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We have incompressible air flowing through a cylindrical pipe, from section A to section B. The process is considered to be stationary.

My theory book says that, because of the frictional losses on the walls of the pipe, the stagnation pressure will eventually decrease from B to A: $$\Delta p_0=p_{0}^{B}-p_{0}^{A} < 0$$

I understand that wall friction actually represents a loss of energy, but I can't see how this loss decreases stagnation pressure.

I'm using these equations: $$p_0^A=p^A+\dfrac{v_A^2}{2c_p}$$ $$p_0^B=p^B+\dfrac{v_B^2}{2c_p}$$

Intuitively, the velocity of the incompressible air will decrease from B to A, but I don't know about the static pressures.

Any help or hint will be greatly appreciated.

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static pressure at B can be larger than at A and can be lower. This is not used for estimating stagnation pressure change. If there is no frictional loss, the fact, that the two total pressures are the same, means the energy is conserved. When this is loss, of course energy at A is larger than that at B in order to conserve the energy.

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  • $\begingroup$ But my book implies something like "friction, therefore loss of energy, therefore loss of stagnation pressure". It's the last statement that I don't understand. $\endgroup$ – Jose Lopez Garcia May 4 '16 at 12:59
  • $\begingroup$ If you review Bernoulli equation, you will see that stagnant pressure is total energy. So the first and second statements are the same. $\endgroup$ – user115350 May 4 '16 at 13:35
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Intuitively, the velocity of the incompressible air will decrease from B to A, but I don't know about the static pressures

Intuition isn’t working for you here. By continuity, the (incompressible, hence same densjty) air has to have the same mass flow at A and B. Since they’re the same area, that means same velocity. Same velocity means same kinetic energy; pressure must drop.

Can we make that more intuitive? Yes. The air is being pushed against a frictional force at constant velocity. That requires a countervailing force to have s net force of zero. That comes from the difference in pressure forces in the two ends of the climb of air.

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