The document you provided refers to a Bernoulli flow problem involving a cylindrical tank. The task is divided into two parts:

Partie 1 : L'orifice est fermé par un bouchon

This part asks you to:

1. Déterminer la pression PB au point B
2. Déduire la valeur de la force moyenne qui s'exerce sur le bouchon

Let's solve these in the context of Bernoulli's equation and the given data.

1) Déterminer la pression PB au point B

When the orifice is closed (bouchon), the system is static, and Bernoulli's equation can be used between points A (surface of the fluid in the reservoir) and B (point where the orifice is sealed). The equation for Bernoulli's principle in this static case simplifies to:

$P_A + \frac{1}{2} \rho v_A^2 + \rho g Z_A = P_B + \frac{1}{2} \rho v_B^2 + \rho g Z_B$

Since the orifice is closed, the velocity at point B is zero (i.e., $v_B = 0$) and the fluid velocity at point A is also zero (since it's at rest in the reservoir). So, this reduces to:

$P_A + \rho g Z_A = P_B + \rho g Z_B$

We can isolate $P_B$:

$P_B = P_A + \rho g (Z_A - Z_B)$

Using the given values:

• $P_A = P_{\text{atm}} = 101325 \, \text{Pa}$
• $\rho = 817 \, \text{kg/m}^3$
• $Z_A - Z_B = H = 2.5 \, \text{m}$

Thus,

$P_B = 101325 \, \text{Pa} + 817 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 2.5 \, \text{m}$

Calculating this:

$P_B = 101325 \, \text{Pa} + 20046.75 \, \text{Pa} = 121371.75 \, \text{Pa}$

So, the pressure at point B is approximately 121372 Pa.

2) Déduire la force moyenne qui s'exerce sur le bouchon

The force on the bouchon is given by the pressure difference acting on the cross-sectional area of the orifice. The formula for this force is:

$F_B = (P_A - P_B) \times A_B$

Where:

• $P_A$ and $P_B$ are the pressures at points A and B, respectively.
• $A_B = \pi \left( \frac{D_B}{2} \right)^2$ is the cross-sectional area of the orifice, with $D_B = 14 \, \text{mm} = 0.014 \, \text{m}$.

So:

$A_B = \pi \left( \frac{0.014}{2} \right)^2 = \pi \times (0.007)^2 \approx 1.538 \times 10^{-4} \, \text{m}^2$

Now, calculating the force:

$F_B = (101325 - 121372) \, \text{Pa} \times 1.538 \times 10^{-4} \, \text{m}^2$

$F_B = (-20047) \times 1.538 \times 10^{-4} \approx -3.08 \, \text{N}$

The negative sign indicates that the force is directed towards the inside of the reservoir (the bouchon is being pushed inward).

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Partie 2 : L'orifice est ouvert

Now, the orifice is open, and the liquid flows out of the reservoir. The velocity at the orifice, $V_B$, can be derived using Bernoulli's principle between point A (top of the fluid) and point B (at the orifice).

1. Exprimer $V_A$ en fonction de $V_B$ et $\alpha$ This would require the relationship between the velocities at point A and point B. If $A$ and $B$ are the cross-sectional areas of the reservoir and orifice, respectively, and $\alpha = \frac{S_B}{S_A}$, then:

   $V_A \times S_A = V_B \times S_B$

   Therefore:

   $V_A = \alpha \times V_B$

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Would you like me to continue with the next steps of Partie 2 or provide further clarifications on any part?