Energy Engineering - Advanced Thermodynamics and Thermoeconomics

Full exam

Department of Energy Politecnico d i Milano Author M. Rocco Pag. 1 of 12 Date 10/0 9/2018 Milan, 28th August 201 8 Exam – Advanced Thermodynamic and Thermoeconomics Session of 05 -09-2018 Exercise 1. (7pt) da 21/07/2015 A vapor -compression refrigeration system is designed to cool down a mass flow rate of a fluid for an industrial process. The cycle works with ammonia and it is characterized by the following transformations: • 1-2: vapor enters the compressor (CMP ) as saturated vapor; • 2-3: vapor condensate s in the condenser (COND) and exits as saturated liquid. The heat is rejected to the env ironment at 25 °C and 1 bar (reference conditions); • 3-4: liquid goes through an isenth alpic process in a lamination valve (LAM); • 4-1: the multi -phase flow evaporates in to the evaporator (EVA) and exits as saturated vapor . Evaporations process cool down a l iquid mass flow rate from 15°C to 7°C; Ammonia T p h s ex °C MPa kJ/kg kJ/kg -K kJ/kg 0 25 0.1013 1689.8 7.0732 0.0 1 -10.0 0.2908 1593.9 6.2284 2 141.7 1.1670 1924.8 6.5699 3 30.0 1.1670 484.9 1.9596 4 -10.0 0.2908 484.9 2.0142 With ref erence to the properties collected in the table, it is required to: a. Draw the schem e of the system, with all the components, highlighting directions of work, heat and bulk flow streams. Compute the exergy value for each stream. b. Define boundaries of the whol e system and apply the energy and the exergy balances. Derive the analytical expressions for the COP and the functional exergy efficiency η ex,f of the system. Which are the differences among the two performance indicators? c. Apply the exergy balance to all the components of the system, deriving the analytical expressions for the exergy destructions (specific to the mass flow of working fluid) . d. Compute numerical results for questions b and c. Notice that: - Irreversibility occurring during the heat transfer processes must be included in the evalu ation of the exergy destructions ; - State 0 refers to the reference sta te of ammonia ; - Values of entropy includes the effects of both temperature and pressure; Department of Energy Politecnico d i Milano Author M. Rocco Pag. 2 of 12 Date 10/0 9/2018 Exercise 1. Solution a. Draw the scheme of the system, with all of its component s, highlighting directions of work, heat and bulk flow streams. Compute t he exergy value for each stream. b. Define boundaries of the whole system and apply the energy and the exergy balances, deriving the analytical expressions for the COP and the functional exergy efficiency ηex,f of the system. Which are the diffe rences among the two performance indicators? Application of the energy balance to EVA: - Steady state; - Adiabatic and rigid component; - No relevant changes in elevation and speed of fluids; Application of the energy balance to CMP: - Steady state; - Adiabatic component; - No relevant changes in elevation and speed of fluids; Therefore: Functional exergy efficiency: Applicati on of the exergy balance to EVA: - Steady state; - Adiabatic and rigid component; - No relevant changes in elevation and speed of fluids; Properties of the cold fluid are unknown. However, the change in exergy of the cold fluid is eq ual to the exergy of the heat provided to the fluid between the same inlet and outlet temperatures:   0 0 0 1 2 3 4 156.0 385.1 319.8 303.5 i i iex h h T s s ex kJ kg ex kJ kg ex kJ kg ex kJ kg         EVA CMP Q COP W            , , , 4 ,1 ,1 , 4 00 f f IN f OUT r r r EVA r r r i i m h m h h m h h Q m h h                 ,1 , 2 , 2 ,1 00 i r r r r r r i ii m h W W m h h W m h h              r EVA CMP m Q COP W      ,1 , 4rr r hh m    ,1 , 4 , 2 ,1 , 2 ,1 rr rr rr hh hh hh     , , Q EVA ex f CMP Ex W           , , , , 4 ,1 , 00 D EVA f f IN f OUT r r r D EVA i i m ex Ex m ex ex m ex ex Ex          Department of Energy Politecnico d i Milano Author M. Rocco Pag. 3 of 12 Date 10/0 9/2018 Application of the energy balance to CMP: - Steady state; - Adiabatic component; - No relevant changes in elevation and speed of flui ds; Therefore: c. Apply the exergy balance to all the components of the system, deriving