121578813-Process-Dynamics-and-Control-Solutions Chapter 4.pdf

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4.1

a) b) c) d)

iii iii v v

a)

5

b)

10

4.2

c)

d) e)

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.

5 (1 − e − s ) (10s + 1) s From the Final Value Theorem, y(t) = 0 when t→∞ Y (s) =

5 1 (10 s + 1) From the Final Value Theorem, y(t)= 0 when t→∞ Y (s) =

f)

g)

Y (s) =

5 6 2 (10 s + 1) ( s + 9)

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→∞. Solution Manual for Process Dynamics and Control, 2nd edition, Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp.

4-1

By using Simulink-MATLAB, above solutions can be verified: 10

0.5

9

0.45

8

0.4

7

0.35

0.3

y(t)

y(t)

6

5

0.25

4

0.2

3

0.15

2

0.1

1

0.05

0

0

5

10

15

20

25

30

35

40

45

0

50

0

5

10

15

20

25

30

35

40

45

50

time

time

Fig S4.2a. Output for part c) and d)

Fig S4.2b. Output for part e) 0.7

5

0.6 4.5

0.5 4

0.4 3.5

0.3

y(t)

y(t)

3

2.5

0.2

0.1 2

0

1.5

-0.1

1

-0.2

0.5

0

-0.3 0

5

10

15

20

25

30

35

40

45

50

0

2

4

6

8

10

12

14

16

18

20

time

time

Fig S4.2c. Output for part f)

Fig S4.2d. Output for part g)

4.3

a)

The dynamic model of the system is given by dV 1 = ( wi − w) dt ρ dT wi Q = (Ti − T ) + dt Vρ VρC

(2-45) (2-46)

Let the right-hand side of Eq. 2-46 be f(wi,V,T),  ∂f dT = f (wi ,V , T ) =  dt  ∂wi

 ∂f   ∂f  ′  wi′ +   V′+  T  ∂V  s  ∂T  s s

4-2

(1)

 ∂f   ∂wi

 1  = (Ti − T ) s Vρ

w Q 1  dT   ∂f  =−    = − 2i (Ti − T ) − 2  =0 V  dt  s V ρ V ρC  ∂V  s w  ∂f    =− i Vρ  ∂T  s w dT 1 = (Ti − T ) wi′ − i T ′ , dt V ρ Vρ

dT dT ′ = dt dt

Taking Laplace transform and rearranging T ′( s ) (Ti − T ) / wi = Wi′( s )  Vρ    s + 1  wi 

(2)

Laplace transform of Eq. 2-45 gives V ′( s ) =

Wi′( s ) ρs

(3)

 ∂f  If   were not zero, then using (3)  ∂V  s  (Ti − T ) V  ∂f  1  +     wi  ∂V  s s  T ′( s )  wi = Wi′( s )  Vρ    s + 1  wi 

(4)

Appelpolscher guessed the incorrect form (4) instead of the correct form  ∂f  (2) because he forgot that   would vanish.  ∂V  s b)

From Eq. 3, V ′( s ) 1 = Wi′( s ) ρs

4-3

4.4 Y( s ) = G( s )X ( s ) =

G(s)

Interpretation of G(s)

K s( τs + 1 )

U(s)

Interpretation of u(t)

K s (τs + 1)

2nd order process *

1

δ(0)

[ Delta function]

K τs + 1

1st order process

1 s

S(0)

[Unit step function]

K s

Integrator

K τs + 1

K

Simple gain

1 s (τs + 1)

(i.e no dynamics)

1 −t / τ e [Exponential input] τ

1 − e −t / τ

[Step + exponential input]

* nd

2 order or combination of integrator and 1st order process

4.5 a)

dy 1 = -2y1 – 3y2 + 2u1 dt dy 2 = 4y1 – 6y2 + 2u1 + 4u2 dt

2

(1) (2)

Taking Laplace transform of the above equations and rearranging, (2s+2)Y1(s) + 3Y2(s) = 2U1(s)

(3)

-4 Y1(s) + (s+6)Y2(s)=2U1(s) + 4U2(s)

(4)

Solving Eqs. 3 and 4 simultaneously for Y1(s) and Y2(s), 4-4

Y1(s) =

(2 s + 6) U 1 ( s ) − 12 U 2 ( s ) 2( s + 3) U1 ( s ) − 12 U 2 ( s ) = 2( s + 3)( s + 4) 2 s 2 + 14 s + 24

