MATLAB: Thermal Systems

We plotted the heating and cooling of coffee using MATLAB. First, we used the formula given to find the cooling of coffee over 1500. 

C=100, Rth = 0.85
We used bang-bang control to stop once it reached 357. So basically when T< 357 we made the power go to 0.
power = 0 when T = 293
power = 150 when T= 357
so gain = 150-0/ 357-293
gain = 2.3
however the temperature is only going up to 335

We found that as the denominator increases, the slope decreases as well.

C = 2000, Rth = 0.85

C = 1000, Rth = 2

C = 2000, Rth = 2

C = 500, Rth = .4

This was our next program, where we simulated the heating of coffee.

p= 75

We then did this with proportional control.

power = error x gain

power = 0 when T = 293

power = 150 when T= 357gain = 150-0/ 357-293gain = 2.3

The temperature only ended up going to 335 then we increased the gain to 30 which increased the temperature.

Finally, we used proportional control with delay and the consonants p= 150 and gain = 2.3 and we wanted the error to be the target minus the temperature and the power to be dependent on the gain p= gain times error.