Pumping Liquid from a Conical Tank

Why the Cone’s Shape Matters

To pump liquid from a tank, we add the work needed to lift many thin slices of liquid. In a cylindrical tank, each horizontal slice has the same radius. In a conical tank, the slice radius changes with height, so its volume and weight change as well.

The key geometric step is to express the radius of a slice in terms of its height. Similar triangles provide that relationship. Once the slice radius is known, its volume, weight, and lifting distance determine the work integral.

Set Up the Geometry

Consider an upright circular cone with its vertex at the bottom. Let the cone have height \(H\) and top radius \(R\). Measure the vertical coordinate \(y\) upward from the vertex, so the top rim is at \(y=H\). Suppose the liquid fills the cone from the bottom to height \(L\), where \(0\le L\le H\).

At height \(y\), a horizontal slice has radius \(r(y)\). The small cone below that slice and the full tank are similar. Their corresponding radius-to-height ratios are equal:

Therefore the slice radius is

This formula fits the geometry: at the vertex, \(r(0)=0\), and at the rim, \(r(H)=R\). The slice is a thin cylinder of radius \(r(y)\) and thickness \(dy\). Its volume is approximately its circular cross-sectional area times its thickness; in the differential model,

From Slice Weight to Work

Let \(\gamma\) denote the liquid’s weight density, measured in newtons per cubic metre. A slice of volume \(dV\) has weight \(dF=\gamma\,dV\). If the liquid has mass density \(\rho\), then \(\gamma=\rho g\), where \(g\) is gravitational acceleration. Using weight density directly keeps the force calculation clear.

Suppose the liquid is pumped vertically to an outlet at height \(D\), measured from the cone’s vertex. For pumping all the liquid upward to an outlet at or above the liquid surface, take \(D\ge L\). A slice at height \(y\) travels a distance \(D-y\). Its work is its weight multiplied by that lifting distance:

Substituting the slice volume gives the work contribution from height \(y\). Adding contributions from the bottom to the liquid surface produces the integral:

The variable \(y\) must be used consistently: it identifies both the slice’s height and, through \(r(y)\), its radius. The work integral has units of joules because \(\gamma\,dV\) is a force in newtons, \(D-y\) is a distance in metres, and a newton-metre is a joule.

Why the Integral Gives the Total Work

The integral follows from adding the work on thin slices. Divide the liquid into horizontal layers. For a layer near height \(y\) with thickness \(\Delta y\), its approximate volume is \(\pi r(y)^2\Delta y\), its approximate weight is \(\gamma\pi r(y)^2\Delta y\), and its lifting distance is approximately \(D-y\). Thus its approximate work is

Adding these approximations over all layers gives a Riemann sum. As the layer thicknesses tend to zero, the sums approach the integral of the work per unit height. Substituting \(r(y)=(R/H)y\) gives the conical-tank formula above. This reasoning requires the tank geometry and liquid weight density to be as specified, and the lifting distance to be measured to the stated outlet.

The integral can also be evaluated for a constant weight density:

This evaluated expression applies when \(D\) and \(L\) are measured from the same vertex and \(D\ge L\). For pumping to the rim, set \(D=H\).

Common Mistake: Reversing the Similar-Triangle Ratio

The ratio must match corresponding lengths: radius to height in the small cone equals top radius to full height in the tank. A common error is to reverse the full-cone ratio. For example, if \(H=6\) m and \(R=2\) m, the incorrect expression \(r(y)=(H/R)y=3y\) predicts \(r(6)=18\) m at the rim. But the rim radius is \(2\) m, so that cannot describe the tank.

The correct ratio is \(r(y)/y=R/H=2/6\), giving \(r(y)=y/3\) and \(r(6)=2\) m as required. Another error is to lift every slice the full tank height. A slice at height \(y\) travels only \(D-y\), not \(D\). Using the same distance for every slice ignores its starting height and overestimates the work.

Using and Interpreting the Model

The geometry, liquid amount, and outlet location each enter the model in a different place. The ratio \(R/H\) determines how quickly the cone widens; the upper limit \(L\) records how much liquid is present; and \(D-y\) gives the lift for each slice. Check all three before evaluating the integral.

For pumping to the rim, the liquid’s upper slices travel a shorter distance than its lower slices. For an outlet above the rim, \(D\) increases, so every slice travels farther. The integral accounts for these differences slice by slice rather than treating the liquid as one object lifted a single distance.