Physical Applications: Learn It 4

Lumen Learning

Physical Applications: Learn It 4

Work Done in Pumping

Consider the work done to pump water (or some other liquid) out of a tank. Pumping problems are a little more complicated than spring problems because many of the calculations depend on the tank’s shape and size. In addition, instead of being concerned about the work done to move a single mass, we are looking at the work done to move a volume of water, and it takes more work to move the water from the bottom of the tank than it does to move the water from the top of the tank.

Assume a cylindrical tank of radius $4$ m and height $10$ m is filled to a depth of $8$ m. How much work does it take to pump all the water over the top edge of the tank?

The first thing we need to do is define a frame of reference. We let $x$ represent the vertical distance below the top of the tank. That is, we orient the $x\text{-axis}$ vertically, with the origin at the top of the tank and the downward direction being positive.

This figure is a right circular cylinder that is vertical. It represents a tank of water. The radius of the cylinder is 4 m, the height of the cylinder is 10 m. The height of the water inside the cylinder is 8 m. There is also a horizontal line on top of the tank representing the x=0. A line is drawn vertical beside the cylinder with a downward arrow labeled x. — Figure 5. How much work is needed to empty a tank partially filled with water?

Using this coordinate system, the water extends from $x=2$ to $x=10.$

Therefore, we partition the interval $\left[2,10\right]$ and look at the work required to lift each individual “layer” of water.

This figure is a right circular cylinder representing a tank of water. Inside of the cylinder is a layer of water with thickness delta x. The thickness begins at xsub(i-1) and ends at xsubi. — Figure 6. A representative layer of water.

In pumping problems, the force required to lift water to the top of the tank is equal to the weight of the water, overcoming gravity. Given that the weight-density of water is $9800 N/m^3$ , or $62.4 lb/ft^3$ .

Calculating the volume of each layer gives us the weight. In this case, we have,

$V=\pi {(4)}^{2}\text{Δ}x=16\pi \text{Δ}x$

Then, the force needed to lift each layer is,

F = 9800 \cdot 16 π Δ x = 156, 800 π Δ x

$F=9800·16\pi \text{Δ}x=156,800\pi \text{Δ}x$

Note that this step becomes a little more difficult if we have a noncylindrical tank. We look at a noncylindrical tank in the next example.

Based on our choice of coordinate systems, we can use ${x}_{i}^{*}$ as an approximation of the distance the layer must be lifted.

Then the work to lift the $i\text{th}$ layer of water ${W}_{i}$ is approximately,

W_{i} \approx 156, 800 π x_{i}^{*} Δ x

${W}_{i}\approx 156,800\pi {x}_{i}^{*}\text{Δ}x$

Adding the work for each layer, we see the approximate work to empty the tank is given by

W = \underset{i = 1}{∑^{n}} W_{i} \approx \underset{i = 1}{∑^{n}} 156, 800 π x_{i}^{*} Δ x

$W=\underset{i=1}{\overset{n}{\text{∑}}}{W}_{i}\approx \underset{i=1}{\overset{n}{\text{∑}}}156,800\pi {x}_{i}^{*}\text{Δ}x$

This is a Riemann sum, so taking the limit as $n\to \infty ,$ we get

\begin{array}{cc} W & = lim_{n \to \infty} \underset{i = 1}{∑^{n}} 156, 800 π x_{i}^{*} Δ x \\ = 156, 800 π \int_{2}^{10} x d x \\ = 156, 800 π {[\frac{x^{2}}{2}] |}_{2}^{10} \\ = 7, 526, 400 π \\ \approx 23, 644, 883. \end{array}

$\begin{array}{cc}\hfill W& =\underset{n\to \infty }{\text{lim}}\underset{i=1}{\overset{n}{\text{∑}}}156,800\pi {x}_{i}^{*}\text{Δ}x\hfill \\ & =156,800\pi {\displaystyle\int }_{2}^{10}xdx\hfill \\ & =156,800\pi {\left[\frac{{x}^{2}}{2}\right]|}_{2}^{10}\hfill \\ & =7,526,400\pi\hfill \\ & \approx 23,644,883.\hfill \end{array}$

The work required to empty the tank is approximately $23,650,000 J$ .

For pumping problems, the calculations vary depending on the shape of the tank or container.

The following problem-solving strategy lays out a step-by-step process for solving pumping problems.

Problem-Solving Strategy: Solving Pumping Problems

Sketch a picture of the tank and select an appropriate frame of reference.
Calculate the volume of a representative layer of water.
Multiply the volume by the weight-density of water to get the force.
Calculate the distance the layer of water must be lifted.
Multiply the force and distance to get an estimate of the work needed to lift the layer of water.
Sum the work required to lift all the layers. This expression is an estimate of the work required to pump out the desired amount of water, and it is in the form of a Riemann sum.
Take the limit as $n\to \infty$ and evaluate the resulting integral to get the exact work required to pump out the desired amount of water.

We now apply this problem-solving strategy in an example with a noncylindrical tank.

Assume a tank in the shape of an inverted cone, with height $12$ ft and base radius $4$ ft. The tank is full to start with, and water is pumped over the upper edge of the tank until the height of the water remaining in the tank is $4$ ft.

How much work is required to pump out that amount of water?

As we did in the example with the cylindrical tank, we orient the $x\text{-axis}$ vertically, with the origin at the top of the tank and the downward direction being positive (step 1).

This figure is an upside-down cone. The cone has an axis through the center. The top of the cone on the axis is labeled x=0. — Figure 7. A water tank in the shape of an inverted cone.

