Question

7.1. (PS) In the gas purification system for the removal of sulfur compounds shown below, an absorption system designed to operate at feed rates up to 820 gmol/hr is configured to handle even higher feed rates by bypassing the absorption system so that the flow will not exceed the designed feed rate. The requirement for the purified gas (stream 5) is that it contains no more than 1.00 mol% H2S and 0.300 mol% COS. This requirement has been met with a feed stream (stream 1) flowing at a rate of 900 gmol/hr and consisting of 15.0% CO2, 5.00% H2S, and 1.41% COS, with the remainder being CH4. Calculate all the stream flows and the mole fraction of H2S in stream 3. 96 CH N5 N' сн. 15.0 %co soz, L4XcOs Absorption system N" CH co H, $ cos CH Coz HS cos cos 0.3% Hys COS

Answer 1

Ans:- Criven That to a- operate system designed An Absorp mo) $20 to Ihr g feed rates Up is The requirement for the pursed gas (stream 5) more han 1.00 molol Hes and That it contains 0.300 MOL 7.(os. diagrams Feed 1000 mer les <3 820 Ad Sopron 180 toolhr 12. Hes bypass 0:37.cos non in As me thane is sue plusons compound it acts Inext and feed Amount of City Intersing in feed = Amount of Cify leaving products Amount of Chi in food 78.54 X100D no = 785.9 :. (Hip moles Ihr

also inest so, (02 in feed coz in product feed D:15 x1000 - Coin iso mol nollho .. product coz fraction and Chip in product one forcu fraction of coz - 100 y. of His % of His -%of los = 100 -1.5 = 98.7 px 0.987 = 150 + 785.9 p=93519 0 1487 ... product ut flow rate = 448.23 milho mother - 820 mo Can I his Enter Into H is given that only Adl sosprior Column. F ~ 1000 FoR+B 100D = 820 +13 Jiow rate = 180 maith rate of By pass stream = 180

Thank you The frow coming out of Adsorption Corucun. = P-B =948.23 -180 = 708-23 solhr moi 76823 mollhr % of co2 in P 150 X100 948.20 15.8y. % of coz in p = 15.8% % of Cty in p = 785.9 X100 948.23 88.97. % of CH4 in P 82192 please give the Thereup's up for This Answer ig you are any dout's In pur Comments