Why Density Layers Change the Work Model
A pumping-work calculation must account for how much each slice weighs and how far it must be lifted. When a liquid has the same density throughout, every slice can use one density value. In a stratified liquid, density depends on depth: a slice from one layer may weigh more or less than an equal-volume slice from another.
The essential modeling choice is to assign density according to the slice’s starting depth in the tank. Then calculate the work for each layer and add those contributions. The earlier material on building work integrals for pumping liquids provides the slice-and-distance framework; here, the additional task is to use the right density in each part of the tank.
Setting Up the Layered-Density Model
Measure vertical position \(y\) upward from the tank bottom. Suppose the liquid occupies heights from \(y=a\) to \(y=b\), and the outlet is at height \(H\), with \(H\ge b\). Let \(A(y)\) be the horizontal cross-sectional area of the tank at height \(y\). The thin slice between \(y\) and \(y+dy\) has approximate volume \(A(y)\,dy\).
Mass density, denoted by \(\rho\), is mass per unit volume. If a slice lies in a layer of constant density \(\rho_i\), its mass is approximately \(\rho_i A(y)\,dy\), and its weight is approximately \(\rho_i g A(y)\,dy\), where \(g\) is gravitational acceleration. The slice is lifted a distance \(H-y\) to reach the outlet.
This model assumes the liquid is pumped quasistatically, so the calculation counts the work against gravity and ignores losses such as friction and energy carried away by fluid motion. The liquid’s density is assigned by its original layer in the tank. The formula also assumes the outlet is at or above the top of the liquid, so every slice is lifted upward.
Why the Layer Integrals Add
Consider one layer with density \(\rho_i\). Divide its height interval into thin pieces. For a slice of thickness \(\Delta y\) at a representative height \(y_j\), its volume is approximately \(A(y_j)\Delta y\). Its weight is approximately \(\rho_i g A(y_j)\Delta y\), and its lifting distance is \(H-y_j\). Thus its work is approximately \(\rho_i g A(y_j)(H-y_j)\Delta y\).
Adding the slice estimates over that layer gives a Riemann sum. As the thicknesses approach zero, the sum becomes the layer’s work integral. The work for all the liquid is the sum of the work for all layers because the total work is the sum of the work done on the individual slices.
Adding these integrals across all layer intervals gives the stated formula. A boundary between layers does not need a separate work term: a single height has zero thickness and therefore contributes no volume or work.
Worked Example: Two Layers in a Cylindrical Tank
Worked Example: Two Layers in a Cylindrical Tank
A cylindrical tank has horizontal area \(2\,\mathrm{m^2}\) and contains liquid from the bottom, \(y=0\), to \(y=4\,\mathrm{m}\). The lower layer, from \(0\) to \(2\,\mathrm{m}\), has density \(1000\,\mathrm{kg/m^3}\). The upper layer, from \(2\) to \(4\,\mathrm{m}\), has density \(800\,\mathrm{kg/m^3}\). The outlet is at \(H=5\,\mathrm{m}\). Use \(g=9.8\,\mathrm{m/s^2}\).
For both layers, \(A(y)=2\,\mathrm{m^2}\), and a slice at height \(y\) travels \(5-y\) metres. Set up one integral for each density:
An antiderivative of \(5-y\) is \(5y-y^2/2\), whose derivative is \(5-y\). For the lower layer, the integral is \((10-2)-(0-0)=8\). Its work is \(1000(9.8)(2)(8)=156800\,\mathrm{J}\).
For the upper layer, the integral is \((20-8)-(10-2)=12-8=4\). Its work is \(800(9.8)(2)(4)=62720\,\mathrm{J}\). Therefore, \(W=156800+62720=219520\,\mathrm{J}\).
Check: Differentiating \(5y-y^2/2\) gives the integrand \(5-y\), and evaluating it at each layer’s endpoints gives \(8\) and \(4\), respectively. The units are \((\mathrm{kg/m^3})(\mathrm{m/s^2})(\mathrm{m^2})(\mathrm{m})(\mathrm{m})=\mathrm{J}\), as required.
Worked Example: A Tank Whose Area Changes with Height
Worked Example: A Tank Whose Area Changes with Height
Suppose a tank contains liquid from \(y=0\) to \(y=3\,\mathrm{m}\), and its horizontal area is \(A(y)=(1\,\mathrm{m})y\). Thus, when \(y\) is measured in metres, the numerical area in square metres equals the numerical height. The lower layer, \(0\le y\le1.5\,\mathrm{m}\), has density \(1200\,\mathrm{kg/m^3}\); the upper layer, \(1.5\le y\le3\,\mathrm{m}\), has density \(900\,\mathrm{kg/m^3}\). The outlet is at \(4\,\mathrm{m}\), and take \(g=10\,\mathrm{m/s^2}\).
