Chapter 6.6, Problem 23E

To determine

To Calculate: The depth to which a tank full of water can be emptied doing a given limited amount of work.

Answer to Problem 23E

The depth of the water remaining is $2.03m$

Explanation of Solution

Given Information: The dimensions of the tank are

• The width of the tank is $3m$ .
• Height of the tank is $3m$ .
• Length of the tank is $8m$ .
• Height of the spout is $2m$ .
• Density of water is $1000kg/m^3$
• The pump breaks down after doing $4.5\times {10}^{5}J$ of work.

Concept Used:The work done by a force $F$ in moving a particle from $x_1$ to $x_2$ is given by

  $W={{\displaystyle \int }}_{x_1}^{x_2}F\left(x\right)dx$

Calculation: At a depth $h$ from the top of the tank (not spout), its height with respect to the bottom is $3-h$ .

The width $w$ of the tank at any level is proportional to the height at that level.

So, to find the width $w$ at any depth $h$ , equate the ratio of widths and heights to write

  $\frac{w}{3-h}=\frac{3}{3}$

  $\Rightarrow w=3-h$

Now, the mass of an infinitesimally thin layer of water of thickness $dh$ and at a depth $h$ from the top of the tank is

  $dm=density\times volume=\rho dv=\rho \times A\times dh=\rho \cdot (l\times w)\cdot dh$

  $=\rho \times l\times (3-h)\cdot dh=1000\times 8\times (3-h)dh$

  $=8000(3-h)dh$

This infinitesimal mass of water located at the depth $h$ moves up a distance equal to the sum of its depth $h$ and the spout height of $2m$ , that is a total of $\left(h+2\right)meters$ .

So, the work done in pumping this infinitesimally small mass of water out is

  $dW=dm\cdot g\cdot (h+2)=8000(3-h)dh\cdot g\cdot (h+2)=8000g(6+h-h^2)dh$

So, the work needed to pump the water out of the tank up to a depth $h^{*}$ is the integral of the above $dW$ from $h=0$ (depth of the top most layer of water) to $h=h^{*}$ (depth of the last layer of water removed).

Therefore

  $W={{\displaystyle \int }}_0^{h^{*}}dW={{\displaystyle \int }}_0^{h^{*}}8000g(6+h-h^2)dh$

  $=8000g(6h^{*}+\frac{h^{{*}^2}}{2}-\frac{h^{{*}^3}}{3})$ .

Equate the above expression for work $W$ with the given available amount of work to find the value of $h^{*}$ .

  $4.7\times {10}^5=8000\times 9.8\times \left(6h^{*}+\frac{h^{{*}^2}}{2}-\frac{h^{{*}^3}}{3}\right)$

The above $3^{rd}$ degree polynomial has three roots out of which only the root $h*=0.97$ is a valid for the present problem.

So, the depth of water remaining $=intialdepth-h^{*}=3-0.97=2.03m$