Fuel, Air, and Combustion Thermodynamics

Transcription

1 Chapter 3 Fuel, Air, and Combustion Thermodynamics 3.1) What is the molecular weight, enthalpy (kj/kg), and entropy (kj/kg K) of a gas mixture at P = 1000 kpa and T = 500 K, if the mixture contains the following species and mole fractions? a) A table of the given and computed parameters is: i y i M i h o f h i h o f s o i ( y i s o i R u ln(y i ) ) [ kg /kmol] [ MJ /kmol] [ MJ /kmol] [ kg /kmol K] [ kj /kmol K] CO H 2 O N CO The mixture molecular mass is: M = y i M i = (0.10)(44.01) + (0.15)(18.01) + (0.70)(28.01) + (0.05)(28.01) M = kg /kmol The specific mixture enthalpy is: h f = y i h i = (0.10)( ) + (0.15)( ) The mixture enthalpy is: + (0.70)( ) + (0.05)( ) h f = 57.8 MJ /kmol = kj/kmol h = h = m h = 2056 kj /kg The specific mixture entropy is: ( ) P s = R u ln + y i (s o i R u ln(y i )) P 0 ( ) 1000 s = ( 8.314) ln s = kj /kmol K 1

2 2 CHAPTER 3. FUEL, AIR, AND COMBUSTION THERMODYNAMICS The mixture entropy is: s = s m = s = kj /kg K

3 3 3.2) What is the enthalpy h (kj/kg) and entropy s (kj/kg-k) of a mixture of 30% H 2 and 70% CO 2 by volume at a temperature of 3000 K? a) A table of the given and computed parameters is: i y i M i h o f h i h o f s o i ( y i s o i R u ln(y i ) ) [ kg /kmol] [ MJ /kmol] [ MJ /kmol] [ kg /kmol K] [ kj /kmol K] CO H The mixture molecular mass is: M = y i M i = (0.70)(44.01) + (0.30)(2.016) M = kg /kmol The specific mixture enthalpy is: h f = y i h i = (0.70)( ) + (0.30)( ) = 14.8 MJ /kmol = kj/kmol The mixture enthalpy is: h = h = m h = 4516 kj /kg The specific mixture entropy is: ( ) P s = R u ln + y i (s o i R u ln(y i )) P 0 ( ) 2000 s = ( 8.314) ln s = kj /kmol K The mixture entropy is: s = s m = s = kj /kg K

4 4 CHAPTER 3. FUEL, AIR, AND COMBUSTION THERMODYNAMICS 3.3) Using the Gordon and McBride equations, Equations (3.38) and (3.39), calculate the enthalpy h and entropy s o of CO 2 and compare with the gas table values used in Example 3.1. A Matlab program for calculating the enthalpy h and entropy s o of CO 2 is t=1000; % temp in K R= % univ. gas const. a1= ; a2= e-3; a3= e-6; a4= e-9; a5= e-16; a6= e4; a7= ; nondimh=a1+a2/2*t+a3/3*t^2+a4/4*t^3+a5/5*t^4+a6/t h=nondimh*t*r nondims=a1*log(t)+a2*t+a3/2*t^2+a4/3*t^3+a5/4*t^4+a7 s=nondims*r The calculated h = kj/kmol and the entropy s o = kj/kmol-k. The gas table values are h = kj/kmol and the entropy s o = kj/kmol-k. The enthalpy values agree to 4 figures, and the entropy values agree to 3 figures.

5 3.4) Using the program Fuel.m, at what temperature is the specific heat c p of methane CH 4 = 3.0 kj/kg-k? 5

6 6 CHAPTER 3. FUEL, AIR, AND COMBUSTION THERMODYNAMICS 3.5) Why does Equation 3.27 contain y i? a) From the Gibbs equations, we have ( ) s i (T, P ) = s o Pi i (T ) R i ln P 0 Where s o i (T ) is the standard entropy. y i = P i P = P i P0 P 0 P ( ) ( ) Pi P0 ln(y i ) = ln + ln P 0 P ( ) ( ) Pi P ln = ln(y i ) + ln P 0 Substituting back in: ( )) P s i (T, P ) = s o i (T ) R i (ln(y i ) + ln The mole fraction y i originates from the pressure ratio term. P 0 P 0

