February 10, 2012  Tagged with: , , , , , , , ,

*Note: This article is part of a series on a specific research project: Part 1, Part 2
*Note: This article assumes you have rudimentary knowledge of how a simple distillation column operates. Though much of the basics of the simple model for a distillation column will be covered below, it should not be considered as a stand alone reference.

# 3.0 Proof of the Equivalence of Methods 1 and 2

## 3.1.0 Geometric Basis for Equivalence

### 3.1.1 The Rationale of the Method for Comparison

Given that method 1 and method 2 involve plotting the exact same enriching and stripping operation lines, the enriching and stripping operation lines are by definition not parallel for any real physical system that does not involve an azeotrope**, and the combined streams q-line from method 2 also passes through the intersection of said enriching and stripping operation lines, there exists at the point of intersection of these three lines a set of three different means of determining the coordinates of the point based on the intersection of a pair of lines.  Two of these lines, the enriching and stripping operation lines, are present in both methods and are defined by Eq. 1-10 and Eq. 1-45 in method 1.  The third line, the combined streams q-line, is defined by Eq. 2-5 in method 2.  As a matter of geometric consistency, evaluation of the ratio of x-ordinates or y-ordinates for a pair of such intersections should result in unity.  In other words, when taking the ratio of a single ordinate between a pair of definitions for the intersection of the three lines, all variables should cancel exactly.

### 3.1.2 The Points to be Examined

The pair of definitions for the intersection of the three lines will be taken as the intersection of the enriching and stripping operation lines and the intersection of the enriching operation line and the combined streams q-line.  The latter selection was made due to the relative simplicity of the definition of the enriching operation line over that of the stripping operation line.

## 3.2.0 Proof of Equivalence

### 3.2.1 Definition of the Intersecting Lines

The enriching operation line is given by Eq. 1-10 from method 1

Eq. 3-1) $\displaystyle \boxed{y=\frac{R_D}{R_D+1}x+\frac{x_D}{R_D+1}}$

The stripping operation line is given by Eq. 1-45 from method 1

Eq. 3-2) $\displaystyle\boxed{y=\frac{\left(Fq_F+Gq_G+DR_D\right)x_Gx-\left(\left(F\left(q_F-q_G\right)+D\left(q_G+R_D\right)+q_GW\right)x+Wx_G\right)x_W+Wx^2_W}{\left(F\left(q_F-1\right)+G\left(q_G-1\right)+D\left(R_D+1\right)\right)x_G-\left(F\left(q_F-q_G\right)+D\left(q_G+R_D\right)+W\left(q_G-1\right)\right)x_W}}$

The combined streams q-line is given by Eq. 2-5 from method 2

Eq. 3-3) $\displaystyle\boxed{y=\frac{q_M}{q_M-1}x-\frac{x_M}{q_M-1}}$

### 3.2.2 Definition of the Intersection (1) of the Enriching Operation Line and the Combined Streams q-Line

Let the point where the enriching operation line intersects the combined streams q-line be defined as $\displaystyle\left\{x_1,y_1\right\}$

Equating Eq. 3-1 and Eq. 3-3 and rearranging to solve for $x$ evaluated at $x_1$ yields

Eq. 3-4) $\displaystyle\boxed{x_1=\frac{x_D\left(q_M-1\right)+x_M\left(R_D+1\right)}{q_M\left(R_D+1\right)-R_D\left(q_M-1\right)}}$

Which, upon substitution back into Eq. 3-1, yields

Eq. 3-5) $\displaystyle\boxed{y_1=\frac{q_Mx_D+R_Dx_M}{q_M+R_D}}$

### 3.2.3 Definition of the Intersection (2) of the Enriching Operation Line and the Stripping Operation Line

Let the point where the enriching operation line intersects the stripping operation line be defined as $\displaystyle\left\{x_2,y_2\right\}$

Equating Eq. 3-1 and Eq. 3-2 and rearranging to solve for $x$ evaluated at $x_1$ yields