the analytical expressions for the exergy destructions (specific to the mass flow of working fluid) Application to the compressor (CMP): - Steady state; - Adiabatic component; - No relevant changes in elevation and speed of fluids; Application to the condenser (COND): - Steady state; - Adiaba tic and rigid component; - No relevant changes in elevation and speed of fluids; - Heat discharged into the environment at T0; Application to the lamination valve (LAM): - Steady state; - Adiabatic and rigid component; - No relevant changes in elevation and speed of fluids;     0 0 0 , 4 ,1 , , ,1 , 4 , , , 1 0 1 1 EVA r r r D EVA Q EVA EVA r r r ml f ml f ml f T T T Q m ex ex Ex Ex Q m h h T T T                                         ,1 , 2 , 2 ,1 00 i CMP r r r CMP r r r i ii m ex W W m ex ex W m ex ex              ,, , , , 284.1 ln f IN f OUT ml f f IN f OUT TT TK T T   , , r Q EVA ex f CMP m Ex W        0 ,1 , 4 , 1 rr ml f r T hh T m        0 , 4 ,1 , , 2 ,1 , 2 ,1 1 rr ml f rr rr T hh T hh hh           1 2 , 21 ,, 0 W ,CMP r D CMP W ,CMP CMP r D CMP D CMP r Ex m ex ex Ex Ex W m h h ex Ex m             , 1 2 2 1D CMPex ex ex h h         2 3 , 00 ,, 0 10 Q ,COND r D COND Q ,COND COND COND COND Q ,COND D COND D COND r Ex m ex ex Ex Ex Q T T T T Ex ex Ex m                 , 2 3D CONDex ex ex  Department of Energy Politecnico d i Milano Author M. Rocco Pag. 4 of 12 Date 10/0 9/2018 Application to the evaporator (EVA): - Steady state; - No relevant changes in elevation and speed of fluids; - Since properties of the cold fluid are unknown, the exergy of heat transfer is considered; d. Compute numerical results for questions b and c. COP = 3.351 Eta_f = 0.165 ex_d kJ/kg CMP 101. 8 COND 65 .3 LAM 16. 3 EVA 92. 7   3 4 , ,, 0 r D LAM D LAM D LAM r m ex ex Ex ex Ex m     , 3 4D LAMex ex ex        4 1 , , 1 0, , 4 0 1 Q ,EVA r D EVA Q ,EVA EVA ml f EVA D r EVA D EVA r Ex m ex ex E m x E h x Q T T Q ex Ex m h                 ,EVA 0 , 4 14 1 1 D ml fex T T ex h x h e    Department of Energy Politecnico d i Milano Author M. Rocco Pag. 5 of 12 Date 10/0 9/2018 Exercise 2. (9pt) 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 produ ces power exclusively for the grid . The condenser condenses the steam by using a sec ond stream of cooling water. An electric pump close s the cycle and allow the water to enter the boiler again. Flue gases water are re leased to the environ ment without any added cost while the stream number 6 (heated water out of the condenser) is then used for district heating. Thermodynamic data, Physical data and Economic data are given in the below tables It is requested to: a. Calculate the mass flow rate of steam and cooling water. b. Write the exergy balance for each component of the plant and for the whole system, and compute analytically, the exergy destructio n rate, and fraction of exergy des truction and the rationa l exergy efficiency. c. Considering the system at the lower level of aggregation write the t hermoeconomic system of equations for each component and by making the proper assumption for auxiliary equations, derive the analytic al 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. Now you can calculate the specific cost of the product for each component, the relative cost difference and th e exergoeconomic factors making some comments on the obtained results. # T ⁰C P bar h kJ/kg s kJ/kgK 1 600 100 3625.84 6.90 2 42 0.0 8 2576.24 8.23 3 42 0.08 173.85 0.59 4 42.5 100 186.76 0.60 5 20 - 84.01 0.30 6 40 - 167.62 0.57 Fuel,Air mass flow rate [kg/s] 1 Fuel,Air Exergy rate [kW] 150 00 Flue gas Exergy rate [kW] 15 0 Net Power [Kw] 4220 T_0 [⁰C] 20 P_0 [atm] 1 unit cost of Fuel,Air (cF/A)[€/GJ] 1 unit cost for electrical P ump [€/ kWh ] 0.5 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 d i Milano Author M. Rocco Pag. 6 of 12 Date 10/0 9/2018 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. 