Fluid with laminar flow going through a constriction in a pipe has low

AI Thread Summary
A fluid with laminar flow experiences lower pressure in a pipe constriction due to the conservation of mass and energy principles. As the fluid enters a constricted section, its velocity increases, leading to an acceleration that results in a decrease in pressure according to Bernoulli's principle. The particles collide with the pipe walls with less force because their increased velocity reduces the frequency and intensity of these collisions. This relationship between velocity and pressure is critical, as it illustrates how energy conservation dictates pressure changes in fluid dynamics. Ultimately, the dynamics of fluid flow in constrictions highlight the interplay between velocity, pressure, and energy conservation.
Beetroot
Messages
5
Reaction score
0
Hi

What is the fundamental reason why a fluid with laminar flow going through a constriction in a pipe has lower pressure?

Pressure is defined as force per unit area so the fluid particles must be hitting the pipe wall in the constriction with less force.

Beetroot
 
Physics news on Phys.org


Beetroot said:
Hi

What is the fundamental reason why a fluid with laminar flow going through a constriction in a pipe has lower pressure?

Pressure is defined as force per unit area so the fluid particles must be hitting the pipe wall in the constriction with less force.

Beetroot
1. The conservation of mass tells us that the average velocity in the constricted cross-section must be higher than in a non-constricted section.
2. Thus, those fluid particles traveling along, must have experienced an ACCELERATION
3. In the inviscid fluid, this acceleration can only be achieved if the pressure in the non-constricted region is higher than the pressure in the constricted region.
4. Suppose that somehow the pressure in the constricted region were to increase to the level in the non-constricted region (say, a plug was inserted),. What would happen?
Well, the water prior to the constriction might try to squeeze itself through the wall (not likely to happen..), yet in that case, the actual pressure there would go up, leading either to a reversal of the flow direction, or by the collapse of the cause for the pressure increase in the constricted region (for example by expelling the plug).
 


arildno said:
The conservation of mass tells us that the average velocity in the constricted cross-section must be higher than in a non-constricted section.
The conservation of mass flow ... Assuming mass isn't continously accumulating at a point in the pipe, the mass flow along the pipe is constant, so the fluid velocity is inversely proportional to the cross sectional area.

Beetroot said:
What is the fundamental reason why a fluid with laminar flow going through a constriction in a pipe has lower pressure?
Flow doesn't have to be laminar. The ideal case is when there is no viscosity or friction with the pipe walls, so that the pipe doesn't perform any work on the fluid. Otherwise, the pressure decreases with distance traveled in the pipe as the fluid flows towards a low pressure exit point at the end of the pipe, even with a constant diameter pipe.

Pressure is defined as force per unit area so the fluid particles must be hitting the pipe wall in the constriction with less force.
That's another way of looking at it. The total energy of a volume of fluid or gas is related to the speed and mass density of the molecules. Pressure is related to the momentum of the molecules as they collide with the pipe. If no work is done, and if the molecules have net increase of component of speed^2 in the direction of flow, then that corresponds with a net decrease in the component of speed^2 perpendicular to the direction of flow, and vice versa, assuming that temperature hasn't changed.

Although not directly related, here's a link to a web page discussion the relationship between the Kelvin temperature scale, kinetic energy, heat, and potential energy (Van der Waals force):

http://id.mind.net/~zona/mstm/physics/mechanics/energy/heatAndTemperature/gasMoleculeMotion/gasMoleculeMotion.html
 
Last edited by a moderator:


Hi

So the particles in the constriction are hitting the pipe wall with less force, that has been established. But is the reduction in force due to:

1) The particles hitting the wall less often, due to their increased velocity, or
2) The particles hitting the wall with less force because the velocity component in the direction of flow is greater than the axial component.

Beetroot
 


The others said why the velocity changes, but you asked why the pressure changes. Velocity changes due to conservation of mass - pressure changes due to conservation of energy.
 


Beetroot said:
1) The particles hitting the wall less often, due to their increased velocity
Mass flow is constant, so the rate of collisions with the wall is also constant.
The particles hitting the wall with less force because the velocity component in the direction of flow is greater than the axial component.
Correct, the average speed of the molecules is constant in this case, so if there's a net flow in a particiular direction, the velocity perpendicular to that flow is a bit less.
 
Thread 'Gauss' law seems to imply instantaneous electric field propagation'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Thread 'Recovering Hamilton's Equations from Poisson brackets'
The issue : Let me start by copying and pasting the relevant passage from the text, thanks to modern day methods of computing. The trouble is, in equation (4.79), it completely ignores the partial derivative of ##q_i## with respect to time, i.e. it puts ##\partial q_i/\partial t=0##. But ##q_i## is a dynamical variable of ##t##, or ##q_i(t)##. In the derivation of Hamilton's equations from the Hamiltonian, viz. ##H = p_i \dot q_i-L##, nowhere did we assume that ##\partial q_i/\partial...
Back
Top