Energy Engineering - Advanced Thermodynamics and Thermoeconomics

Full exam

Department of Energy Politecnico di Milano Author Pag.1 of 11 Date 11/06/2020 Milan, 5 th June 2020 Exam – Advanced Thermodynamic and Thermoeconomics Exam Simulation – Solutions Exercise 1. (15pt) Let’s consider a simple Energy Conversion System (ECS) enclosed in the control volume depicted in the figure (dashed line). The system is composed by 2 micro-CHP systems (a and b), and one absorption chiller (c), exchanging mass and energy flows between themselves and environment. Two micro-CHP systems receive primary energy from outside of the ECS’s boundary, producing heat and electricity. Electricity is used for the residential buildings, while Heat is used for the absorption chiller to provide cooling load for the buildings. Streams number 8 (flue gases) and 9 (steam leakage) are released to the environment without any additional cost. Table 1 State Ex [MW] 1 Primary Energy 100 2 Primary Energy 100 3 Heat 30 4 Heat 20 5 Cooling Load 20 6 Electricity 40 7 Electricity 20 8 Flue gases 5 9 Steam leakage 5 Table 2 unit cost of Fuel (c1, c2) [�/GJ] 8.333 Z_CHP a (Investment, O&M) [�/s] 0.2 Z_CHP b (Investment, O&M) [�/s] 0.1 25 Z_Absorption Chiller (Investment, O&M) [�/s] 0.1 Table 3 Chemical exergy Specific heat value (c p) @ 65�C Molar mass kJ/kmol kJ/kmol*K kg/kmol C2H6 (ethane) 1493900 57.9 30 CH 4 (methane) 830200 37.7 16 Tamb 25�C Pamb 101325 Pa Absorption Chiller (c) ECS ( Primary Energy) Micro CHP (a) Micro CHP (b) 2 ( Primary Energy) 1 4 (Heat) 3 (Heat) (Cooling Load) 5Electricity Electricity6 7 8 Flue gases Steam leakage9 Department of Energy Politecnico di Milano Author Pag.2 of 11 Date 11/06/2020 With reference to the data provided by the tables, it is required to: a. Write analytically the exergy balance and derive of the rational and functional efficiency of each component. Calculate the exergy destructions, rational and functional exergy efficiencies. Comment on the results with specific focus on the differences between the two efficiency definition. b. Write the Thermoeconomic system of equations for each component and derive the analytical expressions of the cost structure of the products. Highlight the auxiliary equations needed to solve the thermoeconomic system and make sure they are coherent with the “functional” use of the components. c. Solve the system and provide the specific costs of the products for each component [€/MWh]. Then collect all the numerical results into a table giving evidence to the three components of the cost structure for each component. Comment on the obtained results. d. 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. e. The second CHP (component b) is fuelled with a perfect mixture of Ethane (80% molar) and Methane (20% molar). Knowing that the combustion reaction occurs with fuel at 200 kPa and 65°C and air at 200 kPa and environmental temperature compute the chemical and physical exergy of both fuel mixture and air (separately) that enter the CHP. f. Assuming an excess of air (λ) equal to 4.038 compute the mass flow rate [kg/s] of fuel (C 2H6 + CH 4) needed to sustain the process. (Assume air composed by oxygen and nitrogen only) Department of Energy Politecnico di Milano Author Pag.3 of 11 Date 11/06/2020 Exercise 1. (Solution) a. Write analytically the exergy balance and derive of the rational and functional efficiency of each component. Calculate the exergy destructions, rational and functional exergy efficiencies. Comment on the results with specific focus on the differences between the two efficiency definition. Exergy Balance for the components: 1 36 ,, 2 478 , , 34 59 , , )30 )55 )25 DaDa DbDb DcDc a Ex Ex Ex ExEx MW b Ex Ex Ex Ex Ex Ex MW c Ex Ex Ex Ex Ex Ex MW  =++ → =  = +++ → =  + = ++ → =           Rational and functional exergy efficiencies: - CHP – a: 36 1 36 1 30 40 0.7 100 30 40 0.7 100 r f Ex Ex Ex Ex Ex Ex η η + + = = = + + = = =     == -=CHP �=b:== 478 2 472 20 20 50.45 100 20 200.4 100 r f Ex Ex ExEx Ex ExEx η η ++++ == = ++ = = =   == -=Absorption Chiller=�=c:= 59 34 5 34 20 5 0.5 30 20 20 0.4 30 20 r f Ex Ex Ex Ex Ex Ex Ex η η + + = = = + + = = = + +     == = In=CHP �=b and Absorption chiller, the decrease of the functional exergy efficiency is due to the presence of= losses. Specifically, in CHP=�=a, no losses occur and the two exe±gy efficiencies are equal.=Functional= efficiency does not take into account outlet stream that do not result to be a useful product.= b. Write the Thermoeconomic system of equations for each component and derive the analytical expressions of the cost structure of the products. Highlight the auxiliary equations needed to solve the thermoeconomic system and make sure they are coherent with the “functional” use of the components. The system is composed by 9 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 = 9, m: number of components = 3: n-m = 9 - 3 = 6. Auxiliary relations: 1368 2479 1) , 3), 5 ) 0 2) , 4), 6) 0 a b c given c c c cc given c c c c= = = = == = = = = 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, in details:= -=streams 3 and 6: c→-products;= Department of Energy Politecnico di Milano Author Pag.4 of 11 Date 11/06/2020 - streams 4 and 7: co-products; - streams 8 and 9: losses. For each component it is required to write economic cost balance based on Inlet/Outlet paradigm and substituting exergy cost relation as follows; ; ; in componentout inoutC Z C Economic cost balance C c ExExergy cost relation  +=  = ⋅  ∑∑     = = = CHP �=a:= = 1 36 , 1 1 3 36 6 36 , 11 36 36 ( . . ) D CHP a a a D CHP a a a Ex Ex Ex Ex c Ex Z c Ex c Ex aux eq c c c Ex Z c cc Ex Ex Ex Ex − −  =++  ⋅ + =⋅ +⋅  →== =++ ++          == = CHP �=b:= = 2 478 , 22 44778847 8 ,8 22 47 47 ( . . ; 0) D CHP b bb D CHP bb b Ex Ex Ex Ex Ex c Ex Z c Ex c Ex c Exaux eq c c c c Ex ExZ c cc Ex Ex Ex Ex − − = +++   ⋅+=⋅+⋅+⋅   →=== + =++ ++          = = Absorption chiller=-=c:= = = 34 59 , 3 34 4 5 59 9 3 4 5 59 9 34 , 34 9 ,9 5, , 55 ( . . 0) cD ABS c c ab c ab in c D ABS c c in c in c Ex Ex Z Ex Ex Ex c Ex c Ex Z c Ex c Ex c Ex c Ex Z c Ex c Ex c Ex c Ex c Ex Ex aux eq c Ex Ex Z cc c Ex Ex − −  + += + +  ⋅ +⋅ + = ⋅ +⋅ → ⋅ +⋅ + = ⋅ +⋅ ⋅ +⋅ = + →= + =++                   = c. 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 components of the cost structure for each component. Comment on the obtained results. 1 1 3 36 6 22 44778a bc Ex Z c Ex c Ex c Ex Z c Ex c Ex c⋅ + =⋅ +⋅ ⋅+=⋅+⋅+      8 3 34 4 5 5 9c Ex c Ex c Ex Z c Ex c⋅ ⋅ +⋅ + =⋅ +    113 6 224 7 3 4 55 9 () () aa bb ab c c Ex Z c Ex Ex c Ex Z c Ex Ex c Ex c Ex Z c Ex Ex  ⋅ +=⋅ +   → ⋅ +=⋅ +  ⋅ +⋅ + =⋅ ⋅          = = = Now, we have 3 equations and 3 unknowns (c a=, c b, c 5).= = Department of Energy Politecnico di Milano Author Pag.5 of 11 Date 11/06/2020 11 36 22 47 34 5 5 a a b b ab c c Ex Z c Ex Ex c Ex Z c Ex Ex c Ex c Ex Z c Ex  ⋅+ =  +  ⋅+  =  +  ⋅ +⋅ +  =         = In order to compute System solutions: 11 36 22 47 34 5 5 €€ 30 100 720 € 53.14 (30 40) €€ 30 100 450 € 86.25 (20 20) € €€ 53.14 30 86.25 20 360 € 183.96 20 a a b b ab c MW c Ex Z MWhh c MWMWh Ex Ex MW c Ex Z MWhh c MWMWh Ex Ex MWMW c Ex c Ex Z MWhMWhh c MWMWh Ex  ⋅+ ⋅+ === + + ⋅+ ⋅+ === + + ⋅+ ⋅+ ⋅ +⋅ + ===         = c1 [€/MWh ] c2 [€/MWh ] c3 [€/MWh ] c4 [€/MWh ] c5 [€/MWh] c6 [€/MWh] c7 [€/MWh] c8,9 [€/MWh] 30 30 53.14 86.25 183 .96 53.14 86.25 0 Cost structures: - CHP – a (c a):  , 11 36 36 ()() Da a a inlet ExergyDestruction Investment Cost Ex Z c cc Ex Ex Ex Ex = +⋅ + ++      = = -=CHP=�=b=(c b):= =  8 ,b 22 47 47 ()() Db b inletExergyDestruction Losses Investment Cost Ex ExZ ccc Ex Ex Ex Ex + + = +⋅ + ++       = -=ABS �=c (c 5):= = For the absorption chiller, since there are two inlet streams, a weighted average of the economic cost needs= to be computed:= 34 , 34 €€ 53.14 30 86.25 20€ 66.38 (30 20) ab in cc Ex c Ex c Ex Ex MWMW MWhMWh MWMWh ⋅ +⋅ == + ⋅+ ⋅ == +   = The costs structure results to be as follows:=   9 ,c 5, , 55 D c in c in c inlet Investment Cost ExergyDestruction Ex Ex Z cc c Ex Ex + = +⋅ +         Department of Energy Politecnico di Milano Author Pag.6 of 11 Date 11/06/2020 component fuel destruction + losses investment total €/MWh €/MWh €/MWh €/MWh CHP – a 30 12.857 * 10.286 * 53.14 CHP – b 30 45 11.25 86.25 ABS – c 66.38 4* 99.57 6* 18 183.96 *third digit is reported in order to match the exact results d. 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. To perform the design evaluation of the system is necessary to compute the cost of the exergy destructions, the exergoeconomic factors and the relative cost differences. The cost of inefficiencies is computed according to the demand driven approach, so the exergy destruction and losses are valued as extra consumption of the fuel. The formula is: , , ,, () D Lj fuelj Dj LjC c Ex Ex + =⋅+   = = Exergoeconomic factor is computed as:= ,j j j jD Z f ZC= +    = Relative cost difference is computed as:= ,, , , ,, P j F j des loss j inv j j F jfuel j cc c c r cc + −+ = = == In the following table, all the results are collected.= Component = Ex D+L,i = CD+L,i = Zi= CD+L,i +Z i= fi= ±i= = €/h €/h €/h €/h - - CHP � a 30 900 720 1620 0.44 0.77 CHP � b 55 +5 180 0 450 225 0 0.2 0 1.87 ABS � c 25 +5 1992 360 2352 0.1 5 1.77 Comments: - According to the exergy destruction, losses and investment, component c is the worst component of the plant with respect to the other components, even if component b is really close to component c’s values. It is worthwhile to note that the component with higher exergy inefficiencies, component b totalizing 60 MW of destruction+losses, is not the worst component according to thermoeconomics, since the exergy inefficiencies of component c are valued at a higher cost (higher fuel cost). - Exergoeconomic factor indicates whether the capital and O&M expenses (Ż ) is the major source of economic cost increase. The indicator takes values between zero and one: high values indicate (Ż ) is the major cost source and the primary aim is to reduce cost of the products by reduction of 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. Based on such explanation, component c has the lowest value of Exergoeconomic factor and needs to be improved by increasing its efficiency. Same rational can be applied to component b, while component a results to be balanced in terms of investment and exergy destruction costs (f=0.44 close to 0.5). - Relative cost difference: high values of this parameter means there is more room for improvement. Based on that, component b and c are the first candidates to be improved. e. the second CHP (component b) is fuelled with a perfect mixture of Ethane (80% molar) and Methane (20% molar). Knowing that the combustion reaction occurs with fuel at 200 kPa and 65°C and air at 200 kPa and environmental temperature compute the chemical and physical exergy of both fuel mixture and air (separately) entering the CHP. The chemical exergy for fuel and air can be computed as follows: