Explanation Write the combustion reaction of diesel fuel by considering 1 lbmol of diesel fuel as follows. C 12 H 26 + a th [ O 2 + 3.76 N 2 ] → 12 CO 2 + 13 H 2 O + ( 3 .76 × a th ) N 2 (I) Write the expression to calculate the total number of moles in a mixture ( N m ) . N m = N A + N B + N C + ........... + N Z (II) Here, number of moles of substance A is N A , number of moles of substance B is N B , Number of moles of substance C is N C , and the number of moles of substance Z is N Z . Express the relation between the mole fraction ratio and pressure fraction ratio for water vapor. N v N total = P v P total (III) Here, mole number of the water vapor is N v , number of moles of products is N total , partial pressure of the water vapor is P v , and total pressure of the process is P total . Write the expression to calculate the entropy generation during the process ( S gen ) . S gen = S P − S R + Q out T surr (IV) Here, entropy of the product is S P , entropy of the reactant is S R , temperature of the surroundings is T surr . The entropy values are to be calculated at the partial pressure of the components. Express the partial pressure of the component i ( P i ) . P i = y i P total (V) Here, mole fraction of the component i is y i , and total pressure is P total . The mole fraction of the components ( y ) is found by the ratio of number of moles of the component and total number moles of the components. Therefore, express the entropy of the component i ( S i ) . S i = N i s ¯ i ( T , P i ) S i = N i [ s ¯ i ( T , P 0 ) − R u ln ( y i P m ) ] (VI) Here, number of moles of component i is N i , absolute entropy of component i is s ¯ i , temperature is T , absolute pressure is P 0 , gas constant is R u , and pressure of the mixture is P m . Write the expression to calculate the reversible work ( W rev ) . W rev = ∑ N R ( h ¯ f ∘ + h ¯ − h ¯ ∘ − T 0 s ¯ ) R − ∑ N P ( h ¯ f ∘ + h ¯ − h ¯ ∘ − T 0 s ¯ ) P W rev = ∑ N R ( h ¯ f ∘ − T 0 s ¯ ) R − ∑ N P ( h ¯ f ∘ − T 0 s ¯ ) P (VII) Here, enthalpy of vaporization is h ¯ f ∘ , sensible enthalpy at specified state is h ¯ , sensible enthalpy at reference state is h ¯ ∘ , thermodynamic temperature of the surroundings is T 0 , and absolute entropy is s ¯ . Conclusion: Balance for the Oxygen from Combustion reaction Equation (I). a th = 12 + 6.5 = 18.5 Rewrite the complete balanced chemical reaction for combustion as follows: C 12 H 26 + ( 18.5 ) [ O 2 + 3.76 N 2 ] → 12 CO 2 + 13 H 2 O + ( 3 .76 × 18 .5 ) N 2 C 12 H 26 + 18.5 [ O 2 + 3.76 N 2 ] → 12 CO 2 + 13 H 2 O + 69.56 N 2 (VIII) Substitute 12 lbmol for N CO 2 , 13 lbmol for N H 2 O , and 69.56 lbmol for N 2 in Equation (II). N total = 12 lbmol + 13 lbmol + 69 .56 lbmol = 9 4 .56 lbmol From the Table A-4 of “Saturated water-Temperature table”, select the partial pressure of the water vapor ( P v ) at the temperature of 77 ° F as 0.4648 psia . Substitute 13 − N w for N v , in , 94.56 − N w for N total , 0.4648 psia for P v , in , and 14.7 psia for P total in Equation (III). 13 − N w 94.56 − N w = 0.4648 psia 14.7 psia N w = 10 .34 lbmol Write the complete balanced combustion reaction of diesel fuel from Equation (VIII) for the amount of water vapor. C 12 H 26 + 18.5 [ O 2 + 3.76 N 2 ] → 12 CO 2 + 10.34 H 2 O ( l ) + 2.66 H 2 O ( g ) + 69.56 N 2 From table A-26E of “Enthalpy of formation, Gibbs function of formation, and absolute entropy at 25 ° C, 1 bar ”, show the Gibbs function of formation values as in Table (1). Substance C 12 H 26 O 2 N 2 H 2 O ( g ) H 2 O ( g ) CO 2 h ¯ f ° Btu / lbmol − 125 , 190 0 0 − 104 , 040 − 122 , 970 − 169 , 300 The absolute entropy of components are obtained from the tables of “Ideal gas properties of oxygen, nitrogen, water, and carbon dioxide” and values are tabulated using Equations (IV), (V), and (VI). Represent the entropy calculations for the reactants as in Table (1). Substance N i y i s ¯ i o , ( 77 ° F , 1 atm ) R u ln ( y i P m ) s ¯ i C 12 H 26 1 – 148.86 – 148.86 O 2 18.5 0.21 49.00 –3.10 52.10 N 2 69.56 0.79 45.77 –0.47 46.24 Represent the entropy calculations for the products as in Table (2). Substance N i y i s ¯ i , ( T , 1 atm ) R u ln ( y i P m ) s ¯ i CO 2 12 0

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Chapter 15.7, Problem 106RP

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The maximum amount of work potential from the diesel fuel.

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