**3. Mathematic model**

#### **3.1 Equations for flows throughout general trays**

The total mole holdup in the nth tray *Mn* is considered constant, but the imbalance in the input and output flows is taken into account for in the component and heat balance equations as shown in Figure 3.1.

Fig. 3.1. A General *n*th Tray

Total mass balance:

$$\frac{d(M\_n)}{dt} = L\_{n+1} - L\_n + V\_{n-1} - V\_n \tag{3.1}$$

Component balance:

$$\frac{d(M\_n\mathbf{x}\_n)}{dt} = L\_{n+1}\mathbf{x}\_{n+1} - L\_n\mathbf{x}\_n + V\_{n-1}y\_{n-1} - V\_n y\_n \tag{3.2}$$

By differentiating (3.2) and substituting for (3.1), the following expression is obtained:

$$\frac{d\langle \mathbf{x}\_n \rangle}{dt} = \frac{L\_{n+1}\mathbf{x}\_{n+1} + V\_{n-1}\mathbf{y}\_{n-1} - (L\_{n+1} + V\_{n-1})\mathbf{x}\_n - V\_n(\mathbf{y}\_n - \mathbf{x}\_n)}{M\_n} \tag{3.3}$$

Energy balance:

$$\frac{d(M\_n h\_n)}{dt} = h\_{n+1}L\_{n+1} - h\_n L\_n + H\_{n-1}V\_{n-1} - H\_n V\_n \tag{3.4}$$

or

$$\hbar M\_n \frac{d\mathbf{h}\_n}{dt} + \hbar\_n \frac{dM\_n}{dt} = \hbar\_{n+1} \mathbf{L}\_{n+1} + \mathbf{H}\_{n-1} V\_{n-1} - \hbar\_n \mathbf{L}\_n - \mathbf{H}\_n V\_n \tag{3.5}$$

Because the term *<sup>n</sup> dh dt* is approximately zero, substituting for the change of hold up *<sup>n</sup> dM dt* in (3.5), and rearranging the terms, the following expression is obtained:

$$V\_n = \frac{h\_{n+1}L\_{n+1} + H\_{n-1}V\_{n-1} - (L\_{n+1} + V\_{n-1})h\_n}{H\_n - h\_n} \tag{3.6}$$

where, n: tray *n*th; *V*: vapor flow; *L*: liquid flow; *x*: liquid concentration of light component; *y*: vapor concentration of light component; *h*: enthalpy for liquid; *H*: enthalpy for vapor.

#### **3.2 Equations for the feed tray: (Stage n=***f***) (See Figure 3.2)**

Total mass balance:

14 Distillation – Advances from Modeling to Applications

The total mole holdup in the nth tray *Mn* is considered constant, but the imbalance in the input and output flows is taken into account for in the component and heat balance

> 1 1 ( ) *<sup>n</sup> n nn n d M L LV V*

 1 1 1 1 ( ) *n n n n nn n n nn*

11 11 1 1 ( ) ( )() *<sup>n</sup> n n n n n n n nn n*

 1 1 1 1 ( ) *n n n n nn n n nn dMh h L hL H V HV*

 11 11 *n n n n n n n n nn nn dh dM M h h L H V hL HV*

(3.5), and rearranging the terms, the following expression is obtained:

*d x L x V y L V x Vy x*

*n*

By differentiating (3.2) and substituting for (3.1), the following expression is obtained:

*dt* (3.1)

*dMx L x Lx V y Vy dt* (3.2)

*dt <sup>M</sup>* (3.3)

*dt* (3.4)

*dt dt* (3.5)

*dt*

in

*dt* is approximately zero, substituting for the change of hold up *<sup>n</sup> dM*

**3. Mathematic model** 

equations as shown in Figure 3.1.

Fig. 3.1. A General *n*th Tray

Total mass balance:

Component balance:

Energy balance:

Because the term *<sup>n</sup> dh*

or

**3.1 Equations for flows throughout general trays** 

$$\frac{d(M\_f)}{dt} = F + L\_{f+1} + V\_{f-1} - L\_f - V\_f \tag{3.7}$$

Component balance:

$$\begin{split} \frac{d(\mathbf{M}\_f \mathbf{x}\_f)}{dt} &= \mathbf{F} \mathbf{c}\_f + \mathbf{L}\_{f+1} \mathbf{x}\_{f+1} + \mathbf{V}\_{f-1} \mathbf{y}\_{f-1} - \mathbf{L}\_f \mathbf{x}\_f - \mathbf{V}\_f \mathbf{y}\_f \\ \Rightarrow \frac{d \mathbf{x}\_n}{dt} &= \frac{\mathbf{L}\_{n+1} \mathbf{x}\_{n+1} + \mathbf{V}\_{n-1} \mathbf{y}\_{n-1} - (\mathbf{L}\_{n+1} + \mathbf{V}\_{n-1}) \mathbf{x}\_n - \mathbf{V}\_n \mathbf{y}\_n - \mathbf{x}\_n \end{split} \tag{3.8}$$

Energy balance:

$$\begin{split} \frac{d\{M\_f h\_f\}}{dt} &= h\_f F + h\_{n+1} L\_{n+1} + H\_{n-1} V\_{n-1} - h\_n L\_n - H\_n V\_n\\ \Rightarrow V\_n &= \frac{h\_F F + h\_{n+1} L\_{n+1} + H\_{n-1} V\_{n-1} - (L\_{n+1} + V\_{n-1}) h\_n}{H\_n - h\_n} \end{split} \tag{3.9}$$

#### Fig. 3.2. Feed Section

Modeling and Control Simulation for a Condensate Distillation Column 17

The condenser duty *QC* is equal to the latent heat required to condense the overhead vapor

 <sup>2</sup> 32 2

*d M LLVV*

2 2

*B*

33 22 2 2

3 3 22 22

*B B dMh hL HV hL HV*

( ) *BB B hL HV L V h <sup>V</sup>*

3 3 3 2

2 2

*B B*

*dMx Lx Lx Vy Vy dt* (3.18)

*dt* (3.19)

*H h* (3.20)

*dt* (3.17)

( )

( )

( ) *B B*

2

The base of the column has some particular characteristics as follows:

There is re-boiler heat flux *QB* establishing the boil-up vapor flow *VB* .

**3.4 Equations for the bottom section: (Stage n=2) (See Figue 3.4)** 

( ) *Q HV h L V H h C in in out out N N N* (3.16)

to bubble point:

Fig. 3.4. Bottom Section and Re-boiler

*Re-boiler and Column Bottoms (stage n=1)* 

*Bottom Tray (stage n=2)*  Total mass balance:

Component balance:

Energy balance:

#### **3.3 Equations for the top section: (stage n=***N+***1) (See Figure 3.3)**

Fig. 3.3. Top Section and Reflux Drum *Equations for the top tray (stage n=N+1)* 

Total mass balance:

$$\frac{d(M\_{N+1})}{dt} = L + V\_N - L\_{N+1} - V\_{N+1} \tag{3.10}$$

Component balance:

$$\frac{d\left(M\_{N+1}\mathbf{x}\_{N+1}\right)}{dt} = \mathbf{L}\mathbf{x}\_{D} + V\_{N}\mathbf{y}\_{N} - L\_{N+1}\mathbf{x}\_{N+1} - V\_{N+1}\mathbf{y}\_{N+1} \tag{3.11}$$

Energy balance:

$$\frac{d(M\_{N+1}h\_{N+1})}{dt} = h\_D L + H\_N V\_N - h\_{N+1} L\_{N+1} - H\_{N+1} V\_{N+1} \tag{3.12}$$

$$V \Longrightarrow V\_{N+1} = \frac{h\_D L + H\_N V\_N - (L + V\_N) h\_{N+1}}{H\_{N+1} - h\_{N+1}} \tag{3.13}$$

#### **Reflux drum and condenser**

Total mass balance:

$$\frac{d(M\_D)}{dt} = V\_{N+1} - L - D \tag{3.14}$$

Component balance:

$$\frac{d(M\_D \mathbf{x}\_D)}{dt} = V\_{N+1} y\_{N+1} - (L+D)\mathbf{x}\_D \tag{3.15}$$

Energy balance around condenser:

The condenser duty *QC* is equal to the latent heat required to condense the overhead vapor to bubble point:

$$\mathbf{Q}\_{\mathbb{C}} = \mathbf{H}\_{in}\mathbf{V}\_{in} - \mathbf{h}\_{out}\mathbf{L}\_{out} = \mathbf{V}\_{N}(\mathbf{H}\_{N} - \mathbf{h}\_{N}) \tag{3.16}$$

#### **3.4 Equations for the bottom section: (Stage n=2) (See Figue 3.4)**

Fig. 3.4. Bottom Section and Re-boiler

*Bottom Tray (stage n=2)* 

Total mass balance:

16 Distillation – Advances from Modeling to Applications

<sup>1</sup>

1 1

1 1

 

<sup>1</sup>

 1 1 ( ) ( ) *D D*

*N N D*

*N d M V LD*

*dM h hL HV h L H V*

*hL HV L V h <sup>V</sup>*

( ) *<sup>D</sup>*

1

*N*

*d M LV L V*

*NN N*

*D NN N N N N dM x Lx V y L x V y dt* (3.11)

*D NN N N N N*

1 1 *D NN N N* ( )

*N N*

1 1

*dt* (3.10)

11 11

11 11

1

*dt* (3.14)

*dM x V y L Dx dt* (3.15)

*H h* (3.13)

*dt* (3.12)

**3.3 Equations for the top section: (stage n=***N+***1) (See Figure 3.3)** 

( ) *<sup>N</sup>*

( ) *N N*

( ) *N N*

Fig. 3.3. Top Section and Reflux Drum *Equations for the top tray (stage n=N+1)* 

Total mass balance:

Component balance:

Energy balance:

**Reflux drum and condenser** 

Energy balance around condenser:

Total mass balance:

Component balance:

$$\frac{d(M\_2)}{dt} = L\_3 - L\_2 + V\_B - V\_2 \tag{3.17}$$

Component balance:

$$\frac{d(M\_2\mathbf{x}\_2)}{dt} = L\_3\mathbf{x}\_3 - L\_2\mathbf{x}\_2 + V\_B\mathbf{y}\_B - V\_2\mathbf{y}\_2\tag{3.18}$$

Energy balance:

$$\frac{d(M\_B h\_B)}{dt} = h\_3 L\_3 + H\_B V\_B - h\_2 L\_2 - H\_2 V\_2 \tag{3.19}$$

$$\implies V\_2 = \frac{h\_3 L\_3 + H\_B V\_B - (L\_3 + V\_B)h\_2}{H\_2 - h\_2} \tag{3.20}$$

*Re-boiler and Column Bottoms (stage n=1)* 

The base of the column has some particular characteristics as follows:

There is re-boiler heat flux *QB* establishing the boil-up vapor flow *VB* .

Modeling and Control Simulation for a Condensate Distillation Column 19

<sup>1</sup> *MDn F n n n x VV* ( )*y Lx Dx* (3.26)

1 1 ( )( ) ( ) *M n Fn n n n x VV y y Lx x* (3.27)

1 1 ( )( ) ( ) *M n n n n n FF n x V y y Lx x V y y* (3.28)

1 1 ( )( ) ( ) *M n n n n n FF n x V y y Lx x L x x* (3.29)

1 1 ( ) ( )( ) *M n n n Fn n x V y y LL x x* (3.30)

1 211 ( ) *MB F x LLx Vy Bx* (3.31)

*LF F q F* (3.32)

*V FL F F* (3.33)

*DV LVV L N F* (3.34)

*BL V LL V* 2 1 *<sup>F</sup>* (3.35)

*Fc L x V F FF FF y* (3.36)

(3.37)

Composition *<sup>F</sup> x* in the liquid and *<sup>F</sup> y* in the vapor phase of the feed are obtained by solving

*x*

 1 ( 1) *F*

Although the model order is reduced, the representation of the distillation system is still

*F*

*x*

*F*

*y*

*nonlinear* due to the vapor-liquid equilibrium relationship in equation (3.25).

