ADVANCED THERMODYNAMICS ENGINEERING phần 6 ppt

jj. Example 36 temperature? Solution µ JT = (∆h/c p )/∆P = ∆T/∆P = (T e –203)/(40 – 60) = 0.31 K bar –1 . Therefore, T e = –73.2ºC. a. Evaluation of µ JT Recall that dh = c p dT + (v – T(∂v/∂T) P )dP. Since dh ≈ 0 during throttling, we obtain the relation µ JT = (dT/dP) h = –(v – T(∂v/∂T) P )/c P = (T 2 /c P ) ∂(v/T)/∂T, or (156) µ JT = (dT/dP) h = –(1 – Tβ P )/(c P /v). (157) b. Remarks For ideal gases, v = RT/P so that T(∂v/∂T) = v and, hence, µ JT = 0. There is no temperature change due to throttling for ideal gases. For incompressible fluids ∂v/∂T = β P = 0, and µ JT = –v/c P has a negative value. There- fore, liquids generally heat up upon throttling. If v < T(∂v/∂T) P or T β P < 1, µ JT > 0, and vice versa. At the inversion point, µ JT = 0. Inversion occurs when T inv β P = (T/v)(∂v/∂T) P = 1. For a real gas Pv = ZRT, i.e., ∂v/∂T = ZR/P + (RT/P)∂Z/∂T. If (T/v)(∂v/∂T) P = 1, this implies that ZRT/Pv + (RT 2 /Pv)∂Z/∂T = 1, or (∂Z/∂T) P = 0. For cooling to occur, the inlet pressure must be lower than the inversion pressure. Cooling of a gas can also be accomplished using isentropic expansion. We define µ s = (∂T/∂P) s that is related to the temperature decrease due to the work delivered during an isentropic process. Recall that ds = c p dT/T – (∂v/∂T) P dP. For an isentropic process, ds = 0, and µ s = T(∂v/∂T) P /c P = Tvβ P /c P . (158) Dividing Eq.(157) by Eq.(158), µ JT /µ s = 1 – (1/(Tβ P )) (159) Equation (159) provides a relation to indicate relative degree of heating of the isentropic to isenthalpic throttling process. For substances that expand upon heating β P > 0 and, hence, µ s > 0. If β P T > 1, then µ JT < µ s , i.e., the isentropic expansion results in greater cooling than isenthalpic expansion for the same pressure ratio. The values of T inv , β P , β T , and P sat can be directly obtained from the state equations, while µ JT , h, u, c v , and other such properties depend both on the equations of state and The experimentally determined value of µ JT for N 2 is 0.31 K bar –1 at–70°C and 5 MPa. If the gas is throttled from 6 MPa and –67ºC to 4 MPa, what is the final exit the ideal gas properties. Therefore, the accuracy of the state equations of state can be directly inferred by comparing the predicted values of T inv , β P , β T , and P sat with ex- perimental data. The entropy change during an adiabatic throttling process equals the entropy generated. Therefore, its value must be positive, since throttling is an inherently irreversible process. We can determine ds by using the following relation: dh = T ds + v dP. Since dh = 0, K L A B µµ µµ JT = 0 µµ µµ JT <0 µµ µµ JT > 0 Inversion Curve (b) P i P e P A P B (a) Exit Figure 20: (a) Illustration of a throttling process. (b) The variation of temperature with respect to pressure during throttling. (∂s/∂P) h = – (v/T). (160) During throttling, dP h < 0 and v/T > 0; hence Eq.(160) dictates that ds h >0. From second law for fixed mass adiabatic system, ds h (= δσ) =-(v/T) dP h > 0. kk. Example 37 Obtain an expression for µ JT for a VW gas in terms of v and T. Solution The Joule Thomson coefficient µ JT = – (∂h/∂p) T /c p , where (A) (∂h/∂p) T = v – T (∂v/∂T) p . (B) Using the VW equation of state dP = R dT/(v – b) + (–RT/(v – b) 2 + 2 a/v 3 )dv. (C) Since dP = 0, (dv/dT) p = (R/(v – b))/(RT/(v – b) 2 – 2 a/v 3 ) (D) Therefore, µ JT = – (1/c p ) (v – ((T R/(v–b))/(RT/(v–b) 2 – 2 a/v 3 ))), i.e., (E) µ JT = – (v/c p )(R T b v 2 – 2 a (v – b) 2 )/(RTv 3 – 2 a (v – b) 2 )). (F) Remarks For many liquids v ≈ b, i.e., µ JT ≈ –v/c p for liquids. Since µ JT < 0, incompressible liquids will heat up upon throttling. In context of Eq.