# Work Done in Raising a Liquid

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?

• There are a lot of things wrong with this, not just your notation (it can be very counterproductive to use $\times$ in your formulas, especially where it can easily be mistaken for the cross product), but also your units (according to your work, you get the units of force as $\frac{m^6}{kg}$, rather than what it actually is, $\frac{kg\cdot m}{s^2}$). The great thing is that energy (work) just depends on final an initial states, I recommend forgetting about forces/integration altogether and just finding the energy of the water after pumping it to the top of the tank and that's your answer.
– user460377
Commented Aug 9, 2017 at 6:00
• @industryfatcat Please show me how you tackle the problem above using calculus or the method you prefer. I'll take note of them.... Thanks! Commented Aug 9, 2017 at 16:01

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).

I think of it as two terms. Lifting the disc of water at some arbitrary distance y (<-2) up to the surface of the water (y=-2) and then lifting the whole lot an extra 2 feet. So the first term is $\rho \int_{-6}^{-2} (y+2) \, \pi \, x^2 dy$ and the second term $2 \,\rho \int_{-6}^{-2} \pi \, x^2 dy$, this just being 2 feet times the density of water times the volume. This gives the 25.09 short tons ft in US units

• I made a mistake in the first term as I was lifting the water to the top of the container not to the water level. I have corrected this to give the right answer. However it would have been simpler just to write the answer as one part $\rho \int_{-6}^{-2} y \, \pi \, x^2 dy$. Commented Aug 10, 2017 at 8:19

Yes, the work done to raise an object weight w a height h is wh. Yes, since the height varies, in order to use that we need to "slice" the hemisphere at different heights. Taking y to be the height, as in your picture, each slice is a circle with radius $r= \sqrt{36- y^2}$ so area $\pi r^2= (36- y^2)\pi$. That "slice" has thickness dy so volume $(36- y^2)\pi dy$. The weight of each slice is $62.4(36- y^2)\pi dy$. The work done lifting that weight a distance y is $62.4y(36- y^2)\pi dy= 62.4(36y- y^3)\pi dy$. The work done lifting all of the water is $62.4\pi\int_2^4 (36y- y^3) dy$

• When I evaluated your integral above using the calculator, I got the answer 30581.5 lb-ft. Then, I converted the 30581.5 lb-ft into ton-ft. $30 \space 581.50 \space lb-ft \times \frac{1 \space kg}{0.4536 \space lbs} \times \frac{1 \space ton}{1 \space 000 \space kg} = 67.42 \space ton-ft$, which is farther than mine. Maybe revisit your answer again? Commented Aug 10, 2017 at 3:59