Why Pumping Work Requires a Slice-by-Slice Model
A liquid in a tank does not all travel the same distance when it is pumped to an outlet. A slice near the outlet may be lifted only a short distance, while a slice farther below must be lifted farther. Since work depends on both the force and the distance through which that force acts, one total force multiplied by one distance generally does not model the whole job.
The useful approach is to divide the liquid into thin horizontal slices. For each slice, determine its volume, weight, and vertical lifting distance. Adding the work for all slices leads to a definite integral.
The model below assumes the liquid is pumped slowly enough that changes in kinetic energy can be neglected. It accounts for work against gravity only; it does not include friction, pump inefficiency, or other losses.
Definitions and the General Slice Model
Let \(y\) measure vertical elevation upward from a chosen horizontal reference level. Suppose the liquid occupies elevations from \(y=a\) to \(y=b\), and the outlet is at elevation \(Y\), with \(Y\ge b\). Let \(A(y)\) be the area of the tank’s horizontal cross-section at elevation \(y\). A thin slice between \(y\) and \(y+dy\) has approximate volume \(A(y)\,dy\), so its weight is \(\gamma A(y)\,dy\). The slice must be lifted a vertical distance \(Y-y\).
Here \(W\) is work in joules. The formula requires the slice dimensions and lifting distances to be measured in compatible units, such as metres, and assumes the outlet is at or above the top of the liquid. The factor \(Y-y\) is a distance, not a signed displacement: under this assumption it is nonnegative for every slice.
Using Depth Below the Liquid Surface
It is often convenient to measure depth downward from the liquid surface instead of measuring elevation upward from a reference level. Let \(d\) denote a slice’s depth below the surface, and let the liquid depth be \(D\). Let \(s\) be the outlet’s height above the liquid surface. A slice at depth \(d\) is lifted \(d+s\): it must travel up \(d\) metres to reach the surface and then another \(s\) metres to reach the outlet. If \(A(d)\) denotes the horizontal area at depth \(d\), the model becomes:
The two formulas describe the same geometry. If the surface has elevation \(y_s\), then a slice at depth \(d\) has elevation \(y=y_s-d\). If the outlet has elevation \(Y=y_s+s\), then \(Y-y=(y_s+s)-(y_s-d)=s+d\). This relation makes the choice of coordinates explicit and prevents a frequent source of errors: mixing a depth measured downward with an elevation measured upward.
Why the Integral Gives the Total Work
The integral follows by adding the work on thin slices and taking a limit. The reasoning is the same variable-force principle used for other lifting problems: a small amount of liquid with weight \(F\), lifted through distance \(L\), requires work approximately \(FL\).
Divide the elevation interval \([a,b]\) into thin subintervals. In a subinterval of width \(\Delta y_i\), choose a sample elevation \(y_i^*\).
The slice volume is approximately \(A(y_i^*)\Delta y_i\), its weight is approximately \(\gamma A(y_i^*)\Delta y_i\), and its lifting distance is \(Y-y_i^*\). Its work is therefore approximately \(\gamma A(y_i^*)(Y-y_i^*)\Delta y_i\).
The total approximate work is the sum of the estimated work for all subintervals.
As the largest subinterval width approaches zero, the sums approach the definite integral of the work per unit vertical thickness.
In symbols, the sum in Step 3 is \(\sum_i \gamma A(y_i^*)(Y-y_i^*)\Delta y_i\). By the definition of a definite integral, its limit is \(\int_a^b \gamma A(y)(Y-y)\,dy\), which proves the stated model when the integrand is integrable. The depth-based form follows by describing the same slices with \(y=y_s-d\), or by applying the same slice argument directly using depth.
Worked Examples
Worked Example: Pumping Water from a Rectangular Tank
A rectangular tank has a horizontal base measuring \(4\) m by \(3\) m. It contains water to a depth of \(2\) m, and the outlet is \(1\) m above the water surface. Use water’s weight density \(\gamma=9800\ \mathrm{N/m^3}\). Find the work required to pump all the water to the outlet.