(F), if b « v, (v–b) 2 ≈ v 2 . Consequently, using Eq.(F), µ JT = –(v/c p ) (RTbv 2 – 2 a v 2 )/(RTv 3 – 2 a v 2 )). (G) Dividing the denominator and numerator by RTv 2 , µ JT = – (v/c p )(b – 2a/RT)/(v –2a/RT) (H) If v » (2a/RT), e.g., for high temperature vapors, µ JT ≈ ((2 a /RT)–b)/c p . (I) At low temperatures, 2a/RT » b, and µ JT >0, i.e., cooling occurs upon throttling. At higher temperatures, µ JT < 0 (i.e., the fluid is heated upon throttling). In the limit T→∞ (when the fluid approaches ideal gas behavior), Eq. (F) implies that µ JT = –b/c p , i.e., µ JT c p /v c ´ = –b/v c ´ = –1/8. This limiting value is based on the real gas equation of state. However, for ideal gases, µ JT = 0, since b→0 for the ideal gas point mass molecules. Recall that b equals the geometrical free volume available for molecules to move around. The corresponding result using the RK equation is µ JT c P /v c ´ ≈ –b/v c ´≈ –0.08664. 2. Temperature Change During Throttling Recall from Chapter 2 that the internal energy of unit mass of any fluid can be changed by frictional process and by performing boundary deformation work (Pdv). For incompressible fluid, dv=0 and hence boundary deformation work is zero; thus "u" can change only with a frictionless process (e.g., flow of liquids through pipes) where the mechanical part of energy “vdP” is converted into internal energy using frictional processes. Thus a combi- nation of both of the processes Pdv and vdP result in temperature change during the throttling process. a. Incompressible Fluid Assume that an incompressible fluid is throttled from a higher to a lower pressure under steady state conditions. Let us follow unit mass as it enters and exits the throttling device. Since dv=0, there is no deformation work and hence u cannot change due to Pdv. However the unit mass is pushed into the throttling device with a pump work of vP i , while the same mass is pushed out with a pump work of vP e . Thus there must have been destruction of mechanical energy from vp i to vP e which is converted into thermal energy (see Chapter 2) resulting in an increase of u and hence an increase of T during throttling. The energy increases by an amount du = –vdP ≈ –v(P e – P i ). (Alternately dh =du+ pdv+ vdp = du +0+ vdp=0.) Also, recall that µ JT ≈ – v/c P < 0. Further throttling is inherently irreversible process and hence entropy always increases in adiabatic throttling process (e.g., increased T causes increased s). b. Ideal Gas Now consider a compressible ideal gas. Visualize the throttling process as a two step procedure. First the specific volume is maintained as though fluid is incompressible and the energy rises by the amount du = –vdP. This leads to gas heating. Now let the volume increase during the second stage (i.e expansion to low pressure). The volume increase cannot change the intermolecular potential energy, since the gas is ideal, but Pdv work is performed. This leads to decrease in the internal energy. The total energy change is du = – vdP – Pdv = – d(Pv), i.e., or du + d(Pv) = dh = 0. For ideal gases, dh = c p dT, and, hence, dT = 0. In