Proving Conservation in a Closed Perfusion Model

Why Conservation Is a Structural Check

In a closed perfusion model, drug can move between the central blood pool and tissues, but there is no dose, metabolism, excretion, or other external source or sink. The amount in any one compartment can therefore rise or fall while the total amount represented by the model remains constant. This is more than an intuitive expectation: it follows directly from how the intercompartmental flows enter the amount equations.

The proof depends on consistent bookkeeping. Every amount must belong to exactly one modeled control volume, and each transfer out of one volume must appear as the same transfer into another. The equations below make those requirements explicit for one central pool connected to $m$ perfused tissues.

Model and Assumptions

Definition: Let $A_c$ be the drug amount in the central blood pool and $A_i$ the drug amount in tissue $i$, for $i=1,\ldots,m$. The total modeled amount is $$ A_{\mathrm{tot}}=A_c+\sum_{i=1}^{m} A_i. $$ For tissue $i$, let $Q_i$ be its blood flow, $C_a$ the arterial concentration leaving the central pool, and $C_{v,i}$ the venous concentration leaving that tissue. The tissue’s net rate of amount change is $Q_i(C_a-C_{v,i})$.

The flow $Q_i$ has units of volume per time, and each concentration has units of amount per volume, so $Q_iC$ has units of amount per time. The model assumes that the central pool supplies the tissues at $C_a$, receives their outflows at $C_{v,i}$, and has total outflow equal to the sum of tissue flows. The equations are

\frac{dA_i}{dt}=Q_i(C_a-C_{v,i}),\qquad i=1,\ldots,m,

\frac{dA_c}{dt}=\sum_{i=1}^{m}Q_iC_{v,i} -\left(\sum_{i=1}^{m}Q_i\right)C_a.

These equations describe a closed model only under specific conditions. There is no dose or other external input, and no elimination or other external loss. The central outflow is exactly the flow delivered to the tissues; a tissue’s venous outflow is exactly the amount returned to the central pool. The compartments also form a nonoverlapping accounting partition: an amount cannot be counted in both the central pool and a tissue. If flows vary over time, the same equations apply at each time provided the matching inflows and outflows use the same instantaneous $Q_i$.

Deriving the Conservation Result

Differentiate the total amount. Since it is a sum of compartment amounts, its rate of change is the sum of their rates. Substituting the model equations gives

\begin{aligned} \frac{dA_{\mathrm{tot}}}{dt} &=\frac{dA_c}{dt}+\sum_{i=1}^{m}\frac{dA_i}{dt}\\ &=\left[\sum_{i=1}^{m}Q_iC_{v,i} -\left(\sum_{i=1}^{m}Q_i\right)C_a\right] +\sum_{i=1}^{m}\left[Q_iC_a-Q_iC_{v,i}\right]\\ &=0. \end{aligned}

For each tissue, the return term $Q_iC_{v,i}$ in the central-pool equation cancels the outflow term $-Q_iC_{v,i}$ in that tissue’s equation. The central-pool outflow $-(\sum_i Q_i)C_a$ cancels the sum of tissue inflows $\sum_i Q_iC_a$. Thus internal transfers change where the drug is located, but not how much is present in the modeled system.

Theorem: If the central-pool and tissue equations include only the matched internal flows shown above, then $dA_{\mathrm{tot}}/dt=0$. Consequently, for any two times $t_0$ and $t$ over which the model applies, $A_{\mathrm{tot}}(t)=A_{\mathrm{tot}}(t_0)$.

The final statement follows by integrating the zero derivative: $A_{\mathrm{tot}}(t)-A_{\mathrm{tot}}(t_0)=\int_{t_0}^{t}0\,ds=0$. Conservation does not require each tissue amount or concentration to be constant. It says that their changes, together with the central-pool change, sum to zero.

Worked Example: Checking the Instantaneous Amount Balance

Consider two tissues with $Q_1=1.2$ L/h and $Q_2=0.8$ L/h. At a particular instant, let $C_a=2.0$ mg/L, $C_{v,1}=1.5$ mg/L, and $C_{v,2}=0.5$ mg/L. Assume there are no external inputs or losses and that the central pool supplies exactly these two tissue flows.

The tissue amount rates are

\frac{dA_1}{dt}=1.2(2.0-1.5)=0.60\ \text{mg/h}, \qquad \frac{dA_2}{dt}=0.8(2.0-0.5)=1.20\ \text{mg/h}.

The central-pool rate is venous return minus arterial outflow:

\frac{dA_c}{dt} =1.2(1.5)+0.8(0.5)-(1.2+0.8)(2.0) =1.8+0.4-4.0 =-1.80\ \text{mg/h}.

Adding the three rates gives $ -1.80+0.60+1.20=0.00$ mg/h. The central pool is losing drug at this instant, while both tissues are gaining it; the total modeled amount is unchanged.

What the Result Does—and Does Not—Say

Conservation concerns amounts, not an unweighted sum of concentrations. If compartment volumes are fixed and $A_j=V_jC_j$, then a change in concentration corresponds to a volume-weighted amount change. In general, $\sum_j C_j$ is not a conserved quantity, even when $\sum_j A_j$ is.

