Energy Engineering - Advanced Thermodynamics and Thermoeconomics

Full exam

Department of Energy Politecnico di Milano Author E. Colombo Pag.1 of 13 Date 26/06/2018 Milan, 21st June 2018 Exam – Advanced Thermodynamic and Thermoeconomics Session of 21-06-2018 Exercise 1. (8pt) Consider a process made by the following sequence: 1. an adiabatic combustion chamber where a complete and isobaric combustion of acetylene with excess air takes place. 2. after combustion, flue gases enter a counter-flow heat exchanger and are cooled at constant pressure heating a flow of water that enter as liquid and exit as superheated steam The main hypotheses and data are below described: - Acetylene , C 2H2 has LHV=1256460 kJ/kmol and ex ch=1269310 kJ/Kmole; - Air and Acetylene enter the combustion chamber @ 353 K and environmental pressure. - The adiabatic flame temperature results 2250 K and is associated to the given excess air. - Flue gasses in the heat exchanger are cooled down @ 673 K, - Water enter the heat exchanger @ 323 K and 500 kPa (h=209.3 kJ/kg, s=0.7027 kJ/kg-K) - The superheated steam exits then @ 873 K and 500 kPa (h=3701 kJ/kg, s=8.3513 kJ/kg-K). - Environmental temperature and Pressure are assumed to be T 0=298 K and p 0=101.3 kPa, - For products P and reactants R useful data are below given: Δh R (353 K) Δh P (2250 K) Δh P (673 K) ex ph,R (353 K) ex ph,P (2250 K) ex ph,P (673 K) kJ/kmol kJ/kmol kJ/kmol kJ/kmol kJ/kmol kJ/kmol C2H2 3101 260 N2 1600 65205 11121 132 45905 3944 CO2 106640 16450 76029 5995 H2O 86162 13164 61636 4697 O2 1622 68771 11647 134 48453 4164 a. (2pt) Write the analytical formulation of the energy balance for the combustion process and derive the analytical expression for the excess air λ. Then calculate it. Do you expect any change in the excess air with the variation of the pressure within the combustion chamber? Please justify your answer [in case you are not able to proceed with this question, use λ=1,478] b. (2pt) Write the exergy balance (per kmol of C 2H2) for the combustion process, identifying all the necessary hypothesis and simplify it accordingly. Derive the analytical formulation for the rational exergy efficiency and then calculate it. Are these values functions of the reactants’ temperature and pressure? Neglect the chemical exergy of products of combustion. c. (2pt) Write the exergy balance to the heat exchanger, evaluate the exergy destructions (per kmol of C 2H2) and the functional exergy efficiency considering water the useful product. And evaluate the mass flow rate of acetylene required to produce 180 kg/h of superheated steam. Neglect the chemical exergy of products of combustion. d. (2pt) In case they were not given, please explain how you could calculate theoretically the chemical exergy for C2H2. Department of Energy Politecnico di Milano Author E. Colombo Pag.2 of 13 Date 26/06/2018 Exercise 1. Solution a. Write the analytical formulation of the energy balance for the combustion process and derive the analytical expression for the excess air λ. Then calculate it. Do you expect any change in the excess air with the variation of the pressure within the combustion chamber? Please justify your answer? [in case you are not able to proceed with this question, use λ=1,478] The combustion reaction can be written as follows:     2 2 2 2 2 2 2 2 2.5 3.76 2 2.5 3.76 1 2.5 C H O N CO H O N O               Assumptions for the application of the energy balance: - The combustion chamber is adiabatic and rigid; - Steady