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]