Worked Example: Why Concentration Changes Must Be Volume-Weighted

Use the amount rates from the preceding example, and suppose the central pool has volume $V_c=3.0$ L while the tissue volumes are $V_1=0.50$ L and $V_2=2.0$ L. For fixed volumes, $dC_j/dt=(dA_j/dt)/V_j$.

The concentration rates are

\frac{dC_c}{dt}=\frac{-1.80}{3.0}=-0.60\ \text{mg/L/h},\quad \frac{dC_1}{dt}=\frac{0.60}{0.50}=1.20\ \text{mg/L/h},\quad \frac{dC_2}{dt}=\frac{1.20}{2.0}=0.60\ \text{mg/L/h}.

Their unweighted sum is $1.20$ mg/L/h, not zero. That is not a failure of conservation: these concentrations have different associated volumes. Weighting each rate by its volume recovers the amount balance:

3.0(-0.60)+0.50(1.20)+2.0(0.60) =-1.80+0.60+1.20=0.00\ \text{mg/h}.

This calculation assumes fixed volumes. If a model uses changing volumes, the relation between amount and concentration must account for that change; conservation should still be formulated for amount.

Assumptions, Diagnostics, and Limitations

The cancellation proof is a useful structural check when assembling equations. Before relying on it, confirm that all terms describe the same physical transfers and use compatible units. In particular, a tissue’s $Q_iC_a$ inflow must match the central pool’s $Q_iC_a$ outflow, and its $Q_iC_{v,i}$ outflow must match the central pool’s return. The total flow leaving the central pool must equal $\sum_i Q_i$.

If summing the equations produces a nonzero term, the model is not closed as written or its bookkeeping is inconsistent. A missing tissue return, an unmatched flow, or a term that represents an external input or loss can explain such a remainder. Check the direction and units of each flux, and check that no blood or tissue amount is included in two states. A nonzero balance is not automatically an algebra error: it may be the correct consequence of a source or sink that the model includes.

The result has limits. It does not establish that the chosen tissue equations accurately describe physiology, nor does it determine tissue concentrations or exchange kinetics. It only proves an accounting identity for the stated model. Conservation can hold in a structurally inappropriate model, and a biologically plausible model can fail the identity if a required amount or boundary flow is omitted.

An amount-based derivation is usually the clearest approach because each state is already an amount and each transfer appears with opposite signs. A concentration-based formulation can also be used, but it requires converting each concentration rate back to an amount rate with the appropriate volume (and accounting for changing volumes where relevant). Summing concentrations directly is not an equivalent shortcut.

Define the system boundary.
List the central pool and each tissue included in the total; decide which amounts are inside that boundary.

Write amount rates as transfers.
For each internal flow, identify the amount leaving one compartment and entering the other.

Sum and simplify.
Cancel matched internal terms. A zero remainder establishes conservation for the closed model; a nonzero remainder identifies a term to interpret or audit.

Common pitfall: “The central concentration falls, so the model must be losing drug.” That conclusion does not follow: concentration can fall as drug moves into tissues. The relevant check is whether the sum of all modeled amount rates is zero, not whether any one concentration or amount is constant.

Worked Example: A General Two-Tissue Balance

Suppose $Q_1=0.75$ L/h and $Q_2=1.25$ L/h, with $C_a=1.0$ mg/L, $C_{v,1}=1.8$ mg/L, and $C_{v,2}=0.4$ mg/L. Treat these as hypothetical instantaneous values in a closed model.

The tissue rates are

\frac{dA_1}{dt}=0.75(1.0-1.8)=-0.60\ \text{mg/h}, \qquad \frac{dA_2}{dt}=1.25(1.0-0.4)=0.75\ \text{mg/h}.

Tissue 1 releases drug at this instant, whereas tissue 2 takes it up. For the central pool,

\frac{dA_c}{dt} =0.75(1.8)+1.25(0.4)-(0.75+1.25)(1.0) =1.35+0.50-2.00 =-0.15\ \text{mg/h}.

The total rate is $-0.60+0.75-0.15=0.00$ mg/h. The different directions of exchange do not affect the result: each tissue’s net exchange is balanced by the corresponding central-pool transfers.

Check Your Understanding

Answer each question, then compare with the solutions below.

In one sentence, state the condition on the total amount rate that defines conservation in this model.
A tissue has $Q=0.5$ L/h, $C_a=3$ mg/L, and $C_v=2$ mg/L. What is its amount rate, and what matching tissue-related terms appear in the central equation?
Why can the sum of compartment concentration rates be nonzero even when total amount is conserved?
If the central pool sends out 2.0 L/h but modeled tissue flows sum to 1.8 L/h, which assumption used in the proof fails?

Solutions. 1. Conservation requires $dA_{\mathrm{tot}}/dt=0$, so total modeled amount is constant over time. 2. The tissue rate is $0.5(3-2)=0.50$ mg/h; the central equation includes $+0.5C_v$ as return and $-0.5C_a$ as outflow, giving $-0.50$ mg/h for those terms, which cancels the tissue rate. 3. Concentrations correspond to amounts through compartment volumes, so amount rates are volume-weighted concentration rates when volumes are fixed; the unweighted sum need not vanish. 4. Central outflow no longer matches the total tissue inflow. The proof leaves an unmatched flow term, so the stated closed-system equations do not satisfy the required bookkeeping.