Edit
This edit is in response to some comments claiming that the wake pressure is the same as the ambient pressure across the entire wake volume.
This is not correct. We know that in the wake, the air stream is moving faster than the stationary ambient air and gradually decelerates to zero speed with respect to the ambient air.
Let,s consider a prismatic strand of the flux of air in the wake equal to a volume of a cube one unit of the surface cross-section by a length equal to the speed of the strand per second $V_{flux}$.
By Newton's second law, we know
the force this cube exerts at its end is
$$F=\frac{dP}{dt}$$
We know the mass is constant because we assumed a stationary state so $P=mV \rightarrow dp=mdV$ therefore the equation becomes
$$F=m \frac{dV}{dt} =m\alpha$$
So we do have a pressure $pressure= \frac{F}{A}$ but we assumed the strand section 1 unit so the $pressure = F$
End of edit
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It depends on factors such as the pressure of downwind flow, and the shape of the nozzle. whether the stream is twisting (it will expand faster and lose pressure).
Usually in small fans, the equi-pressure contour lines are like an oblong onion layered off-center, expanding irregularly and losing energy within 10-20 times the diameter of the fan. Unless the fan is designed to blow farther, e.g has a protective slower stream keeping it centered, or a proper long vent like leaf blowers.
In airplane jets, there are published Engine ingestion and wake hazard areas. there have been examples of pavement been ripped by the pressure of the wake and cause substantial damage to the fuselage. Up to 500 meters aft engine could be dangerous.
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