Pressure variation in a heating channel

gRomK13 · August 6, 2009, 13:18

Hello

I'd like to calculate the pressure variation between the inlet and the outlet of a heating channel. The causes of this variation are well known :

- Friction, because of viscosity and turbulence : IRREVERSIBLE
- Fluid Acceleration, either because of density change (heating for instance!) or because of cross section change : REVERSIBLE
- Gravity, if z coordinate changes : REVERSIBLE

Now I want to give an expression for these different components, but my two methods are not in agreement...

1) First Method: 1D Navier-Stokes

1D Momentum equation:

$\frac{dP}{dz} = - \rho w \frac{dw}{dz} - \rho g - \frac{4 \tau_{w}}{D_{h}}$ ( $\tau_{w}$ is wall friction, thus the whole term is the one responsible for friction pressure drop)

From which we define:

$\left( \frac{dP}{dz} \right) _{acceleration} = - \rho w \frac{dw}{dz} = - G \frac{d}{dz} \left( \frac{G}{\rho} \right)$

where $G = \rho w = \frac{Q}{S}$ is mass velocity (Q is the mass flowrate and S is the cross section)

It comes: $\left( \frac{dP}{dz} \right) _{acceleration} = - \frac{1}{2} G^{2} \frac{d}{dz} \left( \frac{1}{\rho} \right) - \frac{d}{dz} \underbrace{ \left( \frac{1}{2} \frac{G^{2}}{\rho} \right) }_{E_{c}}$

Then integration of a small portion of my channel gives: $\Delta P _{acceleration} = - \frac{1}{2} \int_0^L G^{2} \frac{d}{dz} \left( \frac{1}{\rho} \right) dz\ - \Delta E_{c}$

- When density is constant (no heating) and cross section varies: First term is zero, so that leaves us with $-\Delta E_{c}$ only.

- When density changes but cross section is constant: G is also constant so that both terms are finally equal, and we obtain for the pressure drop $: -2 \Delta E_{c}$

2) Second Method: Generalized Bernoulli's law

Let's define hydraulic head, which represents the energy necessary to the fluid motion through piping:

$h = P + \frac{1}{2} \rho w^{2} + \rho g z$

Generalized Bernoulli's law yields in:

$\Delta P = \Delta h - \Delta \left( \frac{1}{2} \rho w^{2} \right) - \Delta \left( \rho g z \right)$

We recognize:

a) $\Delta h$ : Head loss due to friction
b) $- \Delta \left( \rho g z \right)$ : Pressure drop due to gravity

But then, what is $- \Delta \left( \frac{1}{2} \rho w^{2} \right) = - \Delta E_{c}$ ?? It is indeed pressure drop due to acceleration of the fluid, in the case where the channel is not heated. For the heating channel, the factor 2 is missing!!

CONCLUSION: My opinion is that Bernoulli's law is applicable only for incompressible flows. Yet, when the fluid is heated, $\rho$ changes so that we cannot really speak about incompressibility.

Do you see any mistake in my formulas or reasoning? Thank you for reading me.

Jérôme

August 6, 2009, 13:18	Pressure variation in a heating channel	#1
gRomK13 New Member Join Date: Jul 2009 Location: near Marseille, France Posts: 7 Rep Power: 17	Hello I'd like to calculate the pressure variation between the inlet and the outlet of a heating channel. The causes of this variation are well known : - Friction, because of viscosity and turbulence : IRREVERSIBLE - Fluid Acceleration, either because of density change (heating for instance!) or because of cross section change : REVERSIBLE - Gravity, if z coordinate changes : REVERSIBLE Now I want to give an expression for these different components, but my two methods are not in agreement... 1) First Method: 1D Navier-Stokes 1D Momentum equation: $\frac{dP}{dz} = - \rho w \frac{dw}{dz} - \rho g - \frac{4 \tau_{w}}{D_{h}}$ ( $\tau_{w}$ is wall friction, thus the whole term is the one responsible for friction pressure drop) From which we define: $\left( \frac{dP}{dz} \right) _{acceleration} = - \rho w \frac{dw}{dz} = - G \frac{d}{dz} \left( \frac{G}{\rho} \right)$ where $G = \rho w = \frac{Q}{S}$ is mass velocity (Q is the mass flowrate and S is the cross section) It comes: $\left( \frac{dP}{dz} \right) _{acceleration} = - \frac{1}{2} G^{2} \frac{d}{dz} \left( \frac{1}{\rho} \right) - \frac{d}{dz} \underbrace{ \left( \frac{1}{2} \frac{G^{2}}{\rho} \right) }_{E_{c}}$ Then integration of a small portion of my channel gives: $\Delta P _{acceleration} = - \frac{1}{2} \int_0^L G^{2} \frac{d}{dz} \left( \frac{1}{\rho} \right) dz\ - \Delta E_{c}$ - When density is constant (no heating) and cross section varies: First term is zero, so that leaves us with $-\Delta E_{c}$ only. - When density changes but cross section is constant: G is also constant so that both terms are finally equal, and we obtain for the pressure drop $: -2 \Delta E_{c}$ 2) Second Method: Generalized Bernoulli's law Let's define hydraulic head, which represents the energy necessary to the fluid motion through piping: $h = P + \frac{1}{2} \rho w^{2} + \rho g z$ Generalized Bernoulli's law yields in: $\Delta P = \Delta h - \Delta \left( \frac{1}{2} \rho w^{2} \right) - \Delta \left( \rho g z \right)$ We recognize: a) $\Delta h$ : Head loss due to friction b) $- \Delta \left( \rho g z \right)$ : Pressure drop due to gravity But then, what is $- \Delta \left( \frac{1}{2} \rho w^{2} \right) = - \Delta E_{c}$ ?? It is indeed pressure drop due to acceleration of the fluid, in the case where the channel is not heated. For the heating channel, the factor 2 is missing!! CONCLUSION: My opinion is that Bernoulli's law is applicable only for incompressible flows. Yet, when the fluid is heated, $\rho$ changes so that we cannot really speak about incompressibility. Do you see any mistake in my formulas or reasoning? Thank you for reading me. Jérôme

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
Pressure Rise Error	emueller	CFX	0	May 5, 2009 12:08
Pressure drop across a turbulent channel flow	shiv	Main CFD Forum	0	September 21, 2005 01:52
Using ideal gas law to simulate pressure decline	BjÃ¶rn Mattsson	FLUENT	5	September 5, 2005 05:03
what the result is negatif pressure at inlet	chong chee nan	FLUENT	0	December 29, 2001 06:13
Definition of pressure	Ola Nordblom	FLUENT	1	August 16, 2001 16:58