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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:

• $x_1=-0.008585$
• $x_2=10.00000000080*10^5$

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

This result makes me weird because it's the first time I heard about this pressure behaviour. Have I made one/more mistake/s ?

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!

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 $\dfrac{1073.15}{x}-\dfrac{973.15}{10^6}=x-10^6$
and then again as $\dfrac{1073.15}{x}-\dfrac{973.15}{10^6}=x-10^6$

• $x_1=-0.008585$
• $x_2=10.00000000080*10^6$

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

This result makes me weird because it's the first time I heard about this pressure behaviour. Have I made one/more mistake/s ?

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:

• $x_1=-0.008585$
• $x_2=10.00000000080*10^5$

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

This result makes me weird because it's the first time I heard about this pressure behaviour. Have I made one/more mistake/s ?

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):

• $x_1=-0.008585$
• $x_2=10.00000000080*10^6$

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

This result makes me weird because it's the first time I heard about this pressure behaviour. Have I made one/more mistake/s ?

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 $\dfrac{1073.15}{x}-\dfrac{973.15}{10^6}=x-10^6$
and then again as $\dfrac{1073.15}{x}-\dfrac{973.15}{10^6}=x-10^6$

• $x_1=-0.008585$
• $x_2=10.00000000080*10^6$

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

This result makes me weird because it's the first time I heard about this pressure behaviour. Have I made one/more mistake/s ?