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Energy Engineering - Advanced Thermodynamics and Thermoeconomics

Full exam

Department of Energy Politecnico di Milano Author E. Colombo Pag.1 of 12 Date 23/09/2020 Milan, 1 st September 2020 Exam – Advanced Thermodynamic and Thermoeconomics Session of 04-09-2020 Total score: 21 points (exercise 1 roughly 2/3 of the total) Exercise 1. Consider the closed cycle power plant depicted in the figure with a boiler, a turbine, a condenser and a pump. The boiler is fed by fuel & air and it produces steam which enters the turbine. The turbine produces power for the national electric grid. In the condenser, water condenses by using an external stream of cooling water. The pump closes the cycle and allows the water to enter again the boiler. The pump requirements are 42 MW from the national grid, paid at a cost of 40 €/MWh. The water-steam closed cycle has constant mass flow rate of 1.711 ton/s. Flue gases (c p=1.28kJ/kg K; MM=28.26 kg/kmol) are released to the environment without any additional costs. Cooling water is withdrawn from the environment at zero cost. Stream 6 is not used but recollected in an artificial basin at zero additional costs, ready for other cooling-down purposes. The mass flow rate of Fuel is 0.137 ton/s and its LHV is 49000 MJ/ton, flue gases mass flow rate is instead 5.060 ton/s. The net power of the steam turbine is 2 GW. Thermodynamic data, Physical data and Economic data are given in the tables below. Reference conditions: T 0=15°C, P 0=1.013 bar. Reference environment: 0.74 N 2, 0.22 O 2, 0.01 CO 2 , 0.03 H 2O Flue gases Fuel & Air Cooling water 7 8 Stream T P h s Ex [⁰C] [bar] [kJ/kg] [kJ/ kgK] [MW] 1 700.0 150 3839.48 6.96 3142 2 60.1 0.2 2608.95 7.91 568 3 60.1 0.2 251.40 0.83 23 4 63.0 150 276.17 0.86 51 5 15.0 1.013 63.08 0.22 0 6 35.0 1.013 146.73 0.51 135 7 � Fuel & air ? 8 � Flue gases 90.0 1.013 ? unit cost of Fuel,Air (c F/A)[�/MWh] 15 unit cost of for pump operation[�/MWh] 40 Zboiler (Investment, O&M) [�/h] 9000 Zturbine (Investment, O&M) [�/h] 18000 Zcondenser (Investment, O&M) [�/h] 6000 Zpump (Investment, O&M) [�/h] 1000 Department of Energy Politecnico di Milano Author E. Colombo Pag.2 of 12 Date 23/09/2020 With reference to the data provided, please provide the answers to the below questions, point by point: a. Write the exergy balance and compute the exergy destruction of each component. Calculate the rational and functional exergy efficiencies. Comment on the results. b. Highlight the auxiliary equations needed to solve the thermoeconomic system and make sure they are coherent with the “functional” efficiency of each component. Comments on the results. c. Write the thermoeconomic system of equations for each component and derive the analytical expressions of the cost structure of the products. d. Compute the unit cost of the products for each of them [€/MWh]. Please consider that it is not necessary to solve the systems of equations, knowledge of thermoeconomics provides you with easier way to solve it. Collect all the numerical results into a table giving evidence to the three terms of the cost structure for each component. Please extensively comment on the table. e. Compute the real value for chemical exergy of the fuel. The fuel mixture is composed by 20% of methane (CH 4) and 80% of ethane (C 2H6) on volume basis. Then compute the conversion coefficient necessary to correct the value of LHV to calculate the real chemical exergy of the fuel. The chemical exergies of the components are given: ������������������������������������ℎ,������������������������4= 838789 ������������������������ ������������������������������������������������ ������������������������������������������������4= 16 ������������������������ ������������������������������������������������ ������������������������������������ℎ,������������2������������6= 1502481 ������������������������ ������������������������������������������������ ������������������������������������2������������6= 30 ������������������������ ������������������������������������������������ f. Please consider the specific cost ������������ ������������������������������������ Department of Energy Politecnico di Milano Author E. Colombo Pag.3 of 12 Date 23/09/2020 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 results. First of all, it’s necessary to compute the exergy related to the flue gases stream. Physical Exergy is dominant: ������������������������ ������������������������ =������������̇ ������������������������ ∙������������� ������������������������������������������������� −������������ 