Find a negative feedback controller with at least two tunable gains that (1) results in zero steady state error when the input i
s a unit step (1/s). (and show why it works); (2) Gives a settling time of 4 seconds; (3) has 10% overshoot. Use the standard 2nd order approximation. Plot the step response of the system and compare the standard approximation with the plot.
1 answer:
Answer:
Gc(s) =
Explanation:
comparing the standard approximation with the plot attached we can tune the PI gains so that the desired response is obtained. this is because the time requirement of the setting is met while the %OS requirement is not achieved instead a 12% OS is seen from the plot.
attached is the detailed solution and the plot in Matlab
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Answer:
y=0.12 lbmol(water)/lbmol(products)
Explanation:
First we find the humidity of air. Using humidity tables and the temperature we find that is 0.01 g water/L air.
Now we set the equation assuming dry air:
With this equation we have almost all moles that exit the reactor, we are just missing the initial moles of water due to the humidity. So we proceed to calculate it with the ideal gas law:
PV=nRT
Vair=867.7L
With the volume and the fraction of water, we can calculate the mass of water:
0.01 * 867.7=8.677 g of water
Now we calculate the moles of water:
8.677 g / 18 g/mol = 0.48 moles of water
Now we can calculate the total moles of water in the exit of the reactor:
0.48 + 4 = 4.48 moles of water
And finally we just need to sum all moles at the exit of the reactor and divide:
3 moles of CO2 + 2.5 moles of O2+ 4.48 moles of H2O + 28.2 moles of N2
And we have 38.18 moles in total, then:
4.48/38.18=y=0.12 moles of water/moles of products
As the relation moles/moles is equal to lb-moles/lb-moles, we have our fina result:
y=0.12 lbmol(water)/lbmol(products)