Question in CFD

August 6, 2009, 14:08

Hi freinds

I have this equation

$\frac{1}{R}$ $\frac{\partial u}{ \partial \xi}$ + $\frac{u}{( \xi R+ R_{2} )}$ + $\frac{\partial w}{\partial z}$ - $\frac{1}{R}$ $\frac{\partial w}{\partial \xi}$ ( $\xi$ $\frac{\partial R}{ \partial z}$ + $\frac{\partial R_{2}}{\partial z}$ )= 0

which comes from continuty equation

$\frac{\partial u}{\partial r}$ + $\frac{u}{r}$ + $\frac{\partial w}{\partial z}$ = 0

and the radial coordinate transformation

$\xi$ = $\frac{r-R_{2}}{R_{1}- R_{2}}$ = $\frac{r-R_{2}}{R}$

my question now how I can get the first equation? I mean what the steps to get it by using the continuty equ and transformation?

Kind Regards
Shosho

Ananda Himansu · August 6, 2009, 23:02

I will refer to your equations as (1), (2) and (3). You are transforming independent coordinates from (r, z) to (xi, zeta), where xi is given by your eqn (3) and zeta = z. The dependents u and w were previously functions of r & z, and are now functions of xi & zeta. R1 and R2 (and also R1-R2) are functions of z, and thus of zeta, but are independent of r and of xi. Recognize that when transforming your original continuity eqn (2), you must now treat w as a function of xi and zeta, but the derivative is with respect to z while holding r constant. Apply the chain rule for this derivative (dw/dz = dw/dzeta*dzeta/dz + dw/dxi*dxi/dz) and you get your transformed continuity eqn (1), noting that all occurrences of z in (1) are really zeta and that the derivative dw/dzeta is now holding xi constant (not r). The first two terms in (2) get trivially transformed to the first two terms of (1).

August 7, 2009, 09:58

Thanks Ananda Himansu

Dear friends, one of my friends send me this solution but unfortunatlly i didn't understand it . Can you please One explain it to me?.

1. \xi = \xi(r, R_1, R_2)

2. R_1 = R_1(z,t) and R_2 = R_2(z,t), hence (3. & 4. hold):
3. u(r , z , t) = u(\xi , z , t),

4. w(r , z , t) = w(\xi , z , t) so that (5. & 6. hold):

5. u_r = u_{\xi} * \xi_r (from 3.)

6. w_z = w_{\xi} * \xi_z + w_z (from 4.)

formulas 5. and 6. should answer your question!

Kind Regards
Shosho

August 6, 2009, 14:08	Question in CFD	#1
shosho Guest Posts: n/a	Hi freinds I have this equation $\frac{1}{R}$ $\frac{\partial u}{ \partial \xi}$ + $\frac{u}{( \xi R+ R_{2} )}$ + $\frac{\partial w}{\partial z}$ - $\frac{1}{R}$ $\frac{\partial w}{\partial \xi}$ ( $\xi$ $\frac{\partial R}{ \partial z}$ + $\frac{\partial R_{2}}{\partial z}$ )= 0 which comes from continuty equation $\frac{\partial u}{\partial r}$ + $\frac{u}{r}$ + $\frac{\partial w}{\partial z}$ = 0 and the radial coordinate transformation $\xi$ = $\frac{r-R_{2}}{R_{1}- R_{2}}$ = $\frac{r-R_{2}}{R}$ my question now how I can get the first equation? I mean what the steps to get it by using the continuty equ and transformation? Kind Regards Shosho

August 6, 2009, 23:02		#2
Ananda Himansu New Member Ananda Himansu Join Date: Apr 2009 Location: Cleveland, Ohio, USA Posts: 17 Rep Power: 17	I will refer to your equations as (1), (2) and (3). You are transforming independent coordinates from (r, z) to (xi, zeta), where xi is given by your eqn (3) and zeta = z. The dependents u and w were previously functions of r & z, and are now functions of xi & zeta. R1 and R2 (and also R1-R2) are functions of z, and thus of zeta, but are independent of r and of xi. Recognize that when transforming your original continuity eqn (2), you must now treat w as a function of xi and zeta, but the derivative is with respect to z while holding r constant. Apply the chain rule for this derivative (dw/dz = dw/dzetadzeta/dz + dw/dxidxi/dz) and you get your transformed continuity eqn (1), noting that all occurrences of z in (1) are really zeta and that the derivative dw/dzeta is now holding xi constant (not r). The first two terms in (2) get trivially transformed to the first two terms of (1). Last edited by Ananda Himansu; August 6, 2009 at 23:07. Reason: distinguished between dw/dz in eqns (1) and (2)

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
ASME CFD Symposium - Call for Papers	Chris Kleijn	Main CFD Forum	0	September 25, 2001 10:17
ASME CFD Symposium	Chris Kleijn	Main CFD Forum	0	August 22, 2001 06:41
Where do we go from here? CFD in 2001	John C. Chien	Main CFD Forum	36	January 24, 2001 21:10
Is CFD Science or Art ?	John C. Chien	Main CFD Forum	36	October 5, 1999 12:58
salary range	Frank Muldoon	Main CFD Forum	7	August 3, 1998 19:04

August 7, 2009, 09:58		#3
shosho Guest Posts: n/a	Thanks Ananda Himansu Dear friends, one of my friends send me this solution but unfortunatlly i didn't understand it . Can you please One explain it to me?. 1. \xi = \xi(r, R_1, R_2) 2. R_1 = R_1(z,t) and R_2 = R_2(z,t), hence (3. & 4. hold): 3. u(r , z , t) = u(\xi , z , t), 4. w(r , z , t) = w(\xi , z , t) so that (5. & 6. hold): 5. u_r = u_{\xi} * \xi_r (from 3.) 6. w_z = w_{\xi} * \xi_z + w_z (from 4.) formulas 5. and 6. should answer your question! Kind Regards Shosho