problems_13_1_to_13_17

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f t( ) 0dynes=y 0( ) 100cm=For this problem:

k 1816gm

s2

=kM g

27cm

:=g 980.7cm

s2

=M 50gm:=From the solution of problem 2-9:

y t( )and a second integartion givesd y t( )

dtIntegration of this second derivative results in:

d2

y t( )

dt2

gf t( )

M+

k

My t( )=

Solve for the highest derivative:

Md

2y t( )

dt2

M g k y t( ) f t( )+=

The differential equation representing the motion of the bird mobile is, from Problem 2-9:

13-1. Simulation of Bird Mobile of Problem 2-9.

Solutions to Problems 13-1 to 13-17

Chapter 13. Simulation of Process Control Systems

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The period of oscillation is, as in

the solution to problem 2-9:

Period 2

M

k:=

Period 1.043 s=

The number of complete cycles

in 10 seconds is:

10s

Period9.592=

The simulation plot shows the

same result.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes

only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this workbeyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner

is unlawful.

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Period 1.795 s=Period2

g

L

:=g

L3.501 Hz=The frequency of oscillation is:

(Table 2-1.1)x t( ) x 0( ) cosg

Lt

=r2 ig

L=r1. i

g

L=Roots:

s2 g

L+

X s( ) 0=The solution of the differential equation:

x 0( ) 0.1m=L 0.8m:=M 0.5kg:=g 9.807m

s2

=

d2

x t( )

dt2

g

L x t( )=

Substitute and simplify to obtain:

tan ( ) sin ( )=x t( )

L=

For small angles , from the geometry:

M g tan ( ) Md

2x t( )

dt2

=

Application of Newton's Second Law of Motion:

Mgx(t)

L

Mgsin2

2

13-2. Simulation of a Pendulum

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The number of oscillations in

10 s is:

10s

Period5.572=

The simulation plot shows

the same result.


only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this workbeyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owneris unlawful.

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wi 0.194kg

s=wi Ao 2

M 601300 Pa

Rg T 601300Pa po( ):=

Assume the compressor is initially off and comes on after 200 s (five time constants) with the exact

flow required to maintain the initial pressure:

42.895s=

V 2M

Rg T 601300 Pa 500000 Pa

Ao 2 601300 Pa po( ):=

From the linearization of Problem 2-23, we know that the time constant is:

p 0( ) 500000 101300+( )Pa=T 70 273.16+( )K:=Rg 8.314Pa m

3

mole K:=

po 101300Pa:=M 29gm

mole:=Ao 0.785cm2

:=V 1.5m3

:=Problem parameters:

d p t( )

dt

Rg T

V Mwi t( ) wo t( )( )=Substitute and solve for dp(t)/dt:

t( )M

Rg Tp t( )=Ideal gas law, assuming constant temperature:

wo t( ) Ao 2 t( ) p t( ) po( )=Flow through the orifice:

Vd t( )

dtwi t( ) wo t( )=

In Problem 2-23 the mass balance on the tank produced the following equation:

13-3. Simulation of Punctured Air Tank of Problem 2-23.

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As predicted by the linearized

model, the pressure reaches

steady state in about 200 s.



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Run the simulation for 25 hrs (five time constants). Simulate the oven as a step function from an

inital temperature of 535R to 800R.

d T t( )

dt

A

M cvTs t( )

4T t( )

4=

Integrate the differential equation :

5.16 hr=M cv

4 A 535R( )3

:=

By the linearization done in Problem 2-24, the time constant of the turkey is:

0.1718 108

BTU

hr ft2

R4

:=T 0( ) 535R =cv 0.95BTU

lb R:=

0.6:=Ts 800R:=A 3.5ft2

:=M 12lb:=The parameters, given in this problem are:

M cvdT t( )

dt A Ts t( )

4T t( )

4=

From the solution to Problem 2-24, the differential equation obtained from an energy balace on the

turkey is:

13-4. Simulation of the turkey temperature response of Problem 2-24.

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From the response, the time

constant is much less than 5 hr.

