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v08.07.29 © 2008, Michael E. Hanyak, Jr., All Rights Reserved Page 6-21
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v08.07.29 © 2008, Michael E. Hanyak, Jr., All Rights Reserved Page 6-21

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Problem 6.59 in F&R, 3rd Ed.
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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 is a result of completing Step 11 under the "Development of a Mathematical Model" in Chapter 4 of this CinChE Manual.
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Assumptions 5, 6, and 7 are the result of completing Step 14 under the "Development of a Mathematical Model" in Chapter 4 of this CinChE Manual.
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An example vapor-liquid-equilibrium problem for a binary system with the assumption that Raoult's Law applies over the whole range of mole fractions (i.e., from zero to one). As discussed on Page 6-4, chemical components that are sub-critical with respect to temperature have the potential to condense. The critical temperatures for n-butane and n-hexane are 425.17 K and 507.90 K, respectively. At the operating temperature of 120ºC (393.15 K), both chemical components will distribute themselves in the saturated liquid and saturated vapor phases.
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A vapor feed stream at 120°C, 1.0 atm, 150.0 L/s, 50.0 mole% n-butane, and 50.0 mole% n-hexane is cooled and compressed to partially condense the vapor. The vapor and liquid product streams emerge from this process operation at some equilibrium temperature and 340 mm Hg gauge. The vapor product contains 60.0 mole% n-butane, and the barometric pressure is 1.00 atm. What are the equilibrium temperature in °C, the molar composition of the liquid product, the molar flow rate of the n-hexane in the liquid product, and the mass flow rates (lbm/s) of the vapor and liquid product streams?
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v08.07.29 © 2008, Michael E. Hanyak, Jr., All Rights Reserved Page 6-22

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Problem 6.59 in F&R, 3rd Ed.
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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.
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Why are TXY diagrams sketch here? To identify phase equilibrium, we look for key words like saturated, bubble-point, dew-point, and equilibrium in the problem statement. The words "equilibrium temperature" are in the problem statement for the vapor and liquid streams. Thus, we know to sketch a TXY diagram, in order to help us conceptually while solving the problem.
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vapor region
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liquid region
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vap-liq region
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sat'd vap curve
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sat'd liq curve
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Usually, only one TXY diagram is sketched. Why were two TXY diagrams sketched? The first on the left represents the VLE happening in the problem for a total mole fraction of 50% n-butane in the feed stream. The second on the right shows that the equilibrium can also be depicted by focusing on the saturated vapor condition for a vapor fraction of one with a total mole fraction of 60 mol% n-butane in the vapor stream.
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v08.07.29 © 2008, Michael E. Hanyak, Jr., All Rights Reserved Page 6-23