this case, the energy decrease by the Pdv deformation work equals the energy increase due to pumping (=- vdP) c. Real Gas In a real gas the additional work required to overcome the intermolecular attraction forces (or the increase in the molecular intermolecular potential energy, “ipe”) must be ac- counted for. Then the temperature can decrease if increase in “ipe” is significant. Consider the relation dh = du + d(Pv) = du +P dv + v dP. Since du = c v dT + (T (∂P/∂T) v – P) dv, dh =c v dT + (T(∂P/∂T) v – P)dv + Pdv + vdP. (161) The terms on the RHS of this equation, respectively, denote (1) the change in the temperature due to change in the molecular translational, vibrational, and rotational energies, (2) the intermolecular potential energy change, (3) the deformation work, and (4) the work required for pumping. During throttling, dh = 0, the intermolecular potential energy increases, the work deformation is positive, and dP < 0 so that vdP <0. In case if vdP ≈ 0, then it is apparent that the net change in the molecular translational, vibrational, and rotational energies is negative, i.e., dT < 0. Recall (15), i.e., µ JT = – (v –T (∂v/∂T) P )/c p = – v/c p + {T (∂v/∂T) P }/c p . The first term on the RHS of this expression represents the heating effect due to flow work, while the second term accounts for the entire Pdv work and the energy required to overcome the intermolecular forces. Generally, both the terms are important for fluids. 3. Enthalpy Correction Charts The Joule Thomson coefficient µ JT = – (∂h/∂P) T /c P . However, – (∂h/∂P) T = –(∂(h–h o )/∂P) T – (∂h o /∂P) T = –(∂(h–h o )/∂P) T –0 = {∂h C /∂P} T . Therefore, µ JT = {∂h C /∂P} T /c P , and µ JT,R = µ JT c P P c /RT c = (∂h R,C /∂P R ) T R . (162) where h R,c are given in enthalpy correction charts (Appendix, Figure B.3). The behavior of h R,C is illustrated in Figure 22. Its value increases with a decrease in the pressure at a specified value of T R along the curve ABI. Consequently, (∂h R,c /∂P R ) T R < 0, and µ JT < 0. On the other hand along curve ICD (∂h R,c /∂P R ) T R > 0. Point I represents the inversion point. The temperature change during throttling that accompanies a pressure change can be determined using the enthalpy correction charts. Recall that ∆h ≈ 0, i.e., h 2 – h 1 = 0, i.e., (h o2 – h C,2 ) – (h o1 – h C,1 ) = 0. Assuming a constant specific heat and considering ideal gases, this leads to the relation (T 2 – T 1 ) = (h C,1 (T R1 , P R1 ) – h C,2 (T R2 , P R2 )) /c P,o . ll. Example 38 sume that c P,o = 1.039 kJ kg –1 K –1 . Solution We will select pressures of 4.1 and 5.1 MPa (that lie in the vicinity of 4.6 MPa) at T R = 203/126.2 = 1.6, and P R,1 = 41/33.9 = 1.2 and P R,2 = 51/33.9 = 1.5. At these conditions h R,C,1 = 0.667, h R,C,2 = 0.531, h C,1 = 8.314× 126.2= 700 kJ kmole -1 , h C,2 = 557 kJ kmole -1 so that µ JT,R = (0.667 – 0.531)/(1.5 – 1.2)= 0.453. Consequently, (c P,o – c P )/R = (∂h R,C /∂T R ) P R = 1.35 ≈ (∂h R,C /∂T R ) P R = 1.5 = (0.583 – 0.774)/0.2 = –0.955. Now R = (8.314/28.02)=0.297 kJ kg –1 K –1 ,c P =0.955×0.297+1.039=1.322 kJ kg –1 K –1 , i.e., µ JT = µ JT,R RT c /(P c c P ) [h 0 (T) -h(T,P)]/RT c A B I C D P Figure 21: Use of the enthalpy correction charts to determine the inversion conditions. Determine µ JT for N 2 at –70ºC and 4.6 MPa using the enthalpy correction charts. As- = 0.453 × 0.297 (kJ kg –1 K –1 )126.2 K ÷(33.9 bar×1.322 kJ kg –1 K –1 ) = 0.379 K bar –1 . 