Writing heat balance equation for tank 1 dt dh q qa q A 1 − − 1 = 1 1 1 1 R h q = a a R h q 1 = dt dh A R h R h q a 1 1 1 1 1 = − − = writing the balance equation for tank 2 dt dh q q A 2 1 − 2 = 2 dt dh A R h R h 2 2 2 2 1 1 − = writing steady state equations 0 1 1 − − = R h s R h q a s s
0 2 2 1 1 − = R h s R h s writing the equation in terms of deviation variables dt dH A R R Q H a 1 1 1 1 1 1 = − + dt dH A R H R H 2 2 2 2 1 1 − = taking laplace transforms A H s R R R R Q s H s S a 1 1 1 1 2 1 ( ) ( ) = + − -----------(1) and ( ) ( ) ( ) 2 2 2 2 1 1 A s H s R H s R H s − = -----------(2) from (1) we get + + = a a R R R R A s Q s H s 1 1 1 1 1 ( ) ( ) + + + = 1 ( ) ( ) 1 1 1 1 1 1 s R R R R A R R R R Q s H s a a a a
( ) [ ]1 ( ) 1 1 1 1 + + = s R R R R Q s H s a a τ ; a a R R R R A + = 1 1 1 1 τ and from (2 ) we get ( ) [ ] 1 ( ) 1 ( ) 1 2 1 2 1 2 1 + + + = s s R R R R R R Q s H s a a τ τ 2 = R2A2 τ putting the numerical values of parameters + = 1 3 4 3 2 ( ) ( ) 1 s Q s H s 1 ( ) 1 3 4 3 2 ( ) ( ) 2 + + = s s Q s H s 8.1 A step change of magnitude 4 is introduced into a system having the transfer 6.1 4 10 ( ) ( ) 2 + + = X s s s Y s Determine (a) % overshoot (b)Rise time (c)Max value of Y(t) (d)Ultimate value of Y(t) (e) Period of Oscillation.
Given s X s 4 ( ) = ( 6.1 )4 40 ( ) 2 + + = s s s Y s The transfer function is ) )1 4 6.1 (2.0 ) ( 10 .0 25 ( ) ( ) 2 + + × = s s X s Y s = .0 25 4.0 )1 5.2 2 s + s + .0 25 ; 5.0 2 τ = τ = and 2ξ τ = 4.0 4.0 ( 1 ) )5.0(2 4.0 ξ = = < = system is underdamped we find ultimate value of Y(t) 10 4 40 ( 6.1 )4 40 ( ) ( ) 2 0 0 = = + + = = →∞ → → s s s Lt Y t Lt sY s Lt t S S thus B= 10 now, from laplace transform tables + − = − − sin( ) 1 1 ( ) 10 1 2 α φ ξ τ ξt Y t e where ξ ξ φ τ ξ α 2 2 1 , tan 1 − = − = −
(a) Over shoot = − × = − − = .0 84 4.0 exp 1 exp 2 π ξ πξ B A = 0.254 thus % overshoot = 25.4 c)thus, max value of Y(t) = A+B = B(0.254)+B = 2.54+10 = 12.54 e) Period of oscillation = 2 1 2 ξ πτ − = 3.427 b) For rise time, we need to solve r t e t = fort = t + − − − sin( ) 10 1 1 10 1 2 α φ ξ τ ξ = sin(α φ) τ ξτ + − r e t r = 0 = sin( .1 833 .1 1589) 0 5.0 4.0 + = − r e t r τ solving we get tr = 1.082 thus SOLUTION: % Overshoot = 25.4 Rise time = 1.0842 Max Y(t) = 12.54 U(t) Y(t) = 10 Period of oscillation = 3.427 Comment : we see that the Oscillation period is small and the decay ratio also small = system is efficiently under damped.
8.2 The tank system operates at steady state. At t = 0, 10 ft3 of wateris added to tank 1. Determine the maximum deviation in level in both tanks from the ultimate steady state values, and the time at which each maximum occurs. A1 = A2 = 10 ft3 R1 = 0.1ft/cfm R2 = 0.35ft/cfm. As the tanks are non interacting the transfer functions are ( )1 1.0 ( ) 1 ( ) 1 1 + = + = s s K Q s H s τ ( 1)( 5.3 )1 .0 35 ( ) ( 1)( )1 ( ) 1 2 2 2 + + = + + = s s s s R Q s H s τ τ Now, an impulse of t ft is provided 3 ∂( ) = 10 t e s Q s H s − = + = = = 1 1 ( ) 10 ( ) 1 and 5.3 5.4 1 5.3 ( 1)( 5.3 )1 5.3 ( ) 2 2 + + = + + = s s s s H s Now 5.3 .1 871 2 τ = = τ = .1 202 2 5.4 2 = 5.4 = = = τ ξτ ξ thus, this is an ovedamped system
Using fig8.5, for ξ = 2.1 , we see that maximum is attained at = .0 95,t = .1 776min t τ And the maximum value is around .0 325 τ 2 = Y2 (t) = 0.174 = H2(t) = 0.174x3.5 = 0.16ft thus max deviation is H1 will be at t = 0 = H1 = 1 ft max deviation is H2 will be at t = 1.776 min = H2Max = 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 H2Max 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 A1/A2 = 2 ,find R1/R2 . calculate τ for each tank. How long does it take for level in first tank to reach 90% of total change? For the first tank, transfer function 1 1 1 ( ) 1 ( ) + = s R Q s H s τ For the second tank ( ) ( 1)( )1 ( ) 1 2 2 + + = s s R Q s H s τ τ
= ( ) ( ) 1 ( ) 1 2 2 1 2 2 2 + + + = s s R Q s H s τ τ τ τ ( ) 1 1 ; ( ) 1 ( ) 1 2 2 1 2 2 2 + + + = = s s R s H s s Q s τ τ τ τ 1 2 τ ( parameter) = τ τ For 1 2 τ ( parameter) = τ τ = = − + − 1 2 1 2 2 2 ,1 ( ) 1 1 τ τ τ τ ξ t e t for H t R given, t = 1 for 1 2 τ ( parameter) = τ τ ( ) 2 2 2 H (t → ∞) = R 1− )0( = R I R R e = − = − + − 2 1 1 1 2 1 1 2 2 1 2 τ τ τ τ also 1 2 2ξτ = τ +τ 5.0 2 1 1 2 2 1 1 1 2 2 1 2 = = = = = = + = = A A R R τ A R A R τ τ ξ from I τ τ 1 1 1 5.0 1 − − = + e
.1 372 min .0 596 min 5.0 ;1.0 .1 372min 9.0 1( ) .0 94( ) 1( ) ; ( ) 1( ) ( )1 )3.8 ( ) 90% 1 2 2 1 .0 596 .0 596 1 1 1 1 1 1 1 1 1 1 = = = = = = = − → ∞ = − = − + = − − − − t R R thus e t R R e t R e H t R e s s R H s t t t t τ τ τ τ τ Comment : tan . 1 , 2 . 1 2 , sec tan responds more slowly to changes than first k Small values of τ τ indicate the system regains the steady state quickly Also as R > R the ond 8.4 Assuming the flow in the manometer to be laminar function between applied pressure P1 and the manometer reading h. Calculate a) steady sate gain ,b) τ ,c) ξ . Comment on the parameters and their relation to the physical nature of this problem.
