# pumping out a liquid - work

• Mar 11th 2010, 04:37 PM
mybrohshi5
pumping out a liquid - work
A tank in the shape of an inverted right circular cone has height http://webwork2.asu.edu/webwork2_fil...fc95b6a081.png meters and radius http://webwork2.asu.edu/webwork2_fil...0ef0978d41.png meters. It is filled with http://webwork2.asu.edu/webwork2_fil...47b840b231.png meters of hot chocolate.
Find the work required to empty the tank by pumping the hot chocolate over the top of the tank. Note: the density of hot chocolate is http://webwork2.asu.edu/webwork2_fil...61b7b03a91.png

$\displaystyle A = \pi r^2$

$\displaystyle V = \pi r^2 \triangle{x}$

$\displaystyle \frac{r}{h} = \frac{13}{8} \frac {r}{8-x} = \frac{13}{8} r=\frac{13}{8}(8-x)$

$\displaystyle V = \pi (\frac{13}{8})^2 (8-x)^2 \triangle x$

$\displaystyle F = ma = V*p*g$

$\displaystyle F = \pi (\frac{13}{8})^2 (8-x)^2 (1550)(9.8) \triangle x$

$\displaystyle W = \int^8_1 [\pi (\frac{169}{64}) (8-x)^2 (1550)(9.8)] dx$

$\displaystyle W = \frac{2567110\pi}{64} \int^8_1 (8-x)^2 dx$

$\displaystyle W = 14407454.03 J$

Can anyone see where i went wrong?

Thanks for any help :)
• Mar 11th 2010, 05:41 PM
skeeter
Quote:

Originally Posted by mybrohshi5
A tank in the shape of an inverted right circular cone has height http://webwork2.asu.edu/webwork2_fil...fc95b6a081.png meters and radius http://webwork2.asu.edu/webwork2_fil...0ef0978d41.png meters. It is filled with http://webwork2.asu.edu/webwork2_fil...47b840b231.png meters of hot chocolate.
Find the work required to empty the tank by pumping the hot chocolate over the top of the tank. Note: the density of hot chocolate is http://webwork2.asu.edu/webwork2_fil...61b7b03a91.png

let the bottom of the tank be the origin.

one sloping side of the tank runs along the line $\displaystyle y = \frac{8}{13}x$

volume of a representative horizontal cross-section is ...

$\displaystyle \pi x^2 \cdot dy = \frac{169\pi}{64}y^2 \, dy$

weight-density = $\displaystyle \left(1550 \, \frac{kg}{m^3}\right) \cdot \left(9.8 \, \frac{N}{kg}\right) = 15190 \, \frac{N}{m^3}$

lift distance for a representative horizontal cross-section is $\displaystyle (8-y)$

$\displaystyle W = \int_0^7 15190 \cdot (8-y) \cdot \frac{169\pi}{64}y^2 \, dy$

$\displaystyle W = \frac{1283555 \pi}{32}\int_0^7 8y^2-y^3 \, dy$

$\displaystyle W = 3.96 \times 10^7 \, J$