Why does the water go from left to right in this question?

Question

In this problem stated by Brilliant course in "Flow in pipes", the water flows from left to right without explaining anything:

My doubt is if I'm supposed to assume there is some greater force that is making the water flow from left to right like if that tube would be in the middle of two sides were the pressure in the left is greater than in the right? Or it just doesn't matter and if water was put to be in this tube without anything happening at both sides, the water would flow to the right? I wouldn't think so since the pressure in C must be higher than in A, right? But then the very next question I get asked in which point there is a higher pressure and supposedly is A:

Doesn't pressure only depend on the transversal area of the liquid at that point? How is $P_A>P_C$? Brilliant states that:

"If water is flowing from left to right as it was in the previous problem, then the water speeds up as it goes through the nozzle. In order for the water to accelerate while it's in the middle of the nozzle, the pressure behind it must be higher than the pressure in front of it. Thus, the pressure at Point A should be the highest"

Which doesn't make sense to me since that would imply that pressure would depend on the direction of the flow? What is going on?

Answer 1

The 'pressure' in Bernoulli's equation is the pressure at a point in a fluid. Any single fluid element exerts an outward pressure. The total pressure at a point is due to all the surrounding elements. It's not just the force exerted by a flowing liquid on something in front of it, so your assertion that it should increase with decrease in cross-sectional area isn't right.

To get an intuitive 'feel' of this, consider a container of water just lying there. Now you made a hole in it. Put your finger on the hole and keep the water from coming out. Consider how much pressure you would have to apply. Now consider how much you would have to apply if it were a tube with flowing water instead. You can imagine that it would be less. Note that this is just for the 'feel', it can of course be derived mathematically from conservation of energy. Wikipedia is a good read.

(Apologies for the MS Paint Diagram)

As to your specific case:

It is clear from conservation of mass that if the liquid is incompressible, $A_Av_A = A_Cv_C$. Since area decreases, speed must increase from A to C.

Now, if you consider any cross section of fluid, the net force acting on it is the vector sum of all forces from the cross-section behind it and the one ahead. (Each infinitesimally small section of fluid exerts radially outward force and therefore pressure, so the net force due to a sheet of such sections is normal to the sheet as shown in the diagram. The non-perpendicular components of adjacent elements will cancel.) In the part of tube where velocity of the fluid is constant, the forces must be equal. But since we have speed increasing from A to C, there must be a pressure gradient, i.e. the pressure coming from behind must be greater than the one coming from forward!

Others have already shown this pressure variation using conservation of energy. I will try to show it more 'mechanistically'.

On a sheet of thickness $dx$, going through a section where pressure changes by $dP$,

$$ F = ma \implies -AdP = dm\cdot v\frac{dv}{dx} $$ But $Av =$ constant. $Adv +vdA = 0$: $$ -AdP = \rho dV v\frac{dv}{dx} = \rho Av dv \implies dP + \rho vdv = 0 $$

If you integrate both sides, you will get the same result that Bernoulli's equation gives at a constant height! Which makes sense, since Newton's laws are intrinsically mingled with conservation of energy.

Another way to visualise this is to imagine a bubble going along with the liquid flow without affecting its idealised characteristics. It will expand in going from A to C.

Answer 2

In a pipe flow problem like this, the first thing to remember is that mass is conserved. That means at any point in the pipe you can draw a vertical plane perpendicular to the flow direction, and the amount of water passing that plane, e.g. 10 Liters/sec, must also pass at the same rate through any other plane. Pipe regions like C that are smaller in area can accommodate less water. So in order to satisfy 10 L/s (in our example), the water there needs to move faster. Otherwise, the water would have to explode out the side walls of the pipe having nowhere else to go. That established the general principle that for a constant flow rate (L/s), diameter and flow speed are inversely related.

Now it turns out that pressure and flow speed are also inversely related, so higher speed areas will have lower pressure, and lower speed areas will have higher pressure (in most cases). While the previous principle is due to conservation of mass, this principle is due to conservation of energy. The fluid has a finite amount of energy per mL. And in faster section of pipe, its kinetic energy increases. This energy must come from somewhere, and corresponds to a decrease in potential energy (i.e. pressure).

Answer 3

When a particle has constant velocity, the net force on it is zero. If we take a small cubic fluid element in A or C region, where the flow speed is constant, the balance of forces between opposing faces is zero.

In the B region the fluid is increasing its horizontal velocity from left to right. So, the force at the left face of the small cube (pointing to the right) is greater than the force at the right face (pointing to the left). As the pressure is force divided by area we can apply the second law of Newton (considering the density constant): $$(P_{left} - P_{right}) S = -\Delta P S = m\frac{\Delta v}{\Delta t} = \rho (S \Delta x) \frac{\Delta v}{\Delta t} = \rho S \Delta v \frac{\Delta x}{\Delta t} = \rho S v\Delta v $$

When the $\Delta$'s go to zero, $$vdv = d (\frac{1}{2}v^2)\implies -dP = \frac{1}{2}\rho d(v^2)$$

So, the pressure decreases from left to right as the velocity increases.

