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Energy Engineering - Advanced Thermodynamics and Thermoeconomics

Full exam

Department of Energy Politecnico di Milano Author Emanuela Colombo Pag.1 of 12 Date 11/07/2020 Milan, 24 th June 2020 Exam – Advanced Thermodynamic and Thermoeconomics 03-07-2020 Total score: 21 points (exercise 1 roughly 2/3 of the total) Exercise 1. Let’s consider a domestic hot water (DHW) system based on renewable energy. The system consists of three main components. The first one is the solar thermal panel that collects the solar incident power and warms up a mixture of water glycol. Then, there is a heat exchanger that conveys the heat collected by the water-glycol mixture to the water of a secondary circuit that flows in a hot water storage. The storage component has an inlet directly from the main aqueduct and an outlet that is the delivery of hot water to the end-uses. End-uses are in general shower, bath tub and sinks. The system is supposed to work in a steady condition, storage level and mass flow rates are constant. Exergy values related to each stream are reported in Table 1. Consider that the overall costs are charged to each component according to the shares reported in Table 2. Recommendation: keep second digit accuracy when dealing with exergy streams. Consider that T 0 and P 0 are respectively: 288 K and 1 atm. Additional data: Solar thermal panel Heat Exchanger Storage 0 1 2 3 4 5 6 Table 1 State Exergy [kW] 0 Solar Irradiance 15.00 1 Water -Glycol Inlet 916.06 2 Water-Glycol Outlet 918.21 3 Storage Inlet 1.33 4 Storage Outlet 0.16 5 Water from aqueduct 0 6 Hot water to end -use 0.64 Table 2 Total Cost of Investments [€] 15000 Total cost of Operation & maintenance [€/ y] 300 Solar thermal panel [%] 55 Heat Exchanger [%] 15 Storage [%] 30 Interest rate on capital [%] 3 Plant operative lifetime [years] 15 Load factor [ -] 0.4 FIGURE 1. PLANT CONFIGURATION Department of Energy Politecnico di Milano Author Emanuela Colombo Pag.2 of 12 Date 11/07/2020 With reference to the data provided, it is required to: a. Write the exergy balance and compute exergy destruction of each component. Calculate the rational and functional exergy efficiencies. Comment on the obtained results. b. Write all the auxiliary equations needed to characterize the Thermoeconomic system of equation for this plant. Clearly explain the reason why each auxiliary equation is assigned and make sure that they are coherent with the functional use of the components. Then, compute the total cost rate [€/h] for each component of the plant. c. Write the Thermoeconomic system of equations for each component and derive the analytical expressions of the cost structure of the products. d. Solve the system and provide the unit costs of the products for each component [€/kWh]. Then collect all the numerical results into a table giving evidence to the three constituents of the cost structure for each component. Comment on the results obtained. e. The final product of the DHW is the Hot water to end-use (stream 6). Please cross check the value you have obtained for c 6 in point d by solving the Thermoeconomics system at a different level of aggregation. Comment on the results. f. Apply the design evaluation calculating for each component the relative cost difference and the exergoeconomic factor. Collect the data into a table and provide suggestions for reducing the costs of the products. Please make a specific comments for the Solar thermal panel. g. Compute the physical exergy of stream 1 and 2 neglecting any pressure drop in the circuit of water-glycol and considering that T 1 is equal to 45°C and T 2 is equal to 90°C. Quantify the mass flow rate of water- glycol mixture. For physical exergy calculation, consider a perfect liquid mixture which specific heat is c p, water-glycol = 3.13 kJ/kg K. Department of Energy Politecnico di Milano Author Emanuela Colombo Pag.3 of 12 Date 11/07/2020 Exercise 1. (Solution) a. Write the exergy balance and compute exergy destruction of each component. Calculate the rational and functional exergy efficiencies. Comment on the obtained results. Calculation of exergy streams: Exergy Balance for the components: 0 1 2 ,STP,STP 2 4 1 3 , HX, HX 35 46 , solar thermal panel) 15.00 916.06 918.21 12 .85 heat exchanger) 918.21 0.16 916.06 1.33 0.98 storage) DD DD D St Ex Ex Ex ExExkW Ex Ex Ex Ex Ex ExkW Ex Ex Ex Ex Ex += + → = + − = + =++ → = + − − = + = ++ →         ,St 1.33 0 0.16 0.64 0.53 D ExkW = +− − =  Rational and functional exergy efficiencies: - STP: 2 01 21 0 918.21 0.986 15 916.06 918.21 916.06 0.143 15 r f Ex Ex Ex Ex Ex Ex η η = == + + − − = ==     == -=HX: = 13 24 34 21 916.06 1.33 0.999 918.21 0.16 1.33 0.16 0.544 918.21 916.06 r f Ex Ex kW kW Ex Ex Ex Ex kW kW Ex Ex η η + + = == + + − − = == − −     == -=St:= 46 35 6 345 0.16 0.640.602 1.33 0 0.640.547 1.33 0.16 0 r f Ex ExEx Ex Ex Ex Ex Ex η η ++ = == + + === −+ −+     == = b. Write the auxiliary equations needed to characterize the Thermoeconomic system of equation for this plant. Clearly explain the reason why each auxiliary equation is assigned and make sure that they are coherent with the functional use of the components. Then, compute the total cost rate [€/h] for each component of the plant The system is composed by 7 streams and 3 components, therefore, in order to define and close the Thermoeconomic system of equations, (n-m) numbers of auxiliary relations are required: n: Number of exergy flows = 7, m: number of components = 3: n-m = 7 - 3 = 4. Auxiliary equations: 0 1 2 12 3 4 34 5 1) 0 2) considered as a product of Solar ther mal panel 3) considered as a product of Heat Exchn ager 4) 0 c c c Ex c c Ex c − − = = →∆ = →∆ =   = In detail,=������������ 0 is equal to 0 since solar radiation is received at zero cost. Moreover, the product of solar thermal panel is the variation of the exergy content of water-glycol (stream 2 and 1). The same difference results to Department of Energy Politecnico di Milano Author Emanuela Colombo Pag.4 of 12 Date 11/07/2020 be the fuel of the heat exchanger while the exergy difference between stream 3 and 4 is the product of the analysed component. Finally, the fuel of the storage is the exergy difference between stream 3 and 4. Stream 5 is an extra inflow (fuel) and stream 6 is a product. Auxiliary equations have to be consistent with the formulations of functional exergy efficiencies. In this case, all efficiencies result to be coherent with the auxiliary equations listed above. Computation of total cost rate for each component (������������̇ ������������������������������������ ) CRF = capital recovery factor lt=life time i= interest rate ������������������������������������=������������ ( 1+������������) ������������������������ ( 1+������������) ������������������������− 1 =0. 03( 1+0.03) 15 ( 1+0.03) 15 − 1 =0.0841 ������������ � ������������̇ ������������������������������������ = ������������������������������������∙������������������������������������=0.0841 ������������ ∙15000€=1256€ ������������ ������������̇ ������������ &������������ =300€ ������������ ������������̇ ������������ ������������������������ =������������̇ ������������������������������������ + ������������̇ ������������ &������������ =1256€ ������������ +300€ ������������ =1556€ ������������ Let’s express the total costs in €/h starting from the computation of yearly operating hours: ℎ ������������������������ =������������∙8760=0.4∙8760ℎ ������������ = 3504ℎ ������������ ������������̇ ������������ ������������������������ =1556€ ������������ ∙1 3504 ������������ ℎ=0.444€ ℎ The total cost rate is then charged to each component reflecting the data given. ⎩ ⎪ ⎨ ⎪ ⎧������������̇������������������������ ������������ =������������̇ ������������ ������������������������ ∙0.55=0.244€ ℎ ������������̇������������������������ = ������������̇ ������������ ������������������������ ∙0.15=0.067€ ℎ ������������̇������������������������ = ������������̇ ������������ ������������������������ ∙0.30=0.133€ ℎ c. Write the Thermoeconomic system of equations for each component and derive the analytical expressions of the cost structure of the products. For each component it is required to write economic cost balance and substituting exergy cost relation as follows; ; ; in componentout inoutC Z C Economic cost balance C c ExExergy cost relation  +=  = ⋅  ∑∑     STP: Department of Energy Politecnico di Milano Author Emanuela Colombo Pag.5 of 12 Date 11/07/2020 0 1 2 ,STP 0 0 112 2 ,0 D STP in STP Ex Ex Ex Ex c Ex c Ex Z c Ex cc  += +  + +=  →=     == = The cost structure is:=  ,STP STP 12 , , 21 21 D in STP in STP fuel ExergyDestruction Losses Investment Cost Ex Z cc c c Ex Ex Ex Ex + == +⋅ + −−       = = HX:= = 2 4 1 3 , HX 22 4411 33 , 12 D HX in HX Ex Ex Ex Ex Ex c Ex c Ex Z c Ex c Ex c cc  + =++   + += +   →==        The cost structure is:  , HX 3 4 , HX , HX 34 34 () ()() DHX in in fuelExergyDestruction Losses Investment Cost ExZ ccc c Ex Ex Ex Ex + == +⋅ +−−       = St:= = The costs structure results to be as follows:= = ( ) 35 46 , 33 55 4 4 66 3 3 4 55 ,St 345 D St St in Ex Ex Ex Ex Ex c Ex c Ex Z c Ex c Ex c Ex Ex c Ex c Ex Ex Ex  + = ++  + += +  −+ →= −+          The cost structure results to be:   ,St 6, , 66 D St in St in St fuel Investment Cost ExergyDestruction Losses Ex Z cc c EE + =+⋅ +       = d. Solve the system and provide the specific costs of the products for each component. Then collect all the numerical results into a table giving evidence to the three constituents of the cost structure for each component. Comment on the obtained results. 