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I have the following problem where I need to apply Bernoulli's equation to find the velocity at the end of a tube that is attached to a hole of tank (filled with water), like it is illustrated in the picture. Point B is 10 meters under the surface and point C is 30m under the surface.enter image description here My only question is the following: I know that Bernoulli's equation is constant at every point, therefore: $$\frac{1}{2}\rho v_A^2+\rho g.0 + p_{A}=\frac{1}{2}\rho v_B^2+\rho g.(-y_B) + p_{B}=\frac{1}{2}\rho v_C^2+\rho g.(-y_C) + p_{C}$$

I know that $p_A$=$p_C=p_{atm}$. I found in the solutions that $p_B=p_{atm}$ also. Why is that? Shouldn't it be $p_B=p_A+\rho g(-y_B)?$ Or is the assumption wrong?

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  • $\begingroup$ Neither of these is correct. Why do you think either is right? $\endgroup$ – Chet Miller Jan 31 '18 at 18:48
  • $\begingroup$ I corrected a mistake I made. I meant that $p_B=p_A+\rho g(-y_B)$. Is it still incorrect? $\endgroup$ – RicardoP Jan 31 '18 at 19:15
  • $\begingroup$ No. See my answer below. $\endgroup$ – Chet Miller Jan 31 '18 at 20:05
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The pressure at B is a little tricky. Right at point B, the fluid velocity is equal to that within the exit tube running from B to C. But, within the tank, a few tube diameters upstream of exit B, the flow is converging rapidly toward the exit hole. And, along with the flow convergence, the fluid velocity is increasing from a velocity of basically zero a few diameters upstream, to the much higher velocity $v_{BC}$ at the exit. So right at point B, the pressure is equal to $$p_A=p_B+\rho g (-y_b)+\frac{1}{2}\rho v_{BC}^2\tag{1}$$The condition at point C is $$p_A=p_C=p_C+\rho g (-y_C)+\frac{1}{2}\rho v_{BC}^2\tag{2}$$ Or equivalently: $$\rho g (-y_C)+\frac{1}{2}\rho v_{BC}^2=0\tag{3}$$

Interestingly, if we combine Eqns. 1 and 3, we obtain: $$p_B=p_A-\rho g(y_c-y_b)$$ So, as a result of the exit tube discharging at a lower depth than the tank exit, the pressure at point B is actually below atmospheric (i.e., suction). This is similar to what happens with a siphon.

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  • $\begingroup$ thank you! but how did you arrived to (1)? Bernoulli? $\endgroup$ – RicardoP Feb 1 '18 at 0:03
  • $\begingroup$ Yes. The fluid reaching the exit hole from the tank is already traveling at the same velocity as the velocity in the tube, which is the same as the velocity at the exit of the tube. $\endgroup$ – Chet Miller Feb 1 '18 at 0:27

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