Non-isothermal conditions in liquid-gas nitrogen homogeneous mixture

In summary: This will fully incorporate non-isothermal effects into the SS model. In summary, to incorporate non-isothermal phenomena into the SS cavitation theory, one must include the energy equation (equation 3.7) and account for the rate of heat and external forces in the momentum equation (equation 3.6).
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I am studying the cavitation theory proposed by https://www.researchgate.net/profile/Guenter-Schnerr-Professor-Dr-Inghabil/publication/296196752_Physical_and_Numerical_Modeling_of_Unsteady_Cavitation_Dynamics/links/56f6b62308ae81582bf2f940/Physical-and-Numerical-Modeling-of-Unsteady-Cavitation-Dynamics.pdf (SS) and realized it assumes isothermal conditions within liquid-gas homogeneous mixture as well as incompressible flow. This is alright for, say, water. However, for cryogenic fluids (such as liquid nitrogen) this assumption is no longer valid.

I want to incorporate non-isothermal phenomena in the SS cavitation theory. This is how I am approaching it

To describe the fluid mechanics, we need to work with1) The continuity equation

\begin{equation}
\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho v) = 0
\end{equation}

Mass conservation holds anyway once non-isothermal effects are incorporated.2) Momentum equation

\begin{equation}
\frac{\partial}{\partial t}(\rho \vec v) + \nabla \cdot (\rho \vec v \vec v) = -\nabla p + \mu \nabla^2 \vec v + \rho \vec g + \vec S
\end{equation}

Where ##\mu## is the viscosity and ##\vec S## is the surface tension force due to the interface interaction between phases.

Momentum conservation is independent of non-isothermal effects.

Given that the SS model deals with incompressible, isothermal flow the energy equation is not required. However, we want to incorporate non-isothermal effects. To do so, we need to include it

\begin{equation}
\frac{\partial}{\partial t}(E \rho) + \nabla \cdot (\rho E \vec v) = \rho \dot q - \nabla \cdot (p \vec v ) + \rho (\vec f \cdot \vec v) + \text{viscous terms}
\end{equation}

Where ##E## is the total energy ##E := u + \frac{V^2}{2}## and ##u## the internal energy.

Where ##\dot q## is the rate of heat and ##\vec f## refers to external forces.

Given the liquid-gas mixture, the following density and viscosity equations are set to be

$$\rho = \alpha_l \rho_l + (1 - \alpha_l ) \rho_l, \ \ \ \ \mu = \alpha_l \mu_l + (1 - \alpha_l ) \mu_l$$

Where ##\alpha_l, \alpha_v## are the liquid and vapor volume fraction respectively and ##\alpha_l = 1## means there is all liquid and ##\alpha_l = 0## all vapor. It is common practice to use ##\alpha_l + \alpha_v = 1## to eliminate ##\alpha_v##.

A new variable, ##\alpha_l##, has been introduced. Hence, a new equation must be included. The liquid-vapor mass transfer (evaporation and condensation) is governed by the vapor transport equation:

$$\frac{\partial}{\partial t} (\alpha_l \rho) + \nabla \cdot (\alpha_l \rho \vec v) = \dot m^{+} + \dot m^{-}$$

Where ##\dot m^{+}, \dot m^{-}## represent, respectively, the evaporation and condensation mass transfer rates which model the mechanism of cavitation.

Note that for incompressible flow (constant density) the transport equation takes the simple form

$$\frac{\partial}{\partial t} (\alpha_l) + \nabla \cdot (\alpha_l \vec v) = \frac{\dot m^{+} + \dot m^{-}}{\rho}$$

The above are the fundamental equations. Let's now turn to the SS model

Screenshot (45).png

1.png


Where equation 3.5 above is the transport equation.

Questions

  • Does including the energy equation simply incorporate non-isothermal conditions to the LN2 mixture?
  • Regarding incorporating non-isothermal conditions to the SS model: I see that equation 3.7 would take a more complicated final form, given that the density is no longer constant. Is that the only change?

Thank you! :biggrin:
 
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  • #2
Yes, including the energy equation will incorporate non-isothermal conditions to the LN2 mixture. Regarding the SS model, incorporating the energy equation will result in equation 3.7 taking a more complicated form due to the changing density. However, in addition to this change, you will also need to account for the rate of heat and external forces in the momentum equation (equation 3.6).
 

1. What are non-isothermal conditions in a liquid-gas nitrogen homogeneous mixture?

Non-isothermal conditions refer to a situation where the temperature of a system is not constant and changes over time. In a liquid-gas nitrogen homogeneous mixture, this means that the temperature of the mixture is not uniform and can vary throughout the system.

2. How do non-isothermal conditions affect a liquid-gas nitrogen homogeneous mixture?

Non-isothermal conditions can have a significant impact on a liquid-gas nitrogen homogeneous mixture. The temperature variations can cause changes in the density, viscosity, and other physical properties of the mixture. This can also affect the behavior and stability of the mixture.

3. What factors can cause non-isothermal conditions in a liquid-gas nitrogen homogeneous mixture?

Non-isothermal conditions can be caused by various factors, such as external heating or cooling, changes in pressure, or mixing with other substances. These factors can lead to temperature variations within the mixture, resulting in non-isothermal conditions.

4. How can non-isothermal conditions be managed in a liquid-gas nitrogen homogeneous mixture?

Managing non-isothermal conditions in a liquid-gas nitrogen homogeneous mixture requires careful monitoring and control of temperature. This can be achieved through the use of insulation, temperature control systems, and other techniques to maintain a consistent temperature throughout the system.

5. What are the potential applications of understanding non-isothermal conditions in a liquid-gas nitrogen homogeneous mixture?

Understanding non-isothermal conditions in a liquid-gas nitrogen homogeneous mixture is essential for various industries, including cryogenics, food processing, and chemical production. It can also aid in the design and optimization of processes involving these mixtures, leading to more efficient and effective operations.

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