An example vapor-liquid-equilibrium problem with the assumption of a one-condensable system; that is, moist air being cooled and partially condensing out some of the water. As discussed on Pages 6-3 to 6-4, some chemical components can be sub-critical and some super-critical with respect to temperature. The critical temperatures for nitrogen, oxygen, and water are 126.20 K, 154.40 K, and 647.40 K, respectively. At the operating temperature of 80ºC (353.15 K), water is sub-critical, and oxygen and nitrogen are super-critical. Thus, water has the potential to condense, and we will describe its vapor-liquid equilibrium using Raoult's Law. Since oxygen and nitrogen can not condense, they are called non-condensables. However, they are soluble in water, as discussed on Page 6-3 under Henry's Law. This problem is solved two ways. First, we assume that Raoult's Law applies for water and that the oxygen and nitrogen do not appear in the saturated liquid; that is, the liquid is pure water. Second, we assume that Raoult's Law applies for water and that Henry's Law applies for oxygen and nitrogen; that is, they are soluble in water. We will compare the two solutions and see if the first way is a suitable solution.
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Assumptions 1 and 2 are a result of completing the "Development of a Conceptual Model" in Chapter 4 of this CinChE Manual.
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Assumption 3 is a result of completing Step 5 under the "Development of a Mathematical Model" in Chapter 4 of this CinChE Manual.
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Assumption 4 simplifies the problem and makes it easier to solve, even though the pressure of the gas (6.92 atm) is greater than 3 atm. Because volume balances are a misnomer (i.e., they do not exist), we can only write molar or mass material balances. Since we have vapor-liquid equilibrium in the problem and the K-values are expressed as the ratios of mole fractions, we should write molar balances and not mass balances. We will check the validity of the ideal gas assumption later in the next problem of this CinChE manual, when we solve Exercise 6.3-2, Part B.
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Assumptions 5 to 10 are the result of completing Step 14 under the "Development of a Mathematical Model" in Chapter 4 of this CinChE Manual.
See the blue HYSYS manual, Pages C-6 to C-8, for a detailed explanation of the temperature-composition (or TXY) diagram, or consult your course textbook for an explanation. Note that for this one-condensable system, the liquid is pure water. Thus, the saturated liquid curve must be the vertical at a mole fraction of 1.0. Since water is the less volatile component (i.e., has the high boiling temperature) in the gas mixture, the saturated vapor curve has lower temperatures near the zero mole fraction of water and higher temperatures near the pure mole fraction of water.
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DSHF is the degrees of superheat for Stream F. Its definition is as follows: TF = Tdp + DSHF where TF is the temperature of Stream F, Tdp is the dew-point temperature of F.
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vapor region
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vap-liq region
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The exiting Stream T contains both vapor and liquid phases that are in equilibrium, since the feed Stream F is cooled isobarically to condense part of the water component only. Mentally, we picture Stream T as have two parts; that is, a sat'd vapor part and a sat'd liquid part. We show this mental picture as Streams V and L connected by a dotted line, even though in reality only Stream T exists. This dotted line reminds us that we are mentally looking at Stream T as having two parts, where we DO NOT know the mole fractions of the sat'd vapor part. This mental trick helps us to write the material balances by how we draw our system boundary.
As we have done on many past problems, the first ten equations are written using Guidelines 1 to 13 under the "Development of a Mathematical Model," as described in Chapter 4 of this CinChE manual.
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Usually the "Mix V" is selected as the "check" equation, but in this math model it was not, because this equation happens to appear within the "vlevf" function of Eq'n 8 below. As shown in the VLE math model on Page 6-15, this "Mix V" equation appears in its normalized form as the sum of the liquid mole fractions minus the sum of the vapor mole fractions equals zero. Because of this fact, the "nitrogen" equation was selected as the "check" equation, per Guideline 7 of the "Development of a Mathematical Model" in Chapter 4. Also, the "oxygen" equation becomes a "check" equation, since the "vlevf" function below returns the three mole fractions for Stream V.
