Printable View
I have this question that is throwing me for a loop.
A cone shaped tank has a radius of 10-cm at the top and a height of 20-cm. Determine the work needed to pump water (9800kN/m3) to a height of 6-m above the top of the tank. The tank is filles to a level of 15-m.
Any suggestions?? I'm lost...
I would set this up as an integral over the distance from the top of the tank to the water level. If we consider an infinitesimal circular "slice" of water whose surface is a distance y from the top of the tank, the radius of that circular slice is:
$\displaystyle r =10-1/2y$
The volume then is:
$\displaystyle \pi r^2 dy = \pi (10-1/2y)^2 dy$
(We approximate the volume of the cone using a cylinder of height $\displaystyle \Delta y$ and take the limit as this height goes to 0.)
Integrate from $\displaystyle y=5$ to $\displaystyle y=20$ the equation for the amount of work required to move such a "slice" to the desired height . (observe that work depends not only on the volume but also on how far the water has to move which is this case is $\displaystyle y+6$).
We can use similar triangles.
$\displaystyle \frac{x}{y}=\frac{10}{20}; \;\ x=\frac{y}{2}$
The cross-section is a thin cylinder, But this time the radius changes, $\displaystyle V={\pi}r^{2}h={\pi}x^{2}$
The height of the pumping distance is 26-y.
So, we have:
$\displaystyle \int_{0}^{15}[9800{\pi}(\frac{y}{2})^{2}(26-y)]dy$
$\displaystyle 2450{\pi}\int_{0}^{15}[y^{2}(26-y)]dy$
