Process dynamics and Control seborg 2nd ch11 Solutions

123456789 8

13.1

AR = G ( jω) =

=

3 G1 ( jω) G2 ( jω) G3 ( jω)

3 (−ω) 2 + 1 ω (2ω) 2 + 1

=

3 ω2 + 1 ω 4ω 2 + 1

From the statement, we know the period P of the input sinusoid is 0.5 min and, thus, ω=

2π 2π = = 4π rad/min P 0 .5

Substituting the numerical value of the frequency: 3 16π 2 + 1 Aˆ = AR × A = × 2 = 0.12 × 2 = 0.24 1 2 4π 64π + 1 Thus the amplitude of the resulting temperature oscillation is 0.24 degrees.

13.2

First approximate the exponential term as the first two terms in a truncated Taylor series e − θs ≈ 1 − θs

Then G ( jω) = 1 − jω and

ARtwo term = 1 + (−ωθ) 2 = 1 + ω 2 θ 2

φ two term = tan −1 (−ωθ ) = − tan −1 (ωθ ) Solution Manual for Process Dynamics and Control, 2nd edition, Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp

13-1

For a first-order Pade approximation θs 2 e −θs ≈ θs 1+ 2 from which we obtain 1−

ARPade = 1  ωθ  φ Pade = −2 tan −1    2  Both approximations represent the original function well in the low frequency region. At higher frequencies, the Padé approximation matches the amplitude ratio of the time delay element exactly (ARPade = 1), while the two-term approximation introduces amplification (ARtwo term >1). For the phase angle, the high-frequency representations are:

φ two term → −90 1 φ Pade → −180 1

Since the angle of e − jωθ is negative and becomes unbounded as ω → ∞ , we see that the Pade representation also provides the better approximation to the time delay element's phase angle, matching φ of the pure time delay element to a higher frequency than the two-term representation.

13.3

Nominal temperature T =

127 1 F + 119 1 F = 123 1 F 2

1 Aˆ = (127 1 F − 119 1 F) = 4 1 F 2 τ = 4.5 sec ., ω = 2π(1.8 / 60 sec) = 0.189 rad/s

Using Eq. 13-2 with K=1,

(

)

A = Aˆ ω 2 τ 2 + 1 = 4 (0.189) 2 (4.5) 2 + 1 = 5.25 1 F

Actual maximum air temperature = T + A = 128.25 1 F Actual minimum air temperature = T − A = 117.75 1 F

13-2

13.4 Tm′ ( s ) 1 = T ′( s ) 0.2 s + 1

T ′( s ) = (0.2s + 1)Tm′ ( s ) amplitude of T ′ =3.464

(0.2ω) 2 + 1 = 3.467

phase angle of T ′ = ϕ + tan-1(0.2ω) = ϕ + 0.04 Since only the maximum error is required, set ϕ = 0 for the comparison of T ′ and Tm′ . Then Error = Tm′ − T ′ =3.464 sin (0.2t) – 3.467sin(0.2t + 0.04) = 3.464 sin(0.2t) –3.467[sin(0.2t) cos 0.04 + cos(0.2t)sin 0.04] = 0.000 sin(0.2t) − 0.1386 cos(0.2t) Since the maximum absolute value of cos(0.2t) is 1, maximum absolute error = 0.1386

13-3

13.5

a) Bode Diagram

Magnitude (abs)

0

10

-1

10

-2

10

0

Phase (deg)

-45 -90

-135 -180 -2

-1

10

0

10

10

1

10

Frequency (rad/sec)

ω 0.1 1 10

AR (absolute) 4.44 0.69 0.005

ϕ -32.4° -124° -173°

b) Bode Diagram 0

Magnitude (abs)

10

-2

10

0

Phase (deg)

-45 -90 -135 -180 -225 -270 -2

-1

10

0

10

10 Frequency (rad/sec)

ω 0.1 1 10

AR (absolute) 4.42 0.49 0.001

13-4

ϕ -38.2° -169° -257°

1

10

c)

Magnitude (abs)

Bode Diagram

0

10

-1

10

Phase (deg)

0

-45

-90 -2

-1

10

0

10

10

1

10

2

10

Frequency (rad/sec)

ω 0.1 1 10

AR (absolute) 4.48 2.14 0.003

ϕ -22.1° -44.9° -87.6°

d)

Magnitude (abs)

Bode Diagram

0

10

-1

10

0

Phase (deg)

-45 -90 -135 -180 -225 -270 -2

-1

10

0

10

10

1

10

Frequency (rad/sec)

