please solve only b and c UNIVERSITY OF NEW HAVEN Fall 2014 Introduction to Mode
ID: 526421 • Letter: P
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please solve only b and c
UNIVERSITY OF NEW HAVEN Fall 2014 Introduction to Modeling of Engineering Systems EASC 2211 Final Exam 3. (11%) The system below is used to make methanol (cHao) from carbon monoxide and hydrogen. Composition data for streams 1,2 and 3 are shown in the table, along with the flowrate of stream 3. S1 S3 co 2 H2 CH40 mixer S4 S4 a. First consider the mixing step only. Determine Stream S1 S2 S3 the flowrates (mol/s) of streams 1 & 2 mol 120 mol comol 0.30 000 0.18 mol H2 mol 0.00 0.80 0.32 mol CH40/mol 0.00 0.00 000 mol N2/mol 0.70 0.20 0.50 b. hich reactant is the limiting reactant? What is the percent excess for the other reactant? c. If 60 of the Co reacts, what will be the flow of each species in stream 4? moles CO/s moles H2/s moles CH40/s moles N2/sExplanation / Answer
The reaction is CO+2H2----.CH3OH
from overall balance across the mixer, S1+S2= S3
hence moles of CO in S1 + moles of CO in S2= mole of CO in S3
S1*0.3= 120*0.18, S1= 72 moles/sec
S1 contains CO at 0.3 and N2 at 0.7, moles of CO in S1= 72*0.3= 21.6 and moles of N2 at S1= 72*0.7 =50.4 moles
S2 contains H2 and N2 writing hydrogen balance S2*0.8=120*0.32, S2= 48
moles of H2 in S2= 48*0.8= 38.4, N2 in S2= 48-38.4= 9.6
so there are 21.6 moles of CO in S1 and 38.4 moles of H2 in S2.
The reaction is CO+2H2------->CH3OH
molar ratio of reactants CO:H2 ( theoretical)= 1:2
actual CO:H2= 21.6: 38.4= 1 :1.8
So hydrogen is the limiting reactant and CO is excess reactant.
38.4 moles of H2 requires 38.4/2= 19.2 moles of CO.
% excess Co supplied= 100*{(21.6-19.2)/21.6}=11.11%
moles of Co would have reacted= 19.6 moles/sec
when 60% of it reacts, moles Co reacted= 19.6*0.6= 11.76 moles/sec
moles of H2 reacted= 11.76*2= 23.52 moles/sec
Product contains 11.76 mole/s of CH3OH, CO= 21.6-11.76= 9.84 moles/sec, H2= 38.4-23.52= 14.88 mole/sec N2( unreacted)= 120*0.5=60 moles/s
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