Thermo Q: S2-S1=0? Why Doesn't Equation Yield 0?

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In a steady-state device, the change in entropy (S2-S1) is not necessarily equal to zero. The equation s2-s1 = s(degree)(T2) - s(degree)(T1) - R ln(p2/p1) indicates that entropy can change due to variations in temperature and pressure, even when the system is in a steady state. This means that while the system may be stable, the entropy can still be affected by the thermodynamic properties of the working fluid, leading to a non-zero change in entropy. Clarification is needed on the conditions under which entropy change is considered, as steady-state does not imply no entropy change.
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For a device operating at steady state, is the change in entropy (S2-S1) equal to 0?
If it is equal to 0, then why doesn't the s2-s1=s(degree)(T2)-s(degree)(T1) - R ln (p2/p1) equation yield 0?

I'm confused on this point...
Any help greatly appreciated!
 
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eku_girl83 said:
For a device operating at steady state, is the change in entropy (S2-S1) equal to 0?


That's False. Where have you seen it?
 

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