state operation is assumed throughout; - The balance is written per unit of kmol of acetylene; R R P P dE W Q N h N h dt               The energy balance states that R Ph h  [eq1] Enthalpies of reactants and products can be evaluated as follows:     0 2 2 , 1 2.5 1 R R f C H fh T p h h h          2 3.76 O f h h      2N h              0 2 2 , 2 1 2.5 3.76 P P f CO f H O fh T p h h h h h               2 1 2.5 N f h h          2O h   So we obtain                 2 2 2 2 2 2 2 2 2.5 1 3.76 2 2.5 3.76 1 2.5 f C H O N f CO f H O N Oh h h h h h h h h h                               By aggregating the remaining heat of formation we recognize the LHV                 2 2 2 2 2 2 2 2 2 ,C 2 H 2 , 2 2 , 2 2.5 1 3.76 2 2.5 3.76 1 2.5 2 C H f f CO f H O LH C H O N V CO H O N O h h h h h h h h h h                                                  2 2 2 2 2 2 2 2 2 2 2.5 3.76 2 2.5 3.76 1 2.5 C H R O R N R CO P H O P N P O P C H h T h T h LHV T h T h T h T h T                                                 2 2 2 2 2 2 2 2 2 2 2 2 2.5 2.5 3.76 2.5 2.5 2.5 3.76 C H R CO P H O P O P N P O P O R N R C H h T h T h T h T h T h T h T h T LHV                                In the last equation, all the enthalpy differences can be derived from the table, and the only unknown is the excess air:               2 2 2 2 2 2 2 2 2 2 2 2.5 2 2.5 3.76 2.5 2.5 2.5 3.76 C H C H O P CO P H O P N P O P O R N R h h T h T h T h T h T h T h LH T V                          Department of Energy Politecnico di Milano Author E. Colombo Pag.3 of 13 Date 26/06/2018 12 3101 2.5 68771 2 106640 86162 1.478 2.5 3.76 65205 2.5 68771 2.5 162 564 2 2.5 3.76 1600 60                   An increase in temperature of reactants results in an increase in the enthalpies of them: if the adiabatic flame temperature is kept constant, the excess air will decrease and the composition of the products will change. A change in pressure do not affect the combustion process and the composition of the products. As a reminder, recalling the definition of Lower Heating Value for a stoichiometric combustion reaction in air:   2 2 2 2 2 2 2 2.5 3.76 2 2.5 3.76 C H O N CO H O N       , , i f i j f j R P LHV h h            2 2 ,C 2 H 2 2 , 2 2 , 2 2 , 2 2 , 2 2 , 2 2 2 ,C 2 H 2 2 , 2 2 , 2 2 2 ,C 2 H 2 , 2 , 2 2.5 2.5 2 1 C H f O f O N f N CO f CO H O f H O N f N C H f CO f CO H O f H O C H f f CO f H O LHV h h h h h h LHV h h h LHV h h h                       b. Write the exergy balance ((per kmol of C 2H2) for the combustion process, identifying all the necessary hypothesis and simplify it accordingly. Derive the analytical formulation for the rational exergy efficiency and then calculate it. Are these values functions of the reactants’ temperature and pressure? Neglect the chemical exergy of products of combustion. Assumptions for the application of the exergy balance: - The combustion chamber is adiabatic and rigid; - Steady state operation is assumed throughout; - The balance is written per unit of kmol of acetylene;       0 0 , , , 0 Q ,k A A A A W , j des R R P P D comb i j k i dEx Ex m ex Ex ex T p e Ex x T p ex dt                           0 0 , 2 0 , N 2 0 2 2 2 2 , , 2.5 , 3.76 , 1271821.5 R R ch ph R ph O R ph R C H C H ex T p ex ex T p ex T p ex T p kJ kmol                      0 , 0 2 2 , , 874781.1 P P i ph i P C H P ex T p ex T p kJ kmol        , comb 2 2 , 397040.4 0.6878 P D C H ex r R ex ex kJ kmol ex      Differently from the previous question, both temperature and pressure of reactants affects