Assumptions: Cross-sectional area =a Length of mercury in column = L Friction factor = 16/Re (laminar flow) Mass of mercury = mrg Writing a force balance on the mercury Mass X acceleration = pressure force - drag force - gravitational force ( ) 2 ( ) 2 2 1 2 A gh u Ap Af dt d h AL ρ ρ ρ = − − g p h dt dh dt gD d h g L ρ ρ µ 1 2 2 8 + + = At Steady state, g p h s s ρ 1 = = g p H dt dH dt gD d H g L ρ ρ µ 1 2 2 8 + + = = [ ] g p s sH s H s gD s H s g L ρ ρ µ ( ) ( ) ( ) 8 ( ) 2 1 + + = = [ 1] ( ) ( ) 2 3 1 2 1 k s + k s + H s =k p s
= ( ) ( )1 ( ) 2 2 1 3 1 1 + + = k s k s k p s H s Where ; 1 g L k = ; 8 2 gD k ρ µ = ; 1 3 g k ρ = Thus ( ) ( 2 )1 ( ) 2 2 1 1 + + = s s R p s H s τ ξτ Where ; 1 g R ρ = ; 2 g L τ = ; 8 2 ρgD µ ξτ = Now ) ; g L b τ = 1 4 2 1 . 8 ) − = = g L gD gD c ρ µ ρ τ µ ξ Steady state gain ; 1 ( ) 0 g Lt G s R S ρ = = → Comment : a) τ is the time period of a simple pendulum of Length L. b) ξ is inversely proportional to τ , smaller the τ ,the system will tend to move from under damped to over damped characteristics. 8.5 Design a mercury manometer that will measure pressure of upto 2 atm, and give responses that are slightly under damped with ξ = 0.7 Parameter to be decide upon : .a) Length of column of mercury .b) diameter of tube.
Considering hmax to be the maximum height difference to be used ; 9.81*13600 2* .1 01325*105 p1 = ρghmax =hmax = .1 51 ; hmax = m Assuming the separation between the tubes to be 30 cm, We get an additional length of 0.47 m; Which gives us the total length L= 1.5176.47 L = 2 M Now, ξ = 0.7 = 7.0 4 = L g ρgD µ .0 00015 5.1 *10 .0 74*13600* .9 81 2 .9 81 4* 6.1 *10 * 7.0 4 7 3 = = = = − − g L g D ρ µ As can be seen, the values yielded are not proper, with too small a diameter and too large a length. A smaller ξ value and lower measuring range of pressure might be better.
8.6 verify that for a second order system subjected to a step response, [ ] ξ ξ τ ξ ξ τ ξ 2 2 1 2 1 sin 1 tan 1 1 ( ) 1 − − + − = − − − t Y t e t With ξ <1 ( 2 )1 1 1 ( ) 2 2 + + = s s s Y s τ ξτ where s a b s s s s s s =− + − + − = + + = − − τ ξ τ ξ τ ξτ 1 2 1 ( )( ) 2 1 1 2 2 2 s =− a − b − + − = τ ξ τ ξ 1 2 2 ( )( ) 1 ( ) 1 2 2 s s s s s Y s − − = τ − + − = + ( ) ( ) 1 ( ) 1 2 2 s s C s s B s A Y s τ − + − = + ( ) ( ) 1 ( ) 1 2 2 s s C s s B s A Y s τ 0 ( ( 1 ) 1 2 ) ( 2 ) ( 1 ) 1 2 + + = − + + + − + − = A B C A s s s s s s Bs s s Cs s s s
( ) 1 − A s1 + ss − Bs2 − Cs1 = 1 2 1 2 1 1 1 ;1 s s B C s s As ss = A = = + = − 1 2 2 1 s = Bs + Cs = − 1 2 1 2 1 2 1 2 1 1 s s s s s s Cs Cs − = + = + = ( ) 1 1 ( ) 1 ( ) 1 2 2 1 2 2 1 1 2 1 2 1 s s s s s s s s B s s s C − − = − − = = − − = = ( ) ( ) ( ) − − + = − − = 1 2 1 2 1 1 2 2 1 2 2 1 . ( ) 1 1 . 1 1 . 1 1 ( ) s s s s s s s s s s s s s Y s τ 8.6 − + − = − S t s t e s s s e s s s s s Y t 1 2 ( ) 1 ( ) 1 1 1 ( ) 1 2 1 2 1 2 2 1 2 τ 1 2 2 1 τ s s = 1 − − = − S t S t e s e s s s Y t 1 2 2 1 1 2 2 1 1 ( ) 1 ( ) 1 τ [ ] S t S t s e s e s s Y t 1 2 2 1 2 1 ( ) 1 ( ) 1 − − = − [ ] S t S t Y t s e s e 1 2 2 1 2 2 1 ( ) 1 − − = + ξ τ