Eq. 3-6) $\displaystyle\boxed{x_2=\frac{\left(F\left(q_F-1\right)+G\left(q_G-1\right)+D\left(R_D+1\right)\right)x_Dx_G+\left(F\left(q_G-q_F\right)x_D-D\left(q_G+R_D\right)x_D+W\left(x_D-q_Gx_D+x_G+R_Dx_G\right)\right)x_W-\left(R_D+1\right)Wx^2_W}{F\left(R_Dx_G+q_F\left(x_G-x_W\right)+q_Gx_W\right)+\left(q_G+R_D\right)\left(Gx_G-\left(D+W\right)x_W\right)}}$

Which, upon substitution back into Eq. 3-1, yields

Eq. 3-7) $\displaystyle\boxed{y_2=\frac{\left(Fq_F+Gq_G+DR_D\right)x_Dx_G-\left(\left(F\left(q_F-q_G\right)+D\left(q_G+R_D\right)+q_GW\right)x_D-R_DWx_G\right)x_W-R_DWx^2_W}{F\left(R_Dx_G+q_F\left(x_G-x_W\right)+q_Gx_W\right)+\left(q_G+R_D\right)\left(Gx_G-\left(D+W\right)x_W\right)}}$

### 3.2.4 Substitution to Eliminate any Combined Variables

Substitution for combined variables in Eq. 3-4 yields

Eq. 3-8) $\displaystyle x_1=\cfrac{x_D\left(\cfrac{Fq_F}{F+G}+\cfrac{Gq_G}{F+G}\right)-x_D+\left(\cfrac{Fx_F}{F+G}+\cfrac{Gx_G}{F+G}\right)\left(R_D+1\right)}{\left(R_D+1\right)\left(\cfrac{Fq_F}{F+G}+\cfrac{Gq_G}{F+G}\right)-R_D\left(\cfrac{Fq_F}{F+G}+\cfrac{Gq_G}{F+G}\right)+R_D}$

Which simplifies to yield

Eq. 3-9) $\displaystyle x_1=\frac{x_D\left(Fq_F+Gq_G\right)-x_D\left(F+G\right)+\left(R_D+1\right)\left(Fx_F+G_xG\right)}{\left(F+G\right)+\left(Fq_F+Gq_G\right)}$

And, similar substitution for combined variables in Eq. 3-5 yields

Eq. 3-10) $\displaystyle y_1=\cfrac{\left(\cfrac{Fq_F}{F+G}+\cfrac{Gq_G}{F+G}\right)x_D+R_D\left(\cfrac{Fx_F}{F+G}+\cfrac{Gx_G}{F+G}\right)}{R_D+\left(\cfrac{Fq_F}{F+G}+\cfrac{Gq_G}{F+G}\right)}$

Which simplifies to yield

Eq. 3-11) $\displaystyle y_1=\frac{\left(Fq_F+Gq_G\right)x_D+R_D\left(Fx_F+Gx_G\right)}{\left(F+G\right)R_D+\left(Fq_F+Gq_G\right)}$

### 3.2.5 Evaluation of the ratio of y-Ordinates

The ratio of Eq. 3-11 to Eq. 3-7 is first given by

Eq. 3-12) $\displaystyle\frac{y_1}{y_2}=\cfrac{\cfrac{\left(Fq_F+Gq_G\right)x_D+R_D\left(Fx_F+Gx_G\right)}{\left(F+G\right)R_D+\left(Fq_F+Gq_G\right)}}{\cfrac{\left(Fq_F+Gq_G+DR_D\right)x_Dx_G-\left(\left(F\left(q_F-q_G\right)+D\left(q_G+R_D\right)+q_GW\right)x_D-R_DWx_G\right)x_W-R_DWx^2_W}{F\left(R_Dx_G+q_F\left(x_G-x_W\right)+q_Gx_W\right)+\left(q_G+R_D\right)\left(Gx_G-\left(D+W\right)x_W\right)}}$

To  begin simplifying this expression, the denominator must be rearranged and a substitution taking advantage of the mass balance from Eq. 1-1 shown in method 1

Eq. 3-13) $\displaystyle\frac{y_1}{y_2}=\cfrac{\cfrac{\left(Fq_F+Gq_G\right)x_D+R_D\left(Fx_F+Gx_G\right)}{\left(F+G\right)R_D+\left(Fq_F+Gq_G\right)}}{\cfrac{\left(Fq_F+Gq_G+DR_D\right)x_Dx_G-\left(\left(F\left(q_F-q_G\right)+D\left(q_G+R_D\right)+q_GW\right)x_D-R_DWx_G\right)x_W-R_DWx^2_W}{F\left(R_Dx_G+q_F\left(x_G-x_W\right)+q_Gx_W\right)+\left(q_G+R_D\right)\underbrace{\left(Gx_G-\left(D+W\right)x_W\right)}_{Gx_G-\left(F+G\right)x_W}}}$

