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The Second Law of Thermodynamics IV

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The Second Law of Thermodynamics IV The Second Law of Thermodynamics Note: for a reversible process, on p-v diagram 2 2 Pdv = W P ∫ rev 1 ∫ pdv = w 1 1 2 V and If the volume is constant, P 2 2 W=0 Pdv = 0 v1 = v2 ∫ 1 1 V • Similarly, for any reversible adiabatic process, Q=0 T S2 S = S Q = Tds = 0 1 2 rev ∫ Q = 0 S1 S • For any other reversible process T 2 2 1 Q = Tds Q = Tds rev ∫ rev ∫ 1 1 2 S • Clausius Inequality: δQ ≤ 0 ∫ T • Clausius inequality is valid for all thermodynamic cycles, reversible or irreversible. ⎛ δQ ⎞ ⎜ ⎟ = 0 for internally ∫ reversible cycle ⎝ T ⎠rev ⎛ δQ ⎞ ⎜ ⎟ < 0 for irreversible cycle ∫⎝ T ⎠ δQ ⎞ ⎟ represents a property change T ⎠rev δQ ⎞ dS = ⎟ T ⎠rev 2 δQ ⎞ ∆S = S − S = 2 1 ∫ ⎟ 1 T ⎠rev • Consider a cycle composed of a reversible and an irreversible process: irreversible process 2 internally reversible process 1 • The quantity ∆ s = s 2 − s 1 represents the entropy change of the system. 2 • For a reversible process, δQ ∆ s = ∫ entropy transfer with heat 1 T • For an irreversible process 2 δ Q ∆ s = + s sys ∫ gen 1 T • Some entropy is generated during an irreversible process and is always positive quantity. Its value depends on the process, and thus it is not a property of the system. Entropy Change for Pure Substances • The entropy of a pure substance is determined from the tables, just as for any other property. T 3 T P1 2 s1 ≈ s f at T1 s3 =from T1 P2 superheated table 2 1 Saturated liquid- vapor mixture T2 s2 = sf + x 2sfg x 2 s s = s f + xsfg ∆S = m(s2 − s1 ) • T-s diagram of properties: Constant T pressure lines δQ rev = Tds 1 2 2 Q = rev Q = Tds Area under rev ∫ the process s 1 Q rev = T (S 2 − S1 ) • Entropy of a fixed mass can be changed by: 1) Heat transfer 2) Irreversibilities • The entropy of a fixed mass will not change during a process that is internally reversible and adiabatic. • For a reversible, adiabatic process (called Isentropic process) ∆s =0⇒s1 =s2 • If a steam turbine is reversible, and the turbine is insulated (thus the process is reversible and adiabatic) T then ∆ s = 0 1 or s1 = s 2 2 s • The Tds relations: differential form of the first law: δQrev −δwrev = dU second law: δQ rev = Tds TdS= PdV +dU Tds= Pdv+du First Tds relation Tds = du + Pdv h = u + Pv dh = du + Pdv + vdP Tds = dh − vdP Second Tds relation • Entropy changes for Liquids and solids: Tds=du+pdv du pdv ds= + T T • Liquids and solids are incompressible substances dv ≅ 0 du cdT ds = = (c = c = c) T T p v kJ 2 dT T s − s = c = c ln 2 2 1 ∫ kgK 1 T T1 Isentropic process for liquids and solids: T2 s2 − s1 = 0 = cln ⇒ T2 = T1 T1 the isentropic process of an incompressible substance is also isothermal. Note: Tds equations are derived by considering an internally reversible process. An entropy change obtained by integrating these equations is the change for any process. “Entropy is a property and the change in entropy between any two states is independent of the details of the process linking the states.” Entropy change of an Ideal gas Tds equations are used to evaluate the entropy change between two states of an ideal gas. du P ds = + dv T T dh v ds = − dv T T For an ideal gas: du = cvdT dh = c pdT Pv = RT, c p = cv + R dT dv ds = c + R v T v dT dP ds = c − R P T P On integration these equations give, respectively T2 dT v s − s = c + R ln 2 2 1 ∫ v T v T1 1 T2 dT P s − s = c − R ln 2 2 1 ∫ P T P T1 1 To integrate these relations, we must know the temperature dependence of the specific heats. Using Ideal gas Tables: T c (T ) Define so (T ) = ∫ p dT 0 T Where s o ( T ) is the specific entropy at a temperature T and a pressure of 1 atm. Note: The specific entropy is set to zero at the state where the temperature is 0 K and the pressure is 1 atm. Note: Because So depends only on the temperature, it can be tabulated versus temperature, like h and u. Also, T2 dT T2 dT T1 dT c = c − c = so − so ∫ p T ∫ p T ∫ p T 2 1 T1 0 0 then ; o o P2 KJ s2 − s1 = (s2 − s1 ) − R ln P1 