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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.

Sagot :

xero099

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

Work needed to pump all 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]

To learn more on work, we kindly invite to check this verified question: https://brainly.com/question/17290830

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