Y2(s) =

(4 s + 12) U 1 ( s ) − (8s + 8) U 2 ( s ) 4( s + 3) U 1 ( s ) + 8( s + 1) U 2 ( s ) = 2( s + 3)( s + 4) 2 s 2 + 14 s + 24

Therefore,

Y1 ( s ) 1 = U1 (s) s + 4

,

Y1 ( s ) −6 = U 2 ( s ) ( s + 3)( s + 4)

Y2 ( s ) 2 = U1 (s) s + 4

,

Y2 ( s ) 4( s + 1) = U 2 ( s ) ( s + 3)( s + 4)

4.6

The physical model of the CSTR is (Section 2.4.6) V

dc A = q (c Ai − c A ) − Vkc A dt

VρC

(2-66)

dT = wC (Ti − T ) + (− ∆H )Vkc A + UA(Tc − T ) dt

(2-68)

k = ko e-E/RT

(2-63)

where:

These equations can be written as, dc A = f1 (c A , T ) dt

(1)

dT = f 2 (c A , T , Tc ) dt

(2)

Because both equations are nonlinear, linearization is required. After linearization and introduction of deviation variables, we could get an expression for c ′A (s ) / T ′(s ) .

4-5

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)

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   x 0′ +   x1′ = f ( L, x0 , x1 , V , y1 , y 2 ) =   L ′ +  dt  ∂L  s  ∂x1  s  ∂x 0  s

 ∂f  ′  ∂f  ′  ∂f  ′ +  V +  y1 +  ∂y  y 2  ∂V  s  ∂y1  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   x1′ = (a1 + 2a 2 x1 + 3a 3 x1 2 ) x1′ y1′ = g ( x1 ) =   ∂x1  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

We need to use a Taylor series expansion to linearize dh  1 R 1.5  1 1.5Rh 0.5 = wi − h + ( wi − wi ) − (h − h ) dt  ρA ρA ρA  ρA Since the bracketed term is identically zero at steady state,

dh ′ 1 1.5Rh 0.5 = wi′ − h′ dt ρA ρA Rearranging ρ A dh ′ 1 + h′ = wi′ 0.5 dt 1.5 Rh 1.5 Rh 0.5

Hence where

H ′( s ) K = Wi′( s ) τs + 1

1 h h [height ] = = = 0.5 1.5 1.5w [ flowrate] 1.5 Rh 1.5 Rh [mass] = [time] ρA ρAh ρV τ= = = = 0.5 1.5 1.5w [mass / time] 1.5 Rh 1.5 Rh K=

4.9

a)

The model for the system is given by mC

dT = wC (Ti − T ) + h p A p (Tw − T ) dt

mw C w

dTw = hs As (Ts − Tw ) − h p A p (Tw − T ) dt

(2-51)

(2-52)

Assume that m, mw, C, Cw, hp, hs, Ap, As, and w are constant. Rewriting the above equations in terms of deviation variables, and noting that

4-8

dT dT ′ = dt dt

dTw dTw′ = dt dt

dT ′ = wC (Ti′ − T ′) + h p A p (Tw′ − T ′) dt dT ′ m w C w w = hs As (0 − Tw′ ) − h p A p (Tw′ − T ′) dt mC

Taking Laplace transforms and rearranging, (mCs + wC + h p A p )T ′( s ) = wCTi′( s ) + h p A pTw′ ( s )

(1)

(mw C w s + hs As + h p A p )Tw′ ( s ) = h p A p T ′( s )

(2)

Substituting in Eq. 1 for Tw′ (s ) from Eq. 2, (mCs + wC + h p A p )T ′( s ) = wCTi′( s ) + h p A p

h p Ap (mw C w s + hs As + h p A p )

T ′( s )

Therefore,

wC (mw C w s + hs As + h p A p ) T ′( s ) = Ti′( s ) (mCs + wC + h p A p )(mw C w s + hs As + h p Ap ) − (h p Ap ) 2

b)

c)

wC (hs As + h p A p )  T ′( s )  The gain is  =   Ti′( s )  s =0 wC (hs As + h p A p ) + hs As h p A p

No, the gain would be expected to be 1 only if the tank were insulated so that hpAp= 0. For heated tank the gain is not 1 because heat input changes as T changes.

