Van der Waals Equation of State: Entropy Analysis for Water, Study notes of Thermodynamics

Download Van der Waals Equation of State: Entropy Analysis for Water and more Study notes Thermodynamics in PDF only on Docsity! Example 1 Consider the Van der Waals equation of state. Obtain an expression for s = s(u,v) for water assuming the specific heat to be constant and equal to the ideal gas value cv0 = 28 kJ kmole–1. Select the reference condition such that s = cv0 ln (u/cv0 + a/(v cv0)) + R ln (v–b). Plot the entropy vs. volume for an internal energy value of 7000 kJ kmole–1. Solution Since du = cv dT + (T (∂P/∂T)v – P) dv, duT = (T(∂P/∂T)v – P) dv. (A) Using P = RT/(v–b) – a/v2 (B) in Eq. (A), duT = T(R/(v–b)) – (RT/(v–b) – a/v2) = (a/v2) dv Integrating at constant temperature, u(T,v) = –a/v + f(T). (C) In case a = 0, u0(T) = f(T). (D) Eliminating f(T) between Eqs. (C) and (D), u(T,v) = u0(T) – a/v. (E) Assuming constant (ideal gas) specific heats and u = uref,0 at T = Tref, u0 (T) – uref,0 = cv0(T – Tref), (F) Eq. (E) assumes the form u = cv0 (T–Tref) + uref,0 – a/v, i.e., (G) T = {(u–uref,0) + a/v}/cv0 + Tref. (H) Similarly, we can integrate the expression ds = cvdT/T + ∂P/∂T dv at constant temperature to obtain the relation s(T,v) = s0(T,v) – R ln (v/(v–b)). (I) Using the ideal gas relation for s0(T,v), s0(T,v) – sref,0(TRef,vref) = cv0 ln (T/TRef) + R ln (v/vref), s(T,v) = cv0 ln (T/TRef) + R ln (v/vref) + sref,0(T,vref)– R ln (v/(v–b)). (J) Further, using Eqs. (H) and (J) to eliminate the temperature, s =cv0 ln((u–uref,0)/(cv0Tref)+(a/(vcv0Tref) + 1)+R ln((v–b)/vref +sref,0(TRef,vref). (K) Setting , sref,0(Tref ,vref) =cv0 ln(cv0 Tref) + R ln vref, uref,0 = cv0 Tref we obtain the expression s = cv0 ln (u + a/v) + R ln ((v–b)). (L) For ideal gases, a = b =0, and Eq. (K) leads to the relation s0 = cv0 ln {u/(cv0 TRef)} + R ln (v/vRef) (M) This expression leads to a plot of s vs. both u and v for a real or ideal gas. Using the values ā = 5.3 bar m6 kmole–2, b̄ = 0.0305 m3 kmole–1, u = 7000 kJ kmole–1, Tref = 1 K, v ref = 1 m3 kmole–1, c v0 = 28 kJ kmole–1 K–1, u ref,0 = c v0 Tref = 28 kJ kmole–1 in Eq. (K), a plot of s vs. v at u = 7500 kJ kmole–1 is presented in Figure 1. Remarks The relation s =s (u,v) is the entropy fundamental equation. It is somewhat more difficult to manipulate the RK equation and obtain an explicit expression for s = s(u,v) using Eq. (L). a. Example 2 Consider a rigid and insulated 0.4 m3 volume tank filled with 4 kmole (24×1026 molecules) of water, and an internal energy of 7500 kJ kmole–1. Divide the tank into two equal 0.2 m3 parts. Assume that water follows the VW state equation. What is the entropy of each section? Assume that section 1 is slightly compressed to 0.14 m3 while the section 2 is expanded to 0.26 m3, but maintaining the same total volume and total internal energy. What is the entropy change during the process? (Use Figure 1.) Solution The fluid molecules are distributed uniformly in the tank and initially, v̄1 = v̄2 = v̄ = 0.4÷4 = 0.1 m3 kmole–1. Using Figure 1 or Eq. (K) of Example 1, we note that s̄ = 149.83 kJ kmole–1. The state is represented by point D in the figure, and the extensive entropy of each section is initially SD = (S1 = N1 s̄1) = (S2 = N2 s̄2) = 2×149.83 =299.7 kJ. (A) The total entropy of the tank is S = 2SD = 299.7 + 299.7 = 599.4 kJ. After the perturbation, v̄1 = 0.14÷2 = 0.07 m3 kmole–1, u1 = 7500 kJ kmole–1, and v̄2 = 0. 