From Flow Rates to a Changing Water Level
A reservoir can have several flows operating at once: one pipe may add water while one or more outlets remove it. The water level responds not to any one of these rates by itself, but to their combined effect on the volume of water in the reservoir.
To find how quickly a dimension such as water depth is changing, use two ideas in sequence. First, combine the flows to obtain the net volume rate. Then connect that volume rate to the changing dimension through the reservoir’s volume formula.
Choosing Signs for the Flows
Let \(Q_{\text{in},i}\) denote the rate of the \(i\)th inflow and \(Q_{\text{out},j}\) the rate of the \(j\)th outflow. Treat inflow rates as positive contributions to the stored volume and outflow rates as negative contributions. The net rate is
Every rate must be expressed in the same volume-per-time units before you add or subtract it. For example, a rate in liters per minute must be converted if the other rates are in cubic meters per minute.
A positive net rate means the stored volume is increasing at that instant; a negative net rate means it is decreasing. A zero net rate means that the volume is momentarily unchanged, even if water is still entering and leaving.
Why the Volume-Balance Equation Works
Over a time interval, the change in stored water equals the amount that enters minus the amount that leaves. To express this precisely, suppose the flow rates are continuous functions of time. For times \(t_1\) and \(t_2\), the volume balance is
By the Fundamental Theorem of Calculus, differentiating both sides with respect to the upper time endpoint gives the instantaneous balance equation. If the flow rates are constant, the same statement follows directly by multiplying each rate by the elapsed time and then taking the rate of change.
Now suppose the volume is determined by one changing dimension \(x\), so \(V=V(x)\), and \(x=x(t)\). The chain rule gives the connection to the dimension’s rate:
When \(\frac{dV}{dx}\ne 0\), this can be solved for the dimension’s rate:
The denominator describes how much the volume changes per unit change in \(x\). In a tank with a constant horizontal cross-sectional area \(A\), for instance, \(V=Ah\), so \(\frac{dV}{dh}=A\). A net increase of \(A\) cubic units of water per unit time therefore raises the level by one unit of length per unit time.
Worked Example: A Rectangular Pool with Mixed Flow Units
Worked Example: A Rectangular Pool with Mixed Flow Units
A rectangular pool has a fixed base measuring \(10\text{ m}\) by \(6\text{ m}\). Water enters at \(0.80\text{ m}^3/\text{min}\) and \(450\text{ L}/\text{min}\), while an outlet removes water at \(0.50\text{ m}^3/\text{min}\). How quickly is the water depth changing?
Step 1: Convert the flow rates to common units. Since \(1000\text{ L}=1\text{ m}^3\),
Step 2: Find the net volume rate. The inflows total \(0.80+0.45=1.25\text{ m}^3/\text{min}\). Subtract the outflow:
Step 3: Relate volume to depth. The fixed base area is \(10\cdot 6=60\text{ m}^2\), so \(V=60h\). Differentiating with respect to time gives
Therefore,
Check: Multiplying the depth rate by the base area gives \(60\text{ m}^2\cdot 0.0125\text{ m}/\text{min}=0.75\text{ m}^3/\text{min}\), which matches the net inflow. The positive answer is consistent with the pool filling.
Worked Example: Simultaneous Filling and Draining in a Cylindrical Reservoir
Worked Example: Simultaneous Filling and Draining in a Cylindrical Reservoir
An open cylindrical reservoir has a fixed radius of \(4\text{ m}\). Water flows in at \(14\pi\text{ m}^3/\text{min}\), while two outlets remove water at \(5\pi\text{ m}^3/\text{min}\) and \(3\pi\text{ m}^3/\text{min}\). At an instant when the water depth is \(2\text{ m}\), find the rate of change of the depth.
Step 1: Combine the flow rates with signs.
Step 2: Write the water volume in terms of its depth. The reservoir’s radius is fixed; its water forms a cylinder of radius \(4\text{ m}\) and depth \(h\). Thus
Differentiating with respect to time gives
Step 3: Solve for the depth rate.
