Two chemostats (continuous bioreactors) with the same organism, same feed, and the same dilution rate start with different initial glucose concentrations (reactor 1: 10 g L^-1, reactor 2: 0.1 g L^-1). At steady state, how do their residual glucose concentrations compare?

Introduction / Context:
In a chemostat operating at steady state, the specific growth rate equals the dilution rate, and residual substrate depends on kinetic parameters and dilution, not on initial conditions. This question probes understanding of steady-state independence from start-up history.

Given Data / Assumptions:

• Both reactors: identical organism, feed, temperature, pH, and dilution rate D.
• Monod kinetics apply: mu = mu_max * S / (Ks + S).
• Steady state achieved with biomass washout avoided (D < mu_max).

Concept / Approach:
At steady state, mu = D. Solving Monod for S* gives S* = (D * Ks) / (mu_max − D) when S* ≪ feed substrate and D < mu_max. S* is independent of the initial glucose present at t = 0; transients decay as the system converges to the unique steady state defined by D and kinetics.

Step-by-Step Solution:
Write steady-state condition: mu(S*) = D.Solve: D = mu_max * S* / (Ks + S*) ⇒ S* determined solely by D, mu_max, Ks.Recognize initial substrate influences only the transient trajectory, not S*.Therefore, residual glucose is equal in both reactors at steady state.

Verification / Alternative check:
Phase-plane analysis shows a single asymptotically stable steady state for D below washout; numerical simulations with different initial S converge to the same S*.

Why Other Options Are Wrong:
Greater/lesser/always zero: contradicts steady-state dependence on D and kinetics rather than initial conditions; S* is nonzero unless D → 0 or feed is limiting in special cases.

Common Pitfalls:
Confusing batch carryover with chemostat steady state; assuming memory of initial charge persists despite continuous dilution and consumption.

Final Answer:
Glucose in reactor 1 = glucose in reactor 2