Energy Engineering - Advanced Thermodynamics and Thermoeconomics

Full exam

Department of Energy Politecnico di Milano Author E. Colombo Pag. 1 of 10 Date 01/08/2018 Milan, 1st August 201 8 Exam – Advanced Thermodynamic and Thermoeconomics Session of 2 4-07-2018 Exercise 1. The system depicted in the figure is composed by a Burner and a Boiler. The Burner produces flue gases at 1920 K and 3 bar by burning gaseous p ropane with excess air λ equal to 1.446 . Notice that fuel and air enter separately in the burner at the conditions of 70°C and 3 bar. Flue gases exit the Boiler at 420 K (pressure drop is negligible), and the Boiler produces 90 t/h of superheated steam. Properties of inlet/outle t bulk flows in the two components are reported in table 1 and 2. Table 1 Table 2 Isobaric heat capacities of the fuel and the air are respectively equal to 2.401 kJ/kg -K and 2 9.17 kJ/ kmol -K. Considering the data provided in tables 1 and 2, and the e nvironmental reference conditions of 25°C and 1 bar , it is required to: a. Write the chemical combustion reaction. Then write and simplify th e exergy balance for the burner and write the ana lytical expressions for the exergy of its reactants and products streams, including both the physical and the chemical components. b. Derive numerical results (per unit of molar flow of C8H18) for point a . c. Write the analytical expression s for the exergy dest ruction of the burner and for the rational exergy efficiency of the combustion process. Briefly discuss the possible ways to simplify the calculation of the rational exergy efficiency of the combustion process. d. Der ive numerical results (per unit of molar flow of C8H18) for point c . e. Apply t he energy balance to the boil er and determine the molar flow rate of fuel required to prod uce the mass flow rate of steam. hf Δh (1920 - 298K) Δs (1920 - 298K) Δh (420 - 298K) Δs (420 - 298K) ex ch kJ/kmol kJ/kmol kJ /kmolK kJ/kmol kJ /kmolK kJ/kmol C8H18 -208450 5418533 CO2 -393520 86631 93.0660 4842 13.5730 20108 H2O -241820 68623 73.7770 4139 11.6300 8667 O2 0 56186 62.0820 3632 10.2080 3947 N2 0 53267 59.0000 3556 9.9970 691 Unit Water Steam T °C 30 200 p bar 6 6 h kJ/kg 126 2851 s kJ/kgK 0.4366 6.9683 Burner Boiler C 8H 18 Air Flue Gas es @ 420 K @ 70 °C , 3 bar Steam @ 200 °C , 6 bar Water @ 30 °C , 6 bar Flue Gases @ 1920 K Department of Energy Politecnico di Milano Author E. Colombo Pag. 2 of 10 Date 01/08/2018 f. Write and simplify the exergy balance of the boiler. Then write the analytical expressions for th e exergy destructions and for the functional exergy efficiency of the boiler (the exergy of flue gases and the change in exergy of the steam are considered as useful products). Finally derive numerical results. Exercise 1. ( Solution ) a. Write the chemical c ombustion reaction. Then write and simplify th e exergy balance for the burner and write the analytical expressions for the exergy of its reactants and products streams, including both the physical and the chemical components. The general formulation for the chemical reaction with excess air results as follows: The combustion reaction of Octane can be written as follows: Assumptions for the application of the exergy balan ce: - Adiabatic and steady state combustion process; - No work flow exergy; - The balance is written per unit of kmol of C8H18; According to the text, both physical and chemical exergies must be computed for reactants and products: Analytical Fo rmulation     2 2 2 2 2 2 3.76 3.76 1 4 2 4 4 ab b b b b C H a O N aCO H O a N a O                                  8 18 2 2 2 2 2 2 12.5 3.76 8 9 (12.5 3.76) 1 12.5 C H O N CO H O N O                 8 18 2 2 2 2 2 2 18.06 3.76 8 9 67.915 5.563 C H O N CO H O N O             , , , 0 A A A A