| Author | Message | ||
     Stephen Heermann (sheermann) Member Username: sheermann Post Number: 5 Registered: 12-2008 |
I am trying to wrap my brain around the concept of a stream-function gradient versus a stream-function flux. The context is a sand-filled vertical column and assigning a stream-function boundary conditions along the outer wall where the flow, qrr = 0 and qzz can be calculated. (If this was simply a viscous fluid, the vertical velocity would be zero and the dr(psi) would also be zero -- however us ground-water people have been taught to think differently) My (feable) understanding is that that radial stream-function gradient can be computed as follows: dr(psi)=qz*r However the natural boundary condition would be a flux rather than a gradient. I am guessing that I need to do something like divide qz by the hydraulic conductivity terms, Kr*Kz, where Ki = ki*density*g/viscosity and kij is the permeability. However, this is a bit beyond my level of understanding of stream functions as well as a bit beyond my aptitude for this sort of thing. I realize this request is outside of the scope of the users forum. However your understanding is far greater than any academic(ian) I have access and so hope you will be kind enough to help. | ||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 1194 Registered: 06-2003 |
I found a reference on stream function formulation for groundwater, but its only cartesian, so I have to infer the axisymmetric extension. In cartesian geometry, the streamfunction equation is dx((1/Kz)*dx(psi)) + dz((1/Kx)*dz(psi)) = 0 with specific flow rates given by qx = -dz(psi) = -Kx*dx(h) qz = dx(psi) = -Kz*dz(h) Solving this equation with FlexPDE, the Natural BC means the outward normal component of Vector(dx(psi)/Kz, dz(psi)/Kx) = vector(-dz(h), dx(h)), that is, the tangential gradient (not flow rate!) Since the contours of psi are flow paths, any boundary with value(psi)=const will be an impenetrable wall, with only tangential flow possible. A boundary with value(psi)=Q*s, (where s is the path length along the boundary) will be a constant normal flow rate of Q (need to determine the sign from definitions above). The axisymmetric free-flow equation is drr(psi) - dr(psi)/r + dzz(psi) = 0, with Vr = -(1/r)*dz(psi) Vz = (1/r)*dr(psi) I infer from this that the axisymmetric groundwater equation is dr((1/Kz)*dr(psi)) - (1/r/Kz)dr(psi) + dz((1/Kr)*dz(psi)) = 0 with qr = -dz(psi)/r = -Kr*dr(h) qz = dr(psi)/r = -Kz*dz(h) The space between two contours of psi will represent the total amount of water in the cylindrical ring. A wall at r=R0 with value(psi)=const will be an impermeable wall allowing only axial flow. The Natural BC for this equation will mean the outward normal component of vector(dr(psi)/Kz, dz(psi)/Kr) = vector(qz*r, -qr*r) = vector(-r*dz(h), r*dr(h)). In otherwords, the tangential flow in the ring. A lot of this is invented at the keyboard, so I hope I didn't make too many mistakes. |