Question 4.12: Simple Stability Analysis for a Single Compartment with Flui......
Simple Stability Analysis for a Single Compartment with Fluid Volume ∀(t) and Volume-Dependent Inflow and Outflow
Of interest is the equilibrium point, i.e., \rm ∀(t) = ∀_0 = ⊄, and an associated analysis of system stability. Applications include the human-body fluid and a hydro-electric reservoir.
| Concept | Assumptions | Sketch |
| • First-order rate equation \rm d∀/dt = Q_{in} − Q_{out} | • Well-mixed compartment and homogenous fluid |  |
| • Constant properties | ||
| • Uniform inlet/ outlet streams |
• Fluid mass balance in terms of volume, i.e., with m = ρ∀; ρ = const.; and \dot ∀ = ∀/ t = Q:
\rm\frac{d∀}{dt} =Q_{in}-Q_{out}\qquad\qquad\qquad(E.4.12.1)where Q = Q(∀).
• Equilibrium solution:
\rm\frac{d∀}{dt} =0,\qquad i.e.,\qquad Q_{in}=Q_{out} (E.4.12.2a,b)
Equations (E.4.12.2a, b) indicate that the amount of fluid in the compartment doesn’t change, i.e., \rm ∀(t = t_0) = ∀_0 and hence \rm Q_{in}(t_0) = Q_{out}(t_0).
Graph:
Comments:
• Using the plotted inflow/outflow functions as quantitative examples, small deviations from the equilibrium point (∀_0) may lead to the following (see Graph):
Case (i): ∀(t) = ∀_0 + Δ∀, so that with \rm Q_{out}(t_0 + Δt) > Q_{out}(t_0) while \rm Q_{in}(t_0 + Δt) < Q_{in}(t_0). This implies that \rm Q_{in} – Q_{out} < 0 and hence ∀(t) decreases, i.e., ∀(t) retreats back to the equilibrium volume ∀_0.
Case (ii): Perturbation of ∀(t) to ∀_0 − Δ∀ yields \rm Q_{in}(t_0− Δt) > Q_{out}(t_0 − Δt) and d∀/dt > 0, i.e., the fluid volume increases towards ∀_0 again. In summary, the solution is stable.
• Different functions for \rm Q_{in} \,and\,Q_{out} may prevent the solution to revert back to equilibrium, e.g., the compartment totally drains or overflows, and the system is unstable.
