# Homework Help: Fluid Dynamics - Flow through a pipe

1. Oct 7, 2011

### StrongGaia

I have a fluid dynamics homework question I could really use some help solving.

"Consider a straight, circular pipe of inside radius R, through which flow an incompressible fluid. At the pipe entrance, the flow velocity Vo is uniform and parallel to the pipe centreline. The flow is affected by the action of viscosity and the no-slip condition at the inside wall. The velocity profile far downstream of the entrance, V(r), becomes parabolic with radius from the centreline r. Note that V(r) are parallel to the centreline for all r. Assume laminar flow.

Find an expression for V(r), the flow speed distribution, in terms of Vo and R.

The Bernoulli Equation would show a large variation in static pressure across the pipe at the downstream station, but this is not the case. Why is the Bernoulli Equation inappropriate here?"

I'm not great at fluid dynamics, and I'm not sure where to start. I initially thought I could use the Hagenâ€“Poiseuille equation to solve it, but I don't think I have enough information to, not unless I make more assumptions for pipe length and viscosity values, but if I do that then V(r) will have more unknown terms in it than just Vo and R.

Also, I read that Bernoulli's equation assumes a non-viscous flow, which is why it is inappropriate for this question. Is that correct?

2. Oct 7, 2011

### LawrenceC

"Why is the Bernoulli Equation inappropriate here?"

The Bernoulli Equation does not consider viscosity which is what causes the parabolic profile to form.

Do you know what the Navier Stokes equation is? That is where you would start to derive this.

3. Oct 7, 2011

### Cipherflak

Tha navier-stokes equation sounds a bit overkill for this problem, but in lack of a better suggestion atm, I'll fold.

Btw, V(r=0) is greater than Vo, right, and V(r=R)=0. By continuity, the flux Vo*R at the initial transect should be exactly equal to the flux through a transect far down stream. So we need a parabolic profile that is in averge equal to Vo. I think this calls for logical thinking before heavy dynamics.

4. Oct 8, 2011

### LawrenceC

With the Navier Stokes equation, nearly all the terms drop out. The substantial derivative goes away because it is steady state. For axially symmetric flow, axial velocity does not depend on axial coordinate, z. Thus you are left with a Laplacian operator and a pressure term. Pressure term is only a function of z (axial coordinate). Velocity term is only a function of r (radius). Set each to a separation constant and solve. Constants of integration are evaluated via no slip at wall, etc.

5. Oct 8, 2011

### Kurt Peek

Hi,

As Cipherflak says, we can find the answer by logical thinking. It is given that $V(r)$ has a parabolic form, so we posit it has the form $$V(r)=ar^2+br+c$$ where $a$, $b$, and $c$ are constants to be determined.

Since we have three unknown constants, we need three conditions. These conditions are:

1. By the continuity equation, the volumetric flow rate (in m3/s) at the entrance of the pipe must be equal to the volumetric flow rate at the cross-section of the downstream part of the pipe we are considering. Since the flow is uniform at the entrance of the pipe, the former is equal to the cross section of the pipe times $v_0$, i.e.,$\pi R^2 v_0$. The latter is equal to the surface integral of the fully developed velocity profile $v(r)$ over the pipe cross section. Integrating over annular elements $2\pi r dr$, this gives the relation $$\pi R^2 v_0=\int_0^R{v(r) 2 \pi r dr}.$$
2. By the no-slip condition, the velocity is zero at the pipe wall. Mathematically, this is expressed as $$v(R)=0.$$
3. The flow has radial symmetry about the center of the pipe, $r=0$. This can be mathematically expressed as $$v'(0)=0.$$

Now, from condition (3), we immediately see that $b=0$. I'll leave it as an exercise for you to fill in the remaining to conditions to find $a$ and $b$. I obtain for the final velocity profile: $$v(r)=2 v_0 \left(1-\left(\frac{r}{R}\right)^2\right).$$ Thus, the velocity at the center of the pipe is twice as fast as at the entrance, but decays to zero at the edges of the pipe. The term for this type of flow, by the way, is Poiseuille flow (cf. http://books.google.com/books?id=MX...wAA#v=onepage&q=poiseuille equation:&f=false").

Hope this helps!

Cheers,
Kurt

Last edited by a moderator: Apr 26, 2017