In an effort to satisfy local demand for various fuels, a modest-sized blending
ID: 3161899 • Letter: I
Question
In an effort to satisfy local demand for various fuels, a modest-sized blending operation is considering mixing pure ethanol with gasoline to prepare four fuels: El00, which is pure ethanol; E85 which is 85.0 vol% ethanol and 15.0% gasoline; E10 which is 10.0 vol% ethanol and 90.0% gasoline, and pure gasoline (E0). An estimate of the market for these fuels indicates that 2.00% of the demand (by volume) is for El 00, 10.0% for E85, 40.0% for 110. and the remainder for pure gasoline (E0). Draw' and label a flowchart for the overall blending operation, letting V represent the combined volumetric flow rate of the four product fuels, V_G represent the volumetric flow rate of pure gasoline being fed to the process, and V_E is the volumetric flow rate of pure ethanol being fed to the process. If there is no change in volume upon mixing the fuels and if 1.00 times 10^5 liters/day of E10 is produced (in addition to the other fuels, according to market demand), what are the volumetric flowrates of all feed and product streams? Tank trucks are to transport the fuel from the blending operation to service stations in the area. The gross weight of a loaded truck, which has a tare (empty) weight of 1.27 times 10^4 kg, cannot exceed 3.60 times 10^4 kg. Estimate the maximum volume (L) of each fuel that can be loaded onto a truck. Assume the specific gravity of gasoline is 0.726 and the specific gravity of ethanol is 0.789.Explanation / Answer
E10 (40%) market demand = 1.0e+5 Lt/d
E85 (10%) demand = 2.5e+4 Lt/d
E100 (2%) demand = 5.0e+3 Lt/d
E0 (48%) demand = 1.20 e+5 Lt/d
mixing ratio of E10
out put V0 = a Vg + b Ve = 1.0e+5 Lt/d
Vg = 0.9 V0 = 9.0 e+4Lt/d
Ve = 0.1Vo = 1.0e+4 Lts/d
The mixing ratio ramins constant for all fuels i.e 1/9 ethanol and gasoline
For E100 out put V0 = 5.0e+3 Lt/d
Ve = 5.0e+3 Lt/d and Vg =0
For E85
Vo = 2.5 e+4 Lt/d
Vg*0.9 + Ve *0.1 = 25000 Lt/d - mixiing ratio in the blender
Vg *0.15 +085 Ve = 25000 Lt/d - blending of E85
solving the above
Vg = Ve = 25000 Lt/d
E0 - pure gasoline
V0 = 1.2e+5 Lt/d
Vg = V0 = 1.2e+5 Ltd
Ve = 0
c) SG of Gasoline = 0.726
Ethanol = 0.789
Gross weight limit = 3.5 e+4 kg
tare weight = 1.27 e+4 kg
net weight of liquid allowed = 2.23 e+4 kg
E0 density = 0.726 gm/cc = 0.726 kg/Lt
volume of liquid allowed = 2.23e+4 /0.726 = 30716 Lts
E10 densisty = 0.726*0.9 +0.789*0.1 = 0.7323
volume of liquid allowed = 2.23e+4/0.7323 = 30452 Lts
E85 desnity = 0.726*0.15 +0.789*0.85 = 0.7796 gm/cc
volume of liquid allowed = 2.23e+4/0.7796 = 28604 Lt
E100 density 0.789 kg/Lt
volume of liquid allowed = 2.23e+4/0.789 = 28263 Lts
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