Chemostat calculation: A chemostat of liquid volume 2 L is fed at 4 L·h^-1. What is the dilution rate D for this reactor?

Difficulty: Easy

Correct Answer: 2 h^-1

Explanation:


Introduction / Context:
Dilution rate (D) is the defining parameter of a chemostat, connecting feed flow and reactor volume. This item checks correct unit usage and the simple formula for D.


Given Data / Assumptions:

  • Reactor volume V = 2 L.
  • Feed flow F = 4 L·h^-1.
  • Well-mixed, constant volume operation.


Concept / Approach:
The dilution rate is defined as D = F / V with units of time^-1. It represents the fraction of reactor volume replaced per unit time.


Step-by-Step Solution:
Compute D = F / V = 4 / 2 = 2 h^-1.Check units: (L·h^-1)/L = h^-1.Therefore, D = 2 h^-1.


Verification / Alternative check:
Residence time τ = 1 / D = 0.5 h, consistent with a high-throughput chemostat replacing its content twice per hour.


Why Other Options Are Wrong:
“2 litres” and “4 litres per hour” are not rates normalized by volume; they are volume or volumetric flow, not D.

“2 litres per hour” has wrong units; D must be in h^-1.


Common Pitfalls:

  • Forgetting to normalize by reactor volume.
  • Reporting D with incorrect units.


Final Answer:
2 h^-1

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