Under these assumptions, the dynamic model can be expressed by (George, S., 1986):

Condenser (*n*=*N*+2):

Tray n (n=f+2, …, N+1):

Tray above the feed flow (*n=f+1*):

Tray below the feed flow (*n=f*):

Flow rate are assumed as constant molar flows as

Assuming condenser holdup constant

is the relative volatility.

Assuming boiler holdup constant

the flash equations:

And

where,

Tray n (n=2, …, f-1):

Re-boiler (*n=1*):


Total mass balance:

$$\frac{d(M\_B)}{dt} = L\_2 - V\_B - B \tag{3.21}$$

Component balance:

$$\frac{d(M\_B \mathbf{x}\_B)}{dt} = L\_2 \mathbf{x}\_2 - V\_B \mathbf{y}\_B - B \mathbf{x}\_B \tag{3.22}$$

Energy balance:

$$\frac{d(M\_B h\_B)}{dt} = h\_2 L\_2 + Q\_B - h\_B B - H\_B V\_B \tag{3.23}$$

$$\implies V\_B = \frac{h\_2 L\_2 + Q\_B - h\_B B - M\_B \frac{d\mathbf{h}\_B}{dt} - h\_B \frac{d\mathbf{M}\_B}{dt}}{H\_B} \tag{3.24}$$

When all the modeling equations above are resolved, we find out how the flow rate and concentrations of the two product streams (distillate product and bottoms product) change with time, in the presence of changes in the various input variables.

#### **3.5 Simplified model**

To simplify the model, we make the following assumption (Papadouratis, A. *et al.* 1989):

 The relative volatility is constant throughout the column. This means the vaporliquid equilibrium relationship can be expressed by

$$y\_n = \frac{a\mathbf{x}\_n}{1 + (a-1)\mathbf{x}\_n} \tag{3.25}$$

where *<sup>n</sup> x* : liquid composition on *nth* stage; *<sup>n</sup> y* : vapor composition on *nth* stage; and : relative volatility.


Under these assumptions, the dynamic model can be expressed by (George, S., 1986): Condenser (*n*=*N*+2):

$$M\_D \dot{\mathbf{x}}\_n = (V + V\_F)\mathbf{y}\_{n-1} - L\mathbf{x}\_n - D\mathbf{x}\_n \tag{3.26}$$

Tray n (n=f+2, …, N+1):

18 Distillation – Advances from Modeling to Applications

The outflow of liquid from the bottoms *B* is determined externally to be controlled by a

<sup>2</sup> ( ) *<sup>B</sup>*

*d M LVB*

2 2

2 2

*dh dM hL Q hB M h dt dt <sup>V</sup>*

When all the modeling equations above are resolved, we find out how the flow rate and concentrations of the two product streams (distillate product and bottoms product) change

To simplify the model, we make the following assumption (Papadouratis, A. *et al.* 1989):

*y*

mixed (i.e. immediate liquid response, ( 2 3 <sup>2</sup> *dL dL dL dL* ...... *<sup>N</sup>* ).

sections are constant: 1 2 <sup>1</sup> *VV V* ....... *<sup>N</sup>* and 2 3 <sup>2</sup> *LL L* ........ *<sup>N</sup>* .

for the feed tray, *n*=*N*+1 for the top tray and *n*=*N*+2 for the condenser).

*x*

 1 ( 1) *<sup>n</sup> <sup>n</sup>*

where *<sup>n</sup> x* : liquid composition on *nth* stage; *<sup>n</sup> y* : vapor composition on *nth* stage; and

The liquid holdups on each tray, condenser, and the re-boiler are constant and perfectly

The holdup of vapor is negligible throughout the system (i.e. immediate vapor

The molar flow rates of the vapor and liquid through the stripping and rectifying

The column is numbered from bottom (*n*=1 for the re-boiler, *n*=2 for the first tray, *n*=*f*

*n*

*x*

*BB B B*

*B*

*B*

*BB B*

*B B BB*

*dt* (3.21)

*dMx L x V y Bx dt* (3.22)

*dMh hL Q hB HV dt* (3.23)

*B B*

is constant throughout the column. This means the vapor-

*H* (3.24)

(3.25)

:

The holdup is variable and changes in sensible heat cannot be neglected.

( ) *B B*

( ) *B B*

2 2

with time, in the presence of changes in the various input variables.

liquid equilibrium relationship can be expressed by

The overhead vapor is totally condensed in a condenser.

response, 1 2 <sup>1</sup> *dV dV dV dV* ...... *<sup>N</sup>* ).

*B*

bottoms level controller.

Total mass balance:

Component balance:

Energy balance:

**3.5 Simplified model** 

relative volatility.

The relative volatility

$$M\dot{\mathbf{x}}\_n = (V + V\_F)(y\_{n-1} - y\_n) + L(\mathbf{x}\_{n+1} - \mathbf{x}\_n) \tag{3.27}$$

Tray above the feed flow (*n=f+1*):

$$M\dot{\mathbf{x}}\_n = V(y\_{n-1} - y\_n) + L(\mathbf{x}\_{n+1} - \mathbf{x}\_n) + V\_F(y\_F - y\_n) \tag{3.28}$$

Tray below the feed flow (*n=f*):

$$M\dot{\mathbf{x}}\_n = V(y\_{n-1} - y\_n) + L(\mathbf{x}\_{n+1} - \mathbf{x}\_n) + L\_F(\mathbf{x}\_F - \mathbf{x}\_n) \tag{3.29}$$

Tray n (n=2, …, f-1):

$$M\dot{\mathbf{x}}\_n = V(y\_{n-1} - y\_n) + (L + L\_F)(\mathbf{x}\_{n+1} - \mathbf{x}\_n) \tag{3.30}$$

Re-boiler (*n=1*):

$$M\_B \dot{\mathbf{x}}\_1 = (L + L\_F)\mathbf{x}\_2 - V\mathbf{y}\_1 - B\mathbf{x}\_1 \tag{3.31}$$

Flow rate are assumed as constant molar flows as

$$L\_F = q\_F F \tag{3.32}$$

$$V\_F = F - L\_F \tag{3.33}$$

Assuming condenser holdup constant

$$D = V\_N - L = V + V\_F - L \tag{3.34}$$

Assuming boiler holdup constant

$$B = L\_2 - V\_1 = L + L\_F - V \tag{3.35}$$

Composition *<sup>F</sup> x* in the liquid and *<sup>F</sup> y* in the vapor phase of the feed are obtained by solving the flash equations:

$$F\mathcal{L}\_F = L\_F \mathcal{x}\_F + V\_F \mathcal{y}\_F \tag{3.36}$$

And

$$y\_F = \frac{a\mathbf{x}\_F}{1 + (a - 1)\mathbf{x}\_F} \tag{3.37}$$

where,is the relative volatility.

Although the model order is reduced, the representation of the distillation system is still *nonlinear* due to the vapor-liquid equilibrium relationship in equation (3.25).

Modeling and Control Simulation for a Condensate Distillation Column 21

<sup>13</sup> 12 13 14 13 5.8 164.8559( ) 75.6380( ) *x y y x x*

<sup>12</sup> 11 12 13 12 5.8 164.8559( ) 75.6380( ) *x y y x x*

<sup>11</sup> 10 11 12 11 5.8 164.8559( ) 75.6380( ) *x y y x x*

<sup>10</sup> 9 10 11 10 5.8 164.8559( ) 75.6380( ) *x y y x x*

9 8 9 10 <sup>9</sup> <sup>9</sup> 5.8 66.3407( ) 75.6380( ) 98.5152(0.66728 ) *x y y x x y*

8 7 8 9 <sup>8</sup> <sup>8</sup> 5.8 66.3407( ) 75.6380( ) 104.2491(0.26.95 ) *x y y x x x*

7 6 7 8 <sup>7</sup> 5.8 66.3407( ) 179.8871( ) *x y y x x*

6 5 6 7 <sup>6</sup> 5.8 66.3407( ) 179.8871( ) *x y y x x*

5 4 5 6 <sup>5</sup> 5.8 66.3407( ) 179.8871( ) *x y y x x*

4 3 4 5 <sup>4</sup> 5.8 66.3407( ) 179.8871( ) *x y y x x*

3 2 3 4 <sup>3</sup> 5.8 66.3407( ) 179.8871( ) *x y y x x*

2 1 2 3 <sup>2</sup> 5.8 66.3407( ) 179.8871( ) *x y y x x*

31.11 179.8871 110.9235 66.3407 *xxx* 1 2 11 *y*

 5.68 1 4.68 *<sup>n</sup> <sup>n</sup>*

*y*

And vapor-liquid equilibrium on each tray (*n*=1-16):

<sup>13</sup> <sup>12</sup> <sup>13</sup> <sup>14</sup> <sup>13</sup> *x* 28.4234 28.4234 13.0410 13.0410 *y y x x* (4.4)

<sup>12</sup> <sup>11</sup> <sup>12</sup> <sup>13</sup> <sup>12</sup> *x* 28.4234 28.4234 13.0410 13.0410 *y y x x* (4.5)

<sup>11</sup> <sup>10</sup> <sup>11</sup> <sup>12</sup> <sup>11</sup> *x* 28.4234 28.4234 13.0410 13.0410 *y y x x* (4.6)

<sup>10</sup> <sup>9</sup> <sup>10</sup> <sup>11</sup> <sup>10</sup> *x* 28.4234 28.4234 13.0410 13.0410 *y y x x* (4.7)

9 89910 *xyyxx* 11.4381 28.4234 13.0410 13.0410 11.3340 (4.8)

<sup>87889</sup> *xyyxx* 11.4381 11.4381 31.0150 13.0410 4.8440 (4.9)

7 6787 *x* 11.4381 11.4381 31.0150 31.0150 *y y x x* (4.10)

6 5676 *x* 11.4381 11.4381 31.0150 31.0150 *y y x x* (4.11)

5 4565 *x* 11.4380 11.4381 31.0150 31.0150 *y y x x* (4.12)

4 3454 *x* 11.4381 11.4381 31.0150 31.0150 *y y x x* (4.13)

<sup>32343</sup> *x* 11.4381 11.4381 31.0150 31.0150 *y y x x* (4.14)

2 1232 *x* 11.4381 11.4381 31.0150 31.0150 *y y x x* (4.15)

*n*

*x*

*x*

1 121 *x xx* 3.5655 5.7823 2.1325*y* (4.16)

(4.17)

#### **4. Model simulation and analysis**

#### **4.1 Model dynamic equations**

In the process data calculation, we have calculated for the distillation column with 14 trays with the following initial data - equations (2.1), (2.2), (2.3) and (2.4): The feed mass rate of the plant: *Fmass* 15.47619 (tons/hour); The holdup in the column base: 31.11 *MB* (kmole);

The holdup on each tray: *M* 5.80 (kmole); The holdup in the reflux drum: *MD* 13.07 (kmole); The gas percentage in the feed flow: *cF* 38% ; The internal vapor flow *Vf* selected by empirical: *Vf* 28% ; The feed stream (m3/h) with the density *d*F=0.670 (ton/m3);

 15.47619 0.670 *mass F <sup>F</sup> <sup>F</sup> <sup>d</sup>* =23.0988 (m3/h). The calculated stream data is displayed in the table 4.1.


