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Calculus-Based Model Construction · Tutorial 1469 of 1495

Reservoir Storage Under Rainfall And Release

Learn to convert catchment rainfall into inflow, balance it against release, and interpret capacity violations in a storage model.

Advanced 9 min read

What You'll Learn

  • Convert rainfall intensity over a catchment into a runoff volume rate using area and a runoff coefficient.
  • Write a reservoir storage balance with consistent units and signs.
  • Calculate storage over constant and piecewise operating intervals.
  • Use cumulative net inflow to locate when a reservoir reaches capacity.
  • Distinguish an unconstrained forecast from a physical model that includes overflow.
  • Check storage bounds and identify infeasible release plans.

From Rainfall to Reservoir Storage

A reservoir stores water, but rainfall is usually measured as a depth over a catchment rather than as a volume entering the reservoir. To predict storage, first convert that rainfall into a water inflow rate. Then compare inflow with the controlled release rate. Their difference determines whether reservoir volume increases or decreases.

This model is useful for planning: it can show how much water is expected to be stored over a period and whether a proposed operating plan would exceed the reservoir’s capacity. The result is only as reliable as its assumptions. Here we treat runoff as entering the reservoir at the time it is generated, and we neglect evaporation, seepage, and delays in runoff reaching the reservoir.

Converting Rainfall into a Volume Rate

Let \(i(t)\) be rainfall intensity in millimetres per hour over a catchment of area \(A\) square kilometres. Let \(c\) be the runoff coefficient, a dimensionless fraction representing the portion of rainfall that runs off the catchment and reaches the reservoir. In this model, \(0\le c\le 1\). The remaining rainfall is not counted as reservoir inflow.

One millimetre of rain over one square kilometre has volume \(1000\) cubic metres: \(1\) millimetre is \(0.001\) metres, and \(1\) square kilometre is \(1{,}000{,}000\) square metres. Thus the runoff inflow rate \(R(t)\), in cubic metres per hour, is

$$ R(t)=1000cA i(t). $$

This conversion assumes rainfall intensity is uniform over the stated catchment area, or that \(i(t)\) is a suitable average for that area. If the reservoir itself receives direct rainfall, that contribution must be accounted for separately; it is not included in this catchment formula.

Definition: The runoff inflow rate \(R(t)\) is the rate at which catchment runoff enters the reservoir, measured in volume per unit time. Under the assumptions above, \(R(t)=1000cA i(t)\), with \(i(t)\) in millimetres per hour and \(A\) in square kilometres.

The Storage Balance

Let \(V(t)\) be reservoir volume in cubic metres at time \(t\), measured in hours. Let \(q(t)\) be the controlled release rate, also in cubic metres per hour. A positive release carries water out of the reservoir. If runoff and release are the only water transfers being modelled, the net rate of change of storage is inflow minus release.

Model: Before accounting for overflow, reservoir storage satisfies $$ \frac{dV}{dt}=R(t)-q(t)=1000cA i(t)-q(t), $$ with an initial condition \(V(0)=V_0\). The model applies over intervals where the rates are specified and the stated assumptions are reasonable.

A positive value of \(R(t)-q(t)\) means storage is increasing; a negative value means it is decreasing. The initial condition \(V_0\) is essential: rates determine the change in storage, not the starting amount. For continuous rates, integrating the balance from time \(0\) to \(t\) gives

$$ V(t)=V_0+\int_0^t \bigl(R(s)-q(s)\bigr)\,ds. $$

The integral accumulates net inflow over time. Its units are cubic metres because a rate in cubic metres per hour is integrated with respect to hours. For constant rates over a time interval of length \(\Delta t\), the storage change is simply \((R-q)\Delta t\).

Why the Balance Gives the Storage Change

The balance follows from conservation of water. Over a short time interval of length \(\Delta t\), the water entering is approximately \(R(t)\Delta t\), and the water released is approximately \(q(t)\Delta t\). The change in stored volume is therefore approximately their difference:

$$ V(t+\Delta t)-V(t) \approx \bigl(R(t)-q(t)\bigr)\Delta t. $$

Divide by \(\Delta t\) and take the limit as the interval shrinks to zero. When the rates and storage are differentiable as needed for this limit, the result is the differential balance:

$$ \lim_{\Delta t\to 0} \frac{V(t+\Delta t)-V(t)}{\Delta t} = R(t)-q(t), \qquad\text{so}\qquad \frac{dV}{dt}=R(t)-q(t). $$

Integrating this equation recovers the storage formula. By the Fundamental Theorem of Calculus, the derivative of the right-hand side \(V_0+\int_0^t(R(s)-q(s))\,ds\) is \(R(t)-q(t)\), and its value at \(t=0\) is \(V_0\). Thus it has both the required rate of change and the required initial storage.

