Answer:
1.74 × 10^-3 m ( 1.74 mm)
Explanation:
Using Torricelli"s equation to calculate the speed of water at the surface of the hole and assuming that the speed of water at the surface of the tank is zero
v ( speed of water at the surface of the hole) = √ 2g(h1 - h2) where h1 - h2 is the difference in height which 16 m
v = √ 2×9.81×16 = 17.72 m/s
work done in ( m³/s) = 2.5 × 10^ -3 / 60 seconds = 4.2 × 10^-5 m³/s
work done = A × v
4.2 × 10^-5 = A × 17.72
4.2 × 10^-5 / 17.72 = A
2.37 × 10^-6 = A where A is assumed a circle ( πr²)
r = 2.37 × 10^-6 / π = √2.37 × 10^-6 / 3.142 = 0.869 × 10 ^ -3
diameter = 2r = 2 × 0.869 × 10^-7 = 1.74 × 10^-3 m
Answer:
The diameter of the hole is [tex]1.73 x10^{-3}m[/tex]
Explanation:
The velocity at the top of the tank is zero. As depth increases, velocity increases, thus, potential energy is equal to the kinetic energy at the hole level (16-meter depth).
Step 1: We calculate the velocity of water at the hole level
ρgh = ρv²/2
gh = v²/2 (striking the density of water out)
Therefore,
[tex]v=\sqrt{2gh}[/tex]
v² = (2)(9.81 m/s²)(16 m)
∴ v = 17.7 m/s (velocity of water at the hole level)
Step 2: We find the area of the hole itself.
[tex]Area=\frac{Volumetric flow rate}{(velocity)(time)}[/tex]
[tex]Area=\frac{2.50x10^{-3} m^{3}/min }{(17.7 m/s)(60 s)}[/tex]
[tex]Area=2.35x10^{-6} m^{2}[/tex]
Step 3: We calculate the diameter of the hole from its area.
[tex]Area=\frac{\pi D^{2} }{4}[/tex]
[tex]D=2\sqrt{\frac{A}{\pi} }[/tex]
[tex]D=2\sqrt{\frac{2.35x10^{-6} }{3.142} }[/tex]
∵ [tex]D=1.73x10^{-3} m[/tex]