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Note that the "vlevf" function below returns the mole fractions of Stream V; that is, YV,WA, YV,O2, and YV,N2. Once nV is determined, Eq'ns 6 and 7 can be used to find nV,O2 and nV,N2. Thus, we can not also use the "O2" and "N2" balances to find nV,O2 and nV,N2. Therefore, they are not independent.
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The degrees of freedom for the first four numbered equations is as follows: # vars = 6 #eqns = 4 -------------- dof = 2 At this point in the math model, we known one variable: mF. Note that the molecular weights of MWA, MO2, and MN2 are constants, and they are not counted as variables. Thus, we need to write more equations in order to solve the problem.
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Symbol YV with a bar is counted as three variables, because it represents the three mole fractions of Stream V. Symbol ZF with a bar is counted as three variables, because it represents the three mole fractions of Stream F. Since four results are to be returned by function "vlevf", each returned variable is counted as a separate equation, giving a total of four.
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The degrees of freedom for the first eleven numbered equations is as follows: # vars = 17 #eqns = 11 ---------------- dof = 6 At this point in the math model, we known six variables: mF., ZF,WA, ZF,O2, ZF,N2, TV, and PV. Thus, we can solve for all eleven unknown variables in the first eleven numbered equations.
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Eq'ns 12 to 16 are written to satisfy the list of "Finds" quantities on Page 2 of 9.
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A graphical sketch of the "vlevf" function is depicted in the "TXY Diagram for VLE" on Page 2 of this problem solution. In this TXY diagram, knowing TV = 80ºC, PV = 5260 mm Hg, and ZF,WA = 0.1 fixes all other variables, namely Vf, YV,WA, and XL,WA = 1.
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A graphical sketch of the "vlet" function is depicted in the "TXY Diagram for VLE" on Page 2 of this problem solution. In this TXY diagram, knowing PV = 5260 mm Hg, Vf = 1, and ZF,WA = 0.1 fixes Tdp, the dew-point temperature of Stream F. The dew-point is where the first drop of liquid forms while cooling the moist air mixture.
As per the guidelines on Page 4-14 of the "Development of a Mathematical Algorithm," we are to conduct a dimensional consistency analysis on those steps in the math algorithm that do not represent material balances, mixture equations, composition equations, and the energy balance. Since Step 1 is for molecular weight, it does not have to be included in a dimensional consistency analysis. It has been included here ONLY for instructional purposes.
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The algebraic technique to resolve a "3 x 3" SOLVE construct is discussed on Page 4-15 of this CinChE manual.
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(4)
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(3)
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(8) to (11)
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(16)
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(15)
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(14)
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(13)
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Normally, you do not have to show the units check for molecular weight. It has only been included here for instructional purposes in Step 1.
This manual solution was partially automated as indicated by Steps 3 and 5 below. In Steps 3 and 5, the vapor-liquid equilibrium mathematical model based on Raoult's Law, as found on Page 6-15 of this CinChE manual, was prepared as an "EZ Setup" formulation for a ternary system of water, oxygen and nitrogen. The Excel "EZ Setup"/Solver solution for Steps 3 and 5 is given on the next page.
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"EZ Setup"
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"EZ Setup"
T = 80 ºC P = 5260 mm HgP = 5260 mm Hg Vf = 1zWA = 0.100 zWA = 0.100zO2 = 0.189 zO2 = 0.189zN2 = 0.711 zN2 = 0.711
"EZ Setup" Model for Vapor-Liquid Equilibrium using Raoult's Law
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Water is the only condensable component, and oxygen and nitrogen are assumed to be not soluble in liquid water.
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Note that the math model for VLE on Page 6-15 of this CinChE manual contains the distribution coefficient, Kj. The two equations with Kj in that math model were combined to form one equation here, which is know as Raoult's Law. In the TXY diagram on Page 2 of this problem solution, Raoult's Law relates the mole fractions at the end points of a horizontal line that connects the saturated liquid curve to the saturated vapor curve.
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[ Vf, YV,j's ] = vlevf [ TV, PV, ZF,j's ]
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[ Tdp ] = vlet [ PV, Vf = 1, ZF,j's ]
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Example 6.3-2, Part A in F&R, 3rd Ed.
