# Should I expect fluid velocity within a pipe to increase beyond the inlet velocity?

I'm using commercial software to simulate unsteady incompressible flow of air through a 1m long rectangular pipe with cross section 0.2mx1m. I have a velocity inlet condition (u=100m/s, $\nabla p=0$) and a pressure outlet condition (p=0, $\nabla u=0$). For simplicity, I'm running the simulation in 2D on the x-y plane, with the pipe oriented horizontally and with wall boundary conditions on the bottom of the pipe (along y=0) and a symmetry condition at the top of the pipe (at y=0.1). I used a symmetry condition so as to reduce the number of cells in my mesh and thus speed up computation time. I also made sure to bias my mesh toward the wall boundary to ensure proper capturing of the boundary layers.

When I run the simulation to a steady state, I observe that the maximum velocity within the pipe is slightly larger than the inlet velocity ($|u_{max}|\approx 101$). This is counter-intuitive to me. I expect the velocity to be bounded by the inlet velocity. It doesn't make sense to me that the fluid velocity should accelerate at all. Therefore, I ask:

Would the same thing happen in real life? If fluid enters a rectangular pipe of similar dimensions and length at a particular inlet velocity, should I expect a velocity profile within some cross-section of the pipe to exceed the inlet velocity? If so, what causes the fluid velocity to increase within the pipe (from a physics/mechanical engineering perspective)?

Or is there a problem with the way I setup the problem (i.e. a numerical issue that I should be aware of for this problem)?

• If you take a cross-sectional average of the velocity at the outlet you should find that it is equal to the inlet velocity. If not, then mass is not conserved in your simulation. – nluigi Jul 13 '16 at 16:10

The mean velocity of the fluid must be constant, if the fluid is incompressible and the area is constant.

But since you are modelling a viscous fluid, there will be a velocity profile over a cross section of the pipe, with zero velocity at the pipe walls. So the velocity at the center of the pipe will be greater than the mean velocity.

Presumably you specified that the inlet velocity was constant over the whole area of the pipe, so the mean velocity was 100 m/s. Strictly speaking, that velocity profile is inconsistent with the physics of the problem, because the velocity at the pipe wall is zero not 100 m/s, but for many purposes the inconsistency isn't very important. Of course if you were trying to model exactly what happens at the pipe inlet it would be important, and you would need to find a different way to specify the boundary conditions.

• So, you're suggesting that this would not happen in real life? – Paul Jul 12 '16 at 15:12
• In real life, you would not expect the inlet velocity to be perfectly uniform. You would expect that the velocity at the pipe walls would be less than the average velocity, and the velocity at the center would be higher than the average velocity. – Adam Jul 12 '16 at 20:36
• @Adam: But would it be higher than the inlet velocity, assuming it was uniform at the inlet? – Paul Jul 13 '16 at 19:46
• "...assuming it was uniform at the inlet." Are you assuming the flow IS (somehow?) uniform at the inlet, or it was modeled as uniform at the inlet? Your fluid is incompressible. Anywhere you look, the center of the pipe will be going faster than the average flow, and flow on the edges will be less than the average flow. That is true at the inlet, 1 foot in, and 10 feet in. If the cross-sectional area doesn't change, then the area-averaged velocity won't change either. – Adam Jul 13 '16 at 20:43

Paul,

At first glance, it seems to me that the average velocity a good distance down along the pipe should be less than the average velocity at the inlet.

The reason I believe this is because inlets usually act as constrictions. The cross sectional area of flow would be decreased at the inlet. Within the pipe, the cross sectional area of flow would be larger, so the average velocity would be slower.