Balance Equations for Open Systems With Inputs and Outputs

Why Open-System Balances Need Boundary Terms

A reaction network can redistribute material among its species without changing the total amount of a conserved moiety. That statement applies to internal reactions; it does not imply that the total stays constant when material enters or leaves the modeled system. In a pharmacology model, a species may be infused, supplied by an external source, cleared, or removed by an intervention. A correct balance must distinguish those boundary exchanges from internal reaction transfers.

The key bookkeeping question is: which processes move material between modeled species, and which processes move it across the boundary of the modeled system? Internal reactions appear through the stoichiometric matrix and reaction fluxes. Boundary inputs and outputs, along with external production and removal, appear as additional terms. The balance then makes clear whether a quantity is truly constant or changes at the net rate of exchange.

Definition: Let $n(t)$ be the vector of amounts of $m$ modeled species, and let $S$ be their $m \times r$ stoichiometric matrix for $r$ internal reactions. Let $v(t,n)$ be the vector of reaction fluxes. Write $u_{\mathrm{in}}$ and $u_{\mathrm{out}}$ for nonnegative vectors of external input and output rates, and $p_{\mathrm{ext}}$ and $d_{\mathrm{ext}}$ for nonnegative external production and removal rates. The open-system species balance is $$ \frac{dn}{dt}=S v+u_{\mathrm{in}}-u_{\mathrm{out}}+p_{\mathrm{ext}}-d_{\mathrm{ext}}. $$ Each term has units of amount per time. Internal reaction fluxes may depend on the current state; external rates may also vary with time or state.

The signs follow from the meaning of each term: a process adding material to the modeled species has a positive sign, and a process removing it has a negative sign. The labels “input” and “output” refer to crossing the system boundary. “External production” and “external removal” refer to sources or sinks not represented as reactions among the modeled species. A process should be represented once, not both as an internal reaction and as an external term.

This equation is written for amounts. If a model uses concentrations, converting the amount balance to a concentration balance requires accounting for the compartment volume. Here, when a concentration is used to calculate an amount rate, the example will state its volume assumptions explicitly. This tutorial focuses on species accounting, not on deriving balances for changing volumes.

Internal Reactions and Conserved Moieties

For a reaction $j$, column $j$ of $S$ records the stoichiometric change in every modeled species when that reaction proceeds by one unit of extent. The corresponding flux $v_j$ is the rate of that reaction extent. Thus $S v$ combines the contributions of all internal reactions to each species’ rate of change. A negative entry in $S$ consumes a species; a positive entry produces one.

A vector $c$ defines a linear total $M=c^\mathsf{T}n$. Its entries specify how much of the tracked moiety is counted in each species amount. When $c^\mathsf{T}S=0$, every internal reaction has zero net effect on this total. This is the conserved-moiety condition from the earlier material on conserved moieties. In an open system, it cancels the internal terms but does not automatically cancel environmental exchange.

\frac{dM}{dt} =c^\mathsf{T}\frac{dn}{dt} =c^\mathsf{T}S v +c^\mathsf{T}(u_{\mathrm{in}}-u_{\mathrm{out}}+p_{\mathrm{ext}}-d_{\mathrm{ext}}) =c^\mathsf{T}(u_{\mathrm{in}}-u_{\mathrm{out}}+p_{\mathrm{ext}}-d_{\mathrm{ext}}) \quad\text{if }c^\mathsf{T}S=0.

This is the central open-system result: a moiety conserved by internal reactions changes only through its net external exchange. If the weighted input and production equal the weighted output and removal at every time, then $dM/dt=0$, so $M$ remains constant. If those rates do not balance, $M$ changes even though all internal reactions preserve it.

Deriving the Balance From Amount Accounting

The balance can be justified over a short interval $[t,t+\Delta t]$. Each internal reaction changes the amount vector according to its stoichiometric column. If reaction $j$ proceeds by an extent $\Delta \xi_j$, its contribution is $S_{\cdot j}\Delta \xi_j$. Summing over reactions gives $S\Delta \xi$. External inputs and production add amounts; external outputs and removal subtract them. Therefore,

n(t+\Delta t)-n(t) =S\Delta \xi +\int_t^{t+\Delta t} \left(u_{\mathrm{in}}-u_{\mathrm{out}}+p_{\mathrm{ext}}-d_{\mathrm{ext}}\right)\,ds.

