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- 1. 16-1 Chapter 16 CHEMICAL AND PHASE EQUILIBRIUMThe Kp and Equilibrium Composition of Ideal Gases16-1C Because when a reacting system involves heat transfer, the increase-in-entropy principle relationrequires a knowledge of heat transfer between the system and its surroundings, which is impractical. Theequilibrium criteria can be expressed in terms of the properties alone when the Gibbs function is used.16-2C No, the wooden table is NOT in chemical equilibrium with the air. With proper catalyst, it will reachwith the oxygen in the air and burn.16-3C They are ν ν Δν N CC N ν D ⎛ P v PC C PD D ⎞ Kp = ν , K p = e − ΔG*(T ) / RuT and K p = D ⎜ ⎜ ⎟ ⎟ v PA A PB B N ν A N ν B ⎝ N total A B ⎠where Δν = ν C + ν D −ν A −ν B . The first relation is useful in partial pressure calculations, the second indetermining the Kp from gibbs functions, and the last one in equilibrium composition calculations.16-4C (a) This reaction is the reverse of the known CO reaction. The equilibrium constant is then 1/ KP(b) This reaction is the reverse of the known CO reaction at a different pressure. Since pressure has noeffect on the equilibrium constant, 1/ KP(c) This reaction is the same as the known CO reaction multiplied by 2. The quilibirium constant is then 2 KP(d) This is the same as reaction (c) occurring at a different pressure. Since pressure has no effect on theequilibrium constant, 2 KPPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 2. 16-216-5C (a) This reaction is the reverse of the known H2O reaction. The equilibrium constant is then 1/ KP(b) This reaction is the reverse of the known H2O reaction at a different pressure. Since pressure has noeffect on the equilibrium constant, 1/ KP(c) This reaction is the same as the known H2O reaction multiplied by 3. The quilibirium constant is then 3 KP(d) This is the same as reaction (c) occurring at a different pressure. Since pressure has no effect on theequilibrium constant, 3 KP16-6C (a) No, because Kp depends on temperature only.(b) In general, the total mixture pressure affects the mixture composition. The equilibrium constant for thereaction N 2 + O 2 ⇔ 2NO can be expressed as (ν NO −ν N 2 −ν O 2 ) N ν NO ⎛ P ⎞ Kp = NO ⎜ ⎟ ν ν ⎜N ⎟ N NN 2 N OO 2 ⎝ total ⎠ 2 2The value of the exponent in this case is 2-1-1 = 0. Therefore, changing the total mixture pressure willhave no effect on the number of moles of N2, O2 and NO.16-7C (a) The equilibrium constant for the reaction CO + 1 O 2 ⇔ CO 2 can be expressed as 2 ν (ν CO 2 −ν CO −ν O 2 ) N CO 2 CO ⎛ P ⎞ Kp = 2 ⎜ ⎟ ν ν ⎜N ⎟ N CO N OO 2 CO ⎝ total ⎠ 2Judging from the values in Table A-28, the Kp value for this reaction decreases as temperature increases.That is, the indicated reaction will be less complete at higher temperatures. Therefore, the number of molesof CO2 will decrease and the number moles of CO and O2 will increase as the temperature increases.(b) The value of the exponent in this case is 1-1-0.5=-0.5, which is negative. Thus as the pressureincreases, the term in the brackets will decrease. The value of Kp depends on temperature only, andtherefore it will not change with pressure. Then to keep the equation balanced, the number of moles of theproducts (CO2) must increase, and the number of moles of the reactants (CO, O2) must decrease.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 3. 16-316-8C (a) The equilibrium constant for the reaction N 2 ⇔ 2N can be expressed as ν (ν N −ν N 2 ) N NN ⎛ P ⎞ Kp = ⎜ ⎟ ν ⎜ ⎟ N NN 2 ⎝ N total ⎠ 2Judging from the values in Table A-28, the Kp value for this reaction increases as the temperature increases.That is, the indicated reaction will be more complete at higher temperatures. Therefore, the number ofmoles of N will increase and the number moles of N2 will decrease as the temperature increases.(b) The value of the exponent in this case is 2-1 = 1, which is positive. Thus as the pressure increases, theterm in the brackets also increases. The value of Kp depends on temperature only, and therefore it will notchange with pressure. Then to keep the equation balanced, the number of moles of the products (N) mustdecrease, and the number of moles of the reactants (N2) must increase.16-9C The equilibrium constant for the reaction CO + 1 O 2 ⇔ CO 2 can be expressed as 2 ν (ν CO 2 −ν CO −ν O 2 ) N CO 2 CO ⎛ P ⎞ Kp = 2 ⎜ ⎟ ν ν ⎜N ⎟ N CO N OO 2 CO ⎝ total ⎠ 2Adding more N2 (an inert gas) at constant temperature and pressure will increase Ntotal but will have nodirect effect on other terms. Then to keep the equation balanced, the number of moles of the products (CO2)must increase, and the number of moles of the reactants (CO, O2) must decrease.16-10C The values of the equilibrium constants for each dissociation reaction at 3000 K are, from TableA-28, N 2 ⇔ 2N ⇔ ln K p = −22.359 H 2 ⇔ 2H ⇔ ln K p = −3.685 (greater than - 22.359)Thus H2 is more likely to dissociate than N2.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 4. 16-416-11 The mole fractions of the constituents of an ideal gas mixture is given. The Gibbs function of the COin this mixture at the given mixture pressure and temperature is to be determined.Analysis From Tables A-21 and A-26, at 1 atm pressure, [g * (800 K, 1 atm) = g o + Δ h (T ) − Ts o (T ) f ] = −137,150 + (23,844 − 800 × 227.162) − (8669 − 298 × 197.543) 10% CO2 = −244,837 kJ/kmol 60% H2O 30% COThe partial pressure of CO is 10 atm 800 K PCO = y CO P = (0.30)(10 atm) = 3 atmThe Gibbs function of CO at 800 K and 3 atm is g (800 K, 3 atm) = g * (800 K, 1 atm) + Ru T ln PCO = −244,837 kJ/kmol + (8.314 kJ/kmol)(800 K)ln(3 atm) = −237,530 kJ/kmol16-12 The partial pressures of the constituents of an ideal gas mixture is given. The Gibbs function of thenitrogen in this mixture at the given mixture pressure and temperature is to be determined.Analysis The partial pressure of nitrogen is PN2 = 130 kPa = (130 / 101.325) = 1.283 atm N2 ,CO2, NOThe Gibbs function of nitrogen at 298 K and 3 atm is PN2 = 130 kPa g (800 K, 3 atm) = g * (298 K, 1 atm) + Ru T ln PN2 298 K = 0 + (8.314 kJ/kmol)(298 K)ln(1.283 atm) = 617.4 kJ/kmolPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 5. 