W , j des R R R P P P D comb Q ,k i j k i dEx Ex N ex Ex ex T p e Ex x T p ex dt                11 , 1 0 0 , 00 8 18 ,C 8 H18 , ln ln , R R R ph ch ph ch c i p i p i i CH h ch air tabl s air e ex T p ex ex ex ex ex e Tp c T T T c R Tp x                       1 00 0 , , 0 , 1920 298 1920 298 ln ln i i tot P i c P P P ph ch ph c h i i ii h i p h T s T R p x ex ex T p ex ex e T R x x ex x                 Department of Energy Politecnico di Milano Author E. Colombo Pag. 3 of 10 Date 01/08/2018 b. Derive numerical results (per unit of molar flow of C8H18) for point a. c. Write the analytical expression for the exergy destruction of the burner, and the rational exergy efficiency of the combuston process. Briefly discuss the possible ways to simplify the calculation of the rational exergy efficiency of the combustion process . Following the exergy balance writte n in previous section, exergy destruction and rational exergy can be computed as follows: One possible way of simplification is to ignore physical exergy of the reactants and chemical exergy of the products . Let see how the rati onal exergy efficiency varies : d. Derive numerical results (per unit of molar flow of C8H18) for point c. Results for obtained considering both physical and chemical exergy for products and reactants: Simplify calculation in the abovemention ed way it is possible to obtain a new value for exergy of products and reactants thanks to which a new value for exergy destruction and rational efficiency is computed.       11 , 1 0 0 , 3 ,C 8 H18 8 8 18 00 8 18 , ln ln 1 3565 12.5 4.76 2813 , 245818 R R R ph ch ph ch i p i p i C H C H i C ch ch air tables H air ex T p ex ex e Tp cT x ex T T c R kJ kmol kJ kmol Tp ex ex                         8 18 8 18 1 5418533 5418533 , 5664351 CH CH R R Re kJ kmol kJ kmol x T p             1 0 0 8 18 0 , , 0 8 18 8 18 , , 1920 298 1920 298 ln 3897135 ln 1 401 21372 8508 i C H i tot P i ch i i i C P P P ph ch ph ch P P P H ii CH ex T p ex ex ex e p h T s T R kJ kmol p x ex T R x x kJ kmol kJ k x ex mol Tp                       ,, , , , P D comb R R R P P P ex r R ex ex ex T p ex T p ex       , R R R phex T p ex  ,C 8 H18 , 8 18 ch ch ch ch air table C H air s ex ex ex ex      , P P P ph chex T p ex ex      1 00 0 1920 298 1920 298 ln i h i p p h T s R ex T p           , comb 8 18 , 1645844 , 0.709 D C H ex rex kJ kmol       8 18 8 18 , 5418533 , 3897135 R R R P P P CH CH ex T p e kJ kmol kJ kmol x T p   Department of Energy Politecnico di Milano Author E. Colombo Pag. 4 of 10 Date 01/08/2018 Neglecting those contribu tions results in slightly over estimation of rational efficiency . In this specific case (type of fuel, inlet and outlet condition…) the two terms can be neglected. e. Apply the energy balance to the boiler, and determine the molar flow rate of fuel required t o produce the mass flow rate of steam; Assumptions for the application of the energy balance: - Adiabatic and steady state combustion process; - No work flow f. Write and simpl ify the exergy balance of the boiler. Then the analytical expressions for the exergy destructions and the functional exergy efficiency of the boiler (the exergy of flue gases and the change in exergy of the steam are considered as useful products). Finally derive numerical results. Assumptions for the application of the exergy balance: - Steady state and adiabatic process is assumed for the boiler; Boiler Flue Gas es @ 420 K Steam @ 200 °C , 6 bar Water @ 30 °C , 6 bar Flue Gases @ 1920 K ,comb 3 8 , 1521398 0.719 P D C H ex r R ex ex kJ kmol ex      fg w fg w dE W Q H H H H dt                8 18 ,out ,in 8 18 0 1920 298 420 298 5084036 2724 C H fg water water fg CH i i i i water water water water N h m h hh k h h J kmol k h J kg h                     8 