## 2 Materials and Methods
### Thermodynamic Analysis
#### 2.2.2 Energy Analysis  
Similar to the case of the energy analysis, the same assumptions were valid for the exergy analysis of the control volume. The exergy balance can be expressed as Equation (9) as follows:  
\[\sum\dot{E}x_{in}=\sum\dot{E}x_{out}+\dot{E}x_{dest} \tag{9}\]  
\[\dot{E}x_{air}+\dot{E}x_{fuel}=\dot{E}x_{W}+\dot{E}x_{exh}+\dot{E}x_{loss}+ \dot{E}x_{dest} \tag{10}\]  
where \(\sum\dot{E}x_{in}\), \(\sum\dot{E}x_{out}\) and \(\dot{E}x_{dest}\) represents respectively, the total exergy inlet and outlet rates and the exergy destruction (irreversibility) rate of the control volume. \(\dot{E}x_{air}\) is the inlet exergy rate of air, \(\dot{E}x_{fuel}\) is the inlet exergy rate from the fuel, \(\dot{E}x_{W}\) is the exergy rate by work, \(\dot{E}x_{exh}\) is the exergy outlet rate by the exhaust, and \(\dot{E}x_{loss}\) is the exergy loss rate of the control volume. An equivalent assumption was considered for the air inlet rate, and by considering the environmental atmosphere as a reference state, the exergy associated with naturally aspirated air into the engine control volume is zero [47]. The inlet exergy rate from the fuel (\(\dot{E}x_{fuel}\)) to the control volume can be determined as follows in Equation (11), where \(i\) represents the different fuels that composed the blend. \[\dot{E}x_{fuel}\ =\ \sum_{i=1}^{3}\dot{m}_{fuel_{i}}\cdot ex_{fuel_{i}}\ =\ \dot{m}_{fuel_{i}}\cdot LHV_{fuel_{i}}\cdot\varphi_{fuel_{i}} \tag{11}\]  
where \(ex_{fuel}\) is the fuel-specific exergy of the fuel, which can be obtained by multiplying the fuel lower heating value by the chemical exergy factor (\(\varphi\)) of each fuel, which can be obtained through Equation (12) as in [41; 42] for liquid fuels. The accuracy of this expression is estimated to be \(\pm 0.38\%\). \[\varphi=\ 1.0401+0.1728\frac{h}{c}+0.0432\frac{o}{c}+0.2169\frac{S}{c}\left(1\ -\ 2.0628\frac{h}{c}\right) \tag{12}\]  
where \(h\), \(c\), \(o\), and \(s\) are respectively the mass fractions of hydrogen, oxygen, carbon, and sulfur of the fuel [41; 42]. The exergy work rate (\(\dot{E}x_{W}\)) is equal to the energy work rate of the control volume, as shown in Equation (13). \[\dot{E}x_{W}=\dot{E}n_{W} \tag{13}\]  
The exergy rate of exhaust gases (\(\dot{E}x_{exh}\)) is composed of two components that are the physical (thermomechanical) and chemical exergies, which is expressed in Equation (14). \[\dot{E}x_{exh}\ =\ \sum_{i}\dot{n}_{i}(ex_{ph,i}+ex_{ch,i}) \tag{14}\]  
where \(\dot{n}_{i}\) is the molar flow rate of each exhaust gas species, \(ex_{ph,i}\) and \(ex_{ch,i}\) are the specific physical and chemical exergies of each exhaust gas species, respectively. For a mass flow that goes through the volume control, the specific physical exergy rate (\(ex_{ph,i}\)) of the exhaust gas species is obtained by Equation (15) below. \[ex_{ph,i}=(h_{i}-h_{0})-T_{0}\cdot(s_{i}-s_{0}) \tag{15}\]  
where \(s_{i}\) is the entropy of each gaseous species in the engine exhaust. For a gas mixture, the chemical exergy rate (\(ex_{ch,i}\)) of the exhaust gas species can be calculated using Equation (16) as follow. \[ex_{ch,i}=\sum_{i}x_{i}\cdot\varepsilon_{ch,i}+R\cdot T_{0}\cdot\sum_{i}(x_{i} \cdot ln\gamma_{i}\cdot x_{i}) \tag{16}\]  
where \(x_{i}\) is the molar fraction of each species of the exhaust gas, \(\varepsilon_{ch,i}\) is the standard chemical exergy [41; 42], R is the universal gas constant (8.314 kJ/kmol\(\cdot\)K) and \(\gamma_{i}\) is the activity coefficient (\(\gamma_{i}=1\) for ideal gases). The exergy loss rate (\(\dot{E}x_{loss}\)) shown in Equation (17) is considered to be the amount of exergy loss from the control volume, and the lost-exergy rate was assumed to be all heat losses occurring from the measured coolant temperature (\(T_{cool}=353.15\) K) to the environment at reference state temperature, as in [29; 48]. \[\dot{E}x_{loss}=\sum\biggl{(}1-\frac{T_{o}}{T_{cool}}\biggr{)}\dot{E}n_{loss} \tag{17}\]  
Hence, the exergy destruction (\(\dot{E}x_{dest}\)) is obtained from the exergy balance by Equation (18).