7 Dynamics Tutorial Ans

7/25/2019 7 Dynamics Tutorial Ans

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ANSWERS TO PROCESS DYNAMICS TUTORIAL QUESTIONS

1) What is the difference between lumped parameter and distributed parameter dynamicalsystems? Give an example of each.

Answer

Lumped Parameter Systems Distributed Parameter Systems

States change with time only

Described by ODEs

E.g. Stirred tank reactor

States change with both time and a

spatial dimension

Described by PDEs

E.g.: Plug flow reactor

2) Given the following ODE:

)(2)()(

5)(

102

2

tutydt

tdy

dt

tyd=++

a) What is the Laplace transfer function between the outputy(t)and the input u(t)?

b) What is the time domain solution when the input is a unit step?

Answer

a) [ ] )(2)(1510 2 sUsYss =++

b)From (a)[ ]

)(1510

2)(

2 sU

sssY

++=

If U(s)is a unit step, then ssU /1)( =

Therefore[ ]sss

sY1

1510

2)(

2 ++=

A quick inspection of the transfer functions denominator should reveal that the system hascomplex poles and therefore can be written in the standard form:

2

2 2

n

n ns s

+ + 2

. From the Laplace Transform table provided, the time domain solution will be:

( ) ( )

+

= tSin

tty n

n .11

exp12)( 2

2;

=

2

1 1tan

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Compare terms to get the values of the natural frequency and damping factor:

316.010

1==n and 791.0

316.02

5.0

2

10/5=

=

=

n

and substitute into above solution.

3)Using an appropriate diagram, highlight the major features of the response of an oscillatorysystem to a unit step input.

Answer

A typical response of an oscillatory system is given in the diagram below:

The response can be characterised by the following features: (2 marks each)

a) gain: the ratio C/(magnitude of input step)

b)overshoot: is given by the ratio A/C

c) decay ratio: defined as the ratio B/A

d)rise time: the time taken by the response to first reach its final value

e)response time: the time taken by the response to reach and remain within 5% of itsfinal value

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4)Using appropriate diagrams and numerical examples, explain the terms

a)damping factor

b)inverse response

c) time constant

d)dead-time

e)process poles

f) decay ratio

Answer:

a)Damping Factor

The damping factor determines the degree of damping in a potentially oscillatory system.

With a second order system, is represented by the symbol in the following transfer

function:

22

2

2)(

nn

n

sssG

++

=

A plot of the output of this system to a unit step input for various values of shows theimpact it has on output behaviour.

=0.5

=1

=2

b)Inverse Response

An inverse response is one where the output first moves in one direction before settling out to

an equilibrium at another. It is caused by opposing dynamics, and systems with inverse

responses have zeros that are positive real. An example of an inverse response is shown

below

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c)Time constant

The time-constant of a dynamical process determines the speed with which the system will

respond to an input. A system with a smaller time constant will react faster than one with a

larger time constant, as shown in the following diagram

1

1

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with delay

without delay

time-delay

e)Process poles

Process poles are the roots of the denominator of the process transfer function. They

determine whether the output of the system will oscillate, and whether the system is stable.

The presence of complex poles will indicate that the system is an oscillatory one, while poles

with positive real parts indicates that the system is unstable.

f) Decay Ratio

The decay ratio is a measure of how quickly oscillations will die out and is calculated as the

ratio (C/A) where A is the magnitude of the first overshoot and C is the magnitude of the

second overshoot as shown in the diagram below.

AC

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5)Sketch the responses of the following systems to a unit step change in input:

a)+

1

1 10s b)

2

2 22s s c)

2

1 15

5e

s

s

+

d)1

2s e)

( )( )1

1 2 1 3

+ +

s

s s) f)

1

0 7 12s s+ +.

Your sketches should reveal the essential characteristics of each system

Answer:

+

1

1 10s

(negative gain)

2

2 22s s

(unstable - exponential)

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2

1 15

5e

s

s

+

(first-order plus time

delay)

1

2s(ramping)

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( )( )1

1 2 1 3

+ +

s

s s)

(inverse response)

1

0 7 12s s+ +.(oscillatory)

6)A closed loop system has the following block diagram:

K ( )( )1

1 1 3+ +s s)+-R(s) Y(s)

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a)Determine the transfer function between Y(s) and R(s), where Y(s) =L{y(t)} andR(s) =L{r(t)}

b)Find the value of K that will give the fastest non-oscillatory response in Y(s) to a step changein R(s)?

c)What is the corresponding ODE that relates r(t) to y(t)?

