work done on pumping out water from an inverted right circular cone - integral calculus

[ddeyv]

Question: A cistern in the form of an inverted right circular cone is 20 meters deep and 12 meters diameter at the top. If the water is 16 meters deep in the cistern, find the work done in joules in pumping out the water. The water is raised to a point of discharge 10 meter above the top of the cistern.


Answer: 68, 166.75 kJ


I have searched for many of tutorials and this is what I have come up with:

dW = (density of water)(pi)((6y/20)^2)(36-y)dy, integral limits from 0 to 16.



Please help for what I did wrong.

 

[MarkFL]

Hello, and welcome to FMH!

Let's work this problem in general. Let the radius of the top of the inverted conical tank be $R$, the depth be $H$, and the distance above the top of the tank the fluid must be pumped be $A$. Since the gravitational equipotentials are presumably horizontal in our model, it makes sense to consider breaking the volume of fluid into horizontal disks of radius $r$ and thickness $dy$. Let:

$\rho$ = the weight density of the fluid.

Now, if the distance of the disk from the bottom of the tank is $y$, then by similarity we may state:

[MATH]\frac{r}{y}=\frac{R}{H}\implies r=\frac{R}{H}y[/MATH]
And so the volume of an arbitrary disk can be given by:

[MATH]dV=\pi\left(\frac{R}{H}y\right)^2\,dy[/MATH]
Next, we want to determine the weight $w$ of this disk. Using the definition of weight density, we may state:

[MATH]\rho=\frac{w}{dV}\,\therefore\,w=\rho\,dV=\rho\pi\left(\frac{R}{H}y\right)^2\,dy[/MATH]
The distance $d$ over which this disk must be lifted is:

[MATH]d=H-y+A[/MATH]
Thus, using the fact that work is the product of the applied force and the distance through which this force is applied, we find the work done to lift the arbitrary disk is:

[MATH]dW=wd=\rho\pi\left(\frac{R}{H}y\right)^2(H+A-y)\,dy[/MATH]
[MATH]dW=\frac{R^2\rho\pi}{H^2}\left((H+A)y^2-y^3\right)\,dy[/MATH]
Now, if $y_i$ is the initial depth of fluid in the tank, and $y_f$ is the final depth, where $0\le y_f<y_i\le H$, then the total amount of work required to pump out the required amount of fluid is given by:

[MATH]W=\frac{R^2\rho\pi}{H^2}\int_{y_f}^{y_i} (H+A)y^2-y^3\,dy[/MATH]
Applying the FTOC, there results:

[MATH]W=\frac{R^2\rho\pi}{H^2}\left[\frac{H+A}{3}y^3-\frac{1}{4}y^4\right]_{y_f}^{y_i}=\frac{R^2\rho\pi}{H^2}\left(\left(\frac{H+A}{3}y_i^3-\frac{1}{4}y_i^4\right)-\left(\frac{H+A}{3}y_f^3-\frac{1}{4}y_f^4\right)\right)[/MATH]
[MATH]W=\frac{R^2\rho\pi}{12H^2}\left(3(y_f^4-y_i^4)+4(H+A)(y_i^3-y_f^3)\right)[/MATH]
Now, let's plug in the given data:

[MATH]R=6\text{ m},\,H=20\text{ m},\,\rho=1000\frac{\text{kg}}{\text{m}^2}\cdot9.8\frac{\text{m}}{\text{s}^2},\,A=10\text{ m},\,y_i=16\text{ m},\,y_f=0\text{ m}[/MATH]
And we find:

[MATH]W\approx90796.35\text{ kJ}[/MATH]

 

[MarkFL]

This is the same result you would get from your integral, which as far as I can tell is correctly set up.

 

[ddeyv]

This is the same result you would get from your integral, which as far as I can tell is correctly set up.

Thank you for the solution! I really appreciate it.
But I am not getting the answer from the book. So it is possible that the answer from the book is a wrong one? Thank you again, sir!

 

[ddeyv]

Hello! Problem is solved. I have been using the wrong value.

final equation:

dW = (1000)(9.81)(pi)((6y/20)^2)(30-y)dy, from 0 to 16

I have been using (36-y). I am so sorry for my mistake. Thank you!

 

[MarkFL]

Yes, it's odd that I made exactly the same mistake as well ($H+A=30\text{ m}$)...when I put the correct data into my formula, I get:

[MATH]W\approx68097\text{ kJ}[/MATH]
When I use the same value you gave for the gravitational acceleration, I get the value you gave in your first post.