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#### ericandrews

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« on: July 19, 2005, 07:03:53 PM »
Hey there, simple question, regarding a PV diagram for a combustion engine, whats differene between isentropic expansion and adiabatic expansion ? Definitions say isentropic doesn't change in temp and entropy. Why not just call the expansion adiabatic ? (for an ideal combustion chamber)

#### Donaldson Tan

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« Reply #1 on: July 23, 2005, 12:18:54 AM »
an adiabatic process is one whereby there is no transfer of heat between the system and its surroundings, thus Q (heat) = 0

an isoentropic process is one whereby entropy remains constant. if a process is both adiabatic and reversible, then it is considered as isoentropic.

adiabatic expansion can occur in a well-insulated system. neglecting kinetic energy, electrical energy, etc, drop in enthalpy of the system is effectively converted to work. an adiabatic expanion is thus considered to be most expanion. (dH = Q + W, where Q = 0) even in an ideal engine, it's not 100% efficient. this is well-demonstrated in Carnot'e engine.
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#### ppithermo

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« Reply #2 on: September 12, 2005, 05:51:14 PM »
An isentropic expansion is a  process which is both reversible and adiabatic. The entropy of the system remains constant.

An adiabatic expansion is a process where the system does not exchange heat with its envrironment. But in general, it is irreversible.

As a consequence of an irreversible expansion, the entropy of the adaibatic system increases.

#### Juan R.

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« Reply #3 on: September 14, 2005, 04:42:12 AM »
almost all textbooks work just with the elementary equilibrium formulation for closed systems.

In general, from nonequilbrium thermodynamics

dS = diS + dQopen/T + dmatterS

The last term accounts the transfer of matter with exterior and diS is the production of entropy due to irreversible phenomena. The form of production term is standard and known by decades: the product of thermodynamic flows J by thermodynamic forces F or affinities, diS = Sum F·J.

The mass term which does not appear in standard textbooks is easily understood. Take a flask with 50 mL of water at T, it has entropy S. Now add 50 mL of water also at T. The variation of heat (dQ) is zero and T is constant but variation of entropy is NOT zero because entropy is extensive magnitude and now the 100 mL system has entropy 2 S. The usual expression dS = dQ/T appears in textbooks is only valid for closed systems.

When one works with a closed system

dS = diS + dQ/T

since production of entropy is nonnegative. Many textbooks write above expression like

dS >= dQ/T

but if one want compute explicitely the production of entropy one may use the equation instead of the inequality.

For reversible processes, the production is zero

dS = dQ/T