I was studying for some exams when I encountered this question:
A hemispherical tank of radius 6 feet is filled with water to a depth of 4 feet. Find the work done in pumping the water to the top of the tank.
My work:
I visualize the problem like this:
I do know that $Work = Force \times distance .$ The work done by a variable force from $x= a$ to $x = b$ is: $$W = \int_a ^b F(x) dx$$
I'm getting the differential work $dW$: $$dW = dF \times x$$
The force $F$ is equal to specific volume of liquid $\times$ distance $\times$ area or in differential form: $$dF = w \times h \times dA$$
As seen in the figure above: $$dF = 62.4 \times y \times \pi x^2 dy$$
Substituting this newly-found $dF$ to $dW$ described in the figure above:
$$dW = dF \times y $$
$$ dW= 62.4 \times y \times \pi x^2 dy \times y$$
And since we are raising the water from the point $y = -2$ to $y = 0$ to make it appear that the water reaches the top of tank:
$$ W= 62.4 \int_{-2} ^0 y^2 \times \pi x^2 dy $$
Since the equation of the circle (which is the cross-section of the hemisphere) is $x^2 + y^2 = 6^2$ and rearranging it to get $x^2 = 6^2 – y^2$:
$$ W= 62.4 \int_{-2} ^0 y^2 \times \pi (6^2 – y^2) dy $$
Giving the answer $W = 17 \space 564. 77$ lb-ft.
Converting the units lb-ft into ton-ft:
$$17 \space 564.77 \space lb-ft \times \frac{1 \space kg}{0.4536 \space lbs} \times \frac{1 \space ton}{1 \space 000 \space kg} = 38.7 \space ton-ft $$
But my book said that the work done in pumping the water to the top of the tank is 25.1 ton – ft. How do you get the value of 25.1 ton – ft?
Best Answer
Let me begin with the basics just for the sake of being thorough.
Recall somewhere from physics that the amount of work done in raising a body to a certain height is equal to the increase in the body's potential energy.
$$W=mg\Delta h$$ $$W=\rho Vg \Delta h$$ $$\therefore W=\gamma V \Delta h$$ $$\therefore W=\gamma\int \Delta hdV$$
where $\gamma = $ weight per unit volume; $V=$ volume; $\Delta h=$ change in height, i.e. height to which liquid is pumped.
To get the volume of the hemispherical tank, we revolve a quarter circle either in the 3rd or 4th quadrant around the y-axis.
$$V=\int_{y_1}^{y_2} \pi x^2 dy$$ $$V=\pi\int_{y_1}^{y_2}(6^2-y^2) dy$$
Now for the height $\Delta h$, it is the gap between the vessel's rim (maximum height) and the actual instantaneous height of the liquid. Since the quarter circle is positioned at the lower half of the cartesian plane, the maximum height is $0$.
$$\Delta h=0-y=-y$$
But depth of water is just $4$ ft so you only want to get the volume between $$-6\le y \le-2$$
Putting these together,
$$W=62.4\pi \int_{-6}^{-2} -y(36-y^2)dy$$ $$W=50,185 \space \mathrm {lbft.} = 25.1 \space \mathrm {tonft.}$$
Btw, the correct lb to kg conversion is 2.2 lb/kg and kg to ton(US) 907 kg/ton(US).