Thermodynamics Report


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Thermodynamics Coursework CHEMICAL ENGINEERING THERMODYNAMICS (3CET) MOHAMMED AL-SHEREEM (BENG CHEMICAL ENGINEERING FT) ID: 1599611 27/01/2019 Question 1 Plot the p-v-T diagram for oxygen, using an appropriate equation of state – identify critical regions and places on the diagram. Answer: ➢ Choosing and modelling an equation of state: There are various equations of state that could model the behaviour of oxygen gas. With the appropriate comparisons between the ideal gas law, Van der waals, Redlich-Kwong and the PengRobinson equations, it was deduced that the Peng-Robinson is the most suitable as it: ▪ ▪ ▪ ▪ ▪ Is well established within industries (Ashour, et al., 2011) Is suitable for representing real gases such as the case for oxygen (Sandler, 2017) Considers various essential physical properties such as vapour-liquid equilibria and behaviours (Tsai & Chen, 1998). 𝑇 𝑃 𝑉 Includes reduced variables (𝑇𝑟 , 𝑃𝑟 , 𝑉𝑟 ) derived from dimensionless quantities (𝑇 , 𝑃 , 𝑉 ). 𝑐 𝑐 𝑐 Produces accurate results that are close to the real experimental behaviour (Dahm & Donald, 2014). The Peng-Robinson equation is represented in (𝐸𝑞 1) below (Goodwin, et al., 2010): 𝑅𝑇 𝑎𝜑 𝑃= − (𝐸𝑞 1) 𝑉𝑚 − 𝑏 𝑉𝑚 (𝑉𝑚 + 𝑏) + 𝑏(𝑉𝑚 − 𝑏) Where 𝑃 is pressure (𝑏𝑎𝑟), 𝑅 is the ideal gas constant (𝐽/𝑚𝑜𝑙 ∙ 𝐾), 𝑇 is the temperature (𝐾), 𝑉𝑚 is the molar volume (𝑚3 /𝑚𝑜𝑙) and 𝑎, 𝑏 and 𝜑 are shown in (𝐸𝑞 1.1, 1.2 𝑎𝑛𝑑 1.3 ) below. 𝑎 = 0.45724 𝑅 2 𝑇𝑐 2 (𝐸𝑞 1.1) 𝑃𝑐 𝑏 = 0.07780 𝑅𝑇𝑐 (𝐸𝑞 1.2) 𝑃𝑐 𝜑 = [1 + (1 − 𝑇𝑟 0.5 )(0.37464 + 1.54226𝜔 − 0.26992𝜔2 )]2 (𝐸𝑞 1.3) In which 𝑇𝑐 and 𝑃𝑐 are the critical temperature and pressure respectively, 𝑇𝑟 is the reduced temperature and 𝜔 is the acentric factor. ➢ Plotting a pressure-volume-temperature (P-V-T) diagram (𝐸𝑞 1) can be used to plot the (P-V-T) diagram for oxygen. Firstly, a set of thermodynamic parameters must be defined for oxygen as shown in Table 1. Table 1: Thermodynamic parameters for oxygen Thermodynamic parameter Critical Temperature Critical Pressure Ideal gas constant Acentric factor Corresponding value 154.55 (𝐾) 50.8 (𝑏𝑎𝑟) 83.14 (𝑐𝑚3 ∙ 𝑏𝑎𝑟/𝑚𝑜𝑙 ∙ 𝐾) 0.0226 Accordingly, by using MATLAB and (𝐸𝑞 1, 1.1, 1.2 𝑎𝑛𝑑 1.3 ), a set of isotherms where plotted at temperatures below, at and above the critical temperature 𝑇𝑐 as shown in Figure 1. The MATLAB script can be found in Appendix A. The critical (inflection) point where: 𝑇𝑐 = 154.55 𝐾 𝑃𝑐 = 50.8 𝑏𝑎𝑟 𝑉𝑚 𝑐 = 74.7 𝑐𝑚3 /𝑚𝑜𝑙 Figure 1: The oxygen P-V-T diagram as drawn in MATLAB where the critical point is illustrated As can been seen in Figure 1, the isotherms with 𝑇 > 𝑇𝑐 show an inverse relationship between pressure and volume. Alternatively, for the isotherms where 𝑇 < 𝑇𝑐 , the relationship varies along 𝑑𝑃 the different molar volumes. In both cases 𝑑𝑉 ≠ 0, however at the critical point, it is noticeable that 𝑑𝑃 there is an inflection where 𝑑𝑉 = 0. ➢ Adding the oxygen saturation curve In order to identify the different