Why Layers Must Be Modeled Separately
A tank may contain several liquids arranged in horizontal layers. If the liquids have different densities, equal volumes do not have equal weights. In addition, liquid at different heights must be lifted different distances to reach an outlet. The work calculation must account for both effects.
The useful strategy is to divide the liquid into its density layers. For each layer, use its own density and integrate the weight of a thin horizontal slice times the slice’s lift distance. Then add the work for all layers.
We will assume the liquid is at rest, the outlet is at a fixed height, gravity is constant, and losses such as friction and turbulence are ignored. The result is the work required to lift the liquid itself.
Set Up the Vertical Model
Let \(y\) measure height vertically from the bottom of the tank, with \(y=0\) at the bottom. Let the outlet be at height \(H\). A thin horizontal slice at height \(y\), of thickness \(dy\), must be lifted a distance \(H-y\), provided \(y\leq H\).
Write \(A(y)\) for the horizontal cross-sectional area of the tank at height \(y\), measured in square metres. A slice of thickness \(dy\) has volume \(A(y)\,dy\). If the slice belongs to a liquid of mass density \(\rho\), its mass is \(\rho A(y)\,dy\), and its weight is \(\rho g A(y)\,dy\), where \(g\) is the acceleration due to gravity. We use \(g=9.8\ \text{m/s}^2\) in the examples.
If a layer’s density is given as weight per unit volume rather than mass per unit volume, use that weight density directly in place of \(\rho_i g\). Do not multiply a weight density by \(g\) a second time.
Why the Layer Integrals Give the Work
The formula follows from the definition of work as force multiplied by distance for a small displacement, extended to a varying force by integration. The force needed to lift one thin slice is its weight, and the distance through which it moves is \(H-y\).
At height \(y\), a slice of thickness \(dy\) has volume \(A(y)\,dy\).
Within layer \(i\), the slice’s weight is approximately \(\rho_i g A(y)\,dy\).
The slice rises \(H-y\), so its small work contribution is approximately \(\rho_i g A(y)(H-y)\,dy\).
Summing the slice contributions over the layer and taking the limit gives its integral. Adding the separate layers gives the total work.
More formally, partition a layer interval \([a_i,b_i]\) into thin subintervals. In a subinterval of width \(\Delta y\) near \(y_j\), the slice’s work is approximately \(\rho_i g A(y_j)(H-y_j)\Delta y\). The sum of these terms is a Riemann sum for the integral of \(\rho_i g A(y)(H-y)\). In the limit as the widths tend to zero, the sum becomes the layer’s work integral. The layers occupy disjoint height intervals, so their work contributions add.
This reasoning also explains why the density and lift distance both remain inside the model: density can change at an interface, and lift distance changes continuously with height.
Worked Example: Two Layers in a Cylindrical Tank
Worked Example: Two Layers in a Cylindrical Tank
A vertical cylindrical tank has radius \(0.5\ \text{m}\). Its lower liquid layer occupies \(0\leq y\leq1\) and has mass density \(1000\ \text{kg/m}^3\). The upper layer occupies \(1\leq y\leq2\) and has density \(800\ \text{kg/m}^3\). Pump all the liquid to an outlet at height \(H=2.5\ \text{m}\). Find the work, using \(g=9.8\ \text{m/s}^2\).
The tank has constant cross-sectional area
Use a separate integral for each layer:
An antiderivative of \(2.5-y\) is \(2.5y-\frac{y^2}{2}\). For the lower layer,
For the upper layer,
Therefore,
Check the antiderivative: \(\frac{d}{dy}(2.5y-y^2/2)=2.5-y\), the original integrand. The two evaluated integrals are \(2\) and \(1\), so the density-weighted sum is \(2000+800=2800\), as used. The units are \((\text{kg/m}^3)(\text{m/s}^2)(\text{m}^2)(\text{m})(\text{m})=\text{J}\).
