Process Dynamics and Control Solutions

1234567898

3.1 a)

[

1e

− bt

∞

]

sin ωt = ∫ e

− bt

0

∞

sin ωt e dt = ∫ sin ωt e − ( s + b )t dt − st

0

 [− (s + b) sin ωt − ω cos ωt ] = e − ( s + b ) t  ( s + b ) 2 + ω2  0 ω = ( s + b) 2 + ω2

∞

b)

[

1 e

− bt

∞

]

cos ωt = ∫ e 0

− bt

∞

cos ωt e dt = ∫ cos ωt e − ( s + b )t dt − st

0

 [− (s + b) cos ωt + ω sin ωt ] = e − ( s + b ) t  ( s + b) 2 + ω2  0 s+b = ( s + b) 2 + ω2

∞

3.2

a)

The Laplace transform provided is Y ( s) =

4 s + 3s + 4 s 2 + 6 s + 4 4

3

We also know that only sin ωt is an input, where ω = X ( s) =

ω 2 = 2 s +ω s2 + 2 2

( )

2

=

2 . Then

2 s +2 2

Since Y(s) = D-1(s) X(s) where D(s) is the characteristic polynomial (when all initial conditions are zero), Solution Manual for Process Dynamics and Control, 2nd edition, Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp.

3-1

Y (s) =

2 2 2 2 ( s + 3s + 2) ( s + 2) 2

and the original ode was

d2y dy + 3 + 2 y = 2 2 sin 2t 2 dt dt

with y ′(0) = y (0) = 0

b)

This is a unique result.

c)

The solution arguments can be found from Y (s) =

2 2 2 ( s + 1)( s + 2) + ( s 2 + 2)

which in partial fraction form is Y (s) =

α1 α a s + a2 + 2 + 12 s +1 s + 2 s +2

Thus the solution will contain four functions of time e-t

,

e-2t ,

sin 2 t , cos 2 t

3.3

a)

Pulse width is obtained when x(t) = 0 Since x(t) = h – at tω : h − atω = 0

or

tω = h/a

b) h slope = -a

slope = a

x(t)

x(t)

slope = -a

x(t) = hS(t) – atS(t) + a(t -tω) S(t-tω) 3-2

c)

h a ae − stω h e − stω − 1 X ( s) = − 2 + 2 = + s s s s s2

d)

Area under pulse = h tω/2

a)

f(t) = 5 S(t) – 4 S(t-2) – S(t- 6) 1 F (s ) = = ( 5 - 4e -2s - e -6s ) s

3.4

b)

x(t) x1

a a

x4 tr

2tr

3tr

-a

-a

x2

x3

x(t) = x1(t) + x2(t) + x3(t) + x4(t) = at – a(t − tr)S(t − tr) − a(t −2tr)S(t − 2tr) + a(t − 3tr)S(t − 3tr) following Eq. 3-101. Thus X(s) =

[

a 1 − e −tr s − e − 2tr s + e −3tr s 2 s

]

by utilizing the Real Translation Theorem Eq. 3-104.

3-3

3.5

55 55 t S(t) – (t-30) S(t-30) 30 30 20 55 1 55 1 −30 s 20 55 1 T (s) = + − e = + 1 − e −30 s 2 2 2 s 30 s 30 s s 30 s

T(t) = 20 S(t) +

(

)

3.6 a)

X ( s) =

α1 =

α2 =

α3 =

b)

α α α s ( s + 1) = 1 + 2 + 3 ( s + 2)( s + 3)( s + 4) s + 2 s + 3 s + 4

s ( s + 1) ( s + 3)( s + 4) s ( s + 1) ( s + 2)( s + 4) s ( s + 1) ( s + 2)( s + 3)

=1 s = −2

= −6 s = −3

=6 s = −4

X (s) =

1 6 6 − + s+2 s+3 s+4

X (s) =

s +1 s +1 = 2 ( s + 2)( s + 3)( s + 4) ( s + 2)( s + 3)( s + 2 j )( s − 2 j )

X ( s) =

α + jβ 3 α 3 − j β 3 α1 α + 2 + 3 + s+2 s+3 s+2j s+2j

α1 = α2 =

s +1 ( s + 3)( s 2 + 4) s +1 ( s + 2)( s 2 + 4)

