Unformatted text preview: 5.3 The contaminant concentration c increases according to this expression: c(t) = 5 + 0.2t Using deviation variables and Laplace transforming,
c(t ) = 0.2t or C ( s ) = 0 .2 s2 Hence C m ( s) = 1 0 .2 2 10 s + 1 s and applying Eq. 521 cm (t ) = 2(e t /10  1) + 0.2t As soon as cm (t ) 2 ppm the alarm sounds. Therefore, t = 18.4 s (starting from the beginning of the ramp input) The time at which the actual concentration exceeds the limit (t = 10 s) is subtracted from the previous result to obtain the requested t . t = 18.4  10.0 = 8.4 s
2.5 2 1.5 c'm
1 0.5 0 0 2 4 6 8 10 12 14 16 18 20 time Fig S5.3. Concentration response for a ramp input of magnitude 0.2 Kw/min. 53 5.10 a) The dynamic behavior of the liquid level is given by d 2 h dh +A + Bh = C p (t ) dt dt where A= 6 R 2 B= 3g 2L and C = 3 4L Taking the Laplace Transform and assuming initial values = 0 s 2 H ( s ) + AsH ( s ) + BH ( s ) = C P ( s ) or H ( s ) =
C/B P ( s ) 1 2 A s + s +1 B B We want the previous equation to have the form
H ( s ) = K P ( s ) s + 2s + 1
2 2 Hence K = C/B =
1 = B
2 1 2g 2L then = 1 / B = 3g 3 2 L then = 2 R 3g 1/ 2 1/ 2 A 2 = B b) The manometer response oscillates as long as 0 < < 1 or 0 <
3 2 L R 2 3 g 1/ 2 < 1 b) If is larger , then is smaller and the response would be more oscillatory. If is larger, then is larger and the response would be less oscillatory. 512 5.13 a) The solution of a criticallydamped secondorder process to a step change of magnitude M is given by Eq. 550 in text. t y(t) = KM 1  1 + e t / Rearranging
y t = 1  1 + e  t / KM 1 + t t / y = 1 e KM When y/KM = 0.95, the response is 0.05 KM below the steadystate value. KM 0.95KM y 0 ts time t s t / = 1  0.95 = 0.05 1 + e t t ln1 + s  s = ln(0.05) = 3.00 t t Let E = ln1 + s  s + 3 515 and find value of ts that makes E 0 by trialanderror. ts/ 4 5 4.5 4.75 E 0.6094 0.2082 0.2047 0.0008 b) Y(s) = a value of t = 4.75 is ts, the settling time. a a a2 a4 Ka = 1+ 2 + 3 + 2 s s s + 1 (s + 1) 2 s (s + 1)
2 We know that the a3 and a4 terms are exponentials that go to zero for large values of time, leaving a linear response. a2 = lim
s 0 Ka = Ka (s + 1) 2 Define Q(s) = Ka (s + 1) 2 dQ  2 Ka = ds (s + 1) 3 Then a1 =  2 Ka 1 lim 1! s 0 (s + 1) 3 (from Eq. 362) a1 =  2 Ka the longtime response (after transients have died out) is y 2 (t ) = Kat  2 Ka = Ka (t  2) = a (t  2) for K = 1 and we see that the output lags the input by a time equal to 2. 516 2 y x=at yl =a(t2) 0 actual response time 5.17 a) Assume underdamped secondorder model (exhibits overshoot)
K= 1756456 10  6 ft ft = = 0 .2 123456 140  120 gal/min gal/min
11  10 1 = = 0.25 10  6 4 Fraction overshoot = From Fig 5.11, this corresponds (approx) to = 0.4 From Fig. 5.8 , = 0.4 , we note that tp/ 3.5 Since tp = 4 minutes (from problem statement), = 1.14 min G p(s) = 0.2 0 .2 = 2 (1.14) s + 2(0.4)(1.14) s + 1 1.31s + 0.91s + 1
2 2 b) In Chapter 6 we see that a 2ndorder overdamped process model with a numerator term can exhibit overshoot. But if the process is underdamped, it is unique. ...
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 Spring '09
 Dr.MuhammadAlArfaj
 Trigraph, kA, ramp input, time Fig S5.3, criticallydamped secondorder process

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