ω 0.1 1 10

AR (absolute) 4.48 1.36 0.04

13-5

ϕ -33.6° -136° -266°

2

10

e) Bode Diagram

2

10

1

Magnitude (abs)

10

0

10

-1

10

-2

10

-90

Phase (deg)

-120

-150

-180 -1

0

10

1

10

10

Frequency (rad/sec)

ω 0.1 1 10

AR (absolute) 44.6 0.97 0.01

ϕ -117° -169° -179°

f) 10

Magnitude (abs)

10

10

10

Bode Diagram

2

1

0

-1

-90

Phase (deg)

-120

-150

-180 10

-1

10

0

1

10

Frequency (rad/sec)

ω 0.1 1 10

AR (absolute) 44.8 1.36 0.04

13-6

ϕ -112° -135° -158°

13.6

a)

Multiply the AR in Eq. 13-41a by

ω 2 τ a + 1 . Add to the value of ϕ in 2

Eq. 13-41b the term + tan −1 (ωτ a ) .

G ( jω) = K

ω2 τ a + 1 2

(1 − ω 2 τ 2 ) 2 + (0.4ωτ ) 2

 − 0.4ωτ  + tan −1 (ωτ a ) . ∠G ( jω) = tan −1  2 2  1 − ω τ  

b) Bode Diagram

Phase (deg)

Normalized Ampltude ratio (abs)

2

10

0

10

-2

10

-4

10

90 45 0 -45

Ratio = 0 Ratio = 0.1 Ratio = 1 Ratio = 10

-90 -135 -180 -2

10

-1

0

10

10

1

10

ωτ

Figure S13.6. Frequency responses for different ratios τa/τ

13-7

2

10

13.7

Using MATLAB Bode Diagram

5

Magnitude (abs)

10

0

10

-5

10

-10

10

-45

Phase (deg)

-90 -135 -180 -225 -270 -1

10

0

1

10

10

2

10

Frequency (rad/sec)

Figure S13.7. Bode diagram of the third-order transfer function.

The value of ω that yields a -180° phase angle and the value of AR at that frequency are: ω = 0.807 rad/sec AR = 0.202

13-8

13.8

Using MATLAB, Bode Diagram

Magnitude (abs)

G(s) G(s) w ith Pade approx. 0

10

-1

10

0

Phase (deg)

-50 -100 -150 -200 -250 -2

-1

10

0

10

10

10

Frequency (rad/sec)

Figure S13.8. Bode diagram for G(s) and G(s) with Pade approximation.

13.9 ω=2πf

where f is in cycles/min

For the standard thermocouple, using Eq. 13-20b ϕ1 = -tan-1(ωτ1) = tan-1(0.15ω) Phase difference ∆ϕ = ϕ1 – ϕ2 Thus, the phase angle for the unknown unit is ϕ2 = ϕ1 − ∆ϕ and the time constant for the unknown unit is

13-9

1

τ2 =

1 tan(−ϕ 2 ) ω

using Eq. 13-20b . The results are tabulated below f 0.05 0.1 0.2 0.4 0.8 1 2 4

ω 0.31 0.63 1.26 2.51 6.03 6.28 12.57 25.13

ϕ1 -2.7 -5.4 -10.7 -26.6 -37 -43.3 -62 -75.1

∆ϕ 4.5 8.7 16 24.5 26.5 25 16.7 9.2

ϕ2 -7.2 -14.1 -26.7 -45.1 -63.5 -68.3 -78.7 -84.3

τ2 0.4023 0.4000 0.4004 0.3995 0.3992 0.4001 0.3984 0.3988

That the unknown unit is first order is indicated by the fact that ∆ϕ→0 as ω→∞, so that ϕ2→ϕ1→-90° and ϕ2→-90° for ω→∞ implies a first-order system. This is confirmed by the similar values of τ2 calculated for different values of ω, implying that a graph of tan(-ϕ2) versus ω is linear as expected for a first-order system. Then using linear regression or taking the average of above values, τ2 = 0.40 min.

