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 = δQ/T b + δσ. For any reversible process δσ =0 so that dS = δQ/T. For an adiabatic reversible process, δQ = δσ = 0, so that dS = 0. Consequently, the entropy remains unchanged for an adiabatic reversible process. These processes are also known as isentropic processes. b a Figure 22: a. Carnot engine operating between hot gases and the ambient; b. Carnot engine operating between water and the ambient. 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, δσ 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 Q 12 – W 12 = U 2 – U 1 . 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 Q 1 - 2R–2 – W 1–2R–2 = U 2 – U 1 . Integrating Eq. (28) (with δσ= 0) along this path A , S 2 – S 1 = ∫ 1 2 δQ R /T. (33) since δσ = 0. The infinitesimal heat transfer δQ R 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., δQ R = dU + δW R = dU + P dV. Therefore, Eq. (33) may be written in the form KI J FE CD A B G H Figure 23: T-s diagram for the processes indicated in the previous figure. S 2 – S 1 = ∫(dU + PdV)/T. (34) Integrating this relation between the initial and final equilibrium states S 2 – S 1 = TdU PTdU U U V V −− ∫ + ∫ 11 1 2 1 2 . (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 (T 1 , P 1 ) and final (T 2 , P 2 ) 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. S V U S 2 S 1 Path B Path A 2R irrever 2R’ Figure 24: An irreversible process depicted on a U-V-S diagram. An illustration of estimating s by reversible path. Since the entropy is a property, the difference (S 2 – S 1 ) as shown in Eq. (35) is a func- tion of only the initial (U 1 ,V 1 ) and final (U 2 ,V 2 ) 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 T 2 - T 1 can be determined using the ideal gas relation T 2 = P 2 V 2 /(mR) and T 1 = P 1 V 1 /(mR), even though the final state is reached irreversibly, i.e., the functional relation for T 2 - T 1 is unaffected. Likewise, to determine the final functional form for the difference (S 2 –S 1 ), 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) 0 0.2 0.4 0.6 0.8 1 1.2 150 200 250 300 u, kJ/ k s, kJ/kg K R-1 2 v= 0.0022m 3 /k g 0.0 3 0.1 2 Figure 25: Variation of s with u with v as a parameter. 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 (N 1 , N 2 ,…, etc.) or their masses (m 1 , m 2 ,…, 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 N 1 , N 2 ,…, 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 δW = P dV –E elec δq c , (45) where δq c denotes the electrical charge and E elec the voltage. It is not possible to express W = W(V,q c ), since δW is an inexact differential (so that W is not a point function). 1. Ideal Gases Substituting for the enthalpy dh = c p0 (T) dT, Eq. (43) may be written in the form ds = c po dT/T – R dP/P. (46) a. Constant Specific Heats Integrating Eq. (46) from (T ref ,P ref ) to (T,P) s(T,P) – s(T ref ,P ref ) = c po ln(T/T ref ) – R ln(P/P ref ). (47a) Selecting P ref = 1 atmosphere and letting s(T ref ,1) = 0, we have s(T,P) = c p0 ln (T/T ref ) – R ln (P(atm)/1(atm)). (47b) Selecting an arbitrary value for T ref , and applying Eq. (47b) at states 1 and 2, s(T 2 ,P 2 ) – s(T 1 ,P 1 ) = c po ln(T 2 /T 1 ) – R ln(P 2 /P 1 ). (47c) For an isentropic process s 2 = s 1 . Consequently, c