thermodynamic regions, the oxygen saturation curve has to be added to the P-V-T diagram. By using the data for the range of pressures and corresponding molar volumes as shown in Appendix A (NIST, 2018), the saturation curve can be added to the original plot as shown in Figure 2 where the various regions are clearly shown. The compressed liquid region The critical point The superheated vapour region The saturated vapour line The saturated liquid line Figure 2: The different thermodynamic regions for the oxygen saturation curve in the P-V-T diagram ➢ Adding the horizontal equilibrium lines According to Figure 1, the isotherms in which 𝑇 > 𝑇𝑐 represent one root where those in which 𝑇 < 𝑇𝑐 represent real roots. Consequently, Maxwell constructs can be utilized for the isotherms below the critical point in which horizontal equilibrium lines can be obtained. Another method of constructing the horizontal lines is through the calculation of 𝑃𝑠𝑎𝑡 (saturation pressure) for the isotherms with 𝑇 < 𝑇𝑐 . 𝑃𝑠𝑎𝑡 can be obtained from (𝐸𝑞 2) below (Green, 2008). 𝑃𝑆𝑎𝑡 = 𝑒𝑥𝑝 𝐶1+( 𝐶2 )+𝐶3 ln(𝑇)+𝐶4𝑇 𝐶5 𝑇 (𝐸𝑞 1.4) In which the coefficients for oxygen are (Green, 2008): 𝐶1 = 51.245 𝐶2 = −1200.2 𝐶3 = −6.4361 𝐶4 = 2.8405 ∗ 10−2 𝐶5 = 1 Thus, the saturation pressures at 𝑇 = 130 𝐾 and 𝑇 = 145 𝐾 are 𝑃 = 17.54 𝑏𝑎𝑟 and 𝑃 = 34.56 𝑏𝑎𝑟 respectively. Accordingly, these values were added to the original plot, thus obtaining horizontal equilibrium lines as shown in Figure 3. It is also noticeable that the horizontal lines show no change in pressure during the phase changes. Saturated liquid-vapour region at 130 𝐾 (34.56 𝑏𝑎𝑟) Quality region Saturated liquid-vapour region at 130 𝐾 (17.54 𝑏𝑎𝑟) Figure 3: The P-V-T diagram with the horizontal equilibrium lines added at the saturation pressures Question 2 Show the range of temperatures that can be achieved by burning acetylene under various conditions. Adiabatic flame temperature calculation Answer: In the case where a fuel combusts without any heat loss (adiabatic), the final mixture of gases reaches a maximum temperature called the adiabatic flame temperature (Strahle, 1993). The flame temperature depends on various factors and conditions that will be assessed in this section. ➢ Adiabatic combustion at constant pressure Under constant pressure, some of the energy is lost to expansion work, however no heat is lost to the surroundings (Strahle, 1993). At constant pressure, the first law of thermodynamics is as follows ∆𝑈 = 𝑄𝐶𝑜𝑛𝑠𝑡𝑃 − 𝑊𝐶𝑜𝑛𝑠𝑡𝑃 = 𝑄𝐶𝑜𝑛𝑠𝑡𝑃 − 𝑃∆𝑉 (𝐸𝑞 2) Where 𝑈 is the internal energy (𝐽), 𝑄𝐶𝑜𝑛𝑠𝑡𝑃 is the heat at constant pressure (𝐽) and 𝑊𝐶𝑜𝑛𝑠𝑡𝑃 is the work at constant pressure (𝐽). Further, enthalpy, by definition is 𝐻 = 𝑈 + 𝑃𝑉 (𝐸𝑞 2.1) Where 𝐻 is the enthalpy (𝐽), thus, when there is a change in enthalpy at constant pressure, (𝐸𝑞 2.1) becomes ∆𝐻 = ∆𝑈 + 𝑃∆𝑉 + 𝑉∆𝑃 = ∆𝑈 + 𝑃∆𝑉 (𝐸𝑞 2.1.1) For an adiabatic process 𝑄𝐶𝑜𝑛𝑠𝑡𝑃 = 0 𝑎𝑛𝑑 ∆𝐻 = 0 At equilibrium, the energy balance becomes ∑ 𝐻𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 (𝑇𝑖 , 𝑃) = ∑ 𝐻𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠,𝐼𝑛𝑒𝑟𝑡𝑠 (𝑇, 𝑃) (𝐸𝑞2.2) 𝑇𝑓 ∑ 𝐻𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 (𝑇𝑖 , 𝑃) = ∑ 𝑛𝑖 ∫ 𝐶𝑃𝑖 𝑑𝑇 + 𝑄𝑉𝐿 (𝐸𝑞2.2.1) 𝑇𝑖 Where 𝑛𝑖 is the number of moles of the 𝑖 𝑡ℎ species, 𝑇𝑖 is the initial temperature (𝐾), 𝑇𝑓 is the final temperature (𝐾), 𝐶𝑃𝑖 is the specific heat capacity of the 𝑖 𝑡ℎ species (𝐽/𝑚𝑜𝑙 ∙ 𝐾) and 𝑄𝑉𝐿 is the total latent heat required in the case of phase changes. ➢ Adiabatic combustion at constant volume At constant volume, there is no shaft work (no expansion), thus the adiabatic flame temperature is higher than that at constant pressure. From the first law of thermodynamics (Strahle, 1993). 𝑄𝐶𝑜𝑛𝑠𝑡𝑉 = ∆𝑈, 𝑎𝑛𝑑 𝑄𝐶𝑜𝑛𝑠𝑡𝑉 = 𝑄𝐶𝑜𝑛𝑠𝑡𝑃 − 𝑛𝑅𝑇 = ∆𝐻𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠−𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 − ∆𝑛𝑅𝑇 For adiabatic conditions ∆𝐻𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠−𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 − ∆𝑛𝑅𝑇 = 0 Thus, (𝐸𝑞2.2.1) becomes 𝑇𝑓 ∑ 𝐻𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 (𝑇𝑖 , 𝑃) = ∑ 𝑛𝑖 ∫ 𝐶𝑃𝑖 𝑑𝑇 + 𝑄𝑉𝐿 − 𝑅(𝑛𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 𝑇𝑖 − 𝑛𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠,𝐼𝑛𝑒𝑟𝑡𝑠 𝑇𝑓 ) (𝐸𝑞 2.2.2) 𝑇𝑖 Where 𝑛𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 and 𝑛𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠,𝐼𝑛𝑒𝑟𝑡𝑠 is the number of moles of the reactants and products/inerts respectively. ➢ Calculations Before carrying out energy balances, the specific heat capacities must be determined. The heat capacity varies dramatically as combustion results in a large range of temperatures, thus it is necessary to obtain 𝐶𝑃 functions that could represent a more suitable heat capacity at the different temperatures. Consequently, (𝐸𝑞 2.3) is utilised in which the coefficients A, B, C and D are specific for the different gases (Smith, 2005). 𝐶𝑃𝑖 = 𝑅 ∗ [𝐴 + 𝐵𝑇 + 𝐶𝑇 2 + 𝐷𝑇 −2 ] (𝐸𝑞 2.3) Thus, by using the heats of formation, latent heats and sensible heats, different conditions were examined for the resulting final adiabatic temperatures. Combustion in pure oxygen A complete combustion with pure of oxygen was examined in which exact stoichiometric amount were considered. The reaction is as follows. 5 𝐶2 𝐻2 (𝑔) + 𝑂2 (𝑔) → 2𝐶𝑂2 (𝑔) + 𝐻2 𝑂(𝑙) (𝐸𝑞 2.4) 2 The energy balance is shown in Table 2 Table 2: Energy balance for the combustion in pure oxygen at constant pressure and constant volume Constant pressure Constant volume ∆𝐻𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 (𝐽/𝑚𝑜𝑙) −1.3 × 106 𝑄𝑉𝐿 (𝐽/𝑚𝑜𝑙) 4.1 × 104 𝑄𝐶𝑝 (𝐽/𝑚𝑜𝑙) 1.2 × 106 ∆𝑛𝑅𝑇 N/A 𝑇𝑓 (𝐾) 6179 −1.3 × 106 4.1 × 104 1.4 × 106 −1.6 × 105 6690 Where 𝑄𝐶𝑝 is the total sensible heat as per (𝐸𝑞 2.3) and the integral in (𝐸𝑞2.2.1). By examining Table 2, it is clear to see that the temperature reached is unrealistic for both constant volume and pressure. In a real system, the product gases will decompose thermally well