13.10

From the solution to Exercise 5-19, for the two-tank system

H 1′ ( s ) / h1′max 0.01 K = = Q1′i ( s ) 1.32s + 1 τs + 1 H 2′ ( s ) / h2′ max 0.01 K = = 2 Q1′i ( s ) (1.32s + 1) (τs + 1) 2 Q2′ ( s ) 0.1337 0.1337 = = 2 Q1′i ( s ) (1.32s + 1) (τs + 1) 2 and for the one-tank system

′ H ′( s ) / hmax 0.01 K = = Q1′i ( s ) 2.64s + 1 2τs + 1 Q ′( s ) 0.1337 0.1337 = = Q1′i ( s ) 2.64 s + 1 2τs + 1

13-10

For a sinusoidal input q1′i (t ) = A sin ω t , the amplitudes of the heights and flow rates are ′ ] = KA / 4ω 2 τ 2 + 1 Aˆ [h ′ / hmax

(1)

Aˆ [q ′] = 0.1337 A / 4ω 2 τ 2 + 1

(2)

for the one-tank system, and Aˆ [h1′ / h1′max ] = KA / ω 2 τ 2 + 1

(3)

Aˆ [h2′ / h2′ max ] = KA / (ω 2 τ 2 + 1) 2

(4)

Aˆ [q ′2 ] = 0.1337 A / (ω 2 τ 2 + 1) 2

(5)

for the two-tank system. Comparing (1) and (3), for all ω

′ ] Aˆ [h1′ / h1′max ] ≥ Aˆ [h ′ / hmax Hence, for all ω, the first tank of the two-tank system will overflow for a smaller value of A than will the one-tank system. Thus, from the overflow consideration, the one-tank system is better for all ω. However, if A is small enough so that overflow is not a concern, the two-tank system will provide a smaller amplitude in the output flow for those values of ω that satisfy

[ ]

Aˆ [q 2′ ] ≤ Aˆ q ′ or

0.1337 A (ω 2 τ 2 + 1) 2

≤

0.1337 A 4ω 2 τ 2 + 1

or ω ≥ 2 / τ = 1.07 Therefore, the two-tank system provides better damping of a sinusoidal disturbance for ω ≥ 1.07 if and only if

1.32 2 ω 2 + 1 Aˆ [h1′ / h1′max ] ≤ 1 , that is, A ≤ 0.01

13-11

13.11

Using Eqs. 13-48 , 13-20, and 13-24,

2 ω2 τ a + 1 2

AR=

100ω 2 + 1 4ω 2 + 1

φ = tan-1(ωτa) – tan-1(10ω) – tan-1(2ω) The Bode plots shown below indicate that i) ii) iii) iv) v) vi) vii)

AR does not depend on the sign of the zero. AR exhibits resonance for zeros close to origin. All zeros lead to ultimate slope of –1 for AR. A left-plane zero yields an ultimate φ of -90°. A right-plane zero yields an ultimate φ of -270°. Left-plane zeros close to origin can give phase lead at low ω. Left-plane zeros far from the origin lead to a greater lag (i.e., smaller phase angle) than the ultimate value. φu = −90 º with a leftplane zero present. Bode Diagram

1

Magnitude (abs)

10

Case i Case ii(a) Case ii(b) Case iii

0

10

-1

10

-2

10

90

Phase (deg)

0 -90

-180 -270 -2

10

-1

0

10

10 Frequency (rad/sec)

Figure S13.11. Bode plot for each of the four cases of numerator dynamics.

13-12

1

10

13.12 From Eq. 8-14 with τI = 4τD

a)

( 4τ D s + 1 + 4 τ D s 2 ) (2τ D s + 1) 2 = Kc 4τ D s 4τ D s 2

Gc ( s ) = K c

2

 4τ 2 ω 2 + 1  2 D 4τ D ω 2 + 1   Gc ( jω) = K c = Kc 4τ D ω 4τ D ω

From Eq. 8-15 with τI = 4τD and α = 0.1

b)

Gc ( s ) = K c

(4τ D s + 1)(τ D s + 1) 4τ D s (0.1τ D s + 1)

 16τ 2 ω 2 + 1  τ 2 ω 2 + 1  D  D  Gc ( jω) = K c  2 2 4τ D ω 0.01τ D ω + 1 The differences are significant for 0.25 < ωτD < 1 by a maximum of 0.5 Kc at ωτD = 0.5, and for ωτD >10 by an amount increasing with ωτD . 2

10

Series controller with filter (asymptote)

1

AR/Kc

10

0

10

Parallel controller (asymptote) Parallel controller (actual) Series controller (actual)

-1

10 -2 10

-1

10

0

10

1

10

2

10

ωτ D

Figure S13.12. Nominal amplitude ratio for parallel and series controllers.

