Answer to Question #224033 in Chemical Engineering for sbuda

Question

One thousand kilograms per hour of a mixture of benzene (B) and Toluene (T) containing 60% benzene by mass is separated by distillation into two fractions. The mass flowrate of the benzene (B) in the top stream is 500 kg (B) /h and that of toluene in the bottom stream is 485 kg (T) / h. The unit operates at steady state. 3.1.1. Draw a schematic diagram showing all the relevant streams in the distillation unit. (8) 3.1.2. Write the basis for calculation. 

Answer 1

We are given:

• Mass flowrate of inlet = 1000 kg/hr
• Mass composition of inlet = 60% benzene (B)-40% toluene (T)
• Mass flowrate of benzene (B) at the top of column = 500 kg/hr
• Mass flowrate of toluene (T) at the bottom of column = 485 kg/hr

From the given information, we need to establish the material balance for benzene and toluene and determine the flowrate of toluene at the top of the column and the flowrate of benzene at the bottom of the column.

The first step is to draw a diagram to visualize of the process. Since the inlet stream is 60% benzene by mass and 40% toluene by mass, we can express the inlet in terms of the inlet flowrates of benzene and toluene. Also, we can let x as the flowrate of toluene at the top of the column and y as the flowrate of benzene at the bottom. The diagram is shown above

Schematic Diagram of Separation of Benzene-Toluene Mixture

Next, let's establish the material balance for both benzene and toluene and determine the unknowns assuming steady-state.

Material Balance for Benzene

Mass of Benzene in = Mass of Benzene out

600 kg/hr = 500 kg/hr + y

y = 100 kg/hr benzene

Material Balance for Toluene

Mass of Toluene in = Mass of Toluene out

400 kg/hr = 375 kg/hr + x

x=25 kg/hr toluene