- #1
lol_nl
- 40
- 0
Homework Statement
I'm going to model the velocity of a river stream v(y,z) as a function of the location (y,z) in the cross-section of the river. Here y represents the distance from one of the shores and z the depth of the river.
I have done measurements of the depth z on different locations y, so I can find an approximation for a function z(y) and hence find the profile of the river.
I have also done measurements of the velocity on different points in the river, so now I wish to find a function v(y,z) that describes the speed at every point in the cross-section. This will give me a way to find the average speed by integrating over the function for the cross-area.
Homework Equations
Basically, through a number of assumptions and approximations, the problem can be reduced to a set of two differential equations:
\eta_{y}\frac{\partial^{2} v(y,z)}{\partial y^2} + \eta_{z}\frac{\partial^{2} v(y,z)}{\partial z^2} = -\rho g Sin(\alpha)
By assuming by turns \eta_{y} = 0 and \eta_{z} = 0 I can find functions v(y) and v(z) that describe a kind of an average velocity for any given y or given z.
The problem now is the integration of these two into one function v(y,z).
Given boundary conditions:
* v(0,z) = 0
* v(W,z) = 0 -> W represents max width
* v(y,0) = 0 -> (y,0) is a point on the bottom
* \frac{\partial v}{\partial z}[H] = 0 -> change of speed is zero in for the maximum height H.
The Attempt at a Solution
I've found solved the equations for the assumptions \eta_{y} = 0 and \eta_{z} = 0 and using the boundary conditions found:
v(y) = \frac{\rho g Sin[\alpha]}{2\eta_{y}} (W-y)y
v(z) = \frac{\rho g Sin[\alpha]}{2\eta_{z}} (2H-z)z
The question now is whether there is an easier way to find v(y,z) by directly solving it mathematically, possibly by combining solutions for v(y) and v(z)? I can see that a linear combination of the two would work, but I can't see how I can find the general solution of the function.
I have tried to find a solution using Mathematica, but by filling in the boundary conditions it reduced to the function v(y) that is independent of z. Here is what my code in Mathematica more or less looks like:
Code:
DSolve[{a*D[f[y, z], {y, 2}] + b *D[f[y, z], {z, 2}] == -\[Rho]*g*
Sin[\[Alpha]]}, f, {y, z}]
{{f -> Function[{y, z}, -((g y^2 \[Rho] Sin[\[Alpha]])/(2 a)) +
C[1][(Sqrt[-a b] y)/a + z] + C[2][-((Sqrt[-a b] y)/a) + z]]}}
v[y_, z_] := -((g y^2 \[Rho] Sin[\[Alpha]])/(2 a)) +
c1 ((Sqrt[-a b] y)/a + z) + c2 (-((Sqrt[-a b] y)/a) + z);
r1 := v[0, z];
r2 := v[W, z];
r3 := v[y, 0];
r4 := D[v[y, z], z];
{r1, r2, r3, r4}
{c1 z + c2 z,
c2 (-((Sqrt[-a b] W)/a) + z) + c1 ((Sqrt[-a b] W)/a + z) - (
g W^2 \[Rho] Sin[\[Alpha]])/(2 a), (Sqrt[-a b] c1 y)/a - (Sqrt[-a b] c2 y)/
a - (g y^2 \[Rho] Sin[\[Alpha]])/(2 a), c1 + c2}
Clear[c1, c2]
Solve[{r1 == 0, r2 == 0}, {c1, c2}]
Solve[{r3 == 0, r4 == 0}, {c1, c2}]
Solve[{r1 == 0, r3 == 0}, {c1, c2}]
Solve[{r2 == 0, r3 == 0}, {c1, c2}]
{{c1 -> (g W \[Rho] Sin[\[Alpha]])/(4 Sqrt[-a b]),
c2 -> -((g W \[Rho] Sin[\[Alpha]])/(4 Sqrt[-a b]))}}
{{c1 -> (g y \[Rho] Sin[\[Alpha]])/(4 Sqrt[-a b]),
c2 -> -((g y \[Rho] Sin[\[Alpha]])/(4 Sqrt[-a b]))}}
{{c1 -> (g y \[Rho] Sin[\[Alpha]])/(4 Sqrt[-a b]),
c2 -> -((g y \[Rho] Sin[\[Alpha]])/(4 Sqrt[-a b]))}}
{{c1 -> -((-b g W^2 \[Rho] Sin[\[Alpha]] + b g W y \[Rho] Sin[\[Alpha]] +
Sqrt[-a b] g y z \[Rho] Sin[\[Alpha]])/(4 a b z)),
c2 -> -((-Sqrt[-a b] g W^2 \[Rho] Sin[\[Alpha]] +
Sqrt[-a b] g W y \[Rho] Sin[\[Alpha]] + a g y z \[Rho] Sin[\[Alpha]])/(
4 a Sqrt[-a b] z))}}
Clear[c1, c2]
c1 := (g W \[Rho] Sin[\[Alpha]])/(4 Sqrt[-a b]);
c2 := -((g W \[Rho] Sin[\[Alpha]])/(4 Sqrt[-a b]));
Simplify[v[y, z]]
(g (W - y) y \[Rho] Sin[\[Alpha]])/(2 a)
c1 := -((-b g W^2 \[Rho] Sin[\[Alpha]] + b g W y \[Rho] Sin[\[Alpha]] +
Sqrt[-a b] g y z \[Rho] Sin[\[Alpha]])/(4 a b z));
c2 := -((-Sqrt[-a b] g W^2 \[Rho] Sin[\[Alpha]] +
Sqrt[-a b] g W y \[Rho] Sin[\[Alpha]] + a g y z \[Rho] Sin[\[Alpha]])/(
4 a Sqrt[-a b] z));
Simplify[v[y, z]]
(g W (W - y) \[Rho] Sin[\[Alpha]])/(2 a)In both cases I obtain exactly the solution for v(y).
