Entropy Change by Heat Transfer Define Thermal Energy Reservoir (TER) –Constant mass, constant volume –No work - Q only form of energy transfer –T uniform.

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Presentation transcript:

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 T B >T A

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 (P s = 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: 1.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 2.Reversible engines operating between the same TERs have the same  th,rev

Carnot (Ideal) Cycle Internally reversible Interaction with environment reversible QhQh QLQL W in W out T S Reversible work S - constant Reversible heat transfer T - constant

Carnot efficiency Define efficiency: QHQH QLQL W 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 s O (T) values are tabulated for different gases in Tables D

For a Calorically Perfect Gas