Thermo entropy question
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Thermo entropy question

[From: ] [author: ] [Date: 13-08-11] [Hit: ]
and dS is the change in entropy. The equal sign will hold for a reversible process. For a reversible isentropic process, there is no transfer of heat energy and therefore the process is also adiabatic.Therefore, in isentropic process (adiabatic andreversible process.......
I'm doing a problem that has an isentropic compression of an ideal gas. the temperature changed from state 1 to state 2 and when I looked them up in the ideal gas table they have different values for entropy. I thought isentropic mean s1=s2

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From the 2nd law of thermodynamics.
delta Q is less than for equal to T dS

where delta Q is the amount of energy the system gains by heating, T is the temperature of the system, and dS is the change in entropy. The equal sign will hold for a reversible process. For a reversible isentropic process, there is no transfer of heat energy and therefore the process is also adiabatic.
Therefore, in isentropic process (adiabatic and reversible process.)
delta Q = 0 (no heat transfer)
=> dS = 0
s1 =s2

This constant entropy occurred across the component.
Example the inlet and the outlet of the isentropic turbine have same entropy s1=s2. Based on 2nd law.

On the other hand, the entropy that you looked up from ideal gas table, the value is a "state" and that s value depends on T, P, and V.
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keywords: Thermo,question,entropy,Thermo entropy question
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