Hydraulic Networks

Hydraulic networks are made of a concatenation of network elements (see section 6.2.33) filled with an incompressible medium. A network element consists of three nodes: in the corner nodes the temperature and pressure are the unknowns, in the midside node the mass flow is unknown. The corner nodes play the role of crossing points in the network, whereas the midside 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 corner nodes and conservation of momentum in the midside node. Right now, only stationary flow is considered.

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

$$\mathbf{\nabla} \cdot \mathbf{v} = 0$$ (608)

where $\rho$ is the density and $\mathbf{v}$ the velocity vector. Integration over all elements connected to an corner node yields:

$$\sum_{j \in \text{in}} \dot{m}_{ij} = \sum_{j \in \text{out}} \dot{m}_{ij},$$ (609)

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 reduces to the Bernoulli equation. It is obtained by projecting the general momentum equation (substitute Equation (1.535) into Equation (1.334) in [24]) on a flow line. Since a flow line is everywhere locally parallel to the velocity vector, this amounts to a multiplication by:

$$\frac{dx_k}{ds}= \frac{v_k}{\Vert \boldsymbol{v} \Vert }$$ (610)

leading to (where for the gravity $f_k=g z,_k$ with $z$ the coordinate perpendicular to the earch surface was inserted):

$$\rho \left ( \frac{v_k}{\Vert \boldsymbol{v} \Vert } \frac{\parti... ... t_{kl,l} \frac{dx_k}{ds} - p,_k \frac{dx_k}{ds} - \rho g z,_k \frac{dx_k}{ds}.$$ (611)

Since

$$\frac{v_k v_{k,l}v_l}{\Vert \boldsymbol{v} \Vert } = v_k v_{k,l} ... ... (v_k v_k),_l \frac{dx_l}{ds} = \frac{d}{ds} \left( \frac{v_k v_k}{2} \right ),$$ (612)

one obtains:

$$\rho \left [ \frac{\partial \Vert \boldsymbol{v} \Vert }{\partial... ...t ) \right ] = t_{kl,l} \frac{dx_k}{ds} - \frac{dp}{ds} - \rho g \frac{dz}{ds}.$$ (613)

Dividing by $\rho g$ and integration from $s_1$ to $s_2$ yields:

$$\frac{1}{g} \int_{s_1}^{s_2} \frac{\partial \Vert \boldsymbol{v} ... ...}{\underset{1}{\Delta}} \frac{p}{\rho g} - \overset{2}{\underset{1}{\Delta}} z.$$ (614)

Applying this equation for steady state flow within an element with corner nodes i and j reads:

$$z_i + \frac{p_i}{\rho g} + \frac{\dot{m}_{ij}^2}{2 \rho^2 A_i^2 g... ... + \frac{p_j}{\rho g} + \frac{\dot{m}_{ij}^2}{2 \rho^2 A_j^2 g} + \Delta F_i^j,$$ (615)

where

$$\Delta F_i^j := \int_{i}^{j} t_{kl,l} dx_k.$$ (616)

Here, $z$ is the height of the node, $p$ the pressure, $\rho$ the density, $g$ the gravity acceleration, $A$ the cross section in the node and $\Delta F_i^j$ is the head loss across the element. The head loss is positive if the flow runs from i to j, else it is negative (or has to be written on the other side of the equation). The head losses for different types of fluid sections are described in Section 6.5.

Notice that the height of the node is important, therefore, for hydraulic networks the gravity vector must be defined for each element using a *DLOAD card.

The conservation of energy in stationary form requires ([24]):

$$c_p \mathbf{\nabla} \cdot (\rho T \mathbf{v}) = - \mathbf{\nabla} \cdot \mathbf{q} + \rho h^{\theta},$$ (617)

where $\mathbf{q}$ is the external heat flux, $h^{\theta}$ is the body flux per unit of mass, $c_p$ is the specific heat at constant pressure (which, for a fluid, is also the specific heat at constant specific volume, i.e. $c_p=c_v$ [32]) and $T$ is the absolute temperature (in Kelvin). Integration of the energy equation over all elements belonging to end node $i$ yields:

$$c_p (T_i) \sum_{j \in \text{in}} T_j \dot{m}_{ij} - c_p (T_i) T_i... ...\text{out}} \dot{m}_{ij} + \overline{h}(T_i, T) (T - T_i) + m_i h_i^{\theta}=0,$$ (618)

where $\overline{h}(T_i,T)$ is the convection coefficient with the walls. 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:

$$\sum_{j \in \text{in}} c_p (T_m) (T_j-T_i) \dot{m}_{ij} + \overline{h}(T_i, T) (T - T_i) + m_i h_i^{\theta}=0.$$ (619)

where $T_m=(T_i+T_j)/2$.

The calculation of hydraulic 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, i.e. it is allowed to do *HEAT TRANSFER calculations without any solid elements.

To determine appropriate boundary conditions for a hydraulic network the same rules apply as for aerodynamic networks.

Output variables are the mass flow (key MF on the *NODE PRINT or *NODE FILE card), the static pressure (key PN -- network pressure -- on the *NODE PRINT card and PS on the *NODE FILE card) and the total temperature (key NT on the *NODE PRINT card and TT on the *NODE FILE card). Notice that the labels for the *NODE PRINT keyword are more generic in nature, for the *NODE FILE keyword they are more specific. These are the primary variables in the network. Internally, in network nodes, components one to two of the structural displacement field are used for the mass flow and the static pressure, respectively. So their output can also be obtained by requesting U on the *NODE PRINT or *NODE FILE card.

Notice that for liquids the total temperature virtually coincides with the static temperature. Indeed, since

$$T_t-T = v^2/(2c_p),$$ (620)

the difference between total and static temperature for a fluid velocity of 5 m/s and $c_p=4218$ J/(kg.K) (water) amounts to 0.0030 K. This is different from the gases since typical gas velocities are much higher (speed of sound is 340 m/s) and $c_p$ for gases is usually lower.