[ a jb bt j bt a jb bt j bt] je Y t t ( )(cos sin ) ( )(cos sin 2 1 ( ) 1 2 − − + − − + − − = − − ξ τ τ ξ [ ] 2 ( cos sin ) 2 1 ( ) 1 2 jb b bt a bt je Y t t − + − = − − ξ τ τ ξ [ 1 cos( ) sin( )] 1 ( ) 1 2 2 t t e Y t t ξ α ξ α ξ τ ξ − + − = − − [ ] ξ ξ α 2 1− = [ ] − = − ξ ξ φ 2 1 1 tan verified 8.14 From the figure in your text Y(4) for the system response is expressed b) verify that for ξ = ,1 and a step input τ τ t e t Y t − ( ) = 1− 1+
1 1 1 ( ) 2 2 + + = s s s Y s τ τ 2 2 ( )1 ( )1 1 ( ) + + = + τ + τ Bs C B A s s Y s ( 2 )1 1 2 2 2 A τ s + τs + + Bs Cs = 0 2 Aτ + B = 2Aτ + C = 0 A=1; 2 B = τ ;C = 2τ ( )2 1 1 ( )1 ( ) + + + = − s s s Y s τ τ τ τ ( ) 2 1 ( )1 1 ( ) + − + = − s s s Y s τ τ τ τ τ τ τ t t Y t e te − − = − − 1 ( ) 1 τ τ t te t Y t − ( ) =1− 1( + ) proved c) for ξ > ,1 prove that the step response is Y (t) 1 e [ ] cosh( t) sinh( t) t τ α β α ξ = − + −
1 1 2 2 − = − = ξ ξ β τ ξ α Now ( )( ) /1 ( ) 1 2 2 s s B s s Y s − − = τ Where τ ξ τ ξ 1 2 1 − s = − + τ ξ τ ξ 1 2 2 − s = − − from 8.6(a) − − + − − = − ( ) 1 ( ) 1 ( ) 1 ( ) 1 1 1 1 ( ) 1 2 1 2 1 1 2 2 1 2 2 s s s s s s s s s s s s s Y t τ [ ] S t S t s e s e s s Y t 1 2 2 1 2 1 ( ) 1 ( ) 1 − − = − − + − − − − − − = + − − − − − Y t e e e e t t t t τ ξ τ ξ τ ξ τ ξ τ ξ ξ τ ξ ξ ξ τ 2 1 1 1 2 2 2 2 1 1 2 1 ( ) 1 − + − − − − − = + − − − − − − − t t t t t e e e e e e Y t τ ξ τ ξ τ ξ τ ξ τ ξ ξ ξ ξ ξ ξ 1 2 1 2 1 2 2 2 2 1 1 2 1 ( ) 1
+ − − − = + − − − − 1 2 2 ( ) 1 2 t t t t t e e e e Y t e α α α α τ ξ ξ ξ Y (t) 1 e [ ] cosh( t) sinh( t) t τ α β α ξ = − + − 8.7 Verify that for a unit step-input (1) overshoot = − − 2 1 exp ξ πξ (2) Decay ratio = − − 2 1 2 exp ξ πξ For a unit step input the response (ξ<1): − − + − = − − − ξ ξ τ ξ ξ τ ξ 2 2 1 2 1 1 tan 1 ( ) 1 t Sin e Y t t (1) we have to find time t where the maxima occurs = dY/dt = 0 − − + − = − − ξ ξ τ ξ τ ξ ξ τ ξ 2 2 1 2 1 1 tan 1 ) t Sin e dt dY t 0 1 1 tan 2 2 1 = − − − + − − ξ ξ τ ξ τ τ ξ t Cos e t = − − + − ξ ξ τ ξ 2 2 1 1 tan 1 tan t = ξ ξ 2 1− π ξ ξ n t = − 2 1 for maxima = π ξ ξ n t 2 1 2 = −
= 2 1 ξ π − = t t 8.8 Verify that for X(t) =A sin ωt, for a second order system, ( ) ( ) sin( ) 1 ( ) 2 ( ) 2 2 2 ω φ ω ξτ + − + = t t A Y t 2 1 1 ( ) 2 tan ωτ ξωτ φ − = − − ( 2 )1 1 ( ) ( ) 2 2 2 2 + + + = s s s A Y s ω τ ξτ − + − + + + − = ( ) ( ) ( ) ( ) 2 1 1 1 1 1 2 s s D s s C s j B s j A A Y s τ ω ω ω Now as t ,Y (t) A cosωt B sinωt 11 11 → ∞ = + Where 11 1 1 A = A + B ( ) 11 1 1 B = j A − B to determine A , B put s = jω,− jω in the order 1 1 2 ( )( ) 1 2 1 j s j s j A − − − = ω ω ω 2 ( )( ) 1 2 1 j s j s j B + + = ω ω ω − − − + + = ( )( ) 1 ( )( ) 1 2 1 2 1 2 11 s j s j j s j s j A ω ω ω ω ω
