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Unformatted text preview: Sonntag/Borgnakke Study Problems Chapter 14 CHAPTER 14 STUDY PROBLEMS COMBUSTION
• • • • Introduction and definitions Combustion of hydrocarbon fuels The energy equation The entropy equation
SONNTAG • BORGNAKKE • VAN WYLEN FUNDAMENTALS of
Thermodynamics
Sixth Edition Sonntag/Borgnakke 14.1 Study Problems Chapter 14 A combustion equation for JP8 jet fuel A modern jet engine or gas turbine uses JP8 jet fuel, a mixture with the average composition as C13H23.8 and a molecular weight of 180.13. The combustion takes place with 125% theoretical air. What is the airfuel ratio on a mole and mass basis (A/F) and what is the dew point temperature of the products at 100 kPa? Solution: The stoichiometric reaction equation is C13H23.8 + νO2 (O2 + 3.76 N2) → νCO2 CO2 + νH2O H2O + νN2 N2 Where the stoichiometric coefficients for CO2 and H2O comes from the fuel composition. Continuity C: Continuity H: Continuity O: 13 = νCO2 ; 23.8 = 2 νH2O => νH2O = 11.9 => νO2 = 18.95; The stoichiometric (100% theoretical air) oxygen needed thus becomes 2νO2 = 13 × 2 + 11.9 = 37.9 Then the actual combustion equation now has oxygen as νO2 ac = 1.25 × νO2 stoich. = 1.25 × 18.95 = 23.6875 and nitrogen as νN2 = 3.76 νO2 = 3.76 × 23.6875 = 89.065 Now we see the excess oxygen (23.6875 – 18.95 = 4.7375) in the products as C13H23.8 + 1.25 × 18.95 (O2 + 3.76 N2) → 13 CO2 + 11.9 H2O + 4.7375 O2 + 89.065 N2 The A/F ratio on a mole basis is A/F = 23.6875 × (1 + 3.76) / 1 = 112.75 The A/F ratio on a mass basis is A/F = 23.6875 × (31.999 + 3.76 × 28.013) / 180.13 = 18.06 To find the product dew point we need to find the water partial pressure which is the mole fraction time the total pressure (assuming ideal gas mixture). 11.9 yH2O = 13 + 11.9 + 4.7375 + 89.065 = 0.10 => Pv = yH2O Ptot = 0.1 × 100 kPa = 10 kPa = Pg Tdew Tdew = 45.8oC Now look in Table B.1.2: Sonntag/Borgnakke 14.2 Combustion of sulfur Study Problems Chapter 14 Most coals have some amount of sulfur in it. Consider as a part of a combustion process that S is oxidized to SO2 with stoichiometric air. The reactants are supplied at the reference T and P and the products after some heat transfer are at 1000 K. How much energy did the sulfur combustion provide per kmol of sulfur. Combustion equation: The O balance: S + νO2 (O2 + 3.76 N2) → 1 SO2 + νN2 N2 νO2 = 1 => νN2 = 3.76 Now we can do the energy equation. Notice we do not have the SO2 ideal gas tables so we use the heat capacity from Table A.5 (we could have used A.6 to be more accurate). Energy Eq.: HR = HR + ∆HR = HR = HP + Q = HP + ∆HP + Q
° 0 0 0 HR = hf S + hf O2 + 3.76 hf N2 = 0 + 0 + 0 = 0 (S from A.10) ° 0 0 HP = hf SO2 + 3.76 hf N2 = 296 842 + 0 = 296 842 kJ/kmol − ∆HP = Cp M (TP – To) + 3.76 ∆hN2
° ° ° = 0.624 × 64.059 (1000 – 298.15) + 3.76 × 21 463 = 28 055 + 80 701 = 108 756 kJ/kmol Now solve for Q Q = HR – HP – ∆HP = 0 –(–296 842) – 108 756 = 188 086 kJ/kmol Comment: Today we do what we can to take the sulfur out of the fuel (oil or coal) before we burn it. Water in the products together with sulfur dioxide will generate sulfuric acid causing acid rain. This previously caused sections of forrest to die out.
