Filling a Conical Tank

Modelling Water in a Conical Tank

A conical tank does not have the same horizontal cross-sectional area at every height. Near the tip, a small increase in water depth adds relatively little volume. Higher up, the water is wider, so the same increase in depth adds more volume. This changing geometry is why the water level does not generally rise at a constant rate when water flows in at a constant rate.

To find the rate at which the level rises, we first express the water’s volume in terms of its depth. The two essential ingredients are the cone-volume formula and the relationship between the water’s radius and depth.

Relating Radius and Depth

Let \(h\) be the depth of the water and \(r\) the radius of its circular surface. The tank’s full dimensions \(H\) and \(R\) are constant, while \(h\), \(r\), and the water volume \(V\) can change with time.

A vertical cross-section through the tank’s axis contains two similar triangles: the triangle formed by the water and the tip, and the triangle formed by the whole tank. Their corresponding radius-to-height ratios are equal:

Solving for \(r\) gives the water-surface radius as a function of depth:

This relation matters because the radius of the water is not generally the tank’s full radius. It grows in proportion to the water depth.

Writing Volume in Terms of Depth

The volume of a cone with base radius \(r\) and height \(h\) is \(V=\frac{1}{3}\pi r^2h\). Substituting the similar-triangle relation eliminates \(r\), leaving a formula involving the changing depth and the tank’s fixed dimensions:

The cubic dependence on \(h\) reflects the fact that increasing the depth also increases the water’s radius. Both dimensions of each horizontal cross-section change together.

Deriving the Water-Depth Rate

Suppose the net inflow rate at time \(t\) is \(q=\frac{dV}{dt}\), measured in volume per unit time. The tank dimensions \(R\) and \(H\) are fixed. Differentiating the volume relation with respect to time therefore differentiates \(h^3\) by the chain rule:

For \(h>0\), we can solve for the rate of change of depth:

The division by \(h^2\) means this solved form requires positive water depth. The undivided differentiated equation remains \(\frac{dV}{dt}=\frac{\pi R^2}{H^2}h^2\frac{dh}{dt}\). At exactly \(h=0\), that equation cannot give a finite depth rate for a positive inflow: the idealized model has zero water depth at which to evaluate the solved formula.

Worked Examples

Why the Level Rises More Slowly at Greater Depth

For fixed tank dimensions and a fixed positive inflow, the depth-rate formula contains \(h^2\) in the denominator. As the depth increases, the water surface becomes wider. Each additional unit of depth then corresponds to more added volume, so the same volume entering per unit time produces a smaller increase in depth per unit time.

This conclusion concerns the instantaneous rate at a particular depth; it does not say that the depth itself decreases. With positive inflow and \(h>0\), the formula gives \(\frac{dh}{dt}>0\). If \(q\) is a net outflow instead, then \(q<0\), and the depth rate is negative.

A Common Mistake: Using the Tank’s Full Radius

The tank’s top radius \(R\) is not the radius of the water at every depth. For example, in a tank with \(H=6\) m and \(R=3\) m, at water depth \(h=2\) m, similar triangles give:

not \(3\) m. Treating the water as if it had the tank’s full radius at every depth would incorrectly use \(V=3\pi h\). Differentiating that mistaken expression gives \(\frac{dV}{dt}=3\pi\frac{dh}{dt}\). With inflow \(2\text{ m}^3/\text{min}\), this would imply the wrong result \(\frac{dh}{dt}=\frac{2}{3\pi}\text{ m/min}\).

The correction is to use the changing water radius, \(r=(R/H)h\), before differentiating. The correct volume relation for these dimensions is \(V=\frac{\pi}{12}h^3\), which gives the correct rate \(\frac{dh}{dt}=\frac{2}{\pi}\text{ m/min}\) at \(h=2\) m. Substituting that rate into \(\frac{dV}{dt}=\frac{\pi}{4}h^2\frac{dh}{dt}\) yields \(2\text{ m}^3/\text{min}\), as required.

Check the Model Before Using the Formula

The rate formula applies when the water forms a cone similar to the fixed tank and the given volume rate is the net rate of change of water volume. Before calculating, identify the tank’s full dimensions, the water depth at the instant, and the inflow or net outflow rate. Keep length and volume units consistent, and remember that the formula uses a positive water depth.