16-516-13E The equilibrium constant of the reaction H 2 O ⇔ H 2 + 1 O 2 is to be determined using Gibbs 2function.Analysis (a) The Kp value of a reaction at a specified temperature can be determined from the Gibbsfunction data using K p = e −ΔG*( T )/ Ru T or ln K p = − ΔG *(T ) / Ru Twhere H2O ↔ H2 + ½O2 ∗ ∗ ∗ ΔG * (T ) = ν H2 g H2 (T ) +ν O2 g O2 (T ) −ν H2O g H2O (T ) 1440 RAt 1440 R, ∗ ∗ ∗ ΔG * (T ) = ν H2 g H2 (T ) + ν O2 g O2 (T ) −ν H2O g H2O (T ) = ν H2 (h − Ts ) H2 + ν O2 (h − Ts ) O2 −ν H2O (h − Ts ) H2O = ν H2 [(h f + h1440 − h537 ) − Ts ] H2 + ν O2 [(h f + h1440 − h537 ) − Ts ] O2 −ν H2O [(h f + h1440 − h537 ) − Ts ] H2O = 1× (0 + 9956.9 − 3640.3 − 1440 × 38.079) + 0.5 × (0 + 10,532.0 − 3725.1 − 1440 × 56.326) − 1× (−104,040 + 11,933.4 − 4258 − 1440 × 53.428) = 87,632 Btu/lbmolSubstituting, ln K p = −(87,632 Btu/lbmol)/[(1.986 Btu/lbmol.R)(1440 R)] = −30.64or K p = 4.93 × 10 −14 (Table A - 28 : ln K p = −34.97 by interpolation )At 3960 R, ∗ ∗ ∗ ΔG * (T ) = ν H2 g H2 (T ) +ν O2 g O2 (T ) −ν H2O g H2O (T ) = ν H2 (h − Ts ) H2 + ν O2 (h − Ts ) O 2 −ν H2O (h − Ts ) H2O = ν H2 [(h f + h3960 − h537 ) − Ts ] H2 + ν O2 [(h f + h3960 − h537 ) − Ts ] O2 −ν H2O [(h f + h3960 − h537 ) − Ts ] H2O = 1× (0 + 29,370.5 − 3640.3 − 3960 × 45.765) + 0.5 × (0 + 32,441 − 3725.1 − 3960 × 65.032) − 1× (−104,040 + 39,989 − 4258 − 3960 × 64.402) = 53,436 Btu/lbmolSubstituting, ln K p = −(53,436 Btu/lbmol)/[(1.986 Btu/lbmol.R)(3960 R)] = −6.79or K p = 1.125 × 10 −3 (Table A - 28 : ln K p = −6.768)PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 6. 16-616-14 The reaction 2H 2 O ⇔ 2H 2 + O 2 is considered. The mole fractions of the hydrogen and oxygenproduced when this reaction occurs at 4000 K and 10 kPa are to be determined.Assumptions 1 The equilibrium composition consists of H2O, H2, and O2. 2 The constituents of the mixtureare ideal gases.Analysis The stoichiometric and actual reactions in this case areStoichiometric: 2H 2 O ⇔ 2H 2 + O 2 (thus ν H2O = 2, ν H2 = 2, and ν O2 = 1)Actual: 2H 2 O ⎯ xH 2 O + yH 2 + zO 2 ⎯→ 1 3 14 4 2 2 3 react. productsH balance: 4 = 2x + 2 y ⎯ ⎯→ y = 2 − x 2H2O 4000 KO balance: 2 = x + 2z ⎯ ⎯→ z = 1 − 0.5 x 10 kPaTotal number of moles: N total = x + y + z = 3 − 0.5 xThe equilibrium constant relation can be expressed as ν ν (ν H2 +ν O2 −ν H2O ) N H2 N O2 ⎛ P H2 O2 ⎞ Kp = ⎜ ⎟ ν H2O ⎜ ⎟ N H2O ⎝ N total ⎠From Table A-28, ln K p = −0.542 at 4000 K . Since the stoichiometric reaction being considered isdouble this reaction, K p = exp( −2 × 0.542) = 0.3382Substituting, 2 +1− 2 (2 − x) 2 (1 − 0.5 x) ⎛ 10 / 101.325 ⎞ 0.3382 = ⎜ ⎟ x2 ⎝ 3 − 0.5 x ⎠Solving for x, x = 0.4446Then, y = 2 − x = 1.555 z = 1 − 0.5x = 0.7777Therefore, the equilibrium composition of the mixture at 4000 K and 10 kPa is 0.4446 H 2 O + 1.555 H 2 + 0.7777 O 2The mole fractions of hydrogen and oxygen produced are N H2 1.555 1.555 y H2 = = = = 0.560 N total 3 − 0.5 × 0.4446 2.778 N O2 0.7777 y O2 = = = 0.280 N total 2.778PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 7. 16-716-15 The reaction 2H 2 O ⇔ 2H 2 + O 2 is considered. The mole fractions of hydrogen gas produced is tobe determined at 100 kPa and compared to that at 10 kPa.Assumptions 1 The equilibrium composition consists of H2O, H2, and O2. 2 The constituents of the mixtureare ideal gases.Analysis The stoichiometric and actual reactions in this case areStoichiometric: 2H 2 O ⇔ 2H 2 + O 2 (thus ν H2O = 2, ν H2 = 2, and ν O2 = 1)Actual: 2H 2 O ⎯ xH 2 O + yH 2 + zO 2 ⎯→ 1 3 14 4 2 2 3 react. productsH balance: 4 = 2x + 2 y ⎯ ⎯→ y = 2 − x 2H2O 4000 KO balance: 2 = x + 2z ⎯ ⎯→ z = 1 − 0.5 x 100 kPaTotal number of moles: N total = x + y + z = 3 − 0.5 xThe equilibrium constant relation can be expressed as ν ν (ν H2 +ν O2 −ν H2O ) N H2 N O2 ⎛ P H2 O2 ⎞ Kp = ⎜ ⎟ ν H2O ⎜ ⎟ N H2O ⎝ N total ⎠From Table A-28, ln K p = −0.542 at 4000 K . Since the stoichiometric reaction being considered isdouble this reaction, K p = exp( −2 × 0.542) = 0.3382Substituting, 2 +1− 2 (2 − x) 2 (1 − 0.5 x ) ⎛ 100 / 101.325 ⎞ 0.3382 = ⎜ ⎟ x2 ⎝ 3 − 0.5 x ⎠Solving for x, x = 0.8870Then, y = 2 − x = 1.113 z = 1 − 0.5x = 0.5565Therefore, the equilibrium composition of the mixture at 4000 K and 100 kPa is 0.8870 H 2 O + 1.113 H 2 + 0.5565 O 2That is, there are 1.113 kmol of hydrogen gas. The mole number of hydrogen at 10 kPa reaction pressurewas obtained in the previous problem to be 1.555 kmol. Therefore, the amount of hydrogen gas producedhas decreased.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 8. 16-816-16 The reaction 2H 2 O ⇔ 2H 2 + O 2 is considered. The mole number of hydrogen gas produced is to bedetermined if inert nitrogen is mixed with water vapor is to be determined and compared to the case withno inert nitrogen.Assumptions 1 The equilibrium composition consists of H2O, H2, O2, and N2. 2 The constituents of themixture are ideal gases.Analysis The stoichiometric and actual reactions in this case areStoichiometric: 2H 2 O ⇔ 2H 2 + O 2 (thus ν H2O = 2, ν H2 = 2, and ν O2 = 1)Actual: 2H 2 O + 0.5N 2 ⎯ xH 2 O + yH 2 + zO 2 + 0.5N 2 ⎯→ 1 3 14 4 2 2 3 1 3 2 react. products inertH balance: 4 = 2x + 2 y ⎯ ⎯→ y = 2 − x 2H2O, 0.5N2 4000 KO balance: 2 = x + 2z ⎯ ⎯→ z = 1 − 0.5 x 10 kPaTotal number of moles: N total = x + y + z + 0.5 = 3.5 − 0.5 xThe equilibrium constant relation can be expressed as ν ν (ν H2 +ν O2 −ν H2O ) N H2 N O2 ⎛ P H2 O2 ⎞ Kp = ⎜ ⎟ ν H2O ⎜ ⎟ N H2O ⎝ N total ⎠From Table A-28, ln K p = −0.542 at 4000 K . Since the stoichiometric reaction being considered isdouble this reaction, K p = exp(−2 × 0.542) = 0.3382Substituting, 2 +1− 2 (2 − x) 2 (1 − 0.5 x) ⎛ 10 / 101.325 ⎞ 0.3382 = ⎜ ⎟ x2 ⎝ 3.5 − 0.5 x ⎠Solving for x, x = 0.4187Then, y = 2 − x = 1.581 z = 1 − 0.5x = 0.7907Therefore, the equilibrium composition of the mixture at 4000 K and 10 kPa is 0.4187 H 2 O + 1.581 H 2 + 0.7907 O 2That is, there are 1.581 kmol of hydrogen gas. The mole number of hydrogen without inert nitrogen casewas obtained in Prob. 16-14 to be 1.555 kmol. Therefore, the amount of hydrogen gas produced hasincreased.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 9. 