Derivation of pressure differences in a stream.

In summary: This is why your calculation only works for small changes in diameter.In summary, the conversation discusses the attempt to reach Bernoulli's principle using a horizontal pipe system with different cross-sectional areas and velocities. The conversation covers the calculation of forces and pressure differences, and concludes that the approach is only valid for small changes in diameter due to the assumption of localized pressure changes.
  • #1
chingel
307
23

Homework Statement


I tried to reach the Bernoulli principle this way:
Two pipes are connected, one has a cross-sectional area of [itex]S_{1}[/itex] and speed of [itex]v_{1}[/itex]; [itex]S_{2}[/itex] and [itex]v_{2}[/itex] for the other. The pipes are horizontal, the connecting wall between them at the crossing from one pipe to the other is vertical, water flows from the first pipe to the second, the first one is larger.

The Attempt at a Solution


In any second, a mass of fluid [itex]m=S_{2}v_{2}\rho=S_{1}v_{1}\rho[/itex] has gone from the larger pipe to the smaller and its velocity has changed by [itex]dv=v_{2}-v_{1}[/itex]. Therefore the change in momentum is [itex]F=dmv=S_{2}v_{2}\rho(v_{2}-v_{1})[/itex]. This force comes from the differences in pressure between the two regions and since the size of the connection between them is [itex]S_{2}[/itex], because that is the area of the smaller pipe, then the difference in force at the connection is [itex]dF=S_{2}(P_{1}-P_{2})[/itex]. Equating the two forces and doing some simplifying, I get that [itex]\frac{P_{1}}{\rho}+v_{1}v_{2}=\frac{P_{2}}{\rho}+v_{2}^{2}[/itex], which is clearly the wrong answer.

In Wikipedia, there is the derivation using conservation of energy:

http://en.wikipedia.org/wiki/Bernoulli's_principle#Derivations_of_Bernoulli_equation

However, I would like to know where did I go wrong and why did it give the wrong answer.
 
Physics news on Phys.org
  • #2
I think your problem is that you are assuming the pressure would be uniform in each section of pipe. The calculation, if valid at all, considers the pressures either side of the change in width, as though the pressure change were sudden. In practice, the pressure would change over some volume approaching the boundary.
Bernoulli's equation is derived for a different scenario. Flows are laminar, and the cross-section changes slowly enough that the pressure at some point along the pipe is meaningful.
You could try to rescue your approach by considering the area change to be just small delta. I notice that the difference between the equation you get and Bernoulli's is that between a factor of v2 and one of (v1+v2)/2. That might disappear as a second order small quantity.
 
  • #3
Yes good idea, if the force due to the pressure differences were not [itex]F=S_{2}/(P_{1}-P_{2})[/itex], but instead it were [itex]F=S_{2}/(P_{1}-P_{2})*\frac{2v_{2}}{v_{1}+v_{2}}[/itex] then I would get the right answer. But since the extra term is very near one when the speeds are similar, I consider only small changes in v and integrate. Not that [itex]F=S_{2}v_{2}\rho(v_{2}-v_{1})=S_{2}(P_{1}-P_{2})[/itex] as previously, but instead [itex]v_{2}\rho*dv=-dP[/itex], having divided the S2 out. This gives [itex]\frac{dP}{dv}=-v\rho[/itex]. Then [itex]P=-\frac{v^2}{2}\rho+C[/itex], from where [itex]P+\frac{v^2}{2}\rho=C[/itex].

It was wrong to assume the force difference is just the connecting area times the pressure difference, because for example how would I know if the force exerted by the perpendicular wall at the connection point is exactly enough to counteract the larger area the pressure in the first pipe acts on. I assumed that the force difference is just the connecting area times pressure difference, thinking that the extra cross-sectional area the pressure in the larger pipe acts on is exactly counteracted by the perpendicular walls, which might not be the case because the flow there is probably different than further away from the connection.

But now I wonder why exactly is it right for small changes in diameter that the force is the connection size times the pressure difference, if it isn't right for larger changes?
 
  • #4
chingel said:
But now I wonder why exactly is it right for small changes in diameter that the force is the connection size times the pressure difference, if it isn't right for larger changes?
Because you localised the area change right at the constriction, while the pressure change in reality would not be that localised.
 
  • #5


I appreciate your attempt to derive the pressure differences in a stream using the Bernoulli principle. However, it seems that your approach may have overlooked some important factors.

Firstly, the Bernoulli principle states that in a steady flow of an ideal fluid, the sum of the static pressure, dynamic pressure, and gravitational potential energy per unit volume remains constant along a streamline. In your derivation, you have only considered the change in momentum, but have not taken into account the change in energy.

Additionally, the Bernoulli equation is usually applied to a single streamline, rather than the entire fluid flow. In this case, the change in velocity (dv) and the change in pressure (dF) occur along the same streamline, but in opposite directions. Therefore, the equation should be written as S_{2}v_{2}\rho(v_{2}-v_{1})=-S_{2}(P_{1}-P_{2}).

Furthermore, the Bernoulli principle assumes that the fluid flow is incompressible and has a constant density. In your derivation, you have used the same density for both pipes, but in reality, the density may change as the fluid flows from the larger pipe to the smaller one. This would also affect the accuracy of your result.

In summary, while your approach is commendable, it may not accurately represent the pressure differences in the stream. It would be more appropriate to use the energy conservation approach as shown in the Wikipedia derivation, which takes into account all the relevant factors and assumptions. I hope this explanation helps in understanding where you may have gone wrong in your derivation.
 

Related to Derivation of pressure differences in a stream.

What is the derivation of pressure differences in a stream?

The derivation of pressure differences in a stream is a scientific process that explains the relationship between the velocity of a fluid in a stream and the pressure exerted by that fluid. It involves applying the principles of fluid dynamics and Bernoulli's equation to calculate the pressure differences in a stream.

What is Bernoulli's equation and how is it related to pressure differences in a stream?

Bernoulli's equation is a fundamental equation in fluid dynamics that states that the total pressure of a fluid in a closed system remains constant. This means that as the velocity of a fluid in a stream increases, the pressure exerted by that fluid decreases. Bernoulli's equation is used to derive the pressure differences in a stream.

What factors affect the pressure differences in a stream?

The pressure differences in a stream are affected by several factors, including the velocity of the fluid, the density of the fluid, and the shape of the stream. Additionally, external factors such as elevation, gravity, and viscosity can also impact pressure differences in a stream.

How is the velocity of a fluid related to pressure differences in a stream?

The velocity of a fluid in a stream is directly related to the pressure differences in that stream. As the fluid's velocity increases, the pressure exerted by the fluid decreases. This is due to Bernoulli's equation, which states that in a closed system, the total pressure remains constant.

What are some practical applications of understanding the derivation of pressure differences in a stream?

Understanding the derivation of pressure differences in a stream has many practical applications, including designing efficient and safe water systems, calculating the flow rate of fluids through pipes, and predicting the behavior of fluids in natural environments such as rivers and oceans. It is also crucial in fields such as aeronautics and hydraulics.

Similar threads

  • Introductory Physics Homework Help
Replies
23
Views
417
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
941
  • Introductory Physics Homework Help
Replies
3
Views
908
  • Introductory Physics Homework Help
Replies
6
Views
1K
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
823
  • Introductory Physics Homework Help
Replies
14
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
950
  • Introductory Physics Homework Help
Replies
13
Views
911
Back
Top