1
$\begingroup$

I already know the solution to this problem, but its consequences puzzle me.

We have a pipe with a steady flow of fluid in it. At some point the pipe bends, changing direction from $\hat{n}_a$ to $\hat{n}_b$ and, at the same time, changing section area from $A_a$ to $A_b$.

The fluid before the bending has density $\rho_a$, pressure $p_a$ and velocity $\vec{v}_a$ (parallel to $\hat{n}_a$). After the bending, these values change to $\rho_b$, $p_b$ and $\vec{v}_b$ (parallel to $\hat{n}_b$). The question of the problem is: what is the force that the water exerts on the pipe, as a consequence of the change in direction $\hat{n}$ and section area $A$?

My professor derived this result:

$\vec{F} = (p_a A_a + \rho_a v_a^2 A_a) \hat{n}_a - (p_b A_b + \rho_b v_b^2 A_b) \hat{n}_b, \tag{1}$

which I'm not questioning here. Instead, I have two questions about the consequences of this formula.


First question

Let's assume that the fluid is still, so that $\vec{v}_a = \vec{v}_b = 0$, and let's take $A_b = A_a$ and $p_b = p_a$. Then the fluid exerts on the pipe a force $\vec{F} = p_a A_a (\hat{n}_a - \hat{n}_b)$, that is not zero. But nothing is moving! How is this possible? Does the pipe have to bear a force even when the fluid is still? For example, a pipe with an elbow full of still air at atmospheric pressure should experience a force, but of course this doesn't agree with everyday experience.


Second question

Now let's consider a different situation. This time the fluid is moving, but the direction of the pipe doesn't change ($\hat{n}_a = \hat{n}_b = \hat{n}$), only the section changes from $A_a$ to $A_b$, and I will assume that the pipe is getting smaller, that is $A_b \lt A_a$. The force $(1)$ then becomes:

$\vec{F} = (p_a A_a - p_b A_b + \rho_a v_a^2 A_a - \rho_b v_b^2 A_b) \hat{n} . \tag{2}$

In orther to simplify the problem, let's assume that the fluid is incompressible ($\rho_a = \rho_b = \rho$). Then the conservation of mass requires $v_a A_a = v_b A_b$, that is:

$v_b = v_a \frac{A_a}{A_b} . \tag{3}$

Now I can calculate $p_b$ by means of the Bernoulli Theorem, obtaining:

$p_b = p_a - \frac{1}{2} \rho (v_b^2 - v_a^2) = p_a - \frac{1}{2} \rho v_a^2 \left( \frac{A_a^2}{A_b^2} - 1 \right) . \tag{4}$

Now I substitute equations $(3)$ and $(4)$ in equation $(2)$ and obtain (I skip the details of the calculation):

$\vec{F} = (A_a - A_b) \left(p_a - \frac{1}{2} \rho v_a^2 \frac{A_a - A_b}{A_b} \right) \hat{n} . \tag{5}$

But this leads me to a strange conclusion: since the term depending on the speed is negative while the pressure one is positive, this formula means that, when the fluid is moving, the force that it exerts on the narrowing pipe is less than the force exerted when the fluid is still. Moreover, for high enough speeds, the force can even change sign and become antiparallel to $\hat{n}$.

This looks completely counterintuitive to me. Can this be right? Or maybe there is a mistake in my reasoning?

$\endgroup$
  • 1
    $\begingroup$ "We have a pipe with a stationary stream of fluid flowing in it." Pretty pedantic nitpick, but I assume you didn't mean "stationary" here. Stationary fluid flow is a bit of a paradox. $\endgroup$ – JMac Oct 22 '19 at 16:58
  • $\begingroup$ @JMac Sorry, I meant "steady". I'm editing the post. $\endgroup$ – HicHaecHoc Oct 22 '19 at 17:11
  • $\begingroup$ In German I know the Word „stationär“ for fluid flows, i.e. all time derivatives are zero. Does it translate to “steady” or, indeed, to “stationary”? $\endgroup$ – Hartmut Braun Oct 23 '19 at 11:05
  • $\begingroup$ @HartmutBraun I'm not an English speaker either, but I went to Wikipedia and found this: en.m.wikipedia.org/wiki/Fluid_dynamics#Steady_vs_unsteady_flow, so it seems that the more used expression for that is "steady flow". My original expression "stationary stream" was thus doubly confusing. I apologize for that. $\endgroup$ – HicHaecHoc Oct 23 '19 at 12:18
  • 1
    $\begingroup$ @2b-t Yes, but the momentum pointing into the volume is also entering the volume, while the momentum pointing out of the volume is exiting it (thus adding to the volume an opposite momentum). The reason why the sign of momentum is equal to the pressure one in both cases is that $(-1) \cdot (-1)=+1$. $\endgroup$ – HicHaecHoc Oct 27 '19 at 23:00
1
$\begingroup$

After some thinking, I came up on my own with what I think is an answer. I post it here for anyone interested.