values of exergy destruction and exergy efficiency, since physical exergy of ideal gases is a function of both temperature and pressure. Specifically, exergy destructions increase if temperature and pressure increase. c. Write the exergy balance to the heat exchanger, evaluating the exergy destructions (per kmol of C2H2) and the functional exergy efficiency. Neglect the chemical exergy of products of combustion; Before applying exergy balance, it is required to derive the ratio between the mass flow rate of water and the molar flow rate of flue gases. The heat released per molar unit of fuel is equal to:       0 2 2 2250 673 , 2250 673 1073257 fg i C H i P h K K p h K h K kJ kmol               The heat absorbed by a mass unit of water in the heat exchanger is equal to:   , , IN 873 323 , 3491.9 w w w OUT w wh K K p h h kJ kg     Department of Energy Politecnico di Milano Author E. Colombo Pag.4 of 13 Date 26/06/2018 Therefore by applying the energy balance     0 2250 673 , 873 323 , fg w w N h K K p M h K K p        W e do not need to calculate both M and N since, by defining M r N  we get     0 2 2 2250 673 , 307 .4 873 323 , fg w C H w w h K K p r kg kmol h K K p       Assumptions for the application of the exergy balance: - The heat exchanger is adiabatic and rigid; - Steady state operation is assumed throughout; - The balance is written per unit of kmol of acetylene;     0 , HE 2250 673 K, 873 323 K, 0 fg w w D N ex K p M ex K p N ex               0 , HE 2250 673 K, 873 323 K, 0 fg w w D ex K p r ex K p ex              2 2 0 0 0 2250 673 K, 2250 673 83307 , 2. , 5 i C H i P fg K ex K kJ k ex K p ex p m l p o               ,OUT , IN 0 , OUT , IN 873 323 K, 1211.5 w w w w w w w w ex K p h h T s s kJ kg kJ kg            2 2 372352.7 C H w kJ kmol ex r    ,HE 2 2 460719.7 C H D kJ kmol ex      , f 0 873 323 K, 0.4470 2250 673 K, w w P ex F fg r ex K p ex ex ex K p          Evaluate the mass flow rate of acetylene required to produce 180 kg/h of superheated steam. 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 180 26 0.0042 3600 3600 307 w C H C H C H C H w C H w C H kg kmol kg kg h m m MW r h s kg kmol s            Department of Energy Politecnico di Milano Author E. Colombo Pag.5 of 13 Date 26/06/2018 Exercise 2. (7pt) Consider the power plant depicted in the figure with a boiler, a turbine and a condenser . The Boiler is fed by fuel and air and it produces steam which enter in the turbine. The turbine produces power for the grid and gives power to the pump. The condenser condenses the steam by using a second stream of cooling water. The pump close the cycle and allow the water to enter the boiler again. Flue gases water (flow number 6) are released to the environment without any added cost. Thermodynamic data, Physical data and Economic data are given in the below tables It is requested to: a. (1pt) Calculate the mass flow rate of steam and cooling water. b. (1pt) Write the exergy balance for each component of the plant and for the whole system, and compute analytically, the exergy destruction rate, and fraction of exergy destruction, the rational exergy efficiency and the functional exergy efficiency (and make comment on their differences) c. (3pt) Write the Thermoeconomic system of equations for each component and by making the proper assumption for auxiliary equations, derive the analytical cost structure for the product of each components. Then derive the analytical expression for the relative cost difference and the exergoeconomic