Divide by $\Delta t$ and take the limit as $\Delta t$ approaches zero. By definition, $d\xi/dt=v$; the average external rates converge to their instantaneous rates when the rates are well-defined. This yields the differential balance above. Multiplying the finite-interval equation by $c^\mathsf{T}$ also shows directly why internal reactions disappear from a conserved-moiety balance: $c^\mathsf{T}S\Delta \xi=0$. The remaining change is the integral of net external exchange.

The assumptions are bookkeeping assumptions rather than a particular kinetic law: the modeled species and system boundary have been specified; the stoichiometric changes describe the internal reactions being included; and every external source or sink is represented with the correct sign and rate. Reaction kinetics determine $v$, but the balance form does not require mass-action kinetics or any other specific flux law.

Worked Example: Internal Conversion With Feed and Removal

Consider the internal reaction $A\rightarrow B$, with $A$ and $B$ measured in nmol. At a particular instant, the reaction flux is $3.0$ nmol/h. The system receives $4.0$ nmol/h of $A$, has external production of $A$ at $0.5$ nmol/h, removes $A$ externally at $0.2$ nmol/h, and outputs $B$ at $1.0$ nmol/h. Find the instantaneous species rates and the rate of change of the total $A+B$.

With the species ordered as $(A,B)$, the stoichiometric matrix and flux are $$ S=\begin{pmatrix}-1\\1\end{pmatrix}, \qquad v=3.0\ \text{nmol/h}. $$ Thus the internal contribution is $Sv=(-3.0,\ 3.0)^\mathsf{T}$ nmol/h. Adding each external term gives $$ \frac{dA}{dt}=-3.0+4.0+0.5-0.2=1.3\ \text{nmol/h}, \qquad \frac{dB}{dt}=3.0-1.0=2.0\ \text{nmol/h}. $$

For the total $M=A+B$, $c=(1,1)^\mathsf{T}$, and $c^\mathsf{T}S=0$, so the reaction cancels. The net external rate is $4.0+0.5-0.2-1.0=3.3$ nmol/h. The species rates give the same result: $dM/dt=1.3+2.0=3.3$ nmol/h. Internal conversion redistributes amount; the net boundary exchange increases the total.

Applying the Balance to Multiple Reaction Directions

When a reaction is reversible, one convenient bookkeeping choice is to represent the forward and reverse directions as separate nonnegative fluxes with opposite stoichiometric columns. Their net effect can equivalently be written as one signed net flux. Either representation is valid if the flux convention and stoichiometry are used consistently. The external terms remain separate in either case.

Worked Example: Reversible Conversion With Unequal Outputs

Suppose $A\rightleftharpoons B$ is represented by a forward flux of $5.0$ nmol/h and a reverse flux of $2.0$ nmol/h. There is also an external input of $A$ at $1.0$ nmol/h, an output of $A$ at $0.5$ nmol/h, and an output of $B$ at $1.5$ nmol/h. Find each species rate and the total rate for $A+B$.

The two internal reaction columns and fluxes are $$ S=\begin{pmatrix}-1&1\\1&-1\end{pmatrix}, \qquad v=\begin{pmatrix}5.0\\2.0\end{pmatrix}\ \text{nmol/h}. $$ Their combined contribution is $$ Sv=\begin{pmatrix}-5.0+2.0\\5.0-2.0\end{pmatrix} =\begin{pmatrix}-3.0\\3.0\end{pmatrix}\ \text{nmol/h}. $$ Adding the external terms, $dA/dt=-3.0+1.0-0.5=-2.5$ nmol/h and $dB/dt=3.0-1.5=1.5$ nmol/h. Therefore, $d(A+B)/dt=-2.5+1.5=-1.0$ nmol/h. The internal terms cancel in the total; the total declines because outputs of $2.0$ nmol/h exceed the input of $1.0$ nmol/h.

The same result follows from the net external balance, $1.0-0.5-1.5=-1.0$ nmol/h. This agreement is a useful check on signs and bookkeeping.

Inputs and Clearance Expressed as Rates

An external rate need not be specified directly as an amount per time. For example, an infusion may be given as an amount rate, while a well-mixed output stream may be calculated as flow rate times the concentration in that stream. A linear clearance term is another representation of output: clearance multiplied by concentration gives an amount-per-time removal rate. These expressions are alternative ways to calculate a boundary term; they do not change the balance structure.