Using the numerical values of height in metres, the area factor in the integrand is \(y\), and the lifting distance is \(4-y\). Therefore the two contributions are \(12000\int_0^{1.5}y(4-y)\,dy\) and \(9000\int_{1.5}^{3}y(4-y)\,dy\), in joules.
An antiderivative of \(y(4-y)=4y-y^2\) is \(2y^2-y^3/3\), since its derivative is \(4y-y^2\). At \(y=1.5\), this antiderivative equals \(2(1.5)^2-(1.5)^3/3=4.5-1.125=3.375=27/8\). At \(y=3\), it equals \(18-9=9\).
The lower-layer integral is \(27/8\), so its work is \(12000(27/8)=40500\,\mathrm{J}\). The upper-layer integral is \(9-27/8=45/8\), so its work is \(9000(45/8)=50625\,\mathrm{J}\). The total is \(40500+50625=91125\,\mathrm{J}\).
Check: The derivative of \(2y^2-y^3/3\) is \(4y-y^2=y(4-y)\). The evaluated layer integrals are \(27/8\) and \(45/8\), and their weighted contributions add to \(91125\,\mathrm{J}\). The changing area is included through the factor \(y\); replacing it by a constant area would describe a different tank.
Worked Example: Three Layers and a Partially Filled Tank
Worked Example: Three Layers and a Partially Filled Tank
A tank has constant horizontal area \(1.5\,\mathrm{m^2}\). Liquid occupies only the heights from \(y=1\,\mathrm{m}\) to \(y=5\,\mathrm{m}\); the tank below \(1\,\mathrm{m}\) is empty. The layers are \(1\le y\le2\) with density \(1000\,\mathrm{kg/m^3}\), \(2\le y\le4\) with density \(800\,\mathrm{kg/m^3}\), and \(4\le y\le5\) with density \(600\,\mathrm{kg/m^3}\). The outlet is at \(6\,\mathrm{m}\). Take \(g=10\,\mathrm{m/s^2}\).
Use the actual liquid intervals, not the full tank height. The common factor \(gA\) is \(10(1.5)=15\), so $$ W=15\left[ 1000\int_1^2(6-y)\,dy+ 800\int_2^4(6-y)\,dy+ 600\int_4^5(6-y)\,dy \right]. $$ An antiderivative is \(6y-y^2/2\), with derivative \(6-y\).
The first integral is \((12-2)-(6-0.5)=10-5.5=4.5\), giving \(15(1000)(4.5)=67500\,\mathrm{J}\). The second is \((24-8)-(12-2)=16-10=6\), giving \(15(800)(6)=72000\,\mathrm{J}\). The third is \((30-12.5)-(24-8)=17.5-16=1.5\), giving \(15(600)(1.5)=13500\,\mathrm{J}\). Thus \(W=67500+72000+13500=153000\,\mathrm{J}\).
Check: Differentiation confirms the antiderivative, and endpoint evaluation gives \(4.5\), \(6\), and \(1.5\) for the three lifting-distance integrals. Their weighted sum is \(153000\,\mathrm{J}\). No work is included for the empty part of the tank below \(y=1\).
A Common Mistake: Replacing All Layers by an Average Density
In the first example, the two layers have equal thickness and equal area, so their average density by volume is \(900\,\mathrm{kg/m^3}\). That is a valid average for total mass, but using it uniformly to calculate work gives $$ W_{\text{average}} =900(9.8)(2)\int_0^4(5-y)\,dy =17640(12) =211680\,\mathrm{J}. $$ This is not the correct work, which is \(219520\,\mathrm{J}\).
The problem is that the lower layer must be lifted farther, on average, than the upper layer. Work depends on both the weight of a slice and its lifting distance. An average density that preserves total mass need not preserve the sum of weight times distance. The correction is to retain the separate density in each layer integral.
Check Your Understanding
For each question, identify the layer intervals and the relevant density before evaluating an integral.
- A tank has constant area \(3\,\mathrm{m^2}\), liquid layers on \(0\le y\le1\) and \(1\le y\le3\), and an outlet at \(4\,\mathrm{m}\). What factors belong in each layer’s work integrand?
- Why must the density assigned to a slice be the density at its original height in the tank?
- If the tank’s horizontal area doubles with height, which part of the work integrand accounts for that change?
- For a tank containing liquid only from \(y=2\) to \(y=5\), why should the work integral not include the interval from \(0\) to \(2\)?
- Why can a volume-average density give the right total mass but the wrong pumping work?