Every horizontal slice has area \(A(d)=4\cdot3=12\ \mathrm{m^2}\). At depth \(d\), the slice’s lifting distance is \(d+1\) m. Thus
At the upper limit, \(\frac{2^2}{2}+2=2+2=4\); at the lower limit the expression is \(0\). Therefore
Check: the liquid volume is \(12\cdot2=24\ \mathrm{m^3}\), so its total weight is \(9800\cdot24=235200\ \mathrm{N}\). The average lifting distance is \(470400/235200=2\) m. This is consistent with the slices occupying depths from \(0\) to \(2\) m and each being lifted an additional \(1\) m above the surface.
Worked Example: Using Elevation Above the Tank Bottom
A liquid has weight density \(8000\ \mathrm{N/m^3}\). It fills a tank from the bottom at elevation \(y=0\) to \(y=2\) m. The horizontal area at elevation \(y\) is \(A(y)=4-y\) square metres, and the outlet is at elevation \(Y=2.5\) m. Find the pumping work.
A slice at elevation \(y\) has lifting distance \(2.5-y\), so
Expand the product before integrating:
Hence
The work is therefore
Check: the liquid volume is \(\int_0^2(4-y)\,dy=[4y-y^2/2]_0^2=8-2=6\ \mathrm{m^3}\). Its weight is \(8000\cdot6=48000\ \mathrm{N}\). Dividing the calculated work by this weight gives an average lifting distance of \((232000/3)/48000=29/18\) m, which lies between the slices’ lifting distances of \(0.5\) m and \(2.5\) m.
Worked Example: Depth Coordinates with a Varying Cross-Section
A tank contains water to depth \(2\) m. At depth \(d\) below the surface, its horizontal area is \(A(d)=2(4-d)\ \mathrm{m^2}\), and the outlet is \(0.5\) m above the surface. Find the work needed to pump out the water, using \(\gamma=9800\ \mathrm{N/m^3}\).
The lift for a slice at depth \(d\) is \(d+0.5\), so the work integral is
First expand the factors:
Integrating and evaluating gives
Thus
Check: the volume is \(\int_0^2 2(4-d)\,dd=2[4d-d^2/2]_0^2=2(8-2)=12\ \mathrm{m^3}\), giving total weight \(117600\ \mathrm{N}\). The work divided by this weight is \((490000/3)/117600=25/18\) m, consistent with slices lifted between \(0.5\) m and \(2.5\) m.
A Common Coordinate Mistake
Suppose the liquid surface is the depth reference, the outlet is \(1\) m above that surface, and a slice has depth \(d\). A mistaken setup might use \(1-d\) for its lifting distance. That subtracts the depth from the outlet’s height above the surface, even though the slice is already below the surface. For the rectangular tank in the first example, this error gives
This result is not physically reasonable: the bottom slice, at \(d=2\), would be assigned a negative lifting distance. The correct distance is \(1+d\), so the correct work is \(470400\) J. A useful check is to identify the actual elevation of a slice and the outlet, then subtract the slice’s elevation from the outlet’s elevation. With depth \(d\) below the surface and outlet height \(s\) above it, that difference is \(s+d\), not \(s-d\).
Why the Reference Level Matters
The reference level determines the coordinate values, but it does not change the physical lifting distance. If elevations are measured from a different horizontal reference, both the slice elevation and outlet elevation shift by the same amount. Their difference \(Y-y\) stays unchanged. Likewise, depth coordinates remain tied to the liquid surface, so the outlet height \(s\) must be measured from that same surface.
Before integrating, identify the liquid’s vertical interval, the outlet’s level, the area of a horizontal slice, the liquid’s weight density, and the slice’s upward travel distance. Those quantities determine the model. The integral then adds the work for all slices rather than treating the liquid as though it were concentrated at one height.
Check Your Understanding
For each question, identify the slice geometry and lifting distance before evaluating or comparing work.
- A tank’s liquid surface is \(3\) m below an outlet. What is the lifting distance for a slice at depth \(d\) below the surface?
- In elevation coordinates, a slice is at \(y\) and the outlet is at \(Y\). State the condition under which \(Y-y\) is the upward lifting distance for every slice.
- Why does a horizontal slice with area \(A(y)\) and thickness \(dy\) have approximate weight \(\gamma A(y)\,dy\)?
- In depth coordinates, what is wrong with using \(s-d\) as the lifting distance when \(s\) is the outlet’s height above the liquid surface?
- For a tank with constant weight density and area \(A(y)\), write the work integral when liquid occupies elevations \(a\) through \(b\) and is pumped to elevation \(Y\).