13-13

13.13

MATLAB does not allow the addition of transfer functions with different time delays. Hence the denominator time delay needs to be approximated if a MATLAB program is used. However, the use of Mathematica or even Excel to evaluate derived expressions for the AR and angle, using various values of omega, and to make the plots will yield exact results: MATLAB - Padé approximation: Substituting the 1/1 Padé approximation gives: G (s) ≈

K K (θs + 2) = θτs 2 + 2τs + 4  2 − θs  τs +  + 1   2 + θs 

(1)

By using MATLAB, Bode Diagram

1

Magnitude (abs)

10

0

10

-1

10

-2

10

Phase (deg)

0

-45

-90 -2

10

-1

0

10

10

1

10

Frequency (rad/sec)

Figure S13.13. Bode plot by using Padé approximation.

13-14

2

10

13.14

ω = 600

rotations cycles radians rad ×4 × 2π = 15080 min rotation cycle min Aˆ = 0.02 psig

A = 2 psig

AR = Aˆ / A = 0.01 Volume of the pipe connecting the compressor to the reactor is π 3  = 20 ft ×   ft 2 = 0.982 ft 3 4  12  2

V pipe

Two-tank surge system Using the figure and nomenclature in Exercise 2.5, the 0.02 psig variation in Aˆ refers to the pressure before the valve Rc, namely the pressure P2. Hence the transfer function P2′ ( s ) / Pd′ ( s ) is required in order to use the value of AR. Mass balance for the tanks is (referring to the solution for Exercise 2.5.

V1 M dP1 = wa − wb RT1 dt

(1)

V2 M dP2 = wb − wc RT2 dt

(2)

where the ideal-gas assumption has been used. For linear valves,

wa =

Pd − P1 Ra

,

wb =

P1 − P2 Rb

,

wc =

At nominal conditions, Pd = 200 psig wa = wb = wc = 6000 lb/hr = 100 lb/min Pd − P1 = P1 − P2 =

0.1Pd = 10 psig 2

13-15

P2 − Pf Rc

Ra =

Pd − P1 P − P2 10 psig psig = = 0.1 = 1 = Rb wa 100 lb/min lb/min wb

Assume Rc = Ra = Rv Assume T2 = T1 = 300 1 F = 792 1 R Given V1 = V2 = V Then equations (1) and (2) become  VM  dP Rv  1 = Pd − P1 − ( P1 − P2 ) = Pd − 2 P1 + P2   RT  dt  VM  dP Rv  2 = P1 − P2 − ( P2 − Pf ) = P1 − 2 P2 − Pf   RT  dt

Taking deviation variables, Laplace transforming, and noting that Pf′ is zero since Pf is constant, gives 1 1 Pd′ ( s ) − P1′( s ) + P2′ ( s ) 2 2 1 τsP2′ ( s ) = P1′( s ) − P2′ ( s ) 2 τsP1′( s ) =

(3) (4)

where τ=

1  VM  Rv   2  RT 

 1 lb  psig   0.1 = V ft 3  28  2  lb mole  lb/min 

(

)

=

 ft 3 psig  10.731  792 1 R 1 lb mole R  

(1.647 × 10 −4 V ) min

From Eq. 3 P1′( s ) =

1 1 Pd′ ( s ) + P2′ ( s ) 2(τs + 1) 2(τs + 1)

13-16

(

)

Substituting for P1′( s ) into Eq. 4 (τs + 1) P2′( s ) =

1 1 Pd′ ( s ) + P2′( s ) 4(τs + 1) 4(τs + 1)

or

P2′( s ) 1 1 = = 2 2 2 Pd′ ( s ) 4(τs + 1) − 1 4τ s + 8τs + 3 P2′( jω) 1 = 2 2 Pd′ ( jω) (3 − 4ω τ ) + j8ωτ

1

AR =

(3 − 4ω 2 τ 2 ) 2 + 64ω 2 τ 2

1

=

16ω 4 τ 4 + 40ω 2 τ 2 + 9

Setting AR = 0.01 gives 16ω 4 τ 4 + 40ω 2 τ 2 + 9 = 10000 16ω 4 τ 4 + 40ω 2 τ 2 − 9991 = 0

ω2 τ 2 =

τ= V =

)

(

1 − 40 + 40 2 + 4 × 16 × 9991 = 23.77 2 × 16

23.77 4.875 = = 3.233 × 10 − 4 min ω ω τ = 1.963 ft 3 −4 1.647 × 10

Total surge volume Vsurge = 2V = 3.926 ft 3 Letting the connecting pipe provide part of this volume, the volume of 1 each tank = (Vsurge − V pipe ) = 1.472 ft 3 2

