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Thermodynamics [ENGR 251] [Lyes KADEM 2007] CHAPTER VI ENTROPY nd In the previous chapter, we tried to understand the 2 law of thermodynamics from a conceptual perspective. In this chapter, we begin to consider these concepts in a more analytical manner. In the process, we will introduce a new property, ENTROPY, which is defined as follows: Q dS T rev That is, a differential change in entropy corresponds to a differential quantity of heat being transferred divided by the temperature at which is it transferred in a reversible process. I. Clausius Inequality: nd Recall the Kelvin-Plank statement of the 2 Law. “It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work.” nd Clausius proceeded to consider a device, shown below, which violates the 2 Law. Work is produced, but the device receives heat from a single reservoir Thermal reservoir @ TR Q R Reversible Heat Wrev Engine Q System Wsys st Let us apply the 1 Law for the control volume: Esystem Qin Wout The total work output is the sum of that for the system and the heat engine: W T = (W rev + W sys) Entropy 77 Thermodynamics [ENGR 251] [Lyes KADEM 2007] For an infinitesimal part of the engine cycle we write: dE system QR WT Since the cyclic device is a reversible heat engine, it follows from the definition of the thermodynamical temperature scale that: QR/Q = TR/T Then, we can obtain: TR dEsystem Q WT T Consider the operation of the device through a complete engine cycle. For a cycle the energy of the control volume must return to its initial value so that the energy term drops out. Q 0 TR WT T Or Q WT TR T As noted previously, if the total work is positive then this device would violate the Kelvin-Plank nd statement of the 2 Law. It follows that the total work must be zero or negative. Q 0 T Now let us run the engine in reverse. If the path for the cycle is reversed, then the sign of the integral is reversed and the integral will become greater than or equal to zero. This would nd likewise result in a violation of the 2 Law. The only way that the process can be reversed is if the integral gives a value of zero. If the integral of Q/T about a cycle is equal to zero, then the process is said to be reversible In the special case that the cyclic integral is zero, then the quantity under the integral, Q/T, returns to its initial value during the cycle. It behaves like a thermodynamic property. Then, the entropy (S): Q dS T rev Entropy 78 Thermodynamics [ENGR 251] [Lyes KADEM 2007] II. Increase in Entropy Principle Since Q/T is zero for a reversible process and less than zero for an irreversible process, then Q Q T T rev and Q dS T Finally, for any process between two states: Q dS T or 2 Q S2 S1 1 T or 2 Q S2 S1 S generated 1 T A process can only occur if the entropy generated is greater than or equal to zero And this can be considered as a quantative statement of the second law. Example Air is compressed by a 8 kW compressor from P1 to P2. The air temperature is maintained o constant at 25 C during this process as a result of heat transfer to the surrounding medium at o 10 C. Determine the rate of entropy change of the air. State the assumptions made in solving this problem. III. Gibbs equations or Tds Relationships st From the 1 Law for a closed system: dE system Qin Wout Neglecting kinetic and potential energy, this reduces to: dU system Qin Wout From the definition of entropy, S, we have: Qrev T dS Entropy 79 Thermodynamics [ENGR 251] [Lyes KADEM 2007] For a simple compressible substance, we can write a reversible work term as: W = P dV After substitution and rearranging T dS = dU + P dV The alternate form of this relationship is obtained using enthalpy: H=U+PV So that after differentiation, dH = dU +P dV + V dP Replacing the last two terms in the above T-dS relationship: T dS = dH – V dP These T-dS relationships form the basis for evaluation of entropy as a thermodynamic property. IV- Entropy for Liquids & Solids & ideal gases IV.1. Entropy for liquids and solids Normally we expect volume changes to be small as liquids or solids undergo a change in thermodynamic state. Furthermore, Cp Cv C so that: T ds = du + P dv C dT Then, ds = C dT/T or, s2 – s1 = Cavg ln(T2/T1) IV.2. Entropy for ideal gases Using the T-ds relationships, we substitute the ideal gas law for the last terms: T ds = du + P dv = Cv dT + R T dv/v and T ds = dh – v dP = Cp dT – R T dP/P Divide by T and integrate: Entropy 80 Thermodynamics [ENGR 251] [Lyes KADEM 2007] s2 – s1 = Cv,avg ln(T2/T1) + R ln(v2/v1) 1st T-ds relationship and s2 – s1 = Cp,avg ln(T2/T1) – R ln(P2/P1) 2nd T-ds relationship Note that the above expressions may be placed on a molar basis by dividing by the molecular weight of the particular gas: ŝ2 – ŝ 1 = Ĉv,avg ln(T2/T1) + (R/M) ln(v2/v1) and ŝ2 – ŝ 1 = Ĉp,avg ln(T2/T1) – (R/M) ln(P2/P1) IV.2.1. Isentropic Processes (S = 0) of Ideal Gases Isentropic refers to a process which occurs at constant entropy, s. From the definition of entropy, dS = ∫δQ/T│rev we see that for a reversible, adiabatic process ΔS = 0. Let us take a look at the specific heats relationships: Cp = C v + R k = Cp/Cv st From the 1 TdS relationship we developed the following relationship for an ideal gas: s2 – s1 = Cv,avg ln(T2/T1) + R ln(v2/v1) For an isentropic process s = 0 so that after simplifying and rearranging. ln(T2/T1) = (R/ Cv,avg)·ln(v1/v2) then, we have: Cp/Cv = 1 + R/Cv and k - 1 = R/Cv Substitute this result into our simplified T-dS relationship: ln(T2/T1) = (k-1) ln(v1/v2) Raise both sides to the e power: k 1 T2 v1 st T1 v 2 1 Isentropic relationship for an ideal gas Entropy 81 Thermodynamics [ENGR 251] [Lyes KADEM 2007] For a particular ideal gas P1v1/T1 = R = P2v2/T2 so that v2/v1 = P2T1/ P1T2 st Substitute into 1 isentropic relationship k 1 T2 P2 k nd T1 P1 2 Isentropic relationship for an ideal gas k-1 And, P2v2/ P1v1 = (v1/v2) So that k P2 v1 rd P1 v 2 3 Isentropic relationship for an ideal gas Example 3 A 0.5 m insulated rigid tank contains 0.9 kg of carbon dioxide at 100 kPa. Now paddle wheel work is done on the system until the pressure in the tank rises to 120 kPa. Determine the entropy change of carbon dioxide during this process in kJ/K. Assume constant specific heats. Entropy 82

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