CFD Online Discussion Forums - Pressure variation in a heating channel

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