Answer:

a)Y s

R s

K

s s K

( )

( ) ( )=

+ + +3 4 12

b)For fastest non oscillatory response, poles must be real and equal. Therefore find the value ofK that sets 16-4*3*(1+K) = 0, i.e. K=1/3.

c) 3 4 12

2

d y tdt

dy tdt

K y t Kr t ( ) ( ) ( ) ( ) (+ + + = )

7)Step response curves are often used to develop approximate first-order plus time-delaytransfer function models of systems. On having determined the gain and time-delay values,

the time constant is estimated as the time taken for the response to reach 63.2% of the final

change in output. What is the rationale for this?

Answwer

The transfer function of a first-order systems is:

s

K

sU

sYsG

+==

1)(

)()(

The time domain solution depends on the form of the input. If the the input is a unit step change

in u(t), then:

L{u(t)} =L{1} = U(s)= 1/s

Therefore, ignoring the time-delay for the moment,

Y s K

s sK

s s( )

( )

( / )

( / )=

+ =

+11 1

1

which has the time domain solution:

y t K e t( ) /= 1 or Y t K e Y t ss( )/= +1

Thus time tends to infinity, the exponential term will decay to zero and hence y(t) will tend

towards the gain of the process. Further, at time t = ,y(t)is:

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y t K e K( ) .= =1 0 6321

The time constant, , is therefore the time taken for the process to reach 63.2% of its finalchange in value.

8)What is the transfer function between the input U(s)and the output Y(s)of the followingsystem:

s51

1

+

s+11

s51

1

+

+

+

+

-U(s)

Y(s)

)1)(51(

)31(

)(

)(

ss

s

sU

sY

++

=

9)Estimate the parameters of a 1st

order-plus-time-delay model for the system that has thefollowing step response.

8.0

9.0

10.0

11.0

12.0

13.0

14.0

15.0

0 50 100 150 200

Time (secs)

Output

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4.0

4.5

5.0

5.5

6.0

6.5

7.0

7.5

0 50 100 150 200

Time (secs)

Input

Answer

Sketch a smooth curve to approximate the noisy output response, then apply the process reaction

curve method to the smoothed response, the estimated transfer function should be:

s

s

sU

sY

201

)20exp(2

)(

)(

+

=

10) In an isothermal continuous stirred tank reactor, the relationship between theconcentration of component A in the reactor, CA, and its concentration in the feed, CAo, is

described by the following differential equation:

dC t

dt C t C t

A

A A

( )( ) ( ),+ +

=

1 10

A unit step change in CA,0caused the following change in CA

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0 20 40 60 80 100 1200

0.5

1

1.5

2

2.5

3

Time (secs)

Changeinconc.ofA

inrea

ctor

From the plot, determine the constants and .

Answer:

From the plot, the gain of the process is approximately 2.5, while the time constant is

approximately 20 secs.

Comparing this to the above ODE, we have

1

20+

= and1

12 5

+ =

. , ie. 2 equations with 2 unknowns, which can be solved to give

= 8 and = -0.075

11)

Qin

Qout

h

The schematic on the left shows a header tank.

Qin and Qout are volumetric flows, and h is the

level of liquid in the tank.

Develop a dynamic model of the system that

will enable study of how the level will change

when there are changes in Qin and Qout.

State all assumptions that you make.

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The schematic does not provide sufficient information for a mass balance. The flows are

volumetric flows. So, we need a density term () to convert from volume units to mass units. Wealso need to know the cross-sectional area (A) of the tank, so that we can determine the volume

holdup in the tank, and hence the mass holdup.

Given these information, the mass balance of the tank can be written as:

d Ah

dt Q Qin out

( ) =

Since there is no heating effects, density can be assumed constant. Also, since the tank is

cylindrical, it has constant cross-sectional area. Therefore, the ODE becomes:

Adh

dt Q Qin out = and hence

Adh

dt Q Qin out =

The above equation is correct as a mass balance, but is not really in the right form for solution.Note that the flow out, , is determined by the pressure exerted by the liquid, and is given by:Qout

Q kout= h

Therefore, the mass balance should be written as:

Adh

dt Q k hin= and finally as ( )

dh

dt Q k hin= /A

12) The schematic of a level tank system is shown below:

qi

qo

h

qi and qo are volumetric flow rates. The cylindrical tank is open to atmosphere and the outlet

flow rate can be assumed to be proportional to liquid level.

a) Develop a dynamic mass balance for the system, focusing on the relationship between the

level, h, and input flowrate, qi.

b) Determine the Laplace transfer function relating changes in the level, h, to changes in inputflowrate, qi.