First question

This is the case in which the pipe has constant section $A$ and changes direction from $\hat{n}_a$ to $\hat{n}_b$. The fluid inside is still and has pressure $p$.

In this situation the fluid does exert a force $\vec{F} = p A (\hat{n}_a - \hat{n}_b)$, which originates from pressure alone, without any motion. The reason why the existence of this force may be counterintuitive (at least, it was for me), is that in real practical conditions, outside the pipe there is air at atmospheric pressure $p_{atm}$, so this air exerts another force on the pipe, a force which I didn't take into account at first.

The value of this force can be quickly deduced by considering the situation in which also the fluid inside the pipe is air at atmospheric pressure. In this case we know that the total force on the pipe is zero, of course. But our formula says that the air inside is exerting a force equal to $\vec{F} = p_{atm} A (\hat{n}_a - \hat{n}_b)$, so the force that the external air exerts on the pipe must be the opposite: $\vec{F}_{ext} = -p_{atm} A (\hat{n}_a - \hat{n}_b)$.

In conclusion, the total force experienced by the pipe immersed in the atmosphere, when there is a still fluid at pressure $p$ inside it, is:

$\vec{F}_{tot} = \vec{F} + \vec{F}_{ext} = p A (\hat{n}_a - \hat{n}_b) - p_{atm} (\hat{n}_a - \hat{n}_b) = (p - p_{atm}) A (\hat{n}_a - \hat{n}_b).$

This picture reconciles the equations with my intuition.


Second question

This time the fluid is flowing and the pipe doesn't change direction $\hat{n}$, but only changes section from $A_a$ to $A_b$, shrinking ($A_b \lt A_a$). In this situation the force exerted by the fluid on the pipe, in the simple case of incompressible fluid, is given by equation $(5)$ of the question:

$\vec{F} = (A_a - A_b) \left(p_a - \frac{1}{2} \rho v_a^2 \frac{A_a - A_b}{A_b} \right) \hat{n} . \tag{I} $

It puzzled me that, since the pressure term and the speed term have opposite signs, a greater speed leads to a smaller force. Anyway, this is nothing else but a consequence of the usual behavior of pressure, that reduces with increasing speed (a consequence of Bernoulli Theorem).

Indeed, as noted in the question in equation $(4)$, from Bernoulli Theorem we can calculate the drop in the pressure produced by the change in the section:

$p_a - p_b = \frac{1}{2} \rho v_a^2 \left( \frac{A_a^2}{A_b^2} - 1 \right) . \tag{II}$

As the section area shrinks, the fluid's speed increases and the pressure drops, and equation $(\mathrm{II})$ says that the total drop is greater with increasing initial speed $v_a$. Therefore, if we increase $v_a$ keeping $p_a$ constant, the pressure inside the pipe decreases and so does the force exerted on the pipe itself.

Another thing that puzzled me was that with high enough a speed, the force could change sign. If this was the case, the fluid would be "sucking" the pipe instead of "pushing" it, even when there is vacuum outside of the pipe. This would be impossible for an ordinary fluid, and indeed we can prove that this condition is not possible in reality. Let's consider again equation $(\mathrm{I})$: for the force to change sign, we need

$p_a < \frac{1}{2} \rho v_a^2 \frac{A_a - A_b}{A_b} . \tag{III}$

Since $A_a > A_b$, we can write

$\frac{1}{2} \rho v_a^2 \frac{A_a - A_b}{A_b} < \frac{1}{2} \rho v_a^2 \frac{A_a - A_b}{A_b} \cdot \frac{A_a + A_b}{A_b} = \frac{1}{2} \rho v_a^2 \frac{A_a^2 - A_b^2}{A_b^2} = \frac{1}{2} \rho v_a^2 \left( \frac{A_a^2}{A_b^2} - 1 \right) . \tag{IV}$

Putting together inequalities $(\mathrm{III})$ and $(\mathrm{IV})$, we get $p_a \lt \frac{1}{2} \rho v_a^2 \left( \frac{A_a^2}{A_b^2} - 1 \right) , \tag{V}$ that is $p_a - \frac{1}{2} \rho v_a^2 \left( \frac{A_a^2}{A_b^2} - 1 \right) \lt 0 . \tag{VI}$

Now we put together equation $(4)$ from the question with inequality $(\mathrm{VI})$ and obtain: $p_b = p_a - \frac{1}{2} \rho v_a^2 \left( \frac{A_a^2}{A_b^2} - 1 \right) \lt 0 . \tag{VII}$

In conclusion, for the force to change sign, the pressure $p_b$ should be negative, which is impossible in an ordinary real fluid. Therefore, this situation is not possible in reality.

$\endgroup$
  • $\begingroup$ Damn, you did a good job solving your own problem. I immediately saw the answer to question 1 and arrogantly assumed that I fully understood question 2 as well, but then I read your answer and saw what I had overlooked. Thanks for the good content. $\endgroup$ – Duncan Harris Oct 29 '19 at 2:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.