Eg55p7_tutorial_06_solutions.pdf

EG55P7 - Tutorial 6 Cascade control Problem 1. For the paper drying process shown in the figure below, the following information is available: The flow control loop (FIC47) can be represented by a first-order lag with a gain of 4 (m3 h-1)/%CO and a time constant of 0.1 min. The transfer function of the air heater outlet temperature to the fuel flow is a second-order lag with time constants of 2 min and 0.8 min. A change in fuel flow of 1 m3 h-1 causes a change of 2°C in the outlet air temperature. The drier can be represented by a first-order lag with a time constant of 5 min. A change of 1°C in inlet air temperature causes a change in outlet moisture of 0.5 mass percent. The moisture transmitter (MT47) has a range of 0 to 6 mass percent and a negligible time constant.

(a) For the control scheme shown in the figure, draw the block diagram of the moisture control loop showing the transfer functions. Decide on the fail position of the control valve and the controller action, and make sure that the signs in your block diagram correspond to your decisions. (b) Consider a cascade control scheme using an outlet temperature sensor with a range of 120 to 170°C installed on the air line from the heater to the drier and a temperature controller to manipulate the fuel flow set point; the output of the moisture controller (MC47) sets the set point of the outlet air temperature controller. Draw the instrumentation diagram and the block diagram of the cascade control scheme. Show the defined transfer functions on the block diagram, and specify the action of each controller.

Solutions (a) This is a cascade control scheme. The flow control loop can be represented by a first-order lag process, so the transfer function will follow the generic equation G  s  

K , where K is the gain  s 1

and τ the time constant. This process refers essentially to the dynamic response of the valve manipulating the flow. The valve receives a signal from the FC-47 controller (%CO-controller output) and affects accordingly the flow (m3 h-1). Hence we have:

GV 



KV 4  m3h 1 / %CO  V s  1 0.1s  1



The air heater can be represented by a second-order process, so the transfer function will follow the generic equation G  s  

K , where K is the gain and τ the natural period of oscillation  s  2 s  1 2 2

and ζ the damping factor. As is known from process control theory, this second order transfer function can be further proven to be equal to the product of two first order processes

G s 

K

1s  1 2 s  1

with τ1 and τ2 being the equivalent time constants. Moreover, the gain

specifies essentially how much the output changes per unit change in input, or how much the input affects the output. That is, the gain defines the sensitivity relating the output and input variables, which can be defined mathematically as follows: 𝐾=

𝛥 𝑜𝑢𝑡𝑝𝑢𝑡 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝛥 𝑖𝑛𝑝𝑢𝑡 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒

It is given that a change in fuel flow of 1 m3 h-1 causes a change of 2 °C in the outlet air temperature, so the gain of the air heater transfer function will be equal to 2 °C/(m3 h-1). Hence we have:

GH 

KH 2   H ,1s  1 H ,2 s  1  2s  1 0.8s  1



o

C / m3h1



The drier can be represented by a first-order lag process, so the transfer function will again follow the generic equation G  s  

K . It is given that a change of 1°C in inlet air temperature causes a  s 1

change in outlet moisture of 0.5 mass percent. Here is needs to be also considered that an increase in air temperature leads to a decrease in outlet moisture and vice versa. So the gain of the drier transfer function will be equal to -0.5 %Moist./°C. Hence we have:

GD 

KD 0.5  % Moist. / o C  D s  1 5s  1





It is also given that the moisture transmitter has a range of 0 to 6 mass percent and a negligible time constant. This means it can be represented as a zero-order transfer function, essentially just a gain. The transmitter receives an input signal of moisture mass percent (ranging from 0 to 6%) and

generates an output signal (TO-transmitter output) that ranges from 0 to 100%. As such, its gain is equal to:

GMT  s   K MT 

100  0  %TO  16.67 %TO / %Moist.    6  0  %Moist.