α 3 + jβ 3 =

=− s = −2

= s = −3

x(t ) = e−2t − 6e −3t + 6e −4t

and

1 8

2 13

s +1 ( s + 2)( s + 3)( s − 2 j )

3-4

= s = −2 j

1− 2 j − 3 + 11 j = − 40 − 8 j 208

1 2  −3   11  x(t ) = − e − 2t + e −3t + 2  cos 2t + 2  sin 2t 8 13  208   208  1 2 3 11 = − e − 2t + e −3t − cos 2t + sin 2t 8 13 104 104

c)

X ( s) =

α α2 s+4 = 1 + 2 s + 1 ( s + 1) 2 ( s + 1)

(1)

α 2 = ( s + 4) s =−1 = 3 In Eq. 1, substitute any s≠-1 to determine α1. Arbitrarily using s=0, Eq. 1 gives 4 α1 3 = + 1 12 12

or

α1 = 1

1 3 + and x( t ) = e− t + 3te − t 2 s + 1 ( s + 1) 1 1 1 X (s) = 2 = = 2 2 2 s + s +1  1  3 (s + b ) + ω s +  + 2 4  1 3 where b = and ω = 2 2 X (s) =

d)

t

1 2 −2 3 x(t ) = e −bt sin ω t = e sin t ω 2 3 e)

X(s) =

s +1 e −0.5 s s ( s + 2)( s + 3)

To invert, we first ignore the time delay term. Using the Heaviside expansion with the partial fraction expansion, Xˆ ( s ) =

s +1 A B C = + + s ( s + 2)( s + 3) s s + 2 s + 3

Multiply by s and let s → 0

3-5

1 1 = (2)(3) 6 Multiply by (s+2) and let s→ −2 A=

B=

− 2 +1 −1 1 = = (−2)(−2 + 3) (−2)(1) 2

Multiply by (s+3) and let s→-3 C=

− 3 +1 −2 2 = =− (−3)(−3 + 2) (−3)(−1) 3

Then 1 6 1 2 −2 3 Xˆ ( s ) = + + s s+2 s+3 xˆ (t ) =

1 1 − 2 t 2 −3t + e − e 6 2 3

Imposing shift theorem x(t ) = xˆ (t − 0.5) =

1 1 − 2 (t −0.5) 2 −3( t − 0.5) + e − e 6 2 3

for t ≥ 0.5

3.7 a)

Y (s) =

α 6( s + 1) 6 α = 2 = 1 + 22 2 s s ( s + 1) s s

6 s2 6 Y (s) = 2 s α2 = s2

b)

Y (s) =

=6

α1 = 0

s =0

12( s + 2) α1 α 2 s + α 3 = + 2 s s ( s 2 + 9) s +9

3-6

Multiplying both sides by s(s2+9) 12( s + 2) = α 1 ( s 2 + 9) + (α 2 s + α 3 )( s )

or

12 s + 24 = (α 1 + α 2 ) s + α 3 s + 9α 1 2

Equating coefficients of like powers of s, s2: α1 + α2 = 0 s1: α3 = 12 0 s : 9α1 = 24 Solving simultaneously, −8 3  8   − s + 12  8 1  3  Y (s) = + 3s s2 + 9 α1 =

c)

α2 =

α3 =

α2 =

,

,

( s + 2)( s + 3) ( s + 5)( s + 6) ( s + 2)( s + 3) ( s + 4)( s + 6) ( s + 2)( s + 3) ( s + 4)( s + 5)

=1 s = −4

= −6 s = −5

=6 s = −6

Y (s) =

1 6 6 − + s+4 s+5 s+6

Y (s) =

[(s + 1)

=

α 3 = 12

α α α ( s + 2)( s + 3) = 1 + 2 + 3 ( s + 4)( s + 5)( s + 6) s + 4 s + 5 s + 6

Y (s) =

α1 =

d)

8 3

1 2

]

2

+ 1 ( s + 2)

=

1 ( s + 2 s + 2) 2 ( s + 2) 2

α s + α4 α α1 s + α 2 + 2 3 + 5 2 2 s+2 s + 2s + 2 ( s + 2s + 2)

