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[Fluid Dynamics] Bernoulli's Eqn. Problem

by Je m'appelle
Tags: bernoulli, dynamics, fluid
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Je m'appelle
Dec7-10, 09:52 AM
P: 109
1. The problem statement, all variables and given/known data

A bomb located 3 meters below the ground needs a certain pressure in order to pump the water through the pipe (connecting the ground to the bomb) and 12 meters above the ground. The radius of the pipe is 0.01 meters and the radius of the 'mouthpiece' of the tube in the ground is 0.005 meters. Find the pressure of the bomb.

2. Relevant equations

Bernoulli's Equation

[tex]\frac{1}{2}\rho v^2 + p + \rho g h = C [/tex]

Conservation of Mass Equation for a Stationary Fluid

[tex]A_1 v_1 = A_2 v_2 [/tex]

Eqn for the Velocity of an Accelerating Body

[tex]v_2^2 = v_1^2 + 2a\Delta S [/tex]

3. The attempt at a solution

Using Bernoulli's for A or the opening of the pipe at the bomb and for B the ending (or 'mouthpiece' of the pipe) at the ground we get

[tex]\frac{1}{2}\rho v_A^2 + p_A = \frac{1}{2}\rho v_B^2 + p_B + \rho g h_{AB} [/tex]

[tex](p_A - p_B) = \Delta p = \rho (\frac{1}{2} v_B^2 - \frac{1}{2} v_A^2 + g h_{AB}) [/tex]

Using the Conservation Eqn. we get

[tex]A_1 v_A = A_2 v_B [/tex]

[tex]\pi (0.01)^2 v_A = \pi (0.005)^2 v_B[/tex]

[tex]v_A = 0.25 v_B [/tex]

[tex]v_A^2 = 0.625 v_B^2 [/tex]

Substituting the relation between the velocities in our previous eqn. we get

[tex]\Delta p = \rho (0.1875 v_B^2 + g h_{AB}) [/tex]

[tex]\Delta p = \rho (0.1875 v_B^2 + 29.4) [/tex]

Now we have to find the velocity at B, that can easily be found considering that the water flows 12 meters into the air, therefore

[tex]v_2^2 = v_1^2 + 2a\Delta S [/tex]

[tex]0 = v_B^2 - 2g \Delta h [/tex]

[tex]v_B^2 = 2g \Delta h [/tex]

[tex]v_B^2 = 235.2 [/tex]

Back to the previous eqn.

[tex]\Delta p = 10^3 (44.1 + 29.4) [/tex]

[tex]\Delta p = 7.35 \times 10^4 [/tex]

Notice that the power of the pressure is p of B + delta p, or delta p + atmospheric pressure, therefore

[tex]p_A = p_{bomb} = \Delta p + p_B[/tex]

Is this correct?
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