below 6000K , thus this is an invalid calculation as the kinetics must be considered for a realistic result for pure oxygen. Also, due to the final temperature being different in the cases for constant volume and constant pressure, 𝑄𝐶𝑝 is different between them. This difference is present in all conditions. Also, due to the ∆𝑛𝑅𝑇 term in the constant volume calculation, the final temperature is higher than that of constant pressure. Combustion in air Here, a complete combustion is modelled with exact stoichiometries, however the combustion now occurs in air in which the compositions of oxygen and nitrogen gases are 21% and 79% respectively. The resulting temperatures is shown in Table 3. Table 3: Energy balance for the combustion in air at constant pressure and constant volume Constant pressure Constant volume ∆𝐻𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 (𝐽/𝑚𝑜𝑙) −1.3 × 106 𝑄𝑉𝐿 (𝐽/𝑚𝑜𝑙) 4.1 × 104 𝑄𝐶𝑝 (𝐽/𝑚𝑜𝑙) 1.2 × 106 ∆𝑛𝑅𝑇 N/A 𝑇𝑓 (𝐾) 2848 −1.3 × 106 4.1 × 104 1.6 × 106 −3.4 × 105 3408 The energy balance is consistent between Tables 2 and 3 as the reaction is exactly the same fro constant pressure. The only difference is the final temperature due to the presence of significant amount of nitrogen which contributed to the sensible heat. Alternatively, at constant volume, 𝑄𝐶𝑝 and ∆𝑛𝑅𝑇 are different. The temperatures calculated are much more realistic in this model. Combustion in 50% excess air In this case, an excess amount of air is included for the combustion. The resulting temperatures is shown in Table 4. Table 4: Energy balance for the combustion in excess air at constant pressure and constant volume Constant pressure Constant volume ∆𝐻𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 (𝐽/𝑚𝑜𝑙) −1.3 × 106 𝑄𝑉𝐿 (𝐽/𝑚𝑜𝑙) 4.1 × 104 𝑄𝐶𝑝 (𝐽/𝑚𝑜𝑙) 1.2 × 106 ∆𝑛𝑅𝑇 N/A 𝑇𝑓 (𝐾) 2294 −1.3 × 106 4.1 × 104 1.6 × 106 −3.9 × 105 2806 Again the energy balance is consistent in constant pressure, but different in constant volume. The further reduction in temperature is due to the unreacted air (excess). Incomplete combustion in air producing only carbon monoxide and water An incomplete combustion is modelled for the combustion at stoichiometric amounts as shown in (𝐸𝑞 2.5) 3 𝐶2 𝐻2 (𝑔) + 𝑂2 (𝑔) → 2𝐶𝑂(𝑔) + 𝐻2 𝑂(𝑙) (𝐸𝑞 2.5) 2 The energy balance for the combustion is presented in Table 5 Table 5: Energy balance for the incomplete combustion in air (only CO) at constant pressure and constant volume Constant pressure Constant volume ∆𝐻𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 (𝐽/𝑚𝑜𝑙) −7.3 × 105 𝑄𝑉𝐿 (𝐽/𝑚𝑜𝑙) 4.1 × 104 𝑄𝐶𝑝 (𝐽/𝑚𝑜𝑙) 6.9 × 105 ∆𝑛𝑅𝑇 N/A 𝑇𝑓 (𝐾) 2541 −7.3 × 105 4.1 × 104 9.1 × 105 −2.2 × 105 3118 It is clear to see that the ∆𝐻𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 is lower since the reaction now is different. Accordingly the final temperatures are now lower than for the combustion in air (Table 3) since carbon monoxide is produced instead (which has a lower heat of formation). Incomplete combustion in air producing carbon monoxide, soot (solids) and water Finally, the incomplete combustion is modelled in which solid carbon (soot) is also produced, thus altering the stoichiometries. 