4.10

Additional assumptions 1) 2) 3)

perfect mixing in the tank constant density, ρ , and specific heat, C. Ti is constant.

4-9

Energy balance for the tank, ρVC

dT = wC (Ti − T ) + Q − (U + bv) A(T − Ta ) dt

Let the right-hand side be f(T,v), ρVC

dT  ∂f  ′  ∂f  ′ = f (T , v) =   T +  v dt  ∂T  s  ∂v  s

(1)

 ∂f    = − wC (U + bv ) A  ∂T  s  ∂f    = − bA(T − Ta )  ∂v  s Substituting for the partial derivatives in Eq. 1 and noting that ρVC

dT ′ = − [wC + (U + bv ) A]T ′ − bA(T − Ta )v ′ dt

dT dT ′ = dt dt

Taking the Laplace transform and rearranging

[ρVCs + wC + (U + bv ) A] T ′(s) = − bA(T − T )v′ (s) a

 − bA(T − Ta )    wC + (U + bv ) A  T ′( s )  = v ′( s )   ρVC  wC + (U + bv ) A  s + 1  

4.11

a)

Mass balances on surge tanks dm1 = w1 − w2 dt

(1)

dm2 = w2 − w3 dt

(2)

4-10

Ideal gas law m1 RT M m P2V2 = 2 RT M P1V1 =

Flows

(3) (4)

(Ohm's law is I =

E Driving Force = ) R Resistance

1 ( Pc − P1 ) R1 1 w2 = ( P1 − P2 ) R2 1 w3 = ( P2 − Ph ) R3 w1 =

(5) (6) (7)

Degrees of freedom: number of parameters : 8 (V1, V2, M, R, T, R1, R2, R3) number of variables : 9 (m1, m2, w1, w2, w3, P1, P2, Pc, Ph) number of equations : 7 ∴

number of degrees of freedom that must be eliminated = 9 − 7 = 2

Since Pc and Ph are known functions of time (i.e., inputs), NF = 0. b)

Development of model Substitute (3) into (1) :

MV1 dP1 = w1 − w2 RT dt

(8)

Substitute (4) into (2) :

MV2 dP2 = w2 − w3 RT dt

(9)

Substitute (5) and (6) into (8) :

MV1 dP1 1 1 = ( Pc − P1 ) − ( P1 − P2 ) RT dt R1 R2 MV1 dP1 1 1 1 1 = Pc (t ) − ( + ) P1 + P2 RT dt R1 R1 R2 R2

4-11

(10)

Substitute (6) and (7) into (9):

MV2 dP2 1 1 = ( P1 − P2 ) − ( P2 − Ph ) RT dt R2 R3 MV2 dP2 1 1 1 1 = P1 − ( + ) P2 + Ph (t ) RT dt R2 R2 R3 R3 Note that

dP1 = f 1 ( P1 , P2 ) dt

from Eq. 10

dP2 = f 2 ( P1 , P2 ) dt

from Eq. 11

(11)

This is exactly the same situation depicted in Figure 6.13, therefore the two tanks interact. This system has the following characteristics: i) ii) iii) iv) v)

Interacting (Eqs. 10 and 11 interact with each other ) 2nd-order denominator (2 differential equations) Zero-order numerator (See example 4.4 in text) No integrating elements W ′( s ) is not equal to unity. (Cannot be because Gain of 3 Pc′( s ) the units on the two variables are different).

4.12

a)

A

dh = q i − C v h1 / 2 dt

Let f = qi − C v h1 / 2 1 Then f ≈ q i − C v h 1 / 2 + qi − qi − C v h −1 / 2 (h − h ) 2

so A

C dh ′ = q i′ − 1v/ 2 h ′ dt 2h

because

4-12

qi − C v h 1 / 2 ≡ 0

Cv    sA + 2h 1 / 2  H ′( s ) = Qi′ ( s )  

H ' ( s) = Qi′ ( s )

1

Note: Not a standard form

C sA + 1v/ 2 2h

2h 1 / 2 / C v H ' (s) = Qi′ ( s ) 2 Ah 1 / 2 s +1 Cv where K =

b)

and τ =

2 Ah 1 / 2 Cv

Because q = C v h1 / 2 q′ = Cv

C 1 −1 / 2 1 h h ′ = 1v/ 2 h ′ = h ′ 2 K 2h

Q ′( s ) 1 = , H ′( s ) K

∴

and c)

