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\documentclass{article}
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\begin{document}
%\pagestyle{empty}
\setlength{\oddsidemargin}{.2in}
%\Large
\begin{centering}
{\huge\bf\rr First order perturbative calculations for a conducting liquid jet in a solenoid}
\vskip.3in
R. B. Palmer
\vskip.3in
DRAFT 3~~~~10/11/00
\vskip.3in
\end{centering}

\section{\bb Abstract}

A perturbative calculation is given of the behavior of a continuous jet of conducting fluid as it enters and leaves a solenoidal magnetic field. It is assumed that the changes in direction, jet cross section and velocity are small. 

If the jet enters the field along, or close to the axis, then the induced forces are compressive and retarding. The jet slows, suffers an increase in hydrostatic pressure, and increases in diameter. Later, the jet reaccelerates and shrinks. As the jet leaves the field, the hydrostatic pressure becomes negative and cavitation may, or may not occur. 

If the jet enters at an angle to the axis, there are, in addition, deflections and ellipticl deformations of the jet.

Formulae are given for these effects and numerical values given for the example of a solenoidal field with a Gaussian axial profile.

\section{\bb Formulae}

\subsection{Introduction}

The jet is assumed to have an initial radius $r$, and be travelling at a velocity $v$. Changes in radius, shape, direction and velocity are all assumed to be small. The angle between the jet and solenoid axes is also assumed to be small.

In the following formulae, the coordinate system is defined by the jet: $z$ along the direction of motion; $r$ perpendicular to $z$. If the jet is along the axis then $z$ is also the solenoid axis, and the asymuthal field $B_\phi$ = 0.

If the jet is not directed along the solenoid axis, then we also define:
 $y$ in a direction perpendicular to $z$ (the jet axis) and away from the solenoid axis; and $x$ perpendicular to $y$ and $z$. In this case, $B_x=0$.

\subsection{Induced Azymuthal Current}


The magnetic flux inside a radius r:
$$
\Phi \approx \pi r^2 B_z(z)
$$
As a liquid metal jet passes axially down such a field at a velocity $v={dz / dt}$, a circumferential potential will be generated:
$$
V={d\Phi \over dt}\approx \pi r^2 {dB_z \over dz} v
$$
If the metal electrical conductivity $\kappa$ is low enough so that the resulting current has a negligible effect on the field, then the azimuthal current density $i_{\phi}$ will be:
$$
i_{\phi}\approx {V \over 2\pi r}\kappa
$$
{\rr$$
i_{\phi}~\approx~{r~v~\kappa \over 2}~ {dB_z \over dz} 
$$}

\subsection{Induced Axial Currents}

Consider a transverse field component $B_y$

The magnetic flux between transverse positions $-x$ to $x$:
$$
\Phi_y~ =~ 2~x B_y(z)~dz
$$
As a liquid metal jet passes axially down such a field at a velocity $v={dz / dt}$, axial voltage gradients will be generated:
$$
G~=~x~ {dB_y \over dz}~ v
$$
If the metal electrical conductivity $\kappa$ is low enough so that the resulting current has a negligible effect on the field, then the axial current density $i_z$ will be:
$$
i_z~=~ G~\kappa
$$
{\rr $$
i_z~=~x~v~\kappa~ {dB_y \over dz}
$$}


\subsection{Axial force}

The radial field is given by:
$$
B_r\approx {r \over 2}{dB_z \over dz}
$$ 

The induced axial force per unit volume $(dr~r~d\phi dz)$ is:
$$
f_z=-B_r i_{\phi} \approx {r^2 \over 4}~v \kappa~ \({dB_z \over dz} \)^2 
$$ 
and the average force of the disk of radius $r_o$ is:
{\rr $$
<{f_z}> \approx~ -~{r_o^2 \over 8}~v \kappa~ \({dB_z \over dz} \)^2 
$$ }

\subsection{Radial forces and Hydrostatic Pressure}

The induced radial force per unit volume $(dr~r~d\phi dz)$ is:
$$
f_r=-B_z i_{\phi} \approx  {r \over 2 }~v \kappa ~B_z {dB_z \over dz} 
$$
If we assume the effects of the fields are small so that the jet radius and liquid velocities do not vary by large fractions, and if we ignore radial inertia, then the hydrostatic pressures in a jet of outside radius $r_o$, at radius r, will be given by:
$$
p(r,z)=\int_{r_o}^r -{f_r}dr\approx \({r_o^2-r^2 \over 4}\)~v \kappa~ B_z {dB_z \over dz}
$$
This pressure is a function of $z$, and gradients in it will exert axial pressures $f_p$ on the liquid that must be added to the magnetic term $f_z$.
$$
f_p(r,z)= {dp(r,z) \over dz} \approx~-~ \({r_o^2-r^2 \over 4}\)~v \kappa~
 {d \over dz}\( B_z {dB_z \over dz}\)
$$
and the average value of this force is:
$$
<{f_p(r,z)}> \approx ~-~\({r_o^2 \over 8}\)~v \kappa~ {d \over dz}\( B_z {dB_z \over dz}\)
$$
 
\subsection{Axial acceleration}


These forces will then decelerate, or accelerate layers of the fluid, thus inducing differences of liquid velocity as a function of radius:
$$
\rho~{dv \over dt}~=~f
$$
$$
\Delta~ v(r,z)~=~ \int_{z_o}^z{(f_z+f_p)~{1 \over v \rho}~ dz}
$$
The average change in velocity is then:
{\rr $$
<\Delta v>(z)~=~{\kappa \over \rho}~{r_o^2 \over 8}~   \(\int_{z_o}^z{\({dB_z \over dz} \)^2 
~+~{d \over dz}\( B_z {dB_z \over dz}\)~dz} \)
$$}
and the radius as a function of z is:
$$
r(z)~=~\(1~-~{<\Delta v>(z) \over v} \)
$$


\subsection{Transverse Forces and Deflections}

From above, the radial force per unit volume $(dr~r~d\phi dz)$ is:
$$
f_r=-B_z i_{\phi} \approx  {r \over 2 }~v \kappa ~B_z {dB_z \over dz} 
$$
If $B_z$ varies with a transverse distance $y$, then the component of this radial force in the $y$ direction is:
$$
f_y = f_r~\sin(\phi)
$$
and the net deflective force per unit length $dF_y$ is
$$
{dF_y \over dz}~ =~ \int_o^r ~\int_o^{2\pi} 
   {r \over 2 }~v \kappa ~{dB_z \over dy}~r \sin^2 (\phi ) {dB_z \over dz} ~r~ dr d\phi 
$$
$$
= ~{v \kappa \over 2} ~{dB_z \over dy}~ {dB_z \over dz} ~\int_o^{2\pi}{ \sin^2 (\phi)~d\phi}~
\int_o^r {r^3~dr }
 $$$$
= ~{\pi \over 8}v \kappa ~r^4 ~{dB_z \over dy}~ {dB_z \over dz} 
$$
and the change in transverse velocity:
$$
{d v_y \over dz} ~=~{1 \over v}~{d v_y \over dt}
$$$$
=~ {{dF_y \over dz}~dz~\over~v~\rho~\pi~ r^2 ~dz}
$$$$
=~{\kappa~r^2 \over 8~\rho}  ~{dB_z \over dy}~ {dB_z \over dz} 
$$
and the inverse radius of bend is:
{\rr $$
{d \theta \over dz}~=~{\kappa~r^2 \over 8~v~\rho}  ~{dB_z \over dy}~ {dB_z \over dz} 
$$}


\subsection{Transverse Elliptical Distortion}

If the jet is not on the solenoid axis, the axial induced currents interaxting with the transverse fields will generate distorting forces on the jet. These transvers forces per unit volume $dx~ dy~ dz$ are:
$$
f_x~=~ i_z~B_y~
$$$$
=~x~v~\kappa~B_y~ {dB_y \over dz}
$$
Ignoring surface tension, there is no restoring force, so the shape will change. The sides will go in and the height increase. A conservative estimate of the rate of width reduction can be obtained assuming that the only inertial resistance comes from the elements themselves. Then:

$$ {d_2x \over dz^2}~<~{ f_x \over v~\rho } =~~{x~\kappa \over \rho}~B_y~ {dB_y \over dz}$$

In reality the inertial constraint will be greater, because the inward motion of elements on the sides must generate equal outward motions at the top and bottom, and azymuthal motions between the two. Including these effects we can guess:
{\rr
$$ {d_2y \over dz^2}~\approx~{1\over3}~{x~\kappa \over \rho}~B_y~ {dB_y \over dz}$$
}
If this velocity is sufficiently small, then surface tension will restrain it. The total inward force on each side is:
$$
{dF_B \over dz}~\approx~{1\over3}~ v~\kappa~B_y~ {dB_y \over dz}~ \int_{-\pi}^\pi~\int_o^r~ r~\cos(\phi)~dr~r~d\phi 
$$$$
\approx~{1\over3}~{2\over3}~r^3~\kappa~B_y~ {dB_y \over dz}
$$
and this is opposed by the surface tension difference for a finite ellipticity $\epsilon=y/x$:
$$
{dF_T \over dz}~\approx~2~T~\epsilon
$$
where $T$ is the surface tension.

So the equilibrium ellipticity is:
{\rr $$
\epsilon_{equilib}~\approx~{1\over3}~{r^3\over 3~T}~\kappa~B_y~ {dB_y \over dz}
$$}

Again, this is only an order of magnitude calculation since the factor 1/3 is only an estimate for the inertial effects in distorting th ellipse. With the factor removed, however, the equation gives a harder maximum value. 




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{\bb Gaussian Axial Example}

As an example we can consider a field that varies as a Gaussian in $z$:
$$
B_z=B_o~ e^{-({z^2/2\sigma_z^2})}
$$
$$
{dB_z\over dz}=B_o~{-z\over \sigma_z^2}~e^{-({z^2/2\sigma_z^2})}
$$
$$
B_r=~-~B_o~{r~z\over 2 \sigma_z^2}~e^{-({z^2/2\sigma_z^2})}
$$
$$
f_r\approx ~-~{rz \kappa v \over 2 \sigma_z^2}~B_o^2~ e^{-(z^2/\sigma_z^2)}
$$
$$
f_z\approx ~-~{r^2 \kappa v \over 4 \sigma_z^4}~B_o^2~z^2 ~ e^{-(z^2/\sigma_z^2)}
$$
$$
p_r\approx { \kappa v \over 4 \sigma_z^2}~( r_o^2-r^2 )~B_o^2~z~ e^{-(z^2/\sigma_z^2)}
$$
$$
f_p \approx {\kappa v \over 4 \sigma_z^2}~( r_o^2-r^2 )\(1-{2z^2 \over \sigma_z^2}\)~ B_o^2
~ e^{-(z^2/\sigma_z^2)}
$$
\vskip.3in



Defining $R=r/r_o$ and $Z=z/\sigma_z$, the change of velocities starting from points outside the field are:
$$
\Delta v(R,Z)~ \approx~ {\kappa ~ r_o^2 \over 4~ \rho~\sigma_z}~B_o^2~
$$ $$
\left[  
~-~R^2\int_{-\infty}^Z {Z^2e^{-Z^2}dZ} + (1-R^2)\(\int_{-\infty}^Z{e^{-Z^2}dZ} 
- ~2~\int_{-\infty}^Z {Z^2e^{-Z^2}dZ}     \)  
\right]      
$$
and the average velocity change is:
$$
<\Delta v>(Z)~ \approx~ {\kappa ~ r_o^2 \over 8~ \rho~\sigma_z}~B_o^2~\(
 \int_{-\infty}^Z{e^{-Z^2}dZ} 
- \int_{-\infty}^Z {Z^2e^{-Z^2}dZ}      \)
$$

\newpage
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{plots}

The following plots show calculated values of
some of the functions.

\vskip.5in

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\newpage
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{numbers}

Beynond the magnet center (z=0), the radial force is outward. The pressure in the liquid is now negative and cavitation can occur, causing the jet to break up. If it does so, however, the circulating currents stop, reducing further dispersal. 
\vskip.1in
The maximum reduction of surface velocity prior to reaching the center $\Delta v (out)$ is seen to be given by
$$
\Delta v~\approx~ 0.8~{\kappa ~ r_o^2 \over 8~ \rho~\sigma_z}~B_o^2
$$
and at the end it is approximately twice this value.

For three possible liquids, we obtain:
\vskip.2in

\begin{tabular}{|l|cccc|}
\hline
material & conductivity & density & $\Delta v$ (o)& $\Delta v$ (tot)\\
      & ohm m & kg/m$^3$ & m/s & m/s \\
\hline
gallium & 3.8 $10^6$ & 6~$10^3$ & 2.15 & 4.3 \\ 
mercury & 4~$10^6$ & 13.5~$10^3$& .95 & 1.9\\
Pb(81) In(19) & 2.6~$10^6$ & 10.3~$10^3$ & .8 & 1.6 \\
\hline
\end{tabular}
\vskip.2in

Compared with a jet velocity of 20 m/sec, these are all relatively low. Earlier optimizations had called for a jet radius of 1 cm, which gave values of $\Delta v$ 4 times higher, and thus more significant.


\end{document}



%%%%%%%%%%%%%%%%%%%%%%
Elliptical distortion can result from a second order variation of the $B_z$ field.
\vskip.2in

The hydrostatic pressure will restrain the average inward radial forces, and the net transverse force results in a deflection. Any remaining difference in the inward forces at $\phi=0$ and $\phi=\pi$ will result in inward motion of one and  equal outward motion of the other. The driving force for these assymetric motions, per unit of volume, is: 

$$\Delta f_r~= ~ {f_r(\phi=0)~-~f_r(\phi=\pi) \over 2}    $$

$$\Delta f_r~=~  {r^3 \over 4 }~v \kappa ~{d_2 B_z \over dy^2} {dB_z \over dz} $$

Ignoring surface tension, there is no restoring force, so the shape will change at a rate:

$$ {dy \over dz}~=~-{dx \over dz}~=~{\Delta f_r \over v~\rho } $$

$$ =~{r^3 \over 4~\rho }~ \kappa ~{d_2 B_z \over dy^2} {dB_z \over dz} $$

