Understanding the Bernoulli Equation: Solving a Fluid Dynamics Problem

[JulienB]

Hi everybody! I'm training for an upcoming exam, and I'd like to know if I correctly use the Bernoulli equation.

Homework Statement

The problem is kind of difficult to describe without a drawing, please check out the attached jpg to see the situation and the known data.
The questions are:
a) How much time t is needed for a volume V = 1L of fluid to cross the hopper?
b) What is the diameter of d₂?

Homework Equations

Bernoulli equation, equation of continuity, formulas to calculate the flux of a fluid

The Attempt at a Solution

a) I use the Bernoulli equation at points (1) and (0):

P₁ + ρgh₁ + ½ρv₁² = P₀ + ρgh₀ + ½ρv₀²

Now I think that P₁ = 0 (if we take the gauge pressure), ½ρv₁² ≈ 0 (because the diameter at point (1) is much larger than at point (0) (equation of continuity)) and ρgh₀ = 0 (because I choose h₀ = 0).

What I am not sure of is that I assume P₀ = 0 because the fluid is not being stopped at that point, so it should have no pressure. Is that correct? If so, here is where that takes me:

ρgh₁ = ½ρv₀²
⇔ v₀ = √(2gh₁)

We can then use the flux to determine the time it takes for a litre of fluid to cross the hopper:

R = V/t = A₀ ⋅ v₀
⇒ t = V/(¼ ⋅ π ⋅ d₀² ⋅ √(2⋅g⋅h₁)) = 23.5 s

b) I use the Bernoulli equation again at points (0) and (2).

P₀ + ρgh₀ + ½ρv₀² = P₂ + ρgh₂ + ½ρv₂²

I assume P₀ = 0 and P₂ = 0 for the same reason as before (is that right?) and ρgh₂ = 0 (because I now choose h₂ = 0)

⇒ v₂ = √(v₀² + 2⋅g⋅h₀)

The equation of continuity says A₀ ⋅ v₀ = A₁ ⋅ v₁
⇒ A₂ = A₀ ⋅ v₀/v₂
⇒ r = 1.5 mm
⇒ d₂ = 3 mmMy results are plausible, but I am unsure about how to deal with the pressure when there is a flow. Is there no pressure on the right side of the equation when there is a flow? Is that always the case?Thank you very much in advance for your answers.

 

[The Vinh]

you didn't have to assume Po = 0 because it is the fact.

 

[JulienB]

Thank you very much for your answer. I'm still unclear about how is that a fact: does pressure only apply in a "closed" container? Anytime there is a hole or some sort of opening at the bottom of the container (or at a point we are interested in), there is never pressure because the water flows?Julien.

 

[The Vinh]

In your case, the water flows but it is not at steady state. Static pressure only exists when liquid is steady. Assume that i haven't talked about this, only base on logic, how do you manage to solve this because Po = P1 + pGH1 => V2 = 0.

 

[haruspex]

you didn't have to assume Po = 0 because it is the fact.

No, you should not assume P₀ is 0 because it won't be!

In your case, the water flows but it is not at steady state. Static pressure only exists when liquid is steady.

The state only changes relatively slowly (since d₁>>d₀). That means acceleration is negligible, so the usual equations can be used to find the velocity as a function of head height. Only then do we need to consider that head height, and therefore v, change over time.

@JulienB, allow P₀ to be nonzero and consider the flow in the narrow section.

 

[The Vinh]

No, you should not assume P₀ is 0 because it won't be!

The state only changes relatively slowly (since d₁>>d₀). That means acceleration is negligible, so the usual equations can be used to find the velocity as a function of head height. Only then do we need to consider that head height, and therefore v, change over time.

@JulienB, allow P₀ to be nonzero and consider the flow in the narrow section.

Oh, you win, i have made a huge mistake

 

[The Vinh]

JulienB, i think that i have found the problem and here it is : P1 + pGH1 = P0 + 0,5pV^2 ( P1 = P0 = P air ) => pGH1 = 0,5pV^2 ( it still the same but P1 = P0 = P air)
I forgot about air pressure
I hope that it is right

 

[haruspex]

JulienB, i think that i have found the problem and here it is : P1 + pGH1 = P0 + 0,5pV^2 ( P1 = P0 = P air ) => pGH1 = 0,5pV^2 ( it still the same but P1 = P0 = P air)
I forgot about air pressure
I hope that it is right

No, it is not to do with air pressure. P₀ > P₁. Why do you think they should be equal?

 

[The Vinh]

No, it is not to do with air pressure. P₀ > P₁. Why do you think they should be equal?

it always exists air pressure and i knew that Po = P1 + pGH but if i had really done it then the formula would have been like this: Po = Po + 0,5pV^2 then V = 0 , it is so confusing, huh.

 

[JulienB]

Hi everyone and thank you for your answers. It is a bit confusing though :)

@haruspex I don't get why you're stating that P₁ ≠ 0. If we consider gauge pressure (that is, ignoring air pressure), what other factors may cause a pressure at P₁? Maybe I am wrong, but if so I would like to understand why.

Thank you very much again.Julien.

 

[JulienB]

@haruspex What precisely confuses me, is that I have understood that for example in the case of a side hole on a recipient we should consider P = 0 at that exit. In th case of the present problem, the "hole" is located underneath the fluid. Is that why we should consider the pressure not being 0 for P₀? It is kind of unlogical to me that the pressure of the water matters when it can move away, but if you say so I'm ready to believe :)

 

[haruspex]

@haruspex What precisely confuses me, is that I have understood that for example in the case of a side hole on a recipient we should consider P = 0 at that exit. In th case of the present problem, the "hole" is located underneath the fluid. Is that why we should consider the pressure not being 0 for P₀? It is kind of unlogical to me that the pressure of the water matters when it can move away, but if you say so I'm ready to believe :)

Consider the linear velocity of the water in the two sections of pipe. Are they the same? If not, which is greater, and what can account for the acceleration?

