Energy Management : 2009/2010

Transcrição

1 : 009/010 Class# P3 SGCIE Prof. Miguel Águas Prof.ª Tânia Sousa

2 Problem Production of refined vegetable oil The aim is to develop an analysis of Sankey to a factory that produces ton / year of refined vegetable oil, consuming tons / year of fuel oil and 3 GWh / year of electricity. The fuel is consumed in a steam boiler with a thermal efficiency of 85%. The network of steam is poorly insulated, with permanent thermal loss of 50 kw. Moreover, certain valves have problems of sealing, and the leakage of steam in the order of 5% of the produced steam. Electricity is consumed mainly in electric motors with a conversion to mechanical energy efficiency of 90%. It is estimated that the consumption associated with the waste of energy (electrical equipment working out of production) representing 10% of useful energy. Data: LHV = 41. GJ / ton fuel oil ; toe / ton fueloil ; 15 toe / GWh elec a) Represent and quantify the Sankey diagrams for fuel oil and electricity. b) Determine the primary energy consumption based on the coefficients defined SGCIE and the maximum value of specific consumption at the end of the rationalization plan. c) Determine the total amount of degradation between primary energy and useful energy. Class # P3 : SGCIE Slide of 53

3 Produção de óleo vegetal refinado a) Fuel óleo 3000 MWh Centrais térmicas e transporte Fuel óleo 4000 ton Refinação e Transporte Luz Calor Etc Motores eléctricos Calor Caldeira - degradação de 0% Mau isolamento - degradação de 50 kw Fugas de vapor - degradação de 0% 5% Ventiladores Não indicado Class # P3 : SGCIE Slide 3 of 53

4 Produção de óleo vegetal refinado b) Energia primaria = tep/ton x 4000 ton/ano tep/mwh x 3000 MWh/ano = = 4581 tep/ano Consumo especifico = 4581 tep / ton = 115 kgep/ton CE ano 0 = 115 kgep/ton Melhoria de 6% em 6 anos obriga a CE ano 6 <= 115 x 94% = 108 kgep/ton c) A energia primária é de tep. Como 1 tep = 41,9 GJ = x 41,9 = GJ/ano A energia util tem duas componentes: Energia eléctrica = x conversão para GJ x rendimento dos motores elec. =3000 MWh x 3,6 GJ/MWh x 90% = 9.70 GJ/ano Fuel óleo = Fuel x PCI x rendimento da caldeira x (1-%fugas)- + Potência perdida x 8760 h/ano x 3600 s/ano = 4000 ton/ano x 41, GJ/ton x 85% x (1-0,05) kw x 8760 x 3600 x 10-6 GJ/kJ = GJ/ano TOTAL = = GJ/ano A degradação foi então de = GJ/ano Class # P3 : SGCIE Slide 4 of 53

5 Enunciado Replacement of electricity consumption A factory has an annual consumption of toe, broken down into 60% of electricity consumption and 40% of natural gas consumption, with a specific consumption of 150 kgoe/ton. An energy audit found that electric heaters consumption is around 800 MWh and that half of this consumption can be replaced by heat recovery process and the other half to be replaced by heat from burning additional natural gas, with a thermal efficiency of 80 %. Lighting is responsible for 00 MWh, and it was found that 0% of this consumption occurs when daylight was sufficient. The burning of natural gas intended for the production of hot air to the production process, has an energy efficiency of 80%. Currently the company loses 10% of its production due to defects in the final product. In the next 6 years the company expects to implement the solutions described for electricity, as well as eliminate production losses. Data: LHV = 45.1 GJ/ton GN ; toe/ton GN ; GN = kg/nm 3 ; kg /tep; 15 toe/gwh elec ; 0.47 kg /kwh elec a) Represent and quantify the Sankey diagram before changes. b) What was the plant's annual production and carbon intensity? c) The elimination of electrical heating (with additional consumption of natural gas) will be enough to enforce the SGCIE? Class # P3 : SGCIE Slide 5 of 53

6 Substituição de consumo de electricidade a) Gas natural X MWh Centrais térmicas e transporte Gas natural X m3 Rede de transporte Não indicado Fornos - degradação de 0% Luz Calor Etc Ar quente Iluminação desnecessária - degradação de 0% x 00 MWh Quebras Quebras Class # P3 : SGCIE Slide 6 of 53

7 Substituição de consumo de electricidade b) Consumo de energia primária 1800 tep/ano Consumo especifico= 0,15tep/ton = logo x= ton/ano Produção x Consumo de electricidade = 1800 tep/ano x 60% = 1080 tep/ano sendo 0,15 tep/mwh, vem 1080 tep/ano /0,15 tep/mwh = 5.03 MWh/ano Consumo de gás natural = tep/ano x 40% = 70 tep/ano Intensidade carbónica Logo sendo 1,077 tep/ton gn, vem 70 tep/ano /1,077 tep/ton gn = 669 ton gn /ano = [ kg / ano] Emissões de gases de efeito de estufa Consumo total de energia [ tep / ano] 503 MWh/ano 0,47 ton /MWh 70 tep/ano,6837 ton /tep Intens. Carb. + = =,39ton / tep 1800 tep/ano c) Futuro consumo de electricidade = = 43 MWh/ano Futuro consumo de gás natural = 70 + aumento gn Aumento gn x PCI gn x η gn = Calor fornecido pela electricidade Aumento gn = (400 Mwh e /ano x MJ/MWh) / (80% x MJ/ton gn ) = 40 ton gn /ano logo futuro consumo de gás natural = 70 tep/ano +40 ton gn /ano x 1,077 tep/ton gn =763 tep/ano 43 MWh/ano 0,47 ton /MWh 763 tep/ano,6837 ton /tep 403 Intens. Carb. + = = =,41ton / tep 43 MWh/ano x 0,15 tep/mwh A intensidade carbónica aumenta: não valida! O futuro consumo específico será de tep/1.000 ton = 0,139 tep/ton (-7%, valida) Class # P3 : SGCIE Slide 7 of 53