7 7 3.6) A system whose composition is given below is in equilibrium at P = 101 kpa and T = 298 K. What are the enthalpy (kj/kg), specific volume (m 3 /kg), and quality χ of the mixture? a) A table of the given and computed parameters is: i y i M i h f h i x i [ v i ] [ kg /kmol] [ kj /kmol] [ kj /kg] m 3 /kg H 2 O ,826-13, CO , N The mixture molecular mass is: M = y i M i = (0.141)(18.01) + (0.125)(44.01) + (0.734)(28.01) = kg /kmol Sample calculations for CO 2 (i=2): h 2 = h f = 393,522 = 8942 M i kj /kg ( ) Mi x 2 = y 2 = M = v 2 = Ru T M i P = = m3 /kg If all of the water is vapor, the water partial pressure would be P H2O = y i P = 0.141(101) = 14.2 kpa Whoever, at T = 298K, the saturation pressure of the water is 3.17 kpa (Table 3.1), so some fraction of the water will be liquid. The quality of the gas mixture is given in the Appendix: ( 1 X = y H2O X = ) ( ) 1 P 1 1 = P sat ( ) ( ) 1 The enthalpy of vaporization h fg is 44.02/18.01 = 2444 kj/kg (Table 3.1) The term v fg is v g -v f = ( ) = m 3 /kg (Table 3.1) Therefore using equations (3.55) and (3.57) The enthalpy is given by: h = x 1 h 1 + x 2 h 2 + x 3 h 3 (1 X)x 1 h fg h = (0.0888)( 13,424) + (0.1924)( 8942) + 0 ( )(0.0888)(2444) h = 3087 kj /kg The specific volume is given by: v = x 1 v 1 + x 2 v 2 + x 3 v 3 (1 X)x 1 v fg v = (0.0888)(1.362) + (0.1924)(0.557) + (0.7189)(0.875) ( )(0.0888)(1.361) v = m3 /kg

8 8 CHAPTER 3. FUEL, AIR, AND COMBUSTION THERMODYNAMICS 3.7) A four cylinder four stroke 2.8 L port injected spark ignition engine is running at 2000 rpm on a lean (φ =0.9) mixture of octane and standard air (101 kpa, 298 K) If the octane flow rate is 2.5 g/s, what is the mass of fuel entering each cylinder per cycle and the volumetric efficiency? a) The mass of fuel entering each cylinder per cycle for a four stroke engine ( ) ( ) ( ) ( ) m f = ṁ f = 2.5 N n c m f = g b) Since the engine is port injected e v = m a + m f ρ i v d = m f (AF + 1) ρ i v d = m f ( AFs φ + 1) ρ i v d From Table 3.5, the stoichiometric air-fuel ratio for octane is AF s = Assume R = ρ i = P RT = 101 (0.287)(298) = 1.18 kg /m 3 = 1180 g /m 3 Solving for the volumetric efficiency: η v = ( ) ( ( ) 1) (1180) η v =

9 9 3.8) An engine cylinder has a 90 mm bore and a 85 mm stroke, and contains air and residual gases at 350 K and 1 bar. If the engine is to operate on diesel fuel and run lean with an overall equivalence ratio of φ = 0.7, what is the mass of diesel fuel that needs to be injected during the compression stroke? (Assume R of the air-residual gas mixture is 0.29 kj/kg K). a) m = m a + m r We know that f = m r m Substituting we have Since m a = (1 f)m φ = F A F A s Substituting m f = φ(f A s )(m a ) = φ(f A s )(1 f)m Using the ideal gas law m f = P v d RT = P m f = 0.53 g ( π 4 RT ) b 2 s = 100 ( π 4 ) (0.09) 2 (0.085) (0.29)(350)(1000) From Table 3.5, AF s =14.30 or F A s = The mass of injected diesel fuel is: m f = (0.7)(0.0699)( )(0.53) m f = g

10 10 CHAPTER 3. FUEL, AIR, AND COMBUSTION THERMODYNAMICS 3.9) Using the low temperature (T < 1000K) combustion equations, what are the composition, enthalpy, and entropy of the combustion products of methanol, CH 3 OH, at φ = 1.1, T = 1200 K, and P = 101 kpa? Compare with the results from the program ecp.m.

11 3.10) What are the mole fractions of CO 2, H 2 O, CO, N 2 and H 2 produced when methane (CH 4 ) is burned in rich conditions at φ = 1.1, T = 1000 K, and P = 101 kpa? 11

12 12 CHAPTER 3. FUEL, AIR, AND COMBUSTION THERMODYNAMICS 3.11) If a lean (φ = 0.8) mixture of methane CH 4 is burned at a temperature of 1500 K and pressure of 500 kpa, what are the mole fractions of the products, and the product enthalpy, entropy, and specific heat? Use the program ecp.m.

