[Gasification] Underwater gasification?

[Henri Naths]

Somewhere the efficiency ratio changes because, although Vancouver had the misfortune of the fast ferry fiasco, the jet drive system is a popular marine propulsion engine in certain size ships.There seem to be a definite limit to the efficiency as per power to weight ratio.
H.
  ----- Original Message ----- 
  From: Daniel Chisholm 
  To: Discussion of biomass pyrolysis and gasification 
  Sent: Thursday, May 26, 2011 12:52 PM
  Subject: Re: [Gasification] Underwater gasification?


  A Humphrey is a low-compression ratio, low pressure piston engine.  As such it will have a lower thermal efficiency than a higher compression ratio engine, though that does not necessarily mean it is a bad solution.  The elegance of a Humphrey comes from the fact that its piston and its workload happen to be one and the same thing (a column of water), which allows it to avoid high speed machinery, reduction gearboxes, rotary pumps etc, all of which have a dollar and efficiency cost.  (and a side comment for what it's worth - the combustion process in a Humphrey engine would be a lot closer to being a constant volume rather than a constant pressure one).


  With a reaction engine, the lower the speed of the vehicle, the more strongly its propulsive efficiency depends on accelerating a large mass a small amount.  Marine applications are very slow speed, which is why efficient marine propulsion systems use very large propellers, which accelerate enormous columns of water a very small amount.


  This is why a container ship uses large propellers rather than a jet boat type of propulsion.


  Propelling a ship by means of a steam jet would be an even bigger mismatch (an exhaust velocity of hundreds of meters per second, versus a vessel speed of only ten or so m/s), the efficiency would be miniscule.


  Perhaps this might help to understand why a reaction engine's exhaust speed affects its propulsive efficiency:
    a.. The energy that it takes to accelerate a mass jet is proportional to the mass flow rate, times the square of the jet velocity (which is to say - to the kinetic energy that you are delivering to the jet).  
    b.. The thrust that that mass jet produces is proportional to the mass flow rate times the jet velocity (which is to say - to the momentum that you are delivering to the jet)
  If you look to optimize the problem (reduce the amount of energy it takes to produce a given level of thrust), you will see that propulsive efficiency is improved by accelerating a larger amount of mass by a smaller delta-V.



  -- 
  - Daniel
  Fredericton, NB  Canada



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