18 8 18 8 18 25 2725 0.013 5080860 w CH w CH C fg H kmol s kJ kmo mh kg s kJ kg N h l             , 1920 , 3 420 , 3 200 30 , 5 0 fg fg fg water water D boilerN ex K bar ex K bar m ex K bar Ex             , 8 8 18 , 18 ( ) 1920 , 3 3897135 ( ) 1920 , 3 52 () () 221 C fg in CH H fg in fg kJ kmol already comput ex ph K bar N ed E ex ph K ba x ph k r W            1 0 0 8 , 18 0 , 420 298 420 298 420 , 3 ( ) 420 , 3 403 ln 301 8 ) 8 32 ( fg out ph ch ph fg f i C H i f g g out p h T s T R kJ ex K bar ex ex ex N ex ph K bar kmol p Ex ph kW                   ,out ,in 0 ,out ,in 77 200 30 , 6.89 6 19439 water water water wate wat r wa er water ter h h T s s kJ k ex K bar xE kW g        ,, , , 28743 0.45 water P D boiler ex f R fg out fg in x Ex Ex kW E E x x Ex E        Department of Energy Politecnico di Milano Author E. Colombo Pag. 5 of 10 Date 01/08/2018 Exercise 2. Let’s consider a simple Energy Conversion System (ECS) enclosed in the control volume depicted in the figure (dashed line). The system composed by 2 micro -CHP systems (a and b), and one absorp tion 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 Table 2 With reference to the data provided by the t ables, i t is required to: a. Write the exergy balance and compute exergy destruction of each component. b. Derive the number and the type of auxiliary relations required to solve the Thermoeconomic system of equations, properly justifying your choi ces. c. Write one Thermoeconomic system of equations for each component and derive the cost structure for each of them d. Calculate the numerical values of the unit economic costs of each material/energy flow. Collect numerical results in one tabl e. e. Based on the cost structure in point c. compute the economic cost of the exergy destructions , Exergoeconomic factor and relative cost difference for each component . Collect numerical results in one table and comment on the obtained results. f. Solve the system at the highest level of aggregation (1 unique component having different products) and calculate the new structure of the cost. Based on the analytical formulation obtained for the products in point c. please make some comments on thei r difference. Exercise 2. (Solution) a. Write the exergy balance and compute exergy destruction of each component. Exergy Balance for each component : State Ex [MW] 1 Primary Energy 100 2 Primary Energy 100 3 Heat 30 4 Heat 25 5 Cooling Load 25 6 Electricity 40 7 Electricity 20 8 Flue gases 5 9 Steam leakage 5 unit cost of Fuel (c 1,c2) [€/GJ] 8.333 Z_ CHP a (Investment, O&M) [€/s] 0.2 Z_ CHP b (Investment, O&M) [€/s] 0.1 3 Z_ Absorption Chiller (Investment, O&M) [€/s] 0.1 1 3 6 , , 2 4 7 8 , , 3 4 5 9 , , ) 30 ) 50 ) 25 D a D a D b D b D c D c a Ex Ex Ex Ex Ex MW b Ex Ex Ex Ex Ex Ex MW c Ex Ex Ex Ex Ex Ex MW                      Absorption Chiller (c) ECS (Primary Energy ) Micro CHP (a ) Micro CHP (b ) 2 (Primary Energy ) 1 4 (Heat ) 3 (Heat ) (Cooling Load ) 5 Electricity Electricity 6 7 8 Flue gases Steam leakage 9 Department of Energy Politecnico di Milano Author E. Colombo Pag. 6 of 10 Date 01/08/2018 b. Derive the number and the type of auxiliary relations required to solve th e Thermoeconomic system of equations, properly justifying your choices. The system composed by 9 streams and 3 components, therefore, in order to define and close Thermoeconomic system of equations, (n-m) numbers of auxiliary relations are required: n: N umber of exergy flows = 9, m: number of components = 3 : n-m = 9 - 3 = 6 Auxiliary relations : c. Write one Thermoeconomic system of equations for each component and derive the cost structure for each of the m. For e ach component it is required to write economic cost balance based on Inlet/Outlet paradigm and substituting exergy