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Work and Variable Forces · Tutorial 463 of 998

Pumping Liquid from a Cylindrical Tank

Build and evaluate pumping-work integrals for cylindrical tanks by tracking the height and weight of each liquid slice.

Intermediate 9 min read

What You'll Learn

  • Find the volume and weight of a thin horizontal slice in a cylindrical tank.
  • Set the distance each slice moves using a consistent vertical coordinate.
  • Compute work when the outlet is above the rim or below it.
  • Interpret signed work when some liquid must descend to the outlet.
  • Check pumping-work calculations using units and antiderivatives.

Why the outlet height matters

A tank can hold the same amount of liquid and require different amounts of work to empty, depending on where the liquid must go. Liquid near the bottom usually has farther to travel than liquid near the top. If the outlet is below the rim, some liquid may even be higher than the outlet and move downward on its way out.

For a vertical cylindrical tank, every horizontal slice has the same area. This makes the slice volume simple to find; the key is to use the correct distance from each slice to the outlet. We will calculate the ideal work against gravity, assuming the liquid has constant density and ignoring friction, turbulence, and other losses.

Model assumptions: The tank is vertical with constant radius, the liquid has uniform weight density, and each slice is transferred to an outlet at a fixed height. Work is measured against gravity. A positive contribution means lifting a slice; a negative contribution means the slice descends.

Describe a horizontal slice

Set \(y=0\) at the bottom of the tank and measure height upward. Let \(R\) be the tank radius, \(a\) the liquid depth, and \(H\) the outlet height in the same coordinate system. The tank rim is at height \(h\), so the outlet is above the rim when \(H>h\) and below it when \(H<h\).

A thin horizontal slice at height \(y\), with thickness \(dy\), has circular cross-sectional area \(\pi R^2\). Its volume is area times thickness. If the liquid’s weight density is \(\gamma\), measured in newtons per cubic metre, the slice’s weight is its volume multiplied by \(\gamma\).

$$ dV=\pi R^2\,dy, \qquad dF=\gamma\,dV=\gamma\pi R^2\,dy. $$

The slice must reach height \(H\). Its signed vertical displacement is \(H-y\): this is positive when the outlet is above the slice and negative when the outlet is below it. Work against gravity is weight times signed vertical displacement.

Slice-work formula: For a slice at height \(y\), the signed work against gravity is \(dW=\gamma\pi R^2(H-y)\,dy\). The sign records whether that slice rises or descends on its way to the outlet.

Build the integral

Adding the work contributions from all slices between the bottom \(y=0\) and the liquid surface \(y=a\) gives the ideal signed work:

$$ W=\int_0^a \gamma\pi R^2(H-y)\,dy. $$

Here \(R\), \(H\), and \(\gamma\) are constant. Integrating gives a convenient formula for a partially or fully filled cylindrical tank:

$$ W=\gamma\pi R^2\left[Hy-\frac{y^2}{2}\right]_0^a =\gamma\pi R^2\left(Ha-\frac{a^2}{2}\right). $$

For water, a commonly used approximate weight density is \(\gamma=9800\ \text{N/m}^3\). If a problem gives mass density \(\rho\) instead, convert it to weight density using \(\gamma=\rho g\), where \(g\) is gravitational acceleration. Keep all heights and radii in metres when using \(\gamma\) in newtons per cubic metre.

Why the slice integral gives the work

The integral is not merely a memorized formula. Divide the liquid into thin horizontal layers. For a layer of thickness \(\Delta y_i\) near height \(y_i\), its approximate volume is \(\pi R^2\Delta y_i\), its weight is approximately \(\gamma\pi R^2\Delta y_i\), and its signed displacement is approximately \(H-y_i\).

$$ \Delta W_i\approx \gamma\pi R^2(H-y_i)\Delta y_i. $$

Summing these approximations over the layers estimates the total work. As the layer thicknesses shrink, the sums approach the definite integral of the slice-work expression. This is the same principle used to build work integrals for variable forces: add the small contributions and take the limit.

$$ W=\lim_{\max \Delta y_i\to 0} \sum_i \gamma\pi R^2(H-y_i)\Delta y_i =\int_0^a\gamma\pi R^2(H-y)\,dy. $$

Worked Example: Pumping a full tank to an outlet above the rim

Worked Example: Pumping a full tank to an outlet above the rim

A cylindrical tank has radius \(2\ \text{m}\) and height \(5\ \text{m}\). It is full of water. Find the ideal work against gravity to pump all the water to an outlet \(1\ \text{m}\) above the rim. Use \(\gamma=9800\ \text{N/m}^3\).

The liquid depth is \(a=5\ \text{m}\). Since the bottom is \(y=0\), the rim is at height \(5\ \text{m}\), and the outlet is at \(H=6\ \text{m}\). Each slice has area \(\pi R^2=\pi(2)^2=4\pi\ \text{m}^2\). Therefore,

$$ W=\int_0^5 9800(4\pi)(6-y)\,dy =39200\pi\int_0^5(6-y)\,dy. $$

Evaluate the integral:

$$ W=39200\pi\left[6y-\frac{y^2}{2}\right]_0^5 =39200\pi\left(30-\frac{25}{2}\right) =39200\pi(17.5) =686000\pi\ \text{J}. $$

Numerically, \(W\approx 2{,}155{,}132\ \text{J}\), or about \(2.16\times10^6\ \text{J}\) to three significant figures. The units check: weight density times area times distance times slice thickness has units \((\text{N/m}^3)(\text{m}^2)(\text{m})(\text{m})=\text{N}\cdot\text{m}=\text{J}\).