Worked Example: Constant Rainfall and Release

Worked Example: Constant Rainfall and Release

A catchment has area \(A=2.5\) square kilometres and runoff coefficient \(c=0.4\). Rain falls uniformly at \(12\) millimetres per hour for \(6\) hours. The controlled release is \(8500\) cubic metres per hour. Initial storage is \(1{,}200{,}000\) cubic metres, and capacity is \(1{,}500{,}000\) cubic metres. Find the predicted storage after six hours and check capacity.

First convert rainfall to runoff inflow:

$$ R=1000(0.4)(2.5)(12) =1000(1)(12) =12{,}000\ \text{m}^3/\text{h}. $$

The net rate is inflow minus release:

$$ R-q=12{,}000-8{,}500 =3{,}500\ \text{m}^3/\text{h}. $$

Over six hours, the net storage increase is

$$ (3{,}500\ \text{m}^3/\text{h})(6\ \text{h}) =21{,}000\ \text{m}^3. $$

Therefore the predicted final storage is

$$ V(6)=1{,}200{,}000+21{,}000 =1{,}221{,}000\ \text{m}^3. $$

Check: with constant rates, the model predicts \(V(t)=1{,}200{,}000+3{,}500t\), so its derivative is \(3{,}500\ \text{m}^3/\text{h}=R-q\), as required. Since storage increases throughout this interval, its maximum is the final value \(1{,}221{,}000\ \text{m}^3\). This is less than the capacity \(1{,}500{,}000\ \text{m}^3\), so the forecast stays within capacity.

Piecewise Rates and Operating Regimes

Rainfall and release often change at known times. In that case, calculate the net rate on each interval and carry the ending storage of one interval forward as the initial storage of the next. The storage remains continuous at a change in rate: a rate can change suddenly, but that change does not instantaneously add or remove a finite volume of water.

Worked Example: Rainfall and Release Change at a Set Time

A \(1.5\)-square-kilometre catchment has \(c=0.6\). For the first two hours, rainfall intensity is \(8\) millimetres per hour and release is \(4000\) cubic metres per hour. For the next three hours, rainfall intensity is \(3\) millimetres per hour and release is \(3500\) cubic metres per hour. Initial storage is \(80{,}000\) cubic metres, and capacity is \(100{,}000\) cubic metres.

The rainfall conversion factor is

$$ 1000cA=1000(0.6)(1.5)=900. $$

For the first interval, the inflow rate is \(900(8)=7200\ \text{m}^3/\text{h}\). The net rate is \(7200-4000=3200\ \text{m}^3/\text{h}\), so the storage increase is \(3200(2)=6400\ \text{m}^3\). Storage at hour \(2\) is therefore

$$ V(2)=80{,}000+6{,}400=86{,}400\ \text{m}^3. $$

For the second interval, the inflow rate is \(900(3)=2700\ \text{m}^3/\text{h}\). The net rate is \(2700-3500=-800\ \text{m}^3/\text{h}\). Over three hours the change is \((-800)(3)=-2400\ \text{m}^3\), giving

$$ V(5)=86{,}400-2{,}400=84{,}000\ \text{m}^3. $$

Check each interval: storage rises from \(80{,}000\) to \(86{,}400\ \text{m}^3\) in the first interval, then falls to \(84{,}000\ \text{m}^3\) in the second. The maximum is \(86{,}400\ \text{m}^3\), which is below the \(100{,}000\ \text{m}^3\) capacity. The endpoint calculation alone would not always identify the maximum; here the signs of the net rates show that the switch time is the maximum.

Capacity Checks and Overflow

Let \(V_{\max}\) denote the reservoir’s physical capacity. A forecast that uses only \(dV/dt=R-q\) should be checked over the whole time interval, not just at its endpoint. For constant net rates, storage is monotone on each interval, so checking the interval endpoints is sufficient. With changing rates, a maximum may also occur where the net rate changes sign. If the unconstrained forecast exceeds \(V_{\max}\), it signals that the proposed rates cannot describe storage without some additional mechanism, such as overflow or a changed release policy.

