Why Horizontal Layers Model Pumping Work
When a tank is pumped out, the fluid does not all travel the same distance. A layer near the bottom must be lifted farther than a layer near the top. The tank’s shape also affects how much fluid lies in each thin layer. To calculate the work, we account for both effects: the weight of each layer and the distance that layer must be lifted.
The method applies to tanks of many shapes. At each height, describe the horizontal cross-sectional area of the tank. A thin layer at that height has volume approximately equal to its area multiplied by its thickness. Its weight follows from the fluid’s weight density, and its lifting distance follows from the height of the outlet.
Building the Integral from One Layer
Consider a thin horizontal layer centered at height \(y\), with thickness \(\Delta y\). If the cross-sectional area changes little across this thin layer, its volume is approximately \(A(y)\Delta y\). Multiplying by the weight density gives an approximate layer weight of \(\gamma A(y)\Delta y\).
The layer must be lifted from height \(y\) to height \(s\), so its lifting distance is \(s-y\). The work for the layer is approximately its weight multiplied by this distance:
Summing these contributions over all layers gives an approximation to the total work. As the layers become thinner, the sum approaches the definite integral in the definition. This reasoning requires a fixed outlet height, constant fluid weight density, and a cross-sectional area that is integrable over the fluid’s height range. If the area is continuous or piecewise continuous, the usual Riemann-sum interpretation applies.
The units provide an immediate check. If \(A(y)\) is measured in square metres, \(dy\) in metres, and \(\gamma\) in newtons per cubic metre, then \(\gamma A(y)\,dy\) is in newtons. Multiplying by the distance \(s-y\), measured in metres, gives newton-metres, or joules.
Worked Example: A Rectangular Tank
Worked Example: A Rectangular Tank
A rectangular tank is \(4\) m long and \(3\) m wide. It contains water to a depth of \(4\) m, and the outlet is \(2\) m above the tank’s \(4\)-m-high rim. Use the water weight density \(\gamma=9800\ \text{N/m}^3\). Find the work required to pump out the water.
Set the bottom of the tank at \(y=0\). The water occupies \(0\le y\le4\), and the outlet is at \(s=4+2=6\) m. Each horizontal cross section is a rectangle, so its area is
A layer at height \(y\) must be lifted \(6-y\) metres. Therefore,
An antiderivative of \(6-y\) is \(6y-\frac{y^2}{2}\). Evaluating at the endpoints,
Check the antiderivative: \(\frac{d}{dy}(6y-\frac{y^2}{2})=6-y\). The units are \((\text{N/m}^3)(\text{m}^2)(\text{m})(\text{m})=\text{N}\cdot\text{m}=\text{J}\), as required. The result is positive because every layer is lifted to a point above it.
Tank Geometry Determines the Layer Area
The lifting-distance factor \(s-y\) is determined by the layer and the outlet heights. The tank’s shape enters through \(A(y)\). If the horizontal cross section changes with height, the area must be expressed as a function of \(y\) before the work integral is evaluated.
Worked Example: A Triangular-Prism Tank
Worked Example: A Triangular-Prism Tank
A tank is \(5\) m long. Its triangular end has height \(3\) m and top width \(4\) m, with the triangle’s point at the bottom. The tank is full of water, and water is pumped to an outlet \(1\) m above the top. Find the pumping work using \(\gamma=9800\ \text{N/m}^3\).
Let \(y=0\) at the bottom point and \(y=3\) at the top. By similar triangles, the width at height \(y\) is proportional to \(y\). In particular, the ratio of width to height is \(4/3\), so the width is \(\frac{4}{3}y\) metres. A horizontal slice is a rectangle whose length is \(5\) m, giving
The outlet height is \(s=3+1=4\) m, and the water occupies \(0\le y\le3\). Hence
An antiderivative of \(4y-y^2\) is \(2y^2-\frac{y^3}{3}\). Therefore,
The derivative check is \(\frac{d}{dy}(2y^2-\frac{y^3}{3})=4y-y^2\). As an additional check, the water volume is the triangular end area times the tank length: \((\frac12)(4)(3)(5)=30\ \text{m}^3\), so its weight is \(294{,}000\) N. The work \(588{,}000\) J corresponds to an average lifting distance of \(588{,}000/294{,}000=2\) m, consistent with layers spanning heights \(0\) to \(3\) m and being lifted to height \(4\) m.