Worked Example: A Tank With Changing Cross-Sectional Area
Worked Example: A Tank With Changing Cross-Sectional Area
A tank has horizontal cross-sectional area \(A(y)=2+y\ \text{m}^2\) for \(0\leq y\leq2\). A lower layer of density \(900\ \text{kg/m}^3\) fills \(0\leq y\leq1\), and an upper layer of density \(1200\ \text{kg/m}^3\) fills \(1\leq y\leq2\). The outlet is at \(H=3\ \text{m}\). Find the work using \(g=9.8\ \text{m/s}^2\).
The work is
First expand the integrand:
An antiderivative is \(F(y)=6y+\frac{y^2}{2}-\frac{y^3}{3}\). For the lower layer,
For the upper layer, evaluate both endpoints:
Substituting these values,
Check: \(F'(y)=6+y-y^2=(2+y)(3-y)\), so the antiderivative is correct. The layer contributions before multiplication by \(g\) are \(900(37/6)=5550\) and \(1200(31/6)=6200\); their sum is \(11750\). The area, slice thickness, and lift distance contribute \(\text{m}^4\), which combined with density and \(g\) gives joules.
Worked Example: Three Layers With Different Lift Distances
Worked Example: Three Layers With Different Lift Distances
A tank has constant cross-sectional area \(3\ \text{m}^2\). Its layers occupy \(0\leq y\leq0.5\), \(0.5\leq y\leq1.5\), and \(1.5\leq y\leq2.5\), with densities \(1000\), \(850\), and \(700\ \text{kg/m}^3\), respectively. Pump the liquid to an outlet at \(H=3\ \text{m}\), using \(g=9.8\ \text{m/s}^2\).
Set up the three contributions separately:
An antiderivative of \(3-y\) is \(G(y)=3y-y^2/2\), and \(G'(y)=3-y\). The first integral is
The second is
The third is
Thus,
Check: the three lift integrals are \(11/8\), \(2\), and \(1\), and their density-weighted sum is \(1375+1700+700=3775\). Also, \(G'(y)=3-y\). Each term in the original integral has the required units, so the total is in joules.
Common Pitfall: Averaging the Densities First
A tempting shortcut is to replace different layer densities with their arithmetic mean and then model the whole tank as one liquid. In general, this is wrong because the layers do not have the same average lift distance.
For the two-layer cylindrical example, the layers have equal height, so their arithmetic mean density is \((1000+800)/2=900\ \text{kg/m}^3\). Treating the entire height from \(0\) to \(2\) as density \(900\) would give
But the correct result is \(6860\pi\ \text{J}\). The lower, denser layer has a greater average lift distance than the upper layer, so a simple average of the densities does not preserve the work. The correction is to retain the separate density and height interval for each layer, as in the layer-by-layer integral.
Modeling Checks
Before evaluating a layered-liquid work integral, check that the model matches the physical arrangement:
- Measure \(y\) from a clearly specified reference height, and express the outlet height \(H\) in the same coordinate.
- Use the correct density for every layer’s interval; an interface is a point where the density formula changes.
- Use \(H-y\) for the lift distance when the liquid is discharged at the fixed outlet height \(H\).
- Use \(A(y)\) at the slice height, not the tank’s total volume or surface area.
- Keep units consistent, and distinguish mass density from weight density.
If a layer extends above the outlet, the expression \(H-y\) would be negative there and would not represent the positive distance required to raise that liquid to the outlet. The model must then be adjusted to describe the actual pumping path and direction. For the standard setup in this tutorial, all liquid lies at or below the outlet.
Check Your Understanding
For each question, identify how the density layers and lift distances enter the work model.
- A tank has two layers with different densities and a fixed outlet above both. Why should the total work be written as the sum of two integrals?
- In the formula \(W_i=\int_{a_i}^{b_i}\rho_i g A(y)(H-y)\,dy\), what does each factor represent?
- A tank’s cross-sectional area doubles with height. Which part of the work integrand must reflect that change?
- What is wrong with replacing two equal-height layers by a liquid whose density is the arithmetic mean of their densities?
- If a problem gives weight density in newtons per cubic metre, should it be multiplied by \(g\)? Explain.