po ln(T 2 /T 1 ) = R ln(P 2 /P 1 ). (47d) Since R = c po – c vo , applying Eq. (47d) T 2 /T 1 = (P 2 /P 1 ) k/(k-1) , or P 2 /P 1 = (T 2 /T 1 ) (k-1)/k , (47e and f) where k = c p0 /c v0 . Finally, upon substituting for T = Pv/R in Eq. (47e), we obtain the relation Pv k = Constant. (47g) b. Variable Specific Heats Consider an ideal gas that changes state from (T ref ,P ref ) to (T,P). Integrating Eq. (46) and setting s(T ref ,P ref ) = 0 we have s(T,P) = (c (T) / T) p0 T T ref dT ∫ – R ln (P/P ref ). For ideal gases, the first term on the right is a function of temperature alone. Setting P ref = 1 atm, the entropy s(T,P) = s 0 (T) – R ln (P(atm)/1(atm)), where (48a) s 0 (T) = (c (T) / T) p0 T T ref dT ∫ . (48b) If data for the specific heat c p0 (T) are available (Tables A-6F), Eq. (48b) can be read- ily integrated. In general, tables listing s 0 (T) assume that T ref = 0 K(Tables A-7 to A-19). There- fore, s(T,P) = s 0 (T) – R ln (P/1), (49) where the pressure is expressed in units of atm or bars. If P = 1 atm or approximately bar, s(T, 1) = s o (T) which is the entropy of an ideal gas at a pressure of 1 bar and a temperature T. The second term on the RHS of Eq. (49) is a pressure correction. Applying Eq. (49) to states 1 and 2, s(T 2 ,P 2 ) – s(T 1 , P 1 ) =s 0 (T 2 ) –s 0 (T 1 ) – R ln (P 2 /P 1 ). (50a) For an isentropic process s 0 (T 2 ) –s 0 (T 1 ) – R ln (P 2 /P 1 ) = 0. (50b) Therefore, for an isentropic process, if the initial and final pressures, and T 1 are specified, s 0 (T 2 ) can be evaluated. Using the appropriate tables for s 0 (T) (e.g.: Tables A-7 to A-19), T 2 may be determined. For processes for which the volume change ratios are known, it is useful to replace the pressure term in Eq. (50b) using ideal gas law: s 0 (T 2 ) –s 0 (T 1 ) – R ln ((RT 2 /v 2 )/(RT 1 /v 1 )) = 0. Simplifying this relation, s 0 (T 2 ) –s 0 (T 1 ) – R ln (T 2 /T 1 ) + R ln (v 2 /v 1 ) = 0. (50c) For a known volume ratio and temperature T 1 , Eq. (50c) may be used to solve for T 2 itera- tively. In order to avoid the iterative procedure, relative pressures and volumes, P r and v r , may be defined using Eq. (50b) as follows (further details are contained in the Appendix to this chapter) P r (T)= exp(s 0 (T)/R)/exp (s 0 (T ref ´)/R), and (50d) v r = (T /exp (s 0 (T)/R))/(T ref ´/exp(s 0 (T ref ´)/R)), (50e) where T ref ´ is an arbitrarily defined reference temperature. For air, T ref ´ is taken to be 273 K, and P r = 0.00368 exp(s 0 (T)/R) Equations (50b) and (50c) can also be written in the form P 2 /P 1 =P r2 /P r1 , and (50f) v 2 /v 1 = v r2 /v r1 . (50g) The value of v r in SI units is based on the relation v r = 2.87 T/P r Tabulations for P r and v r particularly for solution of isentropic problems were necessary in the past due to the nonavailability of computers. Since their advent, the system properties at the end of isentropic compression or expansion are readily calculated. The isentropic and nonisentropic processes can now be explained as follows. Con- sider a monatomic gas. When an adiabatic reversible compression process occurs in a closed system the work input is converted into a translational energy increase (e.g., due to increased molecular velocity (V x 2 + V y 2 + V z 2 ) because of a force