Respuesta :
Answer:
The specific gravity of the gel is 1470.68
Explanation:
Given:
Volume [tex]V = 4.24[/tex] [tex]m^{3}[/tex]
Diameter of tank [tex]d = 2[/tex] m
Pressure at surface [tex]P _{o} = 1.41[/tex] atm
Height [tex]h = 6565[/tex] ft [tex]= \frac{6565}{3.281} = 2001[/tex] m
Pressure at bottom is given by,
[tex]P_{bottom} = \rho _{w} g h[/tex]
Where [tex]\rho _{w} = 1000 \frac{kg}{m^{3} }[/tex], [tex]g = 9.8\frac{m}{s^{2} }[/tex]
[tex]P_{bottom} = 1000 \times 9.8 \times 2001[/tex]
[tex]P_{bottom} = 19.6 \times 10^{6}[/tex] [tex]\frac{N}{m^{2} }[/tex]
Here volume is 4.24 [tex]m^{3}[/tex]
[tex]\pi r^{2} h = 4.24[/tex]
[tex]h = \frac{4.24}{3.14} = 1.35[/tex] m
From pascal equation,
[tex]P_{bottom} = P_{o} + \rho _{gel} g h[/tex]
Find density of gel from above equation,
[tex]19600000 = 142865.25 + \rho_{gel} \times 9.8 \times 1.35[/tex]
[tex]\rho_{gel} = 1470682.89[/tex] [tex]\frac{kg}{m^{3} }[/tex]
So specific gravity is given by,
ζ = [tex]\frac{\rho _{gel} }{\rho_{w} }[/tex]
ζ = [tex]\frac{1470682.89}{1000}[/tex]
ζ = [tex]1470.68[/tex]
Therefore, the specific gravity of the gel is 1470.68
