FR4.3

ChE 201 Chemical Process Principles and Calculations Fall, 2009

Washington State University Voiland School of Chemical Engineering and Bioengineering Richard L. Zollars

Problem 4.3, Felder and Rousseau A liquid mixture of benzene and toluene contains 55.0% benzene by mass. The mixture is to be partially evaporated to yield a vapor containing 85.0% benzene and a residual liquid containing 10.6% benzene by mass. (a) Suppose the process is to be carried out continuously and at steady-state, with a feed rate of 100.0 kg/h of the 55% mixture. Let m v (kg/h) and m l (kg/h) be the mass flow rates of the vapor and liquid product streams, respectively. Draw and label a process flowchart, then write and solve balances on total mass and on benzene to determine the expected values of m v and m l . For each balance, state which terms of the general balance equation (accumulation = input + generation – output – consumption) you discarded and why you discarded them. (See Example 4.2-2.) (b) Next suppose the process is to be carried out in a closed container that initially contains 100.0 kg of the liquid mixture. Let mv (kg) and ml (kg) be the masses of the final vapor and liquid phases, respectively. Draw and label a process flowchart, then write and solve integral balances on total mass and on benzene to determine mv and ml . For each balance, state which terms of the general balance equation (accumulation = input + generation – output – consumption) you discarded and why you discarded them. (c) Returning to the continuous process, suppose the evaporator is built and started up and the product stream flow rates and compositions are measured. The measured percentage of benzene in the vapor stream is 85% and the product stream flow rates have the values calculated in part (a), but the liquid product stream is found to contain 7% benzene instead of 10.6%. One possible explanation is that a mistake was made in the measurement. Give at least five others. [Think about the assumptions you made in obtaining the solution of part (a).] SOLUTION (a) The process flow diagram would appear as shown below: m v xv,B = 0.850 xv,T = 0.150

m f = 100 kg/h xf,B = 0.550 xf,T = 0.450

m l xl,B = 0.106 xl,T = 0.895

where the mass fractions of toluene in the liquid and vapor product stream have been determined by subtracting the mass fraction of benzene from 1.00. In the general balance equation we can eliminate the accumulation term (since the operation is at steady-state), the generation and consumption terms (since there is no reaction). Thus for the total mass we get 0  in  out 0  m f  m v  m l m v  m l  m f  100

kg h

For benzene we get

0  in  out 0  m f x f , B  m v x v , B  m l xl , B m v xv , B  m l xl , B  m f x f , B  0.55100

0.850 m v  0.106 m l

 55.0

kg h

kg h

Solve the overall balance for m l and substitute this into the second equation to get 0.850 m v  0.106 100.0  m v

  55.0 kg h

There is only one unknown in this equation so solve it to get m v  59.7

kg . Now h

kg . h (b) The process flow diagram for this situation would look like the following: substitute this value into the overall balance to get m l  40.3

mv xv,B = 0.850 xv,T = 0.150

m f = 100 kg xf,B = 0.550 xf,T = 0.450

ml xl,B = 0.106 xl,T = 0.895

Note that the dot over mf, mv, and ml is no longer there (i.e., these quantities are amounts not flow rates) and the units on mf is now kg, not kg/h. For an integral balance on a batch process the general balance equation has the form (Eqn. 4.2-3) initial input  generation  final output  consumption

We again can ignore the generation and accumulation terms since there is no reaction. For the overall balance you get initial input  100 kg  mv  ml

The benzene balance gives initial input  100 kg 0.550   mv xv , B  ml xl , B 55.0 kg  mv 0.850   ml 0.106 

As in part (a) solve the overall balance for ml and substitute this into the benzene balance to get 55.0 kg  0.850 mv 0.850  0.106 100  mv  Solve this for mv to get mv = 59.7 kg. Substitute this into the overall balance to get ml = 40.3 kg. (c) If all of the product flowrates are correct and the product compositions are as stated then the material balances, as written, are not satisfied if the liquid contains only 7% benzene. Possible reasons include: 1) There is a chemical reaction occurring in the system 2) The system is not at steady-state 3) There are other components in the feed. 4) The feed flow rate is not correct. 5) The feed composition is not correct