0 0 112 2 22 4411 33 33 55 4 4 66 STP HX St c Ex c Ex Z c Ex c Ex c Ex Z c Ex c Ex c Ex c Ex Z c Ex c Ex  + +=  + += +  + += +             = = Unknowns=(c 1== c 2, c 4== c 3, c 6).= = = Let�s compute product costs starting from the cost structure of the products:= = Solar thermal panel:= = Department of Energy Politecnico di Milano Author Emanuela Colombo Pag.6 of 12 Date 11/07/2020 () ,STP STP 1 2 00 21 21 € € 12.850.244€ / 00 (918.21 916.06) kW (918.21 916.06) €€ 0 0 0.113 0.113 D ExZkWh cc cc kWh kWhkW Ex Ex Ex Ex kWh kWh = = +⋅ + = + ⋅ += −− −− = ++ =    Heat Exchanger: ( ) , HX 3 4 11 34 34 () € € (0.98) kW 0.067€ / h 0.113 0.113 (1.33 0.16) kW (1.33 0.16) kW ()() €€ 0.113 0.095 0.057 0.265 D HX Ex Z c c cc kWh kWh Ex Ex Ex Ex kWh kWh = = +⋅ + = + ⋅+= −− −− = ++ =     Water storage: () () 3 3 4 55 ,345 ,St 6, , 66 0.265 1.33 0.16 0 0€ 0.265 1.33 0.16 0 € € 0.53 0.133€ /€ € 0.265 0.265(0.265 0.220 0.208) 0.693 0.64 0.64 in St DSt in St in St c Ex Ex c Ex ckWh Ex Ex Ex ExZkW h cc c kWh kWh kW kWkWh kWh EE −+ − +⋅ === −+ −+ = + ⋅ += + ⋅ + = + + =    Cost structure components are collected in the following table. component fuel destruction + losses investment total €/kWh €/kWh €/kWh €/kWh STP 0.000 0.000 0.113 0.113 HX 0.113 0.095 0.057 0.265 St 0.265 0.22 0 0.208 0.69 3 e. The final product of the DHW is the Hot water to end-use (stream 6). Please cross check the value you have obtained for c 6 in point d by solving the Thermoeconomics system at a different level of aggregation. Comment on the results. Whole System: ������������� ̇ ������������ 0+������������̇ ������������ 5=������������̇ ������������ 6+������������̇ ������������ ������������,������������ ������������ ����������������������������=������������ 0������������ ̇ ������������ 0+������������ 5������������̇ ������������ 5 ������������̇ ������������ 0+������������̇ ������������ 5 =0 ������������ 6=������������ ����������������������������+������������ ����������������������������������������̇ ������������ ������������,������������������������������������ ������������̇ ������������ 6 ̇ +������������ ̇ ������������ ������������������������ ������������̇ ������������ 6 ������������ 6=������������ ̇ ������������ ������������������������ ������������̇ ������������ 6=0.444 € ℎ 0.64 ������������������������ =0.694 ∗ * Consider that the value of c 6 is not affected by rounding errors. The cost of the product c 6 might have been calculated with the original data. The reason is that the cost structure of the Whole system misses two terms due to the zero cost of the inlet. So despite the exergy destruction are not zero, the cost of the product are mostly associated with the investment costs. The selected Department of Energy Politecnico di Milano Author Emanuela Colombo Pag.7 of 12 Date 11/07/2020 strategies should be that of reducing the investment cost for the system as much as possible, provided that c0 and c1 are nearly negligible. f. Apply the design evaluation calculating for each component the relative cost difference and the exergoeconomic factor. Collect the data into a table and provide suggestions for reducing the costs of the products. Please make a specific comments for the Solar thermal panel. To perform the design evaluation of the system it is necessary to compute the cost of the exergy destructions, the exergoeconomic factors and the relative cost differences. The exergy destruction and losses cost rate is computed as: ,, P, D L j des loss j jC c Ex ++ = ⋅   = = Exergoeconomic factor is computed as:= , ,, inv j j inv j des loss j c f cc + = + = Relative cost difference is computed as:= ,, , des loss j inv j j fuel j cc r c + + = == In the following table, all the results are collected.