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c) From the transfer function determined in (b), calculate the final value of the level, h, when

the input flow rate undergoes a unit step-change in magnitude.

d) Find the time domain solution of the transfer function determined in (b).

Answer

a) Given these information, the mass balance of the tank can be written as:

oi qq

dt

Ahd=

)(where is the density of the liquid and A is the cross sectional area

of the tank.

Since there is no heating effects, density can be assumed constant. Also, since the tank is

cylindrical, it has constant cross-sectional area. Therefore, the ODE becomes:

oi qqdt

Adh =

The flow out, , is determined by the pressure exerted by the liquid, and can be approximated

linearly by:

oq

Rhqo /= whereRis the resistance to flow

The final ODE is therefore: io Rqqdt

dhAR =+

b) Taking Laplace Transforms of the ODE above, the resulting transfer function between level

and flowrate in is:ARs

R

sQ

sH

o +=

1)(

)(

c) The final value of level to a unit step change in flowrate in isR.

d) The time domain solution to the transfer function in (b) when the input is a unit step is:)]/exp(1[)( ARtRth =

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13) The schematic of a mixing system is shown in the following diagram.

TANK 1 TANK 2

TANK 3

F0

O1

O2

O3

F1

F2

F0,1 F0,2

The feed, at a flowrate of F0, is split into 2 equal streams and fed to two separate tanks. A

substance is added to tank 1 at a flowrate of F1, while another substance is added to tank 2 at a

flowrate of F2. The mixtures from both tanks flow into tank 3, with flowrates O1 and O2

respectively. Further mixing takes place in tank 3 and the product leaves tank 3 with a flowrate

of O3.

Assuming that

- the above flowrates are expressed in mass per unit time

- perfect mixing occurs in all the tanks

- the flow out of a tank is proportional to the level of liquid in that tank

a) Develop the dynamic material balances for each of these tanks

b) If F1=F0,1and F2= F0,2, develop the Laplace transfer function that relates F0to the level in

tank 3

c) Explain why the level in tank 3 will not oscillate when there is a step change in F0.

Answer

a) The material balances for tanks 1-3 are respectively:

)()()()(

11,11

1 tOtFtF

dt

tdhA o +=

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)()()()(

22,22

2 tOtFtFdt

tdhA o +=

)()()()(

3213

3 tOtOtOdt

tdhA +=

WhereAiand hiare the cross-sectional areas and the liquid levels of tanks 1 to 3.

b)

IfF1=F0,1andF2= F0,2, then the material balances for tanks 1 and 2 become (becauseF0splits

equally toF0,1andF0,2) :

)()()(

11

1 tOtFdt

tdhA o = )()(

)(2

22 tOtF

dt

tdhA o =

Taking Laplace Transforms of all three ODEs, and assuming the using of deviation variables:

)()()( 111 sOsFssHA o =

)()()( 222 sOsFssHA o =

)()()()( 32133 sOsOsOsHA +=

Since outflows are proportional to levels in the tanks, i.e. 111 /)()( RsHsO = ,and , then:

222 /)()( RsHsO =

333 /)()( RsHsO =

)()()()( 111111 ssORAsOsFssHA o == which, after rearrangement, will give:

[ ] )()(1 111 sFsOsRA o=+

Similarly, for tank 2

[ ] )()(1 222 sFsOsRA o=+

While for tank 3

[ ] [ )()()(1 213333 sOsORsHsRA +=+ ]

Substituting for the outflows of tanks 1 and 2, we get:

[ ]

++

+=+

sRAsRAsFRsHsRA o

2211

33331

1

1

1)()(1

Thus the transfer function between the level in tank 3 and Fois:

++

+

+=

sRAsRAsRAR

sFsH

o 221133

33

11

11

11

)()(

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c) Because all the poles of the transfer are real, the level in tank 3 will not oscillate.

14)

The diagram on the left show a heated stirred tank.

