Species h2 o2 h2o oh o h reactant 06667 03333 0 0 0 0

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Unformatted text preview: se three before reaction and after combustion are listed below. Species H2 O2 H2O Reactant 0.6667 0.3333 0 Product 0.2915 0.1457 0.5628 As seen in the table, the dissociation is very significant; about 30% of the products is H2. Let’s find out how much fuel is not burned by considering the following stoichiometric reaction: H2 ðgÞ þ 0:5O2 ðgÞ ! X Á H2 þ 0:5X Á O2 þ ð1 À XÞ Á H2 OðgÞ The mole fraction of H2 in the products is xH2 ¼ X X ¼ : X þ 0: 5X þ 1 À X 0: 5X þ 1 With xH2 ¼ 0.2915, we get X ¼ 0.3412. If we assume 66% of fuel is burned, a new ^ estimate based on cp at 1,500 K leads to Tp ¼ 300 K þ 0:66 Á 241:88 kJ=mol $ 3;700 K 0:047 kJ/mol À K ^ that is in much better agreement with the equilibrium result. If we estimate cp at 1,800 K we get Tp ¼ 300 K þ 0:66 Á 241:88 kJ=mole $ 3;514:7 K: 0:04966 kJ/mole À K 40 2 Thermodynamics of Combustion If we include additional species, H, OH, and O in the products, the predicted equilibrium temperature drops to 3,076 K. The table below shows the mole fractions of each species in this case. Species H2 O2 H2O OH O H Reactant 0.6667 0.3333 0 0 0 0 Product 0.1503 0.0510 0.5809 0.1077 0.0330 0.0771 Evidently, the radicals OH, H, and O take some energy to form; note that their values for enthalpy of formation are positive. Because the space shuttle engine operates at 18.94 MPa (2,747 psi, ~186 atm) at 100% power, the pressure needs to be taken into consideration as the combination of radicals occurs faster at higher pressures. The predicted equilibrium temperature at 18.94 MPa is 3,832.4 K and the mole fractions are listed below. Species H2 O2 H2O OH O H Reactant 0.6667 0.3333 0 0 0 0 Product 0.1169 0.0336 0.7051 0.1005 0.0143 0.0296 The energy needed to vaporize liquid H2 and O2 and heat them from their boiling temperatures to 25 C are estimated to be 8.84 kJ/mol and 12.92 kJ/mol (energy ¼ latent heat + sensible energy from boiling point to STP). With H2 þ 0.5O2, the total energy required is then 8.84 þ 0.5·12.92 or about 15.3 kJ/mol. The temperature drop due to this process is about ~15.3 kJ/(0.049 kJ/mol-K) ¼ 148 K. With this, we estimate the space shuttle main engine temperature is 3,832 À 148 K or ~3,675 K. The follow...
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This document was uploaded on 01/20/2014.

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