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.
We are given:
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