Worked Example: Two Tissues Sharing a Blood Pool

Why a Shared Blood Pool Needs Careful Bookkeeping

When two tissues receive blood from the same central pool, each tissue takes up drug from that pool and returns drug to it. The tissues therefore interact through the blood concentration, even if there is no direct exchange between them. Writing each tissue equation separately is straightforward; the key check is whether every internal flow appears once as an outflow and once as an inflow.

This worked example uses amount states, as in the earlier material on control volumes and perfused-tissue mass balances. The purpose is to construct and verify the coupled equations, not to prescribe a universal relationship between tissue amount and venous concentration. A tissue model must provide that relationship.

Definition: Let $A_b$ be drug amount in a shared, well-mixed blood pool, and let $A_1$ and $A_2$ be amounts in tissues 1 and 2. Let $Q_i$ be blood flow from the pool through tissue $i$, $C_b$ the pool concentration, and $C_{vi}$ the concentration leaving tissue $i$ in venous blood. With fixed blood-pool volume $V_b$, $C_b=A_b/V_b$. The tissue amounts change according to inflow minus outflow: $$ \frac{dA_i}{dt}=Q_iC_b-Q_iC_{vi}, \qquad i=1,2. $$

Assembling the Three Equations

Assume the central pool is well mixed, so both tissues receive blood at the same concentration $C_b$. Flow $Q_i$ is constant and the flow through each tissue returns to the same pool. Drug is neither created nor destroyed during tissue exchange. For now, there is no elimination; an external dose input can be added to the blood equation. Concentrations are amount per volume, so multiplying a flow in volume per time by concentration in amount per volume gives an amount rate.

For tissue 1, arterial delivery is $Q_1C_b$, and venous return carries $Q_1C_{v1}$. Tissue 2 has the corresponding delivery and return terms. The blood pool loses the arterial flows sent to both tissues and gains both venous returns. With an input rate $R_{\mathrm{in}}$ into the pool, the equations are:

\begin{aligned} \frac{dA_1}{dt} &= Q_1C_b-Q_1C_{v1},\\ \frac{dA_2}{dt} &= Q_2C_b-Q_2C_{v2},\\ \frac{dA_b}{dt} &= Q_1C_{v1}+Q_2C_{v2}-(Q_1+Q_2)C_b+R_{\mathrm{in}}. \end{aligned}

These balances specify the exchange structure. To obtain a closed system of ordinary differential equations, the tissue model must also specify each venous concentration from the relevant state and parameters. For the numerical examples below, use a deliberately simple well-mixed tissue closure: each tissue has fixed volume $V_i$, and $C_{vi}=A_i/V_i$. This is an illustrative closure, not a universal tissue relation. It gives a three-state ODE system when substituted along with $C_b=A_b/V_b$.

Worked Example: Writing the Coupled ODEs

Consider a hypothetical model with $V_b=3$ L, $V_1=10$ L, $V_2=20$ L, $Q_1=1.2$ L/h, and $Q_2=0.8$ L/h. There is no external input. Apply the stated well-mixed closure and write the equations in terms of amounts.

The concentrations are $C_b=A_b/3$, $C_{v1}=A_1/10$, and $C_{v2}=A_2/20$, each in mg/L if amounts are in mg. Substitution into the tissue balances gives:

\begin{aligned} \frac{dA_1}{dt}&=1.2\left(\frac{A_b}{3}-\frac{A_1}{10}\right),\\ \frac{dA_2}{dt}&=0.8\left(\frac{A_b}{3}-\frac{A_2}{20}\right),\\ \frac{dA_b}{dt}&=1.2\frac{A_1}{10}+0.8\frac{A_2}{20}-(1.2+0.8)\frac{A_b}{3}. \end{aligned}

Each right-hand side has units $(\mathrm{L/h})(\mathrm{mg/L})=\mathrm{mg/h}$, the required units for an amount derivative. The same pool concentration appears in both tissue inflows, while each tissue has its own return concentration.

Why Internal Exchange Cancels

The total modeled amount is $A_{\mathrm{tot}}=A_b+A_1+A_2$. Its rate of change follows by adding the three amount derivatives. This is the central verification: arterial delivery removed from blood must match the delivery added to tissue, and venous return removed from tissue must match the return added to blood.

Add the tissue balances.
The tissue equations contribute $Q_1C_b+Q_2C_b-Q_1C_{v1}-Q_2C_{v2}$.

Add the blood balance.
The blood equation contributes $Q_1C_{v1}+Q_2C_{v2}-(Q_1+Q_2)C_b+R_{\mathrm{in}}$.

Cancel matching internal terms.
Each $Q_iC_{vi}$ and each $Q_iC_b$ occurs once with a positive sign and once with a negative sign. Only the external input remains.

\begin{aligned} \frac{dA_{\mathrm{tot}}}{dt} &=\frac{dA_b}{dt}+\frac{dA_1}{dt}+\frac{dA_2}{dt}\\ &=R_{\mathrm{in}}. \end{aligned}

The cancellation does not depend on the chosen tissue concentration closure. It follows from matching the same physical flow and concentration in the tissue and blood balances. If elimination at rate $E$ is included from the blood pool, the blood equation gains $-E$, and the total balance becomes $dA_{\mathrm{tot}}/dt=R_{\mathrm{in}}-E$. Elimination is an external loss from this three-state system, not an internal exchange.