This is because the time constant

gets smaller with temperature. At

800R it is:

M cv

4 A 800R( )3

:= 1.54 hr=

From the response, the actual time

constant seems to be about 2 hr,

which is more in line with how long it

takes to cook a turkey.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposesonly to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this workbeyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner

is unlawful.

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E 27820BTU

lbmole:=

Rg 1.987BTU

lbmole R:= 55

lb

ft3

:= Cp 0.88BTU

lb R:= A 36ft

2:=

fc 0.8771ft

3

min:=

c 62.4lb

ft3

:= Hr 12000BTU

lbmole:= U 75

BTU

hr ft2

R

:= Vc 1.56ft3

:= Cpc 1BTU

lb R:=

Check that initial conditions are at steady state (derivatives = 0):

rA ko e

E

Rg 678.9 R 0.2068

lbmole

ft3

2

:= rA 0.039lbmole

ft3

min

=

f

VcAi 0.2068

lbmole

ft3

rA 5.87 10

4

lbmole

ft3

min

=

f

VTi 678.9R( )

Hr

Cp

rA

U A

V Cp

678.9 602.7( )R 7.915 10

3

R

min=

fc

Vc

Tci 602.7R( )U A

Vc c Cpc678.9 602.7( )R+ 0.027

R

min=

The following is the Simulink diagram for the reactor:

13-5. Non-isothermal Chemical Reactor of Section 4-2.3

lbmole 453.59mole:=Rearranging the model equations from Section 4-2.3:

d cA t( )

dt

f t( )

V cAi t( ) cA t( )( ) rA t( )= cA 0( ) 0.2068lbmole

ft3=

rA t( ) ko e

E

Rg T t( ) cA

2 t( )=

d T t( )

dt

f t( )

VTi t( ) T t( )( )

Hr

CprA t( )

U A

V CpT t( ) Tc t( )( )= T 0( ) 678.9R =

d Tc t( )

dt

fc t( )

Vc

Tci t( ) Tc t( )( )U A

Vc c CpcT t( ) Tc t( )( )+= Tc 0( ) 602.7R =

Design conditions: cAi 0.5975lbmole

ft3

:= Ti 633.5R:= f 1.3364ft3

min:= Tci 540R:=

Parameters:V 13.46ft

3:= ko 8.33 10

8

ft3

lbmole min:=

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The following are the responses for a 0.25 ft3/min increase in process flow at 1 minute followed by

a 0.1 ft3/min increase in coolant flow at 30 minutes.

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Observe the inverse response in the reactor temperature for the change in process flow.


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The folowing is the Simulink diagram for the mixer:

f2 37.5gal

min=f2 f f1:=

f1 62.5galmin

=f1 fcA2 0.025mole cm

3

cA2 cA1

:=

f1 cA1 f2 cA2+ f cA 0=At the initial steady state:

(assuming constant volume)f1 f2+ f=Total mass balance:

Ah 200gal:=f 100gal

min:=cA2 0.05

mole

cm3

:=cA1 0.01mole

cm3


cA 0( ) 0.025 mole

cm3

=

d cA

t( )

dt

f1

t( ) cA1

t( ) f2

t( ) cA2

t( )+ f t( ) cA

t( )

A h=

The model equation, from the solution to Problem 3-1:

13-6. Mixing Process of Problem 3-1

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The responses to a step increase in f1 from 62.5 to 67.5 GPM at 1 minute:

The concentration response is

typical first-order with a timeconstant of approximately 2 min.



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13-7. Feedback control of composition in mixer of Problem 3-1

Introduce the following blocks from the Chapter 13c public nodels:

Figure 13-4.1, F401Vlv1, control valve with time constant of 1 min, linear, maximum flow of

100 gpm, and initial condition of 37.5%C.O.Figure 13-4.3, F403PI, PI controller with initial condition of 37.5% CO.Figure 13-4.7, F407Trmr, transmitter with 1 min time constant, range of 0 to 1 mole/cm3,and initial condition of 0.025 mole/cm3

The controller was tuned for quarter decay ratio response with a gain of 20%CO/%TO and an

intgral time of 1.5 min.

