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Re: [Phys-L] Carnot (?) efficiency of non-Carnot cycles





On 3/3/2015 11:38 PM, John Denker wrote, in part:
...The Carnot efficiency formula involves two temperatures, namely the temperature of /the/ heat source and /the/ heat sink. ...
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It has always seemed puzzling that the high temperature source in an internal combustion engine is at the temperature of the internal combustion of a reasonably stoichiometric mixture of hydrocarbon vapor and air, in a flame. Flame temperatures are given as about 1000 degC for a Bunsen (yellow flame), 1300 degC for a blow-torch, and 1400 degC for a candle (!!) while the sink temperature is that of hot water regulated to 82 degC.
Using 1273K and 355K as the relevant temperatures, the Carnot efficiency % would be
100(1 - 355/1273) = 72%
- compared with the efficiency usually given for the automobile otto engine, which is 30%.
Even acknowledging that frictional and other drag effects reduce practical efficiency, perhaps it would be better to use the temperature of the flame when mixed with the air compressed at the ignition point to represent the source temperature thermodynamic efficiency more realistically?

Brian Whatcott