# Steady State Flow out of a Tank Through a Vertical Pipe

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1. Feb 16, 2016

### kartmaze

This is not homework, but something work-related. However, I will post it here since it's homework-relatable.

1. The problem statement, all variables and given/known data

Known variables: Qin , A (and obviously the diameter D of the pipe as well as g = 9.81 m/s2).
The pipe outlet and the top of the tank is at atmospheric pressure.
I want h1 to be as small as possible.

What I'm looking for is an expression for h2.

For the moment I'm assuming zero pipe friction.

2. Relevant equations

Torricelli's law (v is the same as U):

Q = U*A

Might be more?

3. The attempt at a solution
In steady state, Qin and Qout are equal (=Q)

I tried solving for h using the two equations above and got

h = Q2 / (2*g*A2)

Is this h h1, h2, the sum of h1 and h2 or neither of these things?

Is this even the right way to approach this problem?

2. Feb 16, 2016

### haruspex

You used h in applying Torricelli's law. What does h need to represent for that law to apply?

3. Feb 16, 2016

### kartmaze

Thanks for replying :)

I think h needs to represent the water level in a tank or equivalent, but it's the hole in the bottom of the pipe in my scenario (amongst other things) that confuses me. Are you implying that the pressure at the pipe outlet will be equal between this scenario and a scenario where the pipe oulet is clogged and Q = 0?

EDIT: When I think of it, my second sentence isn't relevant. When applying Torricelli's law, it's to the inlet of a pipe. So now I'm guessing that h in the equation is h1 . However, I'm not sure how to add the velocity that is driven by gravity (since the pipe is vertical and not horizontal).

4. Feb 16, 2016

### kartmaze

(double post)

5. Feb 16, 2016

### haruspex

I'll try a different hint. At what points are the pressures known to be equal?

6. Feb 16, 2016

### Staff: Mentor

Let's apply Bernoulli's equation at 3 locations:

Point 1: The top of the fluid in the tank where the pressure is atmospheric Pa. If the datum for potential energy is at the outlet of the pipe, what is the $\rho g h$ at this location? What is the velocity at this location?

Point 2: The top of the outlet pipe, at h=h2. Call the pressure P at this location. What is the $\rho g h$ at this location? What is the velocity at this location?

Point 3: The outlet of the pipe. What is there pressure here? What is the $\rho g h$ here? What is the velocity at this location?

From this assessment, write out Bernoulli's equation for points 1 and 3, and then for points 2 and 3.

7. Feb 17, 2016

### kartmaze

My apologies for not understanding much of this. My fluid dynamics knowledge is almost zero, so I'm learning by "Googling" at the moment.

Point 1:
The velocity is zero. And the pressure is atmospheric, so we're only left with the potential energy part of the Bernoulli equation:

$g(h1+h2) = constant$

Point 2:
I think the pressure P at this loacation is $\rho g h$ which is $\rho g h_1$ . I also think the velocity at this location has to be the velocity in the pipe: $Q / A$. Putting this into Bernoulli's equation I get:

$Q^2 / ( 2 A^2 ) + g h_2 + \rho g h_1 / \rho = constant$

which becomes

$Q^2 / ( 2 A^2 ) + g(h_1 + h_2) = constant$

This is where I think I have done something fundamentaly wrong somewhere, since solving for ponts 1 and 2 would give zero flow no matter what.

Point 3:
The velocity should be the same as in point 2, I think: $Q / A$. I'm really unsure about the pressure here. I would guess it's atmospheric. However, $\rho g h$ is $\rho g(h_1 + h_2)$.

...

When I got to this point in this post, I realized that I maybe shouldn't just insert $\rho g h$ in as P in Bernoulli's equation, but instead find an equation for the pressure difference between the two points. So I did that and got:

$P_3 = \rho ( g ( h_1 + h_2 ) - v^2 / 2)$

for both point 1 and 3, and 2 and 3.

If the pressure at P3 is atmospheric, I can set this to zero and get:

$h_2 = (v^2 / 2g) - h_1$

Am I on the right track here, or am I completely off?

EDIT: I see now that I've changed the symbol for velocity from U to v. For clarification purposes, this is the same thing.

EDIT2: To my surprise, this is the exact equation from my original post if h1 is zero. So I guess, given that I'm on the right track, that the answer to my question regarding which h it is is the sum of h1 and h2. :)

Last edited: Feb 17, 2016
8. Feb 17, 2016

### Staff: Mentor

None of this was correct.

Point 1: $p_a+\rho g (h_1+h_2)+0$

Point 2: $P+\rho g h_1+\rho \frac{v^2}{2}$

Point 3: $p_a+0+\rho \frac{v^2}{2}$

So, from Point 1 and Point 3, we have: $p_a+\rho g (h_1+h_2)+0=p_a+0+\rho \frac{v^2}{2}$ or $$v=\sqrt{2g(h_1+h_2)}\tag{1}$$

From Points 2 and 3 we have: $P+\rho g h_1+\rho \frac{v^2}{2}=p_a+0+\rho \frac{v^2}{2}$. This gives:
$$P=p_a-\rho \frac{v^2}{2}$$
This equation indicates that the pressure at the very entrance to the tube is below atmospheric. So, the tube is sucking fluid out of the tank. That's why Eqn 1 has h_2 included in the total height of water responsible for the velocity.

Note that this analysis is valid as long as h1 is large compared to the diameter of the tube. If it is not, then the velocity at the upper surface is non-uniform (disturbed by the presence of the tube), and our equation for the upper surface is no longer valid.

Chet

9. Feb 17, 2016

### kartmaze

Thank you so much for that comprehensive reply, Chet! :-)
That underpressure at point 2 is interesting. I did get the same expression for v as you though (Eqn 1), which I reorganized to an expression for h2, but my way there was a bit sketchy to say the least (and maybe just a fluke). When looking at your setup, things get a lot clearer. I've learned a lot, and I really appreciate your effort :-)

A good thing is also that you mentioned that this analysis only is valid when h1 is large compared to the tube diameter.

Thanks again.