𝐶2 𝐻2 (𝑔) + 𝑂2 (𝑔) → 𝐶𝑂(𝑔) + 𝐻2 𝑂(𝑙) + 𝐶(𝑠) (𝐸𝑞 2.6) Table 6: Energy balance for the incomplete combustion in air (CO and C) at constant pressure and constant volume Constant pressure Constant volume ∆𝐻𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 (𝐽/𝑚𝑜𝑙) −6.2 × 105 𝑄𝑉𝐿 (𝐽/𝑚𝑜𝑙) 4.1 × 104 𝑄𝐶𝑝 (𝐽/𝑚𝑜𝑙) 5.8 × 105 ∆𝑛𝑅𝑇 N/A 𝑇𝑓 (𝐾) 2754 −6.2 × 105 4.1 × 104 7.7 × 105 −2.2 × 105 3364 As seen in Table 6, ∆𝐻𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 is even lower than in Table 5 since less overall chemical compounds are produced (formation of solid carbon produces no energy). Also, it is noticeable that 𝑄𝐶𝑝 is lower in this case as less gaseous products are present, thus requiring less sensible energy, and thus leading to higher adiabatic flame temperatures at the end. Question 3 The pressure of propane vs. density at 400 K can be fitted by the following expression: 𝑃 = 33.258𝜌 − 7.588𝜌2 + 1.0306𝜌3 − 0.0588𝜌4 − 0.0034𝜌5 + 0.0006𝜌6 Where the units of P and  are bar and mol L-1 respectively. Use the Van der Waals equation and the m to calculate the pressure for  = 0 to 12.3 mol L-1, and plot your results. Discuss the comparison of your results to the above expression. Answer: ➢ Virial equation of state (From the question) The virial equations of state are expressions that incorporate temperature, volume and pressure that could illustrate the behaviour of gases during expansion/compression. They are constructed from empirically from experimental data. Consequently, this is the most suitable mathematical expression that would illustrate the behaviour of gaseous propane at 400K (Plischke, 2005). In order to plot the pressure from the expression given in the question, the relationship between molar density and molar volume must be defined as in (Eq 3) where: 𝜌= 1 𝑉𝑚 Where 𝑉𝑚 is the molar volume in 𝐿/𝑚𝑜𝑙. Accordingly a set of molar densities can be obtained from the set of volumes given in the questions. This results in a range of pressures than can be plotted vs volume. ➢ Van der waals equation of state This is the first practical equation of state in which compound specific coefficients are added to the ideal gas equation in order to model a real gas behaviour. The van der Waals equation is shown in (𝐸𝑞 3) (Dahm & Donald, 2014). 𝑃= 𝑅𝑇 𝑎 − (𝐸𝑞 3) 𝑉𝑚 − 𝑏 𝑉𝑚 2 𝐿 Where 𝑎 = 8.779 (𝐿2 ∙ 𝑏𝑎𝑟/𝑚𝑜𝑙 2 ), 𝑏 = 0.08445 (𝑚𝑜𝑙) and 𝑅 = 8.314 ∗ 10−3 (𝐿 ∙ 𝐵𝑎𝑟/𝐾 ∙ 𝑚𝑜𝑙). 𝑎 Moreover, 𝑉 𝑚 2 accounts for the attractive forces and 𝑏 accounts for the fact that molecules have a finite size (Israelachvili, 2011). ➢ Redlich-Kwong equation of state The Redlich-Kwong equation of state was a successful improvement on the van der Waals equation. The improvements were possible due to the empirical findings of the temperature dependency on the attraction parameter with little attention on modifying the parameter b (Tuckerman & Adewumi, 2016). The equation is represented in (𝐸𝑞 3.1) while the expressions for the coefficients 𝑎 and 𝑏 isare shown in (𝐸𝑞 3.1.1) and (𝐸𝑞 3.1.2) respectively. 