factor for all the components. d. (3 pt) Now you can calculate the specific cost of the product for each component, the relative cost difference and the exergoeconomic factors making some comments on the obtained results. Boiler 4 Pump Ẇ pump Turbine Ẇ n et 2 Condenser 1 6 5 3 Fuel,Air Ė x F/A =14700 kW Flue gas Ė x FG = 150 kW Cooling water # T ⁰C P bar h kJ/kg s kJ/kgK 1 520 100 3426.37 6.66 2 0.9 0.08 2335.97 7.46 3 0 0.08 173.84 0.59 4 43 100 188.83 0.61 5 20 - 83.91 0.30 6 35 - 146.63 0.51 Fuel,Air mass flow rate [kg/s] 1 Fuel,Air Exergy rate [kW] 14700 Flue gas Exergy rate [kW] 150 Net Power [Kw] 4208 T_0 [⁰C] 20 P_0 [atm] 1 unit cost of Fuel,Air (cF/A)[€/GJ] 1 Z_boiler (Investment, O&M) [€/s] 0.25 Z_turbine (Investment, O&M) [€/s] 0.5 Z_condenser (Investment, O&M) [€/s] 0.15 Z_pump (Investment, O&M) [€/s] 0.05 Department of Energy Politecnico di Milano Author E. Colombo Pag.6 of 13 Date 26/06/2018 Exercise 2. (Solution) a. Calculate the mass flow rate of steam and cooling water. These can be derived by applying energy balance to turbine and condenser.   1 2 4 3 ( ) ( ) net st W m h h h h       3.91 kg/s stm   2 3 6 5 ( ) ( )st cwm h h m h h      134.89 kg/s cwm   b. Write the exergy balance for each component of the plant and for the whole system, and compute analytically, the exergy destruction rate, and fraction of exergy destruction, the rational exergy efficiency and the functional exergy efficiency (and make comment on their differences) Boiler   1 1 4 / 4 1 , , , , , / 4 / , / 4 1 , / 4 1 0 4 1 0.4015 0.3892 ( ) ( ) ( ) 8829. FG F A FG D boiler ex r boiler ex f boiler F A F A D boiler F A FG D boiler F A st FG Ex Ex Ex Ex Ex Ex Ex Ex Ex Ex Ex Ex Ex Ex Ex Ex Ex Ex Ex m h h T s s Ex                                                3kW        Turbine   2 1 2 , , , , , 1 1 2 , 1 2 , 1 2 0 1 2 4 0.8413 0.8119 ( ) ( ) ( ) ( net pump net net pump D turbine ex r turbine ex f turbine D turbine net pump D turbine st net st Ex W W W Ex Ex W W Ex Ex Ex Ex Ex Ex Ex W W Ex m h h T s s W m h                                                3) 916.44 h kW         Condenser   3 6 3 2 5 3 6 , , , , , 2 5 2 , 2 3 5 6 , 2 3 0 2 3 5 6 0 5 6 0.376 0.02 ( ) ( ) ( ) ( ) ( ) ( D condenser ex r condenser ex f condenser D condenser D condenser st cw Ex Ex Ex Ex Ex Ex Ex Ex Ex Ex Ex Ex Ex Ex Ex Ex Ex m h h T s s m h h T s s                                                ) 368.58 kW         Pump:   4 3 4 3 4 , , , , , 3 , 3 4 , 3 4 0 3 4 4 3 0.7472 0.6940 ( ) ( ) ( ) ( ) 17.95 pump D pump ex r pump ex f pump pump pump D pump pump D pump st st Ex Ex Ex Ex W Ex Ex Ex W W Ex Ex Ex W Ex m h h T s s m h h kW                                                 Whole plant Department of Energy Politecnico di Milano Author E. Colombo Pag.7 of 13 Date 26/06/2018   6 / 5 6 , , , , , / 5 / , / 5 6 , / 5 6 0 5 6 0.3107 0.2863 ( ) ( ) ( ) FG net net F A FG net D plant ex r plant ex f plant F A F A D plant F A FG net D plant F A cw Ex Ex W W Ex Ex Ex Ex W Ex Ex Ex Ex Ex Ex Ex Ex Ex W Ex Ex m h h T s s E                                                   10132.27 FG netx W kW           Component ηex,r ηex,f ExD yD Boiler 40.15 38.92 8829.30 0.87 Turbine 84.13 81.19 916.44 0.09 Condenser 37.60 36.27*** 368.58 0.04 Pump 74.72 69.40 17.95 0.00 Plant 31.07 0.29 10132.27 1 *** not sure of the definition of functional efficiency for a condenser.. c. Write the Thermoeconomic system of equations for each component and by making the proper assumption for auxiliary equations, derive