The CRC Handbook of Mechanical Engineering. Chapter 2. Engineering Thermodynamics (776125), страница 14

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For steady-state operation and negligible effects of kinetic and potential energy,determine the temperature of the combustion products for complete combustion with the theoreticalamount of air, and (b) the rates of entropy generation and exergy destruction, each per kmol of fuel.Solution.

For combustion of liquid octane with the theoretical amount of air, the chemical equation isC 8 H18 (l) + 12.5O 2 + 47N 2 → 8CO 2 + 9H 2 O(g) + 47N 2(a) At steady state, the control volume energy rate balance reduces to read0=00Q˙ cvW˙− cv +n˙ Fn˙ F∑ n (hiof) ∑ n (h+ ∆h −eiRof+ ∆h)ePwhere R denotes reactants, P denotes products, and the symbols for enthalpy have the samesignificance as in Equation 2.78.

Since the reactants enter at 25°C, the corresponding ( ∆h )i termsvanish, and the energy rate equation becomes∑ n ( ∆h ) = ∑ n h − ∑ n hoeiePfiReofePIntroducing coefficients from the reaction equation, this takes the form( )8 ∆hCO2( )− 8 h f( )+ 9 ∆ho© 1999 by CRC Press LLCH 2O( g )( )oCO2+ 9 hf( )+ 47 ∆h( )o=  h fN2( )oH 2O( g )+ 47 ∆h fN2( )oC8H18 ( l )+ 12.5 h f( )oO2+ 47 h fN22-64Section 2Using data from Table 2.9 to evaluate the right side,( )8 ∆hCO2( )+ 9 ∆hH 2O( g )( )+ 47 ∆h= 5,074,630 kJ kmol (fuel)N2Each ∆h term on the left side of this equation depends on the temperature of the products, Tp,which can be solved for iteratively as Tp = 2395 K.(b) The entropy rate balance on a per-mole-of-fuel basis takes the form0=∑j) ()0S˙genQ˙ j Tj+ s F + 12.5sO2 + 47sN 2 − 8sCO2 + 9sH O( g ) + 47sN 2 +2n˙ Fn˙ F(or on rearrangement,S˙genn˙ F()(= 8sCO2 + 9sH O( g ) + 47sN 2 − s F − 12.5sO2 + 47sN 22)The absolute entropy of liquid octane from Table 2.9 is 360.79 kJ/mol · K.

The oxygen andnitrogen in the combustion air enter the reactor as components of an ideal gas mixture at Tref ,pref . With Equation 2.81, where p = pref , and absolute entropy data from Table 2.9,( )sO2 = sOo 2 Tref − R ln yO2= 205.03 − 8.314 ln 0.21 = 218.01 kJ kmol ⋅ K( )sN 2 = sNo 2 Tref − R ln yN 2= 191.5 − 8.314 ln 0.79 = 193.46 kJ kmol ⋅ KThe product gas exits as a gas mixture at 1 atm, 2395 K with the following composition: yCO2= 8/64 = 0.125, yH2O( g ) = 9/64 = 0.1406, yN 2 = 47/64 = 0.7344.

With Equation 2.81, where p= pref , and absolute entropy data at 2395 K from Table A.2,sCO2 = 320.173 − 8.314 ln 0.125 = 337.46 kJ kmol ⋅ KsH2O = 273.986 − 8.314 ln 0.1406 = 290.30 kJ kmol ⋅ KsN 2 = 258.503 − 8.314 ln 0.7344 = 261.07 kJ kmol ⋅ KInserting values, the rate of entropy generation isS˙genn˙ F= 8(337.46) + 9(290.30) + 47(261.07) − 360.79 − 12.5(218.01) − 47(193.46)= 5404 kJ kmol ⋅ KUsing Equation 2.87 and assuming T0 = 298 K, the rate of exergy destruction is E˙ D / n˙ F = 1.61× 106 kJ/kmol.© 1999 by CRC Press LLC2-65Engineering ThermodynamicsGibbs Function of FormationParalleling the approach used for enthalpy, a value of zero is assigned to the Gibbs function of eachstable element at the standard state.

The Gibbs function of formation of a compound equals the changein the Gibbs function for the reaction in which the compound is formed from its elements. Table 2.9provides Gibbs function of formation data of various substances at 298 K and 1 atm.The Gibbs function at a state other than the standard state is found by adding to the Gibbs functionof formation the change in the specific Gibbs function ∆g between the standard state and the state ofinterest:[(g (T , p) = g fo + g (T , p) − g Tref , pref)] = gof+ ∆g(2.82a)where[(∆g = h (T , p) − h Tref , pref)] − [T s (T , p) − T s (Trefref, pref)](2.82b)The Gibbs function of component i in an ideal gas mixture is evaluated at the partial pressure ofcomponent i and the mixture temperature.As an application, the maximum theoretical work that can be developed, per mole of fuel consumed,is evaluated for the control volume of Figure 2.15, where the fuel and oxygen each enter in separatestreams and carbon dioxide and water each exit separately.

