Solutions_F11_ECH157_HW2

Solutions_F11_ECH157 - Y(s = 1 2 −(1 2 s 2 2 2 2 s s 2 s 22 y(t = ∴ 11 2 − cos 2t sin 2t 22 2 1(1 − cos 2t sin 2t 2 Homework Problem Set#2

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Unformatted text preview: Y(s) = 1 2 − (1 2) s 2 +2 +2 2 s s +2 s + 22 y(t) = ∴ 11 2 − cos 2t + sin 2t 22 2 1 (1 − cos 2t ) + sin 2t 2 Homework Problem Set #2: Solution Key y(t) = Problem 1: 3.11 a) Since convergent and oscillatory behavior does not depend on initial dx 2 (0) dx(0) conditions, assume = = x ( 0) = 0 dt dt 2 Laplace transform of the equation gives s 3 X ( s ) + 2 s 2 X ( s ) + 2 sX ( s ) + X ( s ) = 3 1 3 1 3 s ( s + 1)( s + + j )( s + − j) 22 22 Denominator of [sX(s)] contains complex factors so that x(t) is oscillatory, and denominator vanishes at real values of s= −1 and -½ which are all <0 so that x(t) is convergent. See Fig. S3.11a. 2 s 2 X ( s) − X (s) = s −1 2 2 X (s) = = 2 ( s − 1)( s − 1) ( s − 1) 2 ( s + 1) X (s) = b) 3 s 3 = s ( s + 2 s 2 + 2 s + 1) 3 The denominator contains no complex factors; x(t) is not oscillatory. The denominator vanishes at s=1 ≥0; x(t) is divergent. See Fig. S3.11b. c) s 3 X ( s) + X (s) = X (s) = 1 s +1 2 1 = ( s + 1)( s 3 + 1) 2 1 1 3 1 3 ( s + j )( s − j )( s + 1)( s − + j )( s − − j) 22 22 The denominator contains complex factors; x(t) is oscillatory. The denominator vanishes at real s = 0, ½; x(t) is not convergent. See Fig. S3.11c. 3-13 s 2 X ( s ) + sX ( s ) = 4 s 4 4 =2 s ( s + s ) s ( s + 1) X (s) = 2 The denominator of [sX(s)] contains no complex factors; x(t) is not oscillatory. The denominator of [sX(s)] vanishes at s = 0; x(t) is not convergent. See Fig. S3.11d. 3.5 3 2.5 x(t) 2 1.5 1 0.5 0 -0.5 0 1 2 3 4 5 time 6 7 8 9 10 Figure S3.11a. Simulation of X(s) for case a) 700 600 500 400 x(t) d) 300 200 100 0 0 0.5 1 1.5 2 2.5 time 3 3.5 4 4.5 5 Figure S3.11b. Simulation of X(s) for case b) 3-14 80 60 x(t) 40 20 0 -20 -40 0 1 2 3 4 5 time 6 7 8 9 10 Figure S3.11c. Simulation of X(s) for case c) 18 16 14 12 x (t) 10 8 6 4 2 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 time Figure S3.11d. Simulation of X(s) for case d) 3.12 Since the time function in the solution is not a function of initial conditions, we Laplace Transform with x ( 0) = dx(0) =0 dt τ1τ2s2X(s) + (τ1+τ2)sX(s) + X(s) = KU(s) 3-15 Problem 2: 3.13 dx3 + 4 x = et 3 dt a) d 2 x(0) dx(0) = = x(0) = 0 dt 2 dt with Laplace transform of the equation, s 3 X(s) + 4 X(s) = X (s) = = 1 s −1 1 1 = 3 ( s − 1)( s + 4) ( s − 1)( s + 1.59)( s − 0.79 + 1.37 j )( s − 0.79 − 1.37 j ) α 3 + jβ 3 α 3 − jβ 3 α1 α2 + + + s − 1 s + 1.59 s − 0.79 + 1.37 j s − 0.79 − 1.37 j α1 = α2 = 1 ( s + 4) = 3 s =1 1 5 1 ( s − 1)( s − 0.79 + 1.37 j )( s − 0.79 − 1.37 j ) α 3 + jβ 3 = 1 ( s − 1)( s + 1.59)( s − 0.79 − 1.37 j ) =− s = −1.59 1 19.6 = −0.74 − 0.59 j s = 0.79 −1.37 j 1 1 − − 0.074 − 0.059 j − 0.074 + 0.059 j X(s) = 5 + 19.6 + + s − 1 s + 1.59 s − 0.79 + 1.37 j s − 0.79 − 1.37 j 1 1 −1.59t x(t ) = e t − e − 2e 0.79t (0.074 cos 1.37t + 0.059 sin 1.37t ) 5 