+ + − − − + − − + + + = ( )( ) ( ) ( ) 2 2 2 2 2 2 1 1 2 1 2 2 1 2 1 2 2 11 ω ω ω ω ω ω ω ω ω s s j js js s s js js s s A + + + = ( )( ) ( ) 2 2 2 2 2 1 11 1 2 s ω s ω s s A similarly + + − = ( )( ) ( ) 2 2 2 2 2 1 2 11 1 2 ω ω ω ω s s s s B using 1 2 1 2 2 2 1 τ τ ξ = − s + s = s s = 2 2 2 2 2 2 2 2 1 4 2 2(2 )1 τ ξ τ τ ξ − s + s = − = + − + − = 2 4 2 2 4 2 11 2( )1 1 2 2 ξ ω τ ω τ τ ξ τ Aω A 2 2 2 2 3 1 2 2 + − − = τ ξω τ ω τ Aωξ = 2 2 2 1( ( ) ) 2( ) 2 ωτ ξωτ ωξτ − + − A and + − − = 2 2 2 2 2 2 11 1 2 1 τ ξω ω τ ω ω τ τ Aϖ B 2 2 2 2 1( ( ) ) 2( ) 1( ( ) ) ωτ ξωτ ωτ − + − = A
Thus 11 2 11 1 ( ) 2 tan ωτ ωτξ φ − − = = B A And, 2 2 2 ((1− (ωτ ) ) + 2( ξωυ) = A ANew (using A + B = ANew 2 2 11 11 Thus, ( ) ((1 ( ) ) 2( ) ( ) 2 2 2 ω φ ωτ ξωυ + − + = Sin t A Y t proved 8.9) If a second- order system is over damped, it is more difficult to determine the parameters ξ &τ experimentally. One method for determining the parameters from a step response has been suggested by R.c Olderboung and H.Sartarius (The dynamics of Automatic controls,ASME,P7.8,1948),as described below. (a) Show that the unit step response for the over damped case may be written in the form. 1 2 1 2 2 1 ( ) 1 r r r e r e s t tr tr − = − Where r1 and r2 are the roots of 2 1 0 2 2 τ s + ξτs + = (b) Show that s(t) has an inflection point at ( ) ln( / ) 2 1 2 1 r r r r t i − = © Show that the slope of the step response at the inflection point ( ) ( ) 1 i t t s t dt d s i = − Where, i tr tr i s t r e r e 1 2 1 2 1 ( ) =− = −
( ) 1 2 1 1 1 2 r r r r r r − =− (d) Show that the value of step response at the inflection point is ( ) 1 ( ) 1 1 2 1 1 2 i i s t r r r r s t = + and that hence 1 2 1 1 1 ( ) ( ) 1 s t r r s t i i − =− − (e) on a typical sketch of a unit step response show distances equal to ( ) 1 & ( ) ( ) 1 1 1 i i i s t s t s t − (f) Relate & 1 & 2 ξ τ to r r (a) ( )( ) 1 2 1 1 2 )1 1 ( ) 1 2 2 2 2 2 2 2 s r s r s s s s G s − − = + + = + + = τ τ τ ξ τ τ ξτ = ( )( ) 1 ( ) 1 2 2 s s r s r Y s − − = τ 1 2 1 2 ( )( ) 1 s r C s r B s A s s r s r − + − = + − − 1 ( )( ) ( ) ( ) 1 2 2 1 = A s − r s − r + Bs s − r + cs s − r Put s= 0 = Ar1r2 =1 ; 2 A = τ Put ( ) 1 ( ) ;1 1 1 2 1 1 1 2 r r r s r Br r r B − = = − = = ( ) 1 ( ) ;1 2 2 1 2 2 2 1 r r r s r Cr r r C − = = − = = − − + − − = + ( )( ) 1 ( )( ) 1 1 ( ) 1 1 2 1 2 2 1 2 2 2 s r r r s r r r r s r Y s τ τ
− + − = + ( ) ( ) 1 ( ) 1 1 2 2 2 1 2 2 1 2 r r r e r r r e Y t tr tr τ τ [ ] − − = − tr tr r e r e r r Y t 2 1 1 2 1 2 1 ( ) 1 ( ) ( ) 1 1 2 1 2 2 1 r r r e r e Y t r t tr − − = − φ (b) For inflection point , 0 & 0 2 3 2 2 = = dt d s dt d s 1 2 1 2 ( ) 2 2 r r r r e e dt ds tr tr − − = − 0 ( ) 1 2 1 2 2 1 2 2 2 2 = − − = − r r r r r e r e dt d s r t r t i tr tr r r t e r r u e r e ( ) 1 2 2 1 2 1 1− 2 = = = = 1 2 1 2 ln r r r r t i − = = (c ) ( ) ( ) ' i t t s t dt ds t i = = ] − − = − − 2− 1 1 1 2 1 2 1 2 1 2 1 2 r r r r r r r r r r r r r rr
