Fluid power began with hydraulics and the fluid being water. Water cannot be compressed. Though other fluids can be compressable and some even into gasious states, some remain non-commpressable.
External force or pressure means anything protruding from the outside in being the pushing force creating the pressure. We need non-compressable fluids to make the hydraulics and other fluid powers to work in the ways in which they do today.
Keep in mind that even if the verbage of fluid power were to change, that the principles will continue to work the same as they exist today. To alter the verbage now and in todays society, would only confuse people more in this already confusing world.
Ask yourself this; is it worth the fight to chage laws of theroys for one mans different interiptations? Also, is this why our constituion is so insanely ammended because everyone seems to find a way to change it to how they see it to be?
I think you're over thinking this thing buddy. It's just meant to be used as a basic tool in understanding how fluid power works.
Thanks
Joe
By convention, gauge pressure references atmospheric pressure- for "gauge pressure p in this case is the sum of major and minor losses" to be valid, the duct must discharge to atmosphere. This requires that your Minor losses include the discharge loss, where the duct delivers/returns air to atmospheric conditions (and all kinetic energy/dynamic pressure is lost),where $v_{discharge}=0$, $P_{discharge}=P_{atm}$, and $T_{discharge}=T_{atm}$.
The fan increases the pressure above atmospheric pressure enough to induce a desired flow rate. The manometer measures the difference in pressure between two points- in this case: $P_{manometer} - P_{discharge} = \Delta P$, where $P_{manometer} \approx P_{fan}$.
The pressure difference, $\Delta P$ is a physical measure of flow potential between points. For your problem statement to hold, all of the pressure differential is lost.
These pressure losses are due to Major (friction) losses and Minor (geometry restriction) losses. Without knowing other factors like duct geometry, roughness, velocity, etc. you cannot determine what the Major and Minor losses are, you can only state:
$$P_{gauge}=\Delta P=Major Loss+Minor Loss$$
Assuming uniform roughness and restrictions throughout the duct, moving the manometer closer to the discharge of the duct results in lower gauge pressures.
(Forgive the pitiful diagram!)
Perhaps a more intuitive way to envision your question:
If you wanted to determine losses in a given section of duct, you could place two manometers at chosen locations. Solving for $P_{manometer}$ at both manometers, you can calculate $\Delta P$ between manometers, again, the sum of Major and Minor losses in that section.
Best Answer
May it now be that by definition a free stream has no accelerations? And as derived in the linked page, no acceleration implies no pressure gradients. That with the fact that all the streams considered are connected with the atmosphere at some points, should mean all the stream is at atmospheric pressure.