Energy Engineering - Wind, Hydro and Geothermal Power Generation

Full exam

POWER PRODUCTION FROM RENEWABLE ENERGY AY 2021-22 7 th February 2023 Prof. Silva Time: 1.5 hours Instructions for the examination: 1) Clearly indicate your name on all the files you will deliver. 2) The score refers to exercises done in a comprehensive manner with exact numerical results. Numerical results correct but not accompanied by explanations will not be taken into account. The final score can be normalized according to the average results. 3) Talking with colleagues and / or cheating will cause the cancellation of the exam. 4) All the needed data for the resolution of exercises lies on this paper. It is NOT ALLOWED to use material other than this (e.g. books, clipboard etc.). Exercise 1 (15 points) A geothermal source has the following characteristics at the outlet of the well: temperature 185°C, enthalpy 1100 kJ/kg. CASE A Consider a single flash direct steam cycle cogenerative plant, based on a cooling tower condenser with a condensation pressure of 0.07 bar. The flash pressure is 4.4 bar. Design the lay-out of the plant (i) with all the components (1 point), considering that the cogenerative heat exchanger is fed by the separated liquid from the flash chamber, with an outlet temperature of the geothermal fluid equal to 47°C. Neglecting the presence of non-condensable gases, calculate (ii) the mass flow rate of the geothermal fluid that should be extracted from the well to obtain a gross electric power output of 11.7 MW and that the isentropic efficiency and the organic-electric efficiency of the turbogenerator are respectively equal to 80% and 95% (4 points). Then determine (iii) the vapor fraction at the outlet of the steam turbine (2 points) and (iv) the net electrical efficiency of plant (2 points), considering an electric power consumption of all the auxiliary components of the steam cycle equal to 700 kW and assuming a minimum temperature of reinjection equal to 44°C (c p of the geothermal fluid in the liquid phase equal to 4.4 kJ/kgK). Calculate also (v) the cogenerative thermal power output (1 point), and (vi) the first-law efficiency at nominal conditions (1 point). CASE B Consider an "all electric" plant scheme in which the liquid output from the flash chamber feeds an ORC binary cycle by means of a heat exchanger which cools it to the same exit temperature of the CASE A (for the rest of the system take the same assumptions). Compute (vii) the total net electric power output of the plant, knowing that the second law efficiency of the ORC cycle is 49% and the condensing temperature is the same as the direct steam cycle (4 points). Thermodynamic properties of water at saturation Liquid Vapor P sat [bar] T sat [°C] h liq.sat. [kJ/kg] s liq.sat. [kJ/kgK] v liq.sat. [m3/kg] h vap.sat [kJ/kg] s vap.sat. [kJ/kgK] v vap.sat. [m3/kg] 4.4 147 .1 619. 6 1.8120 0.00100 2741. 9 6.8623 0.4226 0.07 39.0 163.4 0.5591 0.00101 2572.6 8.2767 20.5310 Exercise 2 (15 points) Mechanical and Energy Engineering students The installation of a PV plant for residential applications should be evaluated. The PV plant is installed on the roof of a house (5.5 x 8 m), the azimuth and the tilt angles of the roof are respectively -20° and 25° (the azimuth is determined starting from south direction, positive counterclock-wise). Assuming the PV module and inverter characteristics reported in the following table, determine (i) the number of inverters (2 points) and (ii) the max- imum nominal power installed (AC) (3 points), discussing the results. Single Cell electric characteristics Inverter Power at MMPT @STC 5.17 W Max p ower input in CC 2650 W open circuit voltage @STC 0.66 V Maximum voltage 600V Voltage at MPPT@STC 0.628 V Power point operating voltage 260 -500 V Power coefficient -0.38%/°C Max p ower output in AC 2500 W Module characteristic s N° of cells (series connected) 60 STC (1000 W/m 2 and 25°C) Module size ( l x w) in mm 1660 x 990 NOCT (@800 W/m 2 and T amb 20°C) 45°C Calculate (iii) the yearly electricity produced (5 points) and (iv) the equivalent hours (1 point), assuming only losses due to incidence angle, operating temperature and