cost relation as follows; CHP 1: In order to compute cost structure of the outlet flow of CHP1 ( ca), it is enough to substitute ( Eẋ1) in the above formula for ( ca), with its equivalent expression derived from exergy balance. Therefore, the cost structure of the electric power produced by CHP1 can be expressed as follows CHP 2: same procedure of CHP 1 Absorption Chiller: since there are two inlet streams, a weighted average of the economic cost needs to be computed: Then, the formula related to the computation of c 5 can be rearranged as follows: 1 3 6 8 2 4 7 9 1) , 3) , 5) 0 2) , 4) , 6) 0 a b c given c c c c c given c c c c         ; ; in component out in outC Z C Economic cost balance C c Ex Exergy cost relation         1 3 6 , 1 1 3 3 6 6 1 1 3 6 Da a a a Ex Ex Ex Ex c Ex Z c Ex c Ex c Ex Z c Ex Ex                   1 3 6 , 11 3 6 3 6 , 11 3 6 3 6 () () ( ) ( ) D a a a a Da a a inlet ExergyDestruction Investment Cost c Ex Ex Ex Z c Ex Z c Ex Ex Ex Ex Ex Z c c c Ex Ex Ex Ex              2 4 7 8 , 2 2 4 4 7 7 8 Db b Ex Ex Ex Ex Ex c Ex Z c Ex c Ex c              8 2 2 4 7 bb Ex c Ex Z c Ex Ex          2 4 7 8 , 22 4 7 4 7 8 , b 22 4 7 4 7 () () ( ) ( ) D b b b b D b b inlet ExergyDestruction Investment Cost c Ex Ex Ex Ex Z c Ex Z c Ex Ex Ex Ex Ex Ex Z c c c Ex Ex Ex Ex                3 4 5 9 , 3 3 4 4 5 5 9 Dc c Ex Ex Ex Ex Ex c Ex c Ex Z c Ex c            9 3 4 5 5 a b c Ex c Ex c Ex Z c Ex           34 , 34 ab in c c Ex c Ex c Ex Ex      Department of Energy Politecnico di Milano Author E. Colombo Pag. 7 of 10 Date 01/08/2018 Now, (Eẋ3+Eẋ4) in the above formula should be substituted with its equivalent expression derived from exergy balance. Ther efore, the cost structure of the cooling energy produced by absorption chiller can be expressed as follows: d. Calculate the numerical values of the unit economic costs of each material/energy flow. Collect numerical results in one table . Starting from the cost structure of each component it is possible to compute the cost of outlet for each component Component c_in c_dest c_inv c_out CHPa 30.00 12.86 10.29 53.14 CHPb 30.00 36.67 10.40 77.06 Absorption Chiller 64.02 76. 82 14.40 155.23 It is now possible to determine the cost of each flow. c1 [€/MWh ] c2 [€/MWh ] c3 [€/MWh ] c4 [€/MWh ] c5 [€/MWh] c6 [€/MWh] c7 [€/MWh] c8,9 [€/MWh] 30 30 53.14 77.06 155.23 53.14 77.06 0 e. Based on the cost structure in point c. compute the economic cost of the exergy destructions, Exergoeconomic factor and r elative cost difference for each component. Collect numerical results in one table and comment on the obtained results. The economic cost of th e exergy destruction is computed as follows: Exergoeconomic factor is computed as: = Relative cost difference is computed as: Component Ex D Cost of Ex D f r a 30 900 0.44 0.77 b 55 1650 0.2 2 1.31 c 25 1725 0.1 7 1.42 , 3 4 34 5 55 () in c c a b c c Ex Ex Z c Ex c Ex Z c Ex Ex         , 3 4 , 5 9 , 5 55 9 ,c 5 , , 55 ( ) ( ) in c c in c D c c D c in c in c inlet Investment Cost ExergyDestruction c Ex Ex Z c Ex Ex Ex Z c Ex Ex Ex Ex Z c c c Ex Ex                 , ,, , in j loss j D j Dj j c Ex Ex C Ex     , ,, ,j in j loss j D j j jj jj j j j jD c Ex Ex Z ZZ ff Ex Ex Ex ZC           , , , , , , , , out j in j loss j D j j j in j in j in j in j c c Ex Ex Z r c Ex c Ex         Department of Energy Politecnico di Milano Author E. Colombo Pag. 8 of 10 Date 01/08/2018 Comments: - According to the exergy destruction, component b is most responsible component in total exergy destruction of the system, then component a and c respectively. It is interesting that the order of components according to the economic cost of the exergy destruction is changing: component c has the highest rank and needs to be improved prior to the other