− − = − − − − 2 1 1 1 2 1 2 1 2 2 1 1 2 1 2 1 2 r r r r r r r r r r r r r r r r rr = − − − 2− 1 1 1 2 1 2 1 1 2 ( ) r r r r r r r r r r r = − − = 2 1 1 1 2 1 ( ) r r r t t r i r r r dt ds t Also ( ) ( ) ( 1 2 ) 1 2 2 1 r r r r e e dt ds t r t tr t t i − − = − = − − − = − = 1 ( ) ( ) 2 1 1 2 1 2 1 r r r r r r e dt ds t tr t t i tr r t t t r e r e dt ds t i 1 2 1 2 ( ) = − = − = (d) 1 2 1 2 2 1 1 1 2 1 2 ( ) ( ) 1 1 2 1 r r r r r r s t r r r e r e s t rt i tr i i i − − = + − − = − = 1 2 1 2 2 1 1 ( ) ( ) 1 r r r r r r s t s t i i − − = = + Now
− = = + 1 2 1 1 1 ( ) 1 ( ) r r s t s t i i = = − − − 1 2 ' 1 1 ( ) 1 ( ) s t r r s t i 1 2 1 2 2 1 2 1 ; 1 1 r r r + r = = r r = τ = τ τ ; 2 1 2 = − + = τ ξ r r ξ 1 2 1 2 r + r = −2 rr = − + 1 2 2 1 2 1 r r r r ξ proved. 8.10 Y(0),Y(0.6),Y( ∞) if ( 2 )1 1 25( )1 ( ) 2 + + + = s s s s Y s )1 25 2 25 ( 1 1 ( ) 1 2 + + = + s s s Y s Y(s) impulse response + step response of G(s) Where )1 25 2 25 ( 1 ( ) 2 + + = s s G s
ξ ξ τ ξ ξ τ τ ξ 2 2 1 2 1 sin 1 tan 1 1 ( ) − − + − = − − t Y t e t Y(t) = 1+5.0.3e-t sin (4.899t)-1.02e-t sin(4.899t+1.369) Y(0)= 1-1=0 Y(0.6) = 1+0.561+0.515 Y( ∞) =1 Comment : as we can see ,the system exhibits an inverse response by increasing from zero to more than 1 and as t tend to ∞,will reach the steady state value of 1. 8.11 In the system shown the dev in flow to tank 1 is an impulse of magnitude 5 . A1 = 1 ft2 , A2 = A3 = 2 ft2 , R1 = 1 ft/cfm R2 = 1.5 ft/cfm . (a) Determine H1(s), H2(s), H3(s) Transfer function for tank 1 ( )1 1 ( ) ( ) 1 1 + = Q s s H s τ ( )1 5 ( ) 1 + = s H s
from tank 2, ( 1)(3 )1 5.1 ( ) ( 1)( )1 ( ) 1 2 2 2 + + = + + = s s s s R Q s H s τ τ for tank 3, dt dh q qc A 3 2 − = 3 dt dh q qc const Q A 3 3 2 3 = ( ) = = dt dh A R H 3 3 2 2 = thus, H s s H s R H s A SH s 3 1 ( ) ( ) ( ) ( ) 2 3 2 2 3 3 = = = H s s H s R H s A SH s 3 1 ( ) ( ) ( ) ( ) 2 3 2 2 3 3 = = = ( 1)(3 )1 5.0 ( ) ( ) 3 + + = Q s s s s H s 8.11© ( )1 5 ( ) 1 + = s H s t H t e − ( ) = 5 1 H1 .3( 46) = .0 155A 3 4 1 5.1 ( ) ( ) 2 2 + + = Q s s s H s 3 4 1 5.7 )1( 5 ( ) 2 2 + + = = = s s Q H s
τ = 3 2ξτ = 4 .1 155 2 3 4 2 4 = = = τ ξ from fig 8.5 τ ξ t = .1 155 and = 2 3 .3 46 = = = τ ξ t ( ) .0 265 5.7 τH2 t = X .1 147 .0 265 5.7 ( ) 2 = = τ X H t 3( 4 )1 5.0 ( ) ( ) 2 3 + + = Q s s s s H s 3( 4 )1 5.2 ( ) 5 ( ) 3 2 + + = = = s s s Q s H s τ = 3 3 2 ξ = from fig 8.2 at = ,2 ξ = .1 155 τ t Y(t) =0.54 H3(t) =0.54*2.5 = 1.35
8.12 sketch the response Y(t) if ( 2.1 )1 ( ) 2 2 + + = − s s e Y s S Determine Y(t) for t = 0,1,5, ∞ 2 2 2 2 2 2 2 2 ( )6.0 )8.0( )8.0( 8.0 1 ( 2.1 )1 ( )6.0 )8.0( ( ) + + = + + = + + = − − − s e s e s s e Y s S S S , ( ) 0 ,5 )5( .0 14 ,1 )1( 0 0 )0( 0 ( ) .1 25 sin( (8.0 2)) 2 (6. )2 = ∞ ∞ = = = = = = = = − ≥ − − t Y t Y t Y for t Y Y t e t t t Problem 8.13 The system shown is at steady state at t = 0, with q = 10 cfm A1 = 1ft2 ,A2=1.25ft2 , R1= 1 ft/cfm, R2= 0.8 ft/cfm. a) If flow changes fro 10 to 11 cfm, find H2(s). b) Determine H2(1),H2(4),H2(∞ ) c) Determine the initial levels h1(0),h2(0) in the tanks. d) obtain an expression for H1(s) for unit step change.
