Question

Problem 2: In modeling a combustion process it is required to find enthalpy as a function of temperature. Find linear splines interpolation for the following data. 100 45.2 120 92.9 140 178.8349.4 16 80) 17.2 60 D. 1. 2. 3. Find E(85), and E(105) Plot the data and the splines on the same figure What conclusion can be drawn from the figure?

Answer 1

Please find the MATLAB Script below:

%Script Name: Combustion Process
%------------------------------------------
%Define the T and E vectors
%given in the table
T=[60,80,100,120,140,16];
E=[0.0,17.2,45.2,92.9,178.8,349.4];
% Interpolating linear and spline for E(85)
E85Linear=interp1(T,E,85,'linear');
fprintf("E(85) Linear= %4.2f \n",E85Linear);
E85Spline=interp1(T,E,85,'spline');
fprintf("E(85) Spline= %4.2f \n",E85Spline);
% Interpolating linear and spline for E(105)
E105Linear=interp1(T,E,105,'linear');
fprintf("E(105) Linear= %4.2f \n",E105Linear);
E105Spline=interp1(T,E,105,'spline');
fprintf("E(105) Spline= %4.2f \n",E105Spline);
%Plot the figure
figure
hold on
plot(T,E,'o','LineWidth',4.5);
tempPoints=linspace(16,140,200);
ELinear=interp1(T,E,tempPoints,'linear');
plot(tempPoints,ELinear,':','LineWidth',2.5);
ESpline=interp1(T,E,tempPoints,'spline');
plot(tempPoints,ESpline,'--','LineWidth',2.5);
%Labels for the plot
xlabel('T');
ylabel('E');
legend({'Table Values','Linear','Spline'});
title("Temperature vs Enthalpy");

Output of the Program:

>> CombustionProcess
E(85) Linear= 24.20
E(85) Spline= 23.79
E(105) Linear= 57.13
E(105) Spline= 54.42
>>

Screen-shot

Figure File Edit View Insert Tools Desktop Window Help PLOTS Log In R 4 Find Files S Compare New Open Save Print Temperature vs Enthalpy 350 Table Values Linear Cur Editor-CAMATLABCombustionP CombustionProcess.m x1+ 300 pline % Interpolating 250 13-E10SLinear-interpl 14 - fprintf("E (105) 15-E105Spline-interpl 200 16 17 18- Command Window fprintf ("E (105) S Plot the figure figure u 150 田 100 >> CombustionProcess E (85 Linear- 24.20 E(85) 50 公 Spline= 23.79 E (105) Linear-57.13 E (105) Spline- 54.42 2> 0 50 20 40 60 80 100 120 140

Hope this is helpful. Let me know if you need more information on this.