All entering and exiting streams are at thesame temperature T and pressure p. The reaction is complete:bbCa Hb +  a +  O 2 → aCO 2 + H 2 O42This control volume is similar to idealized devices such as a reversible fuel cell or a van’t Hoff equilibriumbox.FIGURE 2.15 Device for evaluating maximum work.For steady-state operation, the energy rate balance reduces to giveW˙ cv Q˙ cvbb=+ hF +  a +  hO2 − ahCO2 − hH2O42n˙ Fn˙ F© 1999 by CRC Press LLC2-66Section 2where ṅF denotes the molar flow rate of the fuel.

Kinetic and potential energy effects are regarded asnegligible. If heat transfer occurs only at the temperature T, an entropy balance for the control volumetakes the form0=S˙genQ˙ cv n˙ Fbb+ s F +  a +  sO2 − asCO2 − sH2O +T42n˙ FEliminating the heat transfer term from these expressions, an expression for the maximum theoreticalvalue of the work developed per mole of fuel is obtained when the entropy generation term is set to zero: W˙ cv bbbb n˙  int = hF +  a + 4  hO2 − ahCO2 − 2 hH2O (T , p) − T s F +  a + 4  sO2 − asCO2 − 2 sH2O (T , p) F revThis can be written alternatively in terms of the enthalpy of combustion as W˙ cv bb n˙  int = − hRP (T , p) − T s F +  a + 4  sO2 − asCO2 − 2 sH2O (T , p) F (2.83a)revor in terms of Gibbs functions as W˙ cv bb n˙  int = gF +  a + 4  gO2 − agCO2 − 2 gH2O (T , p) F (2.83b)revEquation 2.83b is used in the solution to Example 11.Example 11Hydrogen (H2) and oxygen (O2), each at 25°C, 1 atm, enter a fuel cell operating at steady state, andliquid water exits at the same temperature and pressure.

The hydrogen flow rate is 2 × 10–4 kmol/secand the fuel cell operates isothermally at 25°C. Determine the maximum theoretical power the cell candevelop, in kW.Solution. The overall cell reaction is H2 + 1/2 O2 → H2O(,), and Equations 2.83 are applicable. SelectingEquation 2.83b, and using Gibbs function data from Table 2.9, W˙ cv 1 n˙  int =  gH2 + 2 gO2 − gH2O( l )  (25°C, 1 atm) F rev=0+1(0) − (−237,180) = 237,180 kJ kmol2Then(W˙ )cv intrev© 1999 by CRC Press LLCkJ  kmol   kW =  237,1802 × 10 −4 = 47.4 kWkmols   1kJ s 2-67Engineering ThermodynamicsReaction EquilibriumLet the objective be to determine the equilibrium composition of a system consisting of five gases A,B, C, D, and E, at a temperature T and pressure p, subject to a chemical reaction of the formv A A + v B B ↔ vC C + v D Dwhere the v’s are stoichiometric coefficients.

Component E is assumed to be inert and thus does notappear in the reaction equation. The equation suggests that at equilibrium the tendency of A and B toform C and D is just balanced by the tendency of C and D to form A and B.At equilibrium, the temperature and pressure would be uniform throughout the system. Additionally,the equation of reaction equilibrium must be satisfied:v A µ A + v B µ B = vC µ C + v D µ D(2.84a)where the µ’s are the chemical potentials (Section 2.3, Multicomponent Systems) of A, B, C, and D inthe equilibrium mixture.

In principle, the composition that would be present at equilibrium for a giventemperature and pressure can be determined by solving this equation.For ideal gas mixtures, the solution procedure is simplified by using the equilibrium constant K(T )and the following equation:y vC y vD  p K (T ) = CvA DvB y A y B  pref =vCCvAAn nn nvDDvBBvC + vD − v A − vB p pref  n (2.84b)vC + vD − v A − vBwhere yA, yB, yC, and yD denote the mole fractions of A, B, C, and D in the equilibrium mixture and n= nA + nB + nC + nD + nE, where the n’s denote the molar amounts of the gases in the mixture.