3-7

Multiplying both sides by ( s 2 + 2s + 2) 2 ( s + 2) gives 1 = α1s4 + 4α1s3 + 6α1s2 +4α1s + α2s3 +4α2s2 +6α2s +4α2 + α3s2 +2α3s + α4s + 2α4 + α5s4 + 4α5s3 + 8α5s2 + 8α5s + 4α5 Equating coefficients of like power of s, s4 : α1 + α5 = 0 s3 : 4α1 + α2 + 4α5 = 0 s2 : 6α1 + 4α2 + α3 + 8α5 = 0 s1 : 4α1 + 6α2 + 2α3 + α4 + 8α5 = 0 s0 : 4α2 + 2α4 + 4α5 = 1 Solving simultaneously: α1 = -1/4 Y (s) =

α2 = 0

α3=-1/2

α4=0

− 1 / 4s − 1 / 2s 1/ 4 + 2 + 2 s+2 s + 2 s + 2 ( s + 2 s + 2) 2

3.8

a)

From Eq. 3-100 t  1 1  ∫ f (t * )dt *  = F ( s ) 0  s  t −τ  1 1 we know that 1  ∫ e dτ = 21 e − τ = s ( s + 1) 0  s

[ ]

∴

Laplace transforming yields 2 s ( s + 1) 2 or (s2 + 3s + 1) X(s) = s ( s + 1) s2X(s) + 3X(s) + 2X(s) =

3-8

α5 = ¼

X(s) =

x(t) = 1 − 2te-t − e-2t

and b)

2 s ( s + 1) 2 ( s + 2)

Applying the final Value Theorem lim x(t ) = lim sX ( s ) = lim t →∞

s →0

s →0

2 =2 ( s + 1) ( s + 2) 2

[ Note that Final Value Theorem is applicable here]

3.9 a)

X (s) =

6( s + 2) 6( s + 2) = ( s + 9s + 20)( s + 4) ( s + 4)( s + 5)( s + 4) 2

 6s ( s + 2)  =0 x(0) = lim  s →∞ ( s + 5)( s + 4) 2     6s ( s + 2)  x(∞) = lim =0 s →0 ( s + 5)( s + 4) 2    x(t) is converging (or bounded) because [sX(s)] does not have a limit at s = −4, and s = −5 only, i.e., it has a limit for all real values of s ≥ 0. x(t) is smooth because the denominator of [sX(s)] is a product of real factors only. See Fig. S3.9a. b)

10 s 2 − 3 10 s 2 − 3 X (s) = 2 = ( s − 6 s + 10)( s + 2) ( s − 3 + 2 j ) ( s − 3 − 2 j )( s + 2)

  10s 3 − 3s x(0) = lim  2  = 10 s →∞ ( s − 6 s + 10)( s + 2)   Application of final value theorem is not valid because [sX(s)] does not have a limit for some real s ≥ 0, i.e., at s = 3±2j. For the same reason, x(t) is diverging (unbounded). x(t) is oscillatory because the denominator of [sX(s)] includes complex factors. See Fig. S3.9b.

3-9

16 s + 5 16s + 5 = 2 ( s + 9) ( s + 3 j ) ( s − 3 j )

X (s) =

16s 2 + 5s  x(0) = lim  2  = 16 s →∞  ( s + 9)  Application of final value theorem is not valid because [sX(s)] does not have a limit for real s = 0. This implies that x(t) is not diverging, since divergence occurs only if [sX(s)] does not have a limit for some real value of s>0. x(t) is oscillatory because the denominator of [sX(s)] is a product of complex factors. Since x(t) is oscillatory, it is not converging either. See Fig. S3.9c 0.4

0.35

0.3

0.25

x(t)

0.2

0.15

0.1

0.05

0

-0.05

0

0.5

1

1.5

2

2.5 Time

3

3.5

4

4.5

Figure S3.9a. Simulation of X(s) for case a) 8000

6000

4000

2000

x(t)

c)

0

-2000

-4000

-6000

-8000

-10000

-12000

0

0.5

1

1.5

2

2.5

Time

Figure S3.9b. Simulation of X(s) for case b)

3-10

5

20

15

10

x(t)

5

0

-5

-10

-15

-20

0

0.5

1

1.5

2

2.5 Time

3

3.5

4

4.5

5

Figure S3.9c. Simulation of X(s) for case c)

The Simulink block diagram is shown below. An impulse input should be used to obtain the function’s behavior. In this case note that the impulse input is simulated by a rectangular pulse input of very short duration. (At time t = 0 and t =0.001 with changes of magnitude 1000 and –1000 respectively). The MATLAB command impulse might also be used.

