Fluid speed in a straight pipe

[dara bayat]

Hello everyone

I have a simple question regarding how to calculate the speed of a fluid in a normal pipe with a constant cross-section

If we are given the pressure difference at the two ends of the pipe (P1 and P2) is it possible to calculate the speed of the fluid? (I think I can imagine this problem like an inflated balloon attached to a little pipe)

Using Bernoulli’s equation:

P1/rho+V1^2/2= P2/rho+V2^2/2

Since the channel cross-section “A” does not change we have:

Rho*A*V1= Rho*A*V2 --> V1=V2

Therefore

P1=P2 !?Am I doing something wrong? Should I use another formula or in a different order?Thanks in advance for your helpDara

 

[boneh3ad]

The issue is that if you are using the Bernoulli equation, which assumes that the effects of viscosity are negligible. In that case, then there shouldn't be a pressure difference for a constant velocity, constant area pipe that doesn't change heights appreciably. If you want to calculate the flow rate based on the pressure change, you have to take viscosity into account and use something more like the Hagen-Poiseuille equation, which comes directly from the Navier-Stokes equations.

 

[SteamKing]

Hello everyone

I have a simple question regarding how to calculate the speed of a fluid in a normal pipe with a constant cross-section

If we are given the pressure difference at the two ends of the pipe (P1 and P2) is it possible to calculate the speed of the fluid? (I think I can imagine this problem like an inflated balloon attached to a little pipe)

Using Bernoulli’s equation:

P1/rho+V1^2/2= P2/rho+V2^2/2

Since the channel cross-section “A” does not change we have:

Rho*A*V1= Rho*A*V2 --> V1=V2

Therefore

P1=P2 !?Am I doing something wrong? Should I use another formula or in a different order?Thanks in advance for your helpDara

You can use the Darcy-Weisbach equation, which is for real fluids flowing in real pipes with friction:

https://en.wikipedia.org/wiki/Darcy–Weisbach_equation

 

[dara bayat]

thank you very much

this was quite helpful for me

Dara

 

[boneh3ad]

So this begs the question: do you know which assumptions must be made in order for the Bernoulli equation to be valid? This is an important thing for you to know before you apply such equations.

 

[dara bayat]

So this begs the question: do you know which assumptions must be made in order for the Bernoulli equation to be valid? This is an important thing for you to know before you apply such equations.

Thank you for your question and sorry for the late reply

(I had to read a bit and get confused again before answering :-))These should be the assumptions for the Bernoulli equation:

Inviscid flow (ok in this example)

Constant density (respected in this example)

Along a streamline (respected in this example)

Applied in an inertial reference frame (respected in this example)

Steady flow (NOT respected in this example)

Can I conclude that to use Bernoulli’s equation I need to know the speed and not the Pressure?Also now I have another question regarding the Bernoulli’s Equation

As I have read in the more elementary books the Bernoulli equation is a constant on a streamline

Bernoulli=constant

That is why on a streamline we can write (Bernoulli1= Bernoulli2) and find, for example, the Venturi effect formula.

However I do not completely grasp this when the streamline separates to two parts (a bifurcating pipe): Bernoulli1 = Bernoulli2 = Bernoulli3 = constant

Could you elaborate a bit on this? Or maybe refer me to a good book that explains this a bit more clearly than in the elementary books?Thank you again for your helpDara

 

[boneh3ad]

Well there's your problem, then. You know how to regurgitate the requirements but don't really know what they mean. Also, it is very common for people to say that Bernoulli's equation only holds on a streamline, but that also is not true, as I will show in a moment.

In general, Bernoulli's equation is probably best understood as the following:
$$p + \dfrac{1}{2}\rho v^2 + \rho gz = \textrm{const.},$$
where $p$ is pressure, $\rho$ is density, $v$ is the velocity, $g$ is gravity (or any other body force), and $z$ is height (in the case of gravity). The constant is often called the total pressure, $p_0$, in the case where the gravity term can be neglected (e.g. for very light fluid and small changes in $z$).

Bernoulli's equation really only requires three assumptions:

1. steady flow,
2. incompressible flow, and
3. irrotational flow (inviscid).

These three assumptions are essentially enforcing the fact that the constant in the equation is actually a constant. However, if the flow is, in general, not inviscid, it turns out that all three of those assumptions still apply along a streamline, so this is where the confusion there arises. Bernoulli's equation applies, even in a viscous flow, along a streamline. In an inviscid flow, it can apply to any two points in the fluid.

Now, revisiting your claims, you are correct that the flow is incompressible here (unless the flow is a very fast-moving gas). It does not necessarily follow a streamline, as you claim, but as I just pointed out, that is irrelevant. However, the flow is absolutely steady, in your example. You cited a constant velocity flow through a pipe. All of the time derivatives in that case are zero, so the flow is steady.

However, the flow is not inviscid in this case. Flow moving through a pipe will always come under the influence of viscosity and form, for steady flow through a straight section, a parabolic velocity profile. In this regard, using Bernoulli's equation in a pipe, which is common, is a pretty weak approximation to what is actually occurring. Since the pressure drop in a steady, straight pipe flow comes entirely from viscosity, then it makes perfect sense that Bernoulli's equation would predict zero pressure drop since it is neglecting viscosity. However, it is quite common to make corrections to Bernoulli's equation using empirical relations such as the Darcy-Weisbach equation (as mentioned above). This let's you retain some of that pressure drop that occurs in a real system while still sticking to a Bernoulli-like analysis. However, it is still just an approximation (that can be quite accurate at times). For a laminar flow, you can also get an exact solution, which is the Hagen-Poiseuille equation.

In general, the process for solving an inviscid flow is to get the velocity field first (in this case, through conservation of mass) and then use Bernoulli's equation to solve for the pressure field. You took the correct approach here; you just failed to realize that the pressure drop is due to viscosity, and Bernoull's equation ignores viscosity.

It's been a while since I have dealt with pipe flow in detail (not my typical type of work), so I don't want to accidentally tell you something wrong, but in general, Bernoulli's equation is an even worse performer in the case of a bifurcating pipe since the split itself will induce larger viscous effects (e.g. vortices) that will increase pressure drop beyond what would occur in a straight section. The assumptions you make in that system (e.g. the length of pipe each leg is connected to or the pressure on the discharge end of those legs) will determine the flow rates through each pipe.

The first book I used to learn basic fluid mechanics was https://www.amazon.com/dp/0471675822/?tag=pfamazon01-20. It's rather pricey, but such is life with textbooks these days. There are plenty of books on the subject available, though. If you want something cheap, find something in the Dover series of textbooks. They usually retail for somewhere between $10 and $20 (USD).