13-17

Single-tank system In the figure for the two-tank system, remove the second tank and the valve before it (Rb). Now, Aˆ refers to P1 and AR refers to P1′( s ) / Pd′ ( s ) . Mass balance for the tank is

V1 M dP1 = wa − wc RT1 dt wa =

where

Pd − P1 Ra

,

wc =

P1 − Pf Rc

At nominal conditions Pd − P1 = 0.1 Pd = 20 psig

Ra =

Pd − P1 20 psig psig = = 0.2 wa 100 lb/min lb/min

Assume Rc = Ra = Rv Then Eq. 1 becomes

 V1 M  dP  Rv  1 = Pd − P1 − ( P1 − P7 ) = Pd − 2 P1 + P7  RT1  dt Using deviation variables and taking the Laplace transform

P1′( s ) 1/ 2 = Pd′ ( s ) τs + 1 where

τ=

 1  V1 M  Rv  = (3.294 × 10 −4 V1 ) min 2  RT1 

AR = 0.01= 0.5

ω 2 τ 2 + 1 , τ = 3.315 × 10 −3 min, V1 = 10.06 ft 3

Volume of single tank = (V1 − V pipe ) = 9.084 ft 3 > 4 × 1.472 ft 3 Hence, recommend two surge tanks, each with volume 1.472 ft 3

13-18

13.15 By using MATLAB Nyquist Diagram 4 3

Imaginary Axis

2 1 0 -1 -2 -3 -4 -1

0

1

2

3

4

5

6

4

5

6

Real Axis

Figure S13.15a. Nyquist diagram.

Nyquist Diagram 4 3

Imaginary Axis

2 1 0 -1 -2 -3 -4 -1

0

1

2

3

Real Axis

Figure S13.15b. Nyquist diagram by using Pade approximation.

13-19

The two plots are very different in appearance for large values of ω. The reason for this is the time delay. If the transfer function contains a time delay in addition to poles and zeros, there will be an infinite number of encirclements of the origin. This result is a consequence of the unbounded phase shift for the time delay. A subtle difference in the two plots, but an important one for the Nyquist design methods of Chapter 14, is that the plot in S13.5a “encircles” the -1, 0 point while that in S13.5b passes through it exactly.

13.16

By using MATLAB, Bode Diagram

2

10

Magnitude (abs)

1

10

0

10

-1

10

-2

10

0 Parallel Series w ith filter

Phase (deg)

-45 -90 -135 -180 -225 -270 -2 10

-1

10

0

10

Frequency (rad/sec)

Figure S3.16. Bode plot for Exercise 13.8 Transfer Function multiplied by PID Controller Transfer Function. Two cases: a)Parallel b) Series with Deriv. Filter (α=0.2). .

13-20

Amplitude ratios: Ideal PID controller: AR= 0.246 at ω = 0.80 Series PID controller: AR=0.294 at ω = 0.74 There is 19.5% difference in the AR between the two controllers.

13.17

a)

Method discussed in Section 6.3:

Gˆ 1 ( s ) =

12e −0.3s (8s + 1)(2.2s + 1)

Visual inspection of the frequency responses:

Gˆ 2 ( s ) =

Comparison of three models: Bode plots: Bode Diagram

5

10

G(s) G1(s) G2(s)

Magnitude (abs)

0

10

-5

10

-10

10

0

Phase (deg)

b)

12e −0.4 s (5.64s + 1)(2.85s + 1)

-360

-720

-1

10

0

10

1

10

2

10

Frequency (rad/sec)

Figure S13.17a. Bode plots for the exact and approximate models.

13-21

Impulse responses: Impulse Response 1.4

G(s) G1(s) G2(s)

1.2

1

Amplitude

0.8

0.6

0.4

0.2

0

0

5

10

15

20

25

30

35

40

45

Time (sec)

Figure S13.17b. Impulse responses for the exact and approximate models

13.18

The original transfer function is

G (s) =

10(2s + 1)e −2 s (20 s + 1)(4s + 1)( s + 1)

The approximate transfer function obtained using Section 6.3 is:

13-22

Bode Diagram

1

10

Magnitude (abs)

0

10

-1

10

G(s) G'(s)

-2

10

0 -360 Phase (deg)

-720 -1080 -1440 -1800 -2160 -2520 -2880 -2

10

-1

0

10

10

1

10

Frequency (rad/sec)

Figure S13.18. Bode plots for the exact and approximate models.

As seen in Fig.S13.18, the approximation is good at low frequencies, but not that good at higher frequencies.

13-23