SOLUTIONS MANUAL FOR SELECTED PROBLEMS IN PROCESS SYSTEMS ANALYSIS AND CONTROL DONALD R COUGHANOWR COMPILED BY M.N GOPINATH BTech.,(Chem) CATCH ME AT gopinathchemical@gmail.com Disclaimer: This work is just a compilation from various sources believed to be reliable and I am not responsible for any errors CONTENTS PART 1: SOLUTIONS FOR SELECTED PROBLEMS PART2: LIST OF USEFUL BOOKS PART3: USEFUL WEBSITES PART 1.1 Draw a block diagram for the control system generated when a human being steers an automobile 1.2 From the given figure specify the devices Solution: Inversion by partial fractions: 3.1(a) dx dx + + x = x ( 0) = x ' ( 0) = dt dt  dx  L   = s X ( s ) − sx(0) − x ' (0)  dt   dx  L   = s X ( s ) − x ( 0)  dt  L(x) = X(s) L{1} = 1/s s X ( s ) − sx(0) − x ' (0) + s X ( s ) − x (0) + X ( s ) = = ( s + s + 1) X ( s ) = X ( s) = s s s( s + s + 1) Now, applying partial fractions splitting, we get X ( s) = X ( s) = s +1 − s ( s + s + 1) s +1    − −    2 2 s        3 1  3  s +  +    s +  +  2        L−1 ( X ( s )) = − e X (t ) = − e b) − t − t Cos −2t t− e sin t 2    3  Cos t +   Sin    t     dx dx + + x = x ( 0) = x ' ( 0) = dt dt when the initial conditions are zero, the transformed equation is ( s + s + 1) X ( s ) = s X ( s) = s( s + s + 1) A Bs + C = + s ( s + s + 1) s s + s + = A(s2 + 2s +1) + Bs2 + Cs = A + B(by equating the co − effecient of s ) = A + C (by equating the co − effecients of s ) = A(by equating the co − effecients of const) A+ B = B = −1 C = −2 A A = 1, B = −1, C = −2 s+2 − s s + 2s +  (s + 1) +  L−1{ X ( s )} = L−1  − (s + 1)2  s X ( s) =  1  { X (t )} = − L−1  +   s + (s + 1)  { X (t )} = − e − t (1 + t ) dx dx 3.1 C + + x = x ( 0) = x ' ( 0) = dt dt by Applying laplace transforms, we get = ( s + 3s + 1) X ( s ) = X ( s) = s( s + 3s + 1) s X ( s) = A Bs + C + s s + 3s + 1 = A( s + 3s + 1) + Bs + Cs = A + B(by equating the co − effecient of s ) = A + C (by equating the co − effecients of s) = A(by equating the co − effecients of const) A+ B = B = −1 C = −3 A = −3 A = 1, B = −1, C = −3 s+3  1 L−1{ X ( s )} = L−1  −   s s + 3s +    −1 −1  L { X ( s )} = L  − s  s +      1 L−1{ X ( s )} = L−1  − s  s +    X (t ) = − e − 3t (Cos s+3 3   −       5     3    s+ 5  − 2 3  3      + − s    −       5t + t sinh 2 3.2(a) dx d x + = Cos t; x (0) = x ' (0) = x ''' (0) = dt dt x11 (0) =     5     Applying Laplace transforms, we get s X ( s ) − s x (0) − s x1 (0) − sx '' (0) − x ''' (0) + s X ( s ) − s x (0) − sx ' (0) − x '' (0) = X ( s ) ( s + s ) − ( s + 1) = s s +1 s s +1  s  + 1) + ( s + 1)  s + s X ( s ) =  (  s +1  s + s + s + s + s + s + 2s + = = s ( s + 1)( s + 1) s ( s + 1)( s + 1) s + s + 2s + A B C D Es + F = + + 3+ + s ( s + 1)( s + 1) s s s s +1 s +1 s + s + s + = As ( s + 1)( s + 1) + Bs( s + 1)( s + 1) + c( s + 1)( s + 1) + Ds ( s + 1) + ( Es + F ) s ( s + 1) A+B+E=0 equating the co-efficient of s5 A+B+E+F=0 equating the co-efficient of s4 A+B+C+D+F=0 equating the