Tabulationsof K(T ) for each of several reactions of the form Equation 2.84a are provided in Table 2.11. An applicationof Equation 2.84b is provided in Example 12.Example 12One kmol of CO reacts with the theoretical amount of dry air to form an equilibrium mixture of CO2,CO, O2, and N2 at 2500 K, 1 atm. Determine the amount of CO in the equilibrium mixture, in kmol.Solution. The reaction of CO with the theoretical amount of dry air to form CO2, CO, O2, and N2 isCO +1zO + 1.88N 2 → zCO + O 2 + (1 − z )CO 2 + 1.88N 22 22where z is the amount of CO, in kmol, present in the equilibrium mixture.

The total number of moles n isn=z+z5.76 + z+ (1 − z ) + 1.88 =22At equilibrium CO2 ↔ CO + 1/2 O2; and Equation 2.84b takes the formz( z 2 )1− z12K= p pref  (5.76 + z ) 2 12where p/pref = 1. At 2500 K, Table 2.11 gives K = 0.0363. Solving iteratively, z = 0.175.© 1999 by CRC Press LLC2-68Section 2TABLE 2.11 Logarithms to the Base 10 of the Equilibrium Constant Klog10 KCO2 + H2H2Ο ⇔CO2 ⇔/2O2 + 1/2N2 H2O ⇔⇔H2 + 1/2O2 OH + 1/2H2 CO + 1/2 O2 CO + H2O⇔NO1Temp (K) H2 ⇔ 2H O2 ⇔ 2O N2 ⇔ 2N29850010001200140016001700180019002000210022002300240025002600270028002900300031003200330034003500–71.224–40.316–17.292–13.414–10.630–8.532–7.666–6.896–6.204–5.580–5.016–4.502–4.032–3.600–3.202–2.836–2.494–2.178–1.882–1.606–1.348–1.106–0.878–0.664–0.462–81.208–45.880–19.614–15.208–12.054–9.684–8.706–7.836–7.058–6.356–5.720–5.142–4.614–4.130–3.684–3.272–2.892–2.536–2.206–1.898–1.610–1.340–1.086–0.846–0.620–159.600–92.672–43.056–34.754–28.812–24.350–22.512–20.874–19.410–18.092–16.898–15.810–14.818–13.908–13.070–12.298–11.580–10.914–10.294–9.716–9.174–8.664–8.186–7.736–7.312–15.171–8.783–4.062–3.275–2.712–2.290–2.116–1.962–1.823–1.699–1.586–1.484–1.391–1.305–1.227–1.154–1.087–1.025–0.967–0.913–0.863–0.815–0.771–0.729–0.690–40.048–22.886–10.062–7.899–6.347–5.180–4.699–4.270–3.886–3.540–3.227–2.942–2.682–2.443–2.224–2.021–1.833–1.658–1.495–1.343–1.201–1.067–0.942–0.824–0.712–46.054–26.130–11.280–8.811–7.021–5.677–5.124–4.613–4.190–3.776–3.434–3.091–2.809–2.520–2.270–2.038–1.823–1.624–1.438–1.265–1.103–0.951–0.809–0.674–0.547–45.066–25.025–10.221–7.764–6.014–4.706–4.169–3.693–3.267–2.884–2.539–2.226–1.940–1.679–1.440–1.219–1.015–0.825–0.649–0.485–0.332–0.189–0.054+0.071+0.190Source: Based on data from the JANAF Thermochemical Tables, NSRDS-NBS-37, 1971.© 1999 by CRC Press LLC–5.018–2.139–0.159+0.135+0.333+0.474+0.530+0.577+0.619+0.656+0.688+0.716+0.742+0.764+0.784+0.802+0.818+0.833+0.846+0.858+0.869+0.878+0.888+0.895+0.902Temp(°R)53790018002160252028803060324034203600378039604140432045004680486050405220540055805760594061206300Engineering Thermodynamics2-692.5 Exergy AnalysisThe method of exergy analysis (availability analysis) presented in this section enables the location,cause, and true magnitude of energy resource waste and loss to be determined.

Such information canbe used in the design of new energy-efficient systems and for improving the performance of existingsystems. Exergy analysis also provides insights that elude a purely first-law approach. For example, onthe basis of first-law reasoning alone, the condenser of a power plant may be mistakenly identified asthe component primarily responsible for the plant’s seemingly low overall performance. An exergyanalysis correctly reveals not only that the condenser loss is relatively unimportant (see the last tworows of the Rankine cycle values of Table 2.15), but also that the steam generator is the principal siteof thermodynamic inefficiency owing to combustion and heat transfer irreversibilities within it.When exergy concepts are combined with principles of engineering economy, the result is known asthermoeconomics.

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