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We need a work of 294210 watts to pump the water over the top. [tex]\blacksquare[/tex]
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:
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:
In addition, we know that pressure of the water is entirely hydrostatic:
[tex]p = \rho \cdot g \cdot h[/tex] (4)
Where:
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:
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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