2h 1 / 2 Cv

Q ′( s ) H ′( s ) 1 K = H ′( s ) Qi′( s ) K τs + 1

Q ′( s ) 1 = Qi′ ( s ) τs + 1

For a linear outlflow relation A

dh * = qi − C v h dt

A

dh ′ * = q i′ − C v h ′ dt

A

dh ′ * + C v h ′ = q i′ dt

Note that C v ≠ C v *

A dh′ 1 + h′ = * qi′ * C v dt Cv

or

Multiplying numerator and denominator by h on each side yields Ah dh ′ h + h ′ = * qi′ * C v h dt Cv h

4-13

or

V dh′ h + h′ = qi′ q i dt qi

τ∗ =

V qi

K∗ =

h qi

q.e.d

To put τ and K in comparable terms for the square root outflow form of the transfer function, multiply numerator and denominator of each by h 1/ 2 . K=

2h 1 / 2 h 1 / 2 2h 2h = = = 2K * 1/ 2 1/ 2 Cv h qi Cv h

2 Ah 1 / 2 h 1 / 2 2 Ah 2V τ= = = = 2τ * 1/ 2 1/ 2 Cv h qi Cv h Thus level in the square root outflow TF is twice as sensitive to changes in qi and reacts only ½ as fast (two times more slowly) since τ = 2 τ∗ .

4.13 a)

The nonlinear dynamic model for the tank is:

(

dh 1 = qi − Cv h dt π( 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 =

qi − Cv h π( D − h ) h

Then,  ∂f   ∂f  f (h, qi ) ≈   h′ +   qi′  ∂h  s  ∂qi  s where

4-14

 ∂f  1   =  ∂qi  s π( D − h )h   1 Cv 1 −πD + 2πh   ∂f   = − + q − C h i v    ( π( D − h ) h ) 2  2 h π( D − h ) h  ∂h  s  

)

(

Notice that the second term of last partial derivative is zero from the steady-state relation, and the term π( 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 π( D − h )h   π( D − h ) h  Since

qi = Cv h

   dh′  1 qi 1 1 = − h′ +    qi′ dt  2 h π( D − h )h  π ( D − h ) h   or

dh′ = ah′ + bqi′ dt

where  1q  q 1 a = − i =− i  Vo  2 h π( D − h ) h 

,

  1 b=    π( D − h ) h 

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

4.14

Assumptions 1) Perfectly mixed reactor 2) Constant fluid properties and heat of reaction. a)

Component balance for A, dc A = q (c Ai − c A ) − Vk (T )c A dt Energy balance for the tank,

(1)

V

ρVC

dT = ρqC (Ti − T ) + ( − ∆H )Vk (T )c A dt

(2)

Since a transfer function with respect to cAi is desired, assume the other inputs, namely q and Ti, to be constant. Linearize (1) and (2) and not that

V

dc A dc ′A dT dT ′ = , = , dt dt dt dt

dc ′A 20000 = qc ′Ai − (q + Vk (T ))c ′A − Vc A k (T ) T′ dt T2

ρVC

dT ′ 20000   = − ρqC + ∆HVc A k (T ) T ′ + (−∆H )Vk (T )c′A dt T2  

(3)

(4)

Taking the Laplace transforms and rearranging

[Vs + q + Vk (T )]C ′ (s) = qC ′ (s) − Vc A

Ai

A

k (T )

20000 T ′( s ) T2

(5)

20000   ρVCs + ρqC − (−∆H )Vc A k (T ) T 2  T ′( s ) = (−∆H )Vk (T )C A′ ( s ) (6)

Substituting for C ′A (s ) from Eq. 5 into Eq. 6 and rearranging,

4-16

T ′( s ) = ′ ( s) C Ai

−∆HVk (T ) q 20000  20000  Vs + q + Vk (T )  ρVCs + ρqC − ( −∆H )Vc A k (T ) + ( −∆H )c AV 2 k 2 (T ) 2  T T2  

(7)

c A is obtained from Eq. 1 at steady state, cA =

qc Ai = 0.0011546 mol/cu.ft. q + Vk (T )

Substituting the numerical values of T , ρ, C, ( − ∆H), q, V, c A into Eq. 7 and simplifying,

