# Bernoulli's equation: tube attached to a filled tank

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. 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?

• Neither of these is correct. Why do you think either is right? – Chet Miller Jan 31 '18 at 18:48
• I corrected a mistake I made. I meant that $p_B=p_A+\rho g(-y_B)$. Is it still incorrect? – RicardoP Jan 31 '18 at 19:15
• No. See my answer below. – Chet Miller Jan 31 '18 at 20:05

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.