A fuel gas consists of 75% butane (C4H10), 10% propane (C3H8) and 15% butene (C4H8) by volume. It is to be fed to the combustion chamber in 10% excess air at 25ºC, where it is completely burnt to carbon dioxide and water. The flue gases produced are to be used to generate 5 bar steam from water at 90ºC. Data: Net calorific value (MJ m–3) at 25ºC of: Butane (C4H10) = 111.7 MJ m–3 Butene (C4H8) = 105.2 MJ m–3 Propane (C3H8) = 85.8 MJ m–3 Air is 21% oxygen, 79% nitrogen by volume and 23.3% oxygen and 76.7% nitrogen by mass. Atomic mass of C = 12, O = 16, N=14 and H = 1. I write the question and after the solution ,i just want to know if what i did is ok or i have to do something else, thank you in advance for your help.(also i used a table with enphaly for different temperatures but didnt post it here) Calculate: (i) the net calorific value (CV) per m3 of the fuel/air mix at 25ºC (ii) the net calorific value (CV) per kmol of the fuel/air mix at 25ºC. (i) NCV of fuel in MJ/M3 @25°C = (0.75 x 111.7 + 0.1 x 105.2 + 0.15 x 85.8) MJ/M3 = 107.2 MJ/M3 (ii) (ii) NCV of the fuel will be 2.401MJ/Kg-mol. Weight of 1 kg-mol of the fuel mix = 56.3 kg At atm pressure the volume of 1 kg mol = 22.4m3 And hence the NCV = 107.2 x 22.4 = 2401.28 MJ/Kg-mol. Determine the actual fuel:air ratio (i) by volume (ii) by mass. (i). Fuel gas composition, Butane: C4H10 Propane: C3H8 Butane: C4H8 Oxygen sent is 10% excess Stoichiometric equations (the combustion equation) C4H10 + 6.5 O2 4CO2 + 5H2O C3H8 +5O2 3CO2 + 4H2O C4H8 +6O2 4CO2 + 4H2O Butane : 75% Propane : 10% Butane : 15% 1 Mole of fuel contains 0.75 Moles of Butane 0.1 Moles of Propane 0.15 Moles of Butane Corresponding oxygen requirement for, Butane = 0.75 x 6.5 = 4.875 Moles of O2 Propane = 0.10 x 5 = 0.5 Moles of O2 Butane = 0.15 x 6 = 0.9 Moles of O2 Total = 6.275 Moles of O2 Amount of O2 in 1 mole of air =0.2095 moles Hence, 6.275 moles of O2 Present in 25.95 moles of air Hence fuel to air ratio is 1 mole of fuel needed for 29.95 moles of air for complete combustion. Since the excess air supplied is 10% The actual air intake for 1 mole of fuel Combustion = 32.945 mole / 1 mole of fuel By mass, the amount of fuel intake, for mole Butane = 0.75 x (4 x 12 +10 x 1) = 43.5 Kg Propane = 0.10 x (3 x 12 + 8 x 1) = 4.4 Kg Butane = 0.15 x (4 x 12 +8 x 1) = 8.5 Kg 1 mole of fuel molecular mt = 56.3 kg-mol. Corresponding air intake (with 10% Excess) = 1.1 x 6.275 x 29 kg/kg-mole = 200.1725 kg/kg-mole Hence fuel to air ratio by mass = 56.3/200.1725 = 0.28 Actual air to fuel ratio by mass =3.56 A/F) Mass = 3.56 A/F) Volume = 32.945 Determine the composition of the flue gases by volume (assuming the inlet air is dry): (i) on a wet basis (ii) on a dry basis Composition of fuel gas by volume: (wet basis) (i) Butane, Propane, Butane’s combustion products along with excess oxygen and nitrogen. 1 mole of fuel yields = 0.75 x (4CO2 + 5H2O) + 0.1 x (3CO2 + 4H2O) + 0.15 x (4CO2 + 4H2O) + Excess O2 + N2 = 39 CO2 + 4.75 H2O + 0.64 O2 + 26.03 N2 Excess O2 = (0.21)*32.95 = 6.9195 - 6.275 = 0.645 moles Nitrogen present = 0.79 * 32.95 = 26.03 moles (i) On dry basis, the consistent of the three gases are 3.9 CO2 + 0.645 O2 + 26.03 N2. Determine the ‘furnace efficiency’ if the flue gases leave the boiler at 300ºC. The furnace efficiency is the measure of the furnace combustion efficiency. Furnace efficiency If the flue gases are leaving at 300°C Heating value of the gas = (3.9x26.61+.645x18.33+26.03x17.4+4.75x20.89) = 667.75 MJ/kg-mol Hence the efficiency of the furnace is given by = 667.75/2401 = 27.8% If 5% of the heat available for steam production is lost to the atmosphere, determine the amount of steam raised per hour when the total flow of flue gases is 1400 kmol h–1. Total flow rate of the flue gases = 1400Kmol/hr. 5% of the heat available is lost. The actual amount of the heat available in the flue gases = 2401.28MJ/kg-mol Hence considering the loss of 5% of heat energy the actual heat available for making steam, = 0.95 x 2401.28MJ/kg-mol = 2281.2MJ/kg-mol Steam enthalpy at 5 bar = 2748KJ/kg = 49.464MJ/kg-mol Saturated Water enthalpy at 90 degree Celsius = 385KJ/kg = 6.930MJ/kg-mol Hence the heat requirement for steam generation = 49.5-6.93 = 42.57MJ/kg-mol Flue gas flow rate = 1400 kgmol/hr Hence the steam generation rate = (heat available/heat consumption), = (1400*2281.2)/42.6 =~ 75 tons/hr. Determine the dew point temperature assuming that the flue gas pressure is 1.00 bar and the inlet air: (i) is dry (ii) contains 0.8 kg water per kmol of air at the temperature of the inlet air. The gas is 1.00 bar, (i) Inlet air is dry Flue gas pressure is 1.00 bar If inlet air is dry, the amount of water vapor in the products = 4.75 moles Hence % of vapor by volume = 4.75/3.9+4.75+0.645+26.03 = 13.45% Hence from the charts, the dew points of water vapor in the gas = 53°C (ii) In inlet air contains 0.8kg water/kg-mol of air Actual air intake = 32.945 kg mol Hence water present = 26.356 kg Hence the actual kg-mol of water will be increased by this content proportionately 26.356 kg = 1.46 kg-mol Hence total H2O vapor = 1.46+4.75 = 6.21 kg mol =(6.21/ 3.9+4.75+0.645+26.03 ) +1.46 = 16.88% by volume From charts corresponding dew point = 58°C If the flue gases exiting the boiler are used to preheat the water fed to the boiler from a temperature of 28ºC to 90ºC and assuming: • a mean specific heat capacity for water over this temperature range to be 4.2 kJ kg–1 K–1 • a mean molar heat capacity for the flue gases up to 300ºC to be 31 kJ kmol–1 K–1 • 10% of the heat required to heat the water is lost in the heat exchanger • all water entering the system is converted to steam determine the final outlet temperature of the flue gas and state if the dew point will be reached in both of the cases given in previous question Flue gases used to preheat water From 28°C to 90°C Near specific heat of water = 4.2 kj/kg Water heat capacity of flue gas = 31kj/kg-mol k Heat lost = 10% Water completely became steam Mw x 4.2 x (90-28) = 31 x (-Tf + 300) x 1kg mol 8.4 Mw = 300- Tf Tf = 300-8.4 (Mw) Amount of water Mw/ in kg/kg-mol of the gas if increased proportionately Tf will come down. It is possible to get the dew points of both the previous cases.