Question

(b) Propane must be stripped from an oil stream in a stripping column by using pure steam The pure oil stream (without Propane) has a flow-rate of 150kmol/h and is loaded with 2.5 mol% Propane. The column is set at the temperature of 350 K and pressure of 1.2 atm. The minimum propane amount to be separated is 90% (1) In order to determine the flow of the gas stream, some assumptions are used. What are the assumptions 3 Marks (ii) Calculate the minimum amount of the gas stream required to carry out the stripping 1 Marks Determine the number of stages for the gas stream 1.7 times minimum by the two methods namely the Kremser equation and Mc Cabe Thiele method and exit was composition. The Antoine equation below for propane equilibrium data is 1872.46 Inp' (mm Hg) 15.726 - T(K)-25.16 4 Marks y

Answer 1

i) Assumptions used:

pure steam and oil are immiscible

steam is miscible with propane

Raoults law is applicable (gas behave as ideal gas and liquid as ideal solution)

page1.

page2.

Fabar project Steam + pro pane 0 Propane foil X = 26% X, - male patio g parpane in och = mole of poopane @ more of odp - AL OORS = 0.02564. 1-X 1-0.025 T=300k | p = 1.2atal pure steam X2 = mole of poopane mole of oil = X, + (1-0.9) 90% is stripped) = 0.02564* 0.1 = 0.002564 given, - 1872.46 Hg) = 15.726 - 18 T(K)- 25.16 from geven condition, 1872.46 Inps. - 15.726 -- 350-25-16 mp = 9,9617 bus - ps = 2199.7958 mmity. = 27.894 atm. nated as Y= mx At equilibrium, u= mx can be approximat Y = f xx (from Raoult's law). → Y= 27.894X= 23. 245x ie Y= 23245 *} 1.2

page 3.

minimumtion tt pmpane balance - Les molor flowsate of pure oil = 150kmah Gs - molar flowsate of pare steam to X, + GeY = Leta + GsY, . - Le (X, - *2) - Ge (Y, -Y) - Garmin - La (x,- X2) at minm condition (gas) M.- Y X, and Y, are out e equilibrin 150 10.02564 - 01002564) Y = mx Gsmin - - (23.2458 0.02564 -0) = 3.4614 - 5.8077 kmol /ho. ( 23, 245x0.02564) (Gemin= 5.8077 kmst /ho G = 1.7 Gemin = 9.8730 9 kmother 150 - 15. 1928 Gs 9.87309

The mole fraction of propane at exit gas stream is calculated as 0.2595

Y at Ge=1.76 smin Ls (X, -42) = Ge (Y,-Y₂) >> y = 150(0.0054 -0.00254) to 9.87309 = A-Ls/he - m 0:35058 15-1928 23. 245 = 0.65359 By Kremses equation, NTP = ln xo - Yn ym . 1 Ynty + (1-A) TA] - in 1x - 72/m +(1-4) +A) 1 X2 - Yt/m [10.02564 - 03/23.045 0.002564 - 07/23245 - 065359) +0.65359 in (0.65359) → NTP = 3.331 ~ 4.. By Macabe Theile method, equilibrium linee Y=mx equilibrin line 0.005 0.01 0.015 0.02 0.116225 0.23245 0.348675 0.4649 0.581125 0.27 011 0.025 0128005 0:01 0.015 002 0.025 0.03 0.035 x from graph, no of stages a 4.