co-efficient of s3 A+B+C=0 equating the co-efficient of s2 B+C=2 equating the co-efficient of s A+B+E=0 equating the co-efficient of s2 C=1equating the co-efficient constant C=1 -B=-C+2=1 A=1-B-C=-1 D+F=0 E+F=0D+E=1 D-E=0 2D=1 A=-1; B=1; C=1 D=1/2; E=1/2; F =-1/2  − 1 1 / / 2( s − 1)  + L−1{( s)} = L−1  + + +  s s +1 s2 +  s s 1 / / 2( s − 1)  −1 L−1 {X ( s )} = L−1  + + + +  s s s +1 s2 +   s {X (t )} = −1 + t + t + e −t + Cos t − S int 2 2 d q dq + = t + 2t q(0) = 4; q1 (0) = −2 dt dt applying laplace transforms,we get s 2Q ( s ) − sq(0) − q ' (0) + sQ (( s ) − q(0) = Q ( s )( s + s ) − s + − = 1   + 1 s2  s  2( s + 1) + ( s + 2) s Q( s) = ( s + s) = s + + s + 2s s ( s + 1) 2*3   + Q ( s ) = 4 +  s +  s( s + 1) s ( s + 1) L−1 (Q ( s )) = q(t ) = 4e −t + 2(1 − e −t ) + t 3 therefore q(t ) = + t3 + e −t 2 + s s C= 1 −2 = + s( s + 1) s s + C (t ) = − e − t For KC = 1, response of -ve feed back is 1 − C= = 2+ s ( s + 2) s s + 1 C (t ) = − e −2 t 2 response of +ve feed back is C= s2 C (t ) = t 14.1 Write the characteristics equation and construct Routh array for the control system shown it is stable for (1) Kc= 9.5,(ii) KC =11; (iii) Kc= 12 Characteristics equation Kc =0 ( s + 1)( s + 2)( s + 3) or ( s + 1)(s + 2)(s + 3) + Kc = 1+ ( s + s + 11s + (6 + Kc) = s + s + 11s + (6 + Kc) = Routh array s3 s s 11 (6 + Kc) 6(1 + Kc) For Kc=9.5 = 10-(Kc)= 10-9.5=0.5>0 therefore stable For Kc=11 = 10-(Kc)= 10-11=-1 ⇒ a > Similarly aj a0 = (−1) j ( sum of aoll possible products of j roots) if j = even (−1) j is and the sum is > so a j / a > if j = odd (−1) j is (−1) and the sum is < so aj a0 is again > in both case a j / a > so a j > 0( for j = 1, .n) 14.5 Prove that the converse statement of the problem 14.4 that an unstable root implies that one or more co-efficient will be negative or zero is untrue for all co-efficient ,n>2 Let the converse be true, always Never if we give a counter example we can contradict Routh array s + s + 2s + s3 s2 s −1 s0 System is unstable even when all the coefficient are greater than 0; hence a contradiction, 14.6 Deduce an expression for Routh criterion that will detect the Presence of roots with real parts greater than σ for any rectified σ >0 Characteristic equation a x n + a1 x n −1 + + a n = Routh criteria determines if for any root, real part > Now if we replace x by X such that .x + σ =X Characteristic equation becomes a ( X − σ ) n + a1 ( X − σ ) n −1 + + a n = Hence if we apply Routh criteria, We will actually be looking for roots with real part > σ rather than >0 a x n + a1 x n −1 + a x n −2 + a n = Routh criterion detects if any root α j is greater than zero Is there any x = α , α , .,α j , α n > − − − − − (1) Now we want to detect any root α j > −σ α j> From(1) x = α , α , α j , α n , > implies is there any x = α1 > x = α2 > x =αj >0 x = αn > add σ on both sides is there any x + σ = α1 + σ > x +σ = α2 +σ > x + σ = α j + σ > x +σ = αn +σ > so, Let X = x + σ and