Doehlert Design for Wet Pelletization 
in High Shear Mixer
Author's Data: Housam Binous
Department of Chemical Engineering
National Institute of Applied Sciences and Technology
Tunis, TUNISIA
Email: binoushousam@yahoo.com

Acknowledgement :
This problem was presented in the excellent book by Lewis G. A., D. Mathieu and R. Phan-Than-Luu, Pharmaceutical Experimental Design, Drugs and the Pharmaceutical Sciences, Vol 92, 1999, Marcel Dekker, New York.

Off[General::"spell1"]

Three Factors:
1/ Impeller Speed (rpm)= 
Associated variable: X1, Lower level: 250, Upper level: 400
2/ Amount of binding solution (%)= 
Associated variable: X2, Lower level: 60 Upper level: 70
3/ Pelletization time (min)= 
Associated variable: X3, Lower level: 5 Upper level: 20

One Response: 
Mean size of pellets (mm)

Plan and Experimental Results

TableForm[{{N°,  X1,  X2,  X3,  Speed,  Solution,  
          time,  Size}, {1, 0, 0, 0, 325, 65.0, 12.5, 1594}, {2, 1, 0, 0, 
          400, 65.0, 12.5, 1398}, {3, 0.5, 0.866, 0, 362, 69.3, 12.5, 
          1352}, {4, -0.5, 0.866, 0, 288, 69.3, 12.5, 1466}, {5, -1, 
          0, 0, 250, 65.0, 12.5, 1750}, {6, -0.5, -0.866, 0, 288, 
          60.7, 12.5, 1064}, {7, 0.5, -0.866, 0, 362, 60.7, 12.5, 
          1337}, {8, 0.5, 0.289, 0.816, 362, 66.4, 18.6, 
          1533}, 
{9, -0.5, 0.289, 0.816, 288, 66.4, 
          18.6, 1617}, {10, 0, -0.577, 0.816, 325, 62.1, 18.6, 
          1332}, {11, 0.5, -0.289, -0.816, 326, 63.6, 6.4, 
          1409}, {12, -0.5, -0.289, -0.816, 288, 63.6, 6.4, 
          1560}, {13, 0, 0.577, -0.816, 325, 68.9, 6.4, 
          1431}}]

Responses' vector is labelled y

y = Transpose[{{1, 0, 0, 0, 325, 65.0, 12.5, 1594}, {2, 1, 0, 0, 400, 
              65.0, 12.5, 1398}, {3, 0.5, 0.866, 0, 362, 69.3, 12.5, 
              1352}, {4, -0.5, 0.866, 0, 288, 69.3, 12.5, 
              1466}, {5, -1, 0, 0, 250, 65.0, 12.5, 
              1750}, {6, -0.5, -0.866, 0, 288, 60.7, 12.5, 1064}, {7, 
              0.5, -0.866, 0, 362, 60.7, 12.5, 1337}, {8, 0.5, 0.289, 
              0.816, 362, 66.4, 18.6, 1533}, 
{9, -0.5, 
              0.289, 0.816, 288, 66.4, 18.6, 1617}, {10, 0, -0.577, 
              0.816, 325, 62.1, 18.6, 1332}, {11, 
              0.5, -0.289, -0.816, 326, 63.6, 6.4, 
              1409}, {12, -0.5, -0.289, -0.816, 288, 63.6, 6.4, 
              1560}, {13, 0, 0.577, -0.816, 325, 68.9, 6.4, 
              1431}}][[8]]

Forming M the Computation Matrix

x1 = Transpose[{{1, 0, 0, 0, 325, 65.0, 12.5, 1594}, {2, 1, 0, 0, 
              400, 65.0, 12.5, 1398}, {3, 0.5, 0.866, 0, 362, 69.3, 12.5, 
              1352}, {4, -0.5, 0.866, 0, 288, 69.3, 12.5, 
              1466}, {5, -1, 0, 0, 250, 65.0, 12.5, 
              1750}, {6, -0.5, -0.866, 0, 288, 60.7, 12.5, 1064}, {7, 
              0.5, -0.866, 0, 362, 60.7, 12.5, 1337}, {8, 0.5, 0.289, 
              0.816, 362, 66.4, 18.6, 1533}, 
{9, -0.5, 
              0.289, 0.816, 288, 66.4, 18.6, 1617}, {10, 0, -0.577, 
              0.816, 325, 62.1, 18.6, 1332}, {11, 
              0.5, -0.289, -0.816, 326, 63.6, 6.4, 
              1409}, {12, -0.5, -0.289, -0.816, 288, 63.6, 6.4, 
              1560}, {13, 0, 0.577, -0.816, 325, 68.9, 6.4, 
              1431}}][[2]]