the analytical cost structure for the product of each components. Then derive the analytical expression for the relative cost difference and the exergoeconomic factor for all the components. For each component it is required to derive the analytical cost structure for the product ; Boiler: / 4 1 , / 4 1 F A FG D boiler F A boiler FG Ex Ex Ex Ex Ex C C Z C C C c Ex                           / / 4 4 1F A F A boiler FGc Ex c Ex Z c Ex c           4 4 / / , / 4 / / 4 4 1 FG F A F A in B F A F A F A boiler FG FG Ex c Ex c Ex c Ex Ex c Ex c Ex Z c Ex c Ex                     , , , 1 1 FG D boiler boiler in B in B Ex Ex Z c c c Ex Ex                  ( , , , , 1 1 in B FG D boiler boiler boiler in B in B Ex Ex Z c c r Ex c c Ex                     , , 1 1 1 in B FG D boiler boiler boiler turbine c Ex Ex Z Z f Ex Ex Ex                   Department of Energy Politecnico di Milano Author E. Colombo Pag.8 of 13 Date 26/06/2018 Turbine 1 2 , 1 2 net pump D turbine turbine elec Ex Ex W W Ex C Z C C C c Ex                            1 2 1 2 turbine elec net elec pump turbine elec net pump c Ex Z c Ex c W c W c Ex Ex Z c W W                        ,D turbine turbine elec net pump net pump Ex Z c c c W W W W                     ,D turbine elec turbine turbine net pump net pump Ex c c Z r c W W c W W                        ,D turbine turbine turbine turbine net pump net pump net pump c Ex Z Z f W W W W W W                        Condenser: 2 5 3 6 , 2 5 3 6 D condenser condenser Ex Ex Ex Ex Ex C C Z C C C c Ex                           2 5 c Ex c  5 3 6 condenser Ex Z c Ex c       0 6 2 3 3 condenser Ex cEx Z c Ex         , 6 3 3 3 D condenser condenser Ex Ex Z c c c Ex Ex          , 6 3 3 3 D condenser condenser condenser Ex Ex c c Z r c Ex c Ex              , 6 3 3 3 D condenser condenser condenser condenser c Ex Ex Z Z f Ex Ex Ex                   Pump: Department of Energy Politecnico di Milano Author E. Colombo Pag.9 of 13 Date 26/06/2018 , , , 4 4 4 D pump pump in pump in pump Ex Z c c c Ex Ex        , , 4 , , 4 4 in pump D pump pump pump in pump in pump Ex Z c c r Ex c c Ex          , , 4 4 4 pump D pump pump in pump pump Z Ex Z f c Ex Ex Ex               d. Now you can calculate the specific cost of the product for each component, the relative cost difference and the exergoeconomic factors making some comments on the obtained results. We can write the 4 exergy cost balances for the 4 components: / / 4 4 1 1 1 1 2 2 2 2 5 5 3 3 6 6 3 3 4 4 F A F A boiler FG FG turbine net net pump pump condenser pump pump pump c Ex c Ex Z c Ex c Ex c Ex Z c Ex c W c W c Ex c Ex Z c Ex c Ex c Ex c W Z c Ex                                                         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 flows = 10 m : number of components = 4 n-m = 10 - 4 = 6 Auxiliary relations . 5 6 1 2 1) 0 2) 3) 4) 0 5) 0 6) FG F / A net pump elec c c given c c c c c c c c                   Now substituting auxiliary relations in the economic cost balance: 3 4 , 3 4 pump D pump elec pump Ex W Ex Ex C C Z C C c Ex                        3 3 4 4 3 3 , 3 elec pump pump elec pump in pump pump c Ex c W Z c Ex c Ex c W c Ex W                   Department of Energy Politecnico di Milano Author E. Colombo Pag.10 of 13 Date 26/06/2018 / / 4 4 1 1 2 . 2 3 3 3 3 . 4 4 ( ) F A F A boiler turbine elec net pump condenser elec pump pump c Ex c Ex Z c Ex c Ex Z c Ex c W W c Ex Z c Ex c Ex c W Z c Ex                                               we have 4 equations and 4 unknowns (c,c 4, c 3, and c elec. ). By coupling the third and the fourth equation we get 2 . 