19.6 dx − 12 x = sin 3t with x(0) = 0 dt b) sX (s) − 12 X(s) = X (s) = = 3 s +9 2 3 3 = ( s + 9)( s − 12) ( s + 3 j )( s − 3 j )( s − 12) 2 α3 α 1 + j β1 α 1 − j β 1 + + s+3j s−3j s − 12 3-17 α 1 + j β1 = α3 = = s = −3 j 3 1 4 =− − j − 18 + 72 j 102 102 3 1 = ( s + 9 ) s =12 51 2 − X (s) = x(t ) = − c) 3 ( s − 3 j )( s − 12) 1 4 1 4 1 − j− + j 102 102 + 102 102 + 51 s+3j s −3j s − 12 1 1 (cos 3t + 4 sin 3t ) + e12 t 51 51 d2x dx + 6 + 25 x = e −t 2 dt dt dx(0) = x(0) = 0 dt with s 2 X ( s ) + 6 sX ( s ) + 25X ( s ) + X ( s ) = X ( s) = α1 = 1 or s +1 X( s ) = 1 ( s + 1 )( s + 6 s + 25 ) 2 α α + β2 j α 2 − β2 j 1 = 1+ 2 + ( s + 1)( s + 3 + 4 j )( s + 3 − 4 j ) s + 1 s + 3 + 4 j s + 3 − 4 j 1 ( s + 6 s + 25) = 2 α 2 + jβ 2 = s = −1 1 20 1 ( s + 1)( s + 3 − 4 j ) =− s = −3− 4 j 1 1 − j 40 80 1 1 1 1 1 −− j− − j + X ( s ) = 20 + 40 80 + 40 80 s +1 s + 3 + 4 j s + 3− 4 j x(t ) = d) 1 −t 1 1 e − e −3t ( cos 4t + sin 4t ) 20 20 40 Laplace transforming (assuming initial conditions = 0, since they do not affect results) sY1(s) + Y2(s) = X1(s) (1) sY2(s) – 2Y1(s) + 3 Y2(s) = X2(s) (2) 3-18 From (2), (s+3) Y2(s) = X2(s) + 2Y1(s) Y2 ( s ) = 1 2 X 2 (s) + Y1 ( s ) s+3 s+3 Substitute in Eq.1 sY1(s) + 1 2 X 2 (s) + Y1 ( s ) = X1(s) s+3 s+3 We neglect X2(s) since it is equal to zero. [s(s + 3) + 2]Y1 (s) = ( s + 3) X 1 (s) ( s 2 + 3s + 2)Y1 ( s ) = ( s + 3) X 1 ( s ) Y1 ( s ) = s+3 s+3 X 1 (s) = X 1 (s) ( s + 1)( s + 2) s + 3s + 2 2 1 s +1 s+3 A B C Y1 ( s ) = = + + 2 2 ( s + 1) ( s + 2) ( s + 1) ( s + 1) s+2 Now if x1(t) = e-t then X1(s) = ∴ so that y1(t) will contain e-t/τ , te-t/τ, e–2t functions of time. For Y2(s) Y2 ( s ) = 2 A B C = + + 2 s+2 ( s + 1) ( s + 2) ( s + 1) ( s + 1) 2 so that y2(t) will contain the same functions of time as y1(t) (although different coefficients). 3-19 But it is not possible to get an expression for T ′( s ) / Tc′( s ) from (2) due to the presence of cA in (2). Thus the proposed approach is not feasible because the CSTR is an interacting system. Better approach: After linearization etc., solve for T ′( s ) from (1) and substitute into the ′ linearized version of (2). Then rearrange to obtain the desired, C A ( s ) / Tc′( s ) (See Section 4.3) Problem 3: 4.7 a) The assumption that H is constant is redundant. For equimolal overflow, L0 = L1 = L , V1 = V2 = V dH = L0 + V2 − L1 − V1 = 0 dt , i.e., H is constant. The simplified stage concentration model becomes dx1 = L( x0 − x1 ) + V ( y 2 − y1 ) dt y1 = a0 + a1x1 + a2x12 +a3x13 (1) H b) (2) Let the right-hand side of Eq. 1 be f(L, x0, x1, V, y1, y2) H ∂f ∂f dx1 ∂f = f ( L, x0 , x1 , V , y1 , y 2 ) = L ′ + ′ ∂x x 0 + ∂x x1 ′ dt ∂L s 1 s 0 s ∂f ′ ∂f ′ ∂f ′ + V + ∂y y1 + ∂y y 2 ∂V s 1 s 2 s Substituting for the partial derivatives and noting that H ′ dx1 dx1 = dt dt ′ dx1 ′ ′ ′ ′ = ( x0 − x1 ) L ′ + L x0 − L x1 + ( y 2 − y1 )V ′ + V y 2 − V y1 dt 4-6 (3) Similarly, ∂g 2 ′ ′ y1 = g ( x1 ) = ∂x x1 = (a1 + 2a 2 x1 + 3a 3 x1 ) x1 ′ 1 s c) (4) For constant liquid and vapor flow rates, L ′ = V ′ = 0 Taking Laplace transform