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I'm talking about a:

  • smooth circular pipe without any valve at inlet or outlet
  • without pressure losses due to friction
  • subjected to a constant heat flux from outside on the entire pipe external surface which makes temperature rise from $T_i=973.15[K]$ to a $T_o=1073.15 [K]$
  • in which air flows at pressure of $p_i=10 [bar]$

this doubt was born studying the momentum balance:

$\rho*v^2|_{out}-\rho*v^2|_{in}=p_{out}-p_{in}$

expressed using the perfect gas law $\rho=\dfrac{p}{R^**T}$

and using the conservation law $v=\dfrac{\dot{m}}{\rho*A}$

where $A$ is the cross sectional area of the pipe.

Replacing those in the balance and if I assume mass flow, composition, and cross sectional area constants, I get:

$\dfrac{\dot{m}^2*R^*}{A^2}*\left[\dfrac{T_{out}}{p_{out}}-\dfrac{T_{in}}{p_{in}}\right]=p_{out}-p_{in}$

because the block outside of the parentheses is constant, this bring me to a fractional second order equation, in which $p_{out}$ is the only unknown. The numerical solutions to this equation are (if I put the outside block equals to 1 for simplicity):
$1*\left[\dfrac{1073.15}{x}-\dfrac{973.15}{10^6}\right]=x-10^6$
that I can rewrite as $\dfrac{1073.15}{x}-\dfrac{973.15}{10^6}=x-10^6$
again as $\dfrac{1073.15}{x}-\dfrac{973.15}{10^6}-x+10^6=0$
and definitively as $\dfrac{1073.15-\dfrac{973.15*x}{10^6}-x^2+10^6*x}{x}=0$
denominator simply exclude $x=0$ as possible solution while solving numerator equation bring me this two (and both numerically valid) solutions:

  • $x_1=-0.00107315$
  • $x_2=1000000.0001$

we can easily exclude first result because it has not a physical meaning because can't exist negative pressure, so definitively $p_{out}=x_2=1000000.0001[Pa]=1.0000000001[bar] > p_{in}$

This result makes me weird because it's the first time I heard about this pressure behaviour. I don't understand why!

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I don't know what your x1 and x2 refer to?

The one-dimensional inviscid compressible flow with heat transfer you are referring to is called "Rayleigh flow".

The static pressure will decrease for subsonic flow and increase for supersonic flow when heat is added. As heat is added, the Mach number will approach 1 and the static pressure will approach a value that is a function of the starting static pressure and Mach number, regardless of whether the flow is supersonic or subsonic when heating begins.

The total pressure will always decrease regardless of the starting Mach number. You can read about this in a compressible flow textbook. A popular one is by John Anderson.

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  • $\begingroup$ $x_1$ and $x_2$ are numerical possible solutions to final equation (I "corrected" opening post with this information which I take for granted). I've never seen this equation but admitting you (Rayleigh) are true, why my result bring me to the opposite conclusion ??? $\endgroup$ – mattia.b89 Aug 11 '15 at 7:51
  • $\begingroup$ I'm not a 100% sure but can air be regarded as compressible in the given state? For the given state $Z$ is almost 1. $\endgroup$ – idkfa Aug 11 '15 at 8:17
  • $\begingroup$ Are you talking about Z, the compressibility factor? I'm not sure why you're bringing it up here. For air, I don't think Z deviates much from 1 in the temperatures and pressures the question asker suggested. The compressibility factor just indicates how much a fluid deviates from ideal gas behaviour. Air will be compressible in the given state, it will just be an ideal gas. $\endgroup$ – user1748155 Aug 11 '15 at 16:47
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I don't get your solutions to the equation, mine are: $$ x_1=-0.00107315 \text{Pa}\\ x_2=1000000 \text{Pa}=10\text{bar} $$ which means that the outlet pressure is the same as (or very close to) the inlet pressure as you would expect.

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  • $\begingroup$ you are right! I've corrected opening post with mathematical steps and results anyway things don't change... $\endgroup$ – mattia.b89 Aug 12 '15 at 20:16

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