being applied in a specific direction, say “x” which increases V x ). Thus, the total number of macro-states cannot change. A crude way to interpret is that dS = dU/T + P dV/T so that S generally increases with increased energy U but decreases due to a decrease in volume V. The entropy first increases due to increased U because of work input (the first term on the RHS) but decreases due to the reduced volume (as the second term, due to the intermolecular spacing, is reduced and, consequently, the number of states in which energy can be stored also decreases). The second term counteracts the en- tropy rise due to the increased internal energy, and the entropy is unchanged. l. Example 12 Air is adiabatically and reversibly compressed from P 1 = 1 bar, and T 1 = 300 K to P 2 = 10 bar. Heat is then added at constant volume from a reservoir at 1000 K (T R ) until the air temperature reaches 900 K (T 3 ). During heat addition, about 10% of the added heat is lost to the ambient at 298 K. Determine: The entropy generated σ 12 in kJ kg –1 K –1 for the first process 1–2; The net heat added to the matter; The heat supplied by the reservoir; The entropy generated in an isolated system during the process from (2) to (3). Solution S 2 – S 1 – ∫δQ/T b = σ 12 .(A) Since the process is reversible, σ 12 = 0, (B) which implies that no gradients exist within the system. Therefore, T b = T. (C) Using Eqs. (A), (B), and (C) S 2 – S 1 = ∫δQ/T. (D) Since the process is adiabatic δQ = 0, and S 2 = S 1 or s 2 = s 1 . At state 1, from the air tables (Tables A-7), p r1 = 1.386, u 1 = 214.07, and h 1 = 300.19. Therefore, s 1 = s 0 (T 1 ) – R ln P/1 = 1.702 – 0 = 1.702 kJ kg –1 K –1 . For the isentropic process p r2 (T 2 )/ p r1 (T 1 ) = p 2 /p 1 = 10. Hence, p r2 = p r1 10 = 1.386×10 = 13.86 so that T 2 = 574 K, u 2 = 415 kJ kg –1 , h 2 = 580 kJ kg –1 , and s 2 = s 1 = 1.702 kJ kg –1 K –1 . Temperature gradients can develop inside a system during heat addition from a ther- mal reservoir or heat loss to the ambient, thereby making a process internally irre- versible. In this example, the final states are assumed to be at equilibrium. Applying the First law to the constant volume process, the heat added to the system can be evaluated as follows q 23 = u 3 – u 2 = 674.58 – 415 = 260 kJ kg –1 . If q R denotes the heat supplied by reservoir, the heat added q 23 = 0.9 q R , i.e. q R = 288.88 kJ kg –1 . The heat loss to the ambient is q 0 = 288.88 – 260 = 28.88 kJ kg –1 . Since we must determine the entropy of an isolated system, assuming that there are no gradients outside that system, and selecting the system boundaries to include the res- ervoir at T R and the ambient at T 0 , it follows that s 3 – s 2 – q R /T R – q 0 /T 0 = σ. Now, P 3 /P 2 = T 3 /T 2 = 900÷574 , i.e., P 3 = 15.68 atm, and s 3 = 2.849 – 0.287 ln (15.68÷1) = 2.059 kJ kg –1 K –1 . Therefore, 2.059 – 1.702 – (289/1000) – (–29/298) = σ so that σ = 0.165 kJ kg –1 K –1 . Remarks It is possible to tabulate p r values for a particular gas using Eq. (50c). 