For the flow transmitter no data are given. Assuming that it also has very fast dynamics so that it can be represented as a zero-order transfer function and that it has a measuring range from 0 to 100 m3 h-1, its gain will be:

GFT  s   K FT 

100  0  %TO  1 %TO / m3h1   100  0  m3h1

The respective block diagram considering the above is given below: MC-47 R(s) +

E(s)

%TO -

%

GCM

FC-47 MM(s) + %CO

-

EF(s)

GCF

%

Valve MF(s) %COF

4 0.1s  1

Heater F(s) m3/h

2

 2s  1 0.8s  1

Drier TH(s) o

C

0.5 5s  1

X(s) % Moist.

FT-47 TOF(s) %

1 MT-47

TO(s) %

16.67

Regarding the failure position of the valve, considering it refers to fuel delivery in a combustion chamber it is logical to assume that it must be a Fail-Close (FC) valve. This means that a 0% signal from the respective controller (FC-47) corresponds to a 0% valve opening, hence no flow. Conversely, a 100% signal from the controller corresponds to a fully open valve, hence maximum flow. Regarding the actions of the controllers, we have:  FC-47: Flow higher than Set-point  Valve opening must reduce  Signal to FC valve must reduce  Reverse action  MC-47: Moisture higher than Set-point  Drier temperature must increase  Fuel flow to heater must increase  Set-point to FC-47 must increase  Direct action No data are given for the types of the controllers, but given the actions above the gain (Kc) of FC-47 will be positive, while that of MC-47 will be negative. For example for FC-47, if the flow is higher than the set-point, then EF will be negative. At the same time we saw that the valve must close, so the controller’s output signal must reduce, meaning that in deviation variables it will also be negative. Hence, the gain of the controller must be positive, since both its input and output have the same sign.

(b) This is a three-level cascade control scheme, enhancing the operation of the previous one. The updated instrumentation diagram is shown below:

Paper product

Wet paper

Drier

MT Moist. MC 47 47

Hot air

Tset

T

TT 47

Ambient air

SP

TC 47

Hot gases

Heater

vp

FT 47

Fset

FIC 47

F

Combustion Fuel air

For the new temperature transmitter again it is assumed that it has very fast dynamics, hence it can be represented as a zero-order transfer function. It has a measuring range from 120 to 170 oC, so its gain will be:

GTT  s   KTT 

100  0  %TO  2 %TO / oC   170  120  oC

The updated block diagram considering the above is given below: MC-47 R(s) +

E(s)

%TO -

%

GCM

FC-47

TC-47 MM(s) +

ET(s)

%CO -

%

GCT

MT(s) +

EF(s)

%COT -

%

GCF

Valve MF(s) %COF

4 0.1s  1

Heater F(s) m3/h

2

 2s  1 0.8s  1

Drier TH(s) o

C

0.5 5s  1

X(s) % Moist.

FT-47 TOF(s) %

1 TT-47

TOT(s) %

2 MT-47

TO(s) %

16.67

The failure position of the valve, considering it still refers to the same manipulated variable, remains Fail-Close (FC).

For the actions of the controllers, we now have:  FC-47: Flow higher than Set-point  Valve opening must reduce  Signal to FC valve must reduce  Reverse action  TC-47: Temperature higher than Set-point  Heater temperature must decrease  Fuel flow to heater must decrease  Set-point to FC-47 must decrease  Reverse action  MC-47: Moisture higher than Set-point  Drier temperature must increase  Hot air temperature from heater must increase  Set-point to TC-47 must increase  Direct action Again, given the actions above the gains of FC-47 and TC-47 will be positive, while that of MC-47 will be negative.