Figure S3.9d. Simulink block diagram for cases a), b) and c).

3-11

3.10

a) i)

ii)

iii)

iv)

Y(s) =

2 2 A B C = 2 = 2+ + s s+4 s ( s + 4 s ) s ( s + 4) s

∴

y(t) will contain terms of form: constant, t, e-4t

Y(s) =

2 2 A B C = = + + s ( s + 4 s + 3) s ( s + 1)( s + 3) s s + 1 s + 3

∴

y(t) will contain terms of form: constant, e-t, e-3t

Y (s) =

2 2 A B C = = + + 2 2 s ( s + 2) s+2 s ( s + 4 s + 4) s ( s + 2)

∴

y(t) will contain terms of form: constant, e-2t , te-2t

Y (s) =

2 s ( s + 4 s + 8)

2

2

2

2

s 2 + 4 s + 8 = ( s 2 + 4 s + 4) + (8 − 4) = ( s + 2) 2 + 2 2 2 Y (s) = s[(s + 2) 2 + 2 2 ] ∴ b)

y(t) will contain terms of form: constant, e-2t sin2t, e-2tcos2t A Bs C 2( s + 1) 2( s + 1) = = + 2 + 2 2 2 2 2 s ( s + 4) s ( s + 2 ) s s + 2 s + 22 2( s + 1) 1 A = lim 2 = s → 0 ( s + 4) 2 Y (s) =

2(s+1) = A(s2+4) + Bs(s) + Cs 2s+2 = As2 + 4A + Bs2 + Cs Equating coefficients on like powers of s s2 :

0=A+B

→

B = −A = −

s1 :

2= C

→

s0 :

2 = 4A

→

C=2 1 A= 2

3-12

1 2

∴

Y(s) =

1 2 − (1 2) s 2 + 2 + 2 2 s s +2 s + 22

y(t) =

1 1 2 − cos 2t + sin 2t 2 2 2

y(t) =

1 (1 − cos 2t ) + sin 2t 2

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) 2 2 2 2 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

3-23

Laplace transforming the input function, a constant, Ci ( s ) =

ci s

so that sVC(s) + qC(s) = q

ci s

or

C(s) =

qci ( sV + q ) s

Dividing numerator and denominator by q C(s) =

ci V   s +1 s q 

Use Transform pair #3 in Table 3.1 to invert (τ =V/q) V − t    c(t) = ci 1 − e q     

Using MATLAB, the concentration response is shown in Fig. S3.17. (Consider V = 2 m3, Ci=50 Kg/m3 and q = 0.4 m3/min)

50

45

40

35

c(t)

30

25

20

15

10

5

0

0

5

10

15

20

25

30

Time

Figure S3.17. Concentration response of the reactor effluent stream.

3-24

3.18 a)

If Y(s) =

KAω s( s 2 + ω2 )

and input U(s) =

Aω = 1 {A sin ωt} (s + ω2 ) 2

then the differential equation had to be dy = Ku (t ) dt

b)

Y(s) =

α1 =

with

y(0) = 0

α ω α α s KAω = 1+ 2 2 2 + 2 3 2 2 2 s( s + ω ) s s + ω s +ω

KAω s + ω2 2

= s →0

KA ω

Find α2 and α3 by equating coefficients KAω= α1(s2+ω2) + α2s2+α3ωs KAω = α1s2 + α1ω2 + α2s2 + α3ωs s2 :

0 = α1 + α2

s:

0 = α3 ω

→ α2 = −α1 = → α3 = 0

KAω KA / ω ( KA / ω) s = − 2 2 2 s s( s + ω ) s + ω2 KA (1 − cos ωt ) y(t) = ω

∴ Y(s) =

3-25

− KA ω

c)

A

u(t)

-A

2KA/ω

y(t)

0 Time

i)

We see that y(t) follows behind u(t) by 1/4 cycle = 2π/4= π/2 rad. which is constant for all ω

ii)

The amplitudes of the two sinusoidal quantities are: y : KA/ω u: A Thus their ratio is K/ω, which is a function of frequency.

3-26