Question

In order to meet pollution standards, oil refineries remove trace amounts of sulfur from the crude oil This sulfur is often in the form of HbS and SO. A process has been suggested to convert these compounds into sulfur and water using the following reaction 2HS(g+ SO,(g) 3S(s)+2H,O(g) A schematic of the facility is shown in Figure I on page 2 Due to the difficulty (and cost) of achieving high purity, the input streams (Fi and F2) both contain signifcant concentrations of CO2 The input feeds (Fi and F2) are blended to achieve a stoichiometric ratio of H.S and SO2 and are subsequently combined with a recycle stream and fed into the reactor. The solid sulfur is completely removed in the solids separator while the water is completely removed in a condenser The stream leaving the condenser splits into a purge stream and the recycle stream. The flow rate in Fi is 100 mol'sec and the process achieves a 95% conversion of H2S In addition, the recycle to feed ratio ficant 15 Hl

Can someone help me with these question especially parts d) and e)
thanks

Answer 1

Given that: The formation reaction of sulfur is: 2H,S(g) +SO2(g) +35(s) + 2H2O(g) The process flow sheet is given as, 90.0% H,S 10.0% CO, F FAR Reactor Solids remova Water removal 95.0% SOZ 5.00% CO2 Flow rate of the input stream 1 is given as, F= 100 mol The conversion of H2S is 95%, Xy,s = 0.95. The ratio of recycle, R to feed. F, is given as, (a) The purpose of recycle stream to increase the conversion of reactants H S and SO2 and to recover the traces of unreacted reactants in the product stream by reusing the products. (b) Purge stream is to remove an accumulation of inerts or unwanted material that might otherwise build up in the recycle stream. From the given balanced reaction, 2 moles of H, S needed to react with 1 mole of So, to give the products. And from the given information, the flow rate of stream 1 is F=1000 In stream 1, mole % of H2S is 90.0%. So molar flow rate of H,S in stream 1 is,

Fys = F x 90% =100 mol 90 '5 100 F = 90 mol And, for CO, Fco. = F x10% =100 mol/ 10 Fco. = 10 mol S 100 From stoichiometry, 2 moles of H2S needed to react with 1 mole of SO, Then, molar flow rate of so, will be, 2 mole of H_$+1 mole of so, 90 mol +? F.o. = 90 molx= S2 Fso, = 45 mol But in stream 2, the % of So, is 95%. That means, in 100 moles of Stream 2, the fraction of So, will be 95 moles. Now calculate molar flow rate stream 2 which gives 45 moles of So, for the reaction. For 95%→ 45 mol For 100% F = 45 mol 100 595 S F= 47.37 m And, for CO:

Fco, = F x10% =47.37 mol x 50 S 100 Fco. = 2.37 mol From the material balance: Fo = x +F; =100 mol +47.4 mol F. = 147.37 mo! CO, (From F.) =CO, (From F, )+CO, (From F,) =10 mol +2.37 mol CO, (From F.)=12.37 mol (d) $ The ratio of recycle, R to feed, F, is given as, =15. Substitute the value of F, and get, R =15 R=15x F. =15x147.37 mol R = 2210.55 mol From material balance: FBR = R+F = 2210.55 mol +147.37 mo! Egg =2357.92 md

In reactor the conversion of H2S is given as 95%. Then, the molar flow rate of H2S is: Fys (reacted) = 90 mol 95 5 100 = 85 mol From the reaction stoichiometry, 2 moles of H S needed to react with 1 mole of So, to give 3 moles of sulfur. Now calculate the moles of Sulfur formed from 85.5 moles of H,S. 2 mole of H2S → 3 mole of S 85.5 mol ? S=85,5 mol 3 S=128.25 mol From the reaction stoichiometry, 2 moles of H S needed to react with 1 mole of So, to give 2 moles of H,O. As the ratio is 2:2. The moles of H20 will be same as H,S. Now calculate the moles of H2O formed from 85.5 moles of H,S. 2 mole of H,S+2 mole of S 85.5 mol → W=85.5 mol > W=85,5 mol S 2 From the material balance, stream FAR contains the products sulfur, H,O and unreacted H,S and CO, Then.

Far = unreacted H2S+S+W+CO, (Fed) =[(100–95)%xFx;s]+S+W+CO, (From F,) = 20 mol 160] +12825 mol + 85 s ml +12.37 mal =(4.5+128.25+85.5+2.37) mo Far = 230.62 mo! The purge stream is about removal of unwanted gases. From the process, CO, is the unwanted gas. So, it will be removed as purge stream. And the unreacted H,S will be sent as recycle. Then, CO, (From F.)=CO(In Purge) =12.37 mol The molar flow rate stream which is sent to condenser will be, Condenser input gas = FAR-S = 230.62 mol –128.25 mol =102.37mo! S Condenser output gas = FAR-S-W = 230.62 mol –128.25 mol –85.5 mol S =16.87 mol In that moles of CO2 will be taken as purge. Then, purge percent of gas leaving the condenser is:

Purge gas % of gas purged== -X100 Condenser output gas 12.37 mol s, x100 16.87 mol % of gas purged = 73.3% Single-pass conversion gives the fraction of reactant converted on a single pass through the reactor. The single pass conversion of the rector is 95% of H,S. In single pass conversion, one cannot use the reactant efficiently. One can not reuse the reactant if it is not converted completely. No recycle will be possible in the single pass. (h) Recycling (back feeding of the unreacted reactants) is the best strategy to increase the overall conversion of the process. Continues catalyst regeneration is another strategy to increase the overall conversion.