26÷2 = 0.13 m3 kmole–1, u2 = 7500 kJ kmole–1. The corresponding states are represented by points M and N in Figure 1, i.e., SM= S1 = N1 s̄1 = 2×149.8 = 299.6 kJ, and SN = S2= N2 s̄2 = 2×149.9 = 299.9 kJ. The total entropy after the disturbance S = S1 + S2 = SM + SN = 599.5 kJ. 146 146.5 147 147.5 148 148.5 149 149.5 150 150.5 151 0.04 0.05 0.06 0.07 0.08 0.09 0.1 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.19 0.2 v, m3 / kmole s, k J/ k m ol e K L C M K B H A ND E Liquid Gas y Figure 1: Variation in the entropy vs. volume for water modeled as a VW fluid at u = 7500 kJ kmole–1. Example 5 2.662 kmole of water at are contained in a rigid tank of volume 27 m3 at 320ºC and 140 bar. What are the values of s̄, ū, and ā at this state? If the temperature and volume are maintained constant, what is the most stable equilibrium state, and what are the values of s̄, ū, and ā at this state? Assume that cΡ = 5.96 kJ kg-1 K-1. Solution v̄ = 2.662/27 = 0.0986 m3 kmole-1. Applying the RK equation at 593 K and 140 bar, Psat = 133 bar. Since P > Psat at 593 K, and P < PN = 155 bars the state is metastable. We will use the method described in Chapter 7 to determine the fluid properties. For instance, h̄(320ºC, 0.0986 m3 kmole-1) = 44633 kJ kmole-1 or 2477 kJ kg-1, ū(320C, 0.0986 m3 kmole-1) = h – Pv̄ = 43252 kJ kmole-1 or 2400 kJ kg-1, and s̄ (320C, 0.0986 m3 kmole-1) = 91.34 kJ kmole-1 or 5.07 kJ kg-1. Thereafter, ā = ū – Ts̄ = 43252 – 593×91.34 = –10,912 kJ kmole-1 or –606 kJ kg-1, and ḡ = h̄ – Ts̄ = –9,532 kJ kmole-1. At a specified value of T and V, the value of ā decreases until it reaches a minimum. The fluid at state D transforms into a wet mixture ; but g of liquid = g of vapor. Hence P must beequal to 133 so that ā is minimized, i.e., v̄ = (1–x) vf (320ºC, Pnew) +x vg (320ºC, Pnew) = 0.0986 m3 kmole-1 The pressure Pnew = Psat, which is the saturation pressure at phase equilibrium (that equals 133 bar). Applying the RK equation at 593 K and 133 bar, vf (320ºC, 133 bar) = 0.04275, m3 kmole-1or 0.00237 m3 kg-1, and vg (320ºC, 133 bar) = 0.236 m3 kmole-1or 0.0131 m3 kg-1, i.e., x = (0.0986 – 0.04275)/(0.236 – 0.04275) = 0.289, and ā = ā f (1–x) + ā g×= –11,114 kJ kmole-1. Remarks At equilibrium ā = –11,114 kJ kmole-1, which is lower than the –10,912 kJ kmole-1 value at the meta-stable point at the same temperature and volume. The disturbance creates a wet mixture. State K (593 K, 0.211 m3 kmole-1) is a metastable state which, when disturbed, produces a wet mixture with a higher quality at the same temperature Example 6 Use the relation P = RT/(v–b) – a/(T1/2 v(v+b)) (A) To obtain an expression for P vs. v and T vs. v for water along the spinodal points. Plot P vs. v at 567, 593 and 615 K for water. Obtain an expression for the reduced values PR and TR vs. vR' along the spinodal points, and plot PR and TR vs. vR' and PR vs. TR along these points. Solution At the spinodal points ∂P/∂v = 0. Differentiating Eq. (A), (∂P/∂v = – RT/(v–b)2 + (a/(T1/2 v(v+b))) (1/v + 1/(v+b))) = 0. Thereafter, solving for the temperature, T = ((a(v–b)2/(R v(v+b))) (1/v + 1/(v+b)))2/3. (B) Using Eqs. (A) and (B) we obtain the relation P = (R(v–b))1/3 ((a/(v(v+b))) (1/v + 1/(v+b)))2/3 – (a/(v(v+b)))2/3 (R/((v–b)2 (1/v + 1/(v+b))))1/3 (C) With the values a = 142.59 bar m6 K1/2 kmole-2 and b̄ = 0.0211 m3 kmole-1 in the context of Eq. (B), we obtain a plot of T vs. v̄, and using Eq. (C) we can obtain a plot of P vs. v̄. Using Eq. (C), one can also obtain the spinodal pressure vs. volume for water at the temperatures 567, 593 and 615 K (cf. ). We can