Answer 4

You have conservation of mass and conservation of energy. Conservation of mass means that the flow of the water speeds up. This is because the amount of water flow $ϕ_A$ through area at $A$ has to equal the current at $C$. $$ϕ_A⋅A_A=ϕ_C⋅A_C$$

Area $A$ is bigger than area $C$. This means that $A_A>A_C$. In turn this means that the flow of water at $C$ has to be higher to conserve mass. So: $$ϕ_C>ϕ_A$$

Conservation of energy means the sum of kinetic energy and the potential energy are constant. $$E_{\text{kin}} +E_{\text{pot}} =\text{constant} $$

So the energy at point $C$ and $A$ has to be equal.

$$E_{\text{kin,A}} +E_{\text{pot,A}} =E_{\text{kin,C}} +E_{\text{pot,C}} $$

We know that at point $C$ the velocity of the water is higher than at point $A$ because of mass conservation. So:

$$E_{\text{kin,C}} >E_{\text{kin,A}} $$

This has to mean that the potential energy at point $C$ has to be lower than at point $A$. $$E_{\text{pot,C}} <E_{\text{pot,A}} $$

Because of mass conservation the velocity has to go up. This means the kinetic energy will have to go up And because of energy conservation this means that the potential energy has to go down Which means the pressure in the water has to go down.

Answer 5

I think that you want an explanation in terms of force. However, it turns out that it is inconclusive. More precisely, consider the following momentum balance. Assuming that the velocities are parallel and normal to the tube cross section at $A$ and $C$, I get (assuming stationary flow and an impressible fluid): $$ \rho v_C^2S_C-\rho v_A^2 S_A = -p_CS_C+p_AS_A $$ with $v$ velocity, $\rho$ mass density, $p$ pressure and $S$ cross section area. The LHS is the flux of momentum, namely the outgoing momentum minus the incoming momentum. The RHS is the total force namely the pressure forces at the boundaries.

Recall that by studying the volume flux (or equivalently mass flux if you multiply by $\rho$): $$ v_CS_C-v_A S_A = 0 $$ which is why the fluid speeds up from left to right. This acceleration translates into an accumulation of momentum which must be countered by an unbalanced force. This is why your argument does not work. You are implicitly assuming a constant force, which is not the case here. In fact, the previous argument shows that: $$ F_A>F_C $$ i.e. the fluid is more pushed at the back ($A$) rather than more impeded at the front ($C$) which is why it accelerates. Now you know: $$ p_AS_A>p_CS_C\\ S_A>S_C $$ To conclude that $p_A<p_C$, it would suffice that $F_A\leq F_C$. However, it is the opposite here, so just with the information gathered so far, you cannot compare $p_A$ and $p_C$.

This is why the other answers directly go for the energy balance, i.e. Bernoulli's law. You can apply the same method: $$ \frac{1}{2}\rho v_C^2v_CS_C - \frac{1}{2}\rho v_A^2v_AS_A = -p_Cv_CS_C+p_Av_AS_A $$ Using the volume balance, you can simplify it to: $$ \frac{1}{2}\rho v_C^2 - \frac{1}{2}\rho v_A^2 = -p_C+p_A $$ so that you can conclude that $p_A>p_C$.

Hope this helps.

Answer 6

Since it's given in the question that fluid is moving left to right. There must we external work done as you say

Pc > Pa (in ideal case)

Since it's moving if we use Equation of continuity we can see speed at C is more than speed at A due to decrease in area.

So now if we apply Bernoulli theorem here you would wind that

Pa-Pc > 0

hence from here we can see that pressure at A is more than pressure at C.

So in layman's term if you want to understand we can say that a fluid flows from a region of high pressure to low pressure so if it's given that fluid is moving from left to right point A must have hight pressure than point C it can be created due to a external cause

Answer 7

I also feel a bit puzzled with this question but my explanation goes as follows:

Let's call the larger radius at point A as R and smaller one at C as r. Lets call the distance between A and C as L (assume that vector AC makes zero angle with horizontal). Now think of a cylinder of radius r and length L which is aligned on the line passing through segment AC and it moves from the negative infinity to positive infinity (assuming the extension of this tube extends on this boundaries but this is not really essential).

What happens to the total momentum of the water inside that tube as it moves from -infinity to +infinity? Well, as its center of mass gets closer to A (or C equivalently) the momentum increases, because the water gets faster at one end point. Then this means there exists a force to the right on this tube near our tube of interest, and particularly at when tube extends between A and C we expect the momentum to keep increasing .

Then it seems that pressure difference only arises so as to make the mass conserved since velocity increase was known not by Newton's equation but mass conservation