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Entropy Change by Heat Transfer. Define T hermal E nergy R eservoir (TER) Constant mass, constant volume No work - Q only form of energy transfer T uniform and constant. Entropy Change by Heat Transfer. Consider two TERs at different Ts, in contact but isolated from surroundings
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Entropy Change by Heat Transfer • Define Thermal Energy Reservoir (TER) • Constant mass, constant volume • No work - Q only form of energy transfer • T uniform and constant
Entropy Change by Heat Transfer • Consider two TERs at different Ts, in contact but isolated from surroundings Heat transfer between TERs produces entropy as long as TB>TA
Second Law for Control Mass • Mechanical Energy Reservoir (MER) • CM interacts with a TER and an MER • MER no disorder; provides only reversible work • Overall system isolated
2nd Law No entropy change could occur because: - Isentropic process (Ps = 0) - entropy production cancelled by heat loss Ps - Q/T = 0
Alternative Approach to 2nd Law Clausius • It is impossible to design a cyclic device that raises heat from a lower T to a higher T without affecting its surroundings. (need work) Kelvin-Planck • It is impossible to design a cyclic device that takes heat from a reservoir and converts it to work only (must have waste heat)
Carnot’s Propositions • Corollaries of Clausius and Kelvin-Planck versions of 2nd Law: • It is impossible to construct a heat engine that operates between two TERs that has higher thermal efficiency than a reversible heat engine. th,rev> th,irrev • Reversible engines operating between the same TERs have the same th,rev
Carnot (Ideal) Cycle • Internally reversible • Interaction with environment reversible Qh T Reversible work S - constant Reversible heat transfer T - constant Win Wout QL S
Carnot efficiency • Define efficiency: QH W QL This is the best one can do
Gibbs Equation • State equations relate changes in T.D. variables to each other: e.g., q - w = du • If reversible and pdv work only • In terms of enthalpy: dh = du + d(pv) dh = du + pdv + vdp; Tds = dh -vdp-pdv+pdv Tds = du + pdv Tds = dh - vdp
Unique aspect of Thermodynamics • The Gibbs Equations were derived assuming a reversible process. • However, it consists of state variables only; i.e., changes are path independent. • Proven for reversible processes but applicable to irreversible processes also.
Enthalpy Relations for a Perfect Gas Show yourself: fn (T) fn (p)
Calculating s • Calculate temperature and pressure effects separately sO(T) values are tabulated for different gases in Tables D