Writing mass balances, ( ) ( tan )1 1 1 1 1 2 for k dt dh A R h h q = − − At steady state ( ) S S s s h h R h h q 1 2 1 1 2 2 = − − − Also for tank 2 ( ) dt dh A R h R h h 2 2 2 2 1 1 2 − = − At steady state ( ) 8.0 *10 8 1 8.0 2 1 2 2 = = = = − S s s S h h h h 18 h1S = C) 18 h1S = ft h )0( 8 ft 2 = The equations in terms of deviation variables dt dH Q Q A 1 − 1 = 1 where 1 1 2 1 R H H Q − = dt dH Q Q A 2 1 − 2 = 2 2 2 2 R H Q = 8.2 1 8.0 ( ) ( ) 1 ( ) 2 1 2 1 2 2 1 2 2 2 + + = + + + + = s A R s s s R Q s H s τ τ τ τ
( .8 31( )) ( 8.2 )1 8.0 ( ) 2 2 Ans a s s s H s + + = Step response of a second order system 4.1 2 8.2 2 ;8.2 1 1 2 = = = = = = ξτ ξ τ τ ) 1 ;1 ( ) .0(8.0 22) .0 176 ( ) 2 H t ft from fig t a t = = = = = τ ) 4 ;4 ( ) .0(8.0 78) .0 624 ( ) 2 H t ft from fig t b t = = = = = τ c)t H (t ) 8.0 ft → ∞ = 2 → ∞ = Thus H )1( .0 176 ft 2 = H )4( .0 624 ft 2 = H ( ) 8.0 ft 2 ∞ = 8.13(d) we have ( ) ( ) ( ) ( ) 1 1 1 Q s Q s = A s H s ( ) ( ) ( ) ( ) 1 2 2 2 Q s − Q s = A s H s ( ) ( ) ( ) ( ) ( ) ( ) 2 1 1 2 2 Q s − Q s = A s H s + A s H s − = + ( ) ( ) 1 ( ) ( ) ( ) 2 2 2 1 1 A s H s R Q s A s H s
+ = ( ) 1 2 2 2 H s R τ s − = H s R Q s A sH s H s 2 2 1 1 2 ( ( ) ( ) ( ) τ We have Deg R s A R s R Q s H s 2 1 2 1 2 2 1 2 2 2 ( ) ( ) 1 ( ) = + + + + = τ τ τ τ + − = 1( ) ( ) ( ( ) ( ) 2 2 2 2 1 1 s R Q s A sH s Deg R H s τ = − − + ( ) ( ) 1 1 1 1 2 Q s H s A s Deg τ s − − = = − Deg Deg s Deg A s H s Q s sA H s 2 1 2 1 ( ) 1 1 1 1 ( ) ( ) τ τ + + + + − − = Deg s A R s s Q s sA H s 1 2 1 2 2 2 1 2 1 1 ( ) 1 1 ( ) ( ) τ τ τ τ τ = + + = Deg s A R s Q s sA H s 1 ) ( ) ( ) 1 2 1 1 2 1 τ τ τ + + + + + + = ( ) 1 1( ) ( ) ( ) 1 2 1 2 2 1 2 2 1 2 s A R s R R s Q s H s τ τ τ τ τ
8.14 ( ) 4( 8.0 )1 2 2 4 ( ) 2 + + + = s s s s Y s ( ) 4( 8.0 )1 4 2 ( ) 2 + + + = s s s s Y s 4( 8.0 )1 2 1 ( ) 4 1 2 + + = + s s s Y s 4( 8.0 )1 8 ( ) 2 + + = s s s Y s + 4( 8.0 )1 4 2 s + s + = (step response) + (impulse response) Now,τ = 4 = 2 ; 2ξτ = 8.0 ξ = 2.0 also, 2 2 4 = = τ t impulse response τY(t) = 4*0.63 = 2.52 (from figure) step response = 8*1.15 = 9.2 (from figure) Y(4) = 1.26+9.2 Y(4) =10.46
Q 9.1. Two tank heating process shown in fig. consist of two identical, well stirred tank in series. A flow of heat can enter tank2. At t = 0 , the flow rate of heat to tank2 suddenly increased according to a step function to 1000 Btu/min. and the temp of inlet Ti drops from 60o F to 52o F according to a step function. These changes in heat flow and inlet temp occurs simultaneously. (a) Develop a block diagram that relates the outlet temp of tank2 to inlet temp of tank1 and flow rate to tank2. (b) Obtain an expression for T2’(s) (c) Determine T2(2) and T2(∞) (d) Sketch the response T2’(t) Vs t. Initially Ti = T1 = T2 = 60o F and q=0 W = 250 lb/min Hold up volume of each tank = 5 ft3 Density of the fluid = 50 lb/ft3 Heat Capacity = 1 Btu/lb (o F) Solution: (a) For tank 1 w Ti T1 T2 w q
Input – output = accumulation WC(Ti – To) - WC(T1 – To) = ρ C V dt dT1 -------------------------- (1) At steady state WC(Tis – To) - WC(T1s – To) = 0 ------------------------------------(2) (1) – (2) gives WC(Ti – Tis) - WC(T1 – T1s) = ρ C V dt dT 1 ' WTi’ - WT1 ’ = ρ V dt dT 1 ' Taking Laplace transform WTi(s) = WT1(s) + ρ V s T1(s) Ti s s T s +τ = 1 1 ( ) ( ) 1 , where τ = ρ V / W. From tank 2 q + WC(T1 – To) - WC(T2 – To) = ρ C V dt dT2 -------------------------- (3) At steady state qs + WC(T1s – To) - WC(T2s – To) = 0 ------------------------------------(4) (3) – (4) gives Q ‘ + WC(T1 – T1s) - WC(T2 – T2s) = ρ C V dt dT 2 ' Q ‘ + WCT1 ’ - WCT2 ’ = ρ C V dt dT 2 ' Taking Laplace transform
Q (s) + WC(T1(s) - T2(s)) = ρ C V s T2(s) + + = ( ) ( ) 1 1 ( ) 2 1 T s WC Q s s T s τ , where τ = ρ V / W. (b) τ = 50*5/250 = 1 min WC = 250*1 = 250 Ti(s) = -8/s and Q(s) = 1000/s Now by using above two equations we relate T2 and Ti as below and after taking laplace transform we will get T2(t) ( ) ( ) ( ) ( ) 4( 8 ) 4 1 1 1 1 1 8 1( ) 1 1 ( ) 4 1 8 1( ) 4 ( ) ( ) 1 1 250 ( ) 1 1 ( ) 2 2 2 2 2 2 2 = + − + − + − − + = − + − + = + + + = −t i T t t e s s s s s T s s s T s T s s Q s s T s τ τ (c) T2’(2) = -1.29 T2(2) = T2’(2) + T2s = 60 – 1.29 = 58.71 o F T2’(∞) = -4 T2(∞) = T2’(∞) + T2s = 60 – 4 = 56 o F
