A trough is 5 meters long, 2 meters wide, and 3 meters deep. The vertical cross-section of the trough parallel to an end is shaped like an isoceles triangle (with height 3 meters, and base, on top, of length 2 meters). The trough is full of water (density 1000kgm3 ). Find the amount of work in joules required to empty the trough by pumping the water over the top.

Since the cross section area of the trough ([tex]A[/tex]), in square meters, varies with the height of the water ([tex]h[/tex]), in meters, and considering that pumping system extracts water at constant rate, then the work needed to pump all the water ([tex]W[/tex]), in joules, is:

[tex]W = \int\limits^{V_{max}}_0 {p} \, dV[/tex] (1)

Where:

[tex]p[/tex] - Pressure of the infinitesimal volume, in pascals.
[tex]V[/tex] - Volume, in cubic meters.
[tex]V_{max}[/tex] - Maximum volume allowed by the trough, in cubic meters.

The infinitesimal volume is equivalent to the following expression:

[tex]dV = A\, dh[/tex] (2)

Since the area is directly proportional to the height of the water, we have the following expression:

[tex]A = \frac{A_{max}}{H_{max}}\cdot h[/tex] (3)

Where:

[tex]A_{max}[/tex] - Area of the base of the trough, in square meters.
[tex]H_{max}[/tex] - Maximum height of the water, in meters.

In addition, we know that pressure of the water is entirely hydrostatic:

[tex]p = \rho \cdot g \cdot h[/tex] (4)

Where:

[tex]\rho[/tex] - Density of water, in kilograms per cubic meters.
[tex]g[/tex] - Gravitational acceleration, in meters per square second.

By (2), (3) and (4) in (1):

[tex]W = \frac{\rho\cdot g\cdot W_{max}\cdot L_{max}}{H_{max}} \int\limits^{H_{max}}_{0} {h^{2}} \, dh[/tex] (5)

Where:

[tex]W_{max}[/tex] - Width of the base of the triangle, in meters.
[tex]L_{max}[/tex] - Length of the base of the triangle, in meters.
[tex]H_{max}[/tex] - Maximum height of the triangle, in meters.

The resulting expression is:

[tex]W = \frac{\rho\cdot g\cdot W_{max}\cdot L_{max}\cdot H_{max}^{2}}{3}[/tex] (5b)

If we know that [tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]W_{max} = 2\,m[/tex], [tex]L_{max} = 5\,m[/tex] and [tex]H_{max} = 3\,m[/tex], then the work needed to pump the water is:

[tex]W = \frac{(1000)\cdot (9.807)\cdot (2)\cdot (5)\cdot (3)^{2}}{3}[/tex]

[tex]W = 294210\,W[/tex]

We need a work of 294210 watts to pump the water over the top. [tex]\blacksquare[/tex]

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