Problem 2. Consider the jacketed continuous stirred tank reactor (CSTR) sketched in the figure below. The following information is obtained from testing the reactor and its control system. The transfer function of the reactor temperature to the jacket temperature is a first-order lag with a gain of 0.6 oC/oC and a time constant of 13 min. The transfer function of the jacket temperature to the coolant flow is a first-order lag with a gain of - 2.0 oC/(kg s-1) and a time constant of 2.5 min. The control valve is linear with constant pressure drop and is sized to pass 12 kg s-1 when fully opened. Its time constant is negligible. The reactor temperature transmitter is calibrated for a range of 50 to 100 °C and has a time constant of 1 min. The jacket temperature transmitter is calibrated for a range of 0 to 100 °C, and its time constant is negligible.

(a) Decide on the proper fail position of the control valve and the action of the controller for a simple feedback control loop with the reactor temperature controller manipulating the position of the

coolant valve. Draw the block diagram showing all transfer functions and write the closed-loop transfer function of the reactor temperature to its set point. Pay particular attention to the signs, which must correspond to the fail position of the valve and the controller action. (b) Write the characteristic equation for the single feedback loop. Using the Routh-Hurwitz stability criterion examine its stability characteristics and calculate the gain at the critical stability limit. (c) Design a cascade control system for the reactor temperature with the jacket temperature as the intermediate process variable, specifying the action of both controllers. Draw the complete block diagram for the cascade control system showing all transfer functions and their signs. (d) Assuming a P slave controller with a gain of 2%CO/%TO, write the transfer function for the jacket temperature loop and redraw the block diagram with the jacket temperature loop as a single block. (e) Using the simplified block diagram from part (d), write the characteristic equation of the reactor temperature loop in the cascade control system and use again the Routh-Hurwitz stability criterion to examine its stability characteristics and calculate the gain at the critical stability limit.

Solutions (a) Regarding the failure position of the valve, considering it refers to coolant delivery in a reactor, the safe position to assume is Fail-Open (FO). This means that a 0% signal from the respective controller (TC) corresponds to a 100% valve opening, hence maximum flow. Conversely, a 100% signal from the controller corresponds to a fully closed valve, hence no flow. Regarding the action of the controller, we have:  Temperature higher than Set-point  Coolant flowrate must increase  Valve opening must increase  Signal to FO valve must decrease  Reverse action  Gain positive (error negative) The transfer function of the reactor temperature to the jacket temperature can be represented by a first-order lag process, so the transfer function will follow the generic equation G  s  

K . In  s 1

the given data, it can be seen that the gain is 0.6 oC/oC, logically a positive number, since an increase in the temperature of the coolant (and the jacket) would lead to an increase in the temperature of the reactor. Hence we have:

GR  s  

T s KR 0.6 o o  R  C/ C  R s  1 TJ  s  13s  1





The transfer function of the jacket temperature to the coolant flow can also be represented by a first-order lag process. In the given data, it can be seen that the gain is - 2.0 oC/(kg s-1), logically a negative number, since an increase in the flowrate of the coolant would lead to a decrease in the temperature of the jacket. Hence we have:

GJ  s  

T s KJ 2 o  J  C / kg s 1  J s  1 FC  s  2.5s  1





It is also given that the valve is sized to pass 12 kg s-1 when fully open and has a negligible time constant. This means it can be represented as a zero-order transfer function. The valve, since it is FO, receives a signal from the controller ranging from 0 to 100% and manipulates the flow from 12 to 0 kg s-1. As such, its gain is equal to:

GV  s   KV

FC  s 

12  0  kg s 1     0.12  kg s 1 / %CO  M  s   0  100  %CO

The reactor temperature transmitter from the data given can be represented by a first-order lag process. It has a range from 50 to 100 °, so its gain will be:

K MR 

 0  100  %TO  2 %TO / oC and its transfer function will be:    50  100  oC

GMR 

TO  s  K MR 2   %TO / oC  MR s  1 TR  s  s  1





The respective block diagram considering the above is given below: TC R(s) + %TO -

E(s) %

GC

Valve M(s)