13 ) At what temperature does the saturation pressure P sat of an octane droplet equal 0.5 bar? At that temperature, what is the enthalpy of vaporization h fg? a) Antoine s equation is: [ log 10 (P sat ) = a b ] T + c Solving for T and using coefficients from Table 3.3 T = b a log 10 (P sat ) c Using coefficients from Table T = ( ) = 375 K (octane) log 10 (0.5) T = ( ) = 498 K (tetradecane) log 10 (0.5) This shows the greater volatility of octane relative to tetradecane b) The molar enthalpy of vaporization h fg is: h fg = Ae dt /Tc (1 TTc ) β Using coefficients from Table 3.4: ( h fg = exp ( h fg = exp 694 ) ( ) ( ) = 36.2 MJ /kmol (octane) ) = 53.1 MJ /kmol (tetradecane) The octane value compares well with the value of 36.4 MJ /kmol in Table 3.2. requires about 47% more energy to vaporize. The tetradecane

14 14 CHAPTER 3. FUEL, AIR, AND COMBUSTION THERMODYNAMICS 3.13) Compare the enthalpy of vaporization h fg (MJ/kmol) of nitromethane, methanol, octane, and tetradecane at 400 K. a) The molar enthalpy of vaporization h fg is ( h fg = A exp α T ) ) β (1 TTc T c Using coefficients from Table 3.4 and molecular mass from Table 3.5 For Nitromethane: h fg = (53.33) exp h fg = 0.53 MJ /kg For Methanol: For Octane: h fg = (45.30) exp h fg = 0.95 MJ /kg h fg = (58.46) exp h fg = 0.30 MJ /kg For Tetradecane: h fg = (95.66) exp h fg = 0.31 MJ /kg ( ) ( ) = MJ /kmol = ( ) ( ) = MJ /kmol = ( ) ( ) = MJ /kmol = ( ) ( ) = MJ /kmol = The charge cooling effect of the vaporization of methanol and nitromethane is 2-3 times that of octane and tetradecane on a per kg basis. The charge cooling produces greater volumetric efficiencies by increasing the density of the cylinder gases.

15 3.14) a.) If a rich (φ = 1.1) mixture of diesel fuel is burned at a temperature of 2000 K and pressure of 750 kpa, what are the mole fractions of the products, and the product enthalpy, entropy, specific volume, and specific heat? b.) Repeat the calculation for φ = Discuss the effects of equivalence ratio. 15

16 16 CHAPTER 3. FUEL, AIR, AND COMBUSTION THERMODYNAMICS 3.15) Using the program ecp.m, plot the product equilibrium mole fractions as a function of equivalence ratio (0.5 < φ < 2) resulting from the combustion of methane at 5000 kpa and 2500 K.

17 ) Derive Equation (??) for the species mole fractions of a mixture of air and residual gas. a) The mixture m is composed of the residual mass m r and the premixed fuel-air m fa m = m f + m fa The residual mole fraction y r is y r = n r n fa + n r = 1 n fa n r + 1 Since the residual fraction f = mr /m or 1 f = m = m r + m fa = 1 + m fa m r m r m r m fa m r = 1 f 1 The mole ratio is n fa n r = m fa M fa Mr m r = m fa m r M M = ( ) 1 M f 1 M So y r = [1 + M M ( )] 1 1 f 1 The species mole fractions y i are y i = n i N = n i nfa n fa N + n i nr n r N = ( ni n fa ) y fa + Since the sum of the residual fractions must equal 1 y fa = 1 y r y i = n i n fa y i = n i n r ( ni n r ) y r So the species mole fractions accounting for both the residual gas and the inlet fuel- air y i = (1 y r ) y i + (y r ) y i

18 18 CHAPTER 3. FUEL, AIR, AND COMBUSTION THERMODYNAMICS 3.17) At what equivalence ratio for octane-air mixtures does the carbon to oxygen ratio of the system equal one? Why is this of interest? a) The combustion equation is C 8 H 18 + a s φ (O N 2 ) Products From Table 3.5, a s = for octane. For every mole of octane, there are 8 carbon atoms and 2as /φ oxygen atoms. The carbon/oxygen ratio is: [C] [D] = 8 ( 2 ) a s φ So this ratio is equal to one, i.e. CO formation, φ = 2a s 8 = (2)(12.50) = Therefore for φ > 3.125, there will be solid carbon in the products, since the carbon atoms are in excess of those used to form CO.

19 3.18) At what temperature is the concentration of H 2 a minimum for the combustion of gasoline and air at φ = 1.2 and 4500 kpa? What is that minimum value of H 2? 19

20 20 CHAPTER 3. FUEL, AIR, AND COMBUSTION THERMODYNAMICS 3.19) At what equivalence ratio is the concentration of OH a maximum for the combustion of diesel and air at T = 2500 K and 4500 kpa? What is that maximum value of OH?

21 3.20) At what temperature does the mole fraction of NO reach for the equilibrium products resulting from the combustion of gasoline and air at φ = 1.0 and 5000 kpa? 21

22 22 CHAPTER 3. FUEL, AIR, AND COMBUSTION THERMODYNAMICS 3.21) At what temperature does the mole fraction of CO reach for the equilibrium products resulting from the combustion of methane and air at φ = 1.1 and 3000 kpa?

23 3.22) What is the equilibrium and the frozen specific heat c p of the combustion products of gasoline at a pressure of 2000 kpa and temperature of 2000 K burned at a.) an equivalence ratio of 1.1, and b.) an equivalence ratio of 0.9? 23