The tank starts out full and ends with $4$ ft of water left, so, based on our chosen frame of reference, we need to partition the interval $\left[0,8\right].$ Then, for $i=0,1,2\text{,…},n,$ let $P=\left\{{x}_{i}\right\}$ be a regular partition of the interval $\left[0,8\right],$ and for $i=1,2\text{,…},n,$ choose an arbitrary point ${x}_{i}^{*}\in \left[{x}_{i-1},{x}_{i}\right].$ We can approximate the volume of a layer by using a disk, then use similar triangles to find the radius of the disk (see the following figure).

This figure has two images. The first has the x-axis. Below the axis, on a slant is a line segment extending up to the x-axis. Beside the line segment is a horizontal right circular cylinder. The second image has a triangle. The right triangle mirrors the first image with the hypotenuse the line segment in the first image. The top of the triangle is 4 units. the length of the vertical side is 12 units. The vertical side is also divided into two parts; the first is xsubi, the second is 12-xsubi. It is divided at the level where the first image has the cylinder. — Figure 8. Using similar triangles to express the radius of a disk of water.

From properties of similar triangles, we have

\begin{array}{ccc} \frac{r_{i}}{12 - x_{i}^{*}} & = & \frac{4}{12} = \frac{1}{3} \\ 3 r_{i} & = & 12 - x_{i}^{*} \\ r_{i} & = & \frac{12 - x_{i}^{*}}{3} \\ = & 4 - \frac{x_{i}^{*}}{3} . \end{array}

$\begin{array}{ccc}\hfill \frac{{r}_{i}}{12-{x}_{i}^{*}}& =\hfill & \frac{4}{12}=\frac{1}{3}\hfill \\ \hfill 3{r}_{i}& =\hfill & 12-{x}_{i}^{*}\hfill \\ \hfill {r}_{i}& =\hfill & \frac{12-{x}_{i}^{*}}{3}\hfill \\ & =\hfill & 4-\frac{{x}_{i}^{*}}{3}.\hfill \end{array}$

Then the volume of the disk is

V_{i} = π {(4 - \frac{x_{i}^{*}}{3})}^{2} Δ x (step 2).

${V}_{i}=\pi {(4-\frac{{x}_{i}^{*}}{3})}^{2}\text{Δ}x\text{(step 2).}$

The weight-density of water is $62.4 lb/ft^3$ , so the force needed to lift each layer is approximately

F_{i} \approx 62.4 π {(4 - \frac{x_{i}^{*}}{3})}^{2} Δ x (step 3).

${F}_{i}\approx 62.4\pi {(4-\frac{{x}_{i}^{*}}{3})}^{2}\text{Δ}x\text{(step 3).}$

Based on the diagram, the distance the water must be lifted is approximately ${x}_{i}^{*}$ feet (step 4), so the approximate work needed to lift the layer is

W_{i} \approx 62.4 π x_{i}^{*} {(4 - \frac{x_{i}^{*}}{3})}^{2} Δ x (step 5).

${W}_{i}\approx 62.4\pi {x}_{i}^{*}{(4-\frac{{x}_{i}^{*}}{3})}^{2}\text{Δ}x\text{(step 5).}$

Summing the work required to lift all the layers, we get an approximate value of the total work:

W = \underset{i = 1}{∑^{n}} W_{i} \approx \underset{i = 1}{∑^{n}} 62.4 π x_{i}^{*} {(4 - \frac{x_{i}^{*}}{3})}^{2} Δ x (step 6).

$W=\underset{i=1}{\overset{n}{\text{∑}}}{W}_{i}\approx \underset{i=1}{\overset{n}{\text{∑}}}62.4\pi {x}_{i}^{*}{(4-\frac{{x}_{i}^{*}}{3})}^{2}\text{Δ}x\text{(step 6).}$

Taking the limit as $n\to \infty ,$ we obtain

\begin{array}{cc} W & = lim_{n \to \infty} \underset{i = 1}{∑^{n}} 62.4 π x_{i}^{*} {(4 - \frac{x_{i}^{*}}{3})}^{2} Δ x \\ = \int_{0}^{8} 62.4 π x {(4 - \frac{x}{3})}^{2} d x \\ = 62.4 π \int_{0}^{8} x (16 - \frac{8 x}{3} + \frac{x^{2}}{9}) d x = 62.4 π \int_{0}^{8} (16 x - \frac{8 x^{2}}{3} + \frac{x^{3}}{9}) d x \\ = 62.4 π {[8 x^{2} - \frac{8 x^{3}}{9} + \frac{x^{4}}{36}] |}_{0}^{8} = 10, 649.6 π \approx 33, 456.7 \end{array}

$\begin{array}{cc}\hfill W& =\underset{n\to \infty }{\text{lim}}\underset{i=1}{\overset{n}{\text{∑}}}62.4\pi {x}_{i}^{*}{(4-\frac{{x}_{i}^{*}}{3})}^{2}\text{Δ}x\hfill \\ & ={\displaystyle\int }_{0}^{8}62.4\pi x{(4-\frac{x}{3})}^{2}dx\hfill \\ & =62.4\pi {\displaystyle\int }_{0}^{8}x(16-\frac{8x}{3}+\frac{{x}^{2}}{9})dx=62.4\pi {\displaystyle\int }_{0}^{8}(16x-\frac{8{x}^{2}}{3}+\frac{{x}^{3}}{9})dx\hfill \\ & =62.4\pi {\left[8{x}^{2}-\frac{8{x}^{3}}{9}+\frac{{x}^{4}}{36}\right]|}_{0}^{8}=10,649.6\pi \approx 33,456.7\hfill \end{array}$

It takes approximately $33,450$ ft-lb of work to empty the tank to the desired level.

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “2.5 Physical Applications” here (opens in new window)