- #1

- 1

- 0

For laminar fully developed pipe flow, show the average velocity V = 0.5u(max).

Any help will be greatly appreciated! Thanks!

- Thread starter Student39
- Start date

- #1

- 1

- 0

For laminar fully developed pipe flow, show the average velocity V = 0.5u(max).

Any help will be greatly appreciated! Thanks!

- #2

FredGarvin

Science Advisor

- 5,066

- 8

What does the velocity profile for fully developed laminar flow look like?

- #3

- 8

- 0

u(r)=Umax*(1-r**2/R**2)

Umoy is indeed equal two the half of Umax.

do you need more info about how to get there?

- #4

- 1

- 0

We wand to find average velocity. Now lets start from basics:- Average velocity means average velocity of all the particles in a unit cross-section of the tube. So consider a ring of radius r having a differential are equal to 2*Pi*r*dr. So multiplying this by velocity gives the sum of velocity of all the particles in that differential tube. So if we integrate from 0 to R, we get total velocity of all the particles. and hence dividing the expression by area gives us the average velocity which comes out top be equalt to 0.5 times the Vmax!!!!

Regards,

Vishal

- #5

- 10

- 0

Could you please tell me how to get to the equation for a pipe flow.

u(r)=Umax*(1-r**2/R**2)

Thank you so much in advance.

u(r)=Umax*(1-r**2/R**2)

Umoy is indeed equal two the half of Umax.

do you need more info about how to get there?

- #6

- 757

- 0

you "get it" by solving the Navier-Stokes equations with the appropriate boundary conditions and simplifications (i.e. incompressible,Hello,

Could you please tell me how to get to the equation for a pipe flow.

u(r)=Umax*(1-r**2/R**2)

Thank you so much in advance.

- #7

- 7

- 0

I "read/know" that the velocity profile for laminar flow is parabolic with v=vmax as maximum and v=0 on the edges let us say D=2d. Mathematically I immediately say: the equation of such a parabola is v=vmax•(1-x²/d²) for xε[-d,d]. (This is just the equation of a parabola with max=vmax and roots x=-d and x=d.) No Navier-Stokes required.

If I have a circular pipe, I change x=r, d=R (radius of pipe) and use polar integration to get

Q=vmax•∫∫(1-r²/R²)rdrdθ on intervals rε[0,R] and θε[0,2π] to get Q=(vmax/2)•πR².

Then vavg=Q/A=(vmax/2)•πR²/πR²=vmax/2 (and derive the formula asked at beginning).

So - if I know laminar flow has a parabolic velocity profile, why use NS to find the equation with vmax? I can see using NS to find vmax, but not to get the equation with vmax in it.

BTW: Anybody know of a link with (solved) laminar flow problems for rectangular ducts with a given vmax? Thanks!

- #8

- 545

- 10

Integrate the velocity profile then divide by the length of the integration.

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 3K

- Replies
- 10

- Views
- 5K

- Last Post

- Replies
- 19

- Views
- 13K

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 4

- Views
- 946

- Last Post

- Replies
- 7

- Views
- 2K

- Replies
- 4

- Views
- 2K

- Replies
- 6

- Views
- 17K

- Replies
- 3

- Views
- 958