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Problem 6.59 in F&R, 3rd Ed.
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As we have done on many past problems, the equations are written using Guidelines 1 to 14 under the "Development of a Mathematical Model," as described in Chapter 4 of this CinChE manual.
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Usually "Mix L" is selected as the "check" equation, but in this math model it was not, because this equation happens to appear within the "vlet" function of Eq'n 10 below. As shown in the VLE math model on Page 6-15, this "Mix L" equation appears in its normalized form as the sum of the liquid mole fractions equals one for a vapor fraction of one. Thus, it must be considered as already written as an independent equation in the math model for Problem 6.59. Because of this fact, the "hexane" equation was selected as the "check" equation, per Guideline 7 of the "Development of a Mathematical Model" in Chapter 4.
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The degrees of freedom for the first three numbered equations are as follows: # vars = 7 #eqns = 3 -------------- dof = 4 At this point in the math model, we known three variables: TF, PF, and VF (with a dot). Note that R is a constant, and it is not counted as a variable. Thus, we need to write more equations in order to solve the problem.
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The degrees of freedom for the first nine numbered equations is as follows: # vars = 14 #eqns = 9 ---------------- dof = 5 At this point in the math model, we known three variables: TF, PF, and VF (with a dot). Note that R, MBU, and MHX 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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We known six variables: TF, PF, VF (with a dot), YV,BU, YV,HX, and PV. Thus, we can solve the problem.
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Symbol XL with a bar is counted as two variables, because it represents the two mole fractions of Stream L. Symbol YV with a bar is counted as two variables, because it represents the two mole fractions of Stream V. Since three results are to be returned by function "vlet", each returned variable is counted as a separate equation, giving a total of three.
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As per Guideline 14 of the "Development of a Mathematical Model" in Chapter 4, we are to write first principles for phase equilibrium that apply to the problem. Since Streams V and L are in vapor-liquid equilibrium, we are to select one of the three first principles given at the bottom of Pages 6-15 of this CinChE manual. We are assuming that Raoult's Law and the Antoine equation apply for the assumed condition of VLE. We pick the "vlet" function and apply it to Stream V, because its temperature is one of the "Finds" and its composition is known. Also, Stream V is a saturated vapor, and thus its vapor fraction is one.
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A graphical sketch of the "vlet" function is depicted in the "TXY Diagram for Stream V" on Page 2 of this problem solution. In this TXY diagram, knowing PV, Vf and YV,BU fixes all other variables, namely TV and XL,BU. Note that XL,HX = 1.0 - XL,BU.
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Note that the "vlet" function below returns the mole fractions of Stream L; that is, XL,BU and XL,HX. Once nV and nL are determined, Eq'n 5 can be used to find nL,HX. Thus, we can not also use the "hexane" balance to find nL,HX. Therefore, it is not independent and, it must be used as a "checK" equation. Linear independence must be analyzed using ALL equations written in a possible math model, including the equations that are behind a functional form like that for the "vle" function in this problem. Thus, the equations for the "vle" function are those in the math model on Page 6-15 of the CinChE manual, but written for a binary system and a vapor fraction of one.
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Problem 6.59 in F&R, 3rd Ed.
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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. Note that Steps 2 and 4 are NOT required in a dimensional consistency analysis, since they are equations for the molecular weight. However, they have been included here for instructional purposes ONLY.
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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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6.
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7.
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(3)
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(7)
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(10-12)
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(9)
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(8)
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Note that "mol" and "g-mol" mean the same thing. Normally, you do not have to show the units check for molecular weight. It has only been included here for instructional purposes in Steps 2 and 4.
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6.
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v08.07.29 © 2008, Michael E. Hanyak, Jr., All Rights Reserved Page 6-25

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Problem 6.59 in F&R, 3rd Ed.
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This manual solution was partially automated as indicated by Step 3 below. In Step 3, the vapor-liquid equilibrium math 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 binary system of n-butane and n-hexane. The Excel "EZ Setup"/Solver solution for Step 3 is given on the next page.
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"EZ Setup"

P = 1100 mm HgVf = 1 unitless

zBU = 0.6 mol fraczHX = 0.4 mol frac

T = 56.983 ºCxBU = 0.149479 mol fracxHX = 0.850521 mol frac

yBU = 0.6 mol fracyHX = 0.4 mol frac

PsatBU = 4415.34 mm HgPsatHX = 517.33 mm Hg

kBU = 4.01395 unitlesskHX = 0.4703 unitless

VLET Known Quantities:

VLET Calculated Results:

/* Raoult's Law applied to n-Butane and n-Hexane System */

// Total and Two Component Material Balances 1.0 = Vf + Lf

zBU = Vf * yBU + Lf * xBU zHX = Vf * yHX + Lf * xHX

// Vapor-Liquid Equilibrium using Raoult's Law yBU = kBU * xBU yHX = kHX * xHX

kBU = PsatBU / P kHX = PsatHX / P

// Antoine Equations for the Two Components, F&R, 3rd Ed., Table B.4 log(PsatBU) = 6.82485 - 943.453 / (T + 239.711) // range -78.0 to -0.3 C log(PsatHX) = 6.88555 - 1175.817 / (T + 224.867) // range 13.0 to 69.5 C

// Two mixture equations for the liquid and vapor phases xBU + xHX - yBU - yHX = 0

// Given Information Vf = 1.0 P = 1100 // mm Hg zBU = 0.6 zHX = 1.0 - zBU

TXY Diagram for n-Butane and n-Hexane System at 1100 mm Hg

0 0.2 0.4 0.6 0.8 10

20

40

60

80

100

Tem

pera

ture

, C

Mole Fraction of n-Butane

sat'd vapor

sat'd liquid

Vf = 1

Tdp

Xdp

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"EZ Setup Model for the vlet[ PV, Vf = 1, YV,BU & YV,HX ] Function
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Problem 6.59 in F&R, 3rd Ed.
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If these two equations for n-butane are combined to eliminate KBU, you get the following common expression for Raoult's Law: YBU P = Psat,BU XBU 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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Click here to view this Excel "EZ Setup"/Solver formulation.
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Using the Raoult's VLE model on Page 6-15 as a guide, you can write this "EZ Setup" math model for the binary system of n-butane and n-hexane.

