Pumping Water From Hemispherical Tanks

Why the Shape of the Tank Matters

A tank’s shape determines how much water is present in a thin horizontal layer at each height. In a hemispherical tank, the layers are small near the bottom and grow wider toward the rim. Consequently, the weight being lifted is not constant from one height to another.

To calculate the work of pumping, we model the water as thin horizontal slices. For each slice, we need its area, its weight, and the vertical distance it travels to the outlet. The geometry gives the area; the outlet elevation gives the lifting distance.

Geometry and Coordinate Choice

Consider a hemispherical tank of radius \(R\), with its flat circular opening at the top. Measure height \(y\) vertically upward from the lowest point of the tank. The bottom is at \(y=0\), and the rim is at \(y=R\). Assume the tank is filled with water to height \(h\), where \(0\le h\le R\).

A horizontal cross-section at height \(y\) is a circle. Let \(r(y)\) be its radius. In a vertical cross-section through the center of the tank, the sphere’s center is at height \(R\). The Pythagorean relation between \(r(y)\) and the vertical displacement from the center is therefore

Solving for the cross-sectional radius gives \(r(y)^2=R^2-(y-R)^2=2Ry-y^2\). Thus the area \(A(y)\) of the horizontal slice is

The formula matches the geometry at both ends. At the bottom, \(A(0)=0\). At the rim, \(A(R)=\pi R^2\), the area of the opening.

From Slice Geometry to Work

Let \(\gamma\) denote the weight density of the liquid, measured in newtons per cubic metre. For water, we will use \(\gamma=9800\ \text{N/m}^3\). A slice of thickness \(dy\) at height \(y\) has approximate volume \(A(y)\,dy\), so its weight is \(\gamma A(y)\,dy\).

Let the outlet be at height \(H\) above the bottom. A slice at height \(y\) must rise a vertical distance \(H-y\). The work required to lift it is its weight multiplied by that distance. Adding the work for all slices from the bottom to the water surface gives the pumping-work integral.

The integrand has units \((\text{N/m}^3)(\text{m}^2)(\text{m})=\text{N}\). Multiplying by \(dy\), which has units of metres, gives joules, as required for work.

Deriving a General Formula

The integral can be evaluated for any radius \(R\), water depth \(h\), and outlet height \(H\). First expand the product inside the integral:

Integrating term by term from \(0\) to \(h\) yields the following formula. The derivation also shows how each part depends on the water depth and the outlet elevation.

This expression separates the outlet-height contribution from the terms determined by the tank and water depth. Increasing \(H\) increases the distance every slice is lifted, and therefore increases the work.

Why the Integral Represents the Total Work

The work formula follows by dividing the water into thin horizontal layers. A layer at height \(y\), with thickness \(dy\), has volume approximately \(A(y)\,dy\). Its weight is approximately \(\gamma A(y)\,dy\), and lifting it through \(H-y\) metres requires approximately \(\gamma A(y)(H-y)\,dy\) joules.

For a finite partition of the water column, adding these layer contributions gives a Riemann sum. As the layer thicknesses approach zero, the sums approach the definite integral of the continuous work-per-unit-height expression. This limit accounts for every slice from \(y=0\) to \(y=h\), without treating the varying area or lifting distance as constant.

Worked Example: A Partly Filled Tank Pumped to the Rim

Outlet at the Rim Versus Outlet Above It

When the outlet is at the rim, \(H=R\). Substituting that value into the general formula gives a useful simplification:

An outlet above the rim has \(H>R\), so every slice travels farther than it would to the rim. The general formula applies without changing the slice area; only the distance factor changes from \(R-y\) to \(H-y\).

Worked Example: A Partly Filled Tank Pumped Above the Rim

Worked Example: A Full Tank Pumped Above the Rim

A Common Geometry Mistake

A frequent error is to use \(r(y)^2=R^2-y^2\). That relation incorrectly treats \(y=0\) as the height of the sphere’s center. In this model, \(y=0\) is the bottom of the tank and the center is at height \(R\). Using the incorrect expression would give \(A(0)=\pi R^2\), implying that the tank has a full-width layer at its bottom.

The correction is to measure the vertical displacement from the center as \(y-R\). The sphere equation then gives \(r(y)^2=R^2-(y-R)^2=2Ry-y^2\), and correctly yields zero area at the bottom and \(\pi R^2\) at the rim.

Another modeling check is the lifting distance. It is not the outlet height \(H\) itself: a slice already at height \(y\) rises only \(H-y\). If \(H\ge h\), this distance is nonnegative for every water slice. Use consistent units for \(R,h,H\) and \(\gamma\); with metres and \(\text{N/m}^3\), the resulting work is in joules.