# How much work does it take to pump the oil to the rim of the tank if the conical tank from y=2x is filled to within 2 feet of the top of the olive oil weighing 57(lb)/(ft)^3?

Jan 6, 2018

$W = \frac{1}{16} \pi \rho g {y}^{4} = 5730 l b \cdot f t$, if y =2.0ft .

The height data provided is confusing: Is the oil level 2 ft high or 2 ft from the rim? The answer assumes y= 2ft high.

#### Explanation:

Work is defined as: $W = \int \vec{f} \cdot \mathrm{dv} e c s$

The minimal work done to lift a fluid particle with mass $\delta m$ from the bottom of the cone to height y is:

$\delta W = {\int}_{0}^{y} \left(\delta m\right) g \mathrm{dy} = \delta m g y$

Hence the work to bring a layer of fluid at level y in the cone should be:

$\delta W = \delta m g y = \left(\rho \mathrm{dv}\right) g y = \rho g y \left(\pi {x}^{2}\right) \mathrm{dy}$
where
$\rho =$ density of the fluid; $\mathrm{dv} = A \mathrm{dy} = \pi {x}^{2} \mathrm{dy}$
$x = \frac{y}{2}$ cross-section radius of a fluid layer in the cone

$W = {\int}_{0}^{y} \frac{1}{4} \pi \rho g {y}^{3} \mathrm{dy} = \frac{1}{16} \pi \rho g {\left({y}^{4}\right)}_{0}^{y} = \frac{1}{16} \pi \rho g {y}^{4}$

Let y = 2ft fills when the fluid fills the cone up to the rim

$W = \frac{1}{16} \pi \rho g {\left(2\right)}^{4} = \pi \rho g$

$W = \pi \left(57 \frac{l b}{f {t}^{3}}\right) \left(32 \frac{f t}{\sec} ^ 2\right) \left(f {t}^{4}\right) = 5730 l b \cdot f t$

Basically the work is equal to potential energy of the cone filled with oil. Applying the potential energy concept (mgh) yields the same result.

Note: the data on the fluid level is confusing. Is the fluid 2 ft high or 2 ft from the rim? If the later is the case, the height of the cone is not given. The solution if assumes the general form is:

$W = {\int}_{{y}_{1}}^{{y}_{2}} \frac{1}{4} \pi \rho g {y}^{3} \mathrm{dy} = \frac{1}{16} \pi \rho g {\left({y}^{4}\right)}_{{y}_{1}}^{{y}_{2}} = \frac{1}{16} \pi \rho g \left({y}_{2}^{4} - {y}_{1}^{4}\right)$