Mechanical Engineering - Energy Systems LM

Full exam

Politecnico di Milano School of Industrial Engineering Course Energy Systems proff. S. Consonni, E. Martelli, M. Romano - Academic Year 2022-23 Energy Systems LM – written test of 17 January 2023 page 1 of 3 Written Exam of 17 January 2023 - Problems - Time: 2 hours PLEASE NOTICE 1) Exam is open book, but computers and cell phones are NOT allowed. Talking with colleagues and/or copying will lead to the immediate cancellation of the exam. 2) Answer clearly ONLY to the questions posed by the problem sets. Even if correct, additional considerations and/or calculations will NOT be considered. 3) Fill this sheet with your name and return it together with your solutions. 4) Mark each sheet of the solution with your name and page number. 5) In addition to the points obtained for the solution of each problem, a bonus of max 1 point may be given based on whether the solution of each problem is complete, with clear details and explanations. Problem 1 (15 points) A boiler generates 20 kg/s of saturated steam at 15 bar. The boiler features the following parameters: - The boiler consists of a combustor, a heat exchanger (flue gas – steam), an air preheater, a stack - The temperature of the flue gases at the stack is 120 °C - The ambient air temperature is 10 °C - Air composition = 21% O 2, 79% N 2 - Molar mass of air = 28.84 kg/kmol - The feedwater entering the boiler is at 100 °C, 20 bar - The effectiveness of the air preheater is 80% - The percentage excess of air is 30% - Fuel is pure butane (C 4H10: molar mass=58 kg/kmol; LHV=45.75 MJ/kg) - The specific heat capacity at constant pressure c p of the air is 1.02 kJ/(kg K) - The specific heat capacity at constant pressure c p of the flue gas is 1.10 kJ/(kg K) - Heat losses from the combustor walls = 1 % of the LHV - Energy losses due to incomplete combustion = 1% of the LHV Given the above: 1) Draw the scheme of the boiler with the air preheater. (2 points) 2) Calculate the mass flow rate of fuel. (5 points) 3) Calculate the boiler efficiency. (3 points) 4) Determine the temperature of the preheated air at the air preheater outlet and the flue gas temperature at air preheater inlet (5 points) Politecnico di Milano School of Industrial Engineering Course Energy Systems proff. S. Consonni, E. Martelli, M. Romano - Academic Year 2022-23 Energy Systems LM – written test of 17 January 2023 page 2 of 3 Problem 2 (18 points) The power plant shown in the figure exploits a low temperature geothermal heat source to produce electricity with an organic Rankine cycle (ORC) using R134a as working fluid. Geothermal water is taken from the reservoir at T w,in and is cooled down in the primary heat exchanger (PHE), which includes economizer, evaporator and superheating sections. Superheated R134a fluid from the PHE (stream #6) goes through a control valve before being fed to the turbine (#7). Superheated vapor exiting the turbine (flow #8) is cooled down in a counter-flow regenerator (reg) and then condensed in an air-cooled condenser. Condensed fluid (#1) is pumped (#2), preheated in the regenerator (#3) and fed to the PHE. The following data are given for the process at the design operating point: • Geothermal water: m w = 50 t/h, c p=4.2 kJ/kgK, T w,in = 130 °C • Superheated fluid from the PHE: T 7 = 120 °C • Evaporation temperature: T 5 = 90 °C • Condensing temperature: T cond = 40 °C • Minimum temperature difference in the regenerator = 10 °C • Minimum temperature difference in the evaporator (pinch point) = 5 °C • No pressure drops across the valve and across the heat exchangers (i.e. p 7=p 2; p 1=p 8) • Turbine: isoentropic efficiency ������������ is=80%, mechanical-electric efficiency ������������ me= 96%. • Pump: isoentropic efficiency ������������ is=70%, mechanical-electric efficiency ������������ me= 94%. • Ambient temperature T 0=25 °C • Consider that the temperature increase across the pump due to the combined effects of fluid compressibility and pump fluid-dynamic losses is 2.1 °C. So, T 2 = 42.1 °C. 1) Draw the diagram of the cycle on the given p-h chart and on a qualitative T-s chart, indicating the stream numbers (3 points) 2) Indicate on a table, pressure, temperature, enthalpy and entropy of the 9 streams of the cycle (4 points) Tw,in Tw,out reg 2 1 9 8 7 6 5 4 3 eco sh eva Politecnico di Milano School of Industrial Engineering Course Energy Systems proff. S. Consonni, E. Martelli, M. Romano - Academic Year 2022-23 Energy Systems LM – written test of 17 January 2023 page 3 of 3 3) Calculate net electric power output, net electric efficiency of the cycle and heat recovery efficiency, considering that the minimum temperature of reinjected water (i.e. minimum T w,ou t) is equal to the ambient temperature T 0. (4 points) 4) Calculate the first law efficiency and the second law efficiency of the plant. (3 points) After some years of operation, the geothermal water flow rate is reduced causing a reduction of the working fluid flow rate by 30% compared to the design condition. In such conditions: • the system is controlled with turbine inlet throttling (i.e. p 6 is kept constant by throttling the turbine admission valve) • the turbine keeps working in choked conditions, with the same nondimensional mass flow rate of the design condition) • the PHE is controlled to keep the turbine inlet temperature T 7 constant by-passing a fraction of the geothermal water flow • the condenser fans are controlled to keep the condensing temperature unchanged • the PHE inlet fluid temperature (T 3) is 60 °C. 5) Compute the net electric power output in the new operating condition. (2 points) 6) Assess whether the preheater is expected to be fouled or not. (2 points) Problem 1 PROBLEM 1: DataAir inlet temperature, °C 10 water inlet pressure, bar20 water inlet temperature, °C100 steam pressure (outlet), bar15 stack temperature, °C120 effectiveness of preheater0.8 percentage air excess of combustion30% fuel = C4H10 fuel molecular mass, kg/kmol58 LHV of fuel, MJ/kg45.75 Cp air, kJ/kg/K1.02 Cp flue gases, kJ/kgK1.1 Heat losses from boiler walls, %1.00%of LHV unburned fuel losses, %1.00%of LHV steam mass flow rate, kg/s20 air composition (simplified) 21% O2, 79% N2 Air molecular mass, kg/kmol28.84 Solution2)O2/fuel stoinchiometric molar ratio6.50nC+nH/4wherenC = 4nH = 10moles of C and H atoms iof the fuel molecule air/fuel stoichiometric molar ratio30.95 air/fuel molar ratio40.24 air/fuel stoichiometric mass ratio15.39air/fuel_mass_stoic_ratio= O2/fuel_molar_ratio *(1/0.21) * (air molecular mass / fuel molecular mass) air/fuel mass ratio20.01air/fuel_mass_ratio = stoichiometric_value * (1+0.30) steam outlet enthalpy, kJ/kg2791Saturated steam enthalpy at 5 bar water inlet enthalpy, kJ/kg420.53Liquid water enthalpy at boiler inlet useful thermal power, kW47410Qth = steam flow rate * (Hout- Hin) fuel mass flow rate, kg/s1.120 deriving fuel_flow from the energy balance of whole boiler + air preheater: 3)Boiler efficiency 0.9253 efficiency = thermal_power/(fuel*LHV) 4)air flow rate, kg/s22.407fuel_flow*LHV+fuel_flow*(air/fuel_mass_ratio)*Cp_air*(Tair-25) = FG flow rate, kg/s23.527 = thermal_power + heatlosses + unb_losses +fuel_flow_rate*(air/fuel_mass_ratio + 1)*Cp_fg*(Tstack-25) calculating the inlet preheater temperature of FG(Tout_Air - Tin_AIR) = effectiveness * (Tin_FG - Tin_AIR) definition of effectiveness mair * cp_air * (Tout_AIR-Tin_AIR) = mFG * cpFG * (Tin_FG-Tout_FG) from energy balance of preheater (TairOUT): system