inverter. Assume an average zenith and azimuth angle of the sun equal to 45° and 0° respectively, an average ambient temperature equal to 15°C, an average 700 W/m 2 Direct Normal Irradiance and 80 W/m 2 diffuse horizontal irradiance. Neglect the ground albedo. Equivalent hours of solar radiation are 1800. cos cos cos sin sin cos( ) SZSZSSθ θ β θ β ψψ =+− = where θ S is the incidence angle, θ ZS is the zenith angle, β is the roof tilt, ψ s and ψ are the solar and roof azimuth angles respectively. What would be (v) the electricity produced (3 points) and (vi) the equivalent hours (1 point) by the same number of panels installed on a two axis tracking system? Exercise 2 (15 points) Management Engineering students A steam cycle is coupled with a biomass boiler with mass-flowrate equal to 7 kg/s and biomass composition reported below. The biomass has 50% moisture content (oak with harvest conditions) which results in a boiler efficiency equal to 69% on HHV basis (∆h evaporation equal to 2,44 MJ/kg). Assuming that the steam cycle electric efficiency is equal to 35% (HHV base), what are (i) the net power output of the plant (3 points) and (ii) the net electric efficiency (LHV base) (1 point)? Determine (iii) the stack temperature (4 points) and (iv) the boiler efficiency (LHV) (2 points), knowing that the combustion air flowrate is 22 kg/s, thermal losses from boiler walls are equal to 1% of the fuel HHV thermal input and neglecting ashes losses. Furthermore the un-burned carbon is 0.7% (dry basis), water in the flue gases is 100% in the vapor state, and ambient temperature is 25°C. (c p flue gases 1.11 kJ/kgK, heating value of carbon is 32.8 MJ/kg) A dryer is added before the boiler, exploiting the exhaust gases coming from the boiler in countercurrent with the biomass. Consider a biomass mass flow rate at the dryer inlet equal to the previous case on “as received” basis. Assuming a temperature of the flue gases at the outlet of the dryer equal to 105°C and neglecting the heating of the steam and of the biomass, (v) determine the moisture content of the biomass at the outlet of the dryer (at the inlet of the boiler) (5 points). oak (harvest conditions) %, weight dry basis C 48.80 H 6.09 O 45.00 ash 0.11 LHV, kJ/kg, dry basis 17769 HHV, kJ/kg, dry basis 19107.6 Exercise 1 CASE A Xv is 0,817 h is 2131,1 kJ/kg Dh is 610,8 kJ/kg Dh 488,6 kJ/kg h 2253,2 kJ/kg P el 11700 kW m vap. 25,20 kg/s Xv 0,2264 m geothermal 111,34 kg/s m liq 86,1 kg/s Xv 0,867 Q th 37935 kW hmin 185,3 kJ/kg Q th max 101849 kW Eta el net 10,80% Eta th 37,2% Eta I 48,0% CASE B T m, ln 94,8 °C Eta Lorentz 15,2% Eta binary cycle 7,43% P el net ORC 2816,8 kW P el net tot 13817 kW Exercise 2 - Mechanical and Energy Engineering students Power (single module) 310,2 W cos incidence angle 0,92 22,82845 Voltage 39,7 V 317,6 Incidence radiation 721 W/m^2 MPPT voltage 37,7 V 301,6 temperature 37,5 °C Number of modules 24 mod. power avg 213,1 W Total power DC 7444,8 W average power AC 4825 W Number of inverters 2,809358 electricity 8686 kWh h eq 1237 h Number of inverters 3 2 punti Modules for each inver- ter 8 two axis tracking 1 Power DC (inverter) 2481,6 W radiation 776 W/m2 Voltage 317,6 V temperature 39,3 °C Voltage MPPT 301,6 V power 227,7 W Power AC (inverter) 2341,1 W average AC 5156,5 Power AC 7023 W 3 punti electricity 9282 kWh Exercise 2 - Management Engineering students LHV as received 7664,5 kJ/kg HHV as received 9553,8 kJ/kg boiler thermal input, HHV 66,88 MW boiler thermal input, LHV 53,65 MW steam cycle thermal input, HHV 46,14 MW Net power output 16,15 MW Net electric efficiency, LHV 30,1% air flowrate 22 kg/s Qloss 1,0% Fraction unburnt C 0,7% c_p, flueGas 1,11 kJ/kg -K ambient temperature 25 °C overall losses 20,73 MW m_unc_C 0,012 kg/s energy losses for unc_C 0,39 MW thermal losses 0,67 MW steam content 5,42 kg/s steam losses 13,22 MW stack losses w/o steam 6,45 MW stack temperature 225,4 °C boiler efficiency, LHV 86,0% T stack 105 °C m flue gases 29 kg/s Q dryer 3,875 MW m H2O evaporated 1,588 kg/s m H2O biomass (harvest) 3,5 kg/s m dry biomass 3,5 kg/s m H2O (after dryer) 1,912 kg/s m biomass (after dryer) 5,412 kg/s Moisture content 35,3 %