Table 4.1. Summary of Stream Data

Solving flash equation with the relative volatility (5.68 **)**, 0.26095; 0.66728 *F F x y* .

Reference to equations from (3.28) to (3.39) we can develop a set of nonlinear differential and algebraic equations for the simplified model can be developed as:

$$13.07\dot{\mathbf{x}}\_{16} = 164.8559y\_{15} - 75.6380\mathbf{x}\_{16} - 92.7597\mathbf{x}\_{16}$$

$$\Rightarrow \dot{\mathbf{x}}\_{16} = 12.6133y\_{15} - 12.8863\mathbf{x}\_{16} \tag{4.1}$$

$$\begin{aligned} 5.8\dot{\mathbf{x}}\_{15} &= 164.8559(y\_{14} - y\_{15}) + 75.6380(\mathbf{x}\_{16} - \mathbf{x}\_{15})\\ \implies \dot{\mathbf{x}}\_{15} &= 28.4234y\_{14} - 28.4234y\_{15} + 13.0410\mathbf{x}\_{16} - 13.0410\mathbf{x}\_{15} \end{aligned} \tag{4.2}$$

$$\begin{aligned} 5.8\dot{\mathbf{x}}\_{14} &= 164.8559(y\_{13} - y\_{14}) + 75.6380(\mathbf{x}\_{15} - \mathbf{x}\_{14})\\ \Rightarrow \dot{\mathbf{x}}\_{14} &= 28.4234y\_{13} - 28.4234y\_{14} + 13.0410\mathbf{x}\_{15} - 13.0410\mathbf{x}\_{14} \end{aligned} \tag{4.3}$$

In the process data calculation, we have calculated for the distillation column with 14 trays with the following initial data - equations (2.1), (2.2), (2.3) and (2.4): The feed mass rate of the plant: *Fmass* 15.47619 (tons/hour); The holdup in the column base: 31.11 *MB* (kmole);

The holdup on each tray: *M* 5.80 (kmole); The holdup in the reflux drum: *MD* 13.07 (kmole); The gas percentage in the feed flow: *cF* 38% ; The internal vapor flow *Vf* selected by empirical: *Vf* 28% ; The feed stream (m3/h) with the density *d*F=0.670 (ton/m3);

*<sup>d</sup>* =23.0988 (m3/h). The calculated stream data is displayed in the table 4.1.

rate *<sup>V</sup> Vf* 28 6.4677 0.598 3.8677 58.3 66.3407

rate *<sup>L</sup> Vf*+4 32 7.3916 0.615 4.5458 60.1 75.6380

rate *<sup>D</sup> cF* 38 8.7775 0.576 5.0554 54.5 92.7597

rate *<sup>B</sup>* 100-*cF* 62 14.3213 0.727 10.405 93.8 110.9235

Reference to equations from (3.28) to (3.39) we can develop a set of nonlinear differential

<sup>16</sup> <sup>15</sup> <sup>16</sup> <sup>16</sup> 13.07 164.8559 75.6380 92.7597 *x y x x*

<sup>15</sup> 14 15 16 15 5.8 164.8559( ) 75.6380( ) *x y y x x*

<sup>14</sup> 13 14 15 14 5.8 164.8559( ) 75.6380( ) *x y y x x*

Density (ton/m3)

*cF+4* 42 9.7015 0.591 5.7336 58.2 98.5152

100-*c*<sup>F</sup> 58 13.3973 0.726 9.7264 93.3 104.2491

Mass (ton/h)

Molar (kg/kmol)

5.68 **)**, 0.26095; 0.66728 *F F x y* .

<sup>16</sup> <sup>15</sup> <sup>16</sup> *x* 12.6133 12.8863 *y x* (4.1)

<sup>15</sup> <sup>14</sup> <sup>15</sup> <sup>16</sup> <sup>15</sup> *x* 28.4234 28.4234 13.0410 13.0410 *y y x x* (4.2)

<sup>14</sup> <sup>13</sup> <sup>14</sup> <sup>15</sup> <sup>14</sup> *x* 28.4234 28.4234 13.0410 13.0410 *y y x x* (4.3)

Molar flow (kmole/h)

(m3/h)

**4. Model simulation and analysis** 

**4.1 Model dynamic equations** 

15.47619

*mass F*

Vapor rate in feed *VF*

Liquid rate in feed *LF*

Internal vapor

Internal liquid

Distillate flow

Bottoms flow

Table 4.1. Summary of Stream Data

Solving flash equation with the relative volatility (

and algebraic equations for the simplified model can be developed as:

*<sup>F</sup> <sup>F</sup>*

0.670

Stream Formular % Volume

$$\begin{aligned} 5.8\dot{\mathbf{x}}\_{13} &= 164.8559(y\_{12} - y\_{13}) + 75.6380(\mathbf{x}\_{14} - \mathbf{x}\_{13}) \\ \implies \dot{\mathbf{x}}\_{13} &= 28.4234y\_{12} - 28.4234y\_{13} + 13.0410\mathbf{x}\_{14} - 13.0410\mathbf{x}\_{13} \end{aligned} \tag{4.4}$$

$$5.8\dot{\mathbf{x}}\_{12} = 164.8559(y\_{11} - y\_{12}) + 75.6380(\mathbf{x}\_{13} - \mathbf{x}\_{12})$$

$$\Rightarrow \dot{\mathbf{x}}\_{12} = 28.4234y\_{11} - 28.4234y\_{12} + 13.0410\mathbf{x}\_{13} - 13.0410\mathbf{x}\_{12} \tag{4.5}$$

$$\begin{aligned} \mathbf{5.8\dot{x}\_{11}} &= \mathbf{164.8559(y\_{10} - y\_{11})} + \mathbf{75.6380(x\_{12} - x\_{11})} \\ \Rightarrow \dot{\mathbf{x}}\_{11} &= \mathbf{28.4234}y\_{10} - \mathbf{28.4234}y\_{11} + \mathbf{13.0410x\_{12} - \mathbf{13.0410x\_{11}}} \end{aligned} \tag{4.6}$$

$$\begin{aligned} \mathbf{5.8\dot{x}\_{10}} &= \mathbf{164.8559(y\_9 - y\_{10})} + \mathbf{75.6380(x\_{11} - x\_{10})} \\ \implies \dot{\mathbf{x}}\_{10} &= \mathbf{28.4234}y\_9 - \mathbf{28.4234}y\_{10} + \mathbf{13.0410x\_{11} - \mathbf{13.0410x\_{10}}} \end{aligned} \tag{4.7}$$

$$5.8\dot{x}\_9 = 66.3407(y\_8 - y\_9) + 75.6380(x\_{10} - x\_9) + 98.5152(0.66728 - y\_9)$$

$$\Rightarrow \dot{x}\_9 = 11.4381y\_8 - 28.4234y\_9 - 13.0410x\_9 + 13.0410x\_{10} + 11.3340\tag{4.8}$$

$$\begin{aligned} \text{5.8\dot{x}\_8} &= 66.3407(y\_7 - y\_8) + 75.6380(\mathbf{x}\_9 - \mathbf{x}\_8) + 104.2491(0.26.95 - \mathbf{x}\_8) \\ \Rightarrow \dot{\mathbf{x}}\_8 &= 11.4381y\_7 - 11.4381y\_8 - 31.0150\mathbf{x}\_8 + 13.0410\mathbf{x}\_9 + 4.8440 \end{aligned} \tag{4.9}$$

$$\begin{aligned} 5.8\dot{\mathbf{x}}\_7 &= 66.3407(y\_6 - y\_7) + 179.8871(\mathbf{x}\_8 - \mathbf{x}\_7) \\ \Rightarrow \dot{\mathbf{x}}\_7 &= 11.4381y\_6 - 11.4381y\_7 + 31.0150\mathbf{x}\_8 - 31.0150\mathbf{x}\_7 \end{aligned} \tag{4.10}$$

$$5.8\dot{\mathbf{x}}\_6 = 66.3407(y\_5 - y\_6) + 179.8871(\mathbf{x}\_7 - \mathbf{x}\_6)$$

$$\implies \dot{\mathbf{x}}\_6 = 11.4381y\_5 - 11.4381y\_6 + 31.0150\mathbf{x}\_7 - 31.0150\mathbf{x}\_6\tag{4.11}$$

$$5.8\dot{\mathbf{x}}\_5 = 66.3407(y\_4 - y\_5) + 179.8871(\mathbf{x}\_6 - \mathbf{x}\_5)$$

$$\implies \dot{\mathbf{x}}\_5 = 11.4380y\_4 - 11.4381y\_5 + 31.0150\mathbf{x}\_6 - 31.0150\mathbf{x}\_5\tag{4.12}$$

$$\begin{aligned} 5.8\dot{\mathbf{x}}\_4 &= 66.3407(y\_3 - y\_4) + 179.8871(\mathbf{x}\_5 - \mathbf{x}\_4) \\ \implies \dot{\mathbf{x}}\_4 &= 11.4381y\_3 - 11.4381y\_4 + 31.0150\mathbf{x}\_5 - 31.0150\mathbf{x}\_4 \end{aligned} \tag{4.13}$$

$$\begin{aligned} 5.8\dot{\mathbf{x}}\_3 &= 66.3407(y\_2 - y\_3) + 179.8871(\mathbf{x}\_4 - \mathbf{x}\_3) \\ \Rightarrow \dot{\mathbf{x}}\_3 &= 11.4381y\_2 - 11.4381y\_3 + 31.0150\mathbf{x}\_4 - 31.0150\mathbf{x}\_3 \end{aligned} \tag{4.14}$$

$$\begin{aligned} 5.8\dot{\mathbf{x}}\_2 &= 66.3407(y\_1 - y\_2) + 179.8871(\mathbf{x}\_3 - \mathbf{x}\_2) \\ \implies \dot{\mathbf{x}}\_2 &= 11.4381y\_1 - 11.4381y\_2 + 31.0150\mathbf{x}\_3 - 31.0150\mathbf{x}\_2 \end{aligned} \tag{4.15}$$

$$\begin{aligned} \text{(31.11\dot{x}\_1 = 179.8871x\_2 - 110.9235x\_1 - 66.3407y\_1)}\\ \Rightarrow \dot{x}\_1 = -3.5655x\_1 + 5.7823x\_2 - 2.1325y\_1 \end{aligned} \tag{4.16}$$

And vapor-liquid equilibrium on each tray (*n*=1-16):

$$y\_n = \frac{5.68\chi\_n}{1 + 4.68\chi\_n} \tag{4.17}$$

Modeling and Control Simulation for a Condensate Distillation Column 23

Purity of the Distillate Product (%)

Normal Feed Rate (100%) 96.54 3.75 Reduced Feed Rate (90%) 90.23 0.66 Increased Feed Rate (110%) 97.30 11.66

Table 4.3. Product quality depending on the change of feed rate

Fig. 4.2. Product qualities depending on change of feed rate

Fig. 4.3. Feed flow rate in a sine wave around 5%


Input

Fig. 4.4. Product quality for a sine wave feed rate

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Mole Fraction

(see Figure 4.4 and Table 4.4).

