Setting Up the Capstone Problem

One flowsheet story: reaction, then separation, then design decisions.

capstonereactor designseparations
⏱️ About 18 min

A single good flowsheet can turn scattered calculations into an engineering decision.

💡
The big idea: Before sizing equipment, you need a clear narrative: what streams exist, what changes in each unit, and what numbers must balance.
🎯 By the end, you'll be able to
  • Identify the main streams in an integrated reactor + separator flowsheet
  • State three concrete design questions the flowsheet should answer
  • Perform a quick overall molar balance using stoichiometry and conversion
  • Interpret what “closure” means for a steady-state check
📎 Helpful to know first
  • Real vs Ideal Reactors — A First Look at RTD

The capstone flowsheet (qualitative)

We will follow one simple process story throughout this module:

A liquid-phase reactant A is fed to a reactor where it converts to desired product B (reaction: A → B). The reactor effluent then enters a separator. The separator’s job is to remove an unwanted light component (think: a dissolved byproduct gas or a volatile impurity) so that the liquid product stream meets a purity or emissions target.

Across five lessons, you will repeatedly ask: what must be true at steady state, how big must the reactor be to hit conversion, does the temperature rise stay safe, and what separation duty or recovery is required downstream?

🔑 Design questions this capstone will answer

1) Reactor sizing: what volume (or residence time) is required for a target conversion?

2) Safety check: if the reactor is approximately adiabatic, does the temperature rise stay within an acceptable operating range?

3) Separation: given a simple separator performance model, what fraction is removed or recovered, and what do the outlet stream flow rates look like?

Feed (liquid) A + inert Reactor A → B Effluent Separator remove light Product (liquid B) Vent / byproduct (light component) Goal: purity / emissions
Capstone block-flow diagram used throughout this module (qualitative).

What “steady state” means here

At steady state, what comes in and what goes out must balance after accounting for reaction. In the reactor, moles of A decrease and moles of B increase according to stoichiometry. In the separator, total moles may split into two outlet streams, but conservation still applies: inlet equals the sum of outlets for each conserved quantity (and for each component if you track it).

In this lesson we start with a simple overall molar check using conversion and stoichiometry.

\[ X_A = \frac{F_{A0} - F_A}{F_{A0}} \quad \Rightarrow \quad F_A = F_{A0}(1 - X_A) \]
Conversion definition for reactant A in terms of molar flow rates.
\[ \text{For } A \rightarrow B \text{ (1:1):}\quad F_B = F_{B0} + F_{A0}X_A \]
Stoichiometric link between A consumption and B formation for a 1:1 reaction.
📝 Worked example: A liquid feed contains A and an inert I. The inlet molar flow rates are F<sub>A0</sub> = 100 mol/min and F<sub>I0</sub> = 50 mol/min. The reactor achieves X<sub>A</sub> = 0.40 for the reaction A → B (1:1). Assume no B in the feed. Find outlet flow rates of A, B, and I, and the total outlet molar flow.
  1. Use conversion to find A out: F_A = F_A0(1 - X_A) = 100*(1 - 0.40) = 60 mol/min.
  2. For 1:1 stoichiometry, B formed equals A consumed: A consumed = F_A0*X_A = 100*0.40 = 40 mol/min.
  3. So B out: F_B = 0 + 40 = 40 mol/min.
  4. Inert does not react: F_I = F_I0 = 50 mol/min.
  5. Total outlet molar flow = F_A + F_B + F_I = 60 + 40 + 50 = 150 mol/min.
  6. Check: total in = 100 + 50 = 150 mol/min. For A → B (1:1), total moles stay constant, so closure is expected.
✓ F_A = 60 mol/min, F_B = 40 mol/min, F_I = 50 mol/min; total = 150 mol/min
✏️ Practice: A feed contains F<sub>A0</sub> = 80 mol/min of A and F<sub>I0</sub> = 20 mol/min of inert I. In the reactor, X<sub>A</sub> = 0.25 for A → B (1:1), with no B in the feed. Compute the total outlet molar flow rate.
mol/min
Solution
  1. Compute A out: F_A = 80*(1 - 0.25) = 60 mol/min.
  2. Compute B formed: F_B = 80*0.25 = 20 mol/min.
  3. Inert unchanged: F_I = 20 mol/min.
  4. Total out = 60 + 20 + 20 = 100 mol/min.

Check your understanding

1. In a steady-state reactor with a single reaction, what is the most direct way to connect inlet and outlet flows of A?
Conversion is defined from inlet and outlet flow rates of the reactant; it directly gives F_A from F_A0 and X_A.
2. For the reaction A → B (1:1), if only A and B are present and there are no side reactions, what happens to total molar flow in the reactor?
With 1:1 stoichiometry and no other species changing, total moles are unchanged by reaction (though density/volumetric flow can still change with T).
3. Why is it useful to draw a block-flow diagram before doing detailed calculations?
A clear diagram helps you decide what to balance and where, and prevents missing streams or double-counting.
🔎 Related lessons
Real vs Ideal Reactors — A First Look at RTD · soon
✅ Key takeaways
  • A capstone problem is easiest when the flowsheet story is explicit: reactor then separator.
  • Conversion links inlet and outlet flows for reactant A; stoichiometry gives product formation.
  • Simple molar closure checks can catch setup errors early.
  • A block-flow diagram is a practical tool to prevent missing streams and mismatched assumptions.
➡️ Next, we turn kinetics into a reactor size by computing an Arrhenius-adjusted rate constant and applying a design equation.
Want to test yourself on this? Try the Chemical Aptitude test →