16-916-17E The temperature at which 15 percent of diatomic oxygen dissociates into monatomic oxygen at twopressures is to be determined.Assumptions 1 The equilibrium composition consists of O2 and O. 2 The constituents of the mixture areideal gases.Analysis (a) The stoichiometric and actual reactions can be written asStoichiometric: O 2 ⇔ 2O (thus ν O2 = 1 and ν O = 2)Actual: O 2 ⇔ 0.85O 2 + 0.3O 123 { 4 4 react. prod. O2 ↔ 2OThe equilibrium constant Kp can be determined from 15 % 3 psia ν ν O −ν O2 2 −1 N OO ⎛ P ⎞ 0.3 2 ⎛ 3 / 14.696 ⎞ Kp = ⎜ ⎟ = ⎜ ⎟ = 0.01880 ν O2 ⎜ ⎟ 0.85 ⎝ 0.85 + 0.3 ⎠ N O2 ⎝ N total ⎠and ln K p = −3.974From Table A-28, the temperature corresponding to this lnKp value is T = 3060 K = 5508 R(b) At 100 psia, ν O −ν O2 2 −1 Nν O ⎛ P ⎞ 0.32 ⎛ 100 / 14.696 ⎞ Kp = ν ⎜ O ⎟ = ⎜ ⎟ = 0.6265 N O2 ⎜ N total ⎟ O2 ⎝ ⎠ 0.85 ⎝ 0.85 + 0.3 ⎠ ln K p = −0.4676From Table A-28, the temperature corresponding to this lnKp value is T = 3701 K = 6662 RPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 10. 16-1016-18 The dissociation reaction CO2 ⇔ CO + O is considered. The composition of the products at givenpressure and temperature is to be determined.Assumptions 1 The equilibrium composition consists of CO2, CO, and O. 2 The constituents of the mixtureare ideal gases.Analysis For the stoichiometric reaction CO 2 ⇔ CO + 1 O 2 , from Table A-28, at 2500 K 2 ln K p = −3.331For the oxygen dissociation reaction 0.5O 2 ⇔ O , from Table A-28, at 2500 K, ln K p = −8.509 / 2 = −4.255For the desired stoichiometric reaction CO 2 ⇔ CO + O (thus ν CO2 = 1, ν CO = 1 and ν O = 1) , ln K p = −3.331 − 4.255 = −7.586and K p = exp( −7.586) = 0.0005075 CO2Actual: CO 2 ⎯ xCO 2 + yCO + zO ⎯→ 1 3 1 24 2 4 3 2500 K react. products 1 atmC balance: 1= x+ y ⎯ ⎯→ y = 1 − xO balance: 2 = 2x + y + z ⎯ ⎯→ z = 1 − xTotal number of moles: N total = x + y + z = 2 − xThe equilibrium constant relation can be expressed as ν ν ν CO +ν O −ν CO2 N CO N O ⎛ P ⎞ K p = COν O ⎜ ⎟ ⎜ ⎟ N CO2 ⎝ N total CO2 ⎠Substituting, 1+1−1 (1 − x)(1 − x) ⎛ 1 ⎞ 0.0005075 = ⎜ ⎟ x ⎝ 2− x⎠Solving for x, x = 0.9775Then, y = 1 − x = 0.0225 z = 1 − x = 0.0225Therefore, the equilibrium composition of the mixture at 2500 K and 1 atm is 0.9775 CO 2 + 0.0225 CO + 0.0225 OPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 11. 16-1116-19 The dissociation reaction CO2 ⇔ CO + O is considered. The composition of the products at givenpressure and temperature is to be determined when nitrogen is added to carbon dioxide.Assumptions 1 The equilibrium composition consists of CO2, CO, O, and N2. 2 The constituents of themixture are ideal gases.Analysis For the stoichiometric reaction CO 2 ⇔ CO + 1 O 2 , from Table A-28, at 2500 K 2 ln K p = −3.331For the oxygen dissociation reaction 0.5O 2 ⇔ O , from Table A-28, at 2500 K, ln K p = −8.509 / 2 = −4.255For the desired stoichiometric reaction CO 2 ⇔ CO + O (thus ν CO2 = 1, ν CO = 1 and ν O = 1) , ln K p = −3.331 − 4.255 = −7.586and K p = exp( −7.586) = 0.0005075 CO2, 3N2Actual: CO 2 + 3N 2 ⎯ xCO 2 + yCO + zO + 3N 2 ⎯→ 2500 K 1 3 1 24 2 4 3 { react. products inert 1 atmC balance: 1= x+ y ⎯ ⎯→ y = 1 − xO balance: 2 = 2x + y + z ⎯ ⎯→ z = 1 − xTotal number of moles: N total = x + y + z + 3 = 5 − xThe equilibrium constant relation can be expressed as ν ν ν CO +ν O −ν CO2 N CO N O ⎛ P ⎞ K p = COν O ⎜ ⎟ ⎜ ⎟ N CO2 ⎝ N total CO2 ⎠Substituting, 1+1−1 (1 − x)(1 − x) ⎛ 1 ⎞ 0.0005075 = ⎜ ⎟ x ⎝ 5− x ⎠Solving for x, x = 0.9557Then, y = 1 − x = 0.0443 z = 1 − x = 0.0443Therefore, the equilibrium composition of the mixture at 2500 K and 1 atm is 0.9557 CO 2 + 0.0443 CO + 0.0443 O + 3N 2PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 12. 16-1216-20 It is to be shown that as long as the extent of the reaction, α, for the disassociation reaction X2 ⇔ 2X KPis smaller than one, α is given by α = 4+ KPAssumptions The reaction occurs at the reference temperature.Analysis The stoichiometric and actual reactions can be written asStoichiometric: X 2 ⇔ 2X (thus ν X2 = 1 and ν X = 2)Actual: X 2 ⇔ (1 − α )X 2 + 2αX 1 24 { 4 3 react. prod.The equilibrium constant Kp is given by ν ν X −ν X2 2 −1 N XX ⎛ P ⎞ (2α ) 2 ⎛ 1 ⎞ 4α 2 Kp = ⎜ ⎟ = ⎜ ⎟ = ν X2 ⎜ ⎟ (1 − α ) ⎝ α + 1 ⎠ (1 − α )(1 + α ) N X2 ⎝ N total ⎠Solving this expression for α gives KP α= 4+ KPPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 13. 16-1316-21 A gaseous mixture consisting of methane and nitrogen is heated. The equilibrium composition (bymole fraction) of the resulting mixture is to be determined.Assumptions 1 The equilibrium composition consists of CH4, C, H2, and N2. 2 The constituents of themixture are ideal gases.Analysis The stoichiometric and actual reactions in this case areStoichiometric: CH 4 ⇔ C + 2H 2 (thus ν CH4 = 1, ν C = 1, and ν H2 = 2)Actual: CH 4 + N 2 ⎯ xCH 4 + yC + zH 2 + N 2 ⎯→ 1 3 1 24 { 2 4 3 react. products inertC balance: 1= x+ y ⎯ ⎯→ y = 1 − x CH4, N2H balance: 4 = 4x + 2z ⎯ ⎯→ z = 2 − 2 x 1000 KTotal number of moles: N total = x + y + z + 1 = 4 − 2 x 1 atmThe equilibrium constant relation can be expressed as ν ν CH4 −ν C −ν H2 N CH4 CH4 ⎛ P ⎞ Kp = ⎜ ⎟ ν ν ⎜N ⎟ N CC N H2 H2 ⎝ total ⎠From the problem statement at 1000 K, ln K p = 2.328 . Then, K p = exp(2.328) = 10.257For the reverse reaction that we consider, K p = 1 / 10.257 = 0.09749Substituting, 1−1− 2 x ⎛ 1 ⎞ 0.09749 = ⎜ 2 4 − 2x ⎟ (1 − x)(2 − 2 x) ⎝ ⎠Solving for x, x = 0.02325Then, y = 1 − x = 0.9768 z = 2 − 2x = 1.9535Therefore, the equilibrium composition of the mixture at 1000 K and 1 atm is 0.02325 CH 4 + 0.9768 C + 1.9535 H 2 + 1 N 2The mole fractions are N CH4 0.02325 0.02325 y CH4 = = = = 0.0059 N total 4 − 2 × 0.02325 3.954 NC 0.9768 yC = = = 0.2470 N total 3.954 N H2 1.9535 y H2 = = = 0.4941 N total 3.954 N N2 1 y N2 = = = 0.2529 N total 3.954PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 14. 16-1416-22 The reaction N2 + O2 ⇔ 2NO is considered. The equilibrium mole fraction of NO 1000 K and 1 atmis to be determined.Assumptions 1 The equilibrium composition consists of N2, O2, and NO. 2 The constituents of the mixtureare ideal gases.Analysis The stoichiometric and actual reactions in this case areStoichiometric: N 2 + O 2 ⇔ 2 NO (thus ν N2 = 1, ν O2 = 1, and ν NO = 2)Actual: N 2 + O 2 ⎯ xN 2 + yO 2 + { ⎯→ zNO 14243 products react.N balance: 2 = 2x + z ⎯ ⎯→ z = 2 − 2 x N2, O2O balance: 2 = 2y + z ⎯ ⎯→ y = x 1000 K 1 atmTotal number of moles: N total = x + y + z = 2The equilibrium constant relation can be expressed as ν (ν NO −ν N2 −ν O2 ) N NO NO ⎛ P ⎞ Kp = ⎜ ⎟ ν ν ⎜N ⎟ N N2 N O2 N2 O2 ⎝ total ⎠From Table A-28, at 1000 K, ln K p = −9.388 . Since the stoichiometric reaction being considered is doublethis reaction, K p = exp(−2 × 9.388) = 7.009 × 10 −9Substituting, 2 −1−1 (2 − 2 x) 2 ⎛ 1 ⎞ 7.009 × 10 −9 = ⎜ ⎟ x2 ⎝2⎠Solving for x, x = 0.999958Then, y = x = 0.999958 z = 2 − 2x = 8.4×10-5Therefore, the equilibrium composition of the mixture at 1000 K and 1 atm is 0.999958 N 2 + 0.999958 O 2 + 8.4 × 10 −5 NOThe mole fraction of NO is then N NO 8.4 × 10 −5 y NO = = = 4.2 × 10 −5 (42 parts per million) N total 2PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 15. 