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Purity of the Distillate Product x

**4.2.3 Simulation with a wave change in the feed flow rate by 5%** 

When the input flow rate fluctuates in a *sine wave* by 5% (see Figure 4.3), the purity of the distillate product and the impurity of the bottoms product will also fluctuate in a sine wave

<sup>0</sup> <sup>100</sup> <sup>200</sup> <sup>300</sup> <sup>400</sup> <sup>500</sup> <sup>600</sup> <sup>700</sup> <sup>800</sup> <sup>900</sup> <sup>1000</sup> -0.5

Time

<sup>0</sup> <sup>100</sup> <sup>200</sup> <sup>300</sup> <sup>400</sup> <sup>500</sup> <sup>600</sup> <sup>700</sup> <sup>800</sup> <sup>900</sup> <sup>1000</sup> <sup>0</sup>

Time

<sup>0</sup> <sup>50</sup> <sup>100</sup> <sup>150</sup> <sup>200</sup> <sup>250</sup> <sup>300</sup> <sup>350</sup> <sup>0</sup>

Time

Impurity of the Bottoms Product (%)

*x*

*D*

*xB*

### **4.2 Model simulation with Matlab Simulink**

#### **4.2.1 Simulation without disturbances**

The steady-state solution is determined with dynamic simulation. Figure 4.1 displays the concentration of the light component *xn* at each tray and Table 4.2 shows the steady state values of concentration of *xn* on each tray.

Fig. 4.1. Steady state values of concentration xn on each tray


Table 4.2. Steady state values of concentration *x*n on each tray

If there are no disturbance in operating condition, the system model is to achieve the steady state of product quality that the purity of the distillate product *xD* 0.9654 and the impurity of the bottoms product 0.0375 *Bx* .

#### **4.2.2 Simulation with 10% decreasing and increasing feed flow rate**

When decreasing the feed flow rate by 10%, the quality of the distillate product will get worse while the quality of the bottoms product will get better: the purity of the distillate product reduces from 96.54% to 90.23% while the impurity of the bottoms product reduces from 3.75% to 0.66%.

In contrast, when increasing the feed flow rate by 10%, the quality of the distillate product will be better while the quality of the bottoms product will be worse: the purity of the distillate product increases from 96.54% to 97.30% while the impurity of the bottoms product increases from 3.75% to 11.66%. (See Table 4.3 and Figure 4.2).

The steady-state solution is determined with dynamic simulation. Figure 4.1 displays the concentration of the light component *xn* at each tray and Table 4.2 shows the steady state

*x*

*D*

*xB*

Tray 1 2 3 4 5 6 7 8 *xn* **0.0375** 0.0900 0.1559 0.2120 0.2461 0.2628 0.2701 0.2731 *yn* 0.1812 0.3597 0.5120 0.6044 0.6496 0.6694 0.6776 0.6809 Tray 9 10 11 12 13 14 15 16 *xn* 0.2811 0.3177 0.3963 0.5336 0.7041 0.8449 0.9269 **0.9654**  *yn* 0.6895 0.7256 0.7885 0.8666 0.9311 0.9687 0.9863 0.9937

<sup>0</sup> <sup>50</sup> <sup>100</sup> <sup>150</sup> <sup>200</sup> <sup>250</sup> <sup>300</sup> <sup>350</sup> <sup>0</sup>

Time

If there are no disturbance in operating condition, the system model is to achieve the steady state of product quality that the purity of the distillate product *xD* 0.9654 and the impurity

When decreasing the feed flow rate by 10%, the quality of the distillate product will get worse while the quality of the bottoms product will get better: the purity of the distillate product reduces from 96.54% to 90.23% while the impurity of the bottoms product reduces

In contrast, when increasing the feed flow rate by 10%, the quality of the distillate product will be better while the quality of the bottoms product will be worse: the purity of the distillate product increases from 96.54% to 97.30% while the impurity of the bottoms

**4.2 Model simulation with Matlab Simulink** 

**4.2.1 Simulation without disturbances** 

values of concentration of *xn* on each tray.

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Purity of the Distillate Product x

Fig. 4.1. Steady state values of concentration xn on each tray

Table 4.2. Steady state values of concentration *x*n on each tray

**4.2.2 Simulation with 10% decreasing and increasing feed flow rate** 

product increases from 3.75% to 11.66%. (See Table 4.3 and Figure 4.2).

of the bottoms product 0.0375 *Bx* .

from 3.75% to 0.66%.


Table 4.3. Product quality depending on the change of feed rate

Fig. 4.2. Product qualities depending on change of feed rate

#### **4.2.3 Simulation with a wave change in the feed flow rate by 5%**

When the input flow rate fluctuates in a *sine wave* by 5% (see Figure 4.3), the purity of the distillate product and the impurity of the bottoms product will also fluctuate in a sine wave (see Figure 4.4 and Table 4.4).

Fig. 4.3. Feed flow rate in a sine wave around 5%

Fig. 4.4. Product quality for a sine wave feed rate

Modeling and Control Simulation for a Condensate Distillation Column 25

Linearization for general trays ( *n* 2 15 ÷ ) - ACCUMULATION = INLET – OUTLET:

11 11 1 1

Substituting equation (5.4) into equation (5.5), the following expression is obtained:

12 1 ... *L L L LL F F* ; 2 3 ... *LL L F F* ; 12 1 ... *VV V V <sup>F</sup>* ; 23 1 ... *V V V VV FF N F* ; 5.8 *M M <sup>n</sup>* ; 66.3407 *V V Steady State* ; 75.6380 *L L Steady State* ; 104.2491 *LF* ; 98.5152 *VF* .

1 11 1 1 11 1 1 ( ) ( ) ( ) ( ) *n nn n n nn n nn n n nn n n nn <sup>n</sup>*

In order to obtain a linear approximation to this nonlinear system, this equation may be expanded into a Taylor series about the normal operating point from equation (5.3), and the

Tray above the feed flow (*n=f+1*): <sup>1</sup> <sup>1</sup> ( ) () ( ) *n n n n F nF <sup>n</sup>*

 1 1 <sup>1</sup> 1 1

*K V KV L KV <sup>L</sup> x x y y xx x x x L V M M MM M*

*<sup>n</sup> n nF n n n n n n <sup>n</sup> n n*

*Lx x VKx K x V y y Kx*

Tray below the feed flow (*n=f*): <sup>1</sup> <sup>1</sup> ( ) ( )( ) *n n n n F nF n n*

*n n nn n n F F n nn*

 1 1 <sup>1</sup> 1 1

*K V KV L L <sup>L</sup> x x y y xx x x x L V M MMM M*

*<sup>n</sup> n F n n n n n n <sup>n</sup> n n*

*x*

 1 1 <sup>1</sup> 1 1

*n n nn n n n n n n n n <sup>n</sup> n n*

( )( )

*K V KV L L x x y y xx x x x L V M MM M M L x x V Kx K x M M*

*y Kx V K Vx y Kx V K Vx L x Lx*

*n n n n nn n n*

() ( ) ( ) ( )

() ( ) ( ) ( )

( ) ( ) () ( ) ( )

1 1 1

*y y V x xL y y V*

*M MM*

*y yV x xL x xL*

*M MM*

*n n nn n n*

*M M M M MM*

1 1

(5.5)

(5.6)

and the

(5.7)

and the

(5.8)

*n nn n n n nn n n nn n*

*M x V y L x Vy Lx V VL L xy yx x M MM M*

( ) ( ) *nn n n n n nn nn*

**5.1.2 Material balance relationship in each tray** 

linear approximation equations for general trays are obtained:

linear approximation equations for the tray above the feed flow:

linear approximation equations for the tray above the feed flow:

() ( )

1 1 1

*M M*

( ) ( )( )

*x x*

*n n FF nn n n*

*Lx x Lx VKx K x M M*

1 1 1

where,

*x*

Linearization for special trays:


Table 4.4. Product quality depending on the input sine wave fluctuation

The product quality of this feed rate is not satisfied with *xB* 96% and *xD* 4% .

#### **5. Linearized control model**

#### **5.1 Linear approximation of nonlinear system**

#### **5.1.1 Vapor-Liquid equilibrium relationship in each tray**

$$y\_n = \frac{\alpha \mathbf{x}\_n}{1 + (\alpha - 1)\mathbf{x}\_n} = \frac{5.68 \mathbf{x}\_n}{1 + 4.68 \mathbf{x}\_n} \tag{5.1}$$

In order to obtain a linear mathematical model for a nonlinear system, it is assumed that the variables deviate only slightly from some operating condition (Ogata, K., 2001). If the normal operating condition corresponds to *<sup>n</sup> x* and *<sup>n</sup> y* , then equation (5.1) can be expanded into a Taylor's series as:

$$y\_n = f(\overline{\mathbf{x}}\_n) + \frac{df}{d\mathbf{x}\_n}(\mathbf{x}\_n - \overline{\mathbf{x}}\_n) + \frac{1}{2!} \frac{d^2 f}{d\mathbf{x}\_n^2} (\mathbf{x}\_n - \overline{\mathbf{x}}\_n)^2 + \dots \tag{5.2}$$

where the derivatives 2 2 / , / *n n df dx d f dx* ,… are evaluated at *x x n n* . If the variation *x x n n* is small, the higher-order terms in *x x n n* may be neglected. Then equation (5.2) can be written as:

$$\text{If } y\_n = \overline{y}\_n + K\_n(\mathbf{x}\_n - \overline{\mathbf{x}}\_n) \text{ with } \overline{y}\_n = f(\overline{\mathbf{x}}\_n) \text{ and } K = \frac{df}{d\mathbf{x}\_n}\Big|\_{\mathbf{x}\_n = \overline{\mathbf{x}}\_n} \tag{5.3}$$

From (5.3), equation (5.1) can be written:

$$\text{If } y\_n = \overline{y}\_n + K\_n(\mathbf{x}\_n - \overline{\mathbf{x}}\_n) \text{ with } \overline{y}\_n = \frac{5.68 \overline{\mathbf{x}}\_n}{1 + 4.68 \overline{\mathbf{x}}\_n} \text{ and } K\_n = \frac{5.68}{(1 + 4.68 \overline{\mathbf{x}}\_n)^2} \tag{5.4}$$


Table 5.1. The Concentration *x*n, *y*n and the linearization coefficient *K*<sup>n</sup>

#### **5.1.2 Material balance relationship in each tray**

Linearization for general trays ( *n* 2 15 ÷ ) - ACCUMULATION = INLET – OUTLET:

$$\begin{aligned} M\_n \dot{\mathbf{x}}\_n &= (V\_{n-1} \mathbf{y}\_{n-1} + L\_{n+1} \mathbf{x}\_{n+1}) - (V\_n \mathbf{y}\_n + L\_n \mathbf{x}\_n) \\ \implies \dot{\mathbf{x}}\_n &= \frac{V\_{n-1}}{M\_n} \mathbf{y}\_{n-1} - \frac{V\_n}{M\_n} \mathbf{y}\_n + \frac{L\_{n+1}}{M\_n} \mathbf{x}\_{n+1} - \frac{L\_n}{M\_n} \mathbf{x}\_n \end{aligned} \tag{5.5}$$

where,

24 Distillation – Advances from Modeling to Applications

Max Value 105 96.92 5.53 Min Value 95 95.26 2.06

Table 4.4. Product quality depending on the input sine wave fluctuation

**5. Linearized control model** 

into a Taylor's series as:

can be written as:

From (5.3), equation (5.1) can be written:

( ) *n n nn n y y Kx x* with

Table 5.1. The Concentration *x*n, *y*n and the linearization coefficient *K*<sup>n</sup>

**5.1 Linear approximation of nonlinear system** 

**5.1.1 Vapor-Liquid equilibrium relationship in each tray** 

*y*

The product quality of this feed rate is not satisfied with *xB* 96% and *xD* 4% .

