Nonlinear Models for Regression-More Examples: Chemical ...



Chapter 06.04Nonlinear Models for Regression-More ExamplesChemical EngineeringExample 1Below is given the FT-IR (Fourier Transform Infra Red) data of a 1:1 (by weight) mixture of ethylene carbonate (EC) and dimethyl carbonate (DMC). Absorbance P is given as a function of wavenumber, m.Table 1 Absorbance as a function of wavenumber.Wavenumber, m(cm-1)Absorbance, P(arbitrary unit)804.1840.1591827.3260.0439846.6110.0050869.7530.0073889.0380.0448892.8950.0649900.6090.1204Regress the above data to a second order polynomialP=a0+a1m+a2m2Find the absorbance at m=1000 cm-1SolutionTable 2 shows the summations needed for the calculations of the constants of the regression model.Table 2 Summations for calculating constants of model.iWavenumber, m(cm-1)Absorbance, P(arbitrary unit)m2m3m4m×Pm2×P1804.180.15916.4671×1055.2008×1084.1824×1011127.951.0289×1052827.330.04396.8447×1055.6628×1084.6849×101136.3193.0048×1043846.610.00507.1675×1056.0681×1085.1373×10114.2333.583×1034869.750.00737.5647×1056.5794×1085.7225×10116.3495.522×1035889.040.04487.9039×1057.0269×1086.2471×101139.8283.5409×1046892.900.06497.9726×1057.1187×1086.3563×101157.9485.1742×1047900.610.12048.1110×1057.3048×1086.5787×1011108.439.7655×104i=17?6030.40.44545.2031×1064.4961×1093.8909×1012381.063.2685×105P=a0+a1m+a2m2is the quadratic relationship between the absorbance and the wavenumber where the coefficients a0, a1, a2 are found as followsni=1nmii=1nmi2i=1nmii=1nmi2i=1nmi3i=1nmi2i=1nmi3 i=1nmi4 a0a1a2=i=1nPii=1nmiPii=1nmi2Pin=7i=17mi=6.0304×103i=17mi2=5.2031×106i=17mi3=4.4961×109i=17mi4=3.8909×1012i=17Pi=0.4454i=17miPi=381.06 i=17mi2Pi=3.2685×105We have7.00006.0304×1035.2031×1066.0304×1035.2031×1064.4961×1095.2031×1064.4961×1093.8909×1012a0a1a2=0.4454381.063.2685×105Solve the above system of simultaneous linear equations, we geta0a1a2= 43.985-0.10268-5.9922×10-5The polynomial regression model isP=a0+a1m+a2m2 =43.985-0.10268m-5.9922×10-5m2Figure 1 Second order polynomial regression model for absorbance as a function of wavenumber.To find P where m=1000 cm-1:P=a0+a1m+a2m2 =43.985-0.10268m-5.9922×10-5m2 =43.9855-0.10268×1000-5.9922×10-5×10002 =1.2221Example 2The mechanism of polymer degradation reaction kinetics is suspected to follow Avrami or random nucleation reaction,fα=AT-T0be-ERTwhere fα=-ln1-α, T is the absolute temperature (K), b is the heating rate in K/min, A is the frequency factor with units of rate constant, R is the gas constant (8.314 kJ/kmol-K) and T0 is the activation temperature. Given that T0=338.75 K, b=10 K/min and conversion, α, at different temperatures are as given in table 3. Use the method of least squares to determine the values of AandE.Table 3 Conversion at given different temperaturesTemp K360370380390400410420430440Conversion,α0.10550.20100.34250.51460.67570.80260.89240.95441.00SolutionTo setup the table, we must rewrite equation-ln1-α=AT-T0be-ERTas-bln1-αT-T0=Ae-ERTTaking natural log of both sides of the above equation, we obtainln-bln1-αT-T0=lnA-ERTso that the equation is in the form y=β0+β1x wherey=ln-bln1-αT-T0β0=lnAβ1=-ERx=1T β1=ni=1nxiyi-i=1nxii=1nyini=1nx12-i=1nxi2β0=i=1nyin-β1i=1nxinTable 4 Example on nonlinear exponential problem. iTαxyx2x×y13600.10552.7778×10-3?2.94767.7160×10-6?8.1877×10-323700.20102.7027×10-3?2.63387.3046×10-6?7.1183×10-333800.34252.6316×10-3?2.28626.9252×10-6?6.0163×10-343900.51462.5641×10-3?1.95886.5746×10-6?5.0225×10-354000.67572.5000×10-3?1.69366.2500×10-6?4.2341×10-364100.80262.4390×10-3?1.47965.9488×10-6?3.6088×10-374200.89242.3810×10-3?1.29325.6689×10-6?3.0791×10-384300.95442.3256×10-3?1.08355.4083×10-6?2.5199×10-3∑2.0322×10-2?1.5376×1015.1797×10-5?3.9787×10-2n=8i=18xi=2.0322×10-2i=18yi=-1.5376×101i=18xiyi=-3.9787×10-2i=18xi2=5.1797.10-5β1=8-3.9787×10-2-2.0322×10-2-1.5376×10-185.1797×10-5-2.0322×10-22 =-4.1561×103β0=-1.5376×1018--4.1561×1032.0322×10-28 =8.6352A=eβ0 =e8.6352 =5.6264×103E=-β1R =--4.1561.103×8.3140 =3.4553×104This gives the model as -ln1-α=5.6264×103T-338.7510×e-4.1561×103TFigure 2 Polymer degradation reaction kinetics rate as a function of temperature.Example 3The progress of a homogeneous chemical reaction is followed and it is desired to evaluate the rate constant and the order of the reaction. The rate law expression for the reaction is known to follow the power function form-r=kCn. (1)Use the data provided in the table to obtain n andk.Table 11 Chemical kineticsCA(gmol/l)42.251.451.00.650.250.06-rAgmolls0.3980.2980.2380.1980.1580.0980.048SolutionTaking natural log of both sides of Equation (1), we obtainln-r=lnk+nlnCLet ln-r=zlnC=wlnk=a0 from which k=ea0 (2)n=a1 (3)We getz=a0+a1wThis is a linear relation between z and w, wherea1=i=1mwizi-i=1mwii=1mzimi=1mwi2-i=1mwi2a0=i=1mzim-a1i=1mwim (4a,b)Table 6 Kinetics rate law using power function.iC-rwzw×zw2140.3981.3863?0.92130?1.27721.921822.250.2980.8109?1.2107?0.98180.6576131.450.2380.3716?1.4355?0.53340.13806410.1980.0000?1.61950.00000.0000050.650.158?0.4308?1.84520.79490.1855760.250.098?1.3863?2.32283.22011.921870.060.048?2.8134?3.03668.54317.9153i=17??2.0617?12.3919.765712.7401m=7i=17wi=-2.0617i=17zi=-12.391i=17wizi=9.7657i=17wi2=12.7401From Equation (4a, b)a1 =7×9.7657--2.0617×-12.3917×12.7401--2.06172 =0.50408a0=-12.3917--0.50408-2.06177 =-1.6217From Equations (2) and (3), we obtaink=e-1.6217 =0.19755n=a1 =0.50408Finally, the model of progress of that chemical reaction is-r=0.19755×C0.50408Figure 3 Kinetic chemical reaction rate as a function of concentration. ................
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