Question

Problem One [35 marks] A liquid mixture of benzene and toluene contains 55% benzene by mass, the rest is toluene is to be distilled continuously to produce two streams: the Overhead stream (v) in kg/hr and bottom stream (L.) in kg/hr. The overhead stream was analysed and was found to have 85% by mass benzene and the rest is toluene. The bottom stream was analysed and found to have 89.4% toluene and the rest is benzene. 1) Draw a flowchart for this process and identify/label all streams. 2) Write the general mass balance equation for this process (then state your assumptions) 3) Calculate the overhead and bottom stream flow rates in kg/hr and find the individual benzene mass flow rates in both. (state any assumption in here if you have any)

Answer 1

d) efiven data, Feed a Benzene t toulene Mass traction of benzene CXF) since gt 1A binary Minture Х + X2 = 1 - OSS Mass beection q tollene (I-XF) = 0.45 Mass faction of benzene in overhead stream (²0) = 0185 xi mass flaction q benzene in bottom & team =b-0.894 (XW)=0.106 Mass praction q toulene in overheed streen CI-XD) 1-0.85 Mass traction of boulene in bottom stream 0.894 +) candenses - Reflux drum Feed overheed & trean/ distillate DIX Bentenet toulene xD Reboiles F = Feed How rave D = overheed How rate bottam How CS Scanned with CamScanner rate Wixw Bottom Product

rate a mass in 20 Overall material balance on the proces Rate of Rate q Rate q Mass disappeared genaleted - Rate of mass accumalated Here there is no reaction takes place generation - disapperence =0 We are operated at & tead. State, accumalation 0 = Rare q Mass out Rate of mass in D tw Now component balance - benzene Mars balance an F XF =D&D twaw [ we inow Mass braction X Total Mass - Mass of the component ] ip we are operating Here ou Assumption at steady state I 3 Material balance an toulene FCI-XF) = DC1-XD) + W CI-Xw) 3) NOW we have Equations the given substi tule value in CS Scanned with CamScanner Equations

59.67 kg / h W Feed (P)=100 kg / hy F =D + w 0:55F = 0.85D + oilo 6 0:15 D Now Take, Basis 0.45F 4. . loo =DtW 0:55E100)=0&SD 10:106W solve ① and ② loo = D + wia) - W = (100-D) 55 =0.85b to 106 w 5 5 = 0.85D +0:106 (100-0) 55-0.85 DF10:6-0106D 55-0724 D + 1016 4.4.4 of uuD vi subo titue D ini ។ Too = 59.67 t w .: w = 40.32 kg / hy CS Scanned with CamScanner