Question

3 m3/hr of a mixture containing 75% (by mass) ethanol and 25% (by mass) water and a quantity of a 40% ethanol–60% water (by mass) mixture are blended to produce a mixture containing 60% by mass ethanol. If the density of the 75% ethanol mixture is 877 kg/m3 and the density of the 40% ethanol mixture is 952 kg/m3, what volumetric flowrate of the 40% mixture is needed and what is the mass flowrate of the product solution?

Answer 1

find specific gravity of the 60% mixture

Let X = specific gravity

By interpolation for a straight line with constant m as gradient,

then (75 - 40) / (0.877 - 0.952)

= (60 - 40) / (X - 0.952).

Then 35 / (-0.075) = -466.67

-466.67 = 20 / (X - 0.952). X = -0.0429 x 0.952 = 0.9091.

909.1 kg/m3
Let V = the required volume of the 40 % mixture.

and mixture is 3 m3/hr = 300 gallon /h

Mass of ethanol in 75 % mixture Mass of ethanol in 40 % mixture = Mass of ethanol in 60 % mixture.
Then, 300 (0.877) (0.75) V (0.952) (0.4) = (V 300) (0.9091) (0.6).
Then 197.325 0.3808 V = 0.5455 V 163.638. V = 204.53 gallons.

volumetric flow rate = 2.0453 m3/h