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a b Figure 22: a. Carnot engine operating between hot gases and the ambient; b. Carnot engine operating between water and the ambient. 100´(1 – 298÷1000) » 70 kJ, as illustrated in Figure 22a. Therefore, the quality of energy at 1000 K is 70%. The entropy change of the hot gases is –0.1 kJ K–1 (= –100 ÷ 1000), while the entropy gain for the ambient is 0.1 kJ K–1 (i.e., 30 kJ÷300 K). Alternatively, we can cool the hot gases using water to transfer the 100 kJ of energy. Assume that during this process the water temperature rises by 2 K from 399 K to 401 K (with the average water temperature being 400 K, as shown in Figure 22b). If a Carnot engine is placed between the water at 400 K and the ambient at 300 K, then for the same 100 kJ of heat removed from radiator water, we can extract only 100´(1 – 298÷400) » 25 kJ, and 75 kJ is rejected to the ambient. In this case, the energy quality is only 25% of the extracted heat. Figure 23 illustrates the processes depicted in Figure 22a and b using a T–S diagram. The cy-cle A–B–C–D–A in Figure 23 represents the Carnot engine (CE) of Figure 22a, while area ABJIA and EGKIE represent heat transfer from engine and to hot water, respectively, for Figure 22b, while the area C–D–I–H represents the rejected heat of CE for the first case, and the area D-C-H–K–J–I–D that for the latter case. Since more heat is rejected for the second case, the work potential or the quality of the thermal energy is degraded to a smaller value at the lower temperature. This is due to the irreversible heat transfer or the temperature gradients between hot gases and radiator water (as shown in Figure 22b). In general property gradients cause entropy generation. Now, one might ask about the Maxwell-Boltzmann distribution of molecular veloci-ties. Consider a monatomic gas within a container with rigid adiabatic walls. A “pseudo” tem-perature distribution exists for the monatomic gas. The question is whether with collision and transfer of energy, there can be degradation of energy or generation of entropy. First, tem-perature is a continuum property and the temperature cannot be associated with a group of molecules. Secondly, after frequent collisions, at that location where frequent transfers occur, the intensive state is not altered over a time period much larger than collision time. Thus, no gradient exists and there is no entropy generation. a. Adiabatic Reversible Processes Recall that for any process within a closed or fixed mass system, dS = dQ/Tb + ds. For any reversible process ds =0 so that dS = dQ/T. For an adiabatic reversible process, dQ = ds = 0, so that dS = 0. Consequently, the entropy remains unchanged for an adiabatic reversible process. These processes are also known as isentropic processes. E. ENTROPY EVALUATION The magnitude of heat transfer can be determined through measurements or by ap-plying the First law. Thereupon, in the context of Eq. (28), if the entropy change is known, ds may be determined for a process. The entropy is a property that depends upon the system state and is evaluated at equilibrium. Consider the irreversible process illustrated in Figure 24 involving the sudden com-pression of a gas contained in a piston–cylinder assembly with a large weight. The dashed curve in Figure 24 depicts the accompanying irreversible process. Applying the First law to the process we obtain Q12 – W12 = U2 – U1. The change in state due to an irreversible process can also be achieved through a se-quence of quasiequilibrium processes as described by the path A in Figure 24. Applying the First law to this path, we obtain the relation Q1-2R–2 – W1–2R–2 = U2 – U1. Integrating Eq. (28) (with ds= 0) along this path A , S2 – S1 = ò12 dQR/T. (33) since ds = 0. The infinitesimal heat transfer dQR along the path A is obtained from the First law for a sequence of infinitesimal processes occurring along the reversible path 1–2R–2, i.e., dQR = dU + dWR = dU + P dV. Therefore, Eq. (33) may be written in the form A B E F G D C H I J K Figure 23: T-s diagram for the processes indicated in the previous figure. S S2 S1 U Path B 2R’ irrever V 2R Path A Figure 24: An irreversible process depicted on a U-V-S diagram. An illustration of estimating s by reversible path. S2 – S1 = ò(dU + PdV)/T. (34) Integrating this relation between the initial and final equilibrium states S2 – S1 = òU2 T−1dU+òV2 PT−1dU. (35) The values of pressure and temperature along the path 1–2R–2 in Eq.