Problem 2 The ventilation system of an industrial unit with dimensions 40 x 10 x 4 m, provides 4 air changes per hour. Air in the unit is at 105 kPa and 18°C, and can be considered well mixed. The unit is crossed by a pipe that feeds pure gaseous ammonia to a tank (amoniaco in Spanish). a. Calculate the ammonia leak (g/h) that would result in an ammonia concentration in the unit air equal to 80% of the Spanish legal daily exposure limit (VLA-ED). b. If the ammonia leak is 2.5 m3/h (measured at room conditions), calculate the theoretical ventilation flow rate necessary to maintain the ammonia concentration at 80% of the VLA-ED. c. If at the steady-state conditions of part “b”, the ventilation flow rate drops to half of the value calculated, determine the new steady-state ammonia concentration, and the time (min, measured from the moment when the flowrate drops) that takes this concentration to reach 90% of the steady state value. d. If there is a sudden release of 250 kg of ammonia that in a very short time (assume instaneously) gets well mixed in the air of the unit, determine if the resulting gaseous mixture is flammable. e. If the resulting mixture is flammable, calculate the ventilation flowrate necessary to exit the flammability zone is 3 min. Chemical VLA-ED VLA-EC NOTAS ppm mg/m3 ppm mg/m3 Ammonia 20 14 50 36 VLI Substance Flammability limit (%vol) NFPA class Flash point Autoignition temperature Lower Upper Ammonia 15 28 IIIB 11°C 651°C Answer V = 40 x 10 x 4 = 1600 m3 Q = 1600 m3 x 4 air/h = 6400 m3/h a) 80% of 20 ppm = 16 ppm Fa = Qa . Ca Ca = ya . Ct Ct = = = 0.04 mol/L = 41.9 mol/m3 Ca = (16 x) (41.9) = 6.7xmol/m3 Fa = 6400 m3/h x 6.7xmol/m3 = 4.288 mol/h 4.288 mol/h x 17 g/mol = 72.9 g/h b) F = 2.5 m3/h n = = = 104.77 mol/h Q = = = 156371.77 m3/h c) Q = 1/2 x (156371.77 m3/h) = 78185.88 m3/h Co = = = = 1.34 x10-3 mol/m3 C = 1.34 x10-3 x 90% = 1.2 x10-3 t= V/Q ln F-QoCo/F-QC = = -0.192 Question = What did I do wrong that make the result minus? why is the time in number c minus? what did I do wrong?