The heating mediums flow rate is S[kgmin-1

]. Thetemperature of the bulk liquid it T [degC]. Feed

liquid enters the system at Fi [kgmin-1] and a

temperature of Ti [degC]. This flow is used to

maintain the liquid level in the tank. The heated

liquid leaves the tank with a flowrate of at

Fo[kgmin-1] and a temperature of To[degC].

a) Assuming perfect level control, develop a dynamic energy balance for the system [40%]

b) From (a), obtain the transfer function that describes the effects of changes in heatingmedium flowrate on outlet stream temperature. [10 %]

c) What modifications will you have make to the balance equation(s) if the level is notcontrolled perfectly [50%]

Answer

a) Perfect level control implies that liquid level is constant, which means that:

F F Fi o= = .

There are now several pieces of information to gather, namely:

= latent heat of vaporisation of steam

Cp= heat capacity of the liquid

Assume that heat is transferred to the system purely by condensing steam and that the heatcapacity of the liquid is constant. Assume also that the tank is well stirred, so that the

temperature of the output stream is equal to the temperature of the liquid in the tank.

Using the following general dynamic heat balance equation:

Rate of Energy Accumulation = Rate of Energy Input - Rate of Energy Consumption

Rate of Energy Input = FC T T Sp i a( ) + where Tais the ambient temperature.

Rate of Energy Consumption = FC T Tp o a( )

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Rate of Energy Accumulation = MCdT

dtpo

whereMis the mass of liquid in the tank.

Thus the dynamic model of the stirred tank heating system is:

)()( aopaipp TTFCSTTFCdtdTMC +=

Simplification yields

+= STTFCdt

dTMC oip

op )(

b)

Since the contribution of inlet temperature is not considered, and that Laplace variables have

zero initial values, the required transfer function is obtained from:

=+ STFCdt

dTMC op

op or

SC

Tdt

dT

F

M

p

oo =+

The transfer function is therefore:

sFM

C

sS

sT po

)/(1

)/(

)(

)(

+=

c) If the controller does not provide perfect level control, then a mass balance will have to be

written. That is,

( )dh

dt F k h A

i= /

where h is the level, and A is the cross sectional area of the tank. Since the flow in is being

manipulated by the controller (proportional controller say), we will need to describe this as well.

F k h h Fi c s i= +( ) ,0

where is the value of initial input flow rate; hFi ,0 sis the desired level and kcis the gain of the

controller. The energy balance will then have to be modified to:

)()( aopoaipio

p TTCFSTTCFdt

dTMC += , with = ..hAM

MCdT

dtpo

= F C T T Wi p i i a s, ( ) + F C T To p o o a, ( )

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15) A system is described by the following ODE.

)5(2.0)(1.0)(

=+ tutydt

tdy

a) Determine the Laplace transfer function between the outputy(t)and the input u(t)

b) Sketch the response of the system when the input is a step of magnitude 2.

c) Determine the time domain solution of the ODE when the input is a step of magnitude 2.

Answer

a)( ) 2exp( 5 )

( ) 1 10

Y s s

U s s

=

+

b)

0 10 20 30 40 50 600

0.5

1

1.5

2

2.5

3

3.5

4

Time (units)

Output

c) where is a unit step function.)5()]10/exp(1[4)( = tutty )(tu

16) The concentration of component A (xA) in the top product stream of a distillation columnis related to changes in the reflux flowrate (Re ) according to the ODE:

20 2dx

dtx

A

A+ = .Re

The equilibrium value of xAis 95 weight percent when Re is 10 kgmin-1.

a) stating clearly any assumptions, derive the transfer function between xAand Re

b)find the time domain solution showing the evolution of xAto a step change in Re

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c)what is the value of xA, 20 mins. after Re is reduced to 9 kgmin-1

Answer:

a) state using deviation variables, then transfer function would be:x s

s s

A ( )

Re( )=

+

2

1 20

b)assuming a unit step change, then Re(s)=1/s and x ss sA

( ) .=+

2

1 20

1, which from tables will

give [ ]x t tA ( ) exp( / )= 2 1 20

c)The time constant of the system is 20 mins. Therefore the change in xAdue to a -1 change inRe after 20 mins. would be -1*2*0.632 = -1.264. Hence the final value of xAat this point in

time will be (95-1.264)=93.736 weight percent

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7 Dynamics Tutorial Ans