0�−������������ 0������������� ������������ln������������ ������������������������ ������������0− ������������ ������������������������ ������������������������ ln������������ ������������������������ ������������0� �= =5.06������������ ������������������������ ������������ ∙�1.28 ������������������������ ������������������������ ������������ ( 363.15 ������������−288.15 ������������) −288.15 �������������1.28 ������������������������ ������������������������ ������������ ln363 .15 ������������ 288.15 ������������ −8 .314 28.26 ������������������������ ������������������������ ������������ ln1 .013 1.013 ��= =54 ������������������������ Chemical exergy is dominant, and as a first approximation, chemical exergy for the fuel can be computed as: ������������������������̇ ������������ /������������ =������������̇ ������������������������������������������������ ������������������������������������=0.137������������ ������������������������ ������������ ∙49000������������������������ ������������������������������������ =6713 ������������������������ Boiler / 41, , / 41 1 ,, /4 1 , f, /4 ( ) 6713 (51 3142) 54 3568 0.472 0.464 F Afg D boiler D boiler F Afg FG ex r boiler FA ex boiler FA Ex Ex Ex Ex Ex Ex Ex Ex Ex ExMW Ex Ex Ex Ex Ex Ex Ex η η  +=+ +  = + − − = + − −=  + →= = + →= = +             Turbine 12 , , 12 2 ,, 1 , f,12 ( ) (3142 568) 2000 574 0.817 0.777 net D turbine D turbinenet net ex r turbine net ex turbine Ex Ex W Ex Ex Ex Ex WMW Ex WEx W Ex Ex η η =++  = − −= − − =   + →= = →= =−         Condenser: () 25 36 , ,23 56 36 ,, 25 3 , f, 2 56 ( ) ( ) (568 23) (0 135) 410 0.278 0.053 D condenser D condenser ex r condenser ex condenser Ex Ex Ex Ex Ex Ex Ex Ex Ex ExMW Ex Ex Ex Ex Ex Ex Ex Ex η η + =++  = − + − = − +− =   + →== + →= =+−           We assume that the stream 3 is the unique product and that the difference between the stream 5 and 6 is the cooling power. In other words, 6 is a by-product in the condenser and should bear the same cost of 5. Department of Energy Politecnico di Milano Author E. Colombo Pag.4 of 12 Date 23/09/2020 3 4, , 34 4 ,, 3 4 , f, 3 ( ) (23 51 42) 14 0.785 0.785 pumpD pump D pumppump ex r pump pump ex pump pump Ex W Ex Ex Ex Ex Ex WMW MW Ex Ex W Ex Ex W η η  +=+  = − + = −+ =  →= = + →= = +           b. Highlight the auxiliary equations needed to solve the thermoeconomic system and make sure they are coherent with the “functional” efficiency. Comments on the results. Exergy cost balance composed by 4 equation and 10 unknowns. Therefore, in order to close the system of equations, (n-m) numbers of auxiliary relations are required: n: Number of exergy streams = 10 m: number of components = 4 n-m = 10 - 4 = 6 5 65 12 10 4 6 1) 0 2) 3) 4) 0 5) 6) FG F/ A pump n streams n components c c given c given c cc cc ° −° = − = =  =  =  =  =  =  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 derive the analytical cost structure for the product ; Boiler: / 41, /4 1 F AFG D boiler F AboilerFG Ex Ex Ex Ex Ex C C Z CC C c Ex  +=+ +  ++ =+  = ⋅           = = 44 / / , /4 , boiler 1, , 11 () FA FA in B FA DFG boiler in B in B c Ex c Ex c Ex Ex Ex Ex Z cc c Ex Ex + →= + + =++      = Turbine= = = 12 , 12net D turbine turbineWnetEx Ex W Ex CZ C C C c Ex  =++   +=+  = ⋅       = = Department of Energy Politecnico di Milano Author E. Colombo Pag.5 of 12 Date 23/09/2020 , 11 D turbine turbine Wnet netnet Ex Z c cc WW  =++     Condenser: 25 36 , 2536 D condenser condenser Ex Ex Ex Ex Ex CC Z CC C c Ex  + =++   ++ = +  = ⋅       = = () () 2 2 55 33 66 , 2 2 5 5 6 33 , , 2 5 6 33 ,D condenser D condenser in condD condenserc Ex c Ex c Ex c Ex Ex c Ex c Ex Ex c Ex Ex c Ex Ex Ex c Ex Ex+=++ + −= + +− = +          = ( ) ( ) 22 5 5 6 , 2 56 , 3, , 33 () in cond D cond cond in cond in cond c Ex c Ex Ex c Ex Ex Ex Ex Z cc c Ex Ex +− →= +− =++        = = Pump:= == = d. Compute the unit cost of the products for each component [€/MWh]. Please consider that it is not necessary to solve the systems of equations, knowledge of thermoeconomics provides you with easier way to solve it. Collect all the numerical results into a table giving evidence to the three terms of the cost structure for each component. Please extensively comment on the table. The system is defined in 4 equations and 4 unknowns, but since the aggregated system has only one product, in order to facilitate the computation of the products’ cost, we can compute the value of the cost of W net and compute backward the intermediate products. 