Chapter 6.6, Problem 20E

To determine

To Calculate: The work done needed to pump all the water out of a spherical tank from a spout located at the top.

Answer to Problem 20E

The work needed to empty the spherical tank is $4431.168kJ$

Explanation of Solution

Given Information: The dimensions of the tank are

• The radius of the sphere is $3m$ .
• Height of the spout is $1m$ .
• Density of water is $1000kg/m^3$

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:

  

The mass of an infinitesimally thin disc of water of thickness $dh$ and making an angle $\text{α}$ with the centerof the sphere is

  $dm=density\times volume=\rho dv=\rho \times A\times dh=\rho \times \pi {(r\cos \alpha )}^2\times (rd\alpha \cdot \cos \alpha )$

  $=\rho \pi r^3{\cos }^3\alpha d\alpha$

This infinitesimal mass of discmaking an angle $\text{α}$ with the center of the sphere moves up a distance equal to the sum of its distance from the top of the sphere $=r-rcos\text{α}$ and the spout height $=1m$ , that is a total of $1+r\left(1-\sin \alpha \right)meters$ .

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

  $\begin{array}{l}dW=dm\cdot g\cdot (1+r[1-\sin \alpha ])d\alpha \\ =\rho g\pi r^{3}co{s}^3\alpha (1+r-r\sin \alpha )d\alpha \\ =27000g\pi {\cos }^3\alpha (4-3\sin \alpha )d\alpha \end{array}$

So, the total work needed to pump the entire water out of the tank is the integral of the above

  $dW$ from $\alpha =\frac{-\pi }{2}$ (bottom most layer of water) to $\alpha =\frac{\pi }{2}$ (top most layer of water).

Therefore

  $W={{\displaystyle \int }}_{\frac{-\pi }{2}}^{\frac{\pi }{2}}dW={{\displaystyle \int }}_{\frac{-\pi }{2}}^{\frac{\pi }{2}}27000g\pi {\cos }^3\alpha (4-3\sin \alpha )d\alpha$

  $\begin{array}{l}=27000g\pi {{\displaystyle \int }}_{\frac{-\pi }{2}}^{\frac{\pi }{2}}(4{\cos }^3\alpha -3{\cos }^3\alpha \sin \alpha )d\alpha \\ =27000g\pi \left[{{\displaystyle \int }}_{\frac{-\pi }{2}}^{\frac{\pi }{2}}4{\cos }^3\alpha d\alpha -{{\displaystyle \int }}_{\frac{-\pi }{2}}^{\frac{\pi }{2}}3{\cos }^3\alpha \sin \alpha d\alpha \right]\end{array}$

Use the identity $\cos 3\alpha =4co{s}^3\alpha -3cos\alpha$ .

  $\begin{array}{l}=27000g\pi \left({{\displaystyle \int }}_{\frac{-\pi }{2}}^{\frac{\pi }{2}}(\cos 3\alpha +3\cos \alpha )d\alpha +{{\displaystyle \int }}_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{\cos }^3\alpha d(\cos \alpha )]\right)\\ =27000g\pi {\left[\frac{sin3\alpha }{3}+3sin\alpha +\frac{co{s}^4\alpha }{4}\right]}_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\\ =27000g\pi \left[(\frac{-1-1}{3})+3(1+1)+0\right]\\ =27000\times 9.8\times \pi \times \frac{16}{3}\\ =4431168\end{array}$

Therefore, the total work done is $4431.168kJ$ .