2ndLaw-Part2-Spring2010-9

# 2ndLaw-Part2-Spring2010-9 - Entropy Change by Heat Transfer...

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Unformatted text preview: 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 δPS = dSsystem δPS = d ( STER + SMER + SCM ) δPS = − δQ T + dSCM dSCM = δQ T + δPS dSCM ≥ ∂Q T No entropy change could occur because: - Isentropic process (Ps = 0) € - entropy production cancelled by heat loss δPS - δQ/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 KelvinPlanck 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. ηth,rev> ηth,irrev 2.  Reversible engines operating between the same TERs have the same ηth,rev Carnot (Ideal) Cycle •  Internally reversible •  Interaction with environment reversible T Win QL S Qh Wout Reversible work S - constant Reversible heat transfer T - constant Carnot efficiency •  Define efficiency: ηth = W net , out Q − QL Q =H =1− L QH QH QH Q Q PS = H − l TH TL QH W QL over a reversible cycle QH Q QH T = l⇒ =H TH TL QL TL Carnot efficiency : ∫P S =0 € This is the best one can do Gibbs Equation •  State equations relate changes in T.D. variables to each other; e.g., p=ρRT •  Find a state equation using s: δq - δw = du If reversible and pdv work only Tds = du + pdv In terms of enthalpy: dh = du + d(pv) dh = du + pdv + vdp; Tds = dh -vdp-pdv+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 du p c v dT R Gibbs : ds = + dv = + dv TT T v 2 2 2 dT dv Integrate : ∫1 ds = ∫1 c v + ∫1 R T v Δs = ∫ 2 1 € € ȹ ρ1 ȹ dT cv + R lnȹ ȹ T ȹ ρ 2 Ⱥ Show yourself: ȹ p1 ȹ dT Δs = ∫1 c p + R lnȹ ȹ T ȹ p2 Ⱥ 2 fn (T) fn (p) Calculating Δs •  Calculate temperature and pressure effects separately Δs12 = s − s 0 2 0 1 dT where s = ∫T = 0 c p T 0 T note : s ≡ fn(T, only ) € sO(T) values are tabulated for different gases in Tables D 0 € For a Calorically Perfect Gas ȹ T2 ȹ ȹ p2 ȹ ȹ T2 ȹ ȹ ρ 2 ȹ Δs12 = c p lnȹ ȹ − R lnȹ ȹ = c v lnȹ ȹ − R lnȹ ȹ ȹ T1 Ⱥ ȹ p1 Ⱥ ȹ T1 Ⱥ ȹ ρ1 Ⱥ ȹ p2 ȹ ȹ T2 ȹ ȹ p2 ȹ Δs12 c p ȹ T2 ȹ γ = lnȹ ȹ − lnȹ ȹ = lnȹ ȹ − lnȹ ȹ R R ȹ T1 Ⱥ ȹ p1 Ⱥ γ − 1 ȹ T1 Ⱥ ȹ p1 Ⱥ γ Ⱥ γ Ⱥ ȹ T2 ȹγ −1 ȹ p2 ȹ Δs12 Ⱥ (T2 T1)γ −1 Ⱥ = lnȹ ȹ − lnȹ ȹ = ln Ⱥ R p2 p1 Ⱥ ȹ T1 Ⱥ ȹ p1 Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ € T-s Diagram •  •  •  •  Useful to plot processes Similar to p-v or T-v plots Tracks entropy For perfect gases: p1>p2>p3 • for non-calorically perfect gas use h-s diagram T p1 p2 p3 s T-s Diagram •  •  •  •  •  •  Different “constant processes: Isothermal (T const.) T Isobaric (p const.) Isentropic (s const.) Can only move left if cooling Can move right if: –  heating –  Irreversible (PS>0) p T s s Isentropic Relations •  We saw: ȹ p2 ȹ 0 0 s2 − s1 = s2 − s1 − R lnȹ ȹ ȹ p1 Ⱥ ⇒ ȹ p ȹ s lnȹ ȹ = R ȹ p0 Ⱥ •  Isentropic change from T = 0, 0pO to T, p € • Relative pressure: € € ȹ p ȹ 0 = s − R lnȹ ȹ ȹ p0 Ⱥ 0 ȹ P ȹ ȹ s 0 ȹ Pr = ȹ ȹ = expȹ ȹ ≡ fn (T , only ) ȹ R Ⱥ ȹ P0 Ⱥ • Tabulated in Appendix D Isentropic Relations •  Isentropic process 1 to 2: 0 0 0 ȹ p2 ȹ ȹ s2 − s1 ȹ exp( s2 R) ȹ ȹ = expȹ ȹ = 0 ȹ R Ⱥ exp( s1 R) ȹ p1 Ⱥ s € ȹ p2 ȹ ȹ pr 2 ȹ ȹ ȹ = ȹ ȹ p1 Ⱥ s pr1 Ⱥ ȹ ȹ € Isentropic Relations for Ideal Gases •  We saw: •  Therefore: € ȹ T2 ȹ ȹ p2 ȹ c p lnȹ ȹ = R lnȹ ȹ ȹ T1 Ⱥ S ȹ p1 Ⱥ S ȹ p 2 ȹ ȹ T2 ȹ ȹ ȹ = ȹ ȹ ȹ p1 Ⱥ s ȹ T1 Ⱥ cp R ȹ T2 ȹ = ȹ ȹ ȹ T1 Ⱥ γ γ −1 € ρ2 ρ1 ȹ T2 ȹ p2 T 1 = = ȹ ȹ p1 T2 ȹ T Ⱥ 1 Ⱥ Ⱥ Ⱥ Ⱥ 1 γ γ −1 ȹ T2 ȹ −1 ȹ T2 ȹ = ȹ ȹ ȹ ȹ ȹ T Ⱥ ȹ T Ⱥ 1 1 1 γ −1 = γ −1 Ⱥȹ ȹ γ Ⱥȹ p2 ȹ Ⱥȹ p1 Ⱥ Ⱥ γ −1 = Isentropic Relation for an Ideal Gas p = pvγ = const . ργ dp p γ −1 = γρ (const .) = γ = γRT dρ s ρ we' ll see later (saw ?) that this is the speed of sound squared € Stagnation Conditions •  Defined as the condition that exists when a flow is slowed down reversibly and without loss or gain of energy. V p T V=0 p0 T0 Stagnation Pressure •  Adiabatic, reversible: Adiabatic (h01 = h02), T01 = T02 •  Adiabatic, irreversible: adiabatic 0 Since s2 > s1 Summary •  Isentropic flow (adiabatic and reversible): p0 is constant •  Irreversible but still adiabatic flow: p0 decreases •  T0 is constant even if flow is irreversible •  T0 increases/decreases if heat or work is added/taken away Entropy in an Open System •  Reynolds Transport Theorem: dS d ˆ = ∫CV ρsdV + ∫CS ρs(v rel ⋅ n )dA dt CM dt ˙ ˙s + ∫ δQ dA = d ∫ ρsdV + ∫ ρs(v rel ⋅ n ) dA ˆ P CS T CV CS dt Simplification : uniform flow and Q at constant T ˙ ˙S + ∑heat Q = d ∫ ρsdV + ∑ ms −∑ ms ˙ ˙ P CV transf. T out in dt region steady, one in, one out, one heat transf. region ˙ ˙S + Q = msout − msin ˙ ˙ P T ...
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