𝑃= 𝑅𝑇 𝑎 − (𝐸𝑞 3.1) 𝑉𝑚 − 𝑏 √𝑇𝑉𝑚 (𝑉𝑚 + 𝑏) 𝑎 = 0.42780 𝑅 2 𝑇𝑐 2.5 (𝐸𝑞 3.1.1) 𝑃𝑐 𝑏 = 0.086640 𝑅𝑇𝑐 (𝐸𝑞 3.1.2) 𝑃𝑐 ➢ Plotting the equations of state Figure 4 shows the equations of state plotted for pressure vs molar volume. Similarly, having stated that the virial model is the most accurate of the three, it is possible to it as a reference point, and thus Figure 5 compares the behaviour of van der Waals and Redlich-Kwong and the virial equation of state. Figure 4: A pressure vs volume lot for the three equations of state Figure 5: A comparison between the virial equation of state to the van der Waals and Redlich-Kwong 𝑚𝑜𝑙 ), 𝐿 According to Figure 4, all three models show very similar results up to about 5 ( which is also evident in Figure 5. This is due to the fact that as pressure increases at constant temperature, the compressibility factor 𝑍 = 𝑃𝑉 𝑛𝑅𝑇 increases. The compressibility factor ultimately explains how ideal a real gas is, thus at 𝑍 = 1, a real gas could be modelled as ideal, otherwise it is necessary to consider the gas as a real gas (Rao, 2010). Accordingly, due to the increasing compressibility factor, the van 𝑚𝑜𝑙 ). 𝐿 der Waals and Redlich-Kwong models deviate from the Virial at around 5 ( Moreover, it is noticeable in Figure 4 that the van der Waals model deviates dramatically from both the RedlichKwong and Virial models. This is more clearly illustrated in Figure 5 in which the deviation grows exponentially between the van der Waals and Virial models, whereas the Redlich-Kwong shows a maximum deviation of about 58 (𝑏𝑎𝑟). The increased accuracy of the Redlich-Kwong equation is due to the modification on the temperature reliance of the attractive forces. Therefore, in the case where the virial equation is not available, the use of the Redlich-Kwong equation acceptable to some extent. Finally, there are more complex, yet more accurate models such as the Soave modification of Redlich-Kwong, or even the different forms of the Peng-Robinson equation that could produce more suitable results for real gases (Goodwin, et al., 2010). Question 4 Discuss the use of electrochemical equilibrium measurements in thermodynamics, including relevant derivations and results and illustrative examples where appropriate. Answer: ➢ Electrochemical processes Electrochemical processes occur when electrons flow from one chemical substance to another, with the driving force being the oxidation-reduction reactions (redox). Since all redox reactions occur with both oxidation and reduction, it is common to describe a redox reaction as two half reactions. For example, a typical electrochemical reaction is the redox reaction between Zinc and Bromine as shown in (𝐸𝑞 4.8) (Br ...
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Thermodynamics Report