Department of Energy Politecnico di Milano Author Pag.7 of 11 Date 11/06/2020 ( ) 2626 4426 26 26 26 ,,0 , , ,0 , ln ln( ) ln( ) 1359919.59 0 ch fuel i ch ii i ii ch fuel CH chCH CH chCHCH CH CH CH ch air ex v ex T R v x ex v ex v ex RT v v v vkJ kmol ex = + → = ⋅ +⋅ ++ = →= ∑∑ = The physical exergy for fuel and air can be computed as follows: @ 65@ 200 ,@ 65 0 , 00 @ 65@ 200 , @ 65 @ 65 0 0 , 00 ,0 lnln =() lnln1818.32 l i T Ci p kpa ph fuel i i T Cp i i i T Ci p kpa i PiT C iT Cpi fuel i ph air Tp ex v h T cR Tp Tp v c T T T cRkJ kmol Tp ex T R =°= = ° =°= =°=°   =∆− − =        −−−=      = ∑ ∑ @ 200 0 n1685.56 i p kpa air p kJ kmol p = = Total exergy values for the fuel mixture and air are: ,, ,, 1359919.59 1818.32 1361737.91 0 1685.56 1685.56 fuel ch fuel ph fuel fuel air ch air ph air air kJ ex ex ex kmol kJ ex ex ex kmol =+= + = =+=+ = == f. Assuming an excess of air (λ) equal to 4.038 compute the mass flow rate [kg/s] of fuel (C 2H6 + CH 4) needed to sustain the process. (Assume air composed by oxygen and nitrogen only) The value of lambda allows to compute the number of moles of air per molar unit of fuel starting from the formula of combustion with excess air supply: ( ) ( ) 2 2 2222 3, 763, 76 1 424 4 ab bbb b C H a O N aCO H O a N a O λλλ  ++ + → + ++ +− +    = Since, in this case, we should refer to a mixture of hydrocarbons, the values of=a and b should be weighted according to the fuel mixture composition. ( ) 0.8 2 0.2 1 1.8 ( ) 0.8 6 0.2 4 5.6 ii i ii i a v n atoms C b v n atoms H= ⋅° = ⋅ + ⋅ = = ⋅° = ⋅ + ⋅ = ∑ ∑ == Assuming that air is composed by oxygen and nitrogen only, derived from the combustion reaction, the total= number of moles of air required are:= (1 3.76) 61.51 4 038 4 . air air fuel kmol b va kmol λ λ = + ⋅+ =  = → The value of primary energy ������������̇ ������������ 2 refers to the mixture fuel (C 2H6 + CH 4) and air together. It is required to compute the overall exergy content of the fuel mixture plus the air supply. The procedure is similar to the one expressed previously. First of all, the composition of the new fuel+air mixture is the following: 1 10.016 1 61.51 61.51 61.510.984 1 61.51 fuel mixturefuel fuel airair air n kmol x n kmol x = →= = + = →= = + = Chemical=and physical=exergy=are=computed as follows:= Department of Energy Politecnico di Milano Author Pag.8 of 11 Date 11/06/2020 ( ) ,,0 , ,0 ln ln( ) ln( ) 21553.01 ch fuel air i ch ii i ii fuel ch fuel air ch airfuel fuel air airmix exx ex T R x x x ex x ex RT x x x xkJ kmol + = + = ⋅ +⋅ ++ = ∑∑ ,, ,, 1687.68 / ph fuel air i ph i i fuel ph fuel air ph airmix exx ex x ex x exkJ kmol + =⋅= = ⋅ +⋅ = ∑ Total molar exergy of the mixture is: ,, 23240.69 / mix ph fuel air ch fuel airmixex ex exkJ kmol ++ = += == The molar flow rate of the mixture is computed as:= { 100000 4.303 23240.69 0.016 4.303 0.069 / totmix mix fuel mix fuel fuel totfuel mix kW nkmol s kJ kmol kmol kmol where n x nkmol s kmols == = ⋅= ⋅ =   The mass flow rate of fuel is: 26 26 4 4 27.2 27.2 0.069 1.87 fuel fuel C H C H CH CH fuel fuelfuelfuel fuelfuel fuel fuel kg MM x MM x MM kmol kg kmol kg m MM n kmolss = ⋅ +⋅ = = ⋅= ⋅ =  Department of Energy Politecnico di Milano Author Pag.9 of 11 Date 11/06/2020 Exercise 2. (6pt) Let’s consider the energy statistics for Italy in 2017 (IEA data, values in ktoe). The year after the disposal of a 3 GW Coal plant is planned. In order to maintain the same level of yearly production, the gap of electric energy production is covered for 50% by Natural Gas Combined Cycle power plants (NGCC), and for 50% by a new off-shore wind farm. Using IEA standards and according to the data given in the