Combining Inflows, Outflows, And Internal Transfers

From Individual Compartments to a Combined System

A system may contain several compartments that exchange material with one another while also gaining material from, or losing material to, their surroundings. A compartment's stock can rise or fall because of both external flows and internal transfers. The combined stock, however, changes only when material crosses the boundary of the whole system.

The key is to keep these two kinds of flow distinct. A transfer out of one included compartment is simultaneously a transfer into another included compartment. It changes where the stock is located, but not how much stock the included compartments contain altogether.

Definitions and Sign Convention

Let \(S_i(t)\) denote the stock in compartment \(i\) at time \(t\), for \(i=1,\ldots,n\). A stock is an amount, such as litres, kilograms, or items. Its derivative \(S_i'(t)\) is its instantaneous rate of change, measured in stock units per unit time.

For each compartment, let \(I_i(t)\) be the total external inflow rate and \(O_i(t)\) the total external outflow rate. Both are defined as nonnegative rates. “External” means that the flow crosses the boundary of the full system being modelled, not merely the boundary of one compartment.

For an internal transfer, write \(q_{ij}(t)\) for the rate moving from compartment \(i\) to compartment \(j\), where \(i\ne j\). The direction is part of the notation: \(q_{ij}\) leaves \(i\) and enters \(j\). We take \(q_{ij}\geq 0\) while using that direction convention. A transfer that is absent has rate zero.

With the notation above, the balance for compartment \(i\) is

The first sum collects transfers into \(i\): \(q_{ji}\) comes from compartment \(j\). The second collects transfers out of \(i\): \(q_{ij}\) goes to compartment \(j\). If a model has only some of the possible links, the other terms are simply zero.

Why Internal Transfers Cancel

Define the combined stock of all included compartments by \(S_{\mathrm{tot}}(t)=\sum_{i=1}^{n}S_i(t)\). Since this is a finite sum of differentiable stock functions, its derivative is the sum of their derivatives. Substituting the compartment balances gives

To see the cancellation, consider any one internal transfer from compartment \(a\) to compartment \(b\). It contributes \(-q_{ab}(t)\) to the balance for \(a\), because it leaves \(a\). The same transfer contributes \(+q_{ab}(t)\) to the balance for \(b\), because it enters \(b\). Their sum is \(-q_{ab}(t)+q_{ab}(t)=0\). This reasoning applies to each internal link.

This is the complete cancellation argument: every internal transfer has exactly one source and one destination among the included compartments, so its negative contribution at the source and positive contribution at the destination cancel in the sum. The result does not require the transfer rates to be constant. They may vary with time or depend on the compartment stocks; at each time, the paired terms still cancel.

The boundary of the model matters. If a transfer goes from an included compartment to a compartment left out of the model, it is an outflow from the modelled system, not an internal transfer for that system. If material enters from an omitted compartment, it is an external inflow relative to the chosen system.

Worked Example: Two Compartments with One Internal Transfer

Worked Example: Several Links in a Three-Compartment System

Worked Example: Transfers That Depend on the Stocks

A Common Sign Error

A frequent mistake is to count an internal transfer as an additional gain to the combined stock because it appears as an inflow in the destination compartment. That calculation includes the positive contribution but forgets the equal outflow from the source compartment.

For the three-compartment example, an incorrect total calculation might count the \(10\) kg/min external inflow, the \(4+1+2\) kg/min of internal transfers, and the \(3+4\) kg/min of external outflows:

This is wrong: it treats internal transfers as gains to the combined system. The correction is to include both sides of every transfer. For example, the A-to-B transfer contributes \(-4\) in A’s balance and \(+4\) in B’s balance, giving \(-4+4=0\). Applying the same pairing to all three links leaves only \(10-(3+4)=3\) kg/min, the correct combined rate.

Building and Checking a Balance Model

A reliable construction separates the modelling task into a few checks. These checks do not replace the equations; they help expose missing terms and sign errors.

This structure is useful whenever a model tracks stock in multiple connected parts of a system. Individual compartments can increase or decrease because material moves between them, even while the total changes only through boundary flows. Writing the balances first preserves that distinction and makes the combined result transparent.