1 ( 1) 1 4.68 *n n <sup>n</sup>*

In order to obtain a linear mathematical model for a nonlinear system, it is assumed that the variables deviate only slightly from some operating condition (Ogata, K., 2001). If the normal operating condition corresponds to *<sup>n</sup> x* and *<sup>n</sup> y* , then equation (5.1) can be expanded

where the derivatives 2 2 / , / *n n df dx d f dx* ,… are evaluated at *x x n n* . If the variation *x x n n* is small, the higher-order terms in *x x n n* may be neglected. Then equation (5.2)

( ) *n n nn n y y Kx x* with ( ) *n n y f x* and *x x n n*

Tray 1 2 3 4 5 6 7 8 *xn* **0.0375** 0.0900 0.1559 0.2120 0.2461 0.2628 0.2701 0.2731 *yn* 0.1812 0.3597 0.5120 0.6044 0.6496 0.6694 0.6776 0.6809 *Kn* 4.1106 2.8121 1.8987 1.4312 1.2268 1.1423 1.1081 1.0945 Tray 9 10 11 12 13 14 15 16 *xn* 0.2811 0.3177 0.3963 0.5336 0.7041 0.8449 0.9269 **0.9654**  *yn* 0.6895 0.7256 0.7885 0.8666 0.9311 0.9687 0.9863 0.9937 *Kn* 1.0594 0.9184 0.6970 0.4644 0.3079 0.2314 0.1993 0.1865

*y*

5.68 1 4.68 *<sup>n</sup> <sup>n</sup>*

*x*

*n*

*x*

<sup>1</sup> ( ) ( ) ( ) ... 2! *n n nn n n n n*

*x x*

Feed Flow Rate (%) Distillate Purity (%) Bottoms Impurity (%)

5.68

(5.1)

*dx* (5.3)

(5.4)

*n n*

2

2

*df d f y fx x x x x dx dx* (5.2)

2

*n*

and <sup>2</sup>

*K*

5.68 (1 4.68 ) *<sup>n</sup>*

*n*

*x*

*df <sup>K</sup>*

*x x*

$$L\_1 = L\_2 = \dots = L\_{F-1} = L + L\_F; L\_{F+2} = L\_{F+3} = \dots = L \; \because V\_1 = V\_2 = \dots = V\_{F-1} = V \; \because V\_{F+1} = \overline{V} \; \therefore \; V\_{F-1} = \dots = V \; \because V\_{F+2} = \dots = V \; \because V\_{F-1} = \overline{V} \; \because V\_{F-2} = \dots = V \; \implies V\_{F+1} = \dots = V \; \implies V\_F = \overline{V} \; \therefore \; V\_F = \dots = V \; \implies V\_F = \overline{V} \; \implies V\_F = \overline{V} \; \dots = \overline{V}$$

$$V\_{F+2} = V\_{F+3} = \dots = V\_{N+1} = V + V\_F \; \therefore \; M\_n = M = 5.8 \; \therefore \; V\_{S \; \because S \; \text{int}} = \overline{V} = 66.3407 \; \because V\_{S \; \text{int}} = \overline{V} \; \implies V\_{S \; \text{int}} = \overline{V} \; \implies V\_{S \; \text{int}} = \overline{V} \; \implies V\_{S \; \text{int}} = \overline{V} \; \implies V\_{S \; \text{int}} = \overline{V} \; \implies V\_{S \; \text{int}} = \overline{V} \; \implies V\_{S \; \text{int}} = \overline{V} \; \implies V\_{S \; \text{int}} = \overline{V} \; \implies V\_{S \; \text{int}} = \overline{V} \; \implies V\_{S \; \text{int}} = \overline{V} \; \implies V\_{S \; \text{int}} = \overline{V} \; \implies V\_{S \; \text{int}} = \overline{V} \; \$$

Substituting equation (5.4) into equation (5.5), the following expression is obtained:

$$\dot{\mathbf{x}}\_{n} = \frac{(\overline{y}\_{n-1} - K\_{n-1}\overline{x}\_{n-1})V\_{n-1}}{M\_{n}} + \frac{(K\_{n-1})V\_{n-1}\mathbf{x}\_{n-1}}{M\_{n}} - \frac{(\overline{y}\_{n} - K\_{n}\overline{x}\_{n})V\_{n}}{M\_{n}} - \frac{(K\_{n})V\_{n}\mathbf{x}\_{n}}{M\_{n}} + \frac{L\_{n+1}\mathbf{x}\_{n+1}}{M\_{n}} - \frac{L\_{n}\mathbf{x}\_{n}}{M\_{n}}$$

In order to obtain a linear approximation to this nonlinear system, this equation may be expanded into a Taylor series about the normal operating point from equation (5.3), and the linear approximation equations for general trays are obtained:

$$\begin{split} \dot{\boldsymbol{\chi}}\_{n} - \overline{\boldsymbol{\tilde{\boldsymbol{\chi}}}}\_{n} &= \frac{(\boldsymbol{K}\_{n-1} \overline{\boldsymbol{V}}\_{n})}{M} \boldsymbol{\chi}\_{n-1} - \frac{(\boldsymbol{K}\_{n} \overline{\boldsymbol{V}}\_{n} + \overline{\boldsymbol{L}}\_{n})}{M} \boldsymbol{\chi}\_{n} + \frac{(\overline{\boldsymbol{L}}\_{n})}{M} \boldsymbol{\chi}\_{n+1} + \frac{(\overline{\boldsymbol{\pi}}\_{n+1} - \overline{\boldsymbol{\pi}}\_{n})}{M} \boldsymbol{L} - \frac{(\overline{\boldsymbol{\mathcal{V}}}\_{n} - \overline{\boldsymbol{\mathcal{V}}}\_{n-1})}{M} \boldsymbol{V} \\ &- \frac{\overline{\boldsymbol{L}}\_{n} (\overline{\boldsymbol{\pi}}\_{n+1} - \overline{\boldsymbol{\pi}}\_{n})}{M} + \frac{\overline{\boldsymbol{V}}\_{n} (\boldsymbol{K}\_{n} \overline{\boldsymbol{\pi}}\_{n} - \boldsymbol{K}\_{n-1} \overline{\boldsymbol{\pi}}\_{n-1})}{M} \end{split} \tag{5.6}$$

Linearization for special trays:

Tray above the feed flow (*n=f+1*): <sup>1</sup> <sup>1</sup> ( ) () ( ) *n n n n F nF <sup>n</sup> y y V x xL y y V x M MM* and the linear approximation equations for the tray above the feed flow:

$$\begin{split} \dot{\mathbf{x}}\_{n} - \overline{\dot{\mathbf{x}}}\_{n} &= \frac{\{\mathbf{K}\_{n-1} \overline{\mathbf{V}}\}}{M} \mathbf{x}\_{n-1} - \frac{\{\mathbf{K}\_{n} \overline{\mathbf{V}} + L + \mathbf{K}\_{n} V\_{F}\}}{M} \mathbf{x}\_{n} + \frac{\overline{L}}{M} \mathbf{x}\_{n+1} + \frac{\{\overline{\mathbf{x}}\_{n+1} - \overline{\mathbf{x}}\_{n}\}}{M} L + \frac{\{\overline{\mathbf{y}}\_{n} - \overline{\mathbf{y}}\_{n-1}\}}{M} V \\ &- \frac{\overline{L}(\overline{\mathbf{x}}\_{n+1} - \overline{\mathbf{x}}\_{n})}{M} + \frac{\overline{V}(\mathbf{K}\_{n} \overline{\mathbf{x}}\_{n} - \mathbf{K}\_{n-1} \overline{\mathbf{x}}\_{n-1}) + V\_{F}(\underline{\mathbf{y}}\_{F} - \overline{\mathbf{y}}\_{n} + \mathbf{K}\_{n} \overline{\mathbf{x}}\_{n})}{M} \end{split} \tag{5.7}$$

Tray below the feed flow (*n=f*): <sup>1</sup> <sup>1</sup> ( ) ( )( ) *n n n n F nF n n y yV x xL x xL x x M MM* and the linear approximation equations for the tray above the feed flow:

$$\begin{split} \dot{\overline{\boldsymbol{x}}}\_{n} - \overline{\overline{\boldsymbol{x}}}\_{n} &= \frac{(\boldsymbol{K}\_{n-1} \overline{\boldsymbol{V}})}{M} \mathbf{x}\_{n-1} + \frac{(\boldsymbol{K}\_{n} \overline{\boldsymbol{V}} + \overline{\boldsymbol{L}} + \boldsymbol{L}\_{\text{F}})}{M} \mathbf{x}\_{n} + \frac{\overline{\overline{\boldsymbol{L}}}}{M} \mathbf{x}\_{n+1} + \frac{(\overline{\boldsymbol{x}}\_{n+1} - \overline{\boldsymbol{x}}\_{n})}{M} \boldsymbol{L} - \frac{(\overline{\boldsymbol{y}}\_{n} - \overline{\boldsymbol{y}}\_{n-1})}{M} \boldsymbol{V} \\ &- \frac{\overline{\boldsymbol{L}} (\overline{\boldsymbol{x}}\_{n+1} - \overline{\boldsymbol{x}}\_{n}) - \boldsymbol{L}\_{\text{F}} \mathbf{x}\_{\text{F}}}{M} + \frac{\overline{\boldsymbol{V}} (\boldsymbol{K}\_{n} \overline{\boldsymbol{x}}\_{n} - \boldsymbol{K}\_{n-1} \overline{\boldsymbol{x}}\_{n-1})}{M} \end{split} \tag{5.8}$$

Modeling and Control Simulation for a Condensate Distillation Column 27

12.3312 5.7823 0 0 0 0 0 0 0 0 0 0 0 0 0 0 47.0173 63.1800 31.0150 0 0 0 0 0 0 0 0 0 0 0 0 0 0 32.1649 52.7324 31.0150 0 0 0 0 0 0 0 0 0 0 0 0 0 0 21.7174 47.3852 31.0150 0 0 0 0 0 0 0 0 0 0 0 0 0 0 16.3701 45.0472 31.0150 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 13.0657 43.6895 31.0150 0 0 0 0 0 0 0 0 0 0 0 0 0 0 12.6745 43.5340 13.0410 0 0 0 0 0 0 0 0 0 0 0 0 0 0 12.5189 43.1528 13.0410 0 0 0 0 0 0 0 0 0 0 0 0 0 0 30.1118 39.1451 13.0410 0 0 0 0 0 0 0 0 0 0 0 0 0 0 26.1041 32.8522 13.0410 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 13.1998 21.7926 13.0410 0 0 0 0 0 0 0 0 0 0 0 0 0 0 8.7516 19.6182 13.0410 0 0 0 0 0 0 0 0 0 0 0 0 0 0 6.5772 18.7058 13.0410 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2.5138 12.8843

> 

0.0029 0.0058 0.0114 0.0308 0.0097 0.0263 0.0059 0.0159 0.0029 0.0078 0.0013 0.0034 0.0005 0.0014 0.0014 0.0006 0.0063 0.0015 0.0136 0.0062 0.0237 0.0108 0.0294 0.0135 0.0243 0.0111 0.0141 0.0065 0.0066 0.0030

 

0739 0.0755

 <sup>1000000000000000</sup> 0000000000000001

*B*

0.

