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oksano4ka [1.4K]
3 years ago
11

A large storage tank, open to the atmosphere at the top and filled with water, develops a small hole in its side at a point 10.4

m below the water level. Assume the tank is large so the velocity of the water at the top of the tank is zero. The rate of flow from the leak is 2.53×10^−3 m3/min. Determine the diameter of the hole.
Mathematics
1 answer:
vlabodo [156]3 years ago
4 0

Answer:

Therefore the diameter of the hole is 1.94 \times 10^{-3} m.

Step-by-step explanation:

Bernoulli's equation,

P_1+\frac12 \rho v^2_1+\rho g h_1= P_2+\frac12 \rho v^2_2+\rho g h_2

P₁ = P₂= atmospheric presser

\rho= density

\frac12 \rho v^2_1+\rho g h_1= \frac12 \rho v^2_2+\rho g h_2             [since P₁ = P₂]

\Rightarrow\rho (\frac12 v^2_1+ g h_1)= \rho(\frac12 v^2_2+ g h_2)

\Rightarrow\frac12 v^2_1+ g h_1= \frac12 v^2_2+ g h_2

\Rightarrow\frac12 v^2_2-\frac12 v^2_1=g h_1- g h_2

\Rightarrow v^2_2- v^2_1=2g h                                [h_1-h_2=h]

Here   v_1\approx 0

\Rightarrow v^2_2=2g h

\therefore v_2=\sqrt {2gh

Here g= 9.8 m/s² , h = 10.4 m

The velocity of water that leaves from the hole v_2 = \sqrt {2\times 9.8\times 10.4} m/s

                                                                                  =14.28 m/s.

Given, the rate of flow from the leak is 2.53\times 10^{-3} m^3/min

                                                               =\frac{2.53\times 10^{-3}}{60}  m^3/s

Let the diameter of the hole be d.

Then the cross section area of the hole is =\pi (\frac d2)^2

We know that,

The rate of flow = Cross section area × speed

\Rightarrow \frac{2.53\times 10^{-3}}{60} =\pi (\frac d2)^2\times 14.28

\Rightarrow (\frac d2)^2=\frac{2.53\times 10^{-3}}{60\times 14.28\times \pi}

\Rightarrow d= 1.94 \times 10^{-3}

Therefore the diameter of the hole is 1.94 \times 10^{-3} m.

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