Question

A cubicle vessel of height 1 m is full of water. What will be the minimum work done in taking water out from vessel?

Answer 1

Answer on Question#53117 – Physics / Mechanics – Kinematics – Dynamics

4.905 kJ

Question

A cubicle vessel of height $1\,\mathrm{m}$ is full of water. What will be the minimum work done in taking water out from vessel?

Solution

Consider an infinitely small volume $dV$ of water (elementary volume).

Minimum work to take it out from vessel – transfer to the highest point of the vessel ($dW = g(h - h_i)dm$) and shift horizontally out of vessel (assume that there is no friction with medium (e.g. air), so no contribution to work there).

$g$ – gravitational acceleration, $h$ – height of vessel, $h_i$ – initial height of elementary volume, $dm$ – mass.

Thus, our aim to calculate work over all heights $h_i$. Assume that the lowest point has $h_i = 0\,\mathrm{m}$, accordingly to the data, the highest point has $h_i = 1\,\mathrm{m}$ then (Note: no effect on result). At the same time (in general) we can rewrite $dm = \frac{dm}{dh} dh$, and due to assumption that density of water independent from height, we can replace $\frac{dm}{dh}$ with $\frac{m}{h}$. Finally, we can jot down proper integral:

$$W = \int_{h_{imin}}^{h_{imax}} g(h - h_i) \frac{m}{h} \, dh_i$$

Let us put in actual boundaries and solve it:

$$W = \int_{0}^{1} g(h - h_i) \frac{m}{h} \, dh_i = \frac{mg}{h} \int_{0}^{1} (h - h_i) \, dh_i = mg \int_{0}^{1} dh_i - \frac{mg}{h} \int_{0}^{1} h_i \, dh_i = \\ = m g h_i \left| \frac{1}{0} - \frac{mg}{h} \frac{h_i^2}{2} \right| \frac{1}{0} = mg - \frac{mg}{2h} = mg \left(1 - \frac{1}{2h}\right)$$

There are few numbers from tables we required to know to get numeric answer: (mass of $1\,\mathrm{m}^3$ of water) $m \approx 1000\,\mathrm{kg}$, (gravitational acceleration) $g \approx 9.81\,\mathrm{ms}^{-2}$. Substitute them:

$$W = mg \left(1 - \frac{1}{2h}\right) = 1000 * 9.81 \left(1 - \frac{1}{2 * 1}\right) = 4905\,\mathrm{J} = 4.905\,\mathrm{kJ}$$