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Imagine I have a section of pipe that increases in cross-section at some point. If the gas in the pipe before this section is being displaced with a piston, will the gas increase its pressure and reduce its velocity while going through increasing area, or will it maintain its velocity and pressure at expense of energy delivered by the piston?

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  • $\begingroup$ Consider the expression for pressure losses in a pipe: it has an L/D term so diameter does have an effect Check out ConDi nozzles (Convergent / Divergent)... $\endgroup$
    – Solar Mike
    Feb 8, 2021 at 5:41
  • $\begingroup$ The source of energy input is irrelevant unless you are actually pressurizing the system because the piston moves faster then materials exit the system. In general, it's safe to say that increased velocity equals decreased pressure, and vice versa. And there is no way in a steady state system for velocity to stay the same for varying pipe diameters. $\endgroup$
    – Tiger Guy
    Feb 9, 2021 at 21:49
  • $\begingroup$ i had this realization yesterday. The velocity has to drop in order to maintain the same flow. Therefore to maintain energy the pressure will have to increase to maintain energy. For it to do otherwise would be the opposite of acquiring energy. Sometimes when you look at formulas you lose the intuitive understanding $\endgroup$
    – Francis L.
    Feb 10, 2021 at 2:22

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I'm going to assume the piston is moving subsonically here (since you indicated velocity decreases with increasing area).

The key here is that the end of the pipe is not a closed boundary, so the piston is not compressing the gas or doing work in the sense of $P \Delta V$. As long as the flow is subsonic and not choked via a constriction, pressure will not build up at the piston end because pressure waves will propagate downstream faster than the flow. The piston is still adding energy to the gas, so total (stagnation) pressure would increase, but the static pressure would not. So the piston is imparting kinetic energy to the gas, but I would treat this as a boundary condition, not an energy input, such that $v_{piston} = v_{i, gas}$. From conservation of mass, and neglecting losses, the velocity of the gas everywhere in the straight section of the pipe would equal the piston velocity.

Again neglecting pressure losses in the pipe and assuming an isentropic expansion at the cross-sectional area change (as opposed to a sudden step with pressure loss), the expansion of the subsonic gas would lead to a decrease in velocity and increase in static pressure.

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See the gas law. The pressure x volume / temperature remain constant. For the speed component see the kinetic theory in the wiki article. https://en.wikipedia.org/wiki/Ideal_gas_law

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  • $\begingroup$ you're missing the motion of fluid here. $\endgroup$
    – Tiger Guy
    Feb 9, 2021 at 21:50
  • $\begingroup$ The speed of the gas is a consequence of the pressure difference. So no speed without pressure difference. Assuming that no energy is lost, the pxv / t value remains constant. If the piston is moved back again, for example by an external energy source, the original tooth setting can be reached again. $\endgroup$ Feb 10, 2021 at 5:41

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