° ° Sonntag/Borgnakke 14.3 Study Problems Chapter 14 Heat release by JP8 jet fuel A modern jet engine or gas turbine uses JP8 jet fuel, see Table 14.3. It has a molecular weight of 180.13. The combustion takes place with 125% theoretical air and the fuel is added as a liquid. How much air per kg fuel does it require and what is the specific energy released by the combustion process ( q = −HRP per kg mixture ). If the energy is used as the heat input in a gasturbine with cycle efficiency of 45% how much fuel (kg/s) is needed to have a power output of 10 MW? Solution: The stoichiometric reaction equation is C13H23.8 + νO2 (O2 + 3.76 N2) → 13 CO2 + 11.9 H2O + νN2 N2 Where the stoichiometric coefficients for CO2 and H2O comes from the fuel composition as shown in Study problem 14.1. The oxygen needed becomes νO2 = 13 + 11.9/2 = 18.95; νN2 = 3.76 νO2 = 71.252 The actual combustion equation now has excess oxygen as C13H23.8 + 1.25 × 18.95 (O2 + 3.76 N2) → 13 CO2 + 11.9 H2O + 4.7375 O2 + 89.065 N2 The A/F ratio on a mass basis is AF = 1.25 × 18.95 (31.999 + 3.76 × 28.013) / 180.13 = 18.06 The enthalpy of combustion HRP from Table 14.3 is –42 800 kJ/kg fuel taken for liquid fuel and water as vapor. In order to use that in a cycle calculation we need to convert it to energy per kg mixture. mmix = mfuel + mair = mfuel (1 + AF) We therefore have q = −HRP / (1 + AF) = 42 800 / 19.06 = 2245 kJ/kg For the cycle we then have wnet = η q = 0.45 × 2245 = 1010 kJ/kg . 10 000 . mmix = W/wnet = 1010 = 9.9 kg/s . mmix 9.9 . mfuel = 1 + AF = 19.06 = 0.52 kg/s Sonntag/Borgnakke 14.4 Combustion of nCetane Study Problems Chapter 14 Consider a combustion process with nCetane C16H34, see Table 14.3, in a stoichiometric ratio with air. We notice that there is no entry for this fuel in Table A.10 yet we want to know how much energy we need to supply to a carburetor to vaporize the liquid fuel before we mix it with air. Secondly we also want to know the adiabatic flame temperature assuming the reactants are supplied at the reference P and T with the fuel as a vapor. The stoichiometric reaction equation is C16H34 + νO2 (O2 + 3.76 N2) → 16 CO2 + 17 H2O + νN2 N2 where we have applied the C and H atom balance already. The required oxygen becomes νO2 = 16 + 17/2 = 24.5; The energy equation becomes Energy Eq.: Solve for ∆HP
° ° ° 0 o o ∆HP = HR – HP = –HRP = HV = hf fuel – 16 hf CO2 – 17 hf H2O
° νN2 = 3.76 νO2 = 92.12
° ° HR = HR + ∆HR = HR = HP = HP + ∆HP From Table 14.3 we find for the liquid fuel and water as vapor –HRP = 44 000 kJ/kg. If we look at the fuel vapor and water as vapor we get –HRP = 44 358 kJ/kg so the difference is the fuel enthalpy of vaporization as the other terms cancel out hfg = HVfuel vap – HVfuel liq = hf fuel vap  hf fuel liq = 44 538 – 44 000 = 538 kJ/kg For the energy equation we need the HV per kmol of fuel so kJ kJ ° HV = –HRP = 44 000 kg = 44 000 × (16 ×12.011 + 34 ×1.008) = 9 963 712 kmol − − − ∆HP = 16 ∆hCO2 + 17 ∆hH2O + 92.12 ∆hN2 = 9 963 712 kJ/kmol Now find the temperature for which the enthalpy terms, Table A.9, adds up. We start with the average [79 633 = 9 963 712 /(92.12 + 17 +16) ] for nitrogen giving 2600+ K 2600K: ∆HP = 16 × 128074 + 17 × 104520 + 92.12 × 77963 = 11 007 975 kJ/kmol high 2400K: ∆HP = 16 × 115779 + 17 × 93 741 + 92.12 × 70 640 = 9 953 418 kJ/kmol close Interpolate to get: 9 963 712 – 9 953 418 T = 2400 + 200 11 007 975  9 953 418 = 2402 K
o o ° ° Sonntag/Borgnakke Study Problems Chapter 14 14.5 Combustion of carbon monoxide Assume a steady flow of 0.1 kg/s carbon monoxide CO and a stoichiometric amount of air flowing into a mixing chamber both at reference temperature and pressure. A complete combustion process generates carbon dioxide CO2. What is the product exit temperature assuming an adiabatic process? Solution: The reaction equation is CO + νO2 (O2 + 3.76 N2) → 1 CO2 + νN2 N2 With the carbon balance (νC = 1) done, so the oxygen balance becomes => νO2 = 0.5, νN2 = 1.88 0.5 + νO2 = 1 The energy equation for the whole setup has no Q or W terms so it is Energy Eq.: Solve for ∆HP
