# CFD Research for Air Bearing with Gradient-Depth Recesses
## 2 Principle Model of Guideway without Air Tube
### Structure and Model  
The construction of compound restrictors is wholly concentrated on the guide rail, as shown in Figure 3. Although the guide rail has the length of \(L\), the actual working proportion is determined by the length of the guide sleeve \(L_{q}\). In order to simplify the computational domain of \(S_{\text{a}}\) and \(S_{\text{b}}\), four assumptions can be formatted as follows:  
**Assumption 1**. : _No lateral flow along \(Y\)._  
**Assumption 2**. : _One-dimensional laminar flow along \(X\) in \(S_{b}\)._  
**Assumption 3**. : _No pressure gradient and consistent viscosity \(\mu\) along \(Z\)._  
**Assumption 4**. : _Set status of adiabatic flow \(\kappa\) = 1.4 for compound restrictors and isothermal flow at other positions._  
Figure 2: Working principle of guideway. (**a**) Air tube on guide sleeve. (**b**) Air tube on guide rail (The direction of the arrows are the direction of the airflow). Based on these, the kinematic Equation (1) [1] can be simplified to Equation (2), in which the velocity of air flow along X, Y, and Z is, respectively, \(u\), \(v\), and \(w\). \[\rho\bigg{(}\frac{\partial u}{\partial t}+u\frac{\partial u}{ \partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}\bigg{)} =-\frac{\partial p}{\partial x}+\frac{\partial}{\partial x}\bigg{\{} \bigg{[}\bigg{[}2\frac{\partial u}{\partial x}-\frac{2}{3}\bigg{(}\frac{ \partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{ \partial z}\bigg{)}\bigg{]}\bigg{\}}+\frac{\partial}{\partial y}\bigg{[} \bigg{[}\mu\bigg{(}\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x }\bigg{)}\bigg{]}+\frac{\partial}{\partial z}\bigg{[}\mu\bigg{(}\frac{ \partial u}{\partial z}+\frac{\partial w}{\partial x}\bigg{)}\bigg{]} \tag{1}\] \[\frac{\partial p}{\partial x}=\frac{\partial}{\partial z}\bigg{(}u \frac{\partial u}{\partial z}\bigg{)} \tag{2}\]  
Create the indefinite integral of \(u\). Then, bring in the boundary conditions of \(z\), and integrate \(\rho u\) on the outlet surface along X to obtain the outlet mass flow rate of \(S_{\text{b}}\). \[M_{out}=\int\rho ud2S_{out}=2L_{a}\int_{0}^{h}\rho udz=L_{a}\cdot\frac{1}{2\mu }\frac{dp}{dx}\rho\int_{0}^{h}z(z-h)dz=-\rho\frac{h^{3}L_{a}}{6\mu}\frac{dp}{ dx}=-\frac{h^{3}L_{\rho_{a}}}{6\mu P_{a}}\frac{dp}{dx}p \tag{3}\]  
Separate and integrate \(p\) and \(x\) of Formula (3). Substitute the import and export pressure status again when \(x=0\), \(p=P_{d}\) and when \(x=(L_{b}-L_{b1})/2\), \(p=P_{a}\). \[P_{d}^{2}-p^{2}=-\frac{12\mu P_{a}M_{out}}{h^{3}L_{a}\rho_{a}}(0-x)=\frac{12 \mu P_{a}M_{out}}{h^{3}L_{a}\rho_{a}}x \tag{4}\]  
\[P_{d}^{2}-P_{a}^{2}=-\frac{12\mu P_{a}M_{out}}{h^{3}L_{a}\rho_{a}}(0-\frac{L_{ b}-L_{b1}}{2})=\frac{12\mu P_{a}M_{out}}{h^{3}L_{a}\rho_{a}}\frac{L_{b}-L_{b1}}{2} \tag{5}\]  
Divide Formula (4) by (5) and transform it; the mathematical expression of pressure \(p\) in \(S_{\text{b}}\) can be given. \[p=\bigg{[}P_{d}^{2}-\frac{2x}{L_{b}-L_{b0}}\bigg{(}P_{d}^{2}-P_{a}^{2}\bigg{)} \bigg{]}^{\frac{1}{2}}=P_{d}\bigg{[}1-\frac{2x}{L_{b}-L_{b0}}\bigg{(}1-\frac{P_ {a}^{2}}{P_{d}^{2}}\bigg{)}\bigg{]}^{\frac{1}{2}} \tag{6}\]  
The bearing capacity of the guideway \(W\) is the integration of \(p\) in \(S_{\text{a}}\) and \(S_{\text{b}}\), and then minus the atmospheric pressure on the same area. In Formulas (7) and (8), \(\beta=P_{d}/P_{a}\) and \(P_{d}\) will be determined by \(P_{0}\), \(h\), \(h_{ri}\), and \(d_{ri}\). \[ \begin{array}{rcl}W&=& 2\Bigg{(}L_{a0}\int_{0}^{\dfrac{L_{b}-L_{b1}}{2} }pdx+2\frac{L_{b}-L_{b1}}{2}\frac{L_{a}-L_{a0}}{2}P_{a}\Bigg{)}+L_{a0}L_{b1}P_ {d}-L_{a}L_{b}\cdot P_{a}\\ &=& 2P_{a}(L_{a}-L_{a0})(L_{b}-L_{b1})\bigg{(}\frac{2}{3}\frac{\beta^{ 3}-1}{\beta^{2}-1}-1\bigg{)}+L_{a}L_{b0}P_{a}(\beta-1)\end{array}