Methanol is produced commercially by the following reaction conducted at high pr

Question

Methanol is produced commercially by the following reaction conducted at high pressure and temperature over an appropriate catalyst. The feed contains a stoichiometric mixture of CO and H_2. 35% of the CO and H_2 react in a single pass thru the reactor. 95 % of the methanol is recovered as a liquid in the condenser and the balance is recycled as a gas with all the unreacted CO and H_2. Calculate the molar flow rate and composition (as the mole fraction of each species) for streams 1, 2, 3, and 4. Stream 5 has zero flowrate in Prob. 2.

Explanation / Answer

Step 1 Basis: 99 feed. 33 mol CO and 66 mol H2

Step 2

Feed (F) = Stream 3 (M)

because stream 5= 0

Step 3

Let recycle be R and mole fraction of lethanol in recycle be x

in stream 3

methanol= (33+Rx)* 0.35

unreacted part = (66+R* (1-x))

mole fraction of methanol= x = (33+Rx)* 0.35 / ((33+Rx)* 0.35 + (66+R* (1-x)*.65) )

Step 4

from initial equation we have

M= .95 * (33+Rx)* 0.35 because 95 percent methanol was recovered.

which means 99 = 0.95 * (33+Rx)* 0.35

which means that Rx = 264.7 substitute this value in previous equation we get

solving 2 equations we get 57.6R = 2 * 264*24.95

we get R= 270.6 so x= 264.7/270.6 = .976

methanol mole fraction in recycle = 0.976

methaol mole fraction in output = 1

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