00000

.2409 13.0410 0 0 0

*A*

The matrix B elements:

For n=1, <sup>2</sup> <sup>1</sup> 1,1 1,2

For n=16, <sup>16</sup> <sup>15</sup> 16,1 16,2

And the output matrix C is:

*C*

( ) ( ) , b *B B <sup>x</sup> <sup>y</sup> bL V M M*

For n=2÷15, <sup>1</sup> <sup>1</sup> ,1 ,2

*<sup>x</sup> <sup>y</sup> b Lb V*

( ) ( ) , *D D*

( ) ( ) , *n n n n n n x x y y <sup>b</sup> L b <sup>V</sup> M M*

*M M* , then:

0 0 0 0 14.0322 44.0807 31.0150 0 0 0 0

0 0 0 0 0 0 0 0 0 0 19.8111 26

Reboiler (*n*=1): 2 2 <sup>1</sup> <sup>1</sup> 1 *B BB L x Bx Vy <sup>x</sup> <sup>M</sup> M M* , where, 31.11 *MB* a, *<sup>B</sup>* 110.9235 , and the linear approximation equations for the reboiler:

$$\dot{\mathbf{x}}\_{1} - \overline{\dot{\mathbf{x}}}\_{1} = -\frac{\{\mathbf{K}\_{1}\overline{V} + \mathcal{B}\}}{M\_{\text{B}}} \mathbf{x}\_{1} + \frac{\{\overline{L} + \overline{L}\_{\text{F}}\}}{M\_{\text{B}}} \mathbf{x}\_{2} + \frac{\{\overline{\mathbf{x}}\_{2}\}}{M\_{\text{B}}} L - \frac{\{\overline{y}\_{1}\}}{M\_{\text{B}}} V - \frac{\{\left(\overline{L} + L\_{\text{F}}\right)\overline{\mathbf{x}}\_{2}\}}{M\_{\text{B}}} + \frac{\{\overline{V}\mathbf{K}\_{1}\overline{\mathbf{x}}\_{1}\}}{M\_{\text{B}}} \tag{5.9}$$

Condenser (n=N): 15 15 16 16 16 *D DD V y Lx Dx x M MM* , where, *MD* 13.07 , *<sup>D</sup>* 92.7597 , and the

linear approximation equations for the condenser:

$$\dot{\mathbf{x}}\_{16} - \overline{\dot{\mathbf{x}}}\_{16} = \frac{\{\mathbf{K}\_{15}(\overline{V} + V\_{\mathcal{F}})\}}{M\_{\mathcal{D}}} \mathbf{x}\_{15} - \frac{(\overline{L} + D)}{M\_{\mathcal{D}}} \mathbf{x}\_{16} - \frac{(\overline{\mathbf{x}}\_{16})}{M\_{\mathcal{D}}} L + \frac{(\overline{y}\_{15})}{M\_{\mathcal{D}}} V + \frac{(\overline{L}\overline{\mathbf{x}}\_{16})}{M\_{\mathcal{D}}} - \frac{((\overline{V} + V\_{\mathcal{F}})\mathbf{K}\_{15}\overline{\mathbf{x}}\_{15})}{M\_{\mathcal{D}}} \tag{5.10}$$

As a result, the model is represented in state space in terms of deviation variables:

$$\begin{array}{l} \dot{z}(t) = Az(t) + Bu(t) \\ y(t) = \mathbb{C}\overline{z}(t) \end{array}, \text{where}$$

$$z(t) = \begin{vmatrix} x\_1(t) - \overline{x}\_{1\text{ Standy State}} \\ x\_2(t) - \overline{x}\_{2\text{ Standy State}} \\ \vdots \\ x\_{16}(t) - \overline{x}\_{16\text{ Standy State}} \end{vmatrix},$$

$$u(t) = \begin{vmatrix} L(t) - \overline{L}\_{\text{Standy State}} = dL \\ V(t) - \overline{V}\_{\text{Standy State}} = dV \end{vmatrix}, \quad y(t) = \begin{vmatrix} x\_1(t) - \overline{x}\_{1\text{ Standy State}} = dx\_B \\ x\_{16}(t) - \overline{x}\_{16\text{ Standy State}} = dx\_D \end{vmatrix}$$

The matrix *A* elements:

$$\begin{aligned} \text{For n=1, } & a\_{1,1} = -\frac{(K\_{\overline{L}} \overline{V} + B)}{M\_{B}}, \quad a\_{1,2} = \frac{(\overline{L} + \overline{L}\_{F})}{M\_{B}} \\ \text{For n=2+7, } & a\_{n,n-1} = \frac{(K\_{n-1} \overline{V})}{M}, \quad a\_{n,n} = -\frac{(K\_{n} \overline{V} + \overline{L} + L\_{F})}{M}, \quad a\_{n,n+1} = \frac{(\overline{L} + L\_{F})}{M} \\ \text{For n=8, } & a\_{8,7} = \frac{(K\_{\overline{V}} \overline{V})}{M}, \quad a\_{8,8} = -\frac{(K\_{8} \overline{V} + \overline{L} + L\_{F})}{M}, \quad a\_{8,9} = \frac{(\overline{L})}{M} \\ \text{For n=9, } & a\_{9,8} = \frac{(K\_{8} \overline{V})}{M}, \quad a\_{9,9} = -\frac{(K\_{9} \overline{V} + \overline{L})}{M}, \quad a\_{9,10} = \frac{(\overline{L})}{M} \\ \text{For n=10+15, } & a\_{n,n-1} = \frac{(K\_{n-1} (\overline{V} + V\_{F}))}{M}, \quad a\_{n,n} = -\frac{(K\_{n} (\overline{V} + V\_{F}) + \overline{L})}{M}, \quad a\_{n,n+1} = \frac{(\overline{L})}{M} \\ \text{For n=16, } & a\_{16,15} = \frac{(K\_{15} (\overline{V} + V\_{F}))}{M\_{10}}, \quad a\_{16,16} = -\frac{(\overline{L} + D)}{M\_{10}}, \text{ then:} \end{aligned}$$


The matrix B elements:

26 Distillation – Advances from Modeling to Applications

<sup>1</sup> <sup>2</sup> <sup>1</sup> 2 11

*KV B L L x <sup>y</sup> L L x VK x xx x x L V*

<sup>15</sup> <sup>16</sup> <sup>15</sup> <sup>16</sup> 15 15

As a result, the model is represented in state space in terms of deviation variables:

<sup>1</sup>

*MM M*

 <sup>1</sup> , 1 . , 1 ( ( )) ( ( ) ) ( ) , , *n F nF n n n n n n K V V KV V L L*

*M M M*

() ( ) ( ) , , *n nF <sup>F</sup> n n n n n n K V KV L L L L*

, 1 . , 1

( ) ( ) ( ) , , *K V KV L LF <sup>L</sup>*

() ( ) ( ) , , *KV KV L <sup>L</sup>*

*M M M*

For n=10÷15,

( ( )) ( ) , *<sup>F</sup> D D*

*KVV L D*

*aa a*

*M M* , then:

*M M M*

*aa a*

() () () () () *z t Az t Bu t*

*y t Cz t* , where

1 1 2 2

( )

*xt x xt x*

16 16

( )

*xt x*

*L t L dL xt x dx u t y t V t V dV xt x dx*

*Steady State Steady State*

*Steady State Steady State Steady State B Steady State Steady State D*

( ) ( ) ,

( ) ( ) ( ) , ( ) ( ) ( )

*z t*

( ) () , *<sup>F</sup> B B KV B L L*

*M M*

For n=2÷7,

*aa a*

For n=8, <sup>7</sup> <sup>8</sup> 8,7 8.8 8,9

*aa a*

For n=9, 8 9 9,8 9.9 9,10

For n=16, <sup>15</sup> 16,15 16,16

*a a*

( ( )) ( ) ( ) ( ) ( ) (( ) ) *<sup>F</sup> <sup>F</sup> D D DD D D K VV L D x y Lx V V K x*

( ) ( ) () ( ) (( ) ) ( ) *F F B B BB B B*

*<sup>M</sup> M M* , where, 31.11 *MB* a, *<sup>B</sup>* 110.9235 , and the linear

*M M MM M M* (5.9)

*M M MM M M* (5.10)

1 1 16 16 (5.11)

*M MM* , where, *MD* 13.07 , *<sup>D</sup>* 92.7597 , and the

Reboiler (*n*=1): 2 2 <sup>1</sup> <sup>1</sup> 1

approximation equations for the reboiler:

Condenser (n=N): 15 15 16 16 16

16 16 15 16

The matrix *A* elements:

For n=1, <sup>1</sup> 1,1 1,2

*a a*

linear approximation equations for the condenser:

*x xx x LV*

*x*

*B BB*

*D DD*

*V y Lx Dx*

*L x Bx Vy <sup>x</sup>*

1 1 1 2

$$\begin{aligned} \text{For n=1, } & b\_{1,1} = \frac{(\overline{\mathbf{x}\_2})\_L}{M\_B} L\_\prime \quad \mathbf{b}\_{1,2} = -\frac{(\overline{\mathbf{y}\_1})\_I}{M\_B} V \\\\ \text{For n=2+15, } & b\_{n,1} = \frac{(\overline{\mathbf{x}\_{n+1}} - \overline{\mathbf{x}\_n})\_L}{M} L\_\prime \quad \mathbf{b}\_{n,2} = -\frac{(\overline{\mathbf{y}\_n} - \overline{\mathbf{y}\_{n-1}})\_I}{M} V \\\\ \text{For n=16, } & b\_{16,1} = -\frac{(\overline{\mathbf{x}\_{16}})\_L}{M\_D} L\_\prime \quad \mathbf{b}\_{16,2} \frac{(\overline{\mathbf{y}\_{15}})\_S}{M\_D} V \text{, then:} \\\\ & \begin{vmatrix} 0.0029 & -0.00586 \\ 0.0114 & -0.03085 \\ 0.0097 & -0.0263 \\ 0.0059 & -0.0159 \\ 0.0029 & -0.00786 \\ 0.0013 & -0.00786 \\ 0.0005 & -0.0034 \end{vmatrix} \end{aligned}$$

$$B = \begin{bmatrix} 0.0059 & -0.0159 \\ 0.0029 & -0.0078 \\ 0.0013 & -0.0034 \\ 0.0005 & -0.0014 \\ 0.0014 & -0.0006 \\ 0.0063 & -0.0015 \\ 0.0136 & -0.0062 \\ 0.0237 & -0.0108 \\ 0.0294 & -0.0135 \\ 0.0243 & -0.0111 \\ 0.0141 & -0.0065 \\ 0.0066 & -0.0030 \\ -0.0739 & 0.0755 \end{bmatrix}$$

And the output matrix C is:


Modeling and Control Simulation for a Condensate Distillation Column 29

controller for a distillation column dealing with the disturbance and the model-plant

Adaptive control system is the ability of a controller which can adjust its parameters in such a way as to compensate for the variations in the characteristics of the process. Adaptive control is widely applied in petroleum industries because of the two main reasons: Firstly, most processes are nonlinear and the linearized models are used to design the controllers, so that the controller must change and adapt to the model-plant mismatch; Secondly, most of the processes are non-stationary or their characteristics are changed with time, this leads

The general form of a MRAC is based on an inner-loop Linear Model Reference Controller (LMRC) and an outer adaptive loop shown in Fig. 6.1. In order to eliminate errors between the model, the plant and the controller is asymptotically stable, MRAC will calculate online

Simulation program is constructed using Maltab Simulink with the following data:

*y Cz*

 

*z Az Bu noise*

( ) *<sup>L</sup> t* and *<sup>M</sup>* ( )*t* as detected state error

mismatch as the influence of the plant feed disturbances.

again to adapt the changing control parameters.

the adjustment parameters in gains *L* and *M* by

*e t*( ) when changing *A* , *B* in the process plant.

