This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 5.2 – 1 Solution to problem 5.2 a) For a step change in input of magnitude M: ( ) ( ) 1 (0) t y t KM e y τ = + where 1.5 1 0.5 M Kw = = We need to find both K and τ , and we have information about the initial conditions and two time points. One way of obtaining these parameters would be to substitute the values given in the description of the exercise and solve the system of equations, as follows: ( ) 4 230 0.5 1 80 K e τ = ⋅ + ( ) 30 280 0.5 1 80 K e τ = ⋅ + Dividing these two equations we obtain: 4 30 230 80 1 280 80 1 e e τ τ = Using an iterative calculation (for example with Excel) it is possible to obtain τ . Once this value is available we can calculate K . However, an easier solution can be found by realizing that after 30 min. the process has achieved approximately the steady state, as shown in the following figure: 50 100 150 200 250 300 5 10 15 20 25 30 35 time (min) Temperature (ºC) Since at 30 min. we are at the new steady state, the steady state gain Since at 30 min....
View
Full Document
 Spring '09
 RafiqulGani
 Derivative, exponential term

Click to edit the document details