4. Inversion Curves a. State Equations The inversion conditions can be obtained by several means. For instance, either of the cubic equations of state, the enthalpy correction charts, or empirical state equations can be used. Inversion occurs when 1= Tβ P = (T/v) (∂ v/∂T) P (cf. Eq. (157)) (∂ ln v/∂ ln(T)) P = 1. (163) mm. Example 39 Obtain an expression that describes the inversion curve of a VW gas. Solution We will use Eq. (F) developed in Example 37, namely, µ JT = –(1/c p ) (R T b v 3 – 2 a v (v – b) 2 )/(RTv 3 – 2a (v – b) 2 )). (A) At the inversion, µ JT = 0, so that T inv = 2 a (v – b) 2 /(R b v 2 ). (B) In dimensionless form, T inv,R = (27/4) (v R ´ – (1/8)) 2 /v R ´ 2 (C) If the value of v R ´ is known, T inv,R can be obtained using Eq.(C). The corresponding pressure P inv,R is obtained by applying the normalized VW equation of state, i.e., P R = T R /(v R ´ – (1/8)) – (27/64)/v R ´ 2 . (D) The inversion curve is obtained by assigning a sequence of values for v R ´ and calcu- lating the corresponding values of T R (cf. Eq. (C)), and then determining P R from the state equation. Inversion curves for various state equations are presented in Figure 23. Remarks For a large specific volume, (v – b) 2 ≈ v 2 (1 – 2b/v). Using Eq. (A) one can subse- quently show that µ JT ≈ ((2a/RT)–b)/c p , and T inv /T c ≈ 2a/bR T c = 2 (27/64) (R 2 T c 2 /P c )/((1/8)(RT c /P c )RT c ) = 27/4. (D) b. Enthalpy Charts The enthalpy correction charts (Appendix, Figure B-3) that plot (h o – h)/RT c with respect to P R with T R as a parameter can be used to determine the inversion points. The Joule Thomson coefficient µ JT = 0 when (∂T/∂P) h = 0. The inversion condition can be determined using the relation (∂h/∂P) T = 0, i.e., [∂ {(h o – h)/RT c }/∂P] T = 0. The peak value of (h o – h)/RT c ) with respect to P R at a specified value of T R yields the inversion point. c. Empirical Relations For several gases, such as CO 2 , N 2 , CO, CH 4 , NH 3 , C 3 H 8 , Ar, and C 2 H 4 , the inversion curve is approximately described by the expression P R = 24.21 – 18.54/T R – 0.825 T R 2 . (164) 5. Throttling of Saturated or Subcooled Liquids The cooling or heating of a vapor during throttling is partly due to the destruction of the mechanical part of the energy as well as boundary deformation work that occurs. When a saturated liquid is throttled from (P 1 ,T 1 ) with enthalpy h f1 , the final pressure P 2 < P 1 and temperature T 2 < T 1 . As the pressure decreases, T sat also decreases with decreased enthalpy of saturated liquid to h f2 . Then the difference h f1 - h f2 is used to evaporate a portion of the liquid since h = h f1 . Recall that ln P = A – B/T, (eq. (153)) i.e., ln (P 1 /P 2 ) = B/T 1 (T 1 /T 2 –1) or T 2 – T 1 = – T 1 /(1 +B/(T 1 ln (P 1 /P 2 ))). Therefore, defining the average Joule Thomson coefficient, µ JT = (T 2 – T 1 )/(P 2 – P 1 ) = –T 1 /(P 2 – P 1 ) (1 +B/(T 1 ln (P 1 /P 2 ))), or ≈ (T 1 /(P 2 – P 1 ) –BP 1 /T 1 ) when (P 1 – P 2 ) « P 1 . (A) The quality x 2 at the exit can be determined as follows. Assuming state 1 is a saturated liquid, h 2 = h 1 = h f1 = x 2 h g2 + (1 – x 2 )h f2 = x 2 h fg,2 + h f1 + c l (T 2 – T 1 ), i.e., x 2 h fg,2 = –c l (T 2 – T 1 ), or x 2 = c l (T 1 – T 2 )/h fg,2 (B) Using Eqs. (A) and (B) x 2 = (c l T 1 /h fg,2 )/(1 + B/(T 1 ln(P 1 /P 2 ))). (C) where B = h fg /R. If we assume that h fg,2 = h fg,1 , then x 2 ≈ (c l T 1 /h fg,1 )/(1 + B/(T 1 ln(P 1 /P 2 ))). For water, B ≈ 529 K, while for R134 A, B ≈ 2762 K. 0 1 2 3 4 5 6 7 8 0246810121416 P R T R R K Mille r VW Bert h Figure 22: Inversion curves for various state equations. 