2. Incompressible Liquids For incompressible liquids and solids, the specific volume v is constant. Since u = u(T,v), for incompressible substances it follows that u = u(T). The intermolecular spacing in incompressible liquids is constant and, consequently, the intermolecular potential energy is fixed so that the internal energy varies only as a function of temperature. Since, h = u + Pv, for incompressible liquids h(T,P) = u(T) + P v. Differentiating with respect to the temperature at fixed pressure, c P = (∂h/∂T) P = (∂u/∂T) P . Since u = u(T), it follows that c P = (∂u/∂T) P = du/dT = c v = c, and for incompressible substances du = cdT. (51) The values of c for liquids and solids are tabulated in Tables A-6A and A-6B. Using Eq. (40), ds = du/T + 0, so that ds = cdT/T. (52) Therefore, the entropy is a function of temperature alone. For any substance s = s(T,v) so that if v = constant, s = s(T). Note that Eq. (43) cannot be used since h = h(T,P). Equations (51) and (52) are applied to evaluate the internal energy and entropy of compressed liquids. For example, water at 25ºC and 1 bar exists as compressed liquid, since P > P sat (25ºC). The Steam tables (Tables A-4A) tabulate values of u(T) and s(T) as a function of temperature for saturated water. If the entropy of liquid water is desired at 25ºC and 1 bar, since u(T,P) ≈ u(T,P sat ) = u f (T), and s(T,P) ≈ s(T,P sat ) = s f (T), the respective tabulated values are 104.9 kJ kg –1 K –1 and 0.367 kJ kg –1 K –1 . Likewise, the enthalpy at that state is h = u + P v = 104.9 + 1×100×0.001 = 105 kJ kg –1 . An incompressible substance with constant specific heat is also called a perfect in- compressible substance. For these substances, integration of Eq. (52) between T and reference temperature T ref yields s – s ref = c ln(T/T ref ), (53a) or between two given states s 2 – s 1 = c ln(T 2 /T 1 ). (53b) When an incompressible liquid undergoes an isentropic process, it follows from Eq. (53b) that the process is isothermal. 3. Solids Equation (53), which presumes constant specific heat, is also the relevant entropy equation for incompressible solids. However as T→0, Eq. (53) becomes implausible, forcing us to account for the variation of the specific heat of solids at very low temperatures. At these temperatures c v (T) = 3 R(1 – (1/20)(θ D /T) 2 ), where T » (θ D = 3 R (4 π 4 /5) (T/θ D ) 3 ), (54) where θ D is known as the Debye temperature. A solid that behaves according to Eq. (54) is called a Debye solid. Another pertinent relation is the Dulong–Petit law that states that c v ≈ 3 R. This is based on the presumption that a mole of a substance contains N avag independent oscil- lators vibrating in three directions, with each molecule contributing an amount (3/2)k B T to the energy. Molecules contribute an equal amount of potential energy, i.e., (3/2)k B T. At low tem- peratures, the Dulong–Petit constant specific heat expression leads to erroneous results, and a correction is made using the Einstein function E(T Ein /T), i.e., c v ≈ 3 R E(T Ein /T), where E(T Ein /T) = (T Ein /T) 2 exp(T Ein /T)/(exp(T Ein /T)–1) 2 . Here, T Ein denotes the Einstein temperature. (For many solids, T Ein ≈ 200 K.) As T→0, E(T Ein /T)→T 2 . For coals, c v ≈ 3 R((1/3)E(T Ein,1 /T) + (2/3) E (T Ein,2 /T)), where T Ein,2 denotes the second Einstein temperature. 