T ′( s) 11.38 = C ′Ai ( s) (0.0722s + 1)(50s + 1) b)

 T ′( s)  The gain K of the above transfer function is   ,  C ′Ai ( s )  s =0

K=

0.15766 q c  q c  q  − 3.153 × 106 A2   + 13.84  + 4.364.107 A2  T   1000 T  1000 

(8)

obtained by putting s=0 in Eq. 7 and substituting numerical values for ρ, C, ( − ∆H), V. Evaluating sensitivities, dK K K2 = − dq q 0.15766q

cA   q 2 + 0 . 01384 − 3153 = −6.50 × 10 −4 6 2   10 T  

6 7 dK K 2  q   3.153 × 10 c A × 2  2 × 4.364 × 10 c A  =− + 13 . 84 −     dT 3.153  1000 T3 T3   

= −2.57 × 10 −5

dK dK dc A = × dc Ai dc A dc Ai =

6 −K2   q  3.153 × 10 + 13.84  −  0.15766q   1000 T2 

= 8.87 × 10 −3

4-17

 4.364 × 10 7   q  +   2 T   q + 13840 

4.15

From Example 4.4, system equations are:

dh1′ 1 = qi′ − h1′ dt R1 dh′ 1 1 A2 2 = h1′ − h2′ dt R1 R2 Using state space representation, A1

,

q1′ =

1 h1′ R1

,

q ′2 =

1 h2′ R2

x1 = Ax + Bu y = Cx + Du

where

 h′  x =  1 h2′ 

,

u = qi

and

y = q 2′

then,  dh1′  1   dt  − R A   =  1 1    1  dh2′   R1 A1  dt 

 q ′2 = 0 

 0   1  − R2 A2 

 1 ′  A  h1    +  1     0  h2′   

    qi′   

 h1′  1     R2   h2′ 

Therefore, 1  − R A A=  1 1  1  R1 A1

 0   1  − R2 A2 

 1   A    1  , B=  , C=0   0     

4-18

1   , E=0 R2 

4.16

Applying numerical values, equations for the three-stage absorber are:

dx1 = 0.881 y f − 1.173 x1 + 0.539 x 2 dt dx 2 = 0.634 x1 − 1.173 x 2 + 0.539 x3 dt dx3 = 0.634 x 2 − 1.173 x3 + 0.539 x f dt

yi = 0.72 xi Transforming into a state-space representation form:  dx1  dt    dx 2  dt   dx 3   dt

   0  − 1.173 0.539   =  0.634 − 1.173 0.539       0 0 . 634 − 1 . 173     

0 0   y1  0.72  y  =  0 0.72 0   2    y 3   0 0 0.72

 x1  0.881 x  +  0 y  2    f  x3   0 

 x1  0 x  + 0y  2    f  x3   0 

Therefore, because xf can be neglected in obtaining the desired transfer functions,

0  − 1.173 0.539  A =  0.634 − 1.173 0.539   0 0.634 − 1.173

4-19

0.881 B =  0   0 

0 0  0.72  C= 0 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

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 '+ 2 ( K + S ) s    Ks + S

4-20

  X ' 

Substituting the numerical values for µ m , K s , S and X then:

f 1 ( S ′, X ′) ≈ 0.113S ' + 0.1X ' 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 ' = 0.113S ' + 0.1X ' − 2.25 D' − 0.1X ' dt dS ' −0.113 0.1 = S'− X ' − 9 D ' + 0.1S ' − 0.1S ′f dt 0.5 0.5

Rearranging: dX ' = 0.113S ' − 2.25 D' dt dS ' = −0.126 S ' − 0.2 X ' − 9 D ' − 0.1S ′f dt

Laplace Transforming: sX ' ( s ) = 0.113S ' ( s ) − 2.25 D' ( s )

sS '( s ) = −0.126 S '( s ) − 0.2 X '( s ) − 9 D '( s ) − 0.1S ′f ( s )

4-21

Then, 0.113 2.25 S ' (s) − D' ( s) s s −0.2 9 0.1 S '( s ) = X '( s ) − D '( s ) − S ′f ( s ) s + 0.126 s + 0.126 s + 0.126 X ' (s) =

or

 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, X ' (s) − 1.3005 − 2.25s = 2 D' ( s ) s + 0.126 s + 0.0226

4-22