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To get the "vlet" results, the "T = 80" line near the end of the "EZ Setup" model is replaced with a "Vf = 1" line, and then this model is solved again.
If you apply precision to the "Finds" quantities, the two solutions on Pages 6 and 8 are identical. Thus, the last assumption is valid, and the solution on Page 6 is the easier of the two ways.
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The Reverse Lever Rule is based on the total balance and the component balance for water: nF - nV - nL = 0 nF ZF,WA - nV YV,WA - nL XL,WA = 0 If you divide both equations by nF, you get: 1.0 = Vf + Lf ZF,WA = Vf YV,WA + Lf XL,WA where Vf is the molar vapor fraction, Lf is the molar liquid fraction. If you solve for Lf in the last two equations, you get the Reverse Lever Rule.
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"EZ Setup"
T = 80 ºC P = 5260 mm HgP = 5260 mm Hg Vf = 1zWA = 0.100 zWA = 0.100zO2 = 0.189 zO2 = 0.189zN2 = 0.711 zN2 = 0.711
"EZ Setup" Model for Vapor-Liquid Equilibrium using Raoult's and Henry's Laws
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Water is the only condensable component, and oxygen and nitrogen are soluble in liquid water.
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Note that the math model for VLE on Page 6-15 of this CinChE manual contains the distribution coefficient, Kj. The two equations in that math model with Kj were combined to form one equation here, which is know as Raoult's Law.
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[ Vf, YV,j's ] = vlevf [ TV, PV, ZF,j's ]
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[ Tdp ] = vlet [ PV, Vf = 1, ZF,j's ]
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Example 6.3-2, Part A in F&R, 3rd Ed.
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To get the "vlet" results, the "T = 80" line near the end of the "EZ Setup" model is replaced with a "Vf = 1" line, and then this model is solved again.
"EZ Setup" Direct Solution of the Mathematical Model for Example 6.3-2, Part A The Easiest Way to Solve a One Condensable Vapor-Liquid Equilibrium Problem
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The math model on Page 3 of 9 was used to write this "EZ Setup" model. In the math model, Eq'ns 8 to 11 for the function "vlevf" and Eq'n 16 for the function "vlet" were replaced with Raoult's Law as written for a one-condensable component.
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The "vlevf" equation in the math model: [ Vf, YV ] = vlevf [ TV, PV, ZF ] was replaced with the following two equations: YV,WA * PV = P*WA * XL,WA P*WA = psat [TV, pure H2O ] that is, Raoult's Law and the saturation pressure function that represents the Steam Tables. In the case of the "EZ Setup" model, this function is "pSatW" from Bernhard Spang's Excel IAPWS-IF97 Add-In for the Steam Tables. We can do this replacement because NO iteration is required in the "vlevf" function for a one-condensable system of vapor-liquid equilibrium.
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The "vlet" equation in the math model: [ Tdp ] = vlet [ PV, Vf = 1, ZF ] was replaced with the following two equations: ZF,WA * PV = P*WA * XL,WA Tdp = tsat [P*WA, pure H2O ] that is, Raoult's Law and the saturation temperature function that represents the Steam Tables. In the case of the "EZ Setup" model, this function is "tSatW" from Bernhard Spang's Excel IAPWS-IF97 Add-In for the Steam Tables. We can do this replacement because NO iteration is required in the "vlet" function for a one-condensable system of vapor-liquid equilibrium.
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The oxygen and nitrogen components have be combined into one chemical component, called dry air. The Heuristic Observation for the Mathematical Algorithm on Page 7 of 9 indicates that combining these two components into one will not changed the values for the "Finds" listed on Page 2 of 9.
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Under what condition would the molar liquid fraction be identical in value to the mass liquid fraction?
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The molar liquid fraction will be identical in value to the mass liquid fraction, if the starting mixture is just a pure chemical compound. For a mixture (nc >= 2), the molar liquid fraction does not equal the mass liquid fraction, because the molecular weight of the saturated liquid mixture is different from the molecular weight of the feed mixture.