Check: Differentiating the antiderivative \(6y-y^2/2\) gives \(6-y\), the required distance factor. The outlet is above every slice, so all contributions are positive, as the computed work is.

Worked Example: An outlet below the rim but above the liquid surface

Worked Example: An outlet below the rim but above the liquid surface

A cylindrical tank has radius \(1\ \text{m}\), rim height \(4\ \text{m}\), and water depth \(2\ \text{m}\). The outlet is at height \(3\ \text{m}\), so it is below the rim but above the liquid surface. Find the ideal work. Use \(\gamma=9800\ \text{N/m}^3\).

Here \(R=1\), \(a=2\), and \(H=3\). Every slice lies below the outlet. Its work contribution is positive, although the outlet is lower than the rim.

$$ W=\int_0^2 9800\pi(3-y)\,dy =9800\pi\left[3y-\frac{y^2}{2}\right]_0^2. $$

Substitute the endpoints:

$$ W=9800\pi\left((6-2)-0\right) =39200\pi\ \text{J} \approx 123{,}150\ \text{J}. $$

Thus the work is approximately \(1.23\times10^5\ \text{J}\) to three significant figures. The outlet being below the rim does not, by itself, mean any liquid descends: the liquid surface is only at \(2\ \text{m}\), while the outlet is at \(3\ \text{m}\).

Check: The antiderivative differentiates to \(3-y\), and \(3-y>0\) throughout \(0\le y\le2\). The positive result is consistent with every slice being lifted.

Worked Example: An outlet below the liquid surface

Worked Example: An outlet below the liquid surface

A cylindrical tank of radius \(1.5\ \text{m}\) contains water to a depth of \(3\ \text{m}\). Its rim is at height \(4\ \text{m}\), and the outlet is at height \(H=2\ \text{m}\), below both the rim and the liquid surface. Find the ideal signed work against gravity, using \(\gamma=9800\ \text{N/m}^3\).

The cross-sectional area is \(\pi(1.5)^2=2.25\pi\ \text{m}^2\). Slices below \(y=2\) rise to the outlet; slices above \(y=2\) descend. The signed integral includes both effects:

$$ W=9800(2.25\pi)\int_0^3(2-y)\,dy =22050\pi\left[2y-\frac{y^2}{2}\right]_0^3. $$

Evaluating gives

$$ W=22050\pi\left(6-\frac{9}{2}\right) =33075\pi\ \text{J} \approx 103{,}910\ \text{J}. $$

The positive total means that, for this ideal transfer, the work required to lift the lower slices exceeds the gravitational energy released by the descending upper slices.

Check by splitting the integral: The contribution from \(0\le y\le2\) is \(22050\pi[2y-y^2/2]_0^2=44100\pi\ \text{J}\). The contribution from \(2\le y\le3\) is \(22050\pi[2y-y^2/2]_2^3=-11025\pi\ \text{J}\). Their sum is \(33075\pi\ \text{J}\), matching the result above.

Interpreting signed work

When \(H<a\), the outlet is below the liquid surface, so the integrand \(H-y\) changes sign at \(y=H\), provided \(H\) is above the tank bottom. Slices below the outlet are lifted and contribute positive work; slices above the outlet descend and contribute negative work. The integral gives the net ideal work against gravity for the whole transfer.

A negative net value means that the liquid’s total gravitational potential energy decreases during the modeled transfer. It does not mean that a pump consumes a negative amount of energy. In a real system, descending liquid may release energy, but whether that energy can be recovered depends on the equipment and the flow arrangement. Friction and other losses also require additional energy. If a problem asks specifically for the work to lift only the slices below the outlet, integrate only over those slices; do not count descending slices as negative pump input unless the question is asking for net work against gravity.

For example, the formula \(W=\gamma\pi R^2(Ha-a^2/2)\) can be negative if the outlet is sufficiently low relative to the liquid’s average height. Its sign is therefore informative, not an error to remove.

A common distance error

A frequent mistake is to use the distance from a slice to the rim, \(h-y\), even when the liquid is being pumped to an outlet at a different height. The correct displacement is measured to the actual outlet, so it is \(H-y\).

Pitfall: If the rim is at \(4\ \text{m}\) and the outlet is at \(3\ \text{m}\), a slice at \(y=1\ \text{m}\) travels \(3-1=2\ \text{m}\), not \(4-1=3\ \text{m}\). Use the outlet height in the work integral.

Another sign error is to replace \(H-y\) with \(|H-y|\) without considering what is being calculated. The absolute value gives the distance each slice travels, but it removes the distinction between lifting and descending. For net work against gravity, use the signed displacement \(H-y\). If a problem asks for total distance-weighted work regardless of direction, that is a different model and should be stated explicitly.

Check Your Understanding

Use a bottom-based vertical coordinate and assume uniform liquid density in each question.

  1. A cylindrical tank has radius \(1\ \text{m}\), liquid depth \(2\ \text{m}\), and an outlet at height \(5\ \text{m}\). Write the work integral using weight density \(\gamma\).
  2. For a tank with liquid depth \(a\) and outlet height \(H\), derive the formula for signed work in terms of \(\gamma\), \(R\), \(H\), and \(a\).
  3. When the outlet is below the rim but above the liquid surface, are any slice contributions negative? Explain using \(H-y\).
  4. If \(H<a\), at what height does the slice-work contribution change sign?
  5. Why is replacing \(H-y\) by \(|H-y|\) unsuitable when calculating net work against gravity?

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