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Problem 6.59 in F&R, 3rd Ed.
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"EZ Setup"

nF = 4.64945 mol/s

T_V = 56.983 CP_V = 1100 mm Hg

nV = 3.61743 mol/smV = 0.552992 lbm/s

nL = 1.03202 mol/smL = 0.186515 lbm/s

xL_BU = 0.149479 mol fracxL_HX = 0.850521 mol frac

nL_BU = 0.154265 mol/snL_HX = 0.877751 mol/s

T_V = 57 C

xL_BU = 0.15 mol fracxL_HX = 0.85 mol frac

nL_HX = 0.88 mol/s

mL = 0.19 lbm/smV = 0.55 lbm/s

EZ-Solve Solution

// Problem 6.59 in Felder and Rousseau Textbook, Third Edition

// Total and Component Material Balances

nF - nV - nL = 0 // mol / s

0.50 * nF - 0.60 * nV - nL_BU = 0 // 0.50 * nF - 0.40 * nV - nL_HX = 0 // not independent

// Mixture Equation for Stream L

// nL = nL_BU + nL_HX // equation is already in function vlet

// Ideal Gas Law for Stream F

P_F * V_F = nF * R * (T_F + 273.15) // Composition Equations for Stream L

nL_BU = nL * xL_BU nL_HX = nL * xL_HX

// Mass Flow Rates for Streams V and L

mV = nV * M_V / 453.592 // lbm / s M_V = 0.60 * M_BU + 0.40 * M_HX // g / mol

mL = nL * M_L / 453.592 // lbm / s M_L = xL_BU * M_BU + xL_HX * M_HX // g / mol

// Vapor-Liquid Equilibrium for Streams V and L

T_V = vlet ( P_V, 1.0, 0.60, 0.40, A1,B1,C1, A2,B2,C2 ) // C xL_BU = vlet_x1 ( P_V, 1.0, 0.60, 0.40, A1,B1,C1, A2,B2,C2 ) xL_HX = 1.0 - xL_BU

// Given Information:

T_F = 120 // C P_F = 1 // atm V_F = 150.0 // L / s zF_BU = 0.5 // mol. frac. zF_HX = 0.5 // mol. frac.

Pb = 1 // atm, barometric pressure Pg_V = 340 // mm Hg gauge P_V = Pg_V + Pb*760 // mm Hg absolute

// Physical Constants

R = 0.08206 // L atm / mol K M_BU = 58.12 // g / mol M_HX = 86.17 // g / mol

A1 = 6.82485 // n-butane, Component 1, range of -78.0 to -0.3 C B1 = 943.453 // Antoine constants from Table B.4, F&R, 3rd Ed. C1 = 239.711

A2 = 6.88555 // n-hexane, Component 2, range of 13.0 to 69.5 C B2 = 1175.817 // Antoine constants from Table B.4, F&R, 3rd Ed. C2 = 224.867

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E-Z Solve Direct Solution of the Mathematical Model for Problem 6.59 The Easiest Way to Solve a Binary Vapor-Liquid Equilibrium Problem
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The math model on Page 3 of 10 was used to write this E-Z Solve model. An E-Z Solve user-defined function "vlet" was used to allow for the direct solution of the math model. The E-Z Solve algorithmic code for the function "vlet" is given on the next two pages.
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Problem 6.59 in F&R, 3rd Ed.
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An E-Z Solve user-defined function can only return a single result for a math algorithm. Multiple function calls with slightly different names must be used to get other results from that algorithm. For example, function "vlet" returns temperature in C for only a binary system. Function "vlet_x1" returns mole fraction of Component 1 in the sat'd liquid. Function "vlet_y1" returns mole fraction of Component 1 in the sat'd vapor.
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Click here to view the Excel "EZ Setup"/Solver formulation, which replaces the "E-Z Solve" solution on this page.