This is the Simulink diagram of the loop (the mixer block is the one from Problem 13-5):

The response to a 5 gpm increase in f1 at 1 minute is:

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The outlet flow increases by 5 gpm

at 1 min and then the controller

increases f2 to bring the outlet

concentration back up to the set

point.

The high controller gain rsults in a

very minor deviation of the outlet

concentration from its set point.



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The responses to a 5 ft3/min step increase in process flow are:

The Simulink diagram is given by:

k 1 min1

=k

f cA1 0.5lbmole ft3

V 0.5 lbmole ft3

:=Initial conditions at steady state:

Di 5.5in:=Lp 400ft:=V 150ft3


cA1 2lbmole

ft3

:=f 50ft

3

min:=Design conditions:

cA3 t( ) cA2 t to( )=

cA2 0( ) 0.5 lbmole

ft3

=

d cA2

t( )

dtfV

cA1 t( ) cA2 t( )( ) kcA2 t( )=

The model equations, from the solution to Problem 3-2, are:

13-8. Isothermal reactor of Problem 3-2

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The concentration response shows a

time constant of about 0.75 min, a

dead time of a little over 1 min, and a

steady state change of 0.38 lbmole/ft3.

The values from the linear model are:

V

f k V+:= 0.75 min=

to

Di2

Lp

4f:= to 1.32 min=

cA2

cA1 0.5lbmole

ft3

f k V+

5ft

3

min:=

cA2 0.038lbmole

ft3

=



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The responses to a 0.1 step increase in feed composition are:

The Simulink diagram for this problem is:

z0

0.513=

y0 0.625=z0

V y0 L 0.4+

F:=y0

0.4

1 1( )0.4+:=

At initial steady state:

2.5:=M 500kmole:=Problem parameters:

V 5kmole

s=V F L:=L 5

kmole

s:=F 10

kmole

s:=Design conditions:

y t( ) x t( )

1 1( )x t( )+=

F V L+=

x 0( ) 0.4=d x t( )

dt

1

M

F z t( ) V y t( ) L x t( )( )=

Rearranging the model equation developed in Problem 3-11:kmole 1000mole:=

13-9. Flash drum of Problem 3-11

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These are typical first-order

responses with a time constant

of about 50 s and a gain on x of

about 1 which match the results

of the linear model in the

solution of Problem 3-11.



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The responses to a 20 ft3/min increase in inlet flow are:

The Simulink diagram for the tray is:

As the response is fast, convert time units to s by multiplying the derivative by 60 s/min.

h0 0.136 ft=

h0

fo

0.415 w 2 g

1

1.5

:=fo fi:=fi 30ft

3

min:=Initial steady state conditions:

S 11.2ft2

:=w 3ft:=Problem parameters:

fo t( ) 0.415 w h t( )1.5

2 g=

d h t( )dt

1S

fi t( ) fo t( )( )=

The model equations from the solution to Problem 3-12 are:

13-10. Distillation tray of Problem 3-12

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The first-order response has a time

constant of approximately 2 s and

the steady-state change is about

0.054 ft. The time costant matches

the one from the linerized model from

the solution of Problem 3-12. Usingthe gain from that solution, the

steady-state change in level should

be:

20ft3

min1

331.4ft2

min1

0.06 ft= close!

The students can check the results

for the change in 10 ft3/min.



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From Table 7-1.1, for a series PID controller tuned for quarter decay ratio response:

Kc

Kcu

1.7:= I

Tu

2:= D

Tu

8:= Kc 147

%CO

%TO= I 1.5min= D 0.38 min=

Control valve (from F401Vlv1): Kv 0.0542m

3

min %CO:= v 0.1min:= (Solution of Problem

6-11)

Initial position:f2

Kv

44.28 %CO=

Transmitter (from F407Trmr): T

3min:= Initial output:50 20

70 20

60%TO=

The series PID Controller block is taken from the Public Model Library, F405PIDs

The Simulink block diagram for the blender conentration control loop is:

13.11. Blending tank of Problems 3-18 and 6-11

The diagram for the blender is essentially the same as for Problem 13-6 with slightly different

notation and the following parameter and design values:

%CO %:=

c1 80 kg

m3

:= c2 30 kg

m3

:= c0 50 kg

m3

:= f 4 m

3

min:= V 40m3:= %TO %:=

At the intial steady state: f f1 f2+= f c0 f1 c1 f2 c2+=

f2 fc0 c1

c2 c1:= f1 f f2:= f1 1.6

m3

min= f2 2.4

m3

min=

From the results of Problem 6-11, the ultimate gain and period are:

Kcu 250%CO

%TO:= Tu 3.01min:=

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The responses to a 0.1 m3/min increase in f1 are:

The decay ratio is somewhat greater

than 1/4. Students may adjust the

controller tuning parameters toimprove the response.