Worked Example: Infusion, External Production, and Clearance

A drug amount $n$ is in a compartment of constant volume $V=3.0$ L. At an instant, $n=12$ mg. The compartment receives an infusion of $5.0$ mg/h and an external production rate of $0.4$ mg/h. Its clearance is $0.60$ L/h. Find the instantaneous amount rate.

First calculate the concentration, $C=n/V=12/3.0=4.0$ mg/L. The clearance-mediated removal rate is $Cl\cdot C=(0.60\ \text{L/h})(4.0\ \text{mg/L})=2.4$ mg/h. Thus the amount balance is $$ \frac{dn}{dt}=5.0+0.4-2.4=3.0\ \text{mg/h}. $$ At that instant the drug amount is increasing at $3.0$ mg/h. The calculation uses constant volume so that the stated concentration is $n/V$; clearance enters as a negative external term, not as an internal conversion to another modeled drug species.

Diagnostics, Alternatives, and Common Pitfalls

A balance should be checked both species by species and for any moiety expected to be conserved internally. The following checks often reveal errors before fitting or simulation:

Check signs. Inputs and production should add amount; outputs and removal should subtract it. Confirm that a stated clearance or outflow is not accidentally given a positive sign.
Check the internal contribution. For a proposed conserved moiety, calculate $c^\mathsf{T}S$. If it is zero, internal fluxes must cancel from its total balance.
Check net exchange. Sum weighted external inputs, production, outputs, and removal independently. The result should match the derivative of the weighted species total.
Check process representation. A transfer from one modeled species to another belongs in $Sv$. Counting it again as an external input or output creates a false gain or loss.
Check rate dimensions. All terms added in one species equation must be amount per time. For example, a concentration must be multiplied by an appropriate flow or clearance rate before it can be used as an amount-rate term.

A common mistake is to say, “The reactions conserve $A+B$, so $A+B$ is constant,” even when $A$ is infused and $B$ is removed. The reasoning confuses internal conservation with conservation of the open system. The correction is to apply $c^\mathsf{T}$ to the complete balance: internal reactions cancel when $c^\mathsf{T}S=0$, but the total is constant only when the weighted net external exchange is also zero.

There are two common bookkeeping approaches. A stoichiometric representation is useful when the internal reaction network and its conserved moieties matter: it keeps each reaction’s species changes explicit. An aggregate species balance can instead combine all gains and losses directly for each species; this can be compact for a simple compartment, but it may obscure which internal transfers cancel in a total. These approaches agree when they describe the same processes once, with consistent signs. Neither balance form by itself specifies the kinetics, estimates unknown rates, or guarantees that the chosen system boundary captures every relevant source and sink.

Finally, an unchanged internal moiety does not imply that every species amount is unchanged. Internal reactions may redistribute that moiety, while boundary exchange changes its total. Conversely, a total can be momentarily constant because external gains and losses balance even while individual species are changing. Interpret each equation at the level of the quantity it describes.

Key takeaway: For an open system, write the internal reaction contribution as $Sv$ and list every external input, output, production, and removal separately. For any $c$ satisfying $c^\mathsf{T}S=0$, the total $c^\mathsf{T}n$ changes at exactly its weighted net external exchange rate.

Check Your Understanding

Try each problem, then compare your reasoning with the brief solutions.

For $A\rightarrow B$, let the reaction flux be $2.0$ nmol/h. If $A$ enters at $1.0$ nmol/h and $B$ leaves at $0.5$ nmol/h, find $dA/dt$, $dB/dt$, and $d(A+B)/dt$.
What condition on $c$ ensures that internal reactions do not change $c^\mathsf{T}n$? Does that condition alone guarantee a constant total in an open system?
A modeled species has input $3.0$ mg/h, external production $0.2$ mg/h, and removal $1.1$ mg/h. What is its rate of change due to these terms?
A model reports that a moiety-preserving internal reaction changes the moiety total by $0.8$ nmol/h. Name one bookkeeping check that could identify the error.

Solutions. 1. $dA/dt=-2.0+1.0=-1.0$ nmol/h; $dB/dt=2.0-0.5=1.5$ nmol/h; hence $d(A+B)/dt=0.5$ nmol/h, equal to input minus output. 2. Require $c^\mathsf{T}S=0$. No: weighted net external exchange must also be zero for the total to remain constant. 3. $3.0+0.2-1.1=2.1$ mg/h. 4. Check whether $c^\mathsf{T}S=0$ for the reaction’s stoichiometric column and verify that the reaction was not also counted as an external term.