16-1516-23 Oxygen is heated from a specified state to another state. The amount of heat required is to bedetermined without and with dissociation cases.Assumptions 1 The equilibrium composition consists of O2 and O. 2 The constituents of the mixture areideal gases.Analysis (a) Obtaining oxygen properties from table A-19, an energy balance gives E in − E out = ΔE system 1 24 4 3 1 24 4 3 Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies q in = u 2 − u1 = 57,192 − 6203 O2 = 50,989 kJ/kmol 2200 K(b) The stoichiometric and actual reactions in this case are 1 atmStoichiometric: O 2 ⇔ 2O (thus ν O2 = 1 and ν O = 2)Actual: O2 ⎯ { + { ⎯→ xO 2 yO react. productsO balance: 2 = 2x + y ⎯ ⎯→ y = 2 − 2 xTotal number of moles: N total = x + y = 2 − xThe equilibrium constant relation can be expressed as ν ν O −ν O2 N OO ⎛ P ⎞ Kp = ν ⎜ ⎟ O2 ⎜ ⎟ N O2 ⎝ N total ⎠From Table A-28, at 2200 K, ln K p = −11.827 . Then, K p = exp(−11.827) = 7.305 × 10 −6Substituting, 2 −1 (2 − 2 x) 2 ⎛ 1 ⎞ 7.305 × 10 − 6 = ⎜ ⎟ x ⎝ 2− x⎠Solving for x, x = 0.99865Then, y = 2 − 2x = 0.0027Therefore, the equilibrium composition of the mixture at 2200 K and 1 atm is 0.99865 O 2 + 0.0027 OHence, the oxygen ions are negligible and the result is same as that in part (a), q in = 50,989 kJ/kmolPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 16. 16-1616-24 Air is heated from a specified state to another state. The amount of heat required is to be determinedwithout and with dissociation cases.Assumptions 1 The equilibrium composition consists of O2 and O, and N2. 2 The constituents of themixture are ideal gases.Analysis (a) Obtaining air properties from table A-17, an energy balance gives E in − E out = ΔE system 1 24 4 3 1 24 4 3 Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies q in = u 2 − u1 = 1872.4 − 212.64 O2, 3.76N2 = 1660 kJ/kg 2200 K(b) The stoichiometric and actual reactions in this case are 1 atmStoichiometric: O 2 ⇔ 2O (thus ν O2 = 1 and ν O = 2)Actual: O 2 + 3.76 N 2 ⎯ { + { + 3.76 N 2 ⎯→ xO 2 yO 123 4 4 react. products inertO balance: 2 = 2x + y ⎯ ⎯→ y = 2 − 2 xTotal number of moles: N total = x + y + 3.76 = 5.76 − xThe equilibrium constant relation can be expressed as ν ν O −ν O2 N OO ⎛ P ⎞ Kp = ν ⎜ ⎟ O2 ⎜ ⎟ N O2 ⎝ N total ⎠From Table A-28, at 2200 K, ln K p = −11.827 . Then, K p = exp(−11.827) = 7.305 × 10 −6Substituting, 2 −1 (2 − 2 x) 2 ⎛ 1 ⎞ 7.305 × 10 − 6 = ⎜ ⎟ x ⎝ 5.76 − x ⎠Solving for x, x = 0.99706Then, y = 2 − 2x = 0.00588Therefore, the equilibrium composition of the mixture at 2200 K and 1 atm is 0.99706 O 2 + 0.00588 O + 3.76 N 2Hence, the atomic oxygen is negligible and the result is same as that in part (a), q in = 1660 kJ/kgPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 17. 16-1716-25 The equilibrium constant of the reaction H2 + 1/2O2 ↔ H2O is listed in Table A-28 at differenttemperatures. The data are to be verified at two temperatures using Gibbs function data.Analysis (a) The Kp value of a reaction at a specified temperature can be determined from the Gibbsfunction data using K p = e −ΔG*( T )/ Ru T or ln K p = − ΔG * (T ) / Ru Twhere H2 + ½O2 ↔ H2O ∗ ∗ ∗ ΔG * (T ) = ν H 2O g H 2 O (T ) −ν H 2 g H 2 (T ) −ν O 2 g O 2 (T ) 25ºCAt 25°C, ΔG *(T ) = 1( −228,590) − 1(0) − 0.5(0) = −228,590 kJ / kmolSubstituting, ln K p = −(−228,590 kJ/kmol)/[(8.314 kJ/kmol ⋅ K)(298 K)] = 92.26or K p = 1.12 × 10 40 (Table A - 28: ln K p = 92.21)(b) At 2000 K, ∗ ∗ ∗ ΔG * (T ) = ν H 2 O g H 2 O (T ) −ν H 2 g H 2 (T ) −ν O 2 g O 2 (T ) = ν H 2 O ( h − T s ) H 2 O − ν H 2 ( h − Ts ) H 2 − ν O 2 ( h − Ts ) O 2 = ν H 2 O [(h f + h2000 − h298 ) − Ts ] H 2O −ν H 2 [(h f + h2000 − h298 ) − Ts ] H 2 −ν O 2 [(h f + h2000 − h298 ) − Ts ] O 2 = 1× (−241,820 + 82,593 − 9904 − 2000 × 264.571) − 1× (0 + 61,400 − 8468 − 2000 × 188.297) − 0.5 × (0 + 67,881 − 8682) − 2000 × 268.655) = −135,556 kJ/kmolSubstituting, ln K p = −(−135,556 kJ/kmol)/[(8.314 kJ/kmol ⋅ K)(2000 K)] = 8.152or K p = 3471 (Table A - 28 : ln K p = 8.145)PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 18. 16-1816-26E The equilibrium constant of the reaction H2 + 1/2O2 ↔ H2O is listed in Table A-28 at differenttemperatures. The data are to be verified at two temperatures using Gibbs function data.Analysis (a) The Kp value of a reaction at a specified temperature can be determined from the Gibbsfunction data using K p = e −ΔG*( T )/ Ru T or ln K p = − ΔG * (T ) / Ru Twhere H2 + ½O2 ↔ H2O ∗ ∗ ∗ ΔG * (T ) = ν H 2O g H 2 O (T ) −ν H 2 g H 2 (T ) −ν O 2 g O 2 (T ) 537 RAt 537 R, ΔG *(T ) = 1( −98,350) − 1(0) − 0.5(0) = −98,350 Btu / lbmolSubstituting, ln K p = − ( −98,350 Btu / lbmol) / [(1.986 Btu / lbmol ⋅ R)(537 R)] = 92.22or K p = 1.12 × 10 40 (Table A - 28: ln K p = 92.21)(b) At 3240 R, ∗ ∗ ∗ ΔG * (T ) = ν H 2 O g H 2 O (T ) −ν H 2 g H 2 (T ) −ν O 2 g O 2 (T ) = ν H 2 O ( h − T s ) H 2 O − ν H 2 ( h − Ts ) H 2 − ν O 2 ( h − Ts ) O 2 = ν H 2 O [(h f + h3240 − h537 ) − Ts ] H 2O −ν H 2 [(h f + h3240 − h298 ) − Ts ] H 2 −ν O 2 [(h f + h3240 − h298 ) − Ts ] O 2 = 1× (−104,040 + 31,204.5 − 4258 − 3240 × 61.948) − 1× (0 + 23,484.7 − 3640.3 − 3240 × 44.125) − 0.5 × (0 + 25,972 − 3725.1 − 3240 × 63.224) = −63,385 Btu/lbmolSubstituting, ln K p = −(−63,385 Btu/lbmol)/[(1.986 Btu/lbmol.R)(3240 R)] = 9.85or K p = 1.90 × 104 (Table A - 28: ln K p = 9.83)PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 19. 