Question

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.

Answer 1

We need a work of 294210 watts to pump the water over the top. $\blacksquare$

Work needed to pump all the water over the top

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

$W = \int\limits^{V_{max}}_0 {p} \, dV$ (1)

Where:

• $p$ - Pressure of the infinitesimal volume, in pascals.
• $V$ - Volume, in cubic meters.
• $V_{max}$ - Maximum volume allowed by the trough, in cubic meters.

The infinitesimal volume is equivalent to the following expression:

$dV = A\, dh$ (2)

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

$A = \frac{A_{max}}{H_{max}}\cdot h$ (3)

Where:

• $A_{max}$ - Area of the base of the trough, in square meters.
• $H_{max}$ - Maximum height of the water, in meters.

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

$p = \rho \cdot g \cdot h$ (4)

Where:

• $\rho$ - Density of water, in kilograms per cubic meters.
• $g$ - Gravitational acceleration, in meters per square second.

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

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

Where:

• $W_{max}$ - Width of the base of the triangle, in meters.
• $L_{max}$ - Length of the base of the triangle, in meters.
• $H_{max}$ - Maximum height of the triangle, in meters.

The resulting expression is:

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

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

$W = \frac{(1000)\cdot (9.807)\cdot (2)\cdot (5)\cdot (3)^{2}}{3}$

$W = 294210\,W$

We need a work of 294210 watts to pump the water over the top. $\blacksquare$

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