FUNCTION vlet(P, Vf, z1, z2, A1,B1,C1, A2,B2,C2)

// This EZ-Solve function implements the vapor-liquid equilibriurm (vle)// mathematical algorithm "vlet" for a binary system and Raoult's Law.

// Author: Dr. Michael E. Hanyak, July 29, 2008.

// P - the equilibrium pressure, mm HG.// Vf - the equilibrium vapor fraction, range of 0.0 to 1.0.// z1 - the total composition for Component 1, mole fraction.// z2 - the total composition for Component 2, mole fraction.// A1, B1, C1 - Antoine constants for Component 1.// A2, B2, C2 - Antoine constants for Component 2.//// Antoine Equation is log P = A - B / (T + C)// where P is in mm Hg, T is in C, and log is base 10.

// Function "vlet" does the iteration on temperatue using the Bisection// Method. When the last two temperatures in the iteration are within// 0.00001 C of each other, the equilibrium temperature is then returned // as the value for function "vlet" in degree C.

// First two estimates for the iteration on the equilibrium temperature.// They are the boiling temperatures of each pure compound at P in mm Hg.

T1 = B1 / (A1 - log(P)) - C1 // C T2 = B2 / (A2 - log(P)) - C2 // C

IF ( T1 > T2 ) THEN Tright = T1 Tleft = T2 ELSE Tright = T2 Tleft = T1 ENDIF

// Start the math algorithm for VLET based on Raoult's Law.// Absolute tolerance on the temperature is within 0.00001 C.

Lf = 1.0 - Vf eps = 0.00001

// Iterate on temperature using the Bisection Method.// See "Bisection Method" in http://en.wikipedia.org/

stop = 0 FOR (I = 1, 1000) IF (abs(Tright-Tleft) <= 2*eps) THEN stop = 1 ENDIF

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E-Z Solve User-Defined "vlet" Function in File "vle_alg.msp"
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This E-Z Solve algorithmic code for function "vlet" implements the vapor-liquid equilibrium (VLE) math algorithm found on Page 6-16 of the CinChE manual for any binary system, and it also incorporates the numerical analysis technique called the bisection method.
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Problem 6.59 in F&R, 3rd Ed.
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Click here to view the "EZ Setup" built-in "vleT2" function, which replaces the "E-Z Solve" version given below.

IF ( I = 1 ) THEN Tmid = Tleft ENDIF IF ( I = 2 ) THEN Tmid = Tright ENDIF IF ( I > 2 ) THEN Tmid = (Tleft + Tright) / 2 ENDIF

Psat1 = 10^( A1 - B1/(Tmid + C1) ) // Antoine equation with Psat2 = 10^( A2 - B2/(Tmid + C2) ) // Psat in mm Hg and Tmid in C

K1 = Psat1 / P K2 = Psat2 / P

x1 = z1 / (Vf*K1 + Lf) x2 = z2 / (Vf*K2 + Lf)

y1 = K1 * x1 y2 = K2 * x2

Fmid = (x1 + x2) - (y1 + y2)

IF ( I = 1 ) THEN Fleft = Fmid ENDIF IF ( I = 2 ) THEN Fright = Fmid ENDIF IF ( stop = 1 ) THEN BREAK ENDIF IF ( I > 2 ) THEN IF ( Fleft * Fmid > 0 ) THEN Tleft = Tmid // throw away left half Fleft = Fmid ELSE Tright = Tmid // throw away right half Fright = Fmid ENDIF ENDIF ENDFOR RETURN Tmid // in CEND

Bisection Method

Here, x is T, and a1

and b1 are initial Tright and Tleft

Wikipedia, 7/29/08

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An E-Z Solve user-defined function can ONLY return a single result. To access other results within this function, a copy of the function is created, its name is changed slightly, and another single result is returned. Thus, multiple functional calls would be made in an E-Z Solve model to get multiple results from a specific math algorithm.
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