4. Entropy During Phase Change Consider the case of a boiling liquid. Since the pressure and temperature are generally unchanged during a phase transformation, applying Eq. (43), ds = dh/T – vdP/T = dh/T. (55) Integrating the expression between the saturated liquid and vapor states s g – s f = (h g –h f )/T = h fg /T. (56) Generalizing for any change from phase α to β, s α – s β = h αβ /T (57) m. Example 13 The entropy of water at T tp = 0ºC ,P TP = 0.611 kPa, is arbitrarily set to equal zero, where the subscript tp refers to the triple point. Using this information, determine: s(liquid, 100ºC) assuming c = 4.184 kJ kg –1 K –1 . Compare your results with values tabulated in the Steam tables (Tables A-4). s(sat vapor, 0ºC, 0.611 kPa) assuming h fg = 2501.3 kJ –1 kg –1 K –1 . The entropy generated if the water at 0ºC and 0.611 kPa is mechanically stirred to form vapor at 0ºC in an adiabatic blender. s(393 K, 100 kPa) assuming c p,0 = 2.02 kJ kg –1 K –1 and that steam behaves as an ideal gas. Solution Applying Eq. (53), s(373) – s(273) = 4.184 ln (373/273) = 1.306 kJ kg –1 K –1 . Since s(0ºC) = 0, s(100ºC) =1.306 kJ kg –1 K –1 . From the Table A-4A, s(100ºC) = 1.3069 kJ kg –1 K –1 , which is very close. Applying Eq. (56) to the vaporization process at the triple point, s g – s f = 2501.3÷273 = 9.16 kJ kg –1 K –1 . Since, s f (273 K, 0.611 kPa) = 0, s g (273 K, 0.611 kPa) = 9.16 kJ kg –1 K –1 .(A) ds – δq/T b = δσ. Since δq = 0, ds = δσ. Integrating this expression, s g – s f = σ. (B) Using Eq. (A) and (49b), with s f (0ºC, 0.611 kPa) = 0 s g – s f = 9.16 kJ kg –1 K –1 = σ. (C) s(393 K, 100 kPa) – s(273 K, 0.611) = 2.02 ln (393÷273) – (8.314÷18.02)ln (100÷0.611) = – 1.616 kJ kg –1 K –1 , or s(393, 100 kPa) = 9.16 – 1.616 = 7.54 kJ kg –1 K –1 . Conventional Steam tables (e.g., Table A-4A) yield a value of 7.467 kJ kg –1 K –1 . Figure 26: P–v diagram for water. [...]... u(l,145.5) = 37 7.7 + 5442 =5819.9 kJ kmole–1, and u(l,240 .3) = 5819.9 + 76.5 × (240 .3 – 145.5) = 131 32 kJ kmole–1 In the gaseous state s (g, 240 .3) – s (l,240 .3) = 20,058÷240 .3 = 83. 5 kJ kmole–1 K–1, i.e., s (g,240 .3) = 83. 5 + 142.51 = 226.01 kJ kmole–1 K–1 (point G in Figure 33 ) Therefore, s (g,240 .3) = 226 kJ kmole–1 K–1, and h(g, 240 .3) = 131 32 + 20.058 = 33 ,190 kJ kmole–1 u(g, 240 .3) = 33 ,190 - 8 .31 4 ×... with respect to temperature for cyclopropane C3H6 for which θ = 130 What are the values of the entropy and internal energy at 15 K (cf also Figure 33 ) Solution The molar specific entropy (0 K, 1 bar) = 0 kJ kmole–1 K–1 (i.e., point A in Figure 33 a) Since s = ∫ c s dT / T , s = (1944 T3/ 3) /3 kJ kmole–1 K–4 At 15 K, s (15 K, 1 bar) = (1944 × 1 53/ 130 3)/ 3 = 0.995 kJ kmole–1 K–1 The internal energy u... (145 – 15) + 11.2 = 37 7.7 kJ kmole–1 The liquid entropy may be evaluated as follows: s (l, 145.5) – s(s,145.5) = 5442÷145.5 = 37 .40 kJ kmole–1 K–1 Therefore, s (l, 145.5) = 37 .4 + 66.71 = 104.11 kJ kmole–1 K–1 (point D in Figure 33 ) so that s (l,240 .3) – s(l,145.5) = 76.5 ln(240 .3 145.5) = 38 .4 kJ kmole–1 K–1, and s (l,240 .3) = 38 .4 + 104.11 = 142.51 kJ kmole–1 K–1 (point F in Figure 33 ) Since h = u +... + 0.06 × 254 .3 = 23. 92 kJ K–1, S2 = 0.04 × 221.8 + 0.06 × 2 63. 6 = 24.69 kJ K–1, and S2 – S1 = 24.69 – 23. 