11 Process Yield Steepest Ascent In 6, we described an experiment on a chemical

Example 14-11 Process Yield Steepest Ascent In Example 14-6, we described an experiment on a chemical

  process in which two factors, reaction time (x

  ) and reaction temperature (x 2 ), affect the percent

  conversion or yield (Y). Figure 14-27 shows the 2 design plus fi ve center points used in this study. The engineer found that both factors were important, there was no interaction, and there was no curvature in the response surface. There- fore, the fi rst-order model

  Y =β+β 0 11 x +β 22 + e

  should be appropriate. Now the effect estimate of time is 1.55 hours and the effect estimate of temperature is 0.65°F,

  and because the regression coeffi cients ˆ β 1 and ˆ β 2 are one-half of the corresponding effect estimates, the fi tted fi rst-order

  model is

  y = . 40 44 + 0 775 x 1 0 325 2

  Figure 14-45(a) and (b) show the contour plot and three-dimensional surface plot of this model. Figure 14-45 also

  shows the relationship between the coded variables x 1 and x 2 (that defi ned the high and low levels of the factors) and

  the original variables, time (in minutes), and temperature (in °F).

  Chapter 14Design of Experiments with Several Factors

  2 x (temperature)

  x 2 (temperature)

  150.0 30.00 x 1 (time)

  x 1 (time)

  –1

  Contour plot

  Three-dimensional surface plot

  (a)

  (b)

  FIGURE 14-45 Response surface plots for the fi rst-order model in Example 14-11.

  From examining these plots (or the fi tted model), we see that to move away from the design center—the point

  (x 1 = 0, x 2 = 0)—along the path of steepest ascent, we would move 0.775 unit in the x 1 direction for every 0.325 unit in the x2 direction. Thus, the path of steepest ascent passes through the point (x 1 = 0, x 2 = 0) and has a slope 0.325 0.775.

  The engineer decides to use 5 minutes of reaction time as the basic step size. Now, 5 minutes of reaction time is equiva-

  lent to a step in the coded variable x 1 of Δx 1 = 1. Therefore, the steps along the path of steepest ascent are Δx 1 = 1.0000 and Δx 2 = (0.325 0.775) Δx 1 = 0.42. A change of Δx 2 = 0.42 in the coded variable x 2 is equivalent to about 2°F in the

  original variable temperature. Therefore, the engineer moves along the path of steepest ascent by increasing reaction time by 5 minutes and temperature by 2°F. An actual observation on yield is determined at each point.

  Next Steps: Figure 14-46 shows several points along this path of steepest ascent and the yields actually observed from the process at those points. At points A–D, the observed yield increases steadily, but beyond point D, the yield decreases. Therefore, steepest ascent terminates in the vicinity of 55 minutes of reaction time and 163°F with an observed percent conversion of 67.

  F

  Original region of experimentation

  C Point

  A: 40 minutes, 157 8F, y = 40.5

  B Path of

  Point

  B: 45 minutes, 159 8F, y = 51.3

  A steepest ascent

  Point

  C: 50 minutes, 161 8F, y = 59.6

  Point

  D: 55 minutes, 163 8F, y = 67.1

  Temperature

  Point

  E: 60 minutes, 165 8F, y = 63.6

  Point

  F: 65 minutes, 167 8F, y = 60.7

  Original fitted contours

  30 40 50 60 70 Time

  FIGURE 14-46 Steepest ascent experiment for Example 14-11.

  Section 14-8Response Surface Methods and Designs

  Analysis of a Second-Order Response Surface When the experimenter is relatively close to the optimum, a second-order model is usually required to approximate the response because of curvature in the true response surface. The fi tted second-order model is

  0 ∑ β+ ii x β ii i +

  β ij i xx j

  i = 1 i 1 ij ,

  where ˆ β denotes the least squares estimate of b. In this section, we show how to use this fi tted model to fi nd the optimum set of operating conditions for the x’s and to characterize the nature of the response surface.