0.12

%CO

Jacket FC(s) kg/s

2 2.5s  1

Reactor TJ(s) o

C

0.6 13s  1

TR(s) o

C

TT TO(s) %

2 s 1

(b) The characteristic equation of this loop is (following the rule: 1 + “product of blocks in the loop”): 1 + 𝐺𝑐 (𝑠)𝐺𝑣 (𝑠)𝐺𝐽 (𝑠)𝐺𝑅 (𝑠)𝐺𝑀𝑅 (𝑠) = 0 Assuming that the temperature controller is Proportional only, since no data are given, the above can be expressed in polynomial form to apply the Routh-Hurwitz stability criterion: 1 + 𝐾𝑐 (−0.12) ( 1 2 3 4

32.5 48 A1 B1

−2 0.6 2 ) = 0 ⇒ 32.5𝑠 3 + 48𝑠 2 + 16.5𝑠 + 1 + 0.288𝐾𝑐 = 0 2.5𝑠 + 1 13𝑠 + 1 𝑠 + 1 16.5 1+0.288Kc 0 0

At the critical stability limit all elements of row n=3 are equal to zero, which means: 𝐴1 = 0 ⇒

48.5 ∙ 16.5 − 32.5 ∙ (1 + 0.288𝐾𝑐 ) =0⇒ 48

48.5 ∙ 16.5 = 32.5 ∙ (1 + 0.288𝐾𝑐 ) ⇒ 𝐾𝑐 = 81.14 (%𝐶𝑂/%𝑇𝑂) The gain at the critical stability limit is typically called the ultimate gain (𝐾𝑐𝑢 ). (c) The updated diagram of implementing the cascade control scheme is shown below: TRset

TR 1 TJset TC 2

1

2

vp

The jacket temperature transmitter has a range of 0 to 100 °C and its time constant is negligible, so its gain is equal to:

GMJ  s   K MJ 

100  0  %TO  1 %TO / oC   100  0  oC

The updated block diagram of the cascade control scheme is given below: TC-1 R(s) +

E1(s)

%TO -

%

GC1

Valve

TC-2 M1(s) +

E2(s)

%CO1 -

%

GC2

M2(s) %CO2

0.12

Jacket FC(s)

kg/s

2 2.5s  1

Reactor TJ(s) o

C

0.6 13s  1

TR(s) o

C

TT2 TO2(s) %

1 TT1

TO1(s)

%

2 s 1

The valve obviously remains Fail-Open (FO). Regarding the action of the controllers, we have:  Jacket temperature controller (TC-2): Temperature TJ higher than Set-point  Coolant flowrate must increase  Valve opening must increase  Signal to FO valve must decrease  Reverse action  Gain positive  Reactor temperature controller (TC-1): Temperature TR higher than Set-point  Coolant temperature must decrease  Set-point to TC-2 must decrease  Reverse action  Gain positive

(d) The transfer function for the jacket temperature loop can be found with normal procedures (following the rules: numerator = “product of blocks between input and output”, denominator = 1 + “product of blocks in the loop”), where the secondary loop is treated as an independent system (highlighted below): TC-1

%

GC1

M1(s) +

E2(s)

%CO1 -

%

GC2

M2(s) %CO2

TO2(s) %

0.12

Jacket FC(s)

kg/s

Reactor

2 2.5s  1

0.6 13s  1

TJ(s) o

C

TR(s) o

C

1 TT1

TO1(s)

%

2 s 1

−2 ) 0.48 0.324 2.5𝑠 +1 𝑇𝐽 (𝑠) = 𝑀1 (𝑠) = 𝑀1 (𝑠) = 𝑀 (𝑠) −2 2.5𝑠 + 1 + 0.48 1.69𝑠 + 1 1 )1 1 + 2(−0.12) ( 2.5𝑠 + 1 2(−0.12) (

TC-1 TC-2

% -

) E +1(s)

1

E1(s)

%TO -

Valve

TC-2

TT2

) E +1(s)