(35) are different from those along the dashed line 1–2, except at the initial (T1, P1) and final (T2, P2) states. If the state change is infinitesimal dS = (dU/T) + (P/T) dV, or (36) TdS = dU + PdV, (37) which is also known as the TdS relation. Equation (37) results from a combination of the First and Second laws applied to closed systems. 1.2 1 R-12 0.12 0.03 0.8 v= 0.0022/mkg 0.6 0.4 0.2 0 150 200 250 300 u, kJ/k Figure 25: Variation of s with u with v as a parameter. Since the entropy is a property, the difference (S2– S1) as shown in Eq. (35) is a func-tion of only the initial (U1,V1) and final (U2,V2) states, i.e., for a closed system S = S(U,V). For example, if the initial and the final pressures and volumes are known, the temperature dif-ference T2 - T1 can be determined using the ideal gas relation T2 = P2 V2/(mR) and T1 = P1 V1/(mR), even though the final state is reached irreversibly, i.e., the functional relation for T2 -T1 is unaffected. Likewise, to determine the final functional form for the difference (S2–S1), any reversible path A or B may be selected, since its value being path–independent depends only upon the initial and final states. (This is also apparent from Eq. (36) from which it follows that dS = 0 if dU = dV = 0.) For the processes being discussed, the internal energy change as-sumes the form dU = T dS – P dV. (38) For an infinitesimal process, Eq. (38) represents the change of internal energy between two equilibrium states with the properties U and U+ dU, S and S +dS, and V and V+dV. Recall from Chapter 1 that the higher the energy, the greater the number of ways by which molecules distribute energy. In confirmation, according to Eq.(36), as the internal en-ergy increases in a fixed mass and volume system, the entropy too must increase. Therefore, the entropy is a monatomic function of the internal energy for a given volume and mass. The gradient of the entropy with respect to the internal energy is the inverse of the temperature T–1. If the internal energy is fixed, Eq.(36) implies that as the volume increases, so does the entropy (which confirms the microscopic overview outlined in Chapter 1). This is to be expected, since more quantum states are available due to the increased intermolecular spacing. Upon integrating Eq. (36), functional relation for the entropy is S = S(U,V) + C, or U = U(S,V) + C. (39a and b) The latter is also known as the Gibbs fundamental relation for systems of fixed mat-ter. If the composition of a system is known, it is possible to evaluate the constant C which is a function of the number of moles of the various species (N1, N2,…, etc.) or their masses (m1, m2,…, etc.) that are contained in the closed system of fixed total mass m. If the composition of the system is fixed, i.e., if the number of species moles N1, N2,…, etc. are fixed, then S = S(U,V) which is also known as the fundamental equation in entropy form. On a unit mass basis Eq. (36) may be written in the form ds = du/T + Pdv/T, (40) so that for a closed system of fixed mass s = s (u,v). (41) Figure 25 contains an experimentally–determined relationship between s and u with v as a pa-rameter for the refrigerant R–12. Since dU = dH – d (PV), Eq. (36) assumes the form dS = dH/T – VdP/T. (42) It is apparent from Eq. (42) that S = S(H,P). Writing Eq.(42) on unit mass basis ds = dh/T – v dP/T, i.e., (43) s = s(h,P). (44) Note that only for exact differentials or differentials of properties can one give the functional relation like Eq. (44). On the other hand consider the example of electrical work supplied to a piston–cylinder–weight assembly resulting in gas expansion. In that case. the work dW = P dV –Eelec dqc, (45) where dqc denotes the electrical charge and Eelec the voltage. It is not possible to express W = W(V,qc), since dW is an inexact differential (so that W is not a point function). 1. Ideal Gases Substituting for the enthalpy dh = cp0 (T) dT, Eq. (43) may be written in the form ds = cpo dT/T – R dP/P. (46) a. Constant Specific Heats Integrating Eq. (46) from (Tref,Pref) to (T,P) s(T,P) – s(Tref,Pref) = cpo ln(T/Tref) – R ln(P/Pref). Selecting Pref = 1 atmosphere and letting s(Tref,1) = 0, we have s(T,P) = cp0 ln (T/Tref) – R ln (P(atm)/1(atm)). Selecting an arbitrary value for Tref, and applying Eq. (47b) at states 1 and 2, s(T2,P2) – s(T1,P1) = cpo ln(T2/T1) – R ln(P2/P1). For an isentropic process s2 = s1. Consequently, cpo ln(T2/T1) = R ln(P2/P1). Since R = cpo – cvo, applying Eq. (47d) T2/T1 = (P2/P1)k/(k-1), or P2/P1 = (T2/T1)(k-1)/k, (47a) (47b) (47c) (47d) (47e and f) where k = cp0/cv0. Finally, upon substituting for T = Pv/R in Eq. (47e), we obtain the relation ... - tailieumienphi.vn
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