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 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.

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