apply Routh criteria to det ect any α j + σ > or α j > −σ 14.7 Show that any complex no S1 satisfying S < 1, yields a value of Z= 1+ s that satisfies Re(Z)>0, 1− s Let S=x+iy, x2 + y2 < Z= 1+ s 1− s (1 + x ) + iy (1 − x) + iy (1 − x) − iy (1 − x) + iy (1 − x + (1 = x + − x)iy − y ) + x − 2x + y = − ( x + y ) + 2iy − 2x + (x + y ) Re(Z ) = − (x2 + y ) − 2x + (x + y ) if Re( z ) > then − ( x + y ) > and − 2x + (x + y ) > we have x + y is true therefore x + y < Now + (x + y ) − 2x if x = −1& y = then it is if x = & y = then it is 0 < (1 + ( x + y ) − x) < Re( z ) > example: if s = (0.5 + i0.5) the system is unstable due to the real part   L−1    s − (0.5 + i 0.5)    L−1  = e 0.5t (Cos 0.5t + Sin 0.5t )   s − (0.5 + i0.5)  14.8 For the output C to be stable, we analyze the characteristic equation of the system 1+ 1 × (τ s + 1) = τ I s (τ s + 1)(τ s + 1) τ I s(τ s 2τ + τ s + τ s + 1) + τ s + = τ I τ 1τ s + τ I (τ + τ ) s + (τ I + τ ) s + = Routh Array s3 τ I τ 1τ s2 τ I (τ + τ ) α s s0 α = τ I + τ3 1 τ I (τ + τ )(τ I + τ ) − τ I τ 1τ τ I (τ + τ ) Now (1) τ I τ 2τ > Since τ & τ τ > 0;τ > are process time constant they are definitely +ve (2) τ I (τ + τ ) > (3) α > ⇒ τ I (τ + τ )(τ I + τ ) > τ I τ 1τ τ 1τ I + τ 1τ + τ 2τ I + τ 2τ − τ 1τ > τ I (τ + τ ) > τ 1τ − τ (τ + τ ) τI > τ 1τ −τ τ1 +τ also τ I > 14.9 In the control system shown in fig find the value of Kc for which the system is on the verge of the instability The controller is replaced by a PD controller, for which the transfer function is Kc(1+s) if Kc = 10, determine the range for which the system is stable Characteristics equation 1+ Kc =0 ( s + 1)( s + 2)( s + 3) or ( s + 1)(s + 2)(s + 3) + Kc = ( s + 3s + 2)(s + 3) + Kc = s + s + 11s + (6 + Kc) = Routh array s3 s2 + Kc s  + Kc  3− )   For verge of instability  + Kc  ) 3=   Kc = Characteristics equation + 10(1 + kcs) =0 ( s + 1)3 s + 3s + s (3 + 10 Kcs) + 11 = Routh Array s3 + 10τ D s2 11 s 3(3 + 10τ s ) > 11 for vege 30τ S > τ D > / 30 14.10 (a) Write the characteristics equation for the central system shown (b) Use the routh criteria to determine if the system is stable for Kc=4 © Determine the ultimate value of Kc for which the system is unstable (a) characteristics equation  s +   1 + kc =0     2s =  ( s + 1) ( s + s )(2 s + 1) + kc( s + 2) = 2s + 3s + (1 + kc) s + 2kc = s + 3s + 3s + (1 + kc) = Kc=4Routh array s3 s s − 1/ not stable 3(1 + kc) − 4kc =0 3 − Kc = 0; Kc = For verge of instability 14.11 for the control shown, the characteristics equation is s + s + s + s + (1 + k ) = (a) determine value of k above which the system is unstable (b) Determine the value of k for which the two of the roots are on the imaginary axis, and determine the values of these imaginary roots and remaining roots are real s + s + s + 4s + (1 + k ) = s4 6(1 + k ) s3 4 s s 1+ k 4 − (1 + k ) 1+ k For the system to be unstable  1 + k  41 −    <    1< 1+ k k>4 1+ k < k < −1 k > −1 The system is stable at -1