Felder, Rousseau, Bullard, Elementary Principles of Chemical Processes, 4e an ef
ID: 703821 • Letter: F
Question
Felder, Rousseau, Bullard, Elementary Principles of Chemical Processes, 4e an effort to satisfy local demand for various fuels, a modest-sized blending operation is considering ming pure One calls for preparing four An estimate of the market for these fuels indicates that 5.00% of the ethanol with gasoline using various schemes fuels: E 10°, which is pure ethanol,E85 which is 85.0 vom ethanol and 1S0% gasoline; E10 which 1 0.0 vol% ethanol "nd 90.0% gasoine, and pure gasolne ten demand (by volume) is for E100. 15.0% for Ees. 40.0% for E10. and the remainder for pure gassint ?ethanol and gasoline enter the process. Which of the following process flowcharts best depicts the blending operation? Tn the disgrams, is tne combined volumetric nlow rate of the four product fuels, VG is the volumetric fon rate of pu gasoline being fed to the process, and Ve is the volumetric flow rate of pure ethanol being fed to tre process Ve BienderExplanation / Answer
Part a
Volume of E10 produced = 1.00 x 10^5 liters/day
Total volume of fuel produced
= Volume of E10 produced / fractional market demand for E10
= (1.00 x 10^5 liters/day)/0.40
= 2.50 x 10^5 liters/day
Part b
Volume of E10 produced = 1.00 x 10^5 liters/day
Volumetric flow rate of pure ethanol
= Volume of E10 produced / fractional market demand for pure ethanol
= (1.00 x 10^5 liters/day)/0.05
= 20.00 x 10^5 liters/day
Volumetric flow rate of pure gasoline
= Volume of E10 produced / fractional market demand for pure gasoline
= (1.00 x 10^5 liters/day)/(1 - 0.05 - 0.15 - 0.40)
= 1.003 x 10^5 liters/day
Part C
Max gross weight of truck = 3.6000 x 10^4 kg
Tare weight of truck = 1.2700 x 10^4 kg
Cargo weight limit = (3.6000 - 1.2700) x 10^4
= 2.3300 x 10^4 kg
Part d - scale up
Volume of combined product = 2.5000 x 10^5 L/day
Specific gravity of gasoline = 0.726
Density of gasoline = 0.726 * 1000 = 726 kg/m3
Specific gravity of ethanol = 0.789
Density of ethanol = 0.789 * 1000 = 789 kg/m3
Density of combined product = fractional demand x density of pure ethanol + fractional demand x density of E85 + fractional demand x density of E10 + fractional demand x density of pure gasoline
= ( 0.05 x 789) + (0.15 x 0.85 x 789 + 0.15 x 0.15 x 726) + (0.40 x 0.10 x 789 + 0.40 x 0.9 x 726) + (0.40 x 726)
= 639.2055 kg/m3
Mass of combined product
= 1.00 x 10^5 liters/day x 1m3/1000L x 639.2055 kg/m3
= 0.63925 x 10^5 kg
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