6. Throttling in Closed Systems Consider a large insulated tank that is divided into two sections A and B. Section A consists of high pressure gases at the conditions (P A,1 , T A,1 ) and section B consists of low pressure gases at the state (P B,1 , T B,1 ). If the partition is ruptured, the tank will assume a new equilibrium state. The state change occurs irreversibly and the entropy reaches a maximum value. The new equilibrium state can be obtained either by differentiating S = S A + S B with respect to T A subject to the constraints that U,V and m are fixed, or by applying the energy balance equations. nn. Example 40 volume V B = 3V A . Solution Applying the RK equation to section A 119 = 0.08314×177÷( v A,1 –0.02681) – 15.59÷(177 1/2 v A,1 ( v A,1 +0.02681) bar. (A) Therefore,, v A,1 = 0.0915 m 3 kmole –1 . (Alternatively, we can use the values T R = 1.4 and P R = 3.5 to obtain Z = 0.74 from the appropriate charts. Thereafter, using the relation P v = ZRT, that value of v can be obtained.) Similarly, v B,1 = 0.167 m 3 kmole –1 (for which, P R,B,1 = 1.5 T R,B,1 = 1.2, and Z = 0.68). Using the relations, V A / v A,1 + V B / v B,1 = 8 and V B = 3V A , we obtain the expression V A (1/v A,1 + 3/v B,1 ) = 8, i.e., V A = 0.277 m 3 , V B = 0.831 m 3 so that V = 1.108 m 3 . Thereafter, v 2 = 0.139 m 3 kmole –1 , N A = 3.024 kmole, and N B = 4.976 kmole. Recall that the internal energy u – u o = (3/2 a/bT 1/2 ) ln(v/(v+b)). In section A, u – u o = –1684.7 kJ kmole –1 , i.e., U A,1 – U A,1,o = –5094.8 kJ. Similarly, in section B, u – u o = –1056.8 kJ kmole –1 , i.e., U B,1 – U B,1,o = –5258.6 kJ. Applying the First law to the tank, Q – W = ∆U = 0, so that U A,1 + U B,1 = U 2 , i.e., U 2 = U A,1,o – 5094.8 + U B,1,o – 5258.6 = U A,1,o + U B,1,o –10353.4 kJ. Now, U 2 – U 2,o = U A,1,o + U B,1,o – 10353.4 – U 2o = N(3/2 a/bT 1/2 ) ln (v/(v+b)) = 697799/T 2 1/2 ln(0.139÷(0.139+0.02681) = –123,070/T 2 1/2 kJ. If c vo is a constant, then N A c vo T A1 + N B c vo T B1 – Nc vo T 2 – 10353.4 = –123,070/T 2 1/2 kJ. Therefore, 12.5×(3.024 × 177 + 4.976×151 – 8× T 2 ) = 10353.4 – 123,070/T 2 1/2 kJ, or T 2 = 156 K. Using the RK equation, P 2 = 0.08314×156÷(0.139 – 0.02681)×15.59÷(156 1/2 ×0.139×(0.139+0.0261)) = 61.45 bars Remarks Likewise, for isenthalpic throttling in sssf devices we can use the relation for (h – h o ). Eight kmole of molecular nitrogen is stored in sections A and B of a rigid tank. Sec- tion A corresponds to a pressure P A,1 = 119 bar and temperature T A,1 = 177 K. In section B, P B,1 = 51 bar and T B,1 = 151 K. The partition is suddenly ruptured. Determine the final equilibrium temperature T 2 . Assume that c v = c vo = 12.5 kJ kmole –1 K –1 (c vo does not depend upon the temperature), and that the gas behavior can be described by the RK equation of state P = RT/(v–b) – a/(T 1/2 v(v+b)), where a = 15.59 bar m 6 kmole –2 , b = 0.02681 m 3 kmole –1 , T c = 126.2 K, and P c = 33.9 bar. Assume that the The entropy generated during adiabatic throttling can be determined using a similar procedure. Such calculations are useful in determining the final pressures and temperatures for shock tube experiments. These experiments involve a pressurized gas in a section A that is separated by a diaphragm from section B. During the experiment, this diaphragm is ruptured. 