Pressure in a Fluid-Filled Glass with a Side-Hole Tube: Is Pa = Pb = Pc?

[EMP]

Homework Statement

It's not a homework exercise, it's more of a conceptual question to help my understanding of the subject. Imagine that you have a glass of height H filled with a fluid with density ro and viscosity eta. There is a hole somewhere on a side of the glass, at a height h relative to the top of the fluid. In this side-hole there is a tube attached, uniform, all cross-sections of this tube of length l have the same area. In the tube, which is the thing I'm concerned about (and where the fluid flows out of the glass), the first cross-section is A, the middle of the tube is B, and the end of the tube is C. Okay, my questions are these:

Is the pressure in point A defined by ro*g*h alone?
What is the pressure in point B?
And in C?

Homework Equations

Bernoulli equation -> Pa + ro*g*h + (1/2)*ro*Va^2 = Pb + ro*g*h + (1/2)*ro*Vb^2
Continuity equation -> Sa*Va = Sb*Vb

The Attempt at a Solution

By using Bernoulli's equation between point A and B, for example, I cancel out the potential energy element in both sides of the equation because the height doesn't change. The kinetic energy component of the fluid is also cancelled, because according to the conservation of mass, and given that the cross-section area doesn't change throughout the tube, Va = Vb. That leaves me with Pa = Pb. Is it true that Pa = Pb = Pc? I don't think so, but I can't prove otherwise.
Thanks.

 

[Chestermiller]

See my answer in post # 4 of this thread: https://www.physicsforums.com/threads/pressure-in-piezometers-with-ideal-fluid.896857/#post-5642369

The pressure at point a is not hydrostatic. It is true that Pa = Pb = Pc.