In fluid dynamics Jeffery–Hamel flow is a flow created by a converging or diverging channel with a source or sink of fluid volume at the point of intersection of the two plane walls. It is named after George Barker Jeffery(1915)[1] and Georg Hamel(1917),[2] but it has subsequently been studied by many major scientists such as von Kármán and Levi-Civita,[3] Walter Tollmien,[4] F. Noether,[5] W.R. Dean,[6] Rosenhead,[7] Landau,[8] G.K. Batchelor[9] etc. A complete set of solutions was described by Edward Fraenkel in 1962.[10]
Consider two stationary plane walls with a constant volume flow rate
Q
2\alpha
(r,\theta,z)
r=0
\theta=0
(u,v,w)
z
u=u(r,\theta),v=0,w=0
Then the continuity equation and the incompressible Navier–Stokes equations reduce to
\begin{align} | \partial(ru) |
\partialr |
&=0,\\[6pt] u
\partialu | |
\partialr |
&=-
1 | |
\rho |
\partialp | |
\partialr |
+\nu\left[
1 | |
r |
\partial | \left(r | |
\partialr |
\partialu | |
\partialr |
\right)+
1 | |
r2 |
\partial2u | |
\partial\theta2 |
-
u | |
r2 |
\right]\\[6pt] 0&=-
1 | |
\rhor |
\partialp | |
\partial\theta |
+
2\nu | |
r2 |
\partialu | |
\partial\theta |
\end{align}
The boundary conditions are no-slip condition at both walls and the third condition is derived from the fact that the volume flux injected/sucked at the point of intersection is constant across a surface at any radius.
u(\pm\alpha)=0, Q=
\alpha | |
\int | |
-\alpha |
urd\theta
The first equation tells that
ru
\theta
F(\theta)=
ru | |
\nu |
.
Different authors defines the function differently, for example, Landau[8] defines the function with a factor
6
\theta
1 | |
\rho |
\partialp | |
\partial\theta |
=
2\nu2 | |
r2 |
dF | |
d\theta |
Now letting
p-pinfty | |
\rho |
=
\nu2 | |
r2 |
P(\theta),
the
r
\theta
P=-
1 | |
2 |
(F2+F'')
P'=2F', ⇒ P=2F+C
and substituting this into the previous equation(to eliminate pressure) results in
F''+F2+4F+2C=0
Multiplying by
F'
1 | |
2 |
F'2+
1 | |
3 |
F3+2F2+2CF=D,
1 | |
2 |
F'2+
1 | |
3 |
(F3+6F2+6CF-3D)=0
where
C,D
a,b,c
a+b+c=-6
1 | |
2 |
| ||||
F' |
(F-a)(F-b)(F-c)=0,
1 | |
2 |
| ||||
F' |
(a-F)(F-b)(F-c)=0.
The boundary conditions reduce to
F(\pm\alpha)=0,
Q | |
\nu |
=
\alpha | |
\int | |
-\alpha |
Fd\theta
where
Re=Q/\nu
Q<0
Re
Q>0
Re
Source:[13]
The equation takes the same form as an undamped nonlinear oscillator(with cubic potential) one can pretend that
\theta
F
F'
K.E.+P.E.=0
K.E.=
1 | |
2 |
F'2
P.E.=V(F)
V(F)=- | 1 |
3 |
(a-F)(F-b)(F-c)
where
V\leq0
F=0
\theta=-\alpha
F=0
\theta=\alpha
b,c
a>0
F=0
F=a
F'=0
F''=-dV/dF<0
F=0
0<F<a
F>0
\theta=0
c<b<0<a
F=0
F=a
F=0
F=0
F=b
F=0
F<0
-\alpha\leq\theta\leq\alpha
b\leqF\leqa
\theta=0
For pure outflow, since
F=a
\theta=0
\theta=\sqrt{
3 | |
2 |
and the boundary conditions becomes
\alpha=\sqrt{
3 | |
2 |
The equations can be simplified by standard transformations given for example in Jeffreys.[14]
b,c
a>0
F(\theta)=a- | 3M2 |
2 |
1-\operatorname{cn | |
(M\theta,\kappa)}{1+\operatorname{cn}(M\theta,\kappa)} |
M2=
2 | |
3 |
\sqrt{(a-b)(a-c)},
| |||||
\kappa | + |
a+2 | |
2M2 |
where
\operatorname{sn},\operatorname{cn}
c<b<0<a
F(\theta)=a-6k2m2\operatorname{sn}2(m\theta,k)
m2=
1 | |
6 |
(a-c),
| ||||
k |
.
The limiting condition is obtained by noting that pure outflow is impossible when
F'(\pm\alpha)=0
b=0
\alphac
\begin{align} \alphac&=\sqrt{
3 | |
2 |
where
m2=
3+a | |
3 |
, k2=
1 | \left( | |
2 |
a | |
3+a |
\right)
where
K(k2)
a
\alphac=\sqrt{
3 | |
a |
The corresponding critical Reynolds number or volume flux is given by
\begin{align} Rec=
Qc | |
\nu |
&=2
\alphac | |
\int | |
0 |
(a-6k2m2\operatorname{sn}2m\theta)d\theta,\\ &=
12k2 | |
\sqrt{1-2k2 |
where
E(k2)
a,\left( k2\sim
1 | - | |
2 |
3 | |
2a |
\right)
Re | ||||
|
=12\sqrt{
a | |
3 |
For pure inflow, the implicit solution is given by
\theta=\sqrt{
3 | |
2 |
and the boundary conditions becomes
\alpha=\sqrt{
3 | |
2 |
Pure inflow is possible only when all constants are real
c<b<0<a
F(\theta)=a-6k2m2\operatorname{sn}2(K-m\theta,k)=b+6k2m2\operatorname{cn}2(K-m\theta,k)
m2=
1 | |
6 |
(a-c),
| ||||
k |
where
K(k2)
As Reynolds number increases (
-b
m
\alpha
K
k\sim1
k ≈ 1
\operatorname{sn}t ≈ \tanht, c ≈ b, a ≈ -2b
F(\theta)=b\left\{3\tanh2\left[\sqrt{-
b | |
2 |
It is clear that
F ≈ b
O\left(\sqrt{- | b |
2 |
Q/\nu ≈ 2\alphab
|Re|=O(|b|)
O\left(|Re|1/2\right)