table, it is required to calculate: a. The Total Primary Energy Supply of Italy in 2017 and the new value of the TPES in the year after and comment the obtained values. b. The change in CO 2 emissions due to the new yearly production of energy from NGCC (in Mton) and the installed capacity of the new wind farm (in GW). Assume for the coal, a CO 2 emission factor (k) equal to 0.74 ton/MWh el while for NGCC, 0.39 ton/MWh el. c. Compute the cost of the electricity produced by wind (€/MWh) so that the change in the energy mix results with zero increase total cost of electricity production, assuming a carbon tax of 10€/ton of CO 2. d. If the cost of electricity produced by wind is 85€/MWh, assess the additional overall cost (M€) of the intervention. Additional data: - Natural Gas Combined Cycle plant are assumed to use pure methane with η I=0.55 and LHV ng=50 MJ/kg. - For wind farm assume a load factor of 0.32. - For coal power plant assume η I=0.38 and a load factor of 0.62. - Economic costs of electricity: 67 €/MWh for NGCC, 51 €/MWh for Coal. Department of Energy Politecnico di Milano Author Pag.10 of 11 Date 11/06/2020 Exercise 2. (Solution) a. The Total Primary Energy Supply of Italy in 2017 and the new value of the TPES in the year after and comment the obtained values. The TPES for Italy in 2016 is evaluated through the algebraic sum of the last column of the given table. endogenousstock TPESimports exports bunkers productionchanges = + −− ± = = () []34016 157930 33366 2303 3419 587 153445 TPESktoe ktoe = + − −++ = = = The new value of the TPES:= = , , 3 8760 16294 0.086 3687 0.38 el coalcoal el coal coal Ef GWh E ktoe PEktoe GWh =⋅ ⋅= =⋅= = = ,,, 0.5 16294 0.5 8147 0.086 1274 0.55 el NG el coal el NG NGE EGWh Ektoe PEktoe GWh = ⋅= ⋅= =⋅= = = ,,, 0.5 8147 0.086 701 1 el W el coal el W WE EGWh Ektoe PEktoe GWh = ⋅= =⋅= = = 153445 3687 1274 701 151733 newcoal NG W TPES TPES PE PE PEktoe= − + + = − + += = b. The change in CO2 emissions due to the new yearly production of energy from NGCC (in Mton) and the installed capacity of the new wind farm (in GW). The emissions of CO2 avoided by shutting down the coal plants are: (the emission factor is defined as k) 2,,222 2,,222 2 2,2,2 16294000 0.74 12.10 06 12.1 8147000 0.39 3.2 06 3.2 12.1 3.1 9 CO COAL el coal CO COALCOCO CO NGCC el NGCC CO NGCCCOCO CO CO COAL CO NGCCCO m E ke ton Mton m E ke ton Mton m m mMton − − = ⋅ = ⋅= == ⋅ = ⋅= = →∆ = − = − = == = Known the amount of electricity that has to be produced by the wind farm and given the load factor it is possible to evaluate the nominal=installed capacity=of the=wind=farm.= , , 81472.9 8760 8760 / 0.32 el W el W WEGWh PGW hy f h y = == ⋅⋅ = = c. Compute the cost of the electricity produced by wind (€/MWh) so that the change in the energy mix results with zero increase total cost of electricity production, assuming a carbon tax of 10€/ton of CO2. The theoretical cost of wind is: Department of Energy Politecnico di Milano Author Pag.11 of 11 Date 11/06/2020 , ,2 2,, ,2 2,, , , , ,2 2,, ,2 2, ,, , ,, () € 51 16294 03 1 el COAL el COAL CO CO COAL el NGCC el NGCC CO CO NGCCel WIND th el WIND el COAL el COAL CO CO COAL el NGCC el NGCC CO CO NGCC el WIND th el WIND el WIND thc E cm c E cm c Ec E cm c E cm c E e MWh MWh c+⋅ = +⋅ + +⋅ − +⋅ = ⋅+ = €€ € 0 12.1 06 (67 8147 03 10 3.1 06 ) 8147000 831 06 121 06 546 06 31 06 46€ / 8147000e ton e MWh e ton tonMWhton MWh e e eeMWh ⋅ − ⋅ +⋅ = +− − == d. If the cost of electricity produced by wind is 85€/MWh, assess the additional overall cost (M€) of the intervention. , ,, € C () 8147000 (85 46) 317 06€ 317 € WIND el WIND el WIND thE c cMWhe M MWh ∆= ⋅ − = ⋅ − = =