Fig. 6.1. MRAC block diagram

**Process Plant:** 

#### **5.2 Reduced-order linear model**

The full-order linear model in equation 5-11, which represents a 2 input – 2 output plant can be expressed in the *S* domain as:

$$
\begin{vmatrix} d\mathbf{x}\_D \\ d\mathbf{x}\_B \end{vmatrix} = \frac{1}{1 + \tau\_c s} G(0) \begin{vmatrix} dL \\ dV \end{vmatrix} \tag{5.12}
$$

where *<sup>c</sup>* is the time constant and *G*(0) is the steady state gain

The steady state gain can be directly calculated: <sup>1</sup> *G CA B* (0) or

$$G(0) = \begin{vmatrix} 0.0042 & -0.0060 \\ -0.0050 & 0.0072 \end{vmatrix} \tag{5.13}$$

The time constant *<sup>c</sup>* can be calculated based on some specified assumptions (Skogestad, S., & Morari, M., 1987). The linearized value of *<sup>c</sup>* is given by:

$$\tau\_c = \frac{M\_I}{I\_s \ln S} + \frac{M\_D (1 - \mathbf{x}\_D) \mathbf{x}\_D}{I\_s} + \frac{M\_B (1 - \mathbf{x}\_B) \mathbf{x}\_B}{I\_s} \tag{5.14}$$

where *MI* is the total holdup of liquid inside the column:

$$M\_I = \sum\_{i=1}^{N} M\_i = 5.8 \,\,\text{\*}\,\,14 = 81.2 \,\,\text{(kmole)};$$

*sI* is the "impurity sum": (1 ) (1 ) 7.1021 *s DD BB I D xx B xx* , and *S* is the separation factor: (1 ) 716.1445 (1 ) *D B D B x x <sup>S</sup> x x* .

So that, the time constant *<sup>c</sup>* in equation (5.14) can be determined: 1.9588 ( ) *<sup>c</sup> h* .

As the result, the reduced-order model of the plant is a first order system in equation (5.12):

$$
\begin{vmatrix} d\mathbf{x}\_D \\ d\mathbf{x}\_B \end{vmatrix} = \frac{1}{1 + 1.9588s} \begin{vmatrix} 0.0042 & -0.0060 \\ -0.0050 & 0.0072 \end{vmatrix} d\mathbf{L} \tag{5.15}
$$

or the equivalent reduced-order model in state space:

$$\begin{aligned} \dot{z}\_r(t) &= \begin{vmatrix} -0.5105 & 0 \\ 0 & -0.5105 \end{vmatrix} z\_r(t) + \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} u(t) \\ y\_r(t) &= \begin{vmatrix} 0.0021 & -0.0031 \\ -0.0026 & 0.0037 \end{vmatrix} z\_r(t) \end{aligned} \tag{5.16}$$

#### **6. Control simulation with MRAC**

The reduced-order linear model is then used as the reference model for a model-reference adaptive control (MRAC) system to verify the applicable ability of a conventional adaptive

The full-order linear model in equation 5-11, which represents a 2 input – 2 output plant can

 <sup>1</sup> (0) <sup>1</sup>

(1 ) (1 )

*N I i i*

1

*I D DD B BB <sup>c</sup> ss s*

*sI* is the "impurity sum": (1 ) (1 ) 7.1021 *s DD BB I D xx B xx* , and *S* is the separation

*<sup>c</sup>* in equation (5.14) can be determined:

As the result, the reduced-order model of the plant is a first order system in equation (5.12):

*dx dL*

1 0.0042 0.0060 1 1.9588 0.0050 0.0072

0.0021 0.0031 ( ) ( ) 0.0026 0.0037

*r r*

*r r*

*y t z t*

 0.5105 0 1 0 ( ) () () 0 0.5105 0 1

*z t z t ut*

The reduced-order linear model is then used as the reference model for a model-reference adaptive control (MRAC) system to verify the applicable ability of a conventional adaptive

*M M xx M xx*

0.0042 0.0060 (0) 0.0050 0.0072

*<sup>c</sup>* can be calculated based on some specified assumptions (Skogestad, S.,

*<sup>c</sup>* is given by:

5.8 \* 14 81.2

*M M* (kmole);

*dx <sup>s</sup> dV* (5.12)

*G* (5.13)

*IS I <sup>I</sup>* (5.14)

*dx <sup>s</sup> dV* (5.15)

1.9588 ( ) *<sup>c</sup> h* .

(5.16)

*dx dL G*

*B c*

*D*

 *<sup>c</sup>* is the time constant and *G*(0) is the steady state gain The steady state gain can be directly calculated: <sup>1</sup> *G CA B* (0) or

**5.2 Reduced-order linear model** 

be expressed in the *S* domain as:

& Morari, M., 1987). The linearized value of

(1 ) 716.1445

*x x* .

or the equivalent reduced-order model in state space:

**6. Control simulation with MRAC** 

*D B*

where *MI* is the total holdup of liquid inside the column:

ln

where 

The time constant

factor:

*x x <sup>S</sup>*

So that, the time constant

(1 ) *D B D B* controller for a distillation column dealing with the disturbance and the model-plant mismatch as the influence of the plant feed disturbances.

Adaptive control system is the ability of a controller which can adjust its parameters in such a way as to compensate for the variations in the characteristics of the process. Adaptive control is widely applied in petroleum industries because of the two main reasons: Firstly, most processes are nonlinear and the linearized models are used to design the controllers, so that the controller must change and adapt to the model-plant mismatch; Secondly, most of the processes are non-stationary or their characteristics are changed with time, this leads again to adapt the changing control parameters.

The general form of a MRAC is based on an inner-loop Linear Model Reference Controller (LMRC) and an outer adaptive loop shown in Fig. 6.1. In order to eliminate errors between the model, the plant and the controller is asymptotically stable, MRAC will calculate online the adjustment parameters in gains *L* and *M* by ( ) *<sup>L</sup> t* and *<sup>M</sup>* ( )*t* as detected state error *e t*( ) when changing *A* , *B* in the process plant.

Fig. 6.1. MRAC block diagram

Simulation program is constructed using Maltab Simulink with the following data:

**Process Plant:** 

$$\begin{aligned} \dot{z} &= Az + Bu + moise \\ y &= Cz \end{aligned}$$

Modeling and Control Simulation for a Condensate Distillation Column 31

*<sup>T</sup> Q A P PA m m* . Since matrix *Q* is obviously positive definite, then we

and the plant-model mismatches as the influence of the feed stock disturbances.

*d z e d dt z e P dt c u e d dt u e*

*e Qe dt* and the system is stable with any plant-model mismatches.

 

1 1 1 111

*z d dt z e*

1 1 2 3 1 11

0 / 0 / 2 0 / 0 / 2

It is assumed that the reduced-order linear model in equation (11) can also maintain the similar steady state outputs as the basic nonlinear model. Now this model is used as an MRAC to take the process plant from these steady state outputs ( *xD* 0.9654 and 0.0375 *Bx* ) to the desired targets ( 0.98 1 *xD* and 0 0.02 *Bx* ) amid the disturbances

The design of a new adaptive controller is shown in Figure 6.2 where we install an MRAC and a closed-loop PID (Proportional, Integral, Derivative) controller to eliminate the errors

22 1 2 222

2 2 4 2 22

*u ddt u e*

*c c*

 

*c*

 0 2 *dV <sup>T</sup>*

**Simulation results and analysis:** 

 

between the reference set-points and the outputs.

Fig. 6.2. Adaptive controller with MRAC and PID

always have

**Parameters Adjustment:** 

0.5232 0 0 1.0465

where 1 2 0 0 *A* , 1 2 0 0 *<sup>B</sup>* , 0.004 0.007 0.0011 0.0017 *C* and <sup>1</sup> , <sup>2</sup> , 1 , 2 are changing and dependent on the process dynamics.

#### **Reference Model:**

$$\begin{aligned} \dot{z}\_m &= A\_m z\_m + B\_m \mu\_c \\ y\_m &= C\_m z\_m \end{aligned}$$

$$\text{where } A\_m = \begin{bmatrix} -0.2616 & 0\\ 0 & -0.2616 \end{bmatrix}, B\_m = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}, \text{ C}\_m = \begin{bmatrix} 0.004 & -0.007\\ -0.0011 & 0.0017 \end{bmatrix}$$

#### **State Feedback:**

$$\boldsymbol{\mu} = \boldsymbol{M}\boldsymbol{\mu}\_c - \boldsymbol{L}\boldsymbol{z} \quad \text{where} \quad \boldsymbol{L} = \begin{bmatrix} \boldsymbol{\theta}\_1 & \boldsymbol{0} \\ \boldsymbol{0} & \boldsymbol{\theta}\_2 \end{bmatrix} \text{ and } \boldsymbol{M} = \begin{bmatrix} \boldsymbol{\theta}\_3 & \boldsymbol{0} \\ \boldsymbol{0} & \boldsymbol{\theta}\_4 \end{bmatrix}.$$

#### **Closed Loop:**

$$\dot{z} = (A - BL)z + BM\mu\_c = A\_c(\theta)z + B\_c(\theta)\mu\_c$$

#### **Error Equation:**

$$\begin{aligned} \dot{e} &= z - z\_m = \begin{bmatrix} e\_1 \\ e\_2 \end{bmatrix} \text{ is a vector of state errors,} \\\\ \dot{e} &= \dot{z} - \dot{z}\_m = Az + Bu - A\_m z\_m - B\_m u\_c = A\_m e + (A\_c(\theta) - A\_m)z + (B\_c(\theta) - B\_m)u\_c = A\_m e + \Psi(\theta - \theta^0) \\\\ \text{where} \quad \Psi &= \begin{bmatrix} -\beta\_1 z\_1 & 0 & \beta\_1 u\_{c1} & 0 \\ 0 & -\beta\_2 z\_2 & 0 & \beta\_2 u\_{c2} \end{bmatrix} \end{aligned}$$

#### **Lyapunov Function:**

 <sup>1</sup> 0 0 (, ) ( )( ) <sup>2</sup> *T T V e e Pe* where is an adaptive gain and *P* is a Lyapunov matrix.

#### **Derivative Calculation of Lyapunov Matrix:**

$$\frac{dV}{dt} = -\frac{\mathcal{V}}{2} \varepsilon^T Q \varepsilon + (\theta - \theta^0)^T \left(\frac{d\theta}{dt} + \gamma \Psi^T P e\right) \text{ where } Q = -A\_m^T P - P A\_m \dots$$

For the stability of the system, <sup>0</sup> *dV dt* , we can assign the second item <sup>0</sup> () 0 *T T <sup>d</sup> Pe dt* or *<sup>d</sup> <sup>T</sup> Pe dt* . Then we always have: <sup>2</sup> *dV <sup>T</sup> e Qe dt* . If we select a positive matrix *<sup>P</sup>* <sup>0</sup> , for instance, 1 0 0 2 *P* , then we have

 0.5232 0 0 1.0465 *<sup>T</sup> Q A P PA m m* . Since matrix *Q* is obviously positive definite, then we always have 0 2 *dV <sup>T</sup> e Qe dt* and the system is stable with any plant-model mismatches.

#### **Parameters Adjustment:**

30 Distillation – Advances from Modeling to Applications

0.004 0.007 0.0011 0.0017 *C* and

*Bm* ,

*Cm*

4 0

0.004 0.007 0.0011 0.0017

> 

is an adaptive gain and *P* is a Lyapunov

*dt* , we can assign the second item

 <sup>2</sup> *dV <sup>T</sup> e Qe dt*

*P* , then we have

. If we

 

 <sup>1</sup> , <sup>2</sup> , 1 , 2 are

*<sup>B</sup>* ,

0 *M* .

*z A BL z BMu A z B u* ( ) *cc cc* () ()

 

<sup>0</sup> *e z z Az Bu A z B u A e A A z B B u A e <sup>m</sup> mm mc m c m c m c m* ( () ) ( () ) ( )

where *<sup>T</sup> Q A P PA m m* .

*dt* . Then we always have:

 3

*m mm mc*

*z Az Bu y Cz*

*m mm*

where

0 *A* ,

**Reference Model:** 

**State Feedback:** 

**Closed Loop:** 

**Error Equation:** 

*m*

*ezz*

where

matrix.

 1 2

**Lyapunov Function:** 

 

*e*

*e*

*u Mu Lz c* where

 1

where

0.2616 0 0 0.2616 *Am* ,

2 0

0

changing and dependent on the process dynamics.

0 *L* and

 1

is a vector of state errors,

 

> 

 

or

For the stability of the system, <sup>0</sup> *dV*

 

 

select a positive matrix *<sup>P</sup>* <sup>0</sup> , for instance,

*<sup>d</sup> <sup>T</sup> Pe*

 

*c*

2 2 2 2 0 0

*z u*

*c*

 

 <sup>1</sup> 0 0 (, ) ( )( ) <sup>2</sup> *T T V e e Pe* where

<sup>0</sup> ( ) <sup>2</sup> *dV T TT <sup>d</sup> e Qe Pe*

*dt dt*

 <sup>0</sup> () 0 *T T <sup>d</sup> Pe dt*

0 0

**Derivative Calculation of Lyapunov Matrix:** 

 1 1 1 1

*z u*

2 0

 1

2 0

$$\frac{d\theta}{dt} = -\gamma \begin{bmatrix} -\beta\_1 z\_1 & 0\\ 0 & -\beta\_2 z\_2\\ \beta c\_1 u\_1 & 0\\ 0 & \beta\_2 u\_{2c} \end{bmatrix} \begin{bmatrix} e\_1\\ P \end{bmatrix} \begin{bmatrix} e\_1\\ e\_2 \end{bmatrix} = \begin{bmatrix} d\theta\_1 / \, dt\\ d\theta\_2 / \, dt\\ d\theta\_3 / \, dt\\ d\theta\_4 / \, dt \end{bmatrix} = \begin{bmatrix} \gamma \beta\_1 z\_1 e\_1\\ 2\gamma \beta\_2 z\_2 e\_2\\ -\gamma \beta\_1 u\_{c1} e\_1\\ -2\gamma \beta\_2 u\_{c2} e\_2 \end{bmatrix}$$

#### **Simulation results and analysis:**

It is assumed that the reduced-order linear model in equation (11) can also maintain the similar steady state outputs as the basic nonlinear model. Now this model is used as an MRAC to take the process plant from these steady state outputs ( *xD* 0.9654 and 0.0375 *Bx* ) to the desired targets ( 0.98 1 *xD* and 0 0.02 *Bx* ) amid the disturbances and the plant-model mismatches as the influence of the feed stock disturbances.

The design of a new adaptive controller is shown in Figure 6.2 where we install an MRAC and a closed-loop PID (Proportional, Integral, Derivative) controller to eliminate the errors between the reference set-points and the outputs.

Fig. 6.2. Adaptive controller with MRAC and PID

Modeling and Control Simulation for a Condensate Distillation Column 33

The reduced order linear model is used as the reference model for an MRAC controller. The controller of MRAC and PID theoretically allows the plant outputs tracking the reference set-points to achieve the desired product quality amid the disturbances and the model-plant

In this chapter, the calculation of the mathematical model building and the reduced-order linear adaptive controller is only based on the physical laws from the process. The real system identifications including the experimental production factors, specific designed

The authors would like to thank InTech for providing the opportunity to print this book chapter. Special thanks are also due to the anonymous reviewers and editor, who assisted the author in improving this book chapter significantly. Lastly, the authors would like to thank the Papua New Guinea University of Technology (UNITECH) for their support in the

Franks, R. (1972), Modeling and Simulation in Chemical Engineering, Wiley-Interscience,

George, S. (1986). Chemical Process Control: An Introduction to Theory and Practice,

Kehlen, H. & Ratzsch, M. (1987). Complex Multicomponent Distillation Calculations by

Luyben, W. (1990). Process Modeling, Simulation, and Control for Chemical Engineers,

Marie, E.; Strand, S. & Skogestad S. (2008). Coordinator MPC for Maximizing Plant

Nelson, W. (1985), Petroleum Refinery Engineering, McGraw-Hill, ISBN: 978-0-8247-0599-2,

Ogata, K. (2001). Modern Control Engineering, Prentice Hall, ISBN: 978-0130609076, New

Papadouratis, A.; Doherty, M. & Douglas, J. (1989). Approximate Dynamic Models for

Perry, R. & Green, D. (1984). Perry's Chemical Engineers Handbook, McGraw-Hill, ISBN: 0-

PetroVietnam Gas Company. (1999), Condensate Processing Plant Project – Process

Skogestad, S., & Morari, M. (1987). The Dominant Time Constant for Distillation Columns. Computers & Chemical Engineering, Vol. 11(6), pp. 607-617, ISSN: 0098-1354. Waller, V. (1992).Practical Distillation Control, Van Nostrand Reinhold, *ISBN:* 

Continuous Thermodynamics. Chem. Eng. Sci., Vol. 42(2), pp. 221-232 ISSN: 0009-

Throughput. Computer & Chemical Engineering, Vol. 32(2), pp. 195-204, ISSN:

Chemical Process Systems. Ind. & Eng. Ch. Re., Vol. 28(5), pp. 546-522, ISSN: 0888-

structures, parameters estimation and the system validation are not discussed here.

mismatches as the influence of the feed stock disturbances.

**8. Acknowledgment** 

**9. References** 

2509.

0098-1354.

Singapore.

York.

5885.

471-58626-9, New York.

9780442006013, New York.

preparation of this book chapter.

New York, ISBN: 978-0471275350.

Prentice-Hall, *ISBN*: 9780131286290, New Jersey.

McGraw-Hill, ISBN: 978-0070391598, New York.

Description Document No. 82036-02BM-01, Hanoi.

This controller system was run with different plant-model mismatches, for instance, a plant with 0.50 0 0 0.75 *<sup>A</sup>* , 1.5 0 0 2.5 *B* and an adaptive gain 25 . The operating setpoints for the real outputs are *xDR* 0.99 and 0.01 *BR x* . Then, the reference set-points for the PID controller are *rD* 0.0261 and 0.0275 *Br* since the real steady state outputs are *xD* 0.9654 and 0.0375 *Bx* . Simulation in Figure 6.2 shows that the controlled outputs *xD* and *Bx* are always stable and tracking to the model outputs and the reference set-points (the dotted lines, *rD* and *Br* ) amid the disturbances and the plant-model mismatches (Figure 6.3).

Fig. 6.3. Correlation of Plant Outputs, Model Outputs and Reference Setpoints

#### **7. Conclusion**

A procedure has been introduced to build up a mathematical model and simulation for a condensate distillation column based on the energy balance (*L-V*) structure. The mathematical modeling simulation is accomplished over three phases: the basic nonlinear model, the full order linearized model and the reduced order linear model. Results from the simulations and analysis are helpful for initial steps of a petroleum project feasibility study and design.

The reduced order linear model is used as the reference model for an MRAC controller. The controller of MRAC and PID theoretically allows the plant outputs tracking the reference set-points to achieve the desired product quality amid the disturbances and the model-plant mismatches as the influence of the feed stock disturbances.

In this chapter, the calculation of the mathematical model building and the reduced-order linear adaptive controller is only based on the physical laws from the process. The real system identifications including the experimental production factors, specific designed structures, parameters estimation and the system validation are not discussed here.

#### **8. Acknowledgment**

32 Distillation – Advances from Modeling to Applications

This controller system was run with different plant-model mismatches, for instance, a plant

setpoints for the real outputs are *xDR* 0.99 and 0.01 *BR x* . Then, the reference set-points for the PID controller are *rD* 0.0261 and 0.0275 *Br* since the real steady state outputs are *xD* 0.9654 and 0.0375 *Bx* . Simulation in Figure 6.2 shows that the controlled outputs *xD* and *Bx* are always stable and tracking to the model outputs and the reference set-points (the dotted lines, *rD* and *Br* ) amid the disturbances and the plant-model mismatches (Figure

*B* and an adaptive gain

*x*

*D*

*x*

*B*

25 . The operating

1.5 0 0 2.5

Fig. 6.3. Correlation of Plant Outputs, Model Outputs and Reference Setpoints

A procedure has been introduced to build up a mathematical model and simulation for a condensate distillation column based on the energy balance (*L-V*) structure. The mathematical modeling simulation is accomplished over three phases: the basic nonlinear model, the full order linearized model and the reduced order linear model. Results from the simulations and

20 40 60 80 100 120 140

Time

analysis are helpful for initial steps of a petroleum project feasibility study and design.

with

6.3).

**7. Conclusion** 



*r*

*B*


0

Purity of the Distillate Product x

0.01

0.02

0.03

*r*

*D*

0.50 0 0 0.75

*<sup>A</sup>* ,

The authors would like to thank InTech for providing the opportunity to print this book chapter. Special thanks are also due to the anonymous reviewers and editor, who assisted the author in improving this book chapter significantly. Lastly, the authors would like to thank the Papua New Guinea University of Technology (UNITECH) for their support in the preparation of this book chapter.

#### **9. References**


**2** 

**Energy Conservation in** 

*University of South Africa, Florida Campus* 

Christopher Enweremadu

*South Africa* 

**Ethanol-Water Distillation Column** 

**with Vapour Recompression Heat Pump** 

Ethanol or ethyl alcohol CH3CH2OH, a colorless liquid with characteristic odor and taste; commonly called grain alcohol has been described as one of the most exotic synthetic oxygen-containing organic chemicals because of its unique combination of properties as a solvent, a germicide, a beverage, an antifreeze, a fuel, a depressant, and especially because of its versatility as a chemical intermediate for other organic chemicals. Ethanol could be derived from any material containing simple or complex sugars. The sugar-containing material is fermented after which the liquid mixture of ethanol and water is separated into

Distillation is the most widely used separation operation in chemical and petrochemical industries accounting for around 25-40% of the energy usage. One disadvantage of distillation process is the large energy requirement. Distillation consumes a great deal of energy for providing heat to change liquid to vapour and condense the vapour back to liquid at the condenser. Distillation is carried out in distillation columns which are used for about 95% of liquid separations and the energy use from this process accounts for an estimated 3% of the world energy consumption (Hewitt et al, 1999). It has been estimated that the energy use in distillation is in excess. With rising energy awareness and growing environmental concerns there is a need to reduce the energy use in industry. The potential for energy savings therefore exists and design and operation of energy efficient distillation systems will have a substantial effect on the overall plant energy consumption and

The economic competitiveness of ethanol has been heightened by concerns over prices and availability of crude oil as well as greenhouse gas emissions which have stimulated interest in alternatives to crude oil to provide for automotive power and also by the use of bioethanol in the production of hydrogen for fuel cells. Therefore, there is the need to explore ways of producing ethanol at competitive costs by the use of energy efficient processes. To cope with the high energy demand and improve the benefits from the process, the concept of polygeneration and hydrothermal treatment especially when dealing with small scale ethanol plants is fast gaining interest. However, the analysis of the bioethanol

process shows that distillation is still the most widely used.

**1. Introduction** 

operating costs.

their components using distillation.