Q – 9.2. The two tank heating process shown in fig. consist of two identical , well stirred tanks in series. At steady state Ta = Tb = 60o F. At t = 0 , temp of each stream changes according to a step function Ta’(t) = 10 u(t) Tb’(t) = 20 u(t) (a) Develop a block diagram that relates T2’ , the deviation in the temp of tank2, to Ta’ and Tb’. (b) Obtain an expression for T2’(s) (c) Determine T2(2) W1 = W2 = 250 lb/min V1 = V2 = 10 ft3 ρ1 = ρ2 = 50 lb/ft3 C = 1 Btu/lb (o F) 0.5 0.85 -4 0 T2’(t) t
Solution: (a) For tank1 Ta s s T s 1 1 1 1 ( ) ( ) +τ = , where τ1= ρ V / W1. For tank2 W1C(T1 – To) +W2C(Tb – To) – (W1+W2)C(T2 – To)= ρ C V dt dT2 ------ (1) At steady state W1C(T1s – To) +W2C(Tbs – To) – (W1+W2)C(T2s – To)= 0 -----------------(2) (1) – (2) W1T1 ’ + W2Tb’ - W3T2 ’ = ρ V dt dT 2 ' Taking L.T W1T1(s) + W2Tb(s)- W3T2(s) = ρVs T2(s) W1 Ta T1 T2 W3=W1+W2 Tb W2 W1
[ ( )] 3 2 ( ) 3 1 1 1 ( ) 2 1 T S W T S W W W s T s + b + = τ where τ= ρ V / W3. (b) τ1 = 50*10/250 = 2 min τ = 50*5/250 = 1 min W1/W3 = 1/2 = W2/W3 Ta(s) = 10/s and Tb(s) = 0/s Now by using above two equations we relate T1 and Ta as below and after taking laplace transform we will get T2(t)
( ) ( ) ( ) ( ) 2 2 2 2 2 2 1 2 ( ) 15 5 10 1 2 20 1 15 5 ( ) 1( )(1 2 ) 15 20 ( ) 1 10 1( )(1 2 ) 5 ( ) 1 ( ) 2 1 1( )(1 2 ) ( ) 2 1 ( ) 1 ( ) 2 1 1( ) ( ) 2 1 ( ) t t a b b T t e e s s s T s s s s s T s s s s s s T s s T s s s T s T s s T s s T s T s − − = − − + − + = − + + + = + − + + = + − + + = + − + = (c) T2’(2) = 10.64 o F T2(2) = T2’(2) + T2s = 60 + 10.64 = 70.64 o F Q – 9.3. Heat transfer equipment shown in fig. consist of tow tanks, one nested inside the other. Heat is transferred by convection through the wall of inner tank. 1. Hold up volume of each tank is 1 ft3 2. The cross sectional area for heat transfer is 1 ft2 3. The over all heat transfer coefficient for the flow of heat between the tanks is 10 Btu/(hr)(ft2 )(o F) 4. Heat capacity of fluid in each tank is 2 Btu/(lb)(o F) 5. Density of each fluid is 50 lb/ft3 Initially the temp of feed stream to the outer tank and the contents of the outer tank are equal to 100 o F. Contents of inner tank are initially at 100 o F. the flow of heat to the inner tank (Q) changed according to a step change from 0 to 500 Btu/hr. (a) Obtain an expression for the laplace transform of the temperature of inner tank T(s). (b) Invert T(s) and obtain T for t= 0,5,10, ∞
Solution: (a) For outer tank WC(Ti – To) + hA (T1 – T2)- WC(T2 – To) = ρ C V2 dt dT2 -------------------------- (1) At steady state WC(Tis – To) + hA (T1s – T2s)- WC(T2s – To) = 0 ------------------------------------ (2) (1) – (2) gives WCTi’ + hA (T1’ – T2’)- WCT2’ = ρ C V2 dt dT ' 2 Substituting numerical values 10 Ti’ + 10 ( T1’ – T2’) – 10 T2’ = 50 dt dT ' 2 Taking L.T. Ti(s) + T1(s) – 2T2(s) = 5 s T2(s) Now Ti(s) = 0, since there is no change in temp of feed stream to outer tank. Which gives T s s T s 2 5 1 ( ) ( ) 1 2 + = Q 10 lb/hr T1 T2
For inner tank Q - hA (T1 – T2) = ρ C V1 dt dT1 --------------------- (3) Qs - hA (T1s – T2s) = 0 ------------------------------- (4) (3) – (4) gives Q’ - hA (T1’– T2’ ) = ρ C V1 dt dT ' 1 Taking L.T and putting numerical values Q(s) – 10 T1(s) + 10 T2(s) = 50 s T1(s) Q(s) = 500/s and T2(s) = T1(s) / (2+ 5s) 50 ( ) 2 5 10 ( ) 10 ( ) 500 1 1 1 sT s s T s T s s = + − + + + = − 1 2 5 1 ( ) 5 50 1 s T s s s (25 15 )1 50 2( 5 ) ( ) 1 2 + + + = s s s s T s ( )( ) 50 26.18 50 .3 82 2(2 5 ) ( ) 1 + + + = s s s s T s ( ) ( ) 50 26.18 .5 29 50 .3 82 100 94.71 ( ) 1 + − + = − s s s T s 50 26.18 50 .3 82 ' 1 T (t) 100 - 94.71e - 5.29 e − t − t = and 50 26.18 50 .3 82 1 T (t) 200 - 94.71e - 5.29 e − t − t = For t=0,5,10 and ∞ T(0) = 100 o F T(5) = 134.975 oF T(10) = 155.856 oF T(∞) = 200 oF