4 4 condenser elec pump pump c Ex Z c W Z c Ex           By inserting this last equation into the first equation in the system we get / / 2 . 1F A F A condenser elec pump pump boilerc Ex c Ex Z c W Z Z c Ex                 By using the second equation / / . . ( ) F A F A condenser elec pump pump boiler turbine el ec net pump c Ex Z c W Z Z Z c W W                  . / / elec net F A F A pump boiler condenser turbinec W c Ex Z Z Z Z             / / . F A F A pump boiler condenser turbine elec net c Ex Z Z Z Z c W             Please note that This equation is also coherent with the TE system written for the whole aggregate plants and would also lead to 6 . / / pump boiler condenser turbine DTOT FG elec F A F A net net Z Z Z Z Ex Ex Ex c c c W W                  e. Compute the values making some comments on the obtained results celec. [€/kWh] 0.82531 c [€/kWh] 0.33211 c3 [€/kWh] 59.70660 c4 [€/kWh] 18.18325 Cost structure for each component: Specific cost of the product (a) Cost of fuel for the given component (b) Cost of destruction and Losess (c) Cost of Investment (d) celec. [€/kWh] TOT 0.82531 0.0036 0.00898 0.81274 celec. [€/kWh] 0.82531 0.33211 0.07133 0.42187 c [€/kWh] 0.33211 0.06897 0.10726 0.15588 c3 [€/kWh] 59,70660 0.33211 15.57724 43.79726 c4 [€/kWh] 18.18325 11.05050 3.73956 3.39319 Department of Energy Politecnico di Milano Author E. Colombo Pag.11 of 13 Date 26/06/2018 Relative cost difference [(a)- (b)]/(b) Or [(c)+(d)]/(b) Exergoeconomic factor (d)/[(c)+(d)] TOT 228,2538 0,98908 Boiler 3,81533 0,59238 Turbine 1.48509 0.85537 Condenser 178.78182 0.73764 Pump 0.64547 0.47572 Department of Energy Politecnico di Milano Author E. Colombo Pag.12 of 13 Date 26/06/2018 Exercise 3. (5pt) Consider the energy statistics of Canada for 2015 in the tables below (IEA). The right end side table tis the overall energy balance , while the left end side is the electricity balance Please calculate the a. (1pt) The share of renewable energy in both the primary mix and in the electricity production mix b. (2 pt) primary energy devoted to electricity production and the electric penetration. c. (2 pt) The average efficiency of the electric system in Canada Department of Energy Politecnico di Milano Author E. Colombo Pag.13 of 13 Date 26/06/2018 Exercise 3. Solution a. (1pt) The share of renewable energy in both the primary mix and in the electricity production mix Share in the primary mix Renewable Primary Energy 0.18 TPES   Share in the electricity production mix question 1 prod [GWh] share coal 65977 0.10 oil 8148 0.01 gas 67181 0.10 biofuel 12511 0.02 waste 265 0.00 nuclear 101423 0.15 hydro 380717 0.57 geothermal 0 0.00 solar pv 2895 0.00 solar thermal 0 0.00 wind 26446 0.04 tide 13 0.00 other sources 5275 0.01 total fossil fuels 242729 0.36 total RES 428122 0.64 total 670851 1.00 Share of renewable in the energy production mix = 0.638 b. (2 pt) primary energy devoted to electricity production and the electric penetration Primary energy devoted to electricity production: 96468 ktoe TPES = 270192 ktoe Electric penetration: the ratio between the primary energy devoted to electric energy production and the TPES , 0.357 P i i E EP TPES    c. (2 pt) The average efficiency of the electric system in Canada Average efficiency of the electric system: the ration between the electricity production in ktoe and the primary energy devoted to electricity in ktoe. Average efficiency = 0.6