16-1916-27 The equilibrium constant of the reaction CO + 1/2O2 ↔ CO2 at 298 K and 2000 K are to bedetermined, and compared with the values listed in Table A-28.Analysis (a) The Kp value of a reaction at a specified temperature can be determined from the Gibbsfunction data using K p = e −ΔG*( T )/ Ru T or ln K p = − ΔG * (T ) / Ru Twhere CO + 1 O 2 ⇔ CO 2 2 ∗ ∗ ∗ ΔG * (T ) = ν CO2 g CO2 (T ) −ν CO g CO (T ) −ν O2 g O2 (T ) 298 KAt 298 K, ΔG * (T ) = 1(−394,360) − 1(−137,150) − 0.5(0) = −257,210 kJ/kmolwhere the Gibbs functions are obtained from Table A-26. Substituting, (−257,210 kJ/kmol) ln K p = − = 103.81 (8.314 kJ/kmol ⋅ K)(298 K)From Table A-28: ln K p = 103.76(b) At 2000 K, ∗ ∗ ∗ ΔG * (T ) = ν CO2 g CO2 (T ) −ν CO g CO (T ) −ν O2 g O2 (T ) = ν CO2 (h − Ts ) CO2 −ν CO (h − Ts ) CO −ν O2 (h − Ts ) O2 = 1[(−302,128) − (2000)(309.00)] − 1[(−53,826) − (2000)(258.48)] − 0.5[(59,193) − (2000)(268.53)] = −110,409 kJ/kmolThe enthalpies at 2000 K and entropies at 2000 K and 101.3 kPa (1 atm) are obtained from EES.Substituting, (−110,409 kJ/kmol) ln K p = − = 6.64 (8.314 kJ/kmol ⋅ K)(2000 K)From Table A-28: ln K p = 6.635PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 20. 16-2016-28 EES The effect of varying the percent excess air during the steady-flow combustion of hydrogen isto be studied.Analysis The combustion equation of hydrogen with stoichiometric amount of air is H 2 + 0.5[O 2 + 3.76N 2 ] ⎯ ⎯→ H 2 O + 0.5(3.76) N 2For the incomplete combustion with 100% excess air, the combustion equation is H 2 + (1 + Ex)(0.5)[O 2 + 3.76N 2 ] ⎯ ⎯→ 0.97 H 2 O + a H 2 + b O 2 + c N 2The coefficients are to be determined from the mass balancesHydrogen balance: 2 = 0.97 × 2 + a × 2 ⎯ ⎯→ a = 0.03Oxygen balance: (1 + Ex) × 0.5 × 2 = 0.97 + b × 2Nitrogen balance: (1 + Ex) × 0.5 × 3.76 × 2 = c × 2Solving the above equations, we find the coefficients (Ex = 1, a = 0.03 b = 0.515, c = 3.76) and write thebalanced reaction equation as H 2 + [O 2 + 3.76N 2 ] ⎯ ⎯→ 0.97 H 2 O + 0.03 H 2 + 0.515 O 2 + 3.76 N 2Total moles of products at equilibrium are N tot = 0.97 + 0.03 + 0.515 + 3.76 = 5.275The assumed equilibrium reaction is H 2O ←⎯→ H 2 + 0.5O 2The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K p = e −ΔG*( T )/ Ru T or ln K p = − ΔG *(T ) / Ru Twhere ∗ ∗ ∗ ΔG * (T ) = ν H2 g H2 (Tprod ) + ν O2 g O2 (Tprod ) −ν H2O g H2O (Tprod )and the Gibbs functions are defined as ∗ g H2 (Tprod ) = (h − Tprod s ) H2 ∗ g O2 (Tprod ) = (h − Tprod s ) O2 ∗ g H2O (Tprod ) = (h − Tprod s ) H2OThe equilibrium constant is also given by 1+ 0.5 −1 0.5 ⎛ P ⎞ ab 0.5 ⎛ 1 ⎞ (0.03)(0.515) 0.5 Kp =⎜ ⎜N ⎟ ⎟ 1 =⎜ ⎟ = 0.009664 ⎝ tot ⎠ 0.97 ⎝ 5.275 ⎠ 0.97and ln K p = ln(0.009664) = −4.647The corresponding temperature is obtained solving the above equations using EES to be Tprod = 2600 KThis is the temperature at which 97 percent of H2 will burn into H2O. The copy of EES solution is givennext.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 21. 16-21"Input Data from parametric table:"{PercentEx = 10}Ex = PercentEx/100 "EX = % Excess air/100"P_prod =101.3"[kPa]"R_u=8.314 "[kJ/kmol-K]""The combustion equation of H2 with stoichiometric amount of air isH2 + 0.5(O2 + 3.76N2)=H2O +0.5(3.76)N2""For the incomplete combustion with 100% excess air, the combustion equation isH2 + (1+EX)(0.5)(O2 + 3.76N2)=0.97 H2O +aH2 + bO2+cN2""Specie balance equations give the values of a, b, and c.""H, hydrogen"2 = 0.97*2 + a*2"O, oxygen"(1+Ex)*0.5*2=0.97 + b*2"N, nitrogen"(1+Ex)*0.5*3.76 *2 = c*2N_tot =0.97+a +b +c "Total kilomoles of products at equilibrium""The assumed equilibrium reaction isH2O=H2+0.5O2""The following equations provide the specific Gibbs function (g=h-Ts) foreach H2mponent in the product gases as a function of its temperature, T_prod,at 1 atm pressure, 101.3 kPa"g_H2O=Enthalpy(H2O,T=T_prod )-T_prod *Entropy(H2O,T=T_prod ,P=101.3)g_H2=Enthalpy(H2,T=T_prod )-T_prod *Entropy(H2,T=T_prod ,P=101.3)g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3)"The standard-state Gibbs function is"DELTAG =1*g_H2+0.5*g_O2-1*g_H2O"The equilibrium constant is given by Eq. 15-14."K_P = exp(-DELTAG /(R_u*T_prod ))P=P_prod /101.3"atm""The equilibrium constant is also given by Eq. 15-15.""K_ P = (P/N_tot)^(1+0.5-1)*(a^1*b^0.5)/(0.97^1)"sqrt(P/N_tot )*a *sqrt(b )=K_P *0.97lnK_p = ln(k_P) 2625 ln Kp PercentEx Tprod [%] [K] -5.414 10 2440 2585 -5.165 20 2490 -5.019 30 2520 -4.918 40 2542 2545 -4.844 50 2557 Tprod -4.786 60 2570 -4.739 70 2580 2505 -4.7 80 2589 -4.667 90 2596 -4.639 100 2602 2465 2425 10 20 30 40 50 60 70 80 90 100 PercentExPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 22. 16-2216-29 The equilibrium constant of the reaction CH4 + 2O2 ↔ CO2 + 2H2O at 25°C is to be determined.Analysis The Kp value of a reaction at a specified temperature can be determined from the Gibbs functiondata using K p = e −ΔG*( T )/ Ru T or ln K p = − ΔG *(T ) / Ru T CH4 + 2O2 ↔ CO2 + 2H2Owhere ∗ ∗ ∗ ∗ΔG * (T ) = ν CO 2 g CO 2 (T ) + ν H 2 O g H 2O (T ) −ν CH 4 g CH 4 (T ) −ν O 2 g O 2 (T ) 25°CAt 25°C, ΔG *(T ) = 1( −394,360) + 2( −228,590) − 1( −50,790) − 2(0) = −800,750 kJ / kmolSubstituting, ln K p = −(−800,750 kJ/kmol)/[(8.314 kJ/kmol ⋅ K)(298 K)] = 323.04or K p = 1.96 × 10 140PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 23. 16-2316-30 The equilibrium constant of the reaction CO2 ↔ CO + 1/2O2 is listed in Table A-28 at differenttemperatures. It is to be verified using Gibbs function data.Analysis (a) The Kp value of a reaction at a specified temperature can be determined from the Gibbsfunction data using K p = e − ΔG*(T ) / RuT or ln K p = − ΔG * (T ) / Ru T ∗ ∗ ∗ CO2 ↔ CO + ½O2where ΔG * (T ) = ν CO g CO (T ) + ν O 2 g O 2 (T ) −ν CO 2 g CO 2 (T ) 298 KAt 298 K, ΔG * (T ) = 1(−137,150) + 0.5(0) − 1(−394,360) = 257,210 kJ/kmolSubstituting, ln K p = −(257,210 kJ/kmol)/[(8.314 kJ/kmol ⋅ K)(298 K)] = -103.81or K p = 8.24 × 10 -46 (Table A - 28 : ln K p = −103.76)(b) At 1800 K, ∗ ∗ ∗ ΔG * (T ) = ν CO g CO (T ) + ν O 2 g O 2 (T ) −ν CO 2 g CO 2 (T ) = ν CO ( h − Ts ) CO + ν O 2 ( h − Ts ) O 2 −ν CO 2 ( h − Ts ) CO 2 = ν CO [(h f + h1800 − h298 ) − Ts ] CO + ν O 2 [(h f + h1800 − h298 ) − Ts ] O 2 −ν CO 2 [(h f + h1800 − h298 ) − Ts ] CO 2 = 1× ( −110,530 + 58,191 − 8669 − 1800 × 254.797) + 0.5 × (0 + 60,371 − 8682 − 1800 × 264.701) − 1× ( −393,520 + 88,806 − 9364 − 1800 × 302.884) = 127,240.2 kJ/kmolSubstituting, ln K p = −(127,240.2 kJ/kmol)/[(8.314 kJ/kmol ⋅ K)(1800 K)] = −8.502or K p = 2.03 × 10 -4 (Table A - 28 : ln K p = −8.497)PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 24. 