Net Flow and Gross Flow in Reservoir Models

Why Net Change and Gross Throughput Differ

A reservoir’s volume depends on the balance between what enters and what leaves. If more volume enters than leaves over a time interval, the reservoir’s volume increases. If more leaves than enters, it decreases.

The signed balance is the net flow. The total volume that crosses the reservoir boundary, counting entering and leaving water as positive contributions, is the gross throughput. These quantities answer different questions: net flow describes the change in stored volume, while gross throughput describes the total traffic through the boundary.

For example, if 12 cubic metres enter and 8 cubic metres leave, the reservoir gains 4 cubic metres, but the gross throughput is 20 cubic metres. Counting only the net change would miss most of the movement.

Rates and Sign Convention

Let \(I_j(t)\) denote the rate of the \(j\)th inflow stream and \(O_k(t)\) the rate of the \(k\)th outflow stream. We take each of these physical rates to be nonnegative. Their units might be cubic metres per hour, and time \(t\) might be measured in hours.

With this convention, \(q(t)>0\) means that inflow exceeds outflow at time \(t\), and \(q(t)<0\) means that outflow exceeds inflow. The sign belongs to the balance; the individual physical inflow and outflow rates remain nonnegative.

If there is one inflow and one outflow, the formula simplifies to \(q(t)=I(t)-O(t)\). With several streams, each one must be included in the appropriate sum. The model assumes that the streams listed account for all transfers across the reservoir boundary. If there is an unlisted source or sink, the calculated change will not describe the full volume balance.

Net Volume Change from the Signed Rate

Over a short time interval of length \(\Delta t\), a nearly constant net rate \(q(t)\) produces an approximate volume change \(q(t)\Delta t\). Adding these changes over a partition of the time interval and taking the limit gives the signed-rate integral. When the rates are continuous, the volume function satisfies \(V'(t)=q(t)\), and the Fundamental Theorem of Calculus gives the accumulated change. For piecewise-continuous rates, this accumulation formula still applies, even though the derivative equation need not hold at a jump.

The units provide a useful check. If \(q(t)\) is measured in cubic metres per hour and \(t\) in hours, then \(q(t)\,dt\) has units of cubic metres. Thus its integral is a volume change, not a rate.

Why the Net-Change Formula Works

For continuous rates, the reservoir model says that the instantaneous rate of change of stored volume is \(q(t)\). The Fundamental Theorem of Calculus then relates this derivative to the total change. To see the connection directly, let \(Q(t)=\int_a^t q(s)\,ds\). Since \(q\) is continuous, the Fundamental Theorem gives \(Q'(t)=q(t)\). The model also gives \(V'(t)=q(t)\). Hence \((V-Q)'(t)=0\), so \(V(t)-Q(t)\) is constant. At \(t=a\), \(Q(a)=0\), and therefore \(V(t)-Q(t)=V(a)\). Setting \(t=b\) proves the formula.

For a piecewise-continuous rate, split \([a,b]\) at its finitely many jump points. On each resulting interval, the continuous-rate accumulation formula applies. Adding the changes across those intervals gives the integral over the full interval: both the total volume change and the integral are sums of their corresponding pieces. A single jump point has no effect on the integral. The accumulated volume can therefore have a valid integral formula even when its derivative does not exist at that point.

Gross Throughput: Count Both Directions

To calculate gross throughput, accumulate the inflow and outflow volumes separately, then add them. This is different from integrating their difference. If \(I(t)\) is the total inflow rate and \(O(t)\) is the total outflow rate, the formulas are

Gross throughput counts each transfer across the reservoir boundary in its direction. Consequently, net change alone generally does not determine gross throughput: knowing that the reservoir gained 4 cubic metres does not reveal whether 4 cubic metres entered and none left, or whether 104 entered and 100 left. Separate stream information is needed to calculate gross throughput.

Worked Examples

A Common Mistake

A frequent error is to treat net volume change as gross throughput. In the first example, a mistaken calculation might report gross throughput as \(16.5-6=10.5\text{ m}^3\), because that is the net change. But gross throughput adds both directions: \(16.5+6=22.5\text{ m}^3\). The subtraction answers how much the reservoir’s stored volume changes; the addition answers how much crosses its boundary in total.

Another error is to claim that \(V'(t)=q(t)\) at every time even when the rate jumps. The piecewise example shows why that is not valid at \(t=2\). The correct statement is that the derivative equals the rate on intervals where the rate is continuous; the integral still gives the total change across the full interval.