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 destruct ion, the rational exergy efficiency Boiler Turbine Condenser Pump: Whole plant   1 2 4 3( ) ( ) net stW m h h h h     4.02 kg/s stm  2 3 6 5( ) ( )st cwm h h m h h    115.52 kg/s cwm    1 / 4 1 , , , /4 , / 4 1 , / 4 1 0 4 1 0.4389 () ( ) ( ) 8447.07 FG F A FG D boiler ex r boiler FA 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 m h h T s s Ex kW                             2 1 2 , , , 1 , 1 2 , 1 2 0 1 2 4 3 0.7653 () ( ) ( ) ( ) 1515.63 net pump net pump D turbine ex r turbine D turbine net pump D turbine st net st Ex W W Ex Ex W W Ex Ex Ex Ex Ex W W Ex m h h T s s W m h h kW                              36 2 5 3 6 , , , 25 , 2 3 5 6 , 2 3 0 2 3 5 6 0 5 6 0.61 ( ) ( ) ( ) ( ) ( ) ( ) 261.13 D condenser ex r condenser D condenser D condenser st cw 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 kW                               4 3 4 , , , 3 , 3 4 , 3 4 0 3 4 4 3 0.8622 () ( ) ( ) ( ) 8.85 pump D pump ex r pump pump D pump pump D pump st st Ex Ex W Ex Ex Ex W Ex Ex Ex W Ex m h h T s s m h h kW                        Department of Energy Politecnico d i Milano Author M. Rocco Pag. 7 of 12 Date 10/0 9/2018 Component ηex,r Ex D yD Boiler 43.89 8448.30 0.83 Turbine 76.53 1515.63 0.15 Condenser 61.00 261.13 0.03 Pump 86.22 8.85 0.00 Plant 31.77 10233.91 1 c. Considering the system at the lower level of aggregation write the thermoeconomic system of equations for each component and by mak ing 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 deri ve the analytical cost structure for the product ; Boiler: ( Turbine   6 / 5 6 , , , /5 , / 5 6 , / 5 6 0 5 6 0.3177 () ( ) ( ) 10233.91 FG net F A FG net D plant ex r plant FA D plant F A FG net D plant F A cw FG net Ex Ex W Ex Ex Ex Ex W Ex Ex Ex Ex Ex Ex Ex Ex W Ex Ex m h h T s s Ex W kW                              / 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 FA F A F A boiler FG FG Ex c Ex c Ex c Ex Ex c Ex c Ex Z c Ex c Ex          , ,, 11 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 cc r Ex c c Ex            , , 1 1 1 in B FG D boiler boiler boiler turbine c Ex Ex ZZ f Ex Ex Ex      Department of Energy Politecnico d i Milano Author M. Rocco Pag. 8 of 12 Date 10/0 9/2018 Condenser: Pump: 1 2 , 12 net pump D turbine turbine elec Ex Ex W W Ex C Z C C C c Ex               12 12 turbine elec net elec pump turbine elec net pump c Ex Z c Ex c W c W c Ex Ex Z c WW            ,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 WW c W W              ,D turbine turbine turbine turbine net pump net pump net pump c Ex ZZ f WW W W W W         2 5 3 6 , 2 5 3 6 D condenser condenser Ex Ex Ex Ex Ex C C Z C C C c Ex              25 c Ex c 5 3 3 3 6 2 3 33 condenser condenser Ex Z c Ex c Ex cEx Z c Ex Ex          , 3 3 6 3 6 D condenser condenser Ex Z c c c Ex Ex Ex Ex     , 3 3 6 3 6 () D condenser condenser condenser Ex c c Z r c Ex Ex c Ex Ex        , 3 3 6 3 6 D condenser condenser condenser condenser c Ex ZZ f Ex Ex Ex Ex Ex       Department of Energy Politecnico d i Milano Author M. Rocco Pag. 9 of 12 Date 10/0 9/2018 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 o btained results. We can write the 4 exergy cost balances for the 4 components : Exergy cost balance composed by 4 equation and 10 unknowns. Therefore, in order to close the system of equations, (n -m) numbers of auxiliary relat ions are required: n: Number of exergy flows = 10 m: number of components = 4 n-m = 10 - 4 = 6 Auxiliary relations Now substituting auxiliary relations in the economic cost balance: , ,, 4 44 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 cc r Ex c c Ex      , , 4 4 4 pump D pump pump in pump pump Z Ex Z fc Ex Ex Ex    / / 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 condenser pump pump pump c Ex c Ex Z c Ex c Ex c Ex Z c Ex c W c Ex c Ex Z c Ex c Ex c Ex c W Z c Ex                                  . 5 6 1 2 12 1) 0 2) 3) 4) 0 5) 6) FG F / A net pump elec c c given c c c c c given c c c             / / 4 4 12 1 12 1 12 2 . 12 2 3 3 6 3 3 . 