33 , 3 , 4, , 44 elec pump in pump pump D pump pump in pump in pump c Ex c W c Ex W Ex Z cc c Ex Ex + →= + =++   3 4, 34pump D pump elec pump Ex W Ex Ex CC Z C C c Ex  +=+   ++ =  = ⋅        Department of Energy Politecnico di Milano Author E. Colombo Pag.6 of 12 Date 23/09/2020 ( ) ( ) / / 5566 //556 / 56 €€ 15 6713 0 40 42 (6713 0 135 4 F A F AW pump pump boiler turbine cond pumpfg fg Wn et net F A F AW pump pump system F Apump c Ex c Ex c W Z Z Z Z c Ex c Ex c W MWhMWh c Ex c Ex Ex c W MWhMWh c Ex Ex Ex W − − + + + + ++ = + + ++ + −+ →== +− + +−+                ( ) , € 15.465 2) () 4566 54 (34000) € 15.465 15.465 2000 2000 €€ 15.465 35.724 17 68.189 system D tot FG Wnet system system netnet MWhMWh Z Ex Ex cc c MWh WW MWh MWh = + +  =++= + + =  = ++ =    From the cost structure of the turbine, c 1 can be computed: , 11 12 , 68.189 2000 18000 €€ 46.008 (2000 573) () D turbine turbine Wnet netnet Wnet net turbine net D turbine Ex Z c cc WW cW Z cc MWh MWh W Ex  =++  −  ⋅− →= ===  + +       = So:  Turbine: () , 11 573 18000€€ 46.008 46.00846.008 13.181 9 68.189 2000 2000 D turbineturbine Wnet netnet Ex Z c cc MWh MWh WW  =+ + = + + = ++ =      Condenser: () 22 55 55 , 256 , 3, , 33 46.008 568 0 0 €60.352 568 0 135 ()(411) 6000 € 60.352 60.352 23 23 € 60.352 1078.464 260.870 1399.68 in cond D condcond in cond in cond c Ex c Ex c Ex c MWh Ex Ex Ex ExZ cc c MWh Ex Ex MWh +− ⋅ +− →== = +− +−  = + += + + =  =++ =    € 6 MWh  Pump: 33 , 3 , 4, , 44 1399.686 23 40 42€ 521.120 23 42 14 1000€ 521.120 521.120521.120 143.053 19.608 683.781 51 51 elec pump in pump pump D pump pump in pump in pump c Ex c W c MWh Ex W Ex Z cc c MWh Ex Ex + ⋅+⋅ →= == + + = + + = + += + + =   =  Boiler: Department of Energy Politecnico di Milano Author E. Colombo Pag.7 of 12 Date 23/09/2020 44 / / , /4 , boiler 1, , 11 683.781 51 15 6713 €20.043 51 6713 ()3568 54 9000€ 20.043 20.04320.043 23.105 2.864 46.012 3142 3142 FA FA in B FA Dfg boiler in B in B c Ex c Ex c MWh Ex Ex Ex ExZ cc c MWh Ex Ex + ⋅+⋅ →= == + + ++ =++ = + += + + =     For approximation propagation we have a negligible relative error of: 11 4 1 (46.012 46.008) 10 46.008 boiler allsystem err allsystem cc R c −− − − − − ==< = component cfuel =c des+loss =c inv=c tot= ==€/MWh €/MWh €/MWh €/MWh Boiler 20.043 = 23.105=2.864 = 46.012= Turbine = 46.008 = 13.181=9.000 = 68.189= Condenser = 60.352 = 1078.464=260.87 =1399.686= Pump = 521.120 = 143.053=19.608 = 683.781= Comments: Even without applying the Design Evaluation, some consideration can be derived from the above table recalling the value of the product for each component. - It is evident that the Boiler has the highest cost of Destruction + Investment since the exergy value of the product is very high. Moreover it is possible to anticipate that the value of the relative cost difference is expected to be (due to the strong increase of the cost of the product compared to the cost of the fuel) and that this will be dominated by exergy destruction (as expected in boiler with internal combustion mechanism) - Similarly the condenser is expected to have high relative cost difference, dominated by exergy destruction costs. Indeed, in this component we are destroying exergy for bringing the stream 2 into the right condition to enter the pump (liquid). The condenser is a sink of exergy or a “dissipative” component. - Pump is expected to have a very low relative cost difference as visible by comparing the cost of its product with the cost of the fuel and again value is dominated by exergy destruction. - The turbine cost, is not very much high and the relative contributions of destruction/losses and investment costs are almost equal with a slightly higher contribution of exergy destruction costs. So it would be difficult to give indication and expect outcome after a change of its configuration. One should try, recalculate and see. e. Compute the chemical exergy of the fuel. The fuel mixture is composed by 20% of methane (CH4) and 80% of ethane (C2H6) on volume basis. Then compute the conversion coefficient necessary to correct the value of LHV to calculate the real chemical exergy of the fuel. Methane: ������������������������ ������������ℎ,������������������������4 =838789������������������������ ������������������������������������������������ ������������2������������6������������2 Ethane: ������������������������ ������������ℎ,������������2������������6 =1502481������������������������ ������������������������������������������������ ������������2������������6 Chemical exergy of mixing: Department of Energy Politecnico di Milano Author E. Colombo Pag.8 of 12 Date 23/09/2020 ������������������������ ������������������������������������������������������������������������ =������������ 0������������∙( ������������ ������������������������4 ∙ln( ������������ ������������������������4 ) +������������ ������������2������������6 ∙ln( ������������ ������������2������������6 )) =288.15������������∙8.314������������������������ ������������������������������������������������ ������������ ( 0.2∙ln( 0.2) +0.8∙ln( 0.8) ) =−1199������������������������ ������������������������������������������������ ������������������������������������ Total specific (on volume basis) chemical exergy is: ������������������������ ������������ℎ,������������������������������������ =������������ ������������������������4 ������������������������ ������������������������4 +������������ ������������2������������6 ������������������������ ������������2������������6 +������������������������ ������������������������������������������������������������������������ =( 0.2∙838789+0.8∙1502481+(−1199))������������������������ ������������������������������������������������ ������������������������������������ =1368544������������������������ ������������������������������������������������ ������������������������������������ The mixture molar mass is: ������������������������ ������������������������������������ =������������������������ ������������∙������������������������=�0.2 ������������������������������������ ������������ ������������������������4 ������������������������������������������������ ������������������������������������ ∙16������������������������ ������������������������������������������������ ������������������������4 +0.8������������������������������������ ������������ ������������2������������6 ������������������������������������������������ ������������������������������������ ∙30������������������������ ������������������������������������������������ ������������2������������6 �=27.2������������������������ ������������������������������������������������ ������������������������������������ ������������������������ ������������������������������������ =������������������������ ������������ℎ,������������������������������������ +������������������������ ������������ℎ,������������������������������������ =1368544������������������������ ������������������������������������������������ ������������������������������������ +0������������������������ ������������������������������������������������ ������������������������������������ =1368544������������������������ ������������������������������������������������ ������������������������������������ = =�1368544 ������������������������ ������������������������������������������������ ������������������������������������ 27.2������������������������ ������������������������������������������������ ������������������������������������ �=50314������������������������ ������������������������ ������������������������������������ =50314������������������������ ������������������������������������ ������������������������������������ The chemical exergy conversion coefficient (φ) is: ������������=������������ ������������̇ ������������ ℎ ������������������������������������ = 50314 ������������������������ ������������������������������������ 49000 ������������������������ ������������������������������������ =1.027 f. Please consider the specific cost ������������ ������������������������������������������������ ( ������������̇ ������������������������������������������������ ) and ������������ ������������������������������������������������ ( ������������̇ ������������������������ ������������ ). Then make some comments from the perspective of the owner of the plant. The cost of the power produced by the plant is higher than the price of the electricity paid by the owner of the plant (Pump supply). This means that this power plant is not competitive if we look at a simplified version of the electricity market. There is the need to reduce the cost of the power produced at least below the price of the electricity. Actually, the price at which the power plant owner pays the electricity might be a discounted value/price with respect to the full price of the market. Indeed, for energy-intensive plants, in order to feed auxiliaries, the distribution system operator applies heavy discounts. It may happen that, as in this case, the price of the electricity produced overpasses the average value of the price (contracted with the distribution system operator). In these cases, we should look at the overall clearing price of the market that can be higher than the cost of the power plant. Department of Energy Politecnico