Jeffery–Hamel flow

Formulation

The first equation tells that ${\displaystyle ru}$ is just function of ${\displaystyle \theta }$, the function is defined as

${\displaystyle F(\theta )=\frac{ru}{\nu }.}$

Different authors defines the function differently, for example, Landau defines the function with a factor ${\displaystyle 6}$. But following Whitham, Rosenhead the ${\displaystyle \theta }$ momentum equation becomes

${\displaystyle \frac{1}{\rho }\frac{\mathrm{\partial }p}{\mathrm{\partial }\theta }=\frac{2{\nu }^2}{r^2}\frac{dF}{d\theta }}$

Now letting

${\displaystyle \frac{p-p_{\mathrm{\infty }}}{\rho }=\frac{{\nu }^2}{r^2}P(\theta ),}$

the ${\displaystyle r}$ and ${\displaystyle \theta }$ momentum equations reduce to

${\displaystyle P=-\frac{1}{2}(F^2+F^{″})}$

${\displaystyle P^{\prime }=2F^{\prime },\Rightarrow P=2F+C}$

and substituting this into the previous equation(to eliminate pressure) results in

${\displaystyle F^{″}+F^2+4F+2C=0}$

Multiplying by ${\displaystyle F^{\prime }}$ and integrating once,

${\displaystyle \frac{1}{2}{F}^{\prime 2}+\frac{1}{3}{F}^3+2F^2+2CF=D,}$

${\displaystyle \frac{1}{2}{F}^{\prime 2}+\frac{1}{3}(F^3+6F^2+6CF-3D)=0}$

where ${\displaystyle C,D}$ are constants to be determined from the boundary conditions. The above equation can be re-written conveniently with three other constants ${\displaystyle a,b,c}$ as roots of a cubic polynomial, with only two constants being arbitrary, the third constant is always obtained from other two because sum of the roots is ${\displaystyle a+b+c=-6}$.

${\displaystyle \frac{1}{2}{F}^{\prime 2}+\frac{1}{3}(F-a)(F-b)(F-c)=0,}$

${\displaystyle \frac{1}{2}{F}^{\prime 2}-\frac{1}{3}(a-F)(F-b)(F-c)=0.}$

The boundary conditions reduce to

${\displaystyle F(\pm \alpha )=0,\frac{Q}{\nu }={\int }_{-\alpha }^{\alpha }Fd\theta }$

where ${\displaystyle Re=Q/\nu }$ is the corresponding Reynolds number. The solution can be expressed in terms of elliptic functions. For convergent flow ${\displaystyle Q<0}$, the solution exists for all ${\displaystyle Re}$, but for the divergent flow ${\displaystyle Q>0}$, the solution exists only for a particular range of ${\displaystyle Re}$.

Dynamical interpretation

Source:

The equation takes the same form as an undamped nonlinear oscillator(with cubic potential) one can pretend that ${\displaystyle \theta }$ is time, ${\displaystyle F}$ is displacement and ${\displaystyle F^{\prime }}$ is velocity of a particle with unit mass, then the equation represents the energy equation(${\displaystyle K.E.+P.E.=0}$, where ${\displaystyle K.E.=\frac{1}{2}{F}^{\prime 2}}$ and ${\displaystyle P.E.=V(F)}$ ) with zero total energy, then it is easy to see that the potential energy is

${\displaystyle V(F)=-\frac{1}{3}(a-F)(F-b)(F-c)}$

where ${\displaystyle V\le 0}$ in motion. Since the particle starts at ${\displaystyle F=0}$ for ${\displaystyle \theta =-\alpha }$ and ends at ${\displaystyle F=0}$ for ${\displaystyle \theta =\alpha }$, there are two cases to be considered.

