?
Thanks for bringing math into the equation: it does away with all the guessing and the subjectivity.
I'm still having a problem with your conclusion, however:
I'm still failing to see why situation #2 would mean a cooler smoke.
You're welcome
I can appreciate how 2 may be a bit obtuse; for the uninitiated, the inverse relationship with the inequality doesn't help things. So, I'll try to explain and hope that works. Recall:
∂Q-thinner/∂T-thinner > ∂Q-thicker/∂T-thicker
So, assume the heat transfer (∂Q) is equivalent between the two systems. A quick algebraic manipulation later (shown for completeness):
∂Q-thinner/∂T-thinner > ∂Q-thicker/∂T-thicker
∂Q-thinner/∂Q-thicker > ∂T-thinner/∂T-thicker
since ∂Q-thinner = ∂Q-thicker, then ∂Q-thinner/∂Q-thicker = 1
1 > ∂T-thinner/∂T-thicker
or
∂T-thicker > ∂T-thinner
Since these are the same systems, with only the wall thickness different, then that means: ∂T = T-combustion - T-smoke, with ∂T-combustion between the two systems being equivalent. Tn order to preserve the inequality (a larger temp differential with the thicker system vs. thinner system, the T-smoke-thicker must be > T-smoke-thinner. Algebraically:
∂T-thicker > ∂T-thinner
(T-combustion - T-thicker-smoke exit) > (T-combustion - T-thinner-smoke-exit)
- T-thicker-smoke-exit > -T-thinner-smoke-exit
T-thicker-smoke-exit < T-thinner-smoke-exit
I hope that helps, Olkofri!
To the rest, a few comments:
- While smoking characteristics, and briar, and grain, and ember sizes, and everything else vary wildly between one smoke and the next, they were held constant for this analysis, in order to answer the OP's question
- The temperature differential is likely quite small, and I am sure smoking dynamics can overwhelm what difference there is. Therefore, the many views on this subject, based upon the individuals empirical experiences!
- Net/net: For me, over the long run, I take far more pleasure from smoking a pipe, than thermodynamic calculations
Lee