normalize Equations (B) and (C) so that TR=((0.4275(vR'– 0.08664)2/(vR'(vR'+0.08664))) (1/vR'+1/(vR'+0.08664)))2/3, and (D) PR = TR/(vR' – 0.08664) – 0.4275/(TR1/2 vR' (vR' + 0.08664)) (E) Example 7 Consider the RK equation of state P = RT/(v–b) – a/(T1/2 v(v+b)). (A) Determine the maximum temperature to which water can be superheated at 133 bar without boiling and the temperature to which water vapor can be subcooled at the same pressure without condensation occurring. Assume that Tsat = 593 K at P = 133 bar for water when it is modeled by the RK equation of state. What is the fluid state at 615.001ºC and 133 bar? Solution From Example 12, the temperatures at which ∂P/∂v = 0 are represented by the relation T = ((a (v–b)2/(R v(v+b))) (1/v + 1/(v+b)))2/3. (B) There are two spinodal vapor and liquid volumes at particular temperature, e.g., the states M and N at 133 bar inFigure 3. Using (B) and the RK equation to eliminate the temperature, we obtain the expression P = {R (v–b)}1/3[{a/(v(v+b))}{1/v + 1/(v+b)}]2/3 – {a/(v(v+b))}2/3{R/[(v–b)2(1/v + 1/(v+b))]}1/3. (C) Eq. (C) is useful in btaining the vapor spinodal curves AN and liquid spinodal ON(Figure 3) Using the values R = 0.08314 bar m3 kmole-1 K-1, a = 142.59 bar m6 K1/2 kmole-2, b = 0.0211 m3 kmole-1, P = 133 bar in Eq. (C), vM = 0.0597 m3 kmole-1 (cf. Figure 3), liquid–like spinodal point) and vN = 0.158 m3 kmole-1 (vapor-like spinodal point). and vN = 0.158 m3 kmole-1 (vapor-like spinodal point). Thereafter, applying these results in Eq. (B), for the liquid TSp,l = 615 K and for the vapor TSp,v = 567 K. Therefore, water can be superheated at 133 bar by 615–593 = 22 K and the vapor subcooled by 567–593 = –26 K. Example 8 0 50 100 150 200 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 v, m3/kmole P, b ar s 567 593 615 T3 = 615 K T1= 593 T2 = 567 A N SF G Vapor Liquid Vapor Spinodal Curve Liquid Spinodal Curve B M D Figure 3. Spinodal curves at 567, 593, and 615 K for water modeled according to the RK equation of state. If the pressure is maintained at 133 bar, Tsat = 593 K according to the relation with vapor at state G and liquid at state F. The vapor can be subcooled to 567 K (cf. the spinodal point N) and the liquid superheated to 615 K (cf. spinodal point M) at the same value of Psat = 133 bar. Water boils with the mother liquid phase at 593 K and 133 bar. The vapor embryo is at the same temperature as the liquid, but at a different pressure. Assume that the RK equation applies and that the vapor pressure is 155 bar. (According to the RK equation of state, for water Psat = 133 bar at 593 K.) Determine the bubble size. Assume that the surface tension σ = 10.5×10–6 kN m-1. Solution The bubble pressure is higher than the liquid pressure due to the effect of surface tension. The vapor can exist at a metastable state. The bubble radius rB = 2 σ/(PE – PΜ), i.e., (A) rB = (2×10.5×10–6)/((155–133)×100) = 9.549×10–9 m. Eq. (A) indicates that higher the value of PE, the smaller the bubble size. Remarks Various bubbles have different diameters depending upon the various metastable states of the bubbles. For instance, at 133.1 bar, rB = 2.1 µm. If the mother phase is a vapor at 593 K and 133 bar, the liquid drops exist at a higher pressure in the form of compressed liquid. Consider superheated vapor at 593 K and 50 bar. The lowest pressure in the liquid state at 593 K is 55 bar, which leads to the formation of metastable liquid drops. The associated drop size is rd = 2×10.5×10–6/((55–50)×100) = 0.042 µm