92 = 0.77 kJ K–1 S2 – S1 – Q12/Tb = σ12 Applying the First law, Q12 = U2 – U1 + W12 Therefore, U2 = 0.04 × 438 71 + 0.06 × 26799 = 33 62.8 kJ, U1 = 0.04 × 34 455 + 0.06 × 21815 = 2687.1 kJ, and Q12 = 33 62.8 – 2687.1 + 65 .3 = 741 kJ (E) 31 0K 32 0 K (1) Initial State (2) Final State Figure 29: Illustration... N2 = 0 .36 + 0 .37 6 – 0.188 = 0.548 kmole The entropy increases due to the temperature rise as well as the larger number of moles of gas in the room, while the gas volume remains the same If the room is divided into three equal parts A, B, and C, then VA = VB = VC = V /3 = 9 /3 m3, and likewise SA = SB = SC = S /3, N O2 ,A = = N O2 /3, and UA = UB = UC = U /3 Therefore, U(SA,VA, ) = U(S /3, V /3, N O2 /3, …)... – R ln(4.4÷1) = 234 .1 – 8 .31 4 × ln(4.4÷1) = 221.8 kJ kmole–1 K–1, and, 0 sCO2 (1000K, 4 bar) = sCO2 (1000 K) – R ln(4÷1) = 216.6 kJ kmole–1 K–1 Likewise, 0 sN2 (1200K, 6.6 bar) = sN2 (1200 K) – R ln(6.6÷1) = 279 .3 – 8 .31 4 × ln (6.6÷1) = 2 63. 6 kJ kmole–1 K–1, and sN2 (1000K, 6 bar) = 269.2 – 8 .31 4 × ln 6 = 254 .3 kJ kmole–1 K–1 Using Eqs (C) and (D) S1 = 0.04 × 216.6 + 0.06 × 254 .3 = 23. 92 kJ K–1, S2... Therefore, ˙ σ cv = –100 36 00×24/ (31 0×1000)+0.54×1.610+12.456×1.702– (0.54+ 0.519×24)×1. 735 = 28 .35 kJ K–1 day–1 Per unit mass ˙ σ cv = 28 .35 /70 =0.405 kJ kg–1 day–1 The life span is 10000/ (36 5×0.405) = 68 years Remarks In Chapter 11, the irreversibility due to metabolism will be considered ˙ From Example 12 in Chapter 2 we see that q G∝mb -0 .33 while empirical results sug˙ G (kW/kg) = 0.0 035 52mb-0.26 Part... Example 7, Wmin = – (Ut,2 – Ut,1) + T0 (St,2 – St,1), Wmin = – (mc(T2 – T1) – m usf) + T0 (mc ln (T2/T1) – m usf/Tfreeze), or wmin = Wmin/m = – (4.184×(0–25) 33 5)+298×(4.184×ln(2 73 298) 33 5÷2 73) = 439 .6 – 298 × 1.594 = 35 .41 kJ kg–1 Remarks Figure 32 contains a representation of the process on a T–s diagram The area under the path in the figure represents the reversible heat absorbed from the tank We... ÷ 0.004 = 25 m3 kmole–1 so that s2 / s1 = ( c v0 ln(T2/Tref) + R ln( v2 / vref ))/( c v0 ln(T1/Tref) + R ln( v1 / vref )) Using the values Tref = 2 73 K, vref = 1 m3 kmole–1, c v0 = 20 kJ kmole–1 K–1, and R = 8 .31 4 kJ kmole–1 K–1, T2 = 186 K The chemical work, Wchem,rev = –∫µdN = –(U2 – U1) Now, U2 – U1 = 0.02 × 20 × (186 – 2 73) – 0.004 × 20 × (30 0 – 2 73) = 36 .96 kJ, i.e., Wchem,rev = +36 .96 kJ a Example... at 15 K, u = (1944 × 154 ÷ 130 3 ) ÷ 4 = 11.2 kJ kmole–1 w Example 23 At a pressure of 1 bar, evaluate the entropy and internal energy of cyclopropane C 3 H6 when it exists as (a) saturated solid; (b) saturated liquid; and (c) saturated vapor given that the specific heat cs follows the Debye equation with θD = 130 K and m = 3 when T < 15 K, c s = 28.97 kJ kmole–1 K–1 Figure 33 : Illustration of a: a: P–T . ideal gas. Solution Applying Eq. ( 53) , s (37 3) – s(2 73) = 4.184 ln (37 3/2 73) = 1 .30 6 kJ kg –1 K –1 . Since s(0ºC) = 0, s(100ºC) =1 .30 6 kJ kg –1 K –1 . From the Table A-4A, s(100ºC) = 1 .30 69 kJ kg –1 K –1 ,. s f = 9.16 kJ kg –1 K –1 = σ. (C) s (39 3 K, 100 kPa) – s(2 73 K, 0.611) = 2.02 ln (39 3÷2 73) – (8 .31 4÷18.02)ln (100÷0.611) = – 1.616 kJ kg –1 K –1 , or s (39 3, 100 kPa) = 9.16 – 1.616 = 7.54 kJ. u sf /T freeze ), or w min = W min /m = – (4.184×(0–25) 33 5)+298×(4.184×ln(2 73 298) 33 5÷2 73) = 439 .6 – 298 × 1.594 = 35 .41 kJ kg –1 . Remarks Figure 32 contains a representation of the process on a