Question #344330

5. Suppose that a water tank is shaped like a right circular cone with the tip at the bottom, and has height 10 meters and radius 2 meters at the top. If the tank is full, how much work is required to pump all the water out over the top?

1
Expert's answer
2022-06-01T14:34:51-0400


Let's consider the circular cross-section through the tank at depth h and find its radius using similar triangles:

2r=1010h,\frac{2}{r}=\frac{10}{10-h},

r=2(10h)10=10h5.r=\frac{2(10-h)}{10}=\frac{10-h}{5}.

The aea of the cross-section: A=πr2=π(10h)225.A=\pi r^2=\frac{\pi (10-h)^2}{25}.

The volume of the section of water at depth h:

V=πr2dh=π(10h)225dh.V=\pi r^2dh=\frac{\pi (10-h)^2}{25}dh.


The density of the water: σ=1000kg/m3,\sigma=1000 kg/m^3, so the mass of water in the section is

m=σV=1000π(10h)225dh=40π(10h)2dh.m=\sigma V=1000\frac{\pi (10-h)^2}{25}dh=40\pi (10-h)^2dh.

To raise this mass of water we need to overcome the force of gravity F=mgF=mg , g=9.8m/s2.g=9.8 m/s^2.

F=9.840π(10h)2dh=392π(10h)2dh.F=9.8\cdot 40\pi (10-h)^2dh=392\pi(10-h)^2dh.

We must lift this water a distance h and require a work

W=392π(10h)2hdh.W=392\pi(10-h)^2hdh.

The total work:

W=010392π(10h)2hdh=392π010(10h)2hdh=W=\int_0^{10} 392\pi(10-h)^2hdh=392\pi \int_0^{10} (10-h)^2hdh=

=392π010(100h20h2+h3)dh=392π(100h2220h33+h44)010==392\pi \int_0^{10} (100h-20h^2+h^3)dh=392\pi (100\frac{h ^2}{2}-20\frac{h^3}{3}+\frac{h^4}{4})|_0^{10}=

=392π(500020,0003+10,0004)=392π25003=980,0003π1,026kJ.=392\pi (5000 - \frac{20,000}{3}+\frac{10,000}{4})=392\pi\frac{2500}{3}=\frac{980,000}{3}\pi\approx1,026 kJ.


Answer: 1,026 kJ.


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