Hydrogen is used to reduce 1000 kg/h Fe_2 O_3 to metallic iron according to the

Question

Hydrogen is used to reduce 1000 kg/h Fe_2 O_3 to metallic iron according to the reaction Fe_2 O_3 + 3H_2 rightarrow 2Fe + 3H_2 O The water is condensed and the unreacted hydrogen is recycled. Because the hydrogen in the fresh feed contains 1.0 mol% CO_2 as an impurity, some of the unreacted hydrogen must be purged. Calculate the flow rate and the composition of the purge stream required to limit the CO_2 in the reactor feed to 4.5 mol% if the ratio of recycle to fresh feed is 5:1 on a molar basis.

Explanation / Answer

Let F= Fresh Feed containing CO2 and H2.

Let x= % of CO2 in the purge and 100-x= % of hydrogen in the purge.

Let R= recycle.

At steady state, CO2 entering F*1/100 = P*x/100 , F = Px

R/F= 5, R= 5F

Writing CO2 balance across the reactor,

F*1/100 +R*x/100 = (F+R)*4.5/100 or F+Rx= (F+R)*4.5

F+5Fx = 6F*4.5 ,   F*(1+5x)= 27F, 1+5x= 27, 5x=26 and x= 5.2% and H2= 100-5.2= 94.8%

F= P*5.2, P= F/5.2

Moles of Fe2O3=   mass flow rate/molar mass =1000/160 kg moles/hr =6.25 kg moles/hr

Moles of Fe to be formed = 6.25 kg moles/hr

moles of hydrogen required = 3*6.25= 18.75 kg moles / hr

hydrogen entering the reactor = (F+R)*99.5/100 = (F+5F)+0.995F =5.97F

Hydrogen consumed = 18.75 kg moles/hr

Excess hydrogen = 5.97F -18.75 kg moles/hr

This is leaving as purge and recycle

5.97F- 18.75= (P+R)*0.948= (F/5.2+5F)*0.948

5.97F- 18.75= 5.192F*0.948 =4.922F

5.97F- 4.922F = 18.75

F= 17.89 kg moles/hr

P= F/5.2= 17.89/5.2 = 3.44 kg moles/hr

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