## 3 Calculation Model Notes  
The calculation model is based on the thermodynamic cycle diagram. To achieve, by calculation, results approximately comparable with the directly measured results, polynomials are necessary to be used [10] which are functions of the working temperature to calculate the enthalpy and entropy of the air and gases.  
The polynomial definitions [10] of air, enthalpy and entropy are  
\[i_{a}(T),s_{a}(T)^{p0}=f(T) \tag{5}\]  
\[f(T)=a_{1,a}+a_{2,a}\cdot T+a_{3,a}\cdot T^{2}+a_{4,a}\cdot T^{3}+a_{5,a}\cdot T ^{4}+a_{6,a}\cdot T^{5} \tag{6}\]  
The air enthalpy polynomial coefficients are presented in Table 1 and the air entropy polynomial coefficients are presented in Table 2 [10].  
In the case of gas enthalpy and entropy, the polynomials [10] are defined for an ideal stoichiometric combustion (air excess \(a_{ex}\) is equal to 1).  
\[i_{g,st}(T),s_{g,st}(T)^{p0}=f(T) \tag{7}\]  
\begin{table}
\begin{tabular}{c c c c} \hline \hline \multicolumn{2}{c}{**Domain: T = 200–1000 (K)**} & \multicolumn{2}{c}{**Domain: T = 600–1700 (K)**} \\ \hline
**Symbol** & **Value** & **Symbol** & **Value** \\ \hline \(a_{1,a}\) & \(-4.3213931\) & \(a_{1,a}\) & \(14.058468\) \\ \(a_{2,a}\) & \(1.0473499\) & \(a_{2,a}\) & \(0.9513943\) \\ \(a_{3,a}\) & \(-1.91145951\times 10^{-4}\) & \(a_{3,a}\) & \(-6.735406\times 10^{-6}\) \\ \(a_{4,a}\) & \(3.160497\times 10^{-7}\) & \(a_{4,a}\) & \(1.6270873\times 10^{-7}\) \\ \(a_{5,a}\) & \(-1.3686977\times 10^{-10}\) & \(a_{5,a}\) & \(-9.2577348\times 10^{-11}\) \\ \(a_{6,a}\) & \(1.47654\times 10^{-14}\) & \(a_{6,a}\) & \(1.70236\times 10^{-14}\) \\ \hline \hline \end{tabular}
\end{table}
Table 1: Air enthalpy polynomial coefficients.  
Figure 4: Jet nozzle instrumentation method scheme.  
\begin{table}
\begin{tabular}{c c c c} \hline \hline \multicolumn{2}{c}{**Domain: T = 200–1000 (K)**} & \multicolumn{2}{c}{**Domain: T = 600–1700 (K)**} \\ \hline
**Symbol** & **Value** & **Symbol** & **Value** \\ \hline \(a_{1,a}\) & \(4.941494\) & \(a_{1,a}\) & \(5.6742477\) \\ \(a_{2,a}\) & \(9.64001265\times 10^{-3}\) & \(a_{2,a}\) & \(4.6377175\times 10^{-3}\) \\ \(a_{3,a}\) & \(-1.7785936\times 10^{-5}\) & \(a_{3,a}\) & \(-4.174386\times 10^{-6}\) \\ \(a_{4,a}\) & \(2.1013966\times 10^{-8}\) & \(a_{4,a}\) & \(2.5990432\times 10^{-9}\) \\ \(a_{5,a}\) & \(-1.3263673\times 10^{-11}\) & \(a_{5,a}\) & \(-8.9821609\times 10^{-13}\) \\ \(a_{6,a}\) & \(3.421926\times 10^{-15}\) & \(a_{6,a}\) & \(1.293356\times 10^{-16}\) \\ \hline \hline \end{tabular}
\end{table}
Table 2: Air entropy polynomial coefficients.  
\[f(T)=a_{1,g}+a_{2,g}\cdot T+a_{3,g}\cdot T^{2}+a_{4,g}\cdot T^{3}+a_{5,g}\cdot T^{4 }+a_{6,g}\cdot T^{5} \tag{8}\]  
The gas enthalpy polynomial coefficients are presented in the left part of Table 3 and the gas entropy polynomial coefficients are presented in the right part of Table 3.  
In the case of a real combustion (with excess air), the enthalpy and entropy polynomial equations [10] become  
\[i_{g}(T)=x_{st}\cdot i_{g,st}(T)+x_{a,ex}\cdot i_{a}(T) \tag{9}\]  
\[s_{g}(T)^{p0}=x_{st}\cdot s_{g,st}(T)^{p0}+x_{a,ex}\cdot s_{a}(T)^{p0} \tag{10}\]  
In this case, the mass participations are  
\[x_{st}=(1+\min_{L})/(1+\alpha_{ex}\cdot\min_{L}) \tag{11}\]  
\[x_{a,ex}=((\alpha_{ex}-1)\cdot\min_{L})/(1+\alpha_{ex}\cdot\min_{L}) \tag{12}\]  
The mass participations are determined by applying the energy conservation equation to the combustor chamber [10], where \(\min_{L}\) is the stoichiometric air/fuel ratio.  
\[1\cdot kg(fuel)+\alpha_{ex}\cdot\min_{L}\cdot kg(air)=(1+\min_{L})\cdot kg(ga )_{\alpha_{ex}=1}+(\alpha_{ex}-1)\cdot\min_{L}kg\cdot(air) \tag{13}\]