= = = = component=fuel=destruction + losses=investment =total=C des+loss =C inv=C des+loss+inv =±=f= ==€/MWh €/MWh €/MWh €/MWh €/h €/h €/h - - STP 0.000 0.000 0.113 0.113 0.000 0.244 0.244 na 1.000 HX 0.113 0.095 0.057 0.265 0.112 0.067 0.179 1.345 0.375 St 0.265 0.220 0.208 0.693 0.142 0.133 0.275 1.623 0.484 Comments: Applying the Design Evaluation approach, the first step requires to conjunctly analyse two indicators: - the extensive summation of destruction, losses and investment costs; - the value of relative cost difference. A joint analysis is needed since the destruction + investment cost rate refers to the extensive values but can be affected by high specific costs of the fuel. While relative cost difference represents the margin for improving a component. High values of both exergy destruction and investment costs and r suggest that the component can be further improved. Exergoeconomic factor indicates whether the capital and O&M expenses (Ż ) is the major source of economic cost increase. The range of values for this indicator is between zero and one: high values indicate that investments (Ż ) are the major cost source and the primary aim is to reduce cost of the products by reduction the investment and O&M costs conversely exergy destruction (Ċ D +L ) cost are higher than investments and an increase of efficiency (and investment cost usually) is recommended. - Considering the summation of exergy destruction cost rates and investments, it stands out that the water storage sees the higher value followed by the solar thermal panel and the heat exchanger. Although the solar thermal panel has no destruction costs associated, its investment cost is such that it becomes the second component only after the storage that presents remarkable destruction and investment costs. - Considering the relative cost difference, it cannot be computed for the solar thermal panel since the cost of the fuel is null (solar energy is for free). While Water storage has the highest value of r due to Department of Energy Politecnico di Milano Author Emanuela Colombo Pag.8 of 12 Date 11/07/2020 a remarkable contribution of costs given by investments and destruction terms, despite the high value of the fuel cost. Indeed, almost 2/3 of the product cost is composed by investments and destruction costs - Considering the exergoeconomic factor, the solar thermal panel has a value of 1 since there is no destruction costs associated. Water storage has a f parameter very close to 0.5, so it is suggested not to change the balance between investments and destruction costs. In case of the heat exchanger, since f is lower than 0.5, an increase of investment costs and component efficiency is suggested. g. Compute the physical exergy of stream 1 and 2 neglecting any pressure drop in the circuit of water- glycol and considering that T 1 is equal to 45°C and T 2 is equal to 90°C. Quantify the mass flow rate of water-glycol mixture. Specific physical exergy of stream 1 and 2. Let’s compute the specific physical exergy of stream 1 and 2. We can directly use the specific heat of the water-glycol solution as reported in the text. Glycol is considered as an incompressible fluid. ������������������������ ������������ℎ =������������� ������������,������������������������������������������������������������−������������������������������������������������������������������������ ( ������������������������−������������ 0) −������������ 0�������������������������,������������������������������������������������������������−������������������������������������������������������������������������ ln������������ ������������ ������������0�� ������������������������ ������������ℎ,1 =�3.13������������������������ ������������������������ ������������ ( 318������������−288������������) −288�������������3.13������������������������ ������������������������ ������������ ln318 ������������ 288������������ ��=4.58������������������������ ������������������������ ������������������������ ������������ℎ,2 =�3.13������������������������ ������������������������ ������������ ( 363������������−288������������) −288�������������3.13������������������������ ������������������������ ������������ ln363 ������������ 288������������ ��=26.12������������������������ ������������������������ Chemical exergy is conserved in the transformation between 1 and 2. Besides, mass flow rate of water-glycol is constant between 1 and 2. And in general, the following equations hold. ������������̇ ������������−������������ ������������������������� Substituting, we obtain: ������������������������ 2 ������������������������� ������������ℎ2+������������������������ ������������ℎ�������������������������� ������������ℎ1+������������������������ ������������ℎ�=������������������������ 1 So: ������������������������ ������������ℎ =������������ ������������ ������������ℎ2������������������������ 1−������������������������ ������������ℎ1������������������������ 2 ( ������������������������ 2−������������������������ 1) We can now substitute all the values in the following equation to get the overall value for the chemical exergy of the mixture ������������������������ ������������ℎ,������������������������������������������������������������������������������������ =������������ ������������ ������������ℎ2������������������������ 1−������������������������ ������������ℎ1������������������������ 2 ( ������������������������ 2−������������������������ 1) =26 .12∙918.21−4.58∙916.06 (918.21−916.06) ������������������������ ������������������������ =9173.06������������������������ ������������������������ The mass flow rate, instead, is: ������������̇ ������������−������������ =������������ ������������ 1 ������������������������� ������������ℎ1+������������������������ ������������ℎ�=916 .06 (4.57+9173.06) = 0.100 ������������������������ ������������ Alternatively (one single passage solution without Chemical Exergy calculation): ������������̇������������−������������ =������������ ������������ 2−������������������������ 1 ������������������������� ������������ℎ2−������������������������ ������������ℎ1�=918 .21−916.06 (26.12−4.6) = 0.100 ������������������������ ������������ Department of Energy Politecnico di Milano Author Emanuela Colombo Pag.9 of 12 Date 11/07/2020 Exercise 2. Let’s consider the energy statistics of Denmark in 2017 (IEA data). Table 1: Primary energy balances in ktoe Table 2: Electricity and Heat production in GWh and TJ respectively Department of Energy Politecnico di Milano Author Emanuela Colombo Pag.10 of 12 Date 11/07/2020 With reference to data provided in Table 1 and Table 2 it is required to: a. Compute the overall value of Total Final Consumption (TFC) and the Electric Penetration b. Consider now natural gas only (both in table 1 and 2) and focus on the production of Heat occurring via CHP and Heat plants. Primary energy (in table 1) used by heat plants (traditional natural gas fuelled boilers) is converted into heat with an average efficiency equal to 0.9 (η boilers,NG = 0.9). Taking into consideration the total heat produced by natural gas as reported in table 2, please quantify the shares of the gross heat production from Natural Gas produced by heat plants and CHP plants. c. Consider now coal only (both in table 1 and 2) and focus on the production of Heat and electricity occurring via CHP. Taking also into consideration that all the gross electricity and heat production (in table 2) is produced via CHP, calculate the efficiency of the coal fuelled CHP plants. Then quantify the amount of primary energy that is saved in this configuration with respect to complete separate production (assuming the same gross production for heat and electricity). Consider only production from coal and use as reference values the following efficiency η boilers,coal = 0.9; η el,coal = 0.40. d. Calculate the TFC for oil products and make some consideration about country energy security with reference to the value you find for this specific source; what would happen if the whole imports of crude oil and oil products may stop overnight? Department of Energy Politecnico di Milano Author Emanuela Colombo Pag.11 of 12 Date 11/07/2020 Exercise 2 - Solution a. Compute the overall value of Total Final Consumption (TFC) and the Electric Penetration The Total Final Consumption is evaluated as the balance of Total Primary Energy Supply plus/minus Transfer and Statistical differences plus/minus contributions from energy conversion process minus Energy industries own uses and Losses (pay attention to the signs!): (17007 26 279 5 856 64 3 157 1123 793) 13887 TFC TPES transferslosses ktoe ktoe statistical conversion own differences processes used = ± ± ± −− = = + − −− + +− − − = = The electric penetration can be computed as follow; considering the sum of Primary Energy devoted to both Electricity and CHP plants and the value of TPES: ( & ),electricity CHP i iPE EP TPES =∑ = ( & ),( ),( ), (1337 1446 66 442 2435) 5726 electricity CHP ielectricity iCHP i i iiPEPEPEktoe ktoe =+ = + ++ + = ∑ ∑∑ = = ( & ), 57260.34 17006 electricity CHP i i PEktoe EP TPESktoe = = = ∑ b. Consider now natural gas only (both in table 1 and 2) and focus on the production of Heat occurring via CHP and Heat plants. Primary energy (in tab 1) used by heat plants (traditional natural gas fuelled boilers) is converted into heat with an average efficiency equal to 0.9 (η boilers,NG = 0.9). Taking into consideration the total heat produced by natural gas as reported in table 2, please quantify the shares of the gross heat production from Natural Gas produced by heat plants and CHP plants Given the values of primary energy devoted to heat plants and the reference value of efficiency it is possible to compute the amount of heat that comes from separate heat production and consequently the relative share to the total. The share of the gross heat production that comes from CHP plants will results as complement to 100%. ,, boiler i boilers i boilersQ PE η = ⋅ , 1 308[ ] 0.9 [ ] 3.6[ ] 11604 0.086 boilers NG heat, NG boilersQ PE ktoeGWh / ktoe TJ / GWh TJ η = ⋅= = ⋅⋅⋅ = = % ������������������������������������������������������������������������������������,������������ =������������ ������������ ������������������������������������������������������������������������,������������ ������������������������������������������������������������������������������������������������,������������ +������������ ������������������������������������,������������ % ������������������������������������,������������ =������������ ������������������������������������ ,������������ ������������������������������������������������������������������������������������������������,������������ +������������ ������������������������������������,������������ =1−% ������������������������������������������������������������������������������������,������������ , , ,, , ,, ,, 11604 %0.56 20640 %1 % 1 0.56 0.44 boilers NG boiler NG boilers NG CHP NG CHP NG CHP NGboilers NG boilers NG CHP NG Q QQ Q QQ == = + ==− =−= + c. Consider now coal only (both in table 1 and 2) and focus on the production of Heat and electricity occurring via CHP. Taking also into consideration that all the gross electricity and heat production (in table 2) is produced via CHP, calculate the efficiency of the coal fuelled CHP plants. Then Department of Energy Politecnico di Milano Author Emanuela Colombo Pag.12 of 12 Date 11/07/2020 quantify the amount of primary energy that is saved in this configuration with respect to complete separate production (assuming the same gross production for heat and electricity). Consider only production from coal and use as reference values the following efficiency ηboilers,coal = 0.9; ηel,coal = 0.40. For this calculation we need to take into account the production of heat and electricity from coal, and the primary energy devoted to coal-fired CHP. TAKE CARE OF UNIT OF MEASUREMENT , , 1 19238 19238[ ] [ ] 0.086[ ] 459.6 3.6 6209 6209[ ] 0.086[ / ] = 534.0 1446 4 heat electricity CHP CHP heat coal electricity coal CHP heat electricity CHP CHP QE PE QTJ TJ GWh / TJ ktoe / GWh ktoe EGWh GWh ktoe GWh ktoe PE ktoe QE PE η η + = ==⋅⋅ = = = ⋅ = + == 59.6 534.0 0.69 1446 + = = To compute the primary energy required in case of separated production=it is necessary to consider the reference values for boilers and electricity plants= = , , ,, , ,, 459.6 510.6 0.9 534.0 1335.0 0.4 heat separated heat boilers electricity separated electricity electricity separated heat coal separated,electricity coal separated tot c Q PE E PE ktoe PEktoe ktoe PEktoe PE η η = =  = = = = =  → 1845.6 oal ktoe = = Now it is possible to quantify the value of Primary Energy Savings (PES) in the case of combined production with respect to separate production as the primary energy that would be devote to coal plants (heat and electricity) in the separate case minus the actual value of primary energy devoted to coal-fired CHP plant. ,,, 1845.6 1446 399.6 separated tot coal CHP coal PES PEPEktoe = − = −= d. Calculate the TFC for oil products and make some consideration about country energy security with reference to the value you find for this specific source; what would happen if the whole imports of crude oil and oil products may stop overnight? Calculations can be done focusing only to the oil products column: ( 1172 1678 324 7 66 14 8946 345) 5340 TFC TPES transferslosses ktoe ktoe statistical conversion own differences processes used = ± ± ± −− = =− − − −− − + − = Oil products that comes from domestic refineries are enough to both balance the negative value of TPES (mostly due to higher export than imports) and to sustain the domestic total final consumption. Looking to Crude Oil balances, it can be seen that imports and exports are almost equal in magnitude so, to face a drop in oil imports, Denmark can reply with an equivalent drop in oil products export affecting only marginally its country energy security.