A tank, shaped like a cone has height 12 meter and base radius 1 meter. It is placed so that the circular part is upward. It is full of water, and we have to pump it all out by a pipe that is always leveled at the surface of the water. Assume that a cubic meter of water weighs 10 000 N , i.e. the density of water is 10000 N m 3 . How much work does it require to pump all water out of the tank

Answer :

Answer:

376966.991 Joules

Explanation:

Given that :

the height = 12  m

Let assume the tank have a thickness  = dh

The radius of the tank by using the concept of similar triangle is :

[tex]\dfrac{1}{r} = \dfrac{12}{h}[/tex]

[tex]r = \dfrac{h}{12}[/tex]

The area of the tank = [tex]\mathbf{\pi r^2}[/tex]

The area of the tank = [tex]\mathbf{\pi( \dfrac{h}{12})^2}[/tex]

The area of the tank = [tex]\mathbf{ \dfrac{\pi}{144}h^2}[/tex]

The volume of the tank is  = area × thickness

= [tex]\mathbf{ \dfrac{\pi}{144}h^2 \ dh}[/tex]

Weight of the element = [tex]\rho_ g * volume[/tex]

where;

[tex]\rho_g[/tex] = density of water ; which is given as 10000 N/m³

So;

Weight of the element = [tex]\mathbf{ 10000 *\dfrac{\pi}{144}h^2 \ dh}[/tex]

Weight of the element = [tex]\mathbf{69.44 \ \pi \ h^2 \ dh}[/tex]

However; the work required to pump this water = weight × height  rise

where the height rise = 12 - h

the work required to pump this water  = [tex]\mathbf{69.44 \ \pi \ h^2 \ dh}[/tex](12 - h)

the work required to pump this water  = [tex]\mathbf{69.44 \pi (12h^2-h^3)dh}[/tex]

We can determine the total workdone by integrating the work required to pump this water

SO;

Workdone = [tex]\mathbf{\int\limits^{12}_0 {69.44 \pi(12h^2-h^3)} dh}[/tex]

= [tex]\mathbf{ 69.44 \pi \int\limits^{12}_0 {(12h^2-h^3)} dh}[/tex]

=  [tex]\mathbf{ 69.44 \pi[ \frac{12h^3}{3}- \frac{h^4}{4}]^{12}}_0} }[/tex]

= [tex]\mathbf{69.44 \pi [ \frac{12^4}{3}-\frac{12^4}{4}]}[/tex]

= [tex]\mathbf{69.44 \pi*12^4 [ \frac{4-3}{12}]}[/tex]

= [tex]\mathbf{69.44 \pi*12^4 *\frac{1}{12}}[/tex]

= 376966.991 Joules

The required work done to pump all the water from the tank would be as follows:

[tex]376966.991[/tex] Joules

Find the Work Done?

Given that,

Height [tex](h)[/tex] [tex]= 12 meter[/tex]

Assuming the thickness [tex]= dh[/tex]

Radius through 'similar triangle concept' would be:

[tex]r = h/12[/tex]

Now,

The tank's area would be [tex]=[/tex] π[tex]r^2[/tex]

[tex]=[/tex] π[tex](h/12)^2[/tex]

[tex]=[/tex] (π[tex]/144)[/tex][tex]h^2[/tex]

The tank's volume [tex]=[/tex] Area × Thickness

[tex]=[/tex] (π[tex]/144)[/tex][tex]h^2[/tex] [tex]dh[/tex]

Now,

The element's weight can be determined through [tex]p_{g}[/tex] × [tex]volume[/tex]

where   [tex]p_{g}[/tex] [tex]= 1000 N[/tex]

So,

Element's weight [tex]= 1000[/tex] × (π[tex]/144)[/tex][tex]h^2[/tex] [tex]dh[/tex]

[tex]= 69.44[/tex] π[tex]h^2 dh[/tex]

So,

The work required to pump all the water can be determined through

[tex]weight[/tex] × [tex]height rise[/tex]

where

[tex]height rise = 12 - h[/tex]

Therefore,

The work required to pump all the water

[tex]= 69.44[/tex] π[tex]h^2 dh[/tex] [tex](12 - h)[/tex]

[tex]= 69.44[/tex] π[tex](12h^2 - h^3)dh[/tex]

Now, after integration, the Total work done

[tex]=[/tex][tex]\int\limits^ \,12[/tex][tex]_{0}[/tex] [tex]69.44[/tex]π[tex](12h^2 - h^3)dh[/tex]  

[tex]= 69.44[/tex] π × [tex]12^4[/tex] × [tex]1/12[/tex]

[tex]=[/tex]   [tex]376966.991[/tex]   Joules

Thus, [tex]376966.991[/tex] Joules is the correct answer.

Learn more about "Work Done" here:

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