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Aerodynamic Networks

Aerodynamic networks are made of a concatenation of fluid elements filled with a compressible medium which can be considered as an ideal gas. An ideal gas satisfies

$\displaystyle p=\rho R \theta,$ (74)

where p is the pressure, $ \rho$ is the density, $ R$ is the specific gas constant and $ \theta$ is the absolute temperature. A fluid element (see section 4.1.17) consists of three nodes: in the end nodes the temperature and pressure are the unknowns, in the middle node the mass flow is unknown. The end nodes play the role of crossing points in the network, whereas the middle nodes represent the flow within one element. To determine these unknowns, three types of equations are available: conservation of mass and conservation of energy in the end nodes and conservation of momentum in the middle node. Right now, only stationary flow is considered.

The stationary form of the conservation of mass for compressible fluids is expressed by:

$\displaystyle \mathbf{\nabla} \cdot ( \rho \mathbf{v}) = 0$ (75)

where $ \mathbf{v}$ the velocity vector. Integration over all elements connected to end node i yields:

$\displaystyle \sum_{j \in \text{in}} \dot{m}_{ij} = \sum_{j \in \text{out}} \dot{m}_{ij},$ (76)

where $ \dot{m}_{ij}$ is the mass flow from node i to node j or vice versa. In the above equation $ \dot{m}_{ij}$ is always positive.

The conservation of momentum or element equations are specific for each type of fluid section attributed to the element and are discussed in Section 4.2 on fluid sections. For an element with end nodes i,j it is generally of the form $ f({p_{tot}}_i, {\theta_{tot}}_i, \dot{m}_{ij},{p_{tot}}_j)=0$ (for positive $ \dot{m}_{ij}$, where p is the total pressure and $ \theta_{tot}$ is the total temperature), although more complex relationships exist. Notice in particular that the temperature pops up in this equation (this is not the case for hydraulic networks).

The conservation of energy in stationary form requires ([16]):

$\displaystyle \mathbf{\nabla} \cdot (\rho h_{tot} \mathbf{v}) = - \mathbf{\nabla} \cdot \mathbf{q} + \rho h^{\theta} + \rho \mathbf{f} \cdot \mathbf{v},$ (77)

where $ \mathbf{q}$ is the external heat flux, $ h^{\theta}$ is the body flux per unit of mass and $ \mathbf{f}$ is the body force per unit of mass. $ h_{tot}$ is the total enthalphy satisfying:

$\displaystyle h_{tot} = c_p \theta + \frac{\mathbf{v} \cdot \mathbf{v}}{2},$ (78)

where $ c_p$ is the specific heat at constant pressure and $ \theta$ is the absolute temperature (in Kelvin). This latter formula only applies if $ c_p$ is considered to be independent of the temperature. This is largely true for a lot of industrial applications. In this connection the reader be reminded of the definition of total temperature and total pressure (also called stagnation temperature and stagnation pressure, respectively):

$\displaystyle \theta_{tot} = \theta + \frac{\mathbf{v} \cdot \mathbf{v}}{2 c_p},$ (79)

and

$\displaystyle \frac{p_{tot}}{p} = \left( \frac{\theta_{tot}}{\theta} \right) ^ {\frac{\kappa}{\kappa-1}},$ (80)

where $ \kappa = c_p/c_v$. $ \theta$ and $ p$ are also called the static temperature and static pressure, respectively.

If the end nodes of the elements are considered to be large chambers, the velocity $ \mathbf{v}$ is zero. In that case, the total quantities reduce to the static ones, and integration of the energy equation over all elements belonging to end node $ i$ yields [13]:

$\displaystyle c_p (\theta_i) \sum_{j \in \text{in}} \theta_j \dot{m}_{ij} - c_p...
...{ij} + \overline{h}(\theta_i, \theta) (\theta - \theta_i) + m_i h_i^{\theta}=0,$ (81)

where $ \overline{h}(\theta_i,\theta)$ is the convection coefficient with the walls. Notice that, although this is not really correct, a slight temperature dependence of $ c_p$ is provided for. If one assumes that all flow entering a node must also leave it and taking for both the $ c_p$ value corresponding to the mean temperature value of the entering flow, one arrives at:

$\displaystyle \sum_{j \in \text{in}} c_p (\theta_m) (\theta_j-\theta_i) \dot{m}_{ij} + \overline{h}(\theta_i, \theta) (\theta - \theta_i) + m_i h_i^{\theta}=0.$ (82)

where $ \theta_m=(\theta_i+\theta_j)/2$.

The calculation of aerodynamic networks is triggered by the *HEAT TRANSFER keyword card. Indeed, such a network frequently produces convective boundary conditions for solid mechanics heat transfer calculations. However, network calculations can also be performed on their own.

A particularly delicate issue in networks is the number of boundary conditions which is necessary to get a unique solution. To avoid ending up with more or less equations than unknowns, the following rules should be obeyed:


next up previous contents
Next: Hydraulic Networks Up: Types of analysis Previous: Diffusion mass transfer in   Contents
guido dhondt 2007-08-09