Pumping Water From Conical Tanks

Why a Conical Tank Requires a Changing Slice

A conical tank narrows toward its bottom, so its horizontal cross-sections do not all have the same radius. Water near the bottom occupies a smaller area than water near the top. To calculate the work needed to pump water out, we must account for both this changing volume and the changing distance each layer travels.

Assume the tank is an upright right circular cone with its vertex at the bottom. Let $H$ be its height and $R$ the radius of its circular top. Measure vertical position $y$ upward from the vertex. Suppose the water reaches height $h$, and the water is pumped to an outlet at height $L$, measured from the same vertex. In the usual setup, the outlet is at or above the water surface, so $L\ge h$.

Model assumptions: The water has constant weight density $\gamma$, the tank’s dimensions are $H$ and $R$, and each layer is lifted vertically to the outlet. Weight density is weight per unit volume, such as pounds per cubic foot or newtons per cubic metre.

Radius and Volume of a Horizontal Layer

At height $y$, a horizontal cross-section of the tank is a circle. Its radius is smaller than $R$ unless it is at the top. The side of the cone forms similar triangles: the small triangle from the vertex to height $y$ has radius $r(y)$ and height $y$; the full triangle has radius $R$ and height $H$. Therefore, their corresponding side ratios are equal.

\frac{r(y)}{y}=\frac{R}{H}, \qquad r(y)=\frac{R}{H}y.

A thin layer at height $y$, with thickness $\Delta y$, is approximately a cylinder of radius $r(y)$ and height $\Delta y$. Its volume is approximately its circular cross-sectional area times its thickness. In the limit as the thickness becomes infinitesimal, the layer’s volume is $dV=\pi r(y)^2\,dy$.

dV = \pi\left(\frac{R}{H}y\right)^2dy = \frac{\pi R^2}{H^2}y^2\,dy.

Multiplying by the water’s weight density gives the layer’s weight. Since the outlet is at height $L$, a layer at height $y$ must be lifted a distance $L-y$. The work on that layer is its weight multiplied by this distance.

Key model: For water occupying heights from $0$ to $h$, the total work is the integral of the work on the horizontal layers. Each layer has weight $\gamma\,dV$ and is lifted $L-y$, so $$ W=\int_0^h \gamma(L-y)\frac{\pi R^2}{H^2}y^2\,dy. $$

Why the Integral Gives the Total Work

Divide the water into thin horizontal layers. A layer near height $y$ has volume approximately $\pi r(y)^2\Delta y$, and its weight is approximately $\gamma\pi r(y)^2\Delta y$. Lifting it to the outlet requires approximately that weight times $L-y$. Adding the work for all layers gives a sum of the form

\sum_i \gamma\pi r(y_i)^2(L-y_i)\Delta y.

As the layer thicknesses shrink, these sums approach the definite integral of the work per unit height over the water’s full depth. Substituting the similar-triangle expression for $r(y)$ gives the stated integral. This also explains why the cross-sectional radius and the lifting distance must both be evaluated at the layer’s own height.

The integrand can be integrated directly. For constant $L$, an antiderivative of $Ly^2-y^3$ is $Ly^3/3-y^4/4$. Thus the work has the useful form

W = \frac{\gamma\pi R^2}{H^2} \left( \frac{Lh^3}{3}-\frac{h^4}{4} \right).

This formula applies when the cone points down, the water occupies $0\le y\le h$, and the outlet height $L$ is at least $h$. For an outlet at the rim, set $L=H$. The formula is also valid for an empty tank, $h=0$, giving zero work.

Worked Applications

Worked Example: Pumping Water to the Rim

An upright conical tank is $12$ ft high and has a top radius of $4$ ft. It contains water to a depth of $9$ ft. The outlet is at the rim. Use the water’s weight density $\gamma=62.4$ lb/ft$^3$ to find the work required to pump out the water.

Here $H=12$ ft, $R=4$ ft, $h=9$ ft, and $L=12$ ft. The layer radius and volume are

r(y)=\frac{4}{12}y=\frac{y}{3}\text{ ft}, \qquad dV=\pi\left(\frac{y}{3}\right)^2dy =\frac{\pi y^2}{9}\,dy.

A layer at height $y$ travels $12-y$ ft. Therefore,

\begin{aligned} W &=62.4\int_0^9(12-y)\frac{\pi y^2}{9}\,dy\\ &=\frac{104\pi}{15}\int_0^9(12y^2-y^3)\,dy\\ &=\frac{104\pi}{15} \left[ 4y^3-\frac{y^4}{4} \right]_0^9\\ &=\frac{104\pi}{15} \left( 4(9^3)-\frac{9^4}{4} \right)\\ &=\frac{104\pi}{15} \left( 2916-\frac{6561}{4} \right)\\ &=\frac{104\pi}{15}\cdot\frac{5103}{4} =\frac{44226\pi}{5}\text{ ft}\cdot\text{lb}. \end{aligned}

This is approximately $27{,}788$ ft·lb. To check the integration, differentiate the antiderivative $4y^3-y^4/4$: its derivative is $12y^2-y^3=y^2(12-y)$, the integrand inside the constant factor. The work is positive, as expected for lifting the water to the rim.