 

[JulienB]

@haruspex Well the equation of continuity (A₁⋅v₁ = A₀⋅v₀) states that the water flows faster by the smallest opening (point (0)). And I guess that that is happening because of a greater pressure at that point? Now the formulas I know for pressure are P = F/A or for an submerged object P = ρgh (probably irrelevant here). Is there another one I don't know, or am I missing something important? I can't say much about the force applying on the fluid at that point.

 

[haruspex]

Well the equation of continuity (A1⋅v1 = A0⋅v0) states that the water flows faster by the smallest opening (point (0)). And I guess that that is happening because of a greater pressure at that point?

Exactly. So, as I suggested in post #5, apply the Bernoulli equation to the flow in the narrow section.

 

[Chestermiller]

As I see it, the pressure at point 2 is atmospheric and the pressure at the very top of the hopper is atmospheric. So, from Bernoulli, $$v_2=\sqrt{2g(h_1+h_2)}=\sqrt{2g(0.115+0.026)}=1.66m/s$$
This means that the pressure immediately below point 0 (just below the convergence) is sub-atmospheric. The gauge pressure there would just be $-\rho g h_2=-\rho g(0.026)$. It seems that there would be a jump change in pressure from just above point 0 to just below point 0, with the pressure above being higher than the pressure below. But, the question is, "how much higher?" It seems to me there is not enough information provided to solve this problem without knowing d2.

Chet

 

[JulienB]

@haruspex so I took P₀ = -ρgh₁ -a little out of despair, I must admit- and got that result:

 

[JulienB]

Oops the "2" in the last square root should obviously not be there, but the result was obtained without including it in the calculations. With that t I now get for next question d₂ ≈ 2.5 mm.

I'm very skeptical about my negative pressure though :/

 

[JulienB]

@Chestermiller Are you saying that the gauge pressure just below point (0) is -pgh₂? I'm afraid I don't really understand why: I thought the pressure at point (0) would be depending on the height of the fluid above that point (h₁) and fail to see the connection with h₂.

 

[Chestermiller]

Here is a more fleshed out analysis of my discussion in post #15.

Let z = h₁= elevation of top surface of liquid in hopper
Let z = 0⁺ = elevation of top of short diameter convergence
Let z = 0^- = elevation of bottom of short diameter convergence
Let z = - h₂ = exit elevation

Bernoulli term at each location:

$z = h_1$: $$0 + \rho g h_1 + 0$$
$z=0^+$:$$p^++0+\frac{1}{2}\rho v_0^2$$
$z=0^-$:$$p^-+0+\frac{1}{2}\rho v_2^2$$
$z=-h_2$:$$0-\rho g h_2+\frac{1}{2}\rho v_2^2$$

All the above Bernoulli terms must be equal to one another.

In addition, the velocities in the two tubular sections must satisfy:
$$v_0d_0^2=v_2d_2^2$$

These equations cannot be solved unless we know d2.

 

[Chestermiller]

@Chestermiller Are you saying that the gauge pressure just below point (0) is -pgh₂? I'm afraid I don't really understand why: I thought the pressure at point (0) would be depending on the height of the fluid above that point (h₁) and fail to see the connection with h₂.

See my post #19 for more details. If you can find something wrong with that, please point it out. Setting the Bernoulli terms at the lowest two elevations equal to one another clearly produces this result.

 

[JulienB]

@Chestermiller I see what you are saying. However, we solved that problem in my physics class, although I don't have the solution sheet anymore. I will ask other students if they still have it, and meanwhile I will try again to solve it myself.

@haruspex What do you think about what Chestermiller said. Do you agree?

Thank you both of us for your answers. :)

 

[haruspex]

. It seems that there would be a jump change in pressure from just above point 0 to just below point 0, with the pressure above being higher than the pressure below. But, the question is, "how much higher?" It seems to me there is not enough information provided to solve this problem without knowing d2.

Chet

The question asks to find d₂, but clearly there is insufficient information because one could just make d₂ smaller and no given information would be contradicted. So my guess is that there is, or was intended to be, some other piece of information provided.

 

[Chestermiller]

@Chestermiller Are you saying that the gauge pressure just below point (0) is -pgh₂? I'm afraid I don't really understand why: I thought the pressure at point (0) would be depending on the height of the fluid above that point (h₁) and fail to see the connection with h₂.

If I use the first and third points (as you allude to), I obtain:

$$p^-=\rho g h_1-\frac{1}{2}\rho v_2^2$$not just $\rho g h_1$
If I use the first and 2nd points, I obtain:$$p^+=\rho g h_1-\frac{1}{2}\rho v_0^2$$again, not just $\rho g h_1$.

 

[Chestermiller]

The question asks to find d₂, but clearly there is insufficient information because one could just make d₂ smaller and no given information would be contradicted. So my guess is that there is, or was intended to be, some other piece of information provided.

I think the person who formulated the problem just messed up and was confused, and then transmitted this incorrect analysis to the class.

Chet

 

[JulienB]

I checked and I didn't omit any information from the problem... Sorry for wasting your time on an unsolvable problem! :(

And thank you for your answers, it did help me very much to comprehend better the Bernoulli equation and the pressure at different points of a fluid.

Julien.

 

[Chestermiller]

I checked and I didn't omit any information from the problem... Sorry for wasting your time on an unsolvable problem! :(

And thank you for your answers, it did help me very much to comprehend better the Bernoulli equation and the pressure at different points of a fluid.

Julien.

You have nothing to apologize for. I think many members learned lots of good stuff from this thread. We should be thanking you.