Methanol, called \"wood alcohol\" because it was historically produced by the de
ID: 538259 • Letter: M
Question
Methanol, called "wood alcohol" because it was historically produced by the destructive 30% distillation of wood, is now produced commercially from the process shown schematically below. The reaction is conducted at high pressure and temperature over a catalyst. CO + 2H_2 rightarrow CH_3OH The feed is a stoichiometric mixture of CO and H_2. 45% of the CO reacts in a single pass thru the reactor. 95% of the methanol is recovered as a liquid in the condenser and the balance is recycled with all of the unreacted CO and H_2. Calculate the molar flow rate and composition (as a mole fraction of each species) for streams 1, 2, 3, and 4. Stream 5 has zero flowrate in Prob. 3Explanation / Answer
Feed to the process= 99 moles/hr, It contains 33 moles of CO and 66 moles of H2.
The reaction is CO+2H2-------->CH3OH
1 mole of CO gives reacts with 2 moles of CH3OH to give 1 mole of CH3OH.
At steady state, mole of CO entering = moles of CH3OH formed= 33 Kmoles/hr
This correspond to 95% of the methanol formed. Hence methanol formed= 33/0.95= 34.73 Kmoles/hr
Hence methanol in the recycle= methanol formed - methanol with drawn= 34.73-33= 1.73 Kmoles/hr
let R= recycle, x= mole fraction of CO in recycle ,
Hence CO entering the reactor ( based on balance across the reactor)= Rx+33
mole of CO converted= 45%, moles of CO remaining =100-45=55%
Moles of CO leaving the reacotr = (Rx+33)*0.55
This is leaving the reactor as unconverted and is recycled
hence (Rx+33)*0.55= Rx
hence Rx+33= Rx/0.55= 1.82Rx
0.82Rx= 33, Rx= 33/0.82= 40 = moles of CO in the recycle/hr
let y= mole fraction of hydrogen in the recycle
Moles of hydrogen entering = (Ry+66)
moles of hydrogen consumed= (Ry+66)*0.45
Hydrogen remaining = (Ry+66)*0.55= moles of hydrogen recycled= Ry
Ry+66= Ry/0.55
1.82Ry-Ry= 66
Ry= 66/0.82= 80.48 = moles of hydrogen recycled/hr
Recycle contains (stream-4)( Kmoles/hr) : CO= 40, H2= 80.48, CH3OH= 1.73
flow rate ( total )= 40+80.48+1.73= 122.21 kmole/hr
Composition : CO= molar flow rate of CO/ total moles= 40/122.41= 0.326, H2= 80.48/122.21 =65.85, CH3OH= 1.73/122.21 =0.014
Feed contains (Stream) kmoles/hr) : CO= 33 and H2 =66
Composition : CO= 33/(33+66)= 0.333, H2= 1-0.333=0.667
Feed to the reactor contains (stream-1) :( Kmole/hr) : CO= 33+40= 73 moles/hr, H2= 80.48+66=146.48, CH3OH= 1.73
Total flow rate ( kmoles/hr)= 73+146.48+1.73= 221.21
Composition : CO=73/221.21= 0.33, H2= 146.48/221.21= 0.66, CH3OH= 1.73/221.21= 0.01
Reactor outlet contains (Stream-2) ( kmoles/hr) : CO= 40, H2= 80.48, CH3OH= 34.73 , total flow rate = 155.21
Composition (Stream-2) : CO= 40/155.21= 0.2577, H2= 80.48/155.21=0.5185, CH3OH= 34.73/155.21= 0.223
Stream-3 is 33 moles/hr of Pure methanol.