Worked Example: An Outlet Above the Tank

A conical tank is $10$ ft high with a top radius of $3$ ft. The water depth is $6$ ft, and the outlet is $2$ ft above the rim. Find the work using $\gamma=62.4$ lb/ft$^3$.

The outlet height measured from the vertex is $L=10+2=12$ ft. Similar triangles give $r(y)=3y/10$, so

dV=\pi\left(\frac{3y}{10}\right)^2dy =\frac{9\pi}{100}y^2\,dy.

Each layer travels $12-y$ ft. Substitution into the work integral gives

\begin{aligned} W &=62.4\cdot\frac{9\pi}{100} \int_0^6 y^2(12-y)\,dy\\ &=5.616\pi \left[ 4y^3-\frac{y^4}{4} \right]_0^6\\ &=5.616\pi \left( 4(6^3)-\frac{6^4}{4} \right)\\ &=5.616\pi(864-324)\\ &=5.616\cdot540\pi\\ &=\frac{75816\pi}{25}\text{ ft}\cdot\text{lb}\\ &\approx 9{,}527\text{ ft}\cdot\text{lb}. \end{aligned}

The antiderivative check is $d(4y^3-y^4/4)/dy=12y^2-y^3$, which agrees with $y^2(12-y)$. The outlet is higher than in the first example, so the layers must be lifted farther. This is reflected by using $L=12$, not the tank height $10$, in the distance $L-y$.

Worked Example: Work in SI Units

A conical tank has height $8$ m and top radius $2$ m. It contains water to a depth of $5$ m, and the outlet is at the top. Use water’s weight density $\gamma=9800$ N/m$^3$. Find the work in joules.

Here $L=H=8$ m, and similar triangles give $r(y)=y/4$ m. Thus the layer volume is $\pi y^2/16\,dy$, and

\begin{aligned} W &=9800\int_0^5(8-y)\frac{\pi y^2}{16}\,dy\\ &=612.5\pi\int_0^5(8y^2-y^3)\,dy\\ &=612.5\pi \left[ \frac{8y^3}{3}-\frac{y^4}{4} \right]_0^5\\ &=612.5\pi \left( \frac{1000}{3}-\frac{625}{4} \right)\\ &=612.5\pi\cdot\frac{2125}{12}\\ &=\frac{2603125\pi}{24}\text{ J}\\ &\approx340{,}748\text{ J}. \end{aligned}

To check, differentiating $8y^3/3-y^4/4$ gives $8y^2-y^3=y^2(8-y)$, as required. For units, the factor $9800$ N/m$^3$, multiplied by a volume in m$^3$, gives newtons; multiplying by a lifting distance in metres gives N·m, or joules.

A Common Modeling Error

A frequent mistake is to use the layer’s height $y$ as its lifting distance. In the second example, that mistake would use $y$ rather than $12-y$, producing

5.616\pi\int_0^6 y^3\,dy = 5.616\pi\left[\frac{y^4}{4}\right]_0^6 = 1819.584\pi\text{ ft}\cdot\text{lb}.

This is not the work to pump the water to the outlet: a layer at $y=6$ ft must travel $12-6=6$ ft, while a layer near the bottom must travel nearly $12$ ft. The correct distance is the outlet height minus the layer height, $L-y$. The corrected calculation uses $\int_0^6 y^2(12-y)\,dy=540$, giving $3032.64\pi$ ft·lb.

Another check is to ensure the radius changes with height. In this tank, $r(6)=3(6)/10=1.8$ ft, not the full top radius $3$ ft. Treating every layer as if it had the top radius would model a cylinder rather than a cone.

Check Your Understanding

For each question, identify the geometry and distance that belong in the work model.

A cone has height $H$ and top radius $R$. What is the radius of a horizontal cross-section at height $y$ above its vertex?
Write the volume $dV$ of a thin horizontal water layer at height $y$, with thickness $dy$.
If the outlet is at height $L$, what distance must a layer at height $y$ be lifted?
For a tank with $H=8$ m, $R=2$ m, water depth $h=5$ m, and an outlet at the rim, write the work integral using weight density $\gamma$.
Explain why replacing $L-y$ with $y$ gives the wrong pumping distance.

Pumping Water From Conical Tanks

What You'll Learn

Why a Conical Tank Requires a Changing Slice

Radius and Volume of a Horizontal Layer

Why the Integral Gives the Total Work