16-2416-31 [Also solved by EES on enclosed CD] Carbon monoxide is burned with 100 percent excess air. Thetemperature at which 97 percent of CO burn to CO2 is to be determined.Assumptions 1 The equilibrium composition consists of CO2, CO, O2, and N2. 2 The constituents of themixture are ideal gases.Analysis Assuming N2 to remain as an inert gas, the stoichiometric and actual reactions can be written asStoichiometric: CO + 1 O 2 ⇔ CO 2 (thus ν CO 2 = 1, ν CO = 1, and ν O 2 = 1 ) 2 2Actual: CO +1(O 2 + 3.76 N 2 ) ⎯→ ⎯ 0.97 CO 2 + 0.03 CO + 0.515O 2 + 3.76 N 2 1 24 144 2444 123 4 3 4 3 4 4 product reactants inertThe equilibrium constant Kp can be determined from ν ( νCO 2 − νCO − νO 2 ) N CO 2 CO ⎛ P ⎞ CO + ½O2 ↔ CO2 Kp = 2 ⎜ ⎜N ⎟ ⎟ νCO ν 97 % N CO N OO 2 ⎝ total ⎠ 2 1 atm 1−1.5 0.97 ⎛ 1 ⎞ = 0.5 ⎜ ⎟ 0.03 × 0.515 ⎝ 0.97 + 0.03 + 0.515 + 3.76 ⎠ = 103.48From Table A-28, the temperature corresponding to this Kp value is T = 2276 KPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 25. 16-2516-32 EES Problem 16-31 is reconsidered. The effect of varying the percent excess air during the steady-flow process from 0 to 200 percent on the temperature at which 97 percent of CO burn into CO2 is to bestudied.Analysis The problem is solved using EES, and the solution is given below."To solve this problem, we need to give EES a guess value for T_prop other thanthe default value of 1. Set the guess value of T_prod to 1000 K by selecting VariableInfromation in the Options menu. Then press F2 or click the Calculator icon.""Input Data from the diagram window:"{PercentEx = 100}Ex = PercentEx/100 "EX = % Excess air/100"P_prod =101.3 [kPa]R_u=8.314 [kJ/kmol-K]"The combustion equation of CO with stoichiometric amount of air isCO + 0.5(O2 + 3.76N2)=CO2 +0.5(3.76)N2""For the incomplete combustion with 100% excess air, the combustion equation isCO + (!+EX)(0.5)(O2 + 3.76N2)=0.97 CO2 +aCO + bO2+cN2""Specie balance equations give the values of a, b, and c.""C, Carbon"1 = 0.97 + a"O, oxygen"1 +(1+Ex)*0.5*2=0.97*2 + a *1 + b*2"N, nitrogen"(1+Ex)*0.5*3.76 *2 = c*2N_tot =0.97+a +b +c "Total kilomoles of products at equilibrium""The assumed equilibrium reaction isCO2=CO+0.5O2""The following equations provide the specific Gibbs function (g=h-Ts) foreach component in the product gases as a function of its temperature, T_prod,at 1 atm pressure, 101.3 kPa"g_CO2=Enthalpy(CO2,T=T_prod )-T_prod *Entropy(CO2,T=T_prod ,P=101.3)g_CO=Enthalpy(CO,T=T_prod )-T_prod *Entropy(CO,T=T_prod ,P=101.3)g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3)"The standard-state Gibbs function is"DELTAG =1*g_CO+0.5*g_O2-1*g_CO2"The equilibrium constant is given by Eq. 15-14."K_P = exp(-DELTAG /(R_u*T_prod ))P=P_prod /101.3"atm""The equilibrium constant is also given by Eq. 15-15.""K_ P = (P/N_tot)^(1+0.5-1)*(a^1*b^0.5)/(0.97^1)"sqrt(P/N_tot )*a *sqrt(b )=K_P *0.97lnK_p = ln(k_P)"Compare the value of lnK_p calculated by EES with the value oflnK_p from table A-28 in the text."PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 26. 16-26 PercentEx Tprod [%] [K] 0 2066 20 2194 40 2230 60 2250 80 2263 100 2273 120 2280 140 2285 160 2290 180 2294 200 2297 2350 2300 2250 Tprod [K] 2200 2150 2100 2050 0 40 80 120 160 200 PercentEx [%]PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 27. 16-2716-33E Carbon monoxide is burned with 100 percent excess air. The temperature at which 97 percent ofCO burn to CO2 is to be determined.Assumptions 1 The equilibrium composition consists of CO2, CO, O2, and N2. 2 The constituents of themixture are ideal gases.Analysis Assuming N2 to remain as an inert gas, the stoichiometric and actual reactions can be written asStoichiometric: CO + 1 O 2 ⇔ CO 2 (thus ν CO 2 = 1, ν CO = 1, and ν O 2 = 1 ) 2 2Actual: CO +1(O 2 + 3.76 N 2 ) ⎯→ ⎯ 0.97 CO 2 + 0.03 CO + 0.515O 2 + 3.76 N 2 1 24 144 2444 123 4 3 4 3 4 4 product reactants inertThe equilibrium constant Kp can be determined from ν ( νCO 2 − νCO − νO 2 ) N CO 2 CO ⎛ P ⎞ Kp = 2 ν ⎜ ⎜N ⎟ ⎟ CO + ½O2 ↔ CO2 νCO N CO N OO2 ⎝ total ⎠ 97 % 2 1−1.5 1 atm 0.97 ⎛ 1 ⎞ = ⎜ ⎟ 0.03 × 0.5150.5 ⎝ 0.97 + 0.03 + 0.515 + 3.76 ⎠ = 103.48From Table A-28, the temperature corresponding to this Kp value is T = 2276 K = 4097 R16-34 Hydrogen is burned with 150 percent theoretical air. The temperature at which 98 percent of H2 willburn to H2O is to be determined.Assumptions 1 The equilibrium composition consists of H2O, H2, O2, and N2. 2 The constituents of themixture are ideal gases.Analysis Assuming N2 to remain as an inert gas, the stoichiometric and actual reactions can be written asStoichiometric: H 2 + 2 O 2 ⇔ H 2 O (thus ν H 2 O = 1, ν H 2 = 1, and ν O 2 = 2 ) 1 1Actual: H 2 + 0.75(O 2 + 3.76 N 2 ) ⎯→ ⎯ 0.98 H 2 O + 0.02 H 2 + 0.26O 2 + 2.82 N 2 1 24 144 2444 123 4 3 4 3 4 4 product reactants inertThe equilibrium constant Kp can be determined from ν (ν H 2 O −ν H 2 −ν O 2 ) N HHO 2O ⎛ P ⎞ H2 Kp = 2 ⎜ ⎟ ν H2 ν O2 ⎜N ⎟ NH NO ⎝ total ⎠ Combustion H2O, H2 2 2 1−1.5 chamber O2, N2 0.98 ⎛ 1 ⎞ = ⎜ ⎟ Air 0.02 × 0.26 0.5 ⎝ 0.98 + 0.02 + 0.26 + 2.82 ⎠ = 194.11From Table A-28, the temperature corresponding to this Kp value is T = 2472 K.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 28. 16-2816-35 Air is heated to a high temperature. The equilibrium composition at that temperature is to bedetermined.Assumptions 1 The equilibrium composition consists of N2, O2, and NO. 2 The constituents of the mixtureare ideal gases.Analysis The stoichiometric and actual reactions in this case areStoichiometric: 1 2 N 2 + 1 O 2 ⇔ NO (thus ν NO = 1, ν N 2 = 1 , and ν O 2 = 1 ) 2 2 2 AIRActual: 3.76 N 2 + O 2 ⎯→ ⎯ x NO + y N 2 + z O 2 2000 K 1 3 14243 2 prod. reactants 2 atmN balance: 7.52 = x + 2y or y = 3.76 - 0.5xO balance: 2 = x + 2z or z = 1 - 0.5xTotal number of moles: Ntotal = x + y + z = x + 4.76- x = 4.76The equilibrium constant relation can be expressed as (ν NO −ν N 2 −ν O 2 ) N ν NO ⎛ P ⎞ Kp = ν NO ν ⎜ ⎜N ⎟ ⎟ N NN 2 N OO 2 2 2 ⎝ total ⎠From Table A-28, ln Kp = -3.931 at 2000 K. Thus Kp = 0.01962. Substituting, 1−1 x ⎛ 2 ⎞ 0.01962 = ⎜ ⎟ (3.76 − 0.5 x) 0.5 (1 − 0.5 x) 0.5 ⎝ 4.76 ⎠Solving for x, x = 0.0376Then, y = 3.76-0.5x = 3.7412 z = 1-0.5x = 0.9812Therefore, the equilibrium composition of the mixture at 2000 K and 2 atm is 0.0376NO + 3.7412N 2 + 0.9812O 2The equilibrium constant for the reactions O 2 ⇔ 2O (ln Kp = -14.622) and N 2 ⇔ 2 N (ln Kp = -41.645)are much smaller than that of the specified reaction (ln Kp = -3.931). Therefore, it is realistic to assume thatno monatomic oxygen or nitrogen will be present in the equilibrium mixture. Also the equilibriumcomposition is in this case is independent of pressure since Δν = 1 − 0.5 − 0.5 = 0 .PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 29. 