di Milano Author E. Colombo Pag.9 of 12 Date 23/09/2020 Exercise 2. Consider the energy statistics of Germany 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 E. Colombo Pag.10 of 12 Date 23/09/2020 With reference to data provided in Table 1 and Table 2 it is required to: a. By using Table 1 , compute the Gross Electricity Production in GWh. b. With reference to IEA energy conversion assumptions quantify the share of renewable energies (as an aggregated value) on the electricity production mix. Because of the national commitment to the Paris agreement, a 30% phase-out of actual production for coal power plants is planned in the following 10 years. Keeping the same gross electricity production of the system, the power sector is expected to react under the following assumptions: - 95% of the electricity production gap will be covered by off-shore wind farms in the Northern Sea. - 5% of the electricity production gap will be covered by increasing the operating hours of existing nuclear power plants. c. With reference to this scenario it is required to compute the new wind power capacity that has to be installed before 2027 and the new value of operating hours for nuclear power plants in that year. Consider that average operating hours for off-shore wind power plants in Northern Sea is 3000 h/y and installed capacity of Nuclear Power plant in 2017 is 12 GW. d. In this scenario, please make some comments on how the country would benefit from a partial phase out of power plants that use this source. Department of Energy Politecnico di Milano Author E. Colombo Pag.11 of 12 Date 23/09/2020 Exercise 2. (Solution) a. By using Table 1, compute the Gross Electricity Production in GWh. The Gross Electricity Production can be retrieved by looking at the balances tables by crossing the Electricity column with the rows of Electricity Plants and CHP Plants. Notice that the value has then to be converted into GWh by means of the appropriate conversion factor. [44253 11442]647616 1 0.086 Gross Electricity ProductionktoeGWh GWh ktoe =+⋅ = = = b. With reference to IEA energy conversion assumptions quantify the share of renewable energies (aggregated value) on the electricity production mix. Being able to quantify the Nuclear electricity production from the value of Primary Energy Supply and using the proper conversion assumption, the overall production of electricity from renewables can be computed as difference from the Gross Electricity Production by subtracting the contribution of Nuclear, Coal, Oil and Natural Gas. , 11 19887 0.3376310 0.0860.086 nu nu ref nu GWhGWh EE PEGWh ktoektoe η =⋅⋅=⋅⋅= = = 647616 76310 252824 5571 87685 225226 res gross nu coal oil ng EE EE EE EE EE EEGWh = − − − − = − − −− = = = 225226 %0.35 647616 res res gross EEGWh EEGWh = = = = =c. With reference to this scenario it is required to compute the new wind power capacity that has to be installed before 2027 and the new value of operating hours for nuclear power plants in that year. Consider that average operating hours for off-shore wind power plants in Northern Sea is 3000 h/y and installed capacity of Nuclear Power plant in 2017 is 12 GW. The electricity production gap that has to be covered in 2027 is equal to 30% of the actual production from Coal. 0.3 252824 0.3 75847 gap coal EE EEGWh = ⋅= ⋅= = = According with the reference scenario the electricity=production gap can=be subdivided as follow:= = 0.05 75847 0.05 3792 nu gap EE EEGWh∆ = ⋅= ⋅= = 0.95 75847 0.95 72055 wind gap EE EEGWh∆ = ⋅= ⋅= = = By using the average operating hours for wind farms it is possible=to quantify the amount of wind capacity that= is necessary to install before the 2027= = , 72055 24 3000 wind wind av wind EE GWh CGW hh ∆ ∆= = = = = On the other way round, by considering the=installed capacity of nuclear power plants it is=possible to quantify= the new value of operating hours of nuclear in 2027= = @ 2027@ 2017 [76310 3792] 80102 nununu EE EE EEGWh GWh = +∆ = + = = Department of Energy Politecnico di Milano Author E. Colombo Pag.12 of 12 Date 23/09/2020 @ 2027 @ 2027 80102 6675 12 nu nu nu EE GWh hh C GW = == == Or we can calculate the variation in the operating hours= = @ 2027 3792 316 12 nu nu nu EE GWh hh CGW ∆ ∆= = = = = d. Make some comments on how the country would benefit from a partial phase out of Coal power plants Coal Primary Energy devoted to Electricity and CHP plants is greater than the value of production. Thus, a partial phase out of coal plant can reduce the exposure of the country from imports and a consequent increase of the national energy security