Which is then rearranged (boxes are used to indicate subsequent cancellations)

Eq. 3-14) $\displaystyle\frac{y_1}{y_2}=\cfrac{\cfrac{\left(Fq_F+Gq_G\right)x_D+R_D\left(Fx_F+Gx_G\right)}{\left(F+G\right)R_D+\left(Fq_F+Gq_G\right)}}{\cfrac{\left(Fq_F+Gq_G\right)x_Dx_G+DR_Dx_Dx_G-\left(\left(Fq_F-\boxed{Fq_G}+DR_D+\left(\boxed{F}+G\right)q_G\right)x_D-R_DWx_G\right)x_W-R_DWx^2_W}{F\left(R_Dx_G+q_F\left(x_G-x_W\right)+q_Gx_W\right)+\left(q_G+R_D\right)\left(Gx_G-\left(F+G\right)x_W\right)}}$

Eq. 3-15) $\displaystyle\hookrightarrow\frac{y_1}{y_2}=\cfrac{\cfrac{\left(Fq_F+Gq_G\right)x_D+R_D\left(Fx_F+Gx_G\right)}{\left(F+G\right)R_D+\left(Fq_F+Gq_G\right)}}{\cfrac{\left(Fq_F+Gq_G\right)x_Dx_G+R_D\left(Dx_Dx_G-Wx^2_W\right)-\left(\left(Fq_F+DR_D+Gq_G\right)x_D-R_DWx_G\right)x_W}{FR_Dx_G+Fq_Fx_G-Fq_Fx_W+\boxed{Fq_Fx_W}+Gq_Gx_G-\boxed{Fq_Gx_W}-Gq_Gx_W+R_DGx_G-FR_Dx_W-GR_Dx_W}}$

Eq. 3-16) $\displaystyle\hookrightarrow\frac{y_1}{y_2}=\cfrac{\cfrac{\left(Fq_F+Gq_G\right)x_D+R_D\left(Fx_F+Gx_G\right)}{\left(F+G\right)R_D+\left(Fq_F+Gq_G\right)}}{\cfrac{\left(Fq_F+Gq_G\right)x_D\boxed{x_G}+R_D\left(Dx_D\right)\boxed{\left(x_G-x_W\right)}+R_DWx_W\boxed{\left(x_G-x_W\right)}-\boxed{\left(Fq_F+Gq_G\right)x_Dx_W}}{\boxed{\left(x_G-x_W\right)}\left(FR_D+Fq_F+Gq_G+R_DG\right)}}$

Eq. 3-17) $\displaystyle\hookrightarrow\frac{y_1}{y_2}=\cfrac{\cfrac{\left(Fq_F+Gq_G\right)x_D+R_D\left(Fx_F+Gx_G\right)}{\boxed{\left(F+G\right)R_D+\left(Fq_F+Gq_G\right)}}}{\cfrac{\left(Fq_F+Gq_G\right)x_D+R_D\left(Dx_D+Wx_W\right)}{\boxed{\left(F+G\right)R_D+\left(Fq_F+Gq_G\right)}}}$

To yield

Eq. 3-18) $\displaystyle\frac{y_1}{y_2}=\frac{\left(Fq_F+Gq_G\right)x_D+R_D\left(Fx_F+Gx_G\right)}{\left(Fq_F+Gq_G\right)x_D+R_D\left(Dx_D+Wx_W\right)}$

Using Eq. 1-2 from method 1 to substitute for the second term in the denominator in Eq. 3-18 yields

Eq. 3-19) $\displaystyle\frac{y_1}{y_2}=\frac{\left(Fq_F+Gq_G\right)x_D+R_D\left(Fx_F+Gx_G\right)}{\left(Fq_F+Gq_G\right)x_D+R_D\left(Fx_F+Gx_G\right)}$

Which trivially reduces to yield unity

Eq. 3-20) $\displaystyle\boxed{\frac{y_1}{y_2}=1}$

This proves that for all cases the intersection of the three lines results in the same exact y-ordinate, but this is not sufficient to prove the intersection of all three lines exists at a coincident point.  To complete the proof, the ratio of the x-ordinates must also be unity.

