Question

Book problem 17.2. Use the Chapman-Enskog equation (Eq. (17.28)) to calculate the diffusivity of r-propanol in air. 17.2. An open circular tank 8 m in diameter contains n-propanol at 25°C exposed to the atmosphere in such a manner that the liquid is covered with a stagnant air film esti- mated to be 5 mm thick. The concentration of propanol beyond the stagnant film is negligible. The vapor pressure of propanol at 25°C is 20 mm Hg. If propanol is worth S1.20 per liter, what is the value of the loss of propanol from this tank in dollars per day? The specific gravity of propanol is 0.80.

Answer 1

The chapman Ensky equation Den 1858 x 10 7222 (+ AD Pondo. Compound A = propanol (Cz H30). B= air. TE 273 +252 298k MA = 60 g/mol MB= 29 9 Imol P latm TAB = (GA+OB) characteristic a length in Å 3.711 RALF approximate relation JA=516 (VC) Ve= Gitical vol OB= (ams) gond 10 2 (218) V = 0.218 41 mot 5.015 Å - 0-218 x1000 cmyaros - 218 5.015 + 3.71) = 4.363) A GAB=

collision integral which is function -23 k= bolt 3 mann's const = 1:38+10 E.= characteristic Binary Parameter EAB = LEA EB)? E = 0.755 Te = critical B = 78.6 (for air) Temperatur ER = 78.6 x K 6. Te of propanol. = 536-9 k GA = 0.75 XTC = 0.75 +536-9 .. EAB= (EAXES x 536-9 x K x 78-6xk EAB = 2.45x1021 KT = 1.38x1043298 -.676 2.4521621

Dag = 1.858* 167 (290) ( ) 1 x 4.3631% 1.675 DAB. = 2.95 xroms 5 mm SEESI. diffusion thick air film h-propanos diffusion of propanol through air Güven concentration of propanot beyond air film is negligible – – PAITO 5mm PAD= 20 mmitg. . layer v Propanol = 25 Just above the popanol layer the Partial pressure will be equal to its vapour pressure.

rady state flux of propanol is given by Plan 101323 IP No= DABP in IP-PAI) . RT e VA Р- fao € 2.95 x10 x 101.325*10° in 8-34 x 298x58103 760 10-20) INA: 6.434 x103 kmol mis = 6-934x10 3 x 60 kg fari's = 00 386 kg/mg d=8m ain for im 0.386 kg je g propanol is diffusing into Area = I x*= I x64 => = 50-265 m for 50-265 m- 50.265 80 386 = 19,402 kg/s is diffusing

criven specific gravity: 0.80 a density = 800 kg/m :. Vot of propanol de lost = 19.402 kg/s 800 kg/m3 = 0.02425 m/s im- 10000 - 0.02425 m3= 24-252 24-25C is lost per second - per day 24.25 X60x60x24 2095416L is lost Given cost = $1.20 per litre => loss inn per day= 251 4.499.25 loss of propano) - 2,514,499.2 $ perday