Q – 10.1. A pneumatic PI controller has an output pressure of 10 psi, when the set point and pen point are together. The set point and pen point are suddenly changed by 0.5 in (i.e. a step change in error is introduced) and the following data are obtained. Determine the actual gain (psig per inch displacement) and the integral time. Soln: e(s) = -0.5/s for a PI controller Y(s)/e(s) = Kc ( 1 + τI -1/s) Y(s) = -0.5Kc ( 1/s + τI -1/s2 ) Y(t) = -0.5Kc ( 1 + τI -1 t ) At t = 0+ y(t) = 8 Y(t) = 8 – 10 = -2 2=0.5Kc Kc = 4 psig/in At t=20 y(t) = 7 Y(t) = 7-10 = -3 3 = 2 ( 1 + τI -1 20 ) τI = 40 sec Q-10.2. a unit-step change in error is introduced into a PID controller. If KC = 10 , τI = 1 and τD = 0.5. plot the response of the controller P(t) Soln: P(s)/e(s) = KC ( 1 + τD s+ 1/ τIs) For a step change in error Time,sec Psig 0 - 10 0 + 8 20 7 60 5 90 3.5
P(s) = (10/s)(1 + 0.5 s + 1/s ) P(s) = 10/s + 5 + 10/s2 P(t) = 10 + 5 δ(t) + 10 t Q – 10.3. An ideal PD controller has the transfer function P/e = KC ( τD s + 1) An actual PD controller has the transfer function P/e = KC ( τD s + 1) / (( τD/β) s + 1) Where β is a large constant in an industrial controller If a unit-step change in error is introduced into a controller having the second transfer function, show that P(t) = KC ( 1 + A exp(-βt/ τD)) Where A is a function of β which you are to determine. For β = 5 and KC =0.5, plot P(t) Vs t/ τD. As show that β ∞, show that the unit step response approaches that for the ideal controller. Soln: P/e = KC ( τD s + 1) / (( τD/β) s + 1) For a step change, e(s) = 1/s P(s) = KC s( τD s + 1) / (( τD/β) s + 1) 10 15 10(1+t) P(t) t
= + − + β τ β τ s s K D D C 1 1 1 1 P(t) = − + − D t D D C K e τ β β τ β τ 1 1 1 = + − − D t C K e τ β 1 (β )1 So, A = β – 1 P(t) = 0.5 ( 1 + 4 exp(-5t/ τD)) As β ∞ then τD/β 0 and P/e = KC ( τD s + 1) / (( τD/β) s + 1) becomes P/e = KC ( τD s + 1) that of ideal PD controller Q – 10.4. a PID controller is at steady state with an output pressure of a psig. The set point and pen point are initially together. At time t=0, the set point is moved away from 2.5 0.5 P(t) t/τD
the pen point at a rate of 0.5 in/min. the motion of the set point is in the direction of lower readings. If the knob settings are KC = 2 psig/in of pen travel τI = 1.25 min τD = 0.4 min plot output pressure Vs time Soln: Given de/dt = -0.5 in/min s e(s) = -0.5 Y(s)/e(s) = KC ( 1 + τD s+ 1/ τIs) Y(s) = -( 1/s + 1/ τIs 2 + τD ) Y(t) = -( 1 + t/1.25 + 0.4 δ(t) ) Y(t) = y(t) – 9 = - ( 1 + t/1.25 + 0.4 δ(t) ) y(t) = 8 – 0.8 t – 0.4 δ(t) Q – 10.5. The input (e) to a PI controller is shown in the fig. Plot the output of the controller if KC = 2 and τI = 0.5 min 9 8 7.6 10 y(t) t
e(t) = 0.5 ( u(t) - u(t-1) - u(t-2) + u(t-3) ) e(s) = (0.5/s) ( 1 – e-s - e -2s + e-3s ) P(s)/e(s) = KC ( 1 + (1 / τI s) ) = 2 ( 1+ 2/s ) P(s) = ( 1/s + 2/s2 ) (1 – e-s - e-2s + e-3s ) P(t) = 1 + 2t 0 ≤ t < 1 = 2 1 ≤ t < 2 = 5 – 2t 2 ≤ t < 3 = 0 3 ≤ t < ∞ Q – 12.1. Determine the transfer function Y(s)/X(s) for the block diagrams shown. Wxpress the results in terms of Ga, Gb and Gc 0 1 2 3 4 t, min 0.5 e -0.5
Soln. (a) Balances at each node (1) = GaX (2) = (1) – Y = GaX – Y (3) = Gb(2) = Gb(GaX – Y) (4) = (3) + X = Gb(GaX – Y) + X Y = Gc(4) = Gc (Gb(GaX – Y) + X) = GaGbGcX – GbGcY + GcX GbGc Gc GaGb X Y + + = 1 ( )1 (b) Balances at each node (1) = X – (4) (2) = Gb(1) = Gb( X – (4)) (5) = GcX/Ga (3) = Gc(2) = GbGc( X – (4)) (4) = (3) + (5) --------------------------- 5 = GbGc( X – (4)) + GcX/Ga Y = Ga(4) From the fifth equation (4) = GbGcX – GbGc(4) + GcX/Ga ----------- 6 GbGc Ga GaGbGc Gc X 1( ) ( ) )4( + + = From the sixth equation 1( ) ( )1 GbGc GaGb Gc X Y + + = Q – 12.2 Find the transfer function y(s)/X(s) of the system shown
Soln: Balance at each node (1) = X – Y ---------(a) (2) = (1) + (3) ----------(b) (3) = G1(2) where G1 = 1/(τ1s + 1) ----------(c) (4) Y = G2(3) where G2 = 0.5/(τ1s/2 + 1) ----------(d) From (d) and (c) Y = (2)G1G2 = G1G2 (X – Y + (3) ) ----------(e) Also from (b) and (c) (3) = G1((1) + (3)) (3)(1 – 1/(τ1s + 1)) = 1/(τ1s + 1) (3) τ1s = 1 (3) = 1/(τ1s ) = (X – Y) / (τ1s) Substitute this in (e) ( ) X Y s s s Y − + + + = 1 1 1 1 1 )1 2 ( 1)( 5.0 τ τ τ 2 1 1 1 2 2 1 + + = X s s Y τ τ Q – 12.3. For the control system shown determine the transfer function C(s)/R(s)