components. Economic cost of the exergy destruction indicates the economic expenses that caused by irreversibilities within the considered component. - Exergoeconomic factor indicates whether the capital and O&M e xpenses ( Ż) is the major source of economic cost increase, or it is the economic cost of exergy destruction ( ĊD). It 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. While low values indicate exergy destruction and losses are primary sources of product cost increment. Therefore, increasing the component efficiency even if it requires higher capital investment is worthy . Base on such explanation, component c has le low value of Exergoeconomic factor and needs to be improved by increasing its efficiency. - Relative cost difference: High value of this parameter means there is more room for improvem ent. Based on that, comp onent c is again the first candidate to be i mproved, followed by component b and a. f. Solve the system at the highest level of aggregation (1 unique component having different products) and calculate the new structure of the cost. Based on the ana lytical formulation obtained for the products in point c. please make some comments on their difference. 1 2 5 6 7 8 9 , 1 1 2 2 5 5 6 6 7 7 8 D tot tot Ex Ex Ex Ex Ex Ex Ex Ex c Ex c Ex Z c Ex c Ex c Ex c                   89 Ex c         9 1 1 2 5 6 7 1 5 6 7 8 9 , 5 6 7 8 9 , 1 5 6 7 5 6 7 tot products D tot tot products D tot abc products Ex c Ex Ex Z c Ex Ex Ex c Ex Ex Ex Ex Ex Ex Z c Ex Ex Ex Ex Ex Ex Z Z Z cc Ex Ex Ex Ex Ex Ex                               Department of Energy Politecnico di Milano Author E. Colombo Pag. 9 of 10 Date 01/08/2018 Exercise 3. Consider the energy statistics of Germany and Italy for 2015 in the table s below (IEA) . Please conside r also that Emissions are 730 Mton for Germany and 331 for Italy, while GDP are 3697 billion USD for Germany and 2060 for Italy) . GERMANY (ktoe) ITALY (ktoe) Please c omment on the following phrases: Are they true or false. Your judgment need s to be supported by quantitative data a. “Germany is more advanced than Italy in r enewable energy penetration ” b. “Energy & CO2 intensity are better in Germany than Italy” [toe/thousand of USD and kg C02/USD] Department of Energy Politecnico di Milano Author E. Colombo Pag. 10 of 10 Date 01/08/2018 c. “Overall efficiency of the Elect ricity Sector is higher in Italy than Germany” Exercise 3. Solution a. “Germany is more advanced than Italy in renewable energy penetration” This is not true since the share of the RES is greater in Italy Share in the prim ary mix for italy Share in the primary mix for Germany b. “Energy and CO2 intensity are better in Germany than Italy” This is not true since the two indicators are smaller for Italy. This means that Italy is able to produce a unit of GDP using less primary energy and emitting less CO2 with respect to Germany . Energy intensity for Italy Energy Intensity for Germ any Co2 intensity for Italy Co2 intensity for Germany c. Overall efficiency of the Electricity Sector is higher in Italy than Germany Average eff iciency of Italy: Average efficiency of Germany: Renewable Primary Energy 3916 8908 14591 0.18 152605 TPES     Renewable Primary Energy 1632 11027 29944 0.14 307795 TPES       TPES 152605 *10 ^ 3 0.074 / 2060 *10 ^ 6 toe kUSD GDP     TPES 307795 *10 ^ 3 0.083 / 3697 *10 ^ 6 toe kUSD GDP     CO2 331*10 ^ 9 0.161 / 2060 *10 ^ 9 kg USD GDP     CO2 730 *10 ^ 9 0.198 / 3697 *10 ^ 9 kg USD GDP   Electricity & CHP production in ktoe - own use - losses __ 15969 8245 1784 1696 0.38 9794 795 6185 3916 8567 2864 667 3727 14171 3302 Electricity tranformation withCHP               45190 9934 4502 2202 0.39 56165 896 1521 23920 1632 10257 5585 6792 410 104 16 8099               