Question

A corn ethanol production plant receives 500,000.0 kg/day corn feedstock at a moisture content of 15.5% (wet basis). If all of this corn is converted into ethanol what is the theoretical volume of ethanol that this facility can produce per day? Assume that the starch content of corn grain is 68.5%. The density of ethanol is 789 kg/m3.

Answer 1

207 m³/day

Step-by-step explanation:

Dry corn feed stock = 500000 × ( 100 - 15.5%) = 500000 × 84.5% = 500000× 0.845 = 422500

Starch yield = 68.5% × 422500 = 289412.5

Glucose yield = 1.11 × 289412.5 = 321247.875 where 1.11 is the scarification factor of starch to glucose

Ethanol yield = 0.51 × 321247.875 = 163836.416 where 0.51 is theoretical yield of ethanol from one mole of glucose

density = mass / volume

volume = mass / density = 163836.416 / 789 = 207 m³ / day