Control ENG HW_Part_28

Control ENG HW_Part_28 - Using fig8.5, for = 1.2 , we see...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Using fig8.5, for 2 . 1 = ξ , we see that maximum is attained at min 776 . 1 , 95 . 0 = = t t τ And the maximum value is around 325 . 0 2 = Y 2 (t) = 0.174 = H 2 (t) = 0.174x3.5 = 0.16ft thus max deviation is H 1 will be at t = 0 = H 1 = 1 ft max deviation is H 2 will be at t = 1.776 min = H 2Max = 0.61 ft. comment : the first tank gets the impulse and hence it max deviation turns out to be higher than the deviations for the second tank. The second tank exhibits an increase response ie the deviation increases, reaches the H 2Max falls off to zero. 8.3 The tank liquid level shown operates at steady state when a step change is made in the flow to tank 1.the transaient response in critically damped, and it takes 1 min for level in second tank to reach 50 % of total change. If A 1 /A 2 = 2 ,find R 1 /R 2 . calculate τ for each tank. How long does it take for level in first tank to reach 90% of total change? For the
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 2

Control ENG HW_Part_28 - Using fig8.5, for = 1.2 , we see...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online