Introduction
A cylindrical tank has the same horizontal cross-sectional area at every height. That simple geometry makes it possible to model the work of pumping water by considering thin horizontal layers: each layer has the same volume for a given thickness, but layers at different heights travel different distances to the outlet.
The central modeling decision is to measure each layer’s lifting distance from its own height to the outlet. Water near the bottom must be lifted farther than water near the top. A partially filled tank contains fewer layers to pump, so the lower integration bound stays at the bottom while the upper bound is the water depth, not necessarily the tank height.
Coordinates and Assumptions
Let a vertical cylindrical tank have radius \(R\) and height \(H\). Measure vertical position \(y\) upward from the bottom of the tank, so the bottom is at \(y=0\) and the rim is at \(y=H\). If the water depth is \(h\), the water occupies heights from \(y=0\) to \(y=h\), where \(0\le h\le H\).
Let the outlet be at elevation \(a\), measured from the same bottom reference. If it is \(s\) metres above the rim, then \(a=H+s\). We assume the water has constant density, gravity is constant, and friction, pipe losses, and changes in kinetic energy are negligible. Work here means the positive energy supplied to raise the water to the outlet.
The horizontal cross-sectional area of the tank is \(A=\pi R^2\). A thin layer of water between heights \(y\) and \(y+dy\) therefore has volume \(dV=A\,dy=\pi R^2\,dy\). Its weight is \(\gamma\,dV\), and, when the outlet is above it, its lifting distance is \(a-y\).
The Work Integral for a Cylindrical Tank
For the usual pumping model, the outlet is at or above the water surface, so \(a\ge h\). Every water layer must then be lifted to the outlet. Multiplying the layer’s weight by its lifting distance gives the work for that layer:
Adding the work for all layers from the bottom to the water surface gives the model.
The integral form keeps the geometry and the changing lifting distance visible. The evaluated formula is convenient for calculation. In either form, the tank height \(H\) determines the rim elevation, but the water depth \(h\) determines the upper integration bound.
Why the Integral Gives the Work
The formula follows by approximating the water with thin horizontal layers. To see why, divide the water depth into \(n\) small intervals of width \(\Delta y=h/n\). A layer near height \(y_i\) has volume approximately \(\pi R^2\Delta y\), weight approximately \(\gamma\pi R^2\Delta y\), and lifting distance approximately \(a-y_i\). Its work is therefore approximately \(\gamma\pi R^2(a-y_i)\Delta y\).
Adding those layer-work estimates gives a sum. As the layers become thinner, the sum approaches the definite integral over the water’s full depth.
This is the same variable-force principle used for work: small amounts of work are added over the distance or, here, over the vertical positions of the water. The force associated with a layer is its weight; the distance is specific to that layer. Evaluating the integral gives
Worked Example: A Full Tank Pumped Above the Rim
Worked Example: A Full Tank Pumped Above the Rim
A cylindrical tank has radius \(2\ \text{m}\) and height \(5\ \text{m}\). It is full, and water is pumped to an outlet \(2\ \text{m}\) above the rim. Find the work in joules.
The water depth is \(h=5\ \text{m}\). The outlet elevation from the bottom is \(a=5+2=7\ \text{m}\). The cross-sectional area is
A layer at height \(y\) travels \(7-y\) metres. Using \(\gamma=9800\ \text{N}/\text{m}^3\), set up and evaluate the work integral:
Check: the bracketed term has units of square metres, so \(\gamma\), area, and this term give \((\text{N}/\text{m}^3)(\text{m}^2)(\text{m}^2)=\text{N}\cdot\text{m}=\text{J}\). The lifting distance is \(7\ \text{m}\) at the bottom and \(2\ \text{m}\) at the top, as expected. The answer is \(882000\pi\ \text{J}\).
Partially Filled Tanks
If a tank is only partly filled, the water still begins at the bottom, but its top is at \(y=h\), with \(h<H\). There is no water in the interval from \(h\) to the rim, so that interval contributes no work. The outlet elevation is still measured from the bottom, and the distance for a layer at \(y\) remains \(a-y\), provided the outlet is at or above the water surface.
Worked Example: A Partially Filled Tank with an Outlet Above the Rim
A tank has radius \(1.5\ \text{m}\), height \(4\ \text{m}\), and water depth \(2.5\ \text{m}\). Its outlet is \(1\ \text{m}\) above the rim. Find the work required to pump out the water.
The outlet elevation is \(a=4+1=5\ \text{m}\), and the area is \(\pi(1.5)^2=2.25\pi\ \text{m}^2\). Since the water depth is \(h=2.5\ \text{m}\), the work is
Check: the lifting distance ranges from \(5\ \text{m}\) for the bottom layer to \(5-2.5=2.5\ \text{m}\) for the water surface. Both are positive, so the model’s outlet condition is satisfied. The tank is not full, and the integration correctly stops at \(2.5\ \text{m}\), not at the \(4\ \text{m}\) rim.
