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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.

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

1 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

2 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

3 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

4 2nd Law No entropy change could occur because: - Isentropic process (P s = 0) - entropy production cancelled by heat loss  Ps -  Q/T = 0

5 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)

6 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

7 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

8 Carnot efficiency Define efficiency: QHQH QLQL W This is the best one can do

9 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

10 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.

11 Enthalpy Relations for a Perfect Gas Show yourself: fn (T)fn (p)

12 Calculating  s Calculate temperature and pressure effects separately s O (T) values are tabulated for different gases in Tables D

13 For a Calorically Perfect Gas


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