Entropy Generation in a Rod: Heat Transfer Considerations

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SUMMARY

The discussion focuses on entropy generation in a rod subjected to a temperature gradient, specifically when one end is in contact with a heat source at temperature T1 and the other with a sink at temperature T2. The temperature distribution along the rod is described by the equation T=T1-(T1-T2)*X/L. Participants debate whether entropy is generated within the rod during heat transfer, concluding that while no irreversibility occurs in the rod itself, entropy increases due to the heat transfer process, as dictated by the second law of thermodynamics. The key takeaway is that while the entropy generation term (Δsgen) may be zero, the total change in entropy (dS) is not, as heat flows from the hot end to the cold end.

PREREQUISITES
  • Understanding of the second law of thermodynamics
  • Familiarity with heat transfer principles
  • Knowledge of entropy concepts and calculations
  • Basic grasp of thermal equilibrium and temperature gradients
NEXT STEPS
  • Study the implications of the second law of thermodynamics on closed systems
  • Explore detailed heat transfer mechanisms in insulated materials
  • Learn about entropy calculations in non-equilibrium thermodynamics
  • Investigate the effects of temperature gradients on material properties
USEFUL FOR

This discussion is beneficial for thermodynamics students, engineers working with heat transfer systems, and researchers interested in entropy and thermal dynamics in materials.

kittu1421
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Hi..
Consider a rod which is insulated on its lateral surface, now this rod is brought in contact with a source at temperature T1 and sink at temperature T2 now a temperature gradient sets up in the rod after steady state is reached temperature at some distance X from the source end is given as
T=T1-(T1-T2)*X/L
now this rod is removed and ends are insulated as well
then my question is as heat flows from high temp. region to low temp. region is any entropy generated within the rod...?
what i feel is it should not because there are no irreversibility in the rod during heat transfer, even the heat transfer across any rod element takes place at temp. difference tending to zero, hence in eqn.
ds=δq/T+Δsgen, Δsgen should be zero so that eqn. becomes
ds=δq/T
check me if i am correct and if i have missed something notify me.
 
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Your argument is that if an isolated system is not at equilibrium, it will not move toward equilibrium with time (since that obviously would increase entropy)? Seriously?! S = ∫dQ/T and you think the heat flow magically stops once the ends are insulated? I draw your attention to the second law of thermodynamics. (as well as the zeroth law).
 
Entropy will increase but that increase will be due to heat transfer from hot end to cold end and not because of some generations due to irreversibility and that's what i have said above in last equation i have taken entropy generatio(Sgen=0) and not change in total entropy equal to zero(dS!=0)
 

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