Worked Example: An Outlet at a Specified Elevation
Worked Example: An Outlet at a Specified Elevation
A cylindrical tank has radius \(1\ \text{m}\) and height \(4\ \text{m}\). It contains water to a depth of \(1.5\ \text{m}\), and the outlet is at elevation \(2.5\ \text{m}\) above the tank bottom. Find the work.
The outlet is above the water surface because \(2.5>1.5\), even though it is below the tank rim. A layer at height \(y\) must be raised \(2.5-y\) metres. The area is \(\pi(1)^2=\pi\ \text{m}^2\). Thus
Check: the bottom layer is raised \(2.5\ \text{m}\), and the top layer \(1.5\ \text{m}\), so every layer travels a positive distance. The elevation \(2.5\ \text{m}\) is measured from the bottom; using the tank height \(4\ \text{m}\) as the outlet elevation would describe a different setup.
When the Outlet Is Below the Water Surface
The standard pumping integral above assumes \(a\ge h\). If \(a<h\), the expression \(a-y\) becomes negative for layers above the outlet. Those layers are not lifted up to that outlet; in an ordinary open tank they may instead flow out by gravity. A negative contribution should not be included as though the pump were doing negative work on that water.
For a problem that asks only for pump work on water below an outlet at height \(a\), and assumes that water above the outlet drains by gravity, the pumped portion is \(0\le y\le a\), provided \(0\le a<h\). Its work is \(\gamma\pi R^2\int_0^a(a-y)\,dy\). This describes only the portion that must be lifted; a real installation may have additional requirements not included in this idealized model. Always identify which water is being pumped before choosing bounds.
A Common Mistake: Giving Every Layer the Same Distance
A common error is to multiply the total water weight by the bottom layer’s lifting distance. Consider a full cylindrical tank with radius \(1\ \text{m}\), height \(3\ \text{m}\), and outlet at elevation \(4\ \text{m}\). The water volume is \(3\pi\ \text{m}^3\), and its weight is \(9800(3\pi)=29400\pi\ \text{N}\). Incorrectly treating all the water as if it were at the bottom gives
But the water at height \(y\) travels \(4-y\) metres, not \(4\) metres. The correct calculation is
Check: the bottom layer travels \(4\ \text{m}\), but the top layer travels only \(1\ \text{m}\). Assigning the bottom layer’s distance to all the water overestimates the work. The integral accounts for the changing distance layer by layer.
Using the Model Reliably
Before evaluating an integral, write down what each height means and verify that the outlet lies at or above the water being pumped. The following sequence keeps the geometry and the work model consistent.
Measure \(y=0\) at the tank bottom. Express the rim, water surface, and outlet elevations from that same reference.
For thickness \(dy\), the cylindrical layer has volume \(\pi R^2dy\) and weight \(\gamma\pi R^2dy\).
A layer at height \(y\) lifted to outlet elevation \(a\) travels \(a-y\).
For water filling from the bottom to depth \(h\), integrate from \(0\) to \(h\), whether the tank is full or partly filled.
The lifting distance should be nonnegative over the pumped water, and the final units should be joules.
Why Cylindrical Geometry Simplifies the Calculation
The radius of a cylinder does not change with height, so every layer of the same thickness has the same volume and weight. The only changing factor in the work integral is the layer’s lifting distance. The integral therefore reduces to the cross-sectional area multiplied by an integral of a linear function.
This structure is useful both for modeling and for checking answers. Increasing the radius increases the work in proportion to \(R^2\), because it increases each layer’s volume. Raising the outlet increases the distance every layer must travel. Increasing the water depth adds more layers and also changes the average lifting distance. These relationships help detect an answer that is inconsistent with the physical setup.
Check Your Understanding
For each question, identify the relevant heights and explain how they determine the work model.
- A tank has height \(6\ \text{m}\), water depth \(4\ \text{m}\), and an outlet \(1\ \text{m}\) above the rim. What are the outlet elevation \(a\) and the correct integration bounds?
- For a layer at height \(y\) and outlet elevation \(a\), what is the layer’s lifting distance? What condition makes this distance nonnegative for all water in a tank filled to depth \(h\)?
- A cylindrical tank has radius \(R\), water depth \(h\), and an outlet at elevation \(a\ge h\). Write the work integral using specific weight \(\gamma\).
- Why is the upper integration bound the water depth rather than the tank height when the tank is partially filled?
- What modeling issue arises if the outlet elevation is below the water surface, and why should negative layer work not automatically be included?
