Question

Determine the maximum volume in gallons​ [gal] of olive oil that can be stored in a closed cylindrical silo with a diameter of 3feet​ [ft] so the total pressure at the bottom of the container will not exceed 26 pound-force per square inch​ [psi]. Assume the height of the tank is sufficient to store the amount of olive oil​required, and the surface pressure is 1 atmosphere​ [atm]. The specific gravity of olive oil is 0.86

Answer 1

Answer:

V=1601gal

Explanation:

Hello! This problem is solved as follows,

First we must raise the equation that defines the pressure at the bottom of the tank with the purpose of finding the height that olive oil reaches.

This is given as the sum due to the atmospheric pressure (1atm = 101.325kPa), and the pressure due to the weight of the olive oil, taking into account the above, the following equation is inferred.

P=Poil+Patm

P=total pressure or absolute pressure=26psi=179213.28Pa

Patm= the atmospheric pressure =101325Pa

Poil=pressure due to the weight of olive oil=0.86αgh

α=density of water=1000kg/m^3

g=gravity=9.81m/s^2

h= height that olive oil reaches

solving

P=Poil+Patm

P=Patm+0.86αgh

$h=\frac{P-Patm}{0.86\alpha g } =\frac{179214.28-101325}{(0.86)(1000)(9.81)} \\h=9.23m$[/tex]

Now we can use the equation that defines the volume of a cylinder.

V=$V=\frac{\pi }{4} D^{2} h$

D=3ft=0.9144m

h=9.23m

solving

$V=\frac{\pi }{4} (0.9144)^{2} 9.23=6.06m^3$

finally we use conversion factors to find the volume in gallons

$V=6.06m^3\frac{1000L}{1m^3} \frac{1gal}{3.785L} =1601gal$