7. Euken Coefficient – Throttling at Constant Volume During the adiabatic expansion of pressurized gases, the following relation applies. du = c v dT +(T ∂P/∂T – P) dv. The Euken coefficient µ E is related to a constant volume throttling process during which du = 0, i.e., µ E = (∂T/∂v) u = –(T ∂P/∂T – P)/c v = – T 2 (∂/∂T(P/T)) u /c v . (165) For an RK fluid, µ E = –(3/2) a/(T 1/2 v (v–b))/c v . (166) This coefficient is always negative, i.e., a specific volume increase is accompanied by a temperature decrease during adiabatic irreversible expansion in a rigid system. The corresponding entropy change is obtained by applying the relation du = Tds – P dv, i.e., (∂s/∂v) u = P/T. (167) Using Eqs. (165) and (167), we obtain the expression µ E = –T 2 (∂ 2 s/∂T∂v) u /c v . For an adiabatic throttling process, ds = δσ. Hence from Eq. (167), δσ = ds u = P/T dv u . Since dv > 0 and P/T > 0, ds > 0. For an equation of state P = RT/(v-b) - a/T n v m , Eq.(165) transforms to µ E = (∂T/∂v) u = - (n+1) a/(c v T n v m ). If c v has a constant value, one can integrate and obtain T (n+1) = (n+1) 2 a/(c v (m-1) v (m-1) )+ C, m ≠ 1. If m=2, n=1, then for constant u, T 2 =(4 a/(c v v))+ C. As per this model, at constant values of u, as v → 0, T→ ∞ and T → (T ig ) (n+1) as v → ∞, since attractive forces become negligible as we approach ideal gas (ig) limit. Hence, adiabatic throttling of a closed system yields, T (n+1) - T ig (n+1) = (n+1) 2 a/{c v (m-1) v (m-1) } Recall from Chapter 6 that a = RT c (n+1) (m+1) 2 v c (m-1) /(4 m). Therefore, (T R (n+1) - T ig,R (n+1) )(c v /R) = (n+1) 2 (m+1) 2 ) ((m 2 -1)/4m) (m-1) /(4 m (m-1) v R ’ (m-1) ). For m=2, n=1 (Berthelot equation), the reduced temperature change with pseudo-reduced volume change at constant energy is provided by the expression (T R (n+1) - T ig,R (n+1) )(c v /R) = (27/16) (1/ v R ´). a. Physical Interpretation Consider a container with two sections A and B. Section A is filled with pressurized gases and the second part B contains a vacuum. If the partition in Section A is instantaneously removed, the gas in Section A expands into Section B. As a result of this process the overall internal energy remains constant, but the intermolecular spacing increases (hence, the term (T∂P/∂T – P)dv > 0). Consequently, thermal portion of the energy c v dT must decrease (i.e., dT < 0) in order to compensate for the increase in the intermolecular potential energy. (In the case of an ideal gas, there is a negligible change in the intermolecular potential energy, since the specific volume is very large. Therefore, (T∂P/∂T – P) dv = 0 and there is no change in temperature.) Note that no net boundary work is performed for a rigid system. L. DEVELOPMENT OF THERMODYNAMIC TABLES It is apparent from the information contained in the Chapters 6 and 7 thus far that a set of expressions can be developed for the thermodynamic properties of a fluid that exists in any phase if a state relation is known. For example, properties for superheated vapors as H 2 O and R134 (Tables A-4 and A-5) can be generated using the real gas state equations. Table 2: Reference conditions and ideal gas properties for a few fluids. Property Steam Freon 22 R134A R152A Ammonia Nitrogen Carbon diox- ide Freon 