of two equations and two unknwns (TfgIN and ToutAir) from definition of effectiveness (knowing that the air stream has the lowest heat capacity m*Cp) by direct substitution, you can find the following equation to find Tin_FG: effectiveness = mair*Cpair*(TairOUT-TairIN)/mair*Cpair*(TfgIN-TairIN) Tin_FG(mair*cpAIR*effectiveness-mFG*cpFG)=mair*Cp_air*effec*Tin_AIR-mFG*cpFG*Tout_FG Tout_AIR, °C 309.8 from the energy balance of the air preheater Tin_FG, °C 384.8 Problem 2 Datamw 50 t/h cp4.2kJ/kgK Tw,in130°C Teva90°C T6120°C Tcond40°C DTmin,reg10°C DTmin,eva5°C eta,is,t0.80 eta,me,t0.96 eta,is,p0.70 eta,me,p0.94 T025°C Part-load: mf' / mf0.70 T3'60°C Solution 1) 6=7 5 2 3 4 1 8 9 0 20 40 60 80 100 120 140 1.01.11.21.31.41.51.61.71.81.9 T, °C s, kJ/kg K 9' 9 8 6=7 5 4 1 2 3 2)Fluid nameHEOS::R134a Dhref0.00 p eva =32.4bar Dsref0.000 h6 = h7 = 477.8 kJ/kg s6 = s7 =1.81kJ/kgK p cond =10.2bar h8,is =450.3kJ/kg h8 = h7 - (h7 - h8,is)*eta,is,t =455.8kJ/kg T8 =73.9°C h1 =256.4kJ/kg s1 =1.19kJ/kgK h2,is =258.3kJ/kg h2 = h1 + (h2,is - h1)/eta,is,p =259.2kJ/kg approximated calculation with tables:rho1 = 1147 kg/m3 Dhp = Dp/rho/eta,is,p = 2.8 kJ/kg h2 = h1 + Dhp = 259.2 kJ/kg T2 = 42.1 �C T9 = T2 + DTmin,reg =52.1°C h9 =432.8kJ/kg h3 = h2 + (h8 - h9) =282.2kJ/kg T3 =57.3°C h4 =342.9kJ/kg h5 =425.4kJ/kg p, eva T, °C h, kJ/kg s, kJ/kgK 1 10.2 40 256.4 1.190 2 32.4 42.1 259.2 1.193 3 32.4 57.3 282.2 1.264 4 32.4 90 342.9 1.439 5 32.4 90 425.4 1.666 6 32.4 120 477.8 1.805 7 10.2 120 477.8 1.805 8 10.2 73.9 455.8 1.821 9 10.2 52.1 432.8 1.753 9' 10.2 40 419.4 1.711 (for the chart) 140.01.2(for the chart) 3)mw =13.89kg/s Tw,eva out =95°C Qeva+sh =2042kW Dheva+sh = h6 - h4 =134.9kJ/kg mf =15.14kg/s Pt = mf * (h7 - h8) * eta,me,t =319.9kW Pp = mf * (h2 - h1) / eta,me,p =44.6kW Pnet = 275.3 kW Qin = mf * (h6 - h3) = 2961.4 kW eta,cyc = 9.30 % Qavailable = mw * cp * (Tw,in - T0) = 6125 kW eta,th = Qin / Qav = 48.35 % 4) eta I = Pnet / Qav = 4.49 % Tml,w = 348.0 K eta,Lorenz = 1 - T0/Tml,w = 14.33 % Wrev,w = Qav * eta,L = 877.6 kW eta II = Pnet / Wrev,w = 31.37 % 5)choked flow with T7'=T7 -> p7' / p7 = mf' / mf =0.700 p7' =22.71bar h7' =490.7kJ/kg s7' =1.86kJ/kgK h8',is =469.6kJ/kg h8' = h7' - (h7' - h8',is)*eta,is,t =473.8kJ/kg mf' =10.60kg/s Pt' =171.3kW Dhp' = Dhp -> Pp' =31.19kW Pnet' = 140.2 kW 6)Regenerator design point: DThe =16.6°C DTce =10.0°C LMTD =13.02°C Q = mf * (h3 - h2) = 348.1 kW UA = Q / LMTD = 26.7 kW/K Regenerator off-design point: h3' =286.5kJ/kg T8' =91.2°C h9' = h8' - (h3' - h2') =446.5kJ/kg T9' =65.0°C DThe' =31.2°C DTce' =22.9°C LMTD' =26.86°C Q' = mf' * (h3' - h2') =289.2kW U'A = Q' / LMTD' =10.8kW/K U' / U = 0.403 In the new off-design condition, the overall heat transfer coefficient reduces, which is reasonable considering that the fluid flow rate reduces. However, the reduction of U (-60%) is much higher than the mass flow rate reduction, which means that the heat exchanger is likely fouled. Note that for a clean heat exchanger, convective heat transfer coefficient 'h' is proportional to Re^n, with n dependent on geometry and fluid dynamic regime (typically in a range 0.65-0.85, so the reduction of U should be proportionally lower than the reduction of m). Politecnico di Milano Department of Energy - School of Industrial Engineering Course Energy Systems proff. S. Consonni, E. Martelli, M. Romano - Academic Year 2022-23 Energy Systems – Theoretical Questions – 17 January 2023 page 1 Written Exam 17 January 2023 Theoretical Questions - Time: 1.00 h PLEASE NOTICE 1) Books, lecture notes, cell phones are strictly forbidden. Using them or talking with colleagues and/or copying will lead to the immediate invalidation of the exam. 2) Answer clearly ONLY to the questions. Even if correct, additional considerations will NOT contribute to the final grade – and you have NOT enough time! 3) Fill this sheet with your name and return it with your answers. 