Respuesta :
Answer:
The velocity of water at the bottom, [tex]v_{b} = 28.63 m/s[/tex]
Given:
Height of water in the tank, h = 12.8 m
Gauge pressure of water, [tex]P_{gauge} = 2.90 atm[/tex]
Solution:
Now,
Atmospheric pressue, [tex]P_{atm} = 1 atm = 1.01\tiems 10^{5} Pa[/tex]
At the top, the absolute pressure, [tex]P_{t} = P_{gauge} + P_{atm} = 2.90 + 1 = 3.90 atm = 3.94\times 10^{5} Pa[/tex]
Now, the pressure at the bottom will be equal to the atmopheric pressure, [tex]P_{b} = 1 atm = 1.01\times 10^{5} Pa[/tex]
The velocity at the top, [tex]v_{top} = 0 m/s[/tex], l;et the bottom velocity, be [tex]v_{b}[/tex].
Now, by Bernoulli's eqn:
[tex]P_{t} + \frac{1}{2}\rho v_{t}^{2} + \rho g h_{t} = P_{b} + \frac{1}{2}\rho v_{b}^{2} + \rho g h_{b} [/tex]
where
[tex]h_{t} - h_{b} = 12.8 m[/tex]
Density of sea water, [tex]\rho = 1030 kg/m^{3}[/tex]
[tex]\sqrt{\frac{2(P_{t} - P_{b} + \rho g(h_{t} - h_{b}))}{\rho}} = v_{b}[/tex]
[tex]\sqrt{\frac{2(3.94\times 10^{5} - 1.01\times 10^{5} + 1030\times 9.8\times 12.8}{1030}} = v_{b}[/tex]
[tex]v_{b} = 28.63 m/s[/tex]