Worked Example: Checking Amount Rates at One Instant

Use the volumes and flows from the first example. At a particular instant, suppose $A_b=30$ mg, $A_1=40$ mg, and $A_2=120$ mg. There is no input or elimination. Calculate the concentrations and all three amount rates.

The concentrations are $C_b=30/3=10$ mg/L, $C_{v1}=40/10=4$ mg/L, and $C_{v2}=120/20=6$ mg/L. Thus:

\begin{aligned} \frac{dA_1}{dt}&=1.2(10-4)=7.2\ \mathrm{mg/h},\\ \frac{dA_2}{dt}&=0.8(10-6)=3.2\ \mathrm{mg/h},\\ \frac{dA_b}{dt}&=1.2(4)+0.8(6)-(1.2+0.8)(10)\\ &=4.8+4.8-20=-10.4\ \mathrm{mg/h}. \end{aligned}

The total rate is $7.2+3.2-10.4=0$ mg/h. The tissues collectively gain 10.4 mg/h, exactly the amount lost by the pool. This instantaneous check establishes conservation for these rates; it does not imply that any concentration remains constant over time.

Adding External Processes Without Losing the Check

A dose entering the blood is added only to the blood balance if that is where the input is delivered. Likewise, an elimination process should appear in the compartment from which drug is eliminated. These terms do not have a matching internal counterpart elsewhere in this model, so they remain in the total balance. The following example combines an input with elimination to make that distinction explicit.

Worked Example: Infusion and Elimination

Return to the state in the preceding example, where $C_b=10$ mg/L and the tissue exchange rates are $7.2$ and $3.2$ mg/h into the tissues. Suppose a hypothetical infusion delivers $R_{\mathrm{in}}=5$ mg/h into blood, and elimination from blood is first-order with rate $E=CL\,C_b$, where $CL=0.2$ L/h.

The elimination rate is $E=(0.2\ \mathrm{L/h})(10\ \mathrm{mg/L})=2$ mg/h. The blood equation therefore has the previous exchange rate $-10.4$ mg/h, plus infusion and minus elimination:

\frac{dA_b}{dt}=-10.4+5-2=-7.4\ \mathrm{mg/h}.

The total rate is $dA_{\mathrm{tot}}/dt=7.2+3.2-7.4=3.0$ mg/h. This agrees with the global balance $R_{\mathrm{in}}-E=5-2=3$ mg/h. Tissue exchange still cancels; the remaining increase is the net external input.

Assumptions, Diagnostics, and Model Choice

The cancellation result is a bookkeeping identity under the stated structure, but interpreting the equations as a model also requires assumptions. The pool is represented by a single concentration, both tissues draw from that same pool, and each venous flow returns to it. Flows and volumes are defined consistently, and there are no unlisted sources or sinks. In the worked closure, tissue amount is represented by one well-mixed concentration per tissue; other tissue models can replace that closure while retaining the flow-balance structure.

A useful diagnostic is to add the right-hand sides before solving the ODEs. With no external processes, the sum must be zero for arbitrary values of $C_b$, $C_{v1}$, and $C_{v2}$. With input and elimination, it must equal input minus elimination. Also check that every flow term has units of amount per time and that each tissue’s outflow is the same flow used as its blood-pool return. If these checks fail, the written model does not implement the stated mass balance.

A common mistake is to subtract $Q_iC_b$ from the blood pool and also add that same delivery to the tissue, but then omit the return $Q_iC_{vi}$ from either the tissue or blood equation. The resulting total derivative contains an unmatched term and can falsely suggest loss or creation of drug during internal exchange. The correction is to trace each physical flow across its two control volumes: it is an outflow from one and an inflow to the other.

Explicitly retaining two tissues preserves their separate flows and venous concentrations. Combining them into one aggregate tissue can be simpler, but generally loses the ability to represent distinct tissue amounts or returns unless the aggregate model is constructed to preserve those features. Conversely, a shared blood pool cannot represent spatial differences in arterial concentration between tissue supply sites. The appropriate level of detail depends on the question and available physiological information; adding states alone does not guarantee better predictions.

Key takeaway: For two tissues sharing one blood pool, write each tissue’s arterial delivery and venous return in both relevant balances with opposite signs. Their exchange cancels in the total amount derivative, leaving only external inputs and losses.

Check Your Understanding

Answer each question, then compare with the solutions below.

Write the shared blood-pool balance for two tissue flows when there is no dose or elimination.
If tissue 1 gains 6 mg/h and tissue 2 loses 1 mg/h, what must the blood-pool rate be in a closed model with no external processes?
For $Q_1=1$ L/h, $C_b=8$ mg/L, and $C_{v1}=3$ mg/L, calculate tissue 1’s rate of change.
With infusion of 4 mg/h and elimination of 1.5 mg/h, what is the total amount rate, regardless of the internal tissue flows?
Name one consequence of representing two tissues by a single aggregate tissue.

Solutions. 1. $dA_b/dt=Q_1C_{v1}+Q_2C_{v2}-(Q_1+Q_2)C_b$. 2. The tissue total rate is $6-1=5$ mg/h, so the blood rate must be $-5$ mg/h for the total to remain constant. 3. $dA_1/dt=Q_1(C_b-C_{v1})=(1)(8-3)=5$ mg/h. 4. $dA_{\mathrm{tot}}/dt=4-1.5=2.5$ mg/h. 5. Aggregation can conceal tissue-specific flow or venous-return differences.