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Assignment on Mass Transfer Coefficient Determination PPT

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ASSIGNMENT ON MASS TRANSFER COEFFICIENT DETERMINATION Problem 1: Wetted Wall Column β–  Air-ammonia mixture containing 4% NH3 by volume is being absorbed counter currently with H2SO4 in a wetted wall column of 20 mm diameter and a length of 1 m. The gas enters the column at a rate of 0.25 kmol/hr. The column has a temperature of 25Β°C and a pressure of 101.325 kN/m2. 85 % of the of the entering NH3 is absorbed. Estimate kG. Ammonia-air H2SO4 Solution y1 = 0.04 (mole fraction of NH3) 0.04 π‘šπ‘œπ‘™ 𝑁𝐻3 π‘Œ1 = = 0.04167 1 βˆ’ 0.04 π‘šπ‘œπ‘™ π‘Žπ‘–π‘Ÿ π‘šπ‘œπ‘™ 𝑁𝐻3 π‘Œ2 = 0.04167 0.15 = 0.00625 π‘šπ‘œπ‘™ π‘Žπ‘–π‘Ÿ 𝑦2 = Rate of air entering = 0.000625 = 0.00621 1.000625 0.25 π‘˜π‘šπ‘œπ‘™ 1 β„Žπ‘Ÿ π‘˜π‘šπ‘œπ‘™ Γ— Γ— 1 βˆ’ 0.04 = 6.667 Γ— 10βˆ’5 β„Žπ‘Ÿ 3600 𝑠 𝑠 Solution π‘…π‘Žπ‘‘π‘’ π‘œπ‘“ 𝑁𝐻3 π‘Žπ‘π‘ π‘œπ‘Ÿπ‘π‘’π‘‘, 𝐸 = 6.667 Γ— = 2.36 Γ— 10βˆ’6 10βˆ’5 π‘˜π‘šπ‘œπ‘™ π‘Žπ‘–π‘Ÿ 𝑠 π‘šπ‘œπ‘™ 𝑁𝐻3 0.04167 βˆ’ 0.00625 π‘šπ‘œπ‘™ π‘Žπ‘–π‘Ÿ π‘˜π‘šπ‘œπ‘™ 𝑁𝐻3 𝑠 π‘˜π‘ π‘š2 π‘˜π‘ = 101.325 0.00621 = 0.629 2 π‘š 𝑃𝐴1 = 101.325 0.04 = 4.053 𝑃𝐴2 π‘ƒπ‘™π‘š = 4.053 βˆ’ 0.629 π‘˜π‘ = 1.838 2 4.053 π‘š 𝑙𝑛 0.629 Solution 𝐸 π‘˜πΊ = 𝐴 βˆ™ π‘ƒπ‘™π‘š 20 𝐴 = πœ‹π·πΏ = πœ‹ π‘š 1000 1m = 0.0628 π‘š2 Substituting, 2.36 Γ— 10βˆ’6 π‘˜π‘šπ‘œπ‘™ βˆ’5 π‘˜πΊ = = 2.04459 Γ— 10 0.0628 βˆ™ 1.838 𝑠 βˆ™ π‘š2 βˆ™ π‘˜π‘ƒ ...
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