The model demonstrating the behaviour of oxygen gas can be illustrated by several equations.
These include the Peng-Robinson equation, Van der Waals, Ideal gas law and the Redlichkwong. Out of the four equations, the Peng-Robinson was found to be the most useful because it
is appropriate in representing real gases such as oxygen (Sandler, 2017). As opposed to the other
methods, the equation is also well established within manufacturing industries (Ashour et,al
2011. According to Tsai & Chen, the method also covers several vital physical characteristics,
hich include vapour-liquid equilibriums (1998). As such, it illustrates reduced variables, which
are a result of deriving dimensionless quantities. For instance, (𝑇𝑟 , 𝑃𝑟 , 𝑉𝑟) has been derived from






(𝑇 , 𝑃 , 𝑉 ). Therefore, Dam & Donald assert that the method guarantees precise results similar to
the actual experimental behaviour (2014).
According to Goodwin, et al, The Peng-Robinson equation can be illustrated as follows (2010)



(𝐸𝑞 1)
𝑉𝑚 − 𝑏 𝑉𝑚 (𝑉𝑚 + 𝑏) + 𝑏(𝑉𝑚 − 𝑏)

In this case, P represents Pressure,
R is the constant representing the ideal gas
T stands for temperature
Vm stands for molar volume
a represents 0.45724
b stands for 0.07780

𝑅2 𝑇𝑐 2



(𝐸𝑞 )

𝜑 = [1 + (1 − 𝑇𝑟 0.5 )(0.37464 + 1.54226𝜔 − 0.26992𝜔2 )]2



Here Tc represents the critical temperature, while P c stands for Pressure. Tr and 𝜔 represesent the
reduced temperature and the acentric factor respectively.
The above equations were then used in plotting the P-V-T Diagram after the identiiction of the
thermodynamic properties of oxygen. Additionally, a MATLAB Code was established to help in
the plotting of isotherms sets for various molar volumes and temperature ranges. The values for
the temperature range was at the critical temperature, above it or below it (𝑇 ≤ Tc ≤ 𝑇).

The following is the data representation in the form of a table
Thermodynamic properties of




The gas constant (R)

83.14 cm3 bar/mol K

Critical temperature (Tc )

154.60 K

Critical Pressure ( Pc )

50.80 bar

Acentric factor (𝜔)


graphical representation of
Oxygen Isotherms on a P-V
diagram ranging on the

temperatures of 110, 140, 154, 180, and 210 are demonstrated as follows.



Oxygen isotherm at 154 K is shown as follows with the plotting of equilibrium lines against the
line of saturation and isotherms of oxygen at the critical point.

The complete P-V-T diagram for the five oxygen isotherms at varied temperatures can be shown
as follows.







C4 (1E-02)








According to Strahle (1993), in scenarios where fuel burns without any loss of heat, the resulting
mixture of gases gets to the optimum temperature known as adiabatic flame temperature.
However, the flam temperature varies, depending on several factors or conditions. The
following are some of the points for consideration.

Adiabatic combustion when the pressure is constant

Strahle further argues that when the pressure is constant, a certain amount of energy is lost to
expansion processes, but there is no heat loss to the surroundings. Therefore, the first law of
thermodynamics applies in the following scenario
∆𝑈 = 𝑄𝐶𝑜𝑛𝑠𝑡𝑃 − 𝑊𝐶𝑜𝑛𝑠𝑡𝑃 = 𝑄𝐶𝑜𝑛𝑠𝑡𝑃 − 𝑃∆𝑉 (𝐸𝑞 2)
Here, 𝑈 represents internal energy
𝑄𝐶𝑜𝑛𝑠𝑡𝑃 Represents the heat at constant pressure



𝑊𝐶𝑜𝑛𝑠𝑡𝑃 Stands for work at constant pressure
Enthalpy (H) is equal to 𝑈 + 𝑃𝑉

Where, 𝐻 refers to the enthalpy. Therefore, any change in the enthalpy when the pressure is
constant results to the following

∆𝐻 = ∆𝑈 + 𝑃∆𝑉 + 𝑉∆𝑃 = ∆𝑈 + 𝑃∆𝑉

The representation of the adiabatic process is

𝑄𝐶𝑜𝑛𝑠𝑡𝑃 = 0 𝑎𝑛𝑑 ∆𝐻 = 0

However, at equilibrium, the balance of energy is

∑ 𝐻𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 (𝑇𝑖 , 𝑃) = ∑ 𝐻𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠,𝐼𝑛𝑒𝑟𝑡𝑠 (𝑇, 𝑃) (𝐸𝑞2.2)


∑ 𝐻𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 (𝑇𝑖 , 𝑃) = ∑ 𝑛𝑖 ∫ 𝐶𝑃𝑖 𝑑𝑇 + 𝑄𝑉𝐿 (𝐸𝑞2.2.1)

In this case, 𝑛𝑖 stands for moles number of 𝑖 𝑡ℎ species,
𝑇𝑖 Stands for initial temperature 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 (𝐾)
𝑇𝑓 Illustrates final temperature



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