Nitrogen mole fraction

First, drop all the repeated equations listed in Table 3.5.3. By substituting Equations 3.5.30 to 3.5.34 into Equations 3.5.19 to 3.5.22, we eliminate the mole fraction variables in line seven. We do not need Equations 3.5.35 to 3.5.39 for the solution, so they can be dropped. Table 3.5.2 lists the specified variables, except for the nitrogen mole fraction, y3 6, which is now unspecified. Table 3.5.7 lists the reduced set of equations. 

Pure methane is completely burned with air. The outlet gases from the burner, which contain no oxygen, are passed through a cooler, where some of the water is removed by condensation. The gases leaving the cooler have a nitrogen mole fraction of 0.8335. Calculate the following ... 

All of the above are gases except the solid with the empirical formula Ci()Hg. The gas stream (oxygen and nitrogen mole fractions of 0.01 and 0.099) is at 537.8°C with a Cp of 7.54 calories/g mole °K. Heat of combustion is —100 kilocalories per mole of oxygen consumed. If Sc is 0.8 and Pr 0.7, find the maximum possible particle temperature. 

Thus 41.4% of the mixture is condensed with a nitrogen mole fraction in the condensate of 0.654 while the vapor in equilibrium with the condensate has a nitrogen mole fraction of 0.886. Note that the liquid and vapor concentrations agree with those shown in Fig. 6.9. 

Further cooling of the original mixture provides only a small enrichment of the nitrogen in the vapor. For example, immediately before total condensation of the mixture at a temperature of 78.9 K, the last trace of vapor will contain a nitrogen mole fraction of no better than 0.94. 

To complete the flash calculation, temperatures must be assumed until Eq. (6.25) is balanced (see Example 6.7). A temperature of 79.01 K satisfies the relation with a nitrogen vapor mole fraction of 0.930. With the aid of Eqs. (6.21) and (6.22) and Td = 79.51 K, we can now determine (following the procedure in Example 6.4) that the product stream leaving the top of the column has a nitrogen mole fraction of 0.908. This permits solution of the overall and component balance around the column represented by... 

Finally, Table 2 shows enthalpy calculations for the system nitrogen-water at 100 atm. in the range 313.5-584.7°K. [See also Figure (4-13).] The mole fraction of nitrogen in the liquid phase is small throughout, but that in the vapor phase varies from essentially unity at the low-temperature end to zero at the high-temperature end. In the liquid phase, the enthalpy is determined primarily by the temperature, but in the vapor phase it is determined by both temperature and composition. 

K-factors for vapor-liquid equilibrium ratios are usually associated with various hydrocarbons and some common impurities as nitrogen, carbon dioxide, and hydrogen sulfide [48]. The K-factor is the equilibrium ratio of the mole fraction of a component in the vapor phase divided by the mole fraction of the same component in the liquid phase. K is generally considered a function of the mixture composition in which a specific component occurs, plus the temperature and pressure of the system at equilibrium. 

Ewald22 studied this system at 150° and 155°K. These temperatures are above the critical temperature of pure nitrogen, 126°K, but he found that they are below the lower critical end point of the mixture. The saturated vapor pressure of the system was 50 atm at 150°K and 57 atm at 155°K. The mole fraction of xenon in the saturated gas (X in Figs. 5 and 9) was 0.035 and 0.045 at these temperatures, respectively. 

For a differential reactor, the change in composition across the reactor will be very small, and the bulk fluid composition may be estimated from the inlet molal flow rates. Assuming that the inlet air is 79% nitrogen and 21% oxygen, the calculations below indicate the bulk fluid mole fractions and partial pressures of the various components of the reaction mixture. 

The difference in mole fractions is most significant in the case of S02 where this difference is 15% of the bulk phase level. This result indicates that external mass transfer limitations are indeed significant, and that this difference should be taken into account in the analysis of kinetic data from this system. Note that there is a difference in nitrogen concentration between the bulk fluid and the external surface because there is a change in the number of moles on reaction, and there is a net molar flux toward... 

Because the vessel is pressurized with pure nitrogen, the number of moles of oxygen remains constant during pressurization, whereas the mole fraction decreases. During depressurization, the composition of the gas within the vessel remains constant, but the total number of moles is reduced. Thus the oxygen mole fraction remains unchanged. 

Assume that the nitrogen contains oxygen with a constant mole fraction of yoxy. For a pressure purging procedure the total moles of oxygen present at the end of the first pressurization is given by the moles initially present plus the moles included with the nitrogen. This amount is... 

Example 2.2 Air has the approximate composition by mole fraction of 0.21 for oxygen and 0.79 for nitrogen. What is the molecular weight of air and what is its corresponding composition in mass fractions ... 

Worked Example 7.11 Hydrogen gas is mixed with a nitrogen bath gas . The overall pressure is p. If the mole fraction of the hydrogen is expressed as 10 per cent, what is its activity ... 

As the molar masses of oxygen, nitrogen and argon are so similar, we can approximate the mole fractions of the gases to their percentage compositions. 

Figure 8. Nitrogen + water—mole fraction solubility at 1 atm nitrogen partial pressure vj. temperature ( ) (10, 15, 16) (0) (12, 13, 14, 171 (values used in the...

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