• First case ${\displaystyle b,c}$ are complex conjugates and ${\displaystyle a>0}$. The particle starts at ${\displaystyle F=0}$ with finite positive velocity and attains ${\displaystyle F=a}$ where its velocity is ${\displaystyle F^{\prime }=0}$ and acceleration is ${\displaystyle F^{″}=-dV/dF<0}$ and returns to ${\displaystyle F=0}$ at final time. The particle motion ${\displaystyle 0<F<a}$ represents pure outflow motion because ${\displaystyle F>0}$ and also it is symmetric about ${\displaystyle \theta =0}$.
• Second case ${\displaystyle c<b<0<a}$, all constants are real. The motion from ${\displaystyle F=0}$ to ${\displaystyle F=a}$ to ${\displaystyle F=0}$ represents a pure symmetric outflow as in the previous case. And the motion ${\displaystyle F=0}$ to ${\displaystyle F=b}$ to ${\displaystyle F=0}$ with ${\displaystyle F<0}$ for all time(${\displaystyle -\alpha \le \theta \le \alpha }$) represents a pure symmetric inflow. But also, the particle may oscillate between ${\displaystyle b\le F\le a}$, representing both inflow and outflow regions and the flow is no longer need to symmetric about ${\displaystyle \theta =0}$.

The rich structure of this dynamical interpretation can be found in Rosenhead(1940).

Pure outflow

For pure outflow, since ${\displaystyle F=a}$ at ${\displaystyle \theta =0}$, integration of governing equation gives

${\displaystyle \theta =\sqrt{\frac{3}{2}}{\int }_F^a\frac{dF}{\sqrt{(a-F)(F-b)(F-c))}}}$

and the boundary conditions becomes

${\displaystyle \alpha =\sqrt{\frac{3}{2}}{\int }_0^a\frac{dF}{\sqrt{(a-F)(F-b)(F-c))}},Re=2\sqrt{\frac{3}{2}}{\int }_0^{\alpha }\frac{FdF}{\sqrt{(a-F)(F-b)(F-c))}}.}$

The equations can be simplified by standard transformations given for example in Jeffreys.

• First case ${\displaystyle b,c}$ are complex conjugates and ${\displaystyle a>0}$ leads to

${\displaystyle F(\theta )=a-\frac{3M^2}{2}\frac{1-\mathrm{cn}(M\theta ,\kappa )}{1+\mathrm{cn}(M\theta ,\kappa )}}$

${\displaystyle M^2=\frac{2}{3}\sqrt{(a-b)(a-c)},{\kappa }^2=\frac{1}{2}+\frac{a+2}{2M^2}}$

where ${\displaystyle \mathrm{sn},\mathrm{cn}}$ are Jacobi elliptic functions.

• Second case ${\displaystyle c<b<0<a}$ leads to

${\displaystyle F(\theta )=a-6k^2{m}^2{\mathrm{sn}}^2(m\theta ,k)}$

${\displaystyle m^2=\frac{1}{6}(a-c),k^2=\frac{a-b}{a-c}.}$

Limiting form

The limiting condition is obtained by noting that pure outflow is impossible when ${\displaystyle F^{\prime }(\pm \alpha )=0}$, which implies ${\displaystyle b=0}$ from the governing equation. Thus beyond this critical conditions, no solution exists. The critical angle ${\displaystyle {\alpha }_c}$ is given by

${\displaystyle \begin{array}{cc}{\alpha }_c& =\sqrt{\frac{3}{2}}{\int }_0^a\frac{dF}{\sqrt{F(a-F)(F+a+6))}},\\ & =\sqrt{\frac{3}{2a}}{\int }_0^1\frac{dt}{\sqrt{t(1-t)\{1+(1+6/a)t\}}},\\ & =\frac{K(k^2)}{m^2}\end{array}}$

where

${\displaystyle m^2=\frac{3+a}{3},k^2=\frac{1}{2}\left(\frac{a}{3+a}\right)}$

where ${\displaystyle K(k^2)}$ is the complete elliptic integral of the first kind. For large values of ${\displaystyle a}$, the critical angle becomes ${\displaystyle {\alpha }_c=\sqrt{\frac{3}{a}}K\left(\frac{1}{2}\right)=\frac{3.211}{\sqrt{a}}}$.

The corresponding critical Reynolds number or volume flux is given by

${\displaystyle \begin{array}{cc}R{e}_c=\frac{Q_c}{\nu }& =2{\int }_0^{{\alpha }_c}(a-6k^2{m}^2{\mathrm{sn}}^{2}m\theta )d\theta ,\\ & =\frac{12k^2}{\sqrt{1-2k^2}}{\int }_0^K{\mathrm{cn}}^{2}tdt,\\ & =\frac{12}{\sqrt{1-2k^2}}[E(k^2)-(1-k^2)K(k^2)]\end{array}}$

where ${\displaystyle E(k^2)}$ is the complete elliptic integral of the second kind. For large values of ${\displaystyle a,\left(k^2\sim \frac{1}{2}-\frac{3}{2a}\right)}$, the critical Reynolds number or volume flux becomes ${\displaystyle R{e}_c=\frac{Q_c}{\nu }=12\sqrt{\frac{a}{3}}\left[E\left(\frac{1}{2}\right)-\frac{1}{2}K\left(\frac{1}{2}\right)\right]=2.934\sqrt{a}}$.

Pure inflow

For pure inflow, the implicit solution is given by

${\displaystyle \theta =\sqrt{\frac{3}{2}}{\int }_b^F\frac{dF}{\sqrt{(a-F)(F-b)(F-c))}}}$

and the boundary conditions becomes

${\displaystyle \alpha =\sqrt{\frac{3}{2}}{\int }_b^0\frac{dF}{\sqrt{(a-F)(F-b)(F-c))}},Re=2\sqrt{\frac{3}{2}}{\int }_{\alpha }^0\frac{FdF}{\sqrt{(a-F)(F-b)(F-c))}}.}$

Pure inflow is possible only when all constants are real ${\displaystyle c<b<0<a}$ and the solution is given by

${\displaystyle F(\theta )=a-6k^2{m}^2{\mathrm{sn}}^2(K-m\theta ,k)=b+6k^2{m}^2{\mathrm{cn}}^2(K-m\theta ,k)}$

${\displaystyle m^2=\frac{1}{6}(a-c),k^2=\frac{a-b}{a-c}}$

where ${\displaystyle K(k^2)}$ is the complete elliptic integral of the first kind.

Limiting form

As Reynolds number increases (${\displaystyle -b}$ becomes larger), the flow tends to become uniform(thus approaching potential flow solution), except for boundary layers near the walls. Since ${\displaystyle m}$ is large and ${\displaystyle \alpha }$ is given, it is clear from the solution that ${\displaystyle K}$ must be large, therefore ${\displaystyle k\sim 1}$. But when ${\displaystyle k\approx 1}$, ${\displaystyle \mathrm{sn}t\approx \tanh t,c\approx b,a\approx -2b}$, the solution becomes

${\displaystyle F(\theta )=b\left\{3{\tanh }^2\left[\sqrt{-\frac{b}{2}}(\alpha -\theta )+{\tanh }^{-1}\sqrt{\frac{2}{3}}\right]-2\right\}.}$

It is clear that ${\displaystyle F\approx b}$ everywhere except in the boundary layer of thickness ${\displaystyle O\left(\sqrt{-\frac{b}{2}}\right)}$. The volume flux is ${\displaystyle Q/\nu \approx 2\alpha b}$ so that ${\displaystyle |Re|=O(|b|)}$ and the boundary layers have classical thickness ${\displaystyle O\left(|Re{|}^{1/2}\right)}$.

Jeffery-Hamel Flow with Stretching Shrinking Surface

Jeffery-Hamel Flow in the channel, with various effects has been studied by researchers as Turkyilmazoglu, Ganji, Saumy Mittal, M. Mohan etc. It was observed that fluid velocity decreases in converging channels but increases significantly in diverging channels for higher Williamson and viscosity parameters. The temperature of the fluid increases slightly with these parameters, while the skin-friction coefficient decreases for both converging and diverging channels. It was also found that heat transfer is enhanced more effectively in diverging channels, especially under shrinking-wall conditions and higher Eckert numbers.