Question

a stream containing 25% mass methanol in water is diluted with a second stream containing 10% mass methanol to form a product containing 17% mass methanol . draw and label flow diagram of this process and calculate the feed rate of the 10% mass solution required to produce 1250 kg/hr

Answer 1

  

Let S1 = 1st stream rate, in kg/hr
Let S2 = 2nd stream rate, in kg/hr
We can write two equations:
For total flow rate:
S1 + S2 = 1250 kg/hr
For methanol flow rate (using decimals instead of percents):
(0.25)(S1) + (0.10)(S2) = (0.17)(1250 kg/hr)
Rearranging the top equation, S1 = 1250 - S2
Substituting this for S1 in the second equation:
(0.25)(1250 - S2) + (0.10)(S2) = (0.17)(1250 kg/hr)
Solving,
S2 = 666.66 kg/hr;
S1 = 1250 kg/hr - 666.66 = 583.33 kg/hr.
To check,
(0.25)(583.33 kg/hr) + (0.10)(666.66 kg/hr) = 212.5 kg/hr methanol
(0.17)(1250 kg/hr) = 212.5 kg/hr methanol.