16-2916-36 Hydrogen is heated to a high temperature at a constant pressure. The percentage of H2 that willdissociate into H is to be determined.Assumptions 1 The equilibrium composition consists of H2 and H. 2 The constituents of the mixture areideal gases.Analysis The stoichiometric and actual reactions can be written asStoichiometric: H 2 ⇔ 2H (thus ν H 2 = 1 and ν H = 2) H2Actual: H2 ⎯ ⎯→ { + { xH 2 yH 3200 K react. prod. 8 atmH balance: 2 = 2x + y or y = 2 - 2xTotal number of moles: Ntotal = x + y = x + 2 - 2x = 2 - xThe equilibrium constant relation can be expressed as ν H −ν H 2 Nν H ⎛ P ⎞ K p = νH ⎜ ⎜ ⎟ ⎟ N HH 2 ⎝ N total ⎠ 2From Table A-28, ln Kp = -2.534 at 3200 K. Thus Kp = 0.07934. Substituting, 2 −1 (2 − 2 x) 2 ⎛ 8 ⎞ 0.07934 = ⎜ ⎟ x ⎝2− x⎠Solving for x, x = 0.95Thus the percentage of H2 which dissociates to H at 3200 K and 8 atm is 1 - 0.95 = 0.05 or 5.0%PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 30. 16-3016-37E A mixture of CO, O2, and N2 is heated to a high temperature at a constant pressure. The equilibriumcomposition is to be determined.Assumptions 1 The equilibrium composition consists of CO2, CO, O2, and N2. 2 The constituents of themixture are ideal gases.Analysis The stoichiometric and actual reactions in this case are 2 COStoichiometric: CO + 1 O 2 ⇔ CO 2 (thus ν CO 2 = 1, ν CO = 1, and ν O 2 = 1 ) 2 2 2 O2 6 N2Actual: 2 CO + 2 O 2 + 6 N 2 ⎯→ ⎯ x CO 2 + y CO + z O 2 + 6 N 2 : 4320 R 1 3 14243 2 4 products reactants inert 3 atmC balance: 2= x+ y ⎯→ ⎯ y = 2− xO balance: 6 = 2x + y + 2z ⎯→ ⎯ z = 2 − 0.5xTotal number of moles: N total = x + y + z + 6 = 10 − 0.5xThe equilibrium constant relation can be expressed as ν (ν CO 2 −ν CO −ν O 2 ) N CO 2 CO ⎛ P ⎞ Kp = 2 ⎜ ⎟ ν ν ⎜N ⎟ N CO N OO 2 CO ⎝ total ⎠ 2From Table A-28, ln K p = 3.860 at T = 4320 R = 2400 K. Thus K p = 47.465. Substituting, 1−1.5 x ⎛ 3 ⎞ 47.465 = ⎜ ⎟ (2 − x)(2 − 0.5 x) 0.5 ⎝ 10 − 0.5 x ⎠Solving for x, x = 1.930Then, y = 2 - x = 0.070 z = 2 - 0.5x = 1.035Therefore, the equilibrium composition of the mixture at 2400 K and 3 atm is 1.930CO 2 + 0.070CO + 1.035O 2 + 6N 2PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 31. 16-3116-38 A mixture of N2, O2, and Ar is heated to a high temperature at a constant pressure. The equilibriumcomposition is to be determined.Assumptions 1 The equilibrium composition consists of N2, O2, Ar, and NO. 2 The constituents of themixture are ideal gases.Analysis The stoichiometric and actual reactions in this case areStoichiometric: 1 N 2 + 1 O 2 ⇔ NO (thus ν NO = 1, ν N 2 = 1 , and ν O 2 = 1 ) 3 N2 2 2 2 2 1 O2Actual: 3 N 2 + O 2 + 01 Ar . ⎯→ ⎯ x NO + y N 2 + z O 2 + 01 Ar . 0.1 Ar 1 3 14243 1 3 2 2 2400 K prod. reactants inert 10 atmN balance: 6 = x + 2y ⎯→ ⎯ y = 3 − 0.5xO balance: 2 = x + 2z ⎯→ ⎯ z = 1 − 0.5xTotal number of moles: N total = x + y + z + 01 = 41 . .The equilibrium constant relation becomes, ν NO (ν NO − ν N 2 − νO2 ) 1−0 .5−0 .5 N NO ⎛ P ⎞ x ⎛ P ⎞ Kp = ν ν ⎜ ⎜N ⎟ ⎟ = ⎜ ⎜N ⎟ ⎟ N NN 2 N OO2 ⎝ total ⎠ y 0.5 z 0.5 ⎝ total ⎠ 2 2From Table A-28, ln K p = −3.019 at 2400 K. Thus K p = 0.04885. Substituting, x 0.04885 = ×1 (3 − 0.5 x)0.5 (1 − 0.5 x)0.5Solving for x, x = 0.0823Then, y = 3 - 0.5x = 2.9589 z = 1 - 0.5x = 0.9589Therefore, the equilibrium composition of the mixture at 2400 K and 10 atm is 0.0823NO + 2.9589N 2 + 0.9589O 2 + 0.1ArPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 32. 16-3216-39 The mole fraction of sodium that ionizes according to the reaction Na ⇔ Na+ + e- at 2000 K and 0.8atm is to be determined.Assumptions All components behave as ideal gases.Analysis The stoichiometric and actual reactions can be written asStoichiometric: Na ⇔ Na + + e - (thus ν Na = 1, ν Na + = 1 and ν e - = 1) Na ⇔ Na+ + e- 2000 KActual: ⎯→ { + y Na + + y e − Na ⎯ x Na 0.8 atm 14243 4 4 react. productsNa balance: 1 = x + y or y = 1 − xTotal number of moles: N total = x + 2 y = 2 − xThe equilibrium constant relation becomes, ν N ν Na N Na e- ⎛ P ⎞ (ν Na + +ν - −ν Na ) e y2 ⎛ P ⎞ 1+1−1 e- Kp = ⎜ ⎟ = ⎜ ⎟ ⎜ ⎟ ⎜N ⎟ N ν Na Na ⎝ N total ⎠ x ⎝ total ⎠Substituting, (1 − x) 2 ⎛ 0.8 ⎞ 0.668 = ⎜ ⎟ x ⎝2− x⎠Solving for x, x = 0.325Thus the fraction of Na which dissociates into Na+ and e- is 1 - 0.325 = 0.675 or 67.5%PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 33. 16-3316-40 Liquid propane enters a combustion chamber. The equilibrium composition of product gases and therate of heat transfer from the combustion chamber are to be determined.Assumptions 1 The equilibrium composition C3H8consists of CO2, H2O, CO, N2, and O2. 2 The 25°C COconstituents of the mixture are ideal gases. Combustion 1200 K CO2Analysis (a) Considering 1 kmol of C3H8, the chamber H2Ostoichiometric combustion equation can be Air O2written as 2 atm N2 12°CC 3 H 8 (l) + a th (O 2 + 3.76 N 2 ) ⎯ ⎯→ 3CO 2 + 4H 2 O + 3.76a th N 2where ath is the stoichiometric coefficient and isdetermined from the O2 balance, 2.5a th = 3 + 2 + 1.5a th ⎯→ ⎯ a th = 5Then the actual combustion equation with 150% excess airand some CO in the products can be written as C3H 8 ( l ) + 12.5( O 2 + 3.76 N 2 ) ⎯→ ⎯ xCO 2 + (3 − x )CO + (9 − 0.5 x )O 2 + 4H 2 O + 47N 2After combustion, there will be no C3 H8 present in the combustion chamber, and H2O will act like an inertgas. The equilibrium equation among CO2, CO, and O2 can be expressed as CO 2 ⇔ CO + 1 O 2 (thus ν CO 2 = 1, ν CO = 1, and ν O 2 = 1 ) 2 2and ν ν (ν CO +ν O 2 −ν CO 2 ) N CO N OO 2 ⎛ P CO ⎞ Kp = 2 ⎜ ⎟ ν CO 2 ⎜N ⎟ N CO ⎝ total ⎠ 2where N total = x + (3 − x ) + (9 − 0.5x ) + 4 + 47 = 63 − 0.5xFrom Table A-28, ln K p = −17.871 at 1200 K. Thus K p = 1.73 × 10 −8 . Substituting, 1.5−1 (3 − x )(9 − 0.5 x ) 0.5 ⎛ 2 ⎞ 1.73 × 10 −8 = ⎜ ⎟ x ⎝ 63 − 0.5 x ⎠Solving for x, x = 2.9999999 ≅ 3.0Therefore, the amount CO in the product gases is negligible, and it can be disregarded with no loss inaccuracy. Then the combustion equation and the equilibrium composition can be expressed as C 3 H 8 ( l) + 12.5(O 2 + 3.76N 2 ) ⎯ → 3CO 2 + 7.5O 2 + 4 H 2 O + 47N 2 ⎯and 3CO 2 + 7.5O 2 + 4H 2 O + 47N 2(b) The heat transfer for this combustion process is determined from the steady-flow energy balance E in − E out = ΔE system on the combustion chamber with W = 0, − Qout = ∑ N (h P o f +h −ho ) − ∑ N (h P R o f + h −ho ) RPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 34. 