Kg.K o o P2 KJ or s2 − s1 = (s2 − s1 ) − Ru ln P1 Kmole.K Entropy change of an Incompressible substance: du = C(T)dT du P for incompressible ds = + dv substance, specific heat T T depends solely on C(T)dT P dT ds = + dv = C(T) temperature, and T T T T2 dT S − S = C(T) ← incompressible 2 1 ∫ T T1 When the specific heat is constant: T2 s2 − s1 = C ln ← incompressible, constant C T1 Isentropic Processes of an Ideal Gas For internally reversible process: 2 dT P s − s = c (T ) − R ln 2 2 1 ∫ p 1 T P1 2 dT V s − s = c (T ) − R ln 2 2 1 ∫ v 1 T V1 For Isentropic process(Reversible, adiabatic process): 2 dT P 0= c (T) −Rln 2 s2 −s1 =0 ∫ p 1 T P1 2 dT V 0= c (T) −Rln 2 ∫ v 1 T V1 T2 P2 0 = c p ln − R ln T1 P1 T2 V2 0 = cv ln + R ln T1 V1 R cv T2 R V2 ⎛ V1 ⎞ ⇒ ln = − ln = ln⎜ ⎟ T1 cv V1 ⎝V2 ⎠ k −1 T2 ⎛ V1 ⎞ ln = ln⎜ ⎟ similarly, T1 ⎝V2 ⎠ k −1 k−1 ⎛ T ⎞ ⎛ V ⎞ ⎛ T ⎞ ⎛ P ⎞ k ⎜ 2 ⎟ = ⎜ 1 ⎟ ⎜ 2 ⎟ = ⎜ 2 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ T ⎟ ⎜ P ⎟ ⎝ T1 ⎠s=constant ⎝V2 ⎠ ⎝ 1 ⎠s=constant ⎝ 1 ⎠ Tds = du + Pdv Ideal gas, Isentropic Process (PV=RT, ds=0) du dv dT dv ds = + P 0 = c + R T T v T v Integrate: Constant = cv ln T + R ln v Pv R use PV = RT ⇒ ln + ln v = Constant R cv ln Pv + ln v k −1 = Constant − ln R = Constant ln(Pv × v k −1 ) = Constant ⇒ Pv k = Constant ← Isentropic process ideal gas, wi th constant specific heats T2 dT P s − s = c − R ln 2 2 1 ∫ p T P T1 1 o o P2 s2 − s1 = s2 − s1 − R ln P1 Isentropic process : s2 − s1 = 0 o o P2 s2 − s1 = R ln P1 o o P2 ⎛ s2 − s1 ⎞ ln = ⎜ ⎟ P1 ⎝ R ⎠ P so − so 2 = exp 2 1 P1 R Relative pressure and relative volume: o P2 exp(s2 / R) = o P1 exp(s1 / R) The quantity exp( s o / R ) is a function of temperature only and is given the symbol Pr, define as the relative pressure, and is tabulated for air and other ideal gases P Isentropic Process P2 r2 = (s1=s2) Ideal gas P P variable specific heats 1 r1 where P = P (T ) and P = P (T ) r1 r 1 r2 r 2 Note: Pr is not truly pressure, and also Pr should not be confused with the reduced pressure of compressibility chart. Similarly a relation between specific volumes and temperatures and for two states having the same entropy can also be developed ⎛ RT ⎞ ⎛ RT ⎞ ⎜ 2 ⎟ ⎜ 2 ⎟ ⎜ ⎟ ⎜ ⎟ v2 ⎝ P2 ⎠ v Pr = or 2 = ⎝ 2 ⎠ v1 ⎛ P1 ⎞ v ⎛ P ⎞ ⎜ ⎟ 1 ⎜ r1 ⎟ ⎜ RT ⎟ ⎜ ⎟ ⎝ 1 ⎠ ⎝ RT1 ⎠ RT The ratio = v r is the relative volume Pr v Isentropic process, v2 r2 = ← Ideal gas variable v v specific heats 1 r1 In general Isentropic process for ideal gas: (s2=s1) Pvk = C ← constant specific heat P ⎫ r2 P2 = ⎪ Pr P1 ⎪ 1 ⎬ ← specific heat is a function of temperature v v r2 = 2 ⎪ v v ⎪ r1 1 ⎭ General, reversible (Polytropic) process for an Ideal gas: n Pv =c, where n is constant P 1 n n Slope=-n P1v1 = P2 v 2 2 n P ⎛ v ⎞ Log V 2 1 P = ⎜ ⎟ 1 ⎜ ⎟ Pv n = c P1 ⎝ v 2 ⎠ 2 2 W = Pdv ∫ V 1 2 c P v − P v = dv = 1 1 2 2 ∫ n 1 v n − 1 Polytropic process on P-v and T-s diagrams P n = ±∞ n=0 Constant pressure n=1 (T=constant) Isothermal process n=k (S=constant) Isentropic process V=constant V T n=k n = ±∞ (v = constant) n=0 (P=constant) n=1 n=1 1<n<k S Isentropic Efficiencies of Steady-Flow Devices • The Isentropic process involves no irreversibilities and serves as the ideal process for adiabatic devices. • The actual process is irreversible and the actual device performance is less than the ideal case. • The more closely the actual process approximates the idealized isentropic process, the better the device will perform. • We define the efficiency (isentropic efficiency) of these devices as a measure of deviation of actual processes from the idealized one. 1) Isentropic efficiency of Turbines w& = h1 − h2 Actual turbine work ηt = Isentropic work P T 1 1 wa h1 − h2a 0 ηt = = <100 0 ws h1 − h2s P2 2a 2s s 2) Isentropic efficiency of Compressors and Pumps T T P2 2a 2s 2a 2s P1 1 1 s s W s η c or η p = W a h 2 s − h1 η c or η p = h 2 a − h1 3) Isentropic efficiency of Nozzles Actual K. E. at Nozzle exit η = N Isentropic K.E. at Nozzle exit 2 P V2a T 1 ηN = 2 1 V2s 2 P2 V2a Note : h1 = h2a + 2a 2 2s s h1 − h2a ηN ≅ h1 − h2s.
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