Answer:
The work done to pump all of the kerosene from the tank to an outlet is [tex]W=45238.9\: J[/tex]
Step-by-step explanation:
The work is defined by:
[tex]W=\int dFdx[/tex] (1)
The force here will be the product between the volume and the kerosene weighing, so we have :
[tex]dF=\pi R^{2}dy*50[/tex]
This force will be in-lbs.
Where R is the radius (3 feet)
Then using (1), we have:
[tex]W=\int \pi R^{2}dy*50(8-y)[/tex]
Here 8-y is a distance at some point of the tank. Now, to get the work done from the base to the top of the tank we will need to take integral from 0 to 8 feet.
[tex]W=\int_{0}^{8} \pi 3^{2}dy*50(8-y)[/tex]
[tex]W=450\pi \int_{0}^{8}dy(8-y)[/tex]
[tex]W=450\pi(\int_{0}^{8} 8dy-\int_{0}^{8} ydy)[/tex]
[tex]W=450\pi(8y|_{0}^{8} -\frac{y^{2}}{2}|_{0}^{8})[/tex]
[tex]W=450\pi(8*8 -\frac{8^{2}}{2})[/tex]
[tex]W=450\pi(64 -\frac{64}{2})[/tex]
Therefore, the work done to pump all of the kerosene from the tank to an outlet is [tex]W=45238.9\: J[/tex]
I hope it helps you!
The work taken to pump all of the kerosene from the tank is 25446.6 pound-force-feet.
Let suppose that kerosene has an uniform density ([tex]\rho[/tex]), in pounds per cubic feet, and the pumping process occurs at steady state, which means that work ([tex]W[/tex]), in pound-force-feet, needed is equal to the gravitational potential energy ([tex]U[/tex]), in pound-force-feet, of the fluid in the right circular tank.
By definition of work and the work-energy theorem we have the following line integral:
[tex]W = \oint \vec F \,\bullet\,d \vec s [/tex]
[tex]W = \frac{\rho\cdot A\cdot g}{g_{c}} \int \limits_{0}^{h} s\,ds[/tex] (1)
The solution of this integral is:
[tex]W = \frac{\rho\cdot A\cdot g\cdot h^{2}}{2\cdot g_{c}} [/tex] (1b)
Where:
The area of the transversal section of the right circular cylindrical tank is determined by the following expression:
[tex]A = \frac{\pi\cdot D^{2}}{4} [/tex] (2)
Where [tex]D[/tex] is the diameter of the tank, in feet.
If we know that [tex]D = 6\,ft[/tex], [tex]\rho = 50\,\frac{lbm}{ft^{3}} [/tex], [tex]g = 32.174\,\frac{ft}{s^{2}} [/tex], [tex]h = 6\,ft[/tex] and [tex]g_{c} = 32.174\,\frac{\frac{lbm\cdot ft}{s^{2}} }{lbf} [/tex], then the work needed to pump all of the kerosene is:
[tex]A = \frac{\pi\cdot (6\,ft)^{2}}{4} [/tex]
[tex]A \approx 28.274\,ft^{2}[/tex]
[tex]W = \frac{\left(50\,\frac{lbm}{ft^{3}} \right)\cdot (28.274\,ft^{2})\cdot \left(32.174\,\frac{ft}{s^{2}} \right)\cdot (6\,ft)^{2}}{2\cdot \left(32.174\,\frac{\frac{lbm\cdot ft}{s^{2}} }{lbf} \right)} [/tex]
[tex]W = 25446.6\,lbf\cdot ft[/tex]
The work taken to pump all of the kerosene from the tank is 25446.6 pound-force-feet. [tex]\blacksquare[/tex]
To learn more on work and energy, we kindly invite to check this verified question: https://brainly.com/question/10644371
