Exercise: Material balance with no formation or consumption

Draw a diagram

List the knowns

Essentially, we know the following:

1. There is no formation or consumption of species \(A\) or \(B\)

2. \(\dot m_{\text{feed}} = \SI{100}{kg/hr}\)

3. \(\dot m_{B_{\text{waste}}} = \SI{65}{kg/hr}\)

4. \(x_{A_{\text{feed}}} = 0.2\)

5. \(\dot m_{A_{\text{product}}} = 0.98 \, \dot m_{A_{\text{feed}}}\)

List the unknowns

We would like to determine

• \(\dot m_{A_{\text{waste}}}\)

• \(\dot m_{B_{\text{product}}}\)

Write down the general equations

We also know that the sum of the mass fractions must be one:

\[x_{A_{\text{feed}}} + x_{B_{\text{feed}}} = 1\]

There are similar relationships for the product and waste streams.

Our general species material balances for species \(A\) or \(B\) are

\[ \begin{align}\begin{aligned}\dot m_{A_{\text{feed}}} + R_{A_{\text{formation}}} = \dot m_{A_{\text{product}}} + \dot m_{A_{\text{waste}}} + R_{A_{\text{consumption}}}\\\dot m_{B_{\text{feed}}} + R_{B_{\text{formation}}} = \dot m_{B_{\text{product}}} + \dot m_{B_{\text{waste}}} + R_{B_{\text{consumption}}}\end{aligned}\end{align} \]

Simplify the equations based on the specifics of the problem

With no formation or consumption, our material balances become

\[ \begin{align}\begin{aligned}\dot m_{A_{\text{feed}}} = \dot m_{A_{\text{product}}} + \dot m_{A_{\text{waste}}}\\\dot m_{B_{\text{feed}}} = \dot m_{B_{\text{product}}} + \dot m_{B_{\text{waste}}}\end{aligned}\end{align} \]

Express our equations in terms of what we know

Substituting definitions for relevant process variables into the above equations gives

\[ \begin{align}\begin{aligned}(\dot m \, x_{A})_{\text{feed}} = \dot m_{A_{\text{product}}} + \dot m_{A_{\text{waste}}}\\(\dot m \, x_{B})_{\text{feed}} = \dot m_{B_{product}} + \dot m_{B_{waste}}\end{aligned}\end{align} \]

Substituting in knowns 4 and our equation for mass fractions, we get

\[ \begin{align}\begin{aligned}(\dot m \, x_{A})_{\text{feed}} = 0.98 \, \dot m_{A_{\text{feed}}} + \dot m_{A_{\text{waste}}}\\\dot m_{\text{feed}} \, (1 - x_{A_{\text{feed}}}) = \dot m_{B_{\text{product}}} + \dot m_{B_{\text{waste}}}\end{aligned}\end{align} \]

Algebraically solve for the unknowns

Rearranging, the above equations gives us

\[ \begin{align}\begin{aligned}\dot m_{A_{\text{waste}}} = 0.02 \, \dot m_{\text{feed}} \, x_{A_{\text{feed}}}\\\dot m_{B_{\text{product}}} = \dot m_{\text{feed}} \, (1 - x_{A_{\text{feed}}}) - \dot m_{B_{\text{waste}}}\end{aligned}\end{align} \]

Find the numerical answers

Finally, substituting knowns 2, 3, and 4 gives our desired results

\[ \begin{align}\begin{aligned}\dot m_{A_{\text{waste}}} = (0.02)(\SI{100}{kg/hr})(0.2) = \SI{0.4}{kg/hr}\\\dot m_{B_{\text{product}}} = (\SI{100}{kg/hr})(1 - 0.2) - \SI{65}{kg/hr} = \SI{15}{kg/hr}\end{aligned}\end{align} \]