## 2 Model Derivation
### Transfer Matrix of the Pipe  
Oscillation of the fluid in the pipe is sufficiently described by Equation (29) only for the case when the ends (flanges) of the pipe do not move in the axial direction. The experiment, which is described in the next section, does not fulfil this condition because the excitation is performed only via axial displacement of the flange. Thus, it is also necessary to derive the transfer matrix of the pipe. Generally, the procedure is the same as the abovementioned procedure for the fluid. The transfer matrix of the pipe includes the material properties and mass at both ends of the pipe. The mass represents flanges, valves and sensors because they influence the dynamic behavior of the pipe the most. The scheme is shown in Figure 3.  
The transfer equation between places 0 and 3, when there is no liquid in the pipe, is:  
\[\left(\begin{array}{c}\bar{u}_{0}\\ \bar{F}_{0}\end{array}\right)=\mathbf{P_{m}}\cdot\mathbf{P_{p}}\cdot\mathbf{P_ {m}}\cdot\left(\begin{array}{c}\bar{u}_{3}\\ \bar{F}_{3}\end{array}\right), \tag{31}\]  
where the index refers to the corresponding place in Figure 3. Matrices \(\mathbf{P_{m}}\) and \(\mathbf{P_{p}}\) are transfer matrices of the mass at the end of pipe and the pipe, respectively. Their derivation follows. The transfer matrix \(\mathbf{P_{m}}\) comes from the equation of motion:  
\[F_{0}-F_{1}=m\cdot\frac{\partial^{2}u}{\partial x^{2}}. \tag{32}\]  
When the flange is ideally stiff, it is possible to consider the same displacement of both flange faces \(u_{0}=u_{1}\). When the Laplace transform is applied on Equation (32), the transfer matrix of the flange can be written in the form:  
\[\mathbf{P_{m}}=\left(\begin{array}{cc}1&0\\ m\cdot s^{2}&1\end{array}\right). \tag{33}\]  
The same procedure can be applied on the flange indexed 2 and 3.  
The transfer matrix of the pipe \(\mathbf{P_{P}}\) comes from a longitudinal oscillation:  
\[\frac{\partial\sigma}{\partial x}=\rho\cdot\frac{\partial^{2}u}{\partial t^{ 2}}. \tag{34}\]  
Equations (4) to (6) describing the Kelvin-Voigt model of the body can be implied into the Equation (34). Then it is possible after the Laplace transform to write the Equation (35) as:  
Figure 3: Scheme of the system.  
\[\bar{u} = C_{1}\cdot\cosh\left(\frac{s\cdot x}{a_{0}}\right)+C_{2}\cdot \sinh\left(\frac{s\cdot x}{a_{0}}\right), \tag{35}\] \[C_{1} = \bar{u}\left(0\right),\] (36) \[C_{2} = \frac{\bar{\sigma}\left(0\right)\cdot a_{0}}{\left(E+b\cdot s \right)\cdot s}, \tag{37}\]  
where Equations (36) and (37) can be obtained from the stress \(\bar{\sigma}\left(0\right)\) and the displacement \(\bar{u}\left(0\right)\) taken for the pipe beginning. Knowing that \(\varepsilon=\frac{\bar{u}u}{\bar{\sigma}x}\) and \(\sigma=\frac{F}{S_{p}}\). Speed of sound \(a_{0}\) in the pipe is a complex number:  
\[a_{0}=\sqrt{\frac{E+b\cdot s}{\rho}}. \tag{38}\]  
The transfer matrix of the pipe comes from the combination of Equations (35) and (7):  
\[\mathbf{P_{p}}=\left(\begin{array}{cc}\cosh\left(\frac{s}{a_{0}}\cdot x \right)&\frac{a_{0}}{E\cdot s}\cdot\sinh\left(\frac{s}{a_{0}}\cdot x\right) \\ \frac{k\cdot s}{a_{0}}\cdot\sinh\left(\frac{s}{a_{0}}\cdot x\right)&\cosh \left(\frac{s}{a_{0}}\cdot x\right)\end{array}\right), \tag{39}\]  
where:  
\[k=S_{p}\cdot\left(E+b\cdot s\right). \tag{40}\]  
In this case, the variable \(x\) means the length of the pipe. Boundary conditions \(\bar{F}_{0}=0\) and \(\bar{F}_{3}=0\) can be implied to Equation (31), then it is possible to write:  
\[\left(\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}\right)=\left(\begin{array}{cccc}0&1&0&0\\ 0&0&0&1\\ \mathbf{P_{m}}\cdot\mathbf{P_{p}}\cdot\mathbf{P_{m}}&-1&0\\ 0&0&-1\end{array}\right)\cdot\left(\begin{array}{c}\bar{u}_{0}\\ \bar{F}_{0}\\ \bar{u}_{3}\\ \bar{F}_{3}\end{array}\right). \tag{41}\]  
When the determinant of the matrix in Equation (41) equals zero, it is possible to find constants \(E\) and \(b\) for the measured eigenvalues \(s\).