Question

A feed stream \(F_1\) containing components A and B is processed in a separation–mixer system as shown. The mole fractions of A and B are \(x_A\) and \(x_B\). If \(F_1 + F_2 = 100\ \text{kg h}^{-1}\), the degrees of freedom of the system is \(\underline{\hspace{1cm}}\). 
 

Hint:

Degrees of freedom = (number of independent variables) – (number of independent equations). Flowsheet problems require counting per stream: 1 flow + (components–1) compositions.

Answer 1

Correct Answer: 6

The flowsheet contains the following unknowns:
Flowrates: \(F_1, F_2, F_3, F_4, F_5, F_6\) (6 unknowns).
Mole fractions: each stream has two components (A and B). For each stream, only one independent mole fraction exists.
Thus, composition unknowns:
\[ 6\ \text{streams} \times 1 = 6 \]
Total unknowns:
\[ 6\ (\text{flows}) + 6\ (\text{compositions}) = 12 \]
Available independent equations:
Mass balance for each unit (separation unit & mixer): 4 equations.
Component balances across units: 2 components → 4 more equations.
Given constraint: \(F_1 + F_2 = 100\ \text{kg h}^{-1}\) gives 1 additional equation.
Total equations:
\[ 4 + 4 + 1 = 9 \]
Therefore, degrees of freedom:
\[ \text{DOF} = 12 - 9 = 6 \]