12 Chemical formula H 2 O CF 3 CH 2 F NH 3 N 2 CO 2 W m 18.015 86.476 102.03 66.05 17.03 28.013 44.01 120.92 P c (bar) 220.9 49.775 40.67 45.20 112.8 33.9 73.9 41.2 T c (K) 647.3 369.15 374.30 386.44 405.5 126.2 304.15 385.0 T ref (K) 273.16 233.14 233.14 233.15 64.143 216.55 233.15 P ref (bar) 0.006113 1.0495 0.512615 .7177 .1253 5.178 .6417 h ref (kJ kg –1 ) 0.01 0.0 0.0 150.3 301.45 0.0 u ref (kJ kg –1 ) 0.0 0.0 0.0 150.4 301.01 –0.04 0.0 0.0 0.0 2.431 3.72 0.0 h fg (kJ kg –1 ) 2501.3 233.18 1389.0 215.188 524.534 169.59 A o 30.54 22.54 16.778 17.229 29.75 28.58 29.1519 26.765 † B o 0.01030 0.1141077 0.2865 0.4757 .025 .00377 –.001573 0.17594 † C o 0.0 130196.35 0 0 –154808. 50208 0.292×10 –6 –.27 10 –7† D o 0.0 –0.329×10 –4 –2.276×10 –4 2.893×10 –4 0.0 0.0 0.5283×10 –5 –0.103×10 –6† E o 1.135×10 –7 6.740×10 –8 1. Procedure for Determining Thermodynamic Properties Thermodynamic properties can be determined, once the state equation, critical constants, and corresponding ideal gas properties are known. Some useful formulas are listed be- low and some thermodynamic data is listed in Table 2. The reference conditions should be specified. For example for water, the reference condition is generally specified as that of the saturated liquid at the triple point. The choice of reference conditions is arbitrary. Here, v R ´ = v/v c ´, v c ´ = RT c /P c v R = v/v c , Z = P R v R ´/T R u c,R = – u Res /(RT c ) = (u o (T) – u(T,P))/RT c , h c,R = (h o (T) – h(T,P))/RT c , s c,R = (s o (T,P) – s(T,P))/R, c P,c R = (c P (T,P) – c P,o (T))/R, c v,c,R = (c v (T,P) – c v,o (T))/R, µ JT,R = µ JT c p /v c ´, g c,R = (g o (T,P) – g(T,P))/RT c , a c,R = (a o (T,P) – a(T,P))/RT c φ = f/P = exp((g(T,P)– g o (T,P))/RT) T sat with g f = g g and T inv with µ JT = 0 The constants are used in the formula c po = A o + B o T + C o T –2 + D o T 2 + E o T 3 . In the Figure 23: Schematic illustration of a method of determining the thermodynamic properties of a material using a P–h diagram. s ref (kJ kg –1 K –1 ) [...]... H2O (25ºC, 1 bar) = 6. 621 kJ kmole–1 K–1 Therefore, ˆ s H2O (25ºC, 1 bar, 0.999) = 6. 621 – 8.314 × ln 0.999 = 6. 629 kJ kmole–1 8 Fugacity a Fugacity and Activity As in Chapter 7 we define the fugacity for a component as d g k = R T d (ln fk(T, P)) (65 ) In analogy with the pure component (Eq (65 )), we can write Eq (62 a) in the form ˆ ˆ ˆ d g k = v k dP = R T d ln ( f k(T, P, Xk)) (66 ) ˆ ˆ ˆ The fugacity... is known, Eq (66 ) may be integrated at a given composition to obtain ˆ k (T, P) Subtracting Eq (66 ) from Eq (65 ) f ˆ ˆ ˆ d( g k – g k) = R T d(ln ( f k(T, P,Xk)/fk(T, P))) = ( v k (T, P,Xk) – v k(T, P))dP (67 ) We introduce the activity of the k–th species in the mixture ˆ ˆ α k = ( f k(T, P, Xk)/fk(T, P)), (68 ) ˆ α k = fugacity of species k in the mixture ÷ fugacity of pure species k (69 ) Note that... or 165 .2 kJ kmole–1 K–1 The values of ho(873 K) and so(873 K, 250 bar) can be obtained using the specific heat relations for an ideal gas, i.e., D Figure 25: Schematic illustration of an entropy calculation starting from a temperature of absolute zero; C: critical point ho = 37 06 kJ kg–1 = 66 782 kJ kmole–1 This corresponds to the point D Similarly, so(873 K, 250 bar) = 6. 47 kJ kg–1 K–1 or 1 16. 6 kJ... 2) solution: V = 1.001 + 16. 625N2 + 56. 