wave tank absorbing problem -- CFD Online Discussion Forums

April 23, 2011, 23:17

wave tank absorbing problem

ruirui389

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Join Date: Mar 2011

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hi. i'm trying to absorb waves in the end of a 2-D wave tank using momentum source method. this is what i've done, and i met some problems.

1. suppose that the velocity field after absorbing is

$u_m = Cu_j, v_m = Cv_j, p_m = (1-C)\rho gz + Cp_j$

where C is a spatial transition coefficient varies from 1 to 0. the subscript j represents the value before absorbing, and m represents the value after absorbing. substracting the momentum equations with source term

$\rho \frac{u_m^{n+1}-u_j^n}{\delta t}+\rho u_j \frac{\partial u_j}{\partial x} + \rho v_j \frac{\partial u_j}{\partial y} = -\frac{\partial p_j}{\partial x} + S_x$

with the momentum equations without source term

$\rho \frac{u_m^{n+1}-u_m^n}{\delta t}+\rho u_m \frac{\partial u_m}{\partial x} + \rho v_m \frac{\partial u_m}{\partial y} = -\frac{\partial p_m}{\partial x},$

i got the horizontal momentum source

$S_x = \rho \frac{1}{\delta_t}(C-1)u_j + \rho[(1-C^2)u_j \frac{\partial u_j}{\partial x} - CC' u_j^2] + \rho(1-C^2)v_j \frac{\partial u_j}{\partial y} + (1-C) \frac{\partial p_j}{\partial x}$

where C' is the horizontal deviation of C.
by the same way i got the vertical source

$S_y = \rho \frac{1}{\delta_t}(C-1)v_j + \rho[(1-C^2)u_j \frac{\partial v_j}{\partial x} - CC' u_j v_j] + \rho(1-C^2)v_j \frac{\partial v_j}{\partial y} + (1-C) \frac{\partial p_j}{\partial y} + (C-1)\rho g$ .

after applying them into FLUENT, i got the problem of water level shown in this image

i defined the region from 80 to 100 is the absorbing area. this picture was taken at t=1s. it seems like the absorbing part didn't absort any wave but also generated a wave! and after a long time, the water level at x = 100 went down to y=3.5m. i really don't know where i am wrong. would u help me please? this problem really disappointed me.

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