The depth happens to be \(2\text{ m}\) at the specified instant, but it does not affect this calculation: the horizontal cross-sectional area of a cylinder is constant at every depth.
Check: Substituting the result into the differentiated volume formula gives \(16\pi\cdot\frac{3}{8}=6\pi\text{ m}^3/\text{min}\), the net flow rate. Since this rate is positive, the water level rises.
Worked Example: Water in a Partially Filled Spherical Reservoir
Worked Example: Water in a Partially Filled Spherical Reservoir
A spherical reservoir has a fixed radius \(R=5\text{ m}\). Water enters at \(15\pi\text{ m}^3/\text{min}\) and leaves through an outlet at \(7\pi\text{ m}^3/\text{min}\). When the water depth is \(h=2\text{ m}\), how quickly is the depth increasing?
Step 1: Find the net volume rate.
Step 2: Use the volume of a spherical cap. The tank’s radius \(R\) is fixed. The water is a cap of depth \(h\), not a complete sphere whose radius changes. The cap volume is
For \(R=5\), this becomes \(V=\frac{\pi h^2(15-h)}{3}\). Differentiate with respect to time, keeping \(R\) constant:
Step 3: Substitute the instantaneous depth and net volume rate. At \(h=2\),
Consequently,
Check: Substituting \(\frac{dh}{dt}=\frac12\text{ m}/\text{min}\) into \(\frac{dV}{dt}=16\pi\frac{dh}{dt}\) gives \(8\pi\text{ m}^3/\text{min}\), exactly the net flow. The positive sign means the water depth is increasing.
A Common Sign Error
In the cylindrical-reservoir example, both outlets remove water. A common mistake is to add their rates to the inflow as if all three flows increased the stored volume. That incorrect calculation would give
This treats the outlets as additional sources of water. The correct signed balance subtracts both outlet rates:
A useful check is to ask whether the answer’s sign agrees with the net flow. If total outflow exceeds total inflow, the volume rate and—when increasing depth increases volume—the depth rate should be negative.
A Reliable Sequence of Steps
Convert all flow rates to the same volume-per-time units before combining them.
Add inflows and subtract outflows to obtain \(\frac{dV}{dt}\).
Write \(V\) in terms of the changing dimension, accounting for which dimensions are fixed.
Use the chain rule to connect \(\frac{dV}{dt}\) to the desired rate.
Verify the result in the volume-rate equation and confirm that its sign makes physical sense.
Why Volume Balance Matters
Flow problems often provide several rates, but the requested quantity may be a length rate: a water-level rate, for example. Volume balance supplies the link between those different types of quantities. The geometry determines how much volume corresponds to a change in the dimension, while the signed flow rates determine how quickly that volume is changing.
The volume formula must describe the water that is actually present. A fixed spherical tank that is only partly filled contains a spherical cap, so its water volume depends on cap depth and the tank’s fixed radius. It is not the volume of a complete sphere with a changing radius. Carefully distinguishing the container’s fixed dimensions from the water’s changing dimensions is essential to building a valid model.
Check Your Understanding
For each question, identify the net volume rate before finding any dimension rate.
- A tank receives \(9\text{ m}^3/\text{min}\) and drains at \(4\text{ m}^3/\text{min}\). What is its net volume rate, and is the stored volume increasing or decreasing?
- A rectangular basin has a fixed base area of \(20\text{ m}^2\). Its net volume rate is \(-3\text{ m}^3/\text{min}\). Find the rate of change of its water depth.
- A cylindrical tank of radius \(3\text{ m}\) receives \(12\pi\text{ m}^3/\text{min}\) and drains at \(3\pi\text{ m}^3/\text{min}\). Find the rate of change of its water depth.
- Why must an outlet rate be subtracted when calculating the net rate of stored volume?
- In a partially filled spherical reservoir, which radius is fixed, and what geometric shape describes the water?
