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MAE 5310: COMBUSTION FUNDAMENTALS

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1st Law (Energy Balance) 2nd Law (Equilibrium Relations) Adiabatic Flame ... motors are clustered in modules located on the end of outrigger booms in order ... – PowerPoint PPT presentation

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Title: MAE 5310: COMBUSTION FUNDAMENTALS


1
MAE 5310 COMBUSTION FUNDAMENTALS
  • Lecture 6 Adiabatic Combustion Equilibrium
  • September 3, 2008
  • Mechanical and Aerospace Engineering Department
  • Florida Institute of Technology
  • D. R. Kirk

2
COMMENTS ON KP
  • The Kp of a reaction depends on temperature only
  • Independent of pressure of the equilibrium
    mixture
  • Not affected by presence of inert gases
  • The Kp of the reverse reaction is 1/ Kp
  • The larger the Kp, the more complete the reaction
  • If Kp gt 1,000 (or ln Kp gt 7) reaction assumed
    complete
  • If Kp lt 0.001 (or ln Kp lt -7) reaction assumed
    not to occur
  • Mixture pressure affects the equilibrium
    composition (but not Kp)
  • Presence of inert gases affects the equilibrium
    composition
  • When stoichiometric coefficients are doubled, the
    value of Kp is squared
  • Free electronic in the equilbrium composition can
    be treated as an ideal gas
  • Equilibrium calculations provide information on
    the equilibrium composition of a reaction, not on
    the reaction rate.

3
ADIABATIC COMBUSTION EQUILIBRIUM
  • Previously we have considered
  • Known Stoichiometry 1st Law (Energy Balance) ?
    Adiabatic Flame Temperature
  • Problems 1-4
  • Known P and T 2nd Law (Equilibrium Relations) ?
    Stoichiometry
  • Problems 5-9
  • Now we can combine these
  • 1st Law (Energy Balance) 2nd Law (Equilibrium
    Relations) ? Adiabatic Flame Temperature
    Stoichiometry
  • Problems 10-14
  • Solution Scheme
  • Guess a TTguess
  • Do equilibrium calculation to solve for species
    concentrations at Tguess
  • Plug into 1st Law
  • We want F(Tguess)0
  • If F(Tguess) gt 0, then initial guess was too high
  • If F(Tguess) lt 0, then initial guess was too low
  • Increment Tguess

4
EXAMPLE MONOPROPELLANT ROCKET PROPULSION
  • Monopropellant rockets are simple propulsion
    systems that rely on chemicals which, when
    energized, decompose
  • Decomposition creates both the fuel and an
    oxidizer (which allows the fuel to burn), which
    then react with each other
  • Because they only use a single propellant,
    monopropellant rockets are quite simple and
    reliable, but not very efficient
  • Mainly used to make small adjustments such as
    attitude control
  • Typical Specific Impulse 100-300 sec
  • Typical Thrust 0.1-100 N

Monopropellant hydrazine (N2H4, see Problem 12)
thrusters that were used for trajectory
correction maneuvers (TCMs) during interplanetary
cruise, thrust vector control (TVC) during VOI,
orbit trim maneuvers during the mapping mission,
and attitude control when the action wheels are
being desaturated. The rocket motors are
clustered in modules located on the end of
outrigger booms in order to increase their moment
arm and thus decrease attitude control propellant
requirements. Twelve 0.9-N (Newton) and four 22-N
rocket motors are used for attitude control, with
thrust being provided by eight 445-N rocket
motors or by the 0.9-N motors for small TCMs.
5
EXAMPLE MONOPROPELLANT ROCKET PROPULSION
  • Rocket propellant chemists have proposed a new,
    high-energy liquid oxidizer, penta-oxygen, O5,
    which is also a monopropellant.
  • Calculate the monopropellant decomposition
    temperature at a chamber pressure of 10 atm. If
    it is assumed the only products are O atoms and
    O2 molecules.
  • Supplemental Information
  • Heat of formation of new oxidizer is estimated to
    be very high 1,025 kJ/mole
  • O5 enters system at 298 K
  • Amount of O2 and O should be calculated for one
    mole of O5 decomposing
  • Solution Technique
  • Solution is iterative
  • In order to calculate the final temperature of
    the mixture, one needs to the composition of the
    mixture at equilibrium but the composition of
    the mixture can only be found if the final
    temperature is known.
  • Strategy
  • Guess final temperature and calculate mixture
    composition using equilibrium concept
  • Once composition is known at guessed temperature,
    adiabatic flame temperature is calculated and
    checked against guessed value
  • Repeat process until guessed and calculated
    temperatures are same
  • Answer lies somewhere between 4,000 and 5,000 K
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