4 4 F A F A boiler turbine elec net condenser DH elec pump pump c Ex c Ex Z c Ex c Ex Z c Ex c W c Ex Z c Ex c Ex c Ex c W Z c Ex                              3 3 4 4 33 , 3 elec pump pump elec pump in pump pump c Ex c W Z c Ex c Ex c W c Ex W          3 4 , 34 pump D pump elec pump Ex W Ex Ex C C Z C C c Ex            Department of Energy Politecnico d i Milano Author M. Rocco Pag. 10 of 12 Date 10/0 9/2018 we have 4 equations and 4 unknowns (c,c4, c3, and celec.). By coupling the third and the fourth equation we get By inserting this last equation into the first equation in the system we get By using the second equation From equation 2 From equation 1 from the last equation e. Compute the values making some comments on the obtained results celec [€/kWh] 0.677 c12 [€/kWh] 0.183 c4 [€/kWh] 4.080 C3 [€/kWh] 1.622 fboiler 0.849 fturbine 0.867 fcondenser 0.919 fpump 0.966 r_boiler 8.83 r_turb 2.69 r_cond 7.87 r_pump 4.70 12 2 6 . 4 4 condenser DH elec pump pump c Ex Z c Ex c W Z c Ex          / / 12 2 6 . 12 1F A F A condenser DH elec pump pump boilerc Ex c Ex Z c Ex c W Z Z c Ex               / / 6 .F A F A condenser DH pump boiler turbine elec net p ump c Ex Z c Ex Z Z Z c W W             / / 6 . 0.833[€ / ] F A F A condenser DH pump boiler turbine elec net pump c Ex Z c Ex Z Z Z c kWh WW          12 12 0.6[€ / ] elec net turbinec W Z c kWh Ex Ex    12 1 4 4 52.732[€ / ] boiler FA FA c Ex Z c Ex c kWh Ex      44 3 3 218.839[€ / ] pump elec pump c Ex Z c W c kWh Ex      Department of Energy Politecnico d i Milano Author M. Rocco Pag. 11 of 12 Date 10/0 9/2018 Exercise 3. (5pt) da febbraio 2016 Let’s consider the en ergy statistics for France in 20 15 (IEA data, values in ktoe). For the next year is planned the disposal of a 3 GW Nuclear Power plant (NU). In order to maintain the same level of yearly production, the gap of electric energy production will be covered for 60% by Natural Gas Combined Cycle power plants (NGCC), and for 40% by a new off -shore wind farm. Using IEA convention, calculate: With reference to the data given in the table, it is required to calculate: a. The Total Primary Energy Supply of France in 2015 and The new value of the TPES in the next year . b. The increase of 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) . c. The cost of C O2 emissions (in €/ton) required to make the cost of electricity production by NGCC equal to the cos t of electricity production by NU , with respect to the substitution considered. Additional data: - Natural Gas Combined Cycle plant are assumed to use pure m ethane with ηI=0.55 and LHV ng=50 MJ/kg . - For wind farm assume a load factor of 0.18. - For nuclear power plant assume ηI=0.33 and a load factor of 0.95 . - Economic costs of electricity: 67 €/MW h for NG CC , 90 € /MWh for NU. Exercise 3. Solution a. The Total Primary Energy S upply of France in 2015 and 2016 The TPES for France in 2013 is evaluated through the algebraic sum of the last column of the given table. The new value of the TPES: endogenous stock TPES imports exports bunkers production changes        137757 148404 32684 1618 5938 586 246507 TPES ktoe ktoe         , , 3 8760 24966 0.086 6506 0.33 el NU NU el NU NU E GW h f GWh E ktoe PE ktoe GWh        ,, , 0.6 14980 0.086 2342 0.55 el NG el NU el NG NG E E GWh E ktoe PE ktoe GWh       Department of Energy Politecnico d i Milano Author M. Rocco Pag. 12 of 12 Date 10/0 9/2018 b. The change in CO2 emissions due to the new production of energy form NGCC in one year (in Mton) and the installed capacity of the new wind farm (in GW) The electric energy produced by the new NGCC plants in one year is equal to (according to IEA conventions): To compute the amount of CO2 emitted by NGCC is firstly necessary to estimate the natural gas needed to produce s uch amount of ener gy : 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 power of the farm. c. Evaluate the cost of CO2 emissions (in €/ton) required to make the cost of electricity production by NU equal to the cost of electricity production by NG. ,, , 0.4 9986 0.086 859 el W el NU W el W E E GWh ktoe PE E ktoe GWh       246507 6506 2342 859 243202 new NU NG W TPES TPES PE PE PE ktoe         , el, 0.6 14980[ ] el NGCC NUE E GWh GWh    ng, el, 1 27236 NGCC NGCC NGCC E E GWh GWh     3 44 ng, , 4 4 1 3600 10 1.961 CH CH NGCC ng NGCC CH ng CH ton Mton GJ m E GWh Mton LHV GJ GWh ton       2 2, ng, 2 44 5.393 16 CO CO NGCC NGCC CO ng kg m m Mton kg    , , 6.3 8760 el W el W W E P GW h y f     , , , , , , , 2, 2, 2, 2, 63.9[€ / ] el NG el NU el NG el NU el NG el NG el NG CO NG CO NG CO i CO NG E c c c E c E c m c ton m      