### 3.2.6 Evaluation of the ratio of x-Ordinates

Starting with the ratio of Eq. 3-11 to Eq. 3-6

Eq. 3-21) $\displaystyle\frac{x_1}{x_2}=\cfrac{\cfrac{\left(Fq_F+Gq_G\right)x_D+R_D\left(Fx_F+Gx_G\right)}{\left(F+G\right)R_D+\left(Fq_F+Gq_G\right)}}{\cfrac{\left(F\left(q_F-1\right)+G\left(q_G-1\right)+D\left(R_D+1\right)\right)x_Dx_G+\left(F\left(q_G-q_F\right)x_D-D\left(q_G+R_D\right)x_D+W\left(x_D-q_Gx_D+x_G+R_Dx_G\right)\right)x_W-\left(R_D+1\right)Wx^2_W}{F\left(R_Dx_G+q_F\left(x_G-x_W\right)+q_Gx_W\right)+\left(q_G+R_D\right)\left(Gx_G-\left(D+W\right)x_W\right)}}$

First, the denominator is simplified to to allow for a cancellation

Eq. 3-22) $\displaystyle\frac{x_1}{x_2}=\cfrac{\cfrac{\left(Fq_F+Gq_G\right)x_D+R_D\left(Fx_F+Gx_G\right)}{\left(F+G\right)R_D+\left(Fq_F+Gq_G\right)}}{\cfrac{\boxed{\left(x_G-x_W\right)}\left(x_D\left(F\left(q_F-1\right)+G\left(q_G-1\right)\right)+\left(R_D+1\right)\left(Dx_D+Wx_W\right)\right)}{\boxed{\left(x_G-x_W\right)}\left(FR_D+Fq_F+Gq_G+GR_D\right)}}$

Eq. 3-23) $\displaystyle\frac{x_1}{x_2}=\cfrac{\cfrac{\left(Fq_F+Gq_G\right)x_D+R_D\left(Fx_F+Gx_G\right)}{\left(F+G\right)R_D+\left(Fq_F+Gq_G\right)}}{\cfrac{\left(x_D\left(F\left(q_F-1\right)+G\left(q_G-1\right)\right)+\left(R_D+1\right)\left(Dx_D+Wx_W\right)\right)}{\left(FR_D+Fq_F+Gq_G+GR_D\right)}}$

Then, Eq. 1-2 from method 1 is used to make a substitution in the denominator

Eq. 3-24) $\displaystyle\frac{x_1}{x_2}=\cfrac{\cfrac{\left(Fq_F+Gq_G\right)x_D+R_D\left(Fx_F+Gx_G\right)}{\left(F+G\right)R_D+\left(Fq_F+Gq_G\right)}}{\cfrac{\left(x_D\left(F\left(q_F-1\right)+G\left(q_G-1\right)\right)+\left(R_D+1\right)\underbrace{\left(Dx_D+Wx_W\right)}_{Fx_F+Gx_G}\right)}{\left(FR_D+Fq_F+Gq_G+GR_D\right)}}$

Eq. 3-25) $\displaystyle\frac{x_1}{x_2}=\cfrac{\cfrac{\left(Fq_F+Gq_G\right)x_D+R_D\left(Fx_F+Gx_G\right)}{\left(F+G\right)R_D+\left(Fq_F+Gq_G\right)}}{\cfrac{\left(x_D\left(F\left(q_F-1\right)+G\left(q_G-1\right)\right)+\left(R_D+1\right)\left(Fx_F+Gx_G\right)\right)}{\left(FR_D+Fq_F+Gq_G+GR_D\right)}}$

Which trivially reduces to yield unity

Eq. 3-26) $\displaystyle\boxed{\frac{x_1}{x_2}=1}$

## 3.3.0 Conclusions of the Proof

Since both intersections coincide at the same point, all three lines must always intersect in all cases.

Part 4 will be up soon.  It concerns the generalization of method 1 to a column with an arbitrary number of feed streams and side streams.

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