Chapter 6.6, Problem 24E

To determine

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

Answer to Problem 24E

The work needed to empty the spherical tank is $2741.785kJ$

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 oil is $900kg/m^3$ .

The sphere is half filled.

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 oil of thickness $dh$ and making an angle $\text{α}$ with the center of 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 disc making 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 oil out is

  $\begin{array}{l}dW=dm\cdot g\cdot (1+r[1-\sin \alpha ])d\alpha \\ =\rho g\pi r^{3}co{s}^3\alpha (1+r-r\sin \alpha )d\alpha \\ =24300g\pi {\cos }^3\alpha (4-3\sin \alpha )d\alpha \end{array}$

And since the tank is half filled, the total work needed to pump the entire oil out of the tank is the integral of the above $dW$ from $\alpha =\frac{-\pi }{2}$ (bottom most layer of oil) to $\alpha =0$ (top most layer of oil).

Therefore

  $W={{\displaystyle \int }}_{\frac{-\pi }{2}}^{0}dW={{\displaystyle \int }}_{\frac{-\pi }{2}}^{0}24300g\pi {\cos }^3\alpha (4-3\sin \alpha )d\alpha$

  $\begin{array}{l}=24300g\pi {{\displaystyle \int }}_{\frac{-\pi }{2}}^0(4{\cos }^3\alpha -3{\cos }^3\alpha \sin \alpha )d\alpha \\ =24300g\pi \left[{{\displaystyle \int }}_{\frac{-\pi }{2}}^{0}4{\cos }^3\alpha d\alpha -{{\displaystyle \int }}_{\frac{-\pi }{2}}^{0}3{\cos }^3\alpha \sin \alpha d\alpha \right]\end{array}$

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

  $\begin{array}{l}=24300g\pi \left({{\displaystyle \int }}_{\frac{-\pi }{2}}^0(\cos 3\alpha +3\cos \alpha )d\alpha +{{\displaystyle \int }}_{\frac{-\pi }{2}}^0{\cos }^3\alpha d(\cos \alpha )]\right)\\ =24300\pi {\left[\frac{sin3\alpha }{3}+3sin\alpha +\frac{co{s}^4\alpha }{4}\right]}_{\frac{-\pi }{2}}^0\\ =24300g\pi \left[(\frac{0-1}{3})+3(0+1)+(1-0)\right]\\ =24300\times 9.8\times \pi \times \frac{11}{3}\\ =2741785.2\end{array}$

Therefore, the total work done is $2741.785kJ$ .