Notice that the concentration can be

controlled very tightly.

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f2 f1:= f1 3m

3

min=

Cv1

f1

h10

:= Cv2

f2

h20

:= Cv1 1.897m

2.5

min=

Cv2 1.897m

2.5

min=

The linearized gains and time constants are:

K1

2 h10

Cv1:= 1

2 A1

h10

Cv1:= K1 1.667 min

m2

= 1 15min=

K2

Cv1

Cv2

h20

h10

:= 2

2 A2 h20

Cv2

:= K2 1= 2 15min=


13-12. Non-interacting tanks in series of Fig. 4-1.1

The model equations developed in Section 4-1.1 are:

d h1 t( )

dt

1

A1

f

i

t( ) f

o

t( ) f

1

t( )( )= h10

2.5m:=

d h2 t( )

dt

1

A2

f1 t( ) f2 t( )( )= h20 2.5m:=

f1 t( ) Cv1 h1 t( )= f2 t( ) Cv2 h2 t( )=

Design conditions: fi 5m

3

min:= fo 2

m3

min:=

Problem parameters: A1 9m2

:= A2 9m2

:=

At initial steady state: f1 fi fo:=

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The responses to a 0.2 m3/min step increase in inlet flow are:

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The responses of the level and

flow for the first tank are first-order

with a time constant of 15 min.

The gains are 1.0 for the flows and

the steady-state changes in level

are about 0.35 m, as predicted bythe linear model:

K1 0.2m

3

min0.333 m=

The responses for the second

tank are second order with the

same steady-state change in

level, meaning that the gain K2 is

unity as predicted by the linear

model.



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13-13. Interacting tanks of Fig. 4-2.1

The model equations developed in Section 4-2.1 are the same as for problem 13-12, except for the

flow betwen the tanks:

f1 t( ) Cv1 h1 t( ) h2 t( )=

Design conditions are the same except for the initial condition in tank 1: h10 5m:=

At the initial steady state: Cv1

f1

h10 h20:= Cv1 1.897

m2.5

min=

It can be shown that the gain of the inlet flow on the level in the second tank is the same as K1 in

Problem 13-12.

K1

1.667min

m2

=

The following is the Simulink diagram for the interacting tanks is series:

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The responses to a 0.2 m3/min step increase in inlet flow are:

The change in the level in the

second tank is the same as in

Problem 13-12. Students maywant to study the effect of

reducing the resistance

between the two tanks by

changing the initial condition on

h1 and recalculating Cv1. For

example, for h10 = 2.6 m,

Cv1

f1

2.6m h20:=

Cv1 9.487

m2.5

min=

The response of the second

tank becomes first-order and

the two tanks behave as a

single tank.



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2 5 min=1 5 min=K2 0=2

V2

fA fB+:=1

V1

fA

:=K2

fB

fA fB+:=

K1 1=K1fA

fA fB+:=This problem is linear with a gains and time constants:

T4 500 K=T2 500 K=T4

fA T2 fB T3+

fA fB+:=T2 T1:=At initial steady state:

T3 500K:=T1 500K:=fB 0m

3

min:=V2 5m

3:=V1 5m

3:=fA 1

m3

min:=

Design conditions:

d T4 t( )

dt

1

V2

fA T2 t( ) fB T3 t( )+ fA fB+( ) T4 t( )=

d T2

t( )

dt

fA

V1

T1 t( ) T2 t( )( )=

The model equations developed in Section 4-1.2 are:

13-14. Non-interacting thermal tanks in series of Fig. 4-1.5

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is unlawful.

