How Does the Bernoulli Equation Determine Axial Pressure in Jet Flows?

[Lucus]

Okay, here's the question:

A jet of incompressible fluid emerges from a horizontal channel into an environment of the same fluid. The momentum of the jet at the exit of the channel is measured to be M. The exit momentum is the source of a downstream flow which spreads gradually with distance.

The fluid outside the jet flow is at rest. Use the Bernoulli equation to determine the axial (x) pressure beyond the boundary of the jet flow. How can that result, and the properties of the boundary layer equations be used to determine the axial pressure gradient inside the jet?

So there's the question. I'm not really sure where to start. Any help would be much appreciated. Thanks!

 

[Clausius2]

Okay, here's the question:
A jet of incompressible fluid emerges from a horizontal channel into an environment of the same fluid. The momentum of the jet at the exit of the channel is measured to be M. The exit momentum is the source of a downstream flow which spreads gradually with distance.
The fluid outside the jet flow is at rest. Use the Bernoulli equation to determine the axial (x) pressure beyond the boundary of the jet flow. How can that result, and the properties of the boundary layer equations be used to determine the axial pressure gradient inside the jet?
So there's the question. I'm not really sure where to start. Any help would be much appreciated. Thanks!

Nice problem, although the statement makes no sense. You can't use Bernoulli equation to determine the axial pressure assuming viscous flow as you are doing. The jet core developed in a non dimensional distance of order $$Re_j^{-1}$$ is dominated by viscous forces. You only can apply Bernouilli to calculate external boundary layer pressure, but as it is an stagnant atmosphere, this pressure is trivially constant $$P_a$$.
Usual calculations of incompressible jets assume negligible pressure gradients, in part because the boundary layer is so thin at large $$Re$$ that transversal pressure gradients are very small, and also because by this argument the external pressure impose an uniform pressure across the symmetry axis.