Qantas A380 uncontained #2 engine failure

[chris weston]

OK I'll stop lurking and head goes above parapet.

If we really, really want to get fundamental over this, bear is of course right, it's to do with the hot gas signature and the energy being released by the combustion process.

The super scripts and subscripts are getting a trifle mangled here - the square should be a triangle and they should carry the standard state symbol that looks like a flying saucer but here goes anyway:

The theoretical maximum energy available from the engine is directly proportionate to the relationship:

 G =  H (c) - T S (system)

- which is one way of writing the second law of thermodynamics. (Yes, yes standard conditions.)

G is measuring the energy ‘available’ from/to a reaction relative to equilibrium - hence the name (Gibbs) free energy and which gives us a handle on the maximum thrust theoretically available.
 H (c) is the enthalpy of combustion of the fuel
T is the temperature in Kelvin
 S (system) is the entropy change of the chemicals in the system

• we have the mother of an exothermic reaction going on here as the fuel burns making H negative which in turn makes G negative
• we have a large positive value for T but the - sign in front of TS makes this contribution negative too, making G negative, making the reaction more likely statistically.
• we also have a large, positive, S (sys) as we are forming lots of gas molecules on the right hand side of the combustion reaction, the - sign in front of TS makes this contribution negative too, making G negative, making the reaction more likely statistically.

If G is negative this implies that the reaction will be spontaneous (on energetic grounds) and will make energy available to do work on shoving the 380 hard.

How this energy gets transferred around the engine is for the engineers and not mere chemists.....