The resposes to a 10 K step increase in inlet temperature are:

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The response for the first tank is

first-order with a unity gain and a

time constant of 5 min, matching

the theoretical model. The

response for the second tank is

second-order also with unity ainfor these conditions.

Students may study the effect of

changing flows fA and fB and

temperature T3 on these

responses.

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13-15. Interacting thermal tanks of Fig. 4-2.4

The model equations developed in Section 4-2.2, ignoring fB, are:

d T1 t( )dt

1

V1

fA T1 t( ) fRT4 t( )+ fA fR+( ) T2 t( )=

d T4 t( )

dt

1

V2

fA fR+( ) T1 t( ) T4 t( )( )=

The design conditions and problem parameters are the same as in Problem 13-14, plus the recycle

flow:

fR 1m

3

min:= (The results for fR = 0 are identical to those of Prob. 13-14)

Note: In the model of Section 4-2.2, the recycle flow is assumed to be 0.2*(fA + fB).


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The temperature responses for a 10 K step increase in inlet temperature are:

The students should study the effect

of the recycle flow on the responses.

As the recycle flow is increased, the

temperatures in the two tanks

approach each other and the two

tanks behave as one perfectly mixed

tank with the combined volume of the

two tanks. They should notice that

increasing the recycle flow does not

appreciably change the time to

steady state, or the gain.


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The Simulink diagram for the reactors is:

cA20 0.767lbmol

ft3

=cA20

f1 cA10

f1 k2 V2+:=

cA10 1.726lbmol

ft3

=cA10

fo cAo

f1 k1 V1+ fR

f1

f1 k2 V2+

:=

cA20

f1 cA10

f1 k2 V2+=

f1 20ft

3

min=f1 fo fR+:=At the initial steady state:

fR 10ft

3

min:=For the base case let:

V2 125ft3

:=V1 125ft3

:=fo 10ft

3

min:=

k2 0.2min1

:=k1 0.2min1

:=cAo 7lbmole

ft3

:=Design conditions from Problem 6-15:

cA2 0( ) cA20=d cA2 t( )

dt

f1

V2

cA1 t( ) cA2 t( )( ) k2 cA2 t( )=

cA1 0( ) cA10=d c

A1t( )

dt1

V1

fo cAo t( ) fRcA2 t( )+ f1 cA1 t( )( ) k1 cA1 t( )=

The model equatios are developed in the solution to Problem 4-9:

13-16. Reactors with recycle of Problems 4-9 and 6-15

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The responses to a 0.5 lbmole/ft3 step increase in inlet concentration with a recycle flow of 10

ft3/min are:

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Students shall study the effect of

changing the recycle flow as indicated

in the statement of the problem.

Notice that the initial steady state

conditions vary with the recycle flow.


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The Simulnk block diagram for the extractor is:

f2 14088.6m

3

min=

f2

Ka V c10 me c20( )2 c20

:=

c20 1.2776 104

kmole

m3

=c20

2

f1

ci c10

( )

Ka V

c10+

Ka me V:=

c10 0.04kmole

m3

=c10 1 Rec( ) ci:=At the initial steady state:

V 25m3

:=Ka 3.646min1

:=me 3.95:=Problem parameters:

Rec 90%:=ci 0.4kmole

m3

:=f1 5m

3

min:=Design conditions:

c2 0( ) c20=d c2 t( )

dtKa c1 t( ) me c2 t( )( )

2 f2 t( )

Vc2 t( )=

c

1

0( ) c

10

=

d c1 t( )

dt

2 f1 t( )

V

c

i

t( ) c

1

t( )( ) Ka

c

1

t( ) m

e

c

2

t( )( )=

The model equations developed in the solution to Problem 4-6 are:

13-17. Extraction process of Problem 4-6

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The responses to a 1000 m3/min step increase in solvent flow are:

Obviously the problem parameters

are unreasonable. The large solvent

flow makes for an almost

instantaneous response of theextract composition. The effect on

the raffinate composition is

negligible, as the extract

composition is essentially zero

under the design conditions.

Ask students to try more

reasonable parameter values:

me 0.95:= Ka 209min1

:=



is unlawful.

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problems_13_1_to_13_17