Entropy generation

  • Thread starter kittu1421
  • Start date
  • #1
10
0

Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
140
14
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).
 
  • #3
10
0
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 thats 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)
 

Related Threads on Entropy generation

Replies
21
Views
2K
Replies
1
Views
844
  • Last Post
Replies
6
Views
969
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
4
Views
717
  • Last Post
Replies
4
Views
637
  • Last Post
Replies
15
Views
5K
  • Last Post
Replies
4
Views
817
  • Last Post
Replies
1
Views
8K
  • Last Post
Replies
2
Views
953
Top