For a model that includes overflow, let \(s(t)\) be the spill rate in cubic metres per hour. The balance becomes

$$ \frac{dV}{dt}=R(t)-q(t)-s(t). $$

In a simple capacity-limited model, spill is zero while storage is below capacity. At capacity, if runoff exceeds the planned release, spill must remove the excess to keep storage from rising above \(V_{\max}\). This assumes the spillway can pass the required rate. If it cannot, or if operators change release in response to conditions, those rules must be included explicitly. Capacity is a constraint on a physically meaningful prediction, not a term that can be ignored when a calculation exceeds it.

Worked Example: Finding When Capacity Is Reached

A \(4\)-square-kilometre catchment has runoff coefficient \(0.5\), with constant rainfall intensity \(20\) millimetres per hour for three hours. Release is \(25{,}000\) cubic metres per hour. The reservoir starts at \(90{,}000\) cubic metres and has capacity \(125{,}000\) cubic metres. Determine whether the unconstrained forecast exceeds capacity and, if a spillway can pass the excess, find the spill volume over the three hours.

The runoff rate is

$$ R=1000(0.5)(4)(20) =2000(20) =40{,}000\ \text{m}^3/\text{h}. $$

The net rate before spill is \(40{,}000-25{,}000=15{,}000\ \text{m}^3/\text{h}\). The unconstrained storage forecast is \(V(t)=90{,}000+15{,}000t\). It reaches capacity when

$$ 90{,}000+15{,}000t=125{,}000, \qquad t=\frac{35{,}000}{15{,}000} =\frac{7}{3}\ \text{h}. $$

Because \(7/3\) hours is less than three hours, the unconstrained forecast exceeds capacity. It would predict \(90{,}000+15{,}000(3)=135{,}000\ \text{m}^3\) at hour \(3\), which is \(10{,}000\ \text{m}^3\) above capacity. If the spillway holds storage at capacity after it is reached, it must pass the net inflow of \(15{,}000\ \text{m}^3/\text{h}\) for the remaining \(3-7/3=2/3\) hour. Thus the spill volume is

$$ (15{,}000\ \text{m}^3/\text{h})(2/3\ \text{h}) =10{,}000\ \text{m}^3. $$

Check the water balance over the full interval: total runoff is \(40{,}000(3)=120{,}000\ \text{m}^3\); total release is \(25{,}000(3)=75{,}000\ \text{m}^3\); and spill is \(10{,}000\ \text{m}^3\). Final storage is \(90{,}000+120{,}000-75{,}000-10{,}000=125{,}000\ \text{m}^3\), exactly the capacity.

Common Errors and Physical Checks

A common unit error is to multiply rainfall intensity by area without converting either millimetres to metres or square kilometres to square metres. For example, writing \(12(2.5)=30\) as the inflow rate for \(12\) millimetres per hour over \(2.5\) square kilometres is wrong: the units are not cubic metres per hour, and it omits the runoff fraction. With \(c=0.4\), the correct calculation is \(1000(0.4)(2.5)(12)=12{,}000\ \text{m}^3/\text{h}\).

Another error is reversing the sign of release. A release is an outflow, so it is subtracted: \(dV/dt=R-q\), not \(R+q\). Check the sign against the situation: if \(R=0\) and \(q>0\), the model should predict decreasing storage. Also check that release is feasible. If a model predicts negative storage, the prescribed release cannot continue as stated; a physically realistic model must limit release or specify what happens when water is exhausted.

Capacity takeaway: First calculate the storage predicted by the stated inflow and release rates. Then compare its maximum over the time interval with \(V_{\max}\). A forecast above capacity requires a model of spill or changed operation; simply reporting the excessive volume is not a physical reservoir prediction.

The storage balance connects rainfall, runoff, release, and accumulated volume in a single conservation equation. Careful unit conversion establishes the inflow rate; the sign convention determines whether each transfer raises or lowers storage; and the capacity check tells whether the resulting forecast is physically admissible under the assumed operating rules.

Check Your Understanding

Use the storage balance and show your unit and sign reasoning.

  1. A \(2\)-square-kilometre catchment has \(c=0.5\) and rainfall intensity \(6\) millimetres per hour. What runoff inflow rate reaches the reservoir?
  2. If inflow is \(9000\ \text{m}^3/\text{h}\) and release is \(11{,}000\ \text{m}^3/\text{h}\), how does storage change over a two-hour interval with constant rates?
  3. Why must storage at the end of one piecewise-rate interval be used as the initial storage for the next interval?
  4. A constant net inflow is positive and the starting storage is below capacity. What calculation determines whether capacity is reached during a specified time interval?
  5. What does an unconstrained forecast above capacity tell you, and what additional feature could make the physical model meaningful?

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