of Eqs. 3 and 4, ′ ′ HsX 1′ ( s ) = L X 0 ( s ) − L X 1 ( s ) + V Y2′ ( s ) − V Y1′( s ) (5) ′ Y1′( s ) = (a1 + 2a 2 x1 + 3a3 x1 ) X 1 ( s ) (6) 2 From Eqs. 5 and 6, the desired transfer functions are L τ X 1′ ( s ) =H ′ X 0 ( s ) τs + 1 Y1′( s ) = ′ X 0 ( s) Y1′( s ) = Y2′( s ) V τ ′ X 1 (s) =H Y2′ ( s ) τs + 1 , (a1 + 2a 2 x1 + 3a 3 x1 ) 2 τs + 1 (a1 + 2a 2 x1 + 3a 3 x1 ) 2 τs + 1 L τ H V τ H where τ= H L + V (a1 + 2a 2 x1 + 3a3 x1 ) 2 4.8 From material balance, d (ρAh) = wi − Rh1.5 dt dh 1 R 1.5 = wi − h dt ρA ρA 4-7 0 0 0.72 0 C= 0.72 0 0 0 0.72 0 D= 0 0 Applying the MATLAB function ss2tf , the transfer functions are: Y1′( s ) 0.6343s 2 + 1.4881s + 0.6560 =3 Y f′ ( s ) s + 3.5190 s 2 + 3.443s + 0.8123 Y2′( s ) 0.4022 s + 0.4717 =3 Y f′ ( s ) s + 3.5190s 2 + 3.443s + 0.8123 Y2′( s ) 0.2550 =3 Y f′ ( s ) s + 3.5190s 2 + 3.443s + 0.8123 Problem 4: 4.17 Dynamic model: dX = µ( S ) X − DX dt dS = −µ ( S ) X / Y X / S − D ( S f − S ) dt Linearization of non-linear terms: (reference point = steady state point) 1. f 1 ( S , X ) = µ( S ) X = µm S X Ks + S f1 (S , X ) ≈ f1 (S , X ) + ∂f1 ∂S (S − S ) + S ,X ∂f 1 ∂X (X − X ) S ,X Putting into deviation form, f 1 ( S ′, X ′) ≈ ∂f 1 ∂S S′ + S ,X ∂f 1 ∂X S ,X µ (K + S ) − µ m S µ m S X′ = m s X S '+ K +S (K s + S ) 2 s 4-20 X ' Substituting the numerical values for µ m , K s , S and X then: 11.25 f 1 ( S ′, X ′) ≈ 0.113S ' + 101X ' 0. 2. f 2 ( D, S , S f ) = D( S f − S ) f 2 ( D' , S ′, S ′f ) ≈ ∂f 2 ∂f D' + 2 ∂D D , S , S f ∂S S' + D ,S ,S f ∂f 2 ∂S f S ′f D ,S ,S f f 2 ( D' , S ′, S ′f ) ≈ ( S f − S ) D' − D S ' + D S ′f f 2 ( D ' , S ′, S ′f ) ≈ 9 D ' − 0.1S ' + 0.1S f ' 3. f 3 ( D, X ) = DX f 3 ( D' , X ' ) ≈ D' X + X ' D = 2.25 D' + 0.1X ' Returning to the dynamic equation: putting them into deviation form by including the linearized terms: dX dX ' = 11.25 S ' + 101X ' − 2.25 D' − 0.1X ' 0.113 0. dt 11.25 10 dS ' −0.113 0.1 = S'− X ' − 9 D ' + 0.1S ' − 0.1S ′f dt 0.5 0.5 Rearranging: dX ' 11.25 = 0.113S ' − 2.25 D' +9.9X' dt dS ' = −22.4 S ' − 20 X ' − 9 D ' − 0.1S ′f 0.126 0.2 dt Laplace Transforming: 11.25 sX ' ( s ) = 0.113S ' ( s ) − 2.25 D ' ( s ) +9.9X'(s) sS '( s ) = −22.4 S '( s ) − 20 X '( s ) − 9 D '( s ) − 0.1S ′f ( s ) 0.126 0.2 4-21 Then, 2.25 D' ( s) s-9.9 s-9.9 −20 0.2 9 0.1 S '( s ) = X '( s ) − D '( s ) − S ′f ( s ) s + 22.4 0.126 s + 22.4 0.126 s + 22.4 0.126 X ' (s) = 11.25 0.113 S ' (s) − Substitution of the expression of S'(s) into the expression of X'(s) to get the or following transfer function: 0.0226 X ' ( s ) 1 + = s ( s + 0.126) =− 1.017 0.0113 2.25 D '( s ) − S ′f ( s ) − D′( s ) s + 0.126 s + 0.126 s Therefore, 1.3005 X ' (s) −151.65 − 2.25s =2 D' ( s ) s + 0.126 s + 0.0226 12.5 3.24 Therefore, the steady state gain is -151.65/3.24=-46.8056. 4-22 Practice Problems ...
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This note was uploaded on 01/28/2012 for the course ECH 157 taught by Professor Palagozu during the Fall '08 term at UC Davis.

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