° ° o o ∆HP = HR – HP = hf CO – hf CO2 HR = HR + ∆HR = HR = HP = HP + ∆HP ° ° ° = –110 527 – (–393 522) = 282 995 kJ/kmol CO Now the energy equation reads − − ∆HP = ∆hCO2 + 1.88 ∆hN2 = 282 995 kJ/kmol CO And we need to find the temperature for which the left hand side adds up to the proper value. The enthalpy terms are from Table A.9 so we have trial and error. LHS2000 = 91 439 + 1.88 × 56 137 = 196 976 too small LHS3000 = 152 853 + 1.88 × 92 715 = 327 157 too large LHS2600 = 128 074 + 1.88 × 77 963 = 274 644 too small LHS2800 = 140 435 + 1.88 × 85 323 = 300 842 too large Now interpolate 282 995 – 274 644 T = 2600 + 200 × 300 842 – 274 644 = 2664 K Sonntag/Borgnakke 14.6 Entropy generation burning propene Study Problems Chapter 14 Propene, C3H6, is burned with air in a steady flow burner. The reactants are supplied at the reference pressure and temperature and the mixture is lean so the adiabatic flame temperature is kept to 1800 K. What is the A/F ratio on a mole basis? What is the entropy generation per kmol fuel, neglecting all the partial pressure corrections? The reaction equation for a mixture with excess air is: C3H6 + νO2 (O2 + 3.76 N2) → 3 H2O + 3 CO2 + 3.76νO2 N2 + (νO2 – 4.5)O2 Energy Eq.: The entropy equation: From table A.9 at reference T ∆HR = ∆hFu + νO2(∆hO2 + 3.76 ∆hN2) = 0 From table A.9 at 1800 K: ∆HP = 3 ∆hH2O + 3 ∆hCO2 + 3.76 νO2 ∆hN2 + (νO2 – 4.5) ∆hO2 = 3 × 62 693 + 3 × 79432 + 3.76 νO2 × 48 979 + (νO2 – 4.5) 51 674 = 193 842 + 235 835 νO2 From table 14.3: HP  HR = HRP = 42.081(–45 780) = –1 926 468 kJ/kmol –1 926 468 + 193 842 + 235 835 νO2 = 0 Solve for νO2: νO2 = 1 926 468 – 193 842 = 7.3468, 235 835 νN2 = 27.624
° ° ° HR = HR + ∆HR = HR = HP = HP + ∆HP SR + Sgen = SP => Sgen = SP – SR = SP – SR
° ° ° ° Now substitute all terms into the energy equation A/F = 4.76 νO2 / 1 = 34.97 [ (A/F) / (A/F)S = 7.3468/4.5 = 1.633 ] Table A.9 contains the entropies at 100 kPa so we get: SP = 3 × 259.452 + 3 × 302.969 + (7.3468 – 4.5) 264.797 + 27.624 × 248.304 = 9300.24 kJ/kmol fuel SR = 267.066 + 7.3468 × 205.148 + 27.624 × 191.609 = 7067.25 kJ/kmol fuel Sgen = 9300.24 – 7067.25 = 2233 kJ/kmol fuel Sonntag/Borgnakke Study Problems Chapter 14 14.7 Combustion of carbon monoxide Assume a steady flow of carbon monoxide CO and a stoichiometric amount of air flowing into a mixing chamber both at reference temperature and pressure. A complete combustion process generates carbon dioxide CO2. How much entropy is generated in the mixing chamber and how much is generated in the combustion process both per kmol CO? Solution: The reaction equation is CO + νO2 (O2 + 3.76 N2) → 1 CO2 + νN2 N2 so the oxygen balance becomes => νO2 = 0.5, νN2 = 1.88 0.5 + νO2 = 1 In the mixing process we will have the same temperature out and the total pressure out equals the inlet pressure. The entropy generation is only caused by the mixing process for which we need the molfractions and recall the air was mixed already. 1 yCO = 1 + 0.5 + 1.88 = 0.2958; yair = 1 – yCO = 0.7042 Now the increase in entropy equals the generation due to the reduction to partial pressures Sgen mix = (S2 – S1)CO + (S2 – S1)air − − = 1( – R ln yCO ) + 0.5 × 4.76 (–R ln yair ) = (–8.3145 ln 0.2958) + 2.38 (–8.3145 ln 0.7042) = 17.07 kJ/Kmol CO K The entropy equation for the whole setup has no Q term so it is Entropy Eq.: SR + Sgen tot = SP The entropy for the products are evaluated at the adiabatic flame temperature, see study problem 14.5 where we found T = 2664 K. For the products we have the molfractions 1 yCO2 = 1 + 1.88 = 0.3472 ; yN2 = 1 – yCO2 = 0.6528 SP = SCO2 + 1.88 SN2 = 326.773 – 8.3145 ln(0.3472) + 1.88[262.488 – 8.3145 ln(0.6528)] = 835.712 kJ/kmol K The CO comes in at reference P and the air is mixed i.e. the oxygen and nitrogen are at their partial pressures SR = SCO + 0.5 SO2 + 1.88 SN2 = 197.651 + 0.5[ 205.148 – 8.3145 ln(0.21) ] + 1.88 [191.609 – 8.3145 ln(0.79) ] = 670.622 kJ/kmol K Sgen tot = SP – SR = 835.712 – 670.622 = 165.089 kJ/kmol K Sgen comb = Sgen tot – Sgen mix = 165.089 – 17.07 = 148.0 kJ/kmol CO K ...
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This note was uploaded on 10/20/2009 for the course MECHENG MEecheng 3 taught by Professor Borgnakke during the Fall '09 term at University of Michigan.
 Fall '09
 Borgnakke
 Combustion

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