Solution Manual: Process Dynamics and Control (Second Edition)
Dale E. Seborg, Thomas F. Edgar, Duncan A. Mellichamp
2004, John Wiley and Sons Inc.
1234567898
1.1
a)
b)
c)
d)
e)
True
True
True
False
True
1.2
QL
Q
T
TC
ON/OFF SWITCH
Controlled variable- T (house interior temperature)
Manipulated variable- Q (heat from the furnace)
Solution Manual for Process Dynamics and Control, 2nd edition,
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp.
1-1
Disturbance variable- QL (heat lost to surroundings); other possible
sources of disturbances are the loss of gas pressure and the outside door
opening.
Specific disturbances include change in outside temperature, change in
outside wind velocity (external heat transfer coefficient), the opening of
doors or windows into the house, the number of people inside (each one
generating and transmitting energy into the surrounding air), and what
other electric lights and appliances of any nature are being used.
1.3
The ordinary kitchen oven (either electric or gas), the water heater, and the
furnace (Ex. 1.2) all work similarly, generally using a feedback control
mechanism and an electronic on-off controller. For example, the oven uses
a thermal element similar to a thermocouple to sense temperature; the
sensor's output is compared to the desired cooking temperature (input via
dial or electronic set-point/display unit); and the gas or electric current is
then turned on or off depending on whether the temperature is below or
above the desired value. Disturbances include the introduction or removal
of food from the oven, etc. A non-electronic household appliance that
utilizes built-in feedback control is the water tank in a toilet. Here, a float
(ball) on a lever arm closes or opens a valve as the water level rises and
falls above the desired maximum level. The float height represents the
sensor; the lever arm acting on the valve stem provides actuation; and the
on-off controller and its set point are built into the mechanical assembly.
1.4
No, a microwave oven typically uses only a timer to operate the oven for a
set (desired) period of time and a power level setting that turns the power
on at its maximum level for a fixed fraction of the so-called duty cycle,
generally several seconds.
Thus setting the Power Level at 6 (60% of full power) and the Cook Time
to 1:30 would result in the oven running for a total of one and one-half
minutes with the power proportioned at 60% (i.e., turned on 100% for 6
seconds and off for 4 seconds, if the fixed duty cycle is 10 seconds long).
This type of control is sometimes referred to as programmed control, as it
utilizes only time as the reference variable .
1-2
The big disadvantage of such an approach is that the operator (here the
cook) has to estimate what settings will achieve the desired food
temperature or will cook the food to the desired state. This can be
dangerous, as many people can attest who have left a bag of popcorn in
the oven too long and set the bag on fire, or embarrassing, as anyone
knows who has served a frozen meal that did not quite thaw out, let alone
cook. What good cooks do is provide a measure of feedback control to the
microwave cooking process, by noting the smell of the cooking food or
opening the door and checking occasionally to make sure it is heating
correctly. However, anyone who has used a microwave oven to cook fish
filets, for example, and blown them all over the oven, learns to be very
conservative in the absence of a true feedback control mechanism. [Note
that more expensive microwaves do come equipped with a temperature
probe that can be inserted into the food and a controller that will turn off
the oven when the temperature first reaches the desired (set point) value.
But even these units will not truly control the temperature.]
1.5
a)
In steering a car, the driver's eyes are the sensor; the drivers hands and the
steering system of the car serve as the actuator; and the driver's brain
constitutes the controller (formulates the control action i.e., turning the
steering wheel to the right when the observed position of the car within its
desired path is too far to the left and vice versa). Turns in the road,
obstructions in the road that must be steered around, etc. represent
disturbances.
b)
In braking and accelerating, a driver has to estimate mentally (on a
practically continuous basis) the distance separating his/her car from the
one just ahead and then apply brakes, coast, or accelerate to keep that
distance close to the desired one. This process represents true feedback
control where the measured variable (distance of separation) is used to
formulate an appropriate control response and then to actuate the
brakes/accelerator according to the driver's best judgment. Feedforward
control comes into the picture when the driver uses information other than
the controlled variable (separation distance) that represents any measure of
disturbance to the ongoing process; included would be observations that
brake lights on preceding vehicle(s) are illuminating, that cars are arriving
at a narrowing of the road, etc. Most good drivers also pay close attention
to the rate of change of separation distance, which should remain close to
zero. Later we will see that use of this variable, the time derivative of the
controlled variable, is just another element in feedback control because a
function of the controlled variable is involved.
1-3
1.6
a)
Feedback Control : Measured variable: y
Manipulated variable: D,R, or B(schematic shows D)
b)
Feedforward Control: Measured variable: F
Manipulated variable: D (shown), R or B
1-4
1.7
Both flow control loops are feedback control systems. In both cases, the
controlled variable (flow) is measured and the controller responds to that
measurement.
1.8
a)
TT
LT
Ta
L
TG
G
R
A
V
E
L
Tp
QL
X
Q(t)
TC
LC
leak
p(T)
F
FILTER
PUMP
ON/OFF
VALVE
HEATER
CITY SUPPLY
Tw , Fw
GAS
AIR
Outputs: Tp, L(level)
Inputs: Q(t), Fw
Disturbances: Tw, Ta
b)
Either Tw or Ta or both can be measured in order to add feedforward
control.
c)
Steady-state energy balance
Q(t ) = UA(T p − Ta ) + k G
(T p − TG )
∆x
1-5
+ Fw ρC (T p − Tw )
Notice that, at steady state, Fw = F (from material balance.)
Here, A is the area of water surface exposed to the atmosphere
ρ is the density of supply water
C is the specific heat of supply water.
The magnitudes of the terms UA(Tp-Ta) and FwρC(Tp-Tw) relative to the
magnitude of Q(t) will determine whether Ta or Tw (or both/neither) is the
important disturbance variable.
d)
Determine which disturbance variable is important as suggested in part c)
and investigate the economic feasibility of using its measurement for
feedforward control
1-6
1234567898
1234567898
5
2.1
a)
Overall mass balance:
d (ρV )
= w1 + w2 − w3
dt
(1)
Energy balance:
C
d 1V 2T3 − Tref 3 
dt
= w1C (T1 − Tref ) + w2C (T2 − Tref )
− w3C (T3 − Tref )
(2)
Because ρ = constant and V = V = constant, Eq. 1 becomes:
w3 = w1 + w2
b)
(3)
From Eq. 2, substituting Eq. 3
1CV
d (T3 − Tref )
dt
= 1CV
dT3
= w1C 2T1 − Tref 3 + w2C 2T2 − Tref 3
dt
− ( w1 + w2 ) C (T3 − Tref )
(4)
Constants C and Tref can be cancelled:
ρV
dT3
= w1T1 + w2T2 − