Bickley jet

In fluid dynamics, Bickley jet is a steady two-dimensional laminar plane jet with large jet Reynolds number emerging into the fluid at rest, named after W. G. Bickley, who gave the analytical solution in 1937,[1] to the problem derived by Schlichting in 1933[2] and the corresponding problem in axisymmetric coordinates is called as Schlichting jet.

The solution is valid only for distances far away from the jet origin.

Consider a steady plane emerging into the same fluid, a type of submerged jets from a narrow slit, which is supposed to be very small (such that the fluid loses memory of the shape and size of the slit far away from the origin, it remembers only the net momentum flux).

Let the velocity be

in Cartesian coordinate and the axis of the jet be

axis with origin at the orifice.

The flow is self-similar for large Reynolds number (the jet is so thin that

varies much more rapidly in the transverse

direction than the streamwise

direction) and can be approximated with boundary layer equations.

is the kinematic viscosity and the pressure is everywhere equal to the outside fluid pressure.

Since the fluid is at rest far away from the center of the jet and because the flow is symmetric about

axis and also since there is no solid boundary and the pressure is constant, the momentum flux

across any plane normal to the

axis must be the same is a constant, where

which also constant for incompressible flow.

The constant momentum flux condition can be obtained by integrating the momentum equation across the jet.

is used to simplify the above equation.

The mass flux

across any cross section normal to the

axis is not constant, because there is a slow entrainment of outer fluid into the jet, and it's a part of the boundary layer solution.

This can be easily verified by integrating the continuity equation across the boundary layer.

[3][4] The self-similar solution is obtained by introducing the transformation

The exact solution is given by

( η ) = α tanh ⁡ α η

is solved from the following equation

ξ = α η = 0.2752

{\displaystyle {\begin{aligned}u&=0.4543\left({\frac {M^{2}}{\nu \rho ^{2}x}}\right)^{1/3}\operatorname {sech} ^{2}\xi ,\\v&=0.5503\left({\frac {M\nu }{\rho x^{2}}}\right)^{1/3}(2\xi \operatorname {sech} ^{2}\xi -\tanh \xi ).\end{aligned}}}

The mass flow rate

from the orifice normal to the jet is[5][6][7]