16-34Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables, (Theh fo of liquid propane is obtained by adding the hfg at 25°C to h fo of gaseous propane). h fo h 285 K h 298 K h1200 K Substance kJ/kmol kJ/kmol kJ/kmol kJ/kmol C3H8 (l) -118,910 --- --- --- O2 0 8696.5 8682 38,447 N2 0 8286.5 8669 36,777 H2O (g) -241,820 --- 9904 44,380 CO2 -393,520 --- 9364 53,848Substituting, − Qout = 3( −393,520 + 53,848 − 9364 ) + 4( −241,820 + 44,380 − 9904 ) + 7.5( 0 + 38,447 − 8682 ) + 47( 0 + 36,777 − 8669 ) − 1( −118,910 + h298 − h298 ) − 12.5( 0 + 8296.5 − 8682 ) − 47( 0 + 8186.5 − 8669 ) = −185,764 kJ / kmol of C3H 8or Qout = 185,764 kJ / kmol of C3H 8The mass flow rate of C3H8 can be expressed in terms of the mole numbers as & m & . 12 kg / min N= = = 0.02727 kmol / min M 44 kg / kmolThus the rate of heat transfer is & & Qout = N × Qout = (0.02727 kmol/min)(185,746 kJ/kmol) = 5066 kJ/minThe equilibrium constant for the reaction 1 2 N 2 + 1 O 2 ⇔ NO is ln Kp = -7.569, which is very small. This 2indicates that the amount of NO formed during this process will be very small, and can be disregarded.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 35. 16-3516-41 EES Problem 16-40 is reconsidered. It is to be investigated if it is realistic to disregard the presenceof NO in the product gases.Analysis The problem is solved using EES, and the solution is given below."To solve this problem, the Gibbs function of the product gases is minimized. Click on theMin/Max icon."For this problem at 1200 K the moles of CO are 0.000 and moles of NO are 0.000, thus wecan disregard both the CO and NO. However, try some product temperatures above 1286 Kand observe the sign change on the Q_out and the amout of CO and NO present as theproduct temperature increases.""The reaction of C3H8(liq) with excess air can be written: C3H8(l) + (1+Ex)A_th (O2+3.76N2) = a C02 + b CO + c H2O + d N2 + e O2 + f NOThe coefficients A_th and EX are the theoretical oxygen and the percent excess air on a decimalbasis. Coefficients a, b, c, d, e, and f are found by minimiming the Gibbs Free Energy at a totalpressure of the product gases P_Prod and the product temperature T_Prod.The equilibrium solution can be found by applying the Law of Mass Action or by minimizing theGibbs function. In this problem, the Gibbs function is directly minimized using the optimizationcapabilities built into EES.To run this program, click on the Min/Max icon. There are six compounds present in the productssubject to four specie balances, so there are two degrees of freedom. Minimize the Gibbsfunction of the product gases with respect to two molar quantities such as coefficients b and f.The equilibrium mole numbers a, b, c, d, e, and f will be determined and displayed in the Solutionwindow."PercentEx = 150 [%]Ex = PercentEx/100 "EX = % Excess air/100"P_prod =2*P_atmT_Prod=1200 [K]m_dot_fuel = 0.5 [kg/s]Fuel$=C3H8T_air = 12+273 "[K]"T_fuel = 25+273 "[K]"P_atm = 101.325 [kPa]R_u=8.314 [kJ/kmol-K]"Theoretical combustion of C3H8 with oxygen:C3H8 + A_th O2 = 3 C02 + 4 H2O "2*A_th = 3*2 + 4*1"Balance the reaction for 1 kmol of C3H8""C3H8(l) + (1+Ex)A_th (O2+3.76N2) = a C02 + b CO + c H2O + d N2 + e O2 + f NO"b_max = 3f_max = (1+Ex)*A_th*3.76*2e_guess=Ex*A_th1*3 = a*1+b*1 "Carbon balance"1*8=c*2"Hydrogen balance"(1+Ex)*A_th*2=a*2+b*1+c*1+e*2+f*1 "Oxygen balance"(1+Ex)*A_th*3.76*2=d*2+f*1 "Nitrogen balance"PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 36. 16-36"Total moles and mole fractions"N_Total=a+b+c+d+e+fy_CO2=a/N_Total; y_CO=b/N_Total; y_H2O=c/N_Total; y_N2=d/N_Total; y_O2=e/N_Total;y_NO=f/N_Total"The following equations provide the specific Gibbs function for each component as a function ofits molar amount"g_CO2=Enthalpy(CO2,T=T_Prod)-T_Prod*Entropy(CO2,T=T_Prod,P=P_Prod*y_CO2)g_CO=Enthalpy(CO,T=T_Prod)-T_Prod*Entropy(CO,T=T_Prod,P=P_Prod*y_CO)g_H2O=Enthalpy(H2O,T=T_Prod)-T_Prod*Entropy(H2O,T=T_Prod,P=P_Prod*y_H2O)g_N2=Enthalpy(N2,T=T_Prod)-T_Prod*Entropy(N2,T=T_Prod,P=P_Prod*y_N2)g_O2=Enthalpy(O2,T=T_Prod)-T_Prod*Entropy(O2,T=T_Prod,P=P_Prod*y_O2)g_NO=Enthalpy(NO,T=T_Prod)-T_Prod*Entropy(NO,T=T_Prod,P=P_Prod*y_NO)"The extensive Gibbs function is the sum of the products of the specific Gibbs function and themolar amount of each substance"Gibbs=a*g_CO2+b*g_CO+c*g_H2O+d*g_N2+e*g_O2+f*g_NO"For the energy balance, we adjust the value of the enthalpy of gaseous propane given by EES:"h_fg_fuel = 15060"[kJ/kmol]" "Table A.27"h_fuel = enthalpy(Fuel$,T=T_fuel)-h_fg_fuel"Energy balance for the combustion process:""C3H8(l) + (1+Ex)A_th (O2+3.76N2) = a C02 + b CO + c H2O + d N2 + e O2 + f NO"HR =Q_out+HPHR=h_fuel+ (1+Ex)*A_th*(enthalpy(O2,T=T_air)+3.76*enthalpy(N2,T=T_air))HP=a*enthalpy(CO2,T=T_prod)+b*enthalpy(CO,T=T_prod)+c*enthalpy(H2O,T=T_prod)+d*enthalpy(N2,T=T_prod)+e*enthalpy(O2,T=T_prod)+f*enthalpy(NO,T=T_prod)"The heat transfer rate is:"Q_dot_out=Q_out/molarmass(Fuel$)*m_dot_fuel "[kW]"SOLUTIONa=3.000 [kmol] g_CO2=-707231 [kJ/kmol] Q_dot_out=2140 [kW]A_th=5 g_H2O=-515974 [kJ/kmol] Q_out=188732 [kJ/kmol_fuel]b=0.000 [kmol] g_N2=-248486 [kJ/kmol] R_u=8.314 [kJ/kmol-K]b_max=3 g_NO=-342270 [kJ/kmol] T_air=285 [K]c=4.000 [kmol] g_O2=-284065 [kJ/kmol] T_fuel=298 [K]d=47.000 [kmol] HP=-330516.747 [kJ/kmol] T_Prod=1200.00 [K]e=7.500 [kmol] HR=-141784.529 [kJ/kmol] y_CO=1.626E-15Ex=1.5 h_fg_fuel=15060 [kJ/kmol] y_CO2=0.04878e_guess=7.5 h_fuel=-118918 [kJ/kmol] y_H2O=0.06504f=0.000 [kmol] m_dot_fuel=0.5 [kg/s] y_N2=0.7642Fuel$=C3H8 N_Total=61.5 [kmol/kmol_fuel] y_NO=7.857E-08f_max=94 PercentEx=150 [%] y_O2=0.122Gibbs=-17994897 [kJ] P_atm=101.3 [kPa]g_CO=-703496 [kJ/kmol] P_prod=202.7 [kPa]PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.

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