092 N23/2 + 119.4 N22, (A) ˆ where V is expressed in units of liters, and N2 in kmole Obtain an expression for v 2 ˆ 1 Determine v 2,X → 0 or v 2 in liter per ˆ 2 ˆ∞ in terms of N2, and determine the value of v ˆ kmole Obtain an expression for v 1 in terms of N2, and one in terms of X1 Solution Differentiating Eq (A), ˆ v 2 = ∂V/∂N2 = 16. 625 + 84.138 N21/2 + 238.8... 2 = 16. 625 l kmole ˆ ˆ Furthermore, V = v 1 N1 + v 2 N2, i.e., ˆ ˆ v 1 = (V – v 2N2)/N1 Since a liter of water corresponds to a 1 kg mass, (C) N1 = 1 kg/18.02 kg kmole–1 = 0.0555 kmole, Therefore, applying Eq (C), ˆ v 1 = 18.02 – 505.4 N23/2 – 2151 .6 N22 in units of l kmole–1 (D) The mole fraction X2 = (1 – X1) = N2/(N2 + N1) = N2/(N2 + 0.055), i.e., N2 = 0.055 (1/X1 –1), and ˆ v 1 = 18.04 6. 625(1/X1–1)3/2 6. 567 (1/X1... where 3 –1 v 1(433 K, 100 kPa) = 8.314 × 423 ÷ 100 = 36 m kmole (from the tables for superheated vapor v 1 = 35.8 m3 kmole–1), and 3 –1 v 2(433 K, 100 kPa) = 8.314 × 423 ÷ 100 = 36 m kmole Vid = 0.3 × 36 + 9.7 × 36 = 10 × 36 = 360 m3 At 298 K, Vid = 0.3 v 1(298 K, 100 kPa) + 9.7 v 2(298 K, 100 kPa) Although water exists as a liquid under these conditions, in the mixture it exists as a vapor Therefore,... 873 K can be determined using the relation uo = ho – RT = 66 782 – 8.314 × 873 = 59,524 kJ kmole–1 The correction or residual factors at 873 K and 250 bar can be obtained From charts we see that at PR = 1.13, TR = 1.35, Z = 0.845 Therefore, (ho – h)/RTc = 0.735, (uo – u)/RTc = 0.5 26, and (so – s)/R = 0.389 Consequently, u = 31 46 kJ kg–1, and s = 6. 29 kJ kg–1 K–1 Thereafter, the enthalpy can be determined... 3.75×10–3 kmole of H2SO4, the total moles are (1000÷18.02) + 3.75 = 59.24 gmole Since M for H2SO4 is 98 kg kmol–1, the total mixture mass is 1000 g + 3.75 × 98 = 1 367 .5 g Therefore, x H2SO4 = 3.75÷59.24 = 0. 063 , Y H2SO4 = 367 .5 ÷ 1 367 .5 = 0.27, and M = 1 367 .5 ÷ 59.24 = 23.08 kg kmole–1 2 Generalized Relations Recall that the internal energy of a mixture containing K components is U = U(S,V,N1, N2, , NK), (2a)... Mixture ˆ We have discussed how to determine the value of g kid for ideal solutions (Eqs (72) ˆ and (62 c)) For non ideal mixtures, if a relation for v k in terms of Xk, T, and P is available, then ˆ a relation for g k(T, P,Xk) can be obtained (Eqs (62 a) and (71) Recall from Eq (62 c) ˆ g kid– g k = R T ln Xk (62 c) Further, from Eq (71) ˆ ˆ RT ln α k = ∫ (v k − vk ) dP , (71) ˆ ˆ and the activity α k = f... loses meaning i Remarks In Chapters 1 and 6 we have discussed the functional form for the intermolecular attraction forces given by the Lennard– Jones empirical potential Φ(l) between a pair of molecules For like pairs of molecules, Φ(l) = 4ε((σ/l)12 – (σ/l )6) (20a) For an unlike molecular pair consisting of species k and j, Φ(l) = 4εkj((σkj/l)12 – (σkj/l )6) , (20b) where εkj = (εkεj)1/2, and the collision . NH 3 N 2 CO 2 W m 18.015 86. 4 76 102.03 66 .05 17.03 28.013 44.01 120.92 P c (bar) 220.9 49.775 40 .67 45.20 112.8 33.9 73.9 41.2 T c (K) 64 7.3 369 .15 374.30 3 86. 44 405.5 1 26. 2 304.15 385.0 T ref (K) 273. 16 233.14. × 177 + 4.9 76 151 – 8× T 2 ) = 10353.4 – 123,070/T 2 1/2 kJ, or T 2 = 1 56 K. Using the RK equation, P 2 = 0.08314×1 56 (0.139 – 0.0 268 1)×15.59÷(1 56 1/2 ×0.139×(0.139+0.0 261 )) = 61 .45 bars Remarks Likewise,. 524.534 169 .59 A o 30.54 22.54 16. 778 17.229 29.75 28.58 29.1519 26. 765 † B o 0.01030 0.1141077 0.2 865 0.4757 .025 .00377 –.001573 0.17594 † C o 0.0 1301 96. 35 0 0 –154808. 50208 0.292×10 6 –.27