1

R(s) +

% -

TC-1 Valve TC-2

Valve TC-2 TC-1 Jacket

Jacket Valve Reactor TC-2

So the block diagram can be redrawn as follows: +1(s) +M E21(s) E2(s)R(s) + M1(s) (s) 0.324 + (s) E2(s)2 M1(s) F M2(s)E2(s) EM (s) F T +JC(s) 0.62 E2(s) R(s) M GC2 C1 2 GC GC2 0.12 GC2 GC 0.12 GC2 o G0.12 C

Reactor Jacket Valve FJ(s) (s) TRT (s) M C 2(s)

C 2 2.5 TC-2 1 % Ckg/s 2.5 s  %CO 1TC-21 kg/s %13 %CO sJacket sValve s11 -C%CO21.69 Valve TC-1 Jacket Reactor Reduced secondary +1(s) +M E21(s) E2(s)R(s) (s) FJ(s) (s) 0.324 + (s) E2(s)TT 2 2 M (s)E2(s) EM (s) F T TT (s) + (s) 0.6loop 2 2 M1(s) F M E2(s) R(s) M C(s) TT TT2+ M1(s) GC2 C1 2 GC2 oJC 2 cascade GC1 GC2 0.12 GC2 oR o 2 GC10.12 G0.12 C1 %CO kg/s kg/s C 2.5 s  1 %TO % %CO % %CO % %CO kg/s C C 2.5 s  1 % %CO % %TO%CO %CO 13 s  1 - 1TO 1  s 1.69 1 -2% 1 2 2 2 TO2(s) TO2(s) TO2(s) 2 (s) 1 1 1 Reduced secondary % % % % cascade loop TT2 TT2 TT2 %CO %TO %TO %CO %TC-1 - 1 %CO 1% - %COkg/s - % 2 % TC-1 - 2 Valve TC-2 TC-2 1

1

TO2(s) % TO1(s)

TT11

%

2 s 1

TO1(s)

2

TT1

TO2(s) % TO1(s)

1

TT11

%

2 s 1

TO1(s)

2

TO2(s) % TO1(s)

TT11

%

2 s 1

TO1(s)

2

TT1

TT1

o o

TO2(s) % TO1(s)

Reactor Jacket

0.6 2 0.12 13 s2.5  1s  1 Reactor Jacket Valve

TCR(s) TJ(s) F

0.6 TT22 0.12 13s2.5  1s  1

TCR(s) TJ(s) F

o o

CC kg/s

o o CC kg/s

0.6 2 13 ss 11 2.5 Reactor Jacket

TTJ(s) R(s)

0.6 2 13ss 11 2.5

TTJ(s) R(s)

1 TT2 TT11

%

2 s 1

TO1(s)

2

TT1

(e) Assuming again a P temperature controller,%the characteristic equations of this loop is now: % % s 1 s 1 1 % s 1 1 + 𝐾𝑐

0.324 0.6 2 = 0 ⇒ 21.97𝑠 3 + 36.66𝑠 2 + 15.69𝑠 + 1 + 0.39𝐾𝑐 = 0 1.69𝑠 + 1 13𝑠 + 1 𝑠 + 1

Applying the Routh-Hurwitz stability criterion: 1 21.97 15.69 2 36.66 1+0.39Kc 3 A1 0 4 B1 0 At the critical stability limit all elements of row n=3 are equal to zero, which means: 𝐴1 = 0 ⇒

Reactor

36.66 ∙ 15.69 − 21.97 ∙ (1 + 0.39𝐾𝑐 ) =0⇒ 36.66

36.66 ∙ 15.69 − 21.97 ∙ (1 + 0.39𝐾𝑐 ) ⇒ 𝐾𝑐 = 64.57 (%𝐶𝑂/%𝑇𝑂) The cascade system is actually less stable, since it ultimate gain is smaller than (b). This effect however is correlated to the probably poor choice of the value of the gain of the inner loop.

oo

CC

oo

CC

0.6 13 s 1 Reactor

0.6 13s  1