## 2 Fundamental Theories  
[PERSON] [8] emphasizes that the compressible MRF solver is designed to conduct calculations using rothalpy instead of enthalpy, so the energy equation is also solved through the iEqn.H file rather than hEqn.H. In addition, they also emphasize that the rothalpy jump is applied at the interface between the stator and the rotor. Therefore, it is necessary to provide a brief summary of the rothalpy in order to understand this solver.  
The definition of the torque is \(\mathbf{\tau}=\mathbf{r}\times\mathbf{F}\), which, assuming a steady state, can be rearranged by Reynolds transform theorem as follows:  
\[\mathbf{\tau}=\left(\mathbf{r}_{2}\times\mathbf{U}_{\theta 2}\dot{m}_{2}-\mathbf{r}_{1}\times \mathbf{U}_{\theta 1}\dot{m}_{1}\right) \tag{1}\]  
meaning the time rate of change in the angular momentum of the working fluid through 1 to 2, around some axis A-A, as depicted in Figure 3, where \(U\) is the fluid's absolute velocity, \(U_{\theta}\) is the fluid's tangential velocity, \(r\) is the radius, and \(\dot{m}\) is the mass flow rate. Therefore, the power \(P\) is derived as follows [13]:  
\[P=\tau\Omega=\dot{m}(U_{R\theta 2}U_{\theta 2}-U_{R\theta 1}U_{\theta 1}) \tag{2}\]  
\begin{table}
\begin{tabular}{c c c} \hline \hline  & **KARI Turbine** & **E\({}^{3}\) LPT** \\ \hline Stages & \(1\) & \(5\) \\ Number of stator vanes & \(56\) & \(72/102/96/114/120\) \\ Number of rotor blades & \(104\) & \(120/122/122/156/110\) \\ RPM & \(17/000\) & \(3208.7\) \\ Inlet total T & \(1673\) K & \(416.7\) K \\ Inlet total p & \(3.0685\) MPa & \(310\) kPa \\ Outlet static p & \(1.15\) MPa & \(65\) kPa \\ \hline \hline \end{tabular}
\end{table}
Table 1: Information on two turbines.  
Figure 2: Two different turbine geometries: (**a**) KARI turbine and (**b**) E\({}^{3}\) LPT.  
where the rotational blade speed is \(U_{Rot}=\Omega r\), and \(\Omega\) is the angular velocity [rad/s]. The specific work is equal to the power divided by the mass flow rate or the change in total enthalpy between the outlet and the inlet, expressed as follows:  
\[\Delta w=\frac{P}{m}=U_{Rot2}U_{\theta 2}-U_{Rot1}U_{\theta 1}=h_{02}-h_{01} \tag{3}\]  
The work conducted on the fluid is positive (\(\Delta w>0\)) when it involves a compressor or pump rotor, and the work carried out on the rotor by the fluid is negative (\(\Delta w<0\)) when it involves a turbine. After rearranging Equation (3), we obtain the following:  
\[h_{01}-U_{Rot1}U_{\theta 1}=h_{02}-U_{Rot2}U_{\theta 2}=Const. \tag{4}\]  
In this way, each side has the same value, which means that it has a constant value along the flow in the rotating body. Therefore,  
\[\therefore i_{0}=h_{0}-U_{Rot}U_{\theta} \tag{5}\]  
or  
\[i_{0}=h+\frac{1}{2}U^{2}-U_{Rot}U_{\theta} \tag{6}\]  
where \(h_{0}=h+\frac{1}{2}U^{2}\). This \(i_{0}\) is defined as the rothalpy, and, in this paper, the symbol \(i_{0}\) is used to emphasize the relationship with the stagnation property. If there is no rotation, the value of the \(U_{Rot}\) is 0, so rothalpy = total enthalpy. In other words, the rothalpy can be understood as the total enthalpy with an additional term by rotation, and this additional term \(U_{Rot}U_{\theta}\) is called a rothalpy jump, because it creates a discontinuous value between the stator and the rotor.  
The rothalpy can be redefined using the static enthalpy and the velocity triangle, as follows:  
\[i_{0}=h+\frac{1}{2}\Big{(}U_{Rel}^{2}-U_{Rot}^{2}\Big{)} \tag{7}\]