21.00/24.00 h Off peak 00.00/07.00 h Off peak 00.00/07.00 h Saturday Saturday Peak 09.30/13.00 h 09.00/14.00 h Peak 18.30/22.00 h 20.00/22.00 h 00.00/09.30 h 00.00/09.00 h Off peak 13.00/18.30 h Off peak 14.00/20.00 h 22.00/24.00 h 22.00/24.00 h Sunday Sunday Off peak 00.00/24.00 h Off peak 00.00/24.00 h Dil Daily cycle Winter legal time Monday Friday 09.30/11.30 h Peak 19.00/21.00 h Full Off peak 08.00/09.30 h 11.30/19.30 h 21.00/22.00 h Summer legal time Monday Friday 10.30/12.30 h Peak 20.00/22.00 h 09.00/ /10 h 20.30/19.30 h Full 12.30/20.00 h 22.00/23.00 h 00.00/08.00 h 00.00/09.00 h Off peak 22.00/24.00 h 23.00/24.00 h Selling tariffs to the final consumer (2008) Class # 3P :: Energy Prices Slide 7 of 53

8 Tariffs in SEP - BTN Exemple: Selling tariffs to final consumers (residential) in BTN ( 20,7 kva e > 2,3 kva) Contracted power (kva) Fixed term for simple tariff ( /month) Fixed term for dual tariff ( /month) ,74 8, ,45 10, ,15 13, ,85 15, ,7 21, ,61 28, ,42 35, ,42 42,14 Dual tariff: Peak (peak + full) Off-peak (off-peak) Energy cost (variable term) /kwh /kwh Peak hours 0,1143 0,1132 Off peak hours 0,0614 Selling tariffs to the final consumer (2008) Application: A house with a 6,9 kva contracted power has a monthly consumption of 200 kwh at peak hours and 100 kwh at offpeak hours. What is the best option in the Regulated Market? Class # 3P :: Energy Prices Slide 8 of 53

9 Problems EXAMPLE 1: Consider a house with an electrical contract of 10,35 kva. The present contract is in simple tariff. The consumption power daily profile is represented in the next figure. The owner is considering to substitute the fridge by a new one. The fridge presents an average power of 200 W, while the new one is expected to consume 100 W. Consider the following simplifications: a) Winter peak schedule b) Constant power consumption of the fridge Questions: a) What is the present electricity monthly bill? b) Is dual tariff, cycle daily, less expensive? c) Considering that the contract has been optimized, what will be the annual saves from the fridge substitution? Class # 3P :: Energy Prices Slide 9 of 53

10 Problems PROBLEM 1: Class # 3P :: Energy Prices Slide 10 of 53

11 Problems SOLUTION PROBLEM 1: a) Daily consumption is Daily consumption = 6x x5+3x9 + 5x1 = 81 kwh/day Monthly consumption: Monthly consumption = 81 x 30 = 2430 kwh/month Present electricity monthly bill: Monthly bill = Fixed term + Variable term = 15, x 0,1143 = 293 /month b) For the daily cycle: Peak hours (8h 22h) = 6x x 9 = 54 kwh/day or 1620 kwh/month Off peak hours = = 27 kwh/day or 810 kwh/month Present electricity monthly bill: Monthly bill = 21, x0, x0,0614= 255 /month There is a benefit with the dual tariff, daily cycle. c) Supposing that, the contract has been changed for the dual tariff, daily cycle. The daily cost saving will be: (considering that there are 14 peak hours and 10 off peak hours) Daily cost saving = (0,2 01) 0,1) x (14x0, x0,0614) 0614) = 0,22 /day The cost saving in one year will be: Monthly cost saving = 0,21x365 = 80 /year Class # 3P :: Energy Prices Slide 11 of 53

12 Energy transformation PROBLEM 02 Imagine you are building yourself a new house and still haven t decided on the heating system you will install. After a brief market research you have narrowed down your options to only two: a heating system based on electrical resistances and a heating system based on a hot water circuit heated by a natural gas boiler. Based on the records you kept form your previous house, you got to the conclusion that you will use the heating system 16 hours per day (between 8:00 to and 24:00), only on week days and only during 5 winter months, with an average power of 2kW. a) Considering that you have an electrical contract of 10.35kVA with dual tariff on a daily cycle, determine which of these two solutions is best (assume an 80% efficiency for the hot water boiler, a LHV of 40MJ/m 3 and an average price for natural gas of /m 3 ). b) Comment on the above solution bearing in mind what you have learned about Useful, Final and Primary Energy. Class # 3P :: Energy Prices Slide 12 of 53

13 Energy transformation SOLUTION P02 The heating system runs 16 hours per day. According to the existing tariff plans, this corresponds to 14 hours on peak and 2 hours off-peak (dual electricity tariff). The annual cost of the first solution is therefore given by: (14 h *2kW * /kwh +2h *2kW *.0614 /kwh) *5days/week *4weeks/month *5months/year = /year As for the Nt NaturalGas solution, lti the annual amount of final energy consumed is given by (80% efficiency): i 16 h *2kW *5days/week *4weeks/month *5months/year / 0.8 = 4000 kwh/year Considering a lower heating value of 9054 kcal/m 3, this yields: 4000 kwh/year * 860 kcal/kwh / 9054 kcal/m 3 = 380 m 3 /year Which, considering the current tariff results in: 380 m 3 /year * /m 3 = /year Answer: The best solution is the natural gas boiler. The fixed terms are not taken into account, as it would be necessary to allocate a certain percentage of its value to the heating use, which in this case cannot be done. However, considering that the total fixed value is assignedtoheating g the final result remains the same, as the electricity fixed costs are higher than the ones for natural gas (4.27 /month). Class # 3P :: Energy Prices Slide 13 of 53

14 Energy transformation SOLUTION P02 Comments: In this problem the natural gas solution proves to be the best. Additionally, if we calculate the price of natural gas per kwh of final energy, we obtain a value of /kwh, which is lower than the electricity prices. However, considering that a typical natural gas heating system will have an efficiency of approximately 80%, the price per kwh of useful energy will go up to /kwh, surpassing the off-peak value for a dual tariff electricity contract ( /kwh), where we would consider a 100% efficient electric resistance system. This means that, assuming the maximum power usage (10.35 kw) during all the off-peak period (10h per day), the monthly fee using electricity would become lower than the one using natural gas: Electricity: kw *10h *30days/month *.0614 /kwh /month = /month Natural Gas: kwh * 30 days/month * /kwh /month = /month This scenario, although theoretical, alerts to fact that an energy source that is more Primary Energy consuming should always be more expensive than others (which does not happen in this case), as the failure to do so may lead to a unwanted solution from an environmentally integrated perspective. Unlike the previous calculations, the fixed terms are now considered, as it is assumed that all power is allocated to the heating system. Otherwise the price difference would increase furthermore, emphasizing the above remarks. Class # 3P :: Energy Prices Slide 14 of 53