KING FAHD UNIVERSITY CHEMICAL ENGINEERING COURSE NOTES (Process Control)-sol-hw-4

# KING FAHD UNIVERSITY CHEMICAL ENGINEERING COURSE NOTES (Process Control)-sol-hw-4

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Unformatted text preview: 4.2 a) b) c) 5 10 10 s (10 s + 1) From the Final Value Theorem, y(t) = 10 when t Y (s) = y(t) = 10(1-e-t/10) , then y(10) = 6.32 = 63.2% of the final value. d) e) 5 (1 - e - s ) Y (s) = (10s + 1) s From the Final Value Theorem, y(t) = 0 when t 5 1 (10 s + 1) From the Final Value Theorem, y(t)= 0 when t Y (s) = Y (s) = 5 6 2 (10 s + 1) ( s + 9) f) g) then y(t) = 0.33e-0.1t - 0.33cos(3t) + 0.011sin(3t) The sinusoidal input produces a sinusoidal output and y(t) does not have a limit when t. By using Simulink-MATLAB, above solutions can be verified: 10 0.5 9 0.45 8 0.4 7 0.35 6 0.3 y(t) y(t) 5 0.25 4 0.2 3 0.15 2 0.1 1 0.05 0 0 5 10 15 20 25 time 30 35 40 45 50 0 0 5 10 15 20 25 time 30 35 40 45 50 Fig S4.2a. Output for part c) and d) 0.7 5 Fig S4.2b. Output for part e) 0.6 4.5 0.5 4 0.4 3.5 0.3 3 y(t) 0.2 y(t) 2.5 0.1 2 1.5 0 1 -0.1 0.5 -0.2 0 -0.3 0 5 10 15 20 25 time 30 35 40 45 50 0 2 4 6 8 10 time 12 14 16 18 20 Fig S4.2c. Output for part f) Fig S4.2d. Output for part g) 4.7 a) The assumption that H is constant is redundant. For equimolal overflow, L0 = L1 = L , V1 = V2 = V , i.e., H is constant. dH = L0 + V2 - L1 - V1 = 0 dt The simplified stage concentration model becomes H dx1 = L( x0 - x1 ) + V ( y 2 - y1 ) dt y1 = a0 + a1x1 + a2x12 +a3x13 (1) (2) b) Let the right-hand side of Eq. 1 be f(L, x0, x1, V, y1, y2) f f dx1 f x0 + x1 = f ( L, x0 , x1 , V , y1 , y 2 ) = L + x dt x1 s L s 0 s H 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 (3) 4-6 Similarly, g 2 y1 = g ( x1 ) = x x1 = (a1 + 2a 2 x1 + 3a 3 x1 ) x1 1 s c) 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 ) (4) (5) (6) Y1( s ) = (a1 + 2a 2 x1 + 3a3 x1 ) X 1 ( s ) 2 From Eqs. 5 and 6, the desired transfer functions are L X 1 ( s ) = H X 0 ( s ) s + 1 Y1( s ) = X 0 ( s) Y1( s ) = Y2( s ) 2 , L H V X 1 (s) = H Y2 ( s ) s + 1 (a1 + 2a 2 x1 + 3a 3 x1 ) s + 1 2 (a1 + 2a 2 x1 + 3a 3 x1 ) s + 1 V H where = H L + V (a1 + 2a 2 x1 + 3a3 x1 ) 2 4.13 a) The nonlinear dynamic model for the tank is: dh 1 = qi - Cv h dt ( D - h)h ( ) (1) (corrected nonlinear ODE; model in first printing of book is incorrect) To linearize Eq. 1 about the operating point (h = h , qi = qi ) , let f = Then, f f f (h, qi ) h + qi h s qi s where qi - Cv h ( D - h ) h 4-14 f 1 = qi s ( D - h )h 1 Cv 1 -D + 2h f =- + qi - Cv h ( ( D - h ) h ) 2 2 h ( D - h ) h h s ( ) Notice that the second term of last partial derivative is zero from the steady-state relation, and the term ( D - h )h is finite for all 0 < h < D . Consequently, the linearized model of the process, after substitution of deviation variables is, dh 1 Cv 1 1 = - h + qi dt 2 h ( D - h )h ( D - h ) h Since qi = Cv h dh 1 qi 1 1 = - h + ( D - h )h qi dt 2 h ( D - h )h or where 1q qi 1 a = - i = -V 2 h ( D - h ) h o , 1 b= ( D - h ) h dh = ah + bqi dt Vo = volume at the initial steady state b) Taking Laplace transform and rearranging s h( s ) = ah( s ) + bqi( s ) Therefore h( s ) b = qi( s ) ( s - a ) or h( s ) (-b / a ) = qi( s ) (-1/ a ) s + 1 Notice that the time constant is equal to the residence time at the initial steady state. 4-15 ...
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