Worked Example: A Tank with Sloping Sides
Worked Example: A Tank with Sloping Sides
A \(3\)-m-long tank has a trapezoidal vertical end. Its width is \(2\) m at the bottom and increases linearly to \(5\) m at a height of \(4\) m. Water fills the tank to height \(3\) m, and the outlet is at height \(5\) m. Find the work required to pump out the water, using \(\gamma=9800\ \text{N/m}^3\).
At height \(y\), the width has increased from its bottom value by a fraction \(y/4\) of the total increase \(5-2=3\) m. Thus the width is \(2+\frac34y\) metres. Multiplying by the tank length gives the horizontal area:
The outlet is at \(s=5\) m, and the fluid occupies \(0\le y\le3\). The work integral is
Expand the integrand before integrating:
An antiderivative is \(30y+\frac{21}{8}y^2-\frac34y^3\). Evaluating,
Differentiating \(30y+\frac{21}{8}y^2-\frac34y^3\) gives \(30+\frac{21}{4}y-\frac94y^2\), the expanded integrand. The volume is \(\int_0^3(6+\frac94y)\,dy=18+\frac{81}{8}=\frac{225}{8}\ \text{m}^3\), so the water weighs \(9800(\frac{225}{8})=275{,}625\) N. The calculated work divided by this weight is about \(3.32\) m, which is a plausible average lifting distance for water lifted from heights between \(0\) and \(3\) m to an outlet at \(5\) m.
A Common Error: Measuring from the Water Surface
The lifting distance is measured from each layer to the outlet, not from the layer to the water surface. For the rectangular-tank example, the outlet is at height \(6\) m. A layer at height \(y\) therefore travels \(6-y\) metres.
A mistaken calculation might use \(4-y\), the distance to the water surface, and write
This is only half the correct result because it ignores the extra \(2\) m that every layer must travel from the rim to the outlet. The correct distance is \(6-y\), giving \(1{,}881{,}600\) J. A reliable setup names the vertical coordinate origin and the outlet height before writing the distance factor.
Setting Up a Pumping Model Reliably
Set a clear height origin, commonly the tank bottom, and use it consistently for fluid levels and outlet height.
Express the slice area \(A(y)\) using the tank’s dimensions at height \(y\).
A slice has weight \(\gamma A(y)\,dy\) and, for an outlet at height \(s\), travels \(s-y\).
Use the actual lower and upper fluid heights as bounds, then check units and whether the computed work is positive.
The method’s strength is that it separates the geometry from the lifting. The tank shape determines the area function; the outlet location determines the distance function. Combining those quantities produces the work contribution of each layer and, through integration, the total pumping work.
Check Your Understanding
For each question, identify the slice geometry and lifting distance before evaluating or interpreting an integral.
- A tank has constant horizontal area \(A_0\), fluid height \(h\), and outlet height \(s>h\). Write the work integral using weight density \(\gamma\).
- A tank is \(6\) m long. Its horizontal width at height \(y\) is \(1+y\) metres, and fluid occupies \(0\le y\le2\). If the outlet is at \(4\) m, write the pumping-work integral using \(\gamma\).
- Why is the lifting distance for a layer at height \(y\) equal to \(s-y\), rather than \(s-h\), when the fluid surface is at height \(h\)?
- For a tank with horizontal area \(A(y)=10\ \text{m}^2\) throughout \(0\le y\le3\), explain how the work model changes if the outlet height increases while the fluid and tank remain unchanged.
- What units result from multiplying weight density, horizontal area, slice thickness, and lifting distance?
