# on fully developed duct flow

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 November 19, 2001, 22:21 on fully developed duct flow #1 yfyap Guest   Posts: n/a hi there, i'm interested in solving for u,v,w,P for fully developed flow only in a straight duct. the problem is formulated in primitve variables and all derivatives with respect to z(streamwise direction) are dropped except for the streamwise pressure gradient. it seems, by doing these, u and v are independent of w, is this true? if this were true, then, can we solve for u,v,P using the continuity equation ,x and y momentum equations first? these equations would resemble those for cavity driven flow except all those velocities at the boundaries remain zero. then, would it possible to get trivial solution? after attaining the correct u,v,P, w is obtained from z-momentum equation. thanks. regards, yfyap

 November 20, 2001, 01:10 Re: on fully developed duct flow #2 S.P.Asok Guest   Posts: n/a I keen to go through experts' views on this.Regards

 November 22, 2001, 15:09 Re: on fully developed duct flow #3 I. Dotsikas Guest   Posts: n/a Fully developed duct flow means that your velocity w (in the main direction) varies only across ; and not along the duct :dw/dz = 0 and dw/dx <>0 ; dw/dy<>0 ; dP/dz = 0. All other derivatives of the x and y momentum equations are zero!!! You get the trivial solution u= v= dP/dx = dP/dy = 0 best regards I. Dotsikas

 November 22, 2001, 21:43 Re: on fully developed duct flow #4 yfyap Guest   Posts: n/a dear I. Dotsikas, from the physics of the problem, we know that u(x,y), v(x,y), w(x,y) and P can be written as P = P1(z) + P2(x,y), the flow along z is driven by constant pressure gradient, so, dP/dz = dP1/dz = constant, is this right? i feel that the momentum equations should look like udu/dx + vdu/dy = -1/rho * dP/dx + nu * [ d^2u/dx^2 + d^2u/dy^2 ] ---(1) udv/dx + vdv/dy = -1/rho * dP/dy + nu * [ d^2v/dx^2 + d^2v/dy^2 ] ---(2) udw/dx + vdw/dy = -1/rho * dP/dz + nu * [ d^2w/dx^2 + d^2w/dy^2 ] ---(3) and the continuity equation: du/dx + dv/dy = 0 ----(4) from (1) to (4), it seems (3) is uncoupled from (1), (2) and (4). furthermore, i feel that solely solving (1), (2) and (4) with zero velocity at the boundaries, would render a trivial solution. please explain what is meant by the statement "All other derivatives of the x and y momentum equations are zero!!!", i do not quite get it. thanks. regards, yfyap

 November 23, 2001, 11:58 Re: on fully developed duct flow #5 andy Guest   Posts: n/a Before writing off the W equation consider what gives rise to your dp/dz. If you want something interesting to happen in the plane of the duct then may I suggest making your diffusion coefficient variable (e.g. a turbulence model), varying the density (e.g. extracting heat from the walls) or adding a transverse pressure gradient (e.g. curving the duct). If you want to investigate such parabolic approachs further a good place to start might be the old genmix program: http://www.cham.co.uk/website/new/genmix/code.htm and the large body of associated literature.

 December 11, 2001, 07:49 Re: on fully developed duct flow #6 vassilis sakalis Guest   Posts: n/a in the fully developed region of a straight duct, the axial pressure gradient remains constant. As we see from the n-s equation, cross section velocities are independent of the axial velocity. If someone wants to solve such a problem, the only way is to set initial the axial pressure gradient and to geuess cross sections pressure gradients dp/dx and dp/dy with the other varriables zero (eg:inlet velocities). Then solve the axial n-s equation to obtain axial velocity and solve the cross n-s equations to obtain cross velocities which are not zero because of nonzero pres.gradients. By using a pressure correction technique you can correct the obtained cross section velocities and pressures from the continuity equation. Those corrected values will be set as initial values to solve the n-s equation again till convergence. At the end of thenumerical procedure you will see that the cross section velocities have very low values, but their sign depend especially on the sign of the guess cross section pressure gradients and they are independent of the axial velocity. This problem has not unique solution. The best approach of such a problem is to solve the developing n-s equation in order to obtain the fully developed valuues of the velocities. In this case, the cross section velocities are coupled with axial velocitie by both momentum and continuity equations

 November 7, 2012, 10:59 thermal fully developed #7 New Member   Ayoub Join Date: Nov 2012 Posts: 7 Rep Power: 5 Hi I need to reach fully develop flow in microchannel(D=100 micrometer,L=1mm) q=500,Re=100 to 1000,Tin=300,working fluid is water. i can not reach thermal fully developed,i don't know why even when i apply Le=.05 Re D Pr would you please show me a solution?

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