4) MARK EACH SHEET OF YOUR ANSWERS with YOUR NAME AND PAGE/SHEET NUMBER. 5) WRITING AND SHEET/PAGE SEQUENCE MUST BE LEFT TO RIGHT AS USED IN WESTERN COUNTRIES. 6) Wherever possible, support your statements with clear drawings and/or graphical representation 7) ORGANIZE YOUR ANSWERS CLEARLY: ANSWER TO 1.a; ANSWER TO 1.b, ETC. FIRST NAME…………………………………….......……FAMILY NAME………………………………………………….. Question 1 (17 points) A Combined Cycle comprises a gas turbine which discharges flue gases at 500°C and a steam cycle with a single evaporation pressure level at 90 bar and condensation at 0.1 bar. The liquid exiting the condenser goes first to the low-temperature economizer, which brings water to 90°C; such water goes to the deaerator operating at 2 bar. The deaerator uses steam generated (at 2 bar) in a dedicated evaporator of the HRSG. The liquid exiting the deaerator goes to the high-temperature economizer, which brings the liquid to a temperature 10°C lower than the evaporation temperature (i.e. ∆T subcooling is 10°C). Superheated steam exits the superheater at 470°C (i.e. ∆T approach is 30°C). For the sake of simplicity, assume that all pressure losses are negligible A) Draw the plant scheme (3 points) B) Draw the T-Q diagram of the HRSG and the T-Q diagram of the dearator (3 points) C) Discuss the criteria adopted to select the value of the pinch point temperature difference (∆T pp) (3points) What are advantages / disadvantages of adopting a very high or very low value ? Is there an optimum value ? D) Motivate whether it is more efficient to generate the steam required by the deaerator in the HRSG or to extract the steam from the steam turbine (3 points) After some years, a supplementary firing burner is installed at the inlet of the HRSG (i.e. fuel injection at the gas turbine outlet to increase the gas temperature at the HRSG inlet). When the supplementary firing is turned on, the temperature of the flue gases is increased from 500°C to 600°C and the heat recovery steam cycle operates in “off-design” conditions. Politecnico di Milano Department of Energy - School of Industrial Engineering Course Energy Systems proff. S. Consonni, E. Martelli, M. Romano - Academic Year 2022-23 Energy Systems – Theoretical Questions – 17 January 2023 page 2 E) Discuss the variations expected for the steam cycle temperatures, pressures and mass flow rates and power output (3 points) F) What is the expected impact of supplementary firing. on the efficiency of the Combined Cycle ? (2 points) Question 2 (17 points) Consider a heat pump cycle with an intermediate pressure level with recompression of the vapor formed at the intermediate pressure (i.e. two stage throttling system with intermediate gas removal). The heat pump is used to make available warm water at 60°C for residential heating, sucking heat from ambient air. At design conditions, ambient temperature is assumed -5°C A) Draw the plant scheme (2 points) B) Draw the T-s and P-h diagram (3 points) C) Draw the T-Q diagram of the evaporator and of the condenser, with illustrative, reasonable values for the inlet / outlet temperatures of the hot and cold stream (3 points) D) Discuss advantages and disadvantages of adopting the intermediate pressure level (3 points) E) Discuss how the intermediate pressure level could be determined. Is there an optimum value ? (3 points) F) Discuss how heat pump performances (net absorbed electric power and COP) would change in mild climate, e.g. when ambient temperature is +5°C rather than -5°C (3 points) GIVE CONCISE ANSWERS CLEARLY SPECIFYING WHICH ANSWER YOU ARE ADDRESSING: A), B), C), …