Interpolation

To demonstrate how the interp1 function works, let us create an x-y table with the following commands; x = 0:5; y = [0,20,60,68,77,110]; To tabulate the output, we can use [x',y'] The result is ans = 0 01 20 2 603 68 4 775 110 Suppose we want to find the corresponding value for 1.5 or interpolate for 1.5. Using y_new = interp1(x,Y,x_new) syntax, let us type in: y_new=interp1(x,y,1.5) y_new =40

The table we created above has only 6 elements in it and suppose we need a more detailed table. In order to do that, instead of a single new x value, we can define an array of new x values, the interp1 function returns an array of new y values: new_x = 0:0.2:5; new_y = interp1(x,y,new_x); Let's display this table [new_x',new_y'] The result is ans = 0 00.2000 4.0000 0.4000 8.00000.6000 12.0000 0.8000 16.00001.0000 20.0000 1.2000 28.00001.4000 36.0000 1.6000 44.00001.8000 52.0000 2.0000 60.00002.2000 61.6000 2.4000 63.20002.6000 64.8000 2.8000 66.40003.0000 68.0000 3.2000 69.80003.4000 71.6000 3.6000 73.40003.8000 75.2000 4.0000 77.00004.2000 83.6000 4.4000 90.20004.6000 96.8000 4.8000 103.40005.0000 110.0000

Using the table below, find the internal energy of steam at 215 ˚C and the temperature if the internal energy is 2600 kJ/kg (use linear interpolation).

First let us enter the temperature and energy values temperature = [100, 150, 200, 250, 300, 400, 500];energy = [2506.7, 2582.8, 2658.1, 2733.7, 2810.4, 2967.9, 3131.6];[temperature',energy'] returns ans = 1.0e+003 *0.1000 2.5067 0.1500 2.58280.2000 2.6581 0.2500 2.73370.3000 2.8104 0.4000 2.96790.5000 3.1316 issue the following for the first question: new_energy = interp1(temperature,energy,215) returns new_energy = 2.6808e+003 Now, type in the following for the second question: new_temperature = interp1(energy,temperature,2600) returns new_temperature = 161.4210