Welcome to Westonci.ca, the Q&A platform where your questions are met with detailed answers from experienced experts. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
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
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.