x2 = Transpose[{{1, 0, 0, 0, 325, 65.0, 12.5, 1594}, {2, 1, 0, 0, 
              400, 65.0, 12.5, 1398}, {3, 0.5, 0.866, 0, 362, 69.3, 12.5, 
              1352}, {4, -0.5, 0.866, 0, 288, 69.3, 12.5, 
              1466}, {5, -1, 0, 0, 250, 65.0, 12.5, 
              1750}, {6, -0.5, -0.866, 0, 288, 60.7, 12.5, 1064}, {7, 
              0.5, -0.866, 0, 362, 60.7, 12.5, 1337}, {8, 0.5, 0.289, 
              0.816, 362, 66.4, 18.6, 1533}, 
{9, -0.5, 
              0.289, 0.816, 288, 66.4, 18.6, 1617}, {10, 0, -0.577, 
              0.816, 325, 62.1, 18.6, 1332}, {11, 
              0.5, -0.289, -0.816, 326, 63.6, 6.4, 
              1409}, {12, -0.5, -0.289, -0.816, 288, 63.6, 6.4, 
              1560}, {13, 0, 0.577, -0.816, 325, 68.9, 6.4, 
              1431}}][[3]]

x3 = Transpose[{{1, 0, 0, 0, 325, 65.0, 12.5, 1594}, {2, 1, 0, 0, 
              400, 65.0, 12.5, 1398}, {3, 0.5, 0.866, 0, 362, 69.3, 12.5, 
              1352}, {4, -0.5, 0.866, 0, 288, 69.3, 12.5, 
              1466}, {5, -1, 0, 0, 250, 65.0, 12.5, 
              1750}, {6, -0.5, -0.866, 0, 288, 60.7, 12.5, 1064}, {7, 
              0.5, -0.866, 0, 362, 60.7, 12.5, 1337}, {8, 0.5, 0.289, 
              0.816, 362, 66.4, 18.6, 1533}, 
{9, -0.5, 
              0.289, 0.816, 288, 66.4, 18.6, 1617}, {10, 0, -0.577, 
              0.816, 325, 62.1, 18.6, 1332}, {11, 
              0.5, -0.289, -0.816, 326, 63.6, 6.4, 
              1409}, {12, -0.5, -0.289, -0.816, 288, 63.6, 6.4, 
              1560}, {13, 0, 0.577, -0.816, 325, 68.9, 6.4, 
              1431}}][[4]]

x1 x1

x2 x2

x3 x3

x1 x2

x1 x3

x2 x3

M = 
        Transpose[{{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}, x1, x2, x3, 
            x1 x1, x2 x2,  x3 x3,  x1 x2,  x1 x3, 
            x2 x3}];

M // TableForm

Determination of the quadratic model by multilinear regression 
of the mean particle size:

\(TraditionalForm\`\(y\&^\)\) = 1594 - 97.5 x\(TraditionalForm\`\_1\) + 117.6 x\(TraditionalForm\`\_\(\(2\)\(\ \)\)\)+ 16.7 x\(TraditionalForm\`\(\(\_3\)\(\ \)\)\)- 20.0 (((x\(\_1\)))^2) - 379.0 (((x\(\_2\)))^2) - 70.8 (((x\(\_3\)))^2)
- 223.4 x\(\_1\) x\(\_2\) + 120.1 x\(\_1\)x\(\_3\) + 83.0 x\(\_2\)x\(\_3\)

Results are in agreement with the model obtained by Lewis G. A., D. Mathieu and R. Phan-Than-Luu in Pharmaceutical Experimental Design, Drugs and the Pharmaceutical Sciences, Vol 92, 1999, Marcel Dekker, New York.

a = Inverse[Transpose[M] . M] . Transpose[M] . y

Contour plots of the obtained mean size versus impeller speed (x1) and pelletization time (x3) for x2 (binding solution) =±0.5 and 0. We get the same figures as plotted in Figure 5.14 page 229 of the book by Lewis G. A., D. Mathieu and R. Phan-Than-Luu in Pharmaceutical Experimental Design, Drugs and the Pharmaceutical Sciences, Vol 92, 1999, Marcel Dekker, New York.

yp = 
      1594 - 97.5 X1 - 20.  (X1^2) + 117.6 X2 - 223.4 X1 X2 - 
        379.  (X2^2) + 16.7 X3 + 120.1 X1 X3 + 83.  X2 X3 - 
        70.8 (X3^2)

X2 = 0.5;

ContourPlot[yp, {X1, -1, 1}, {X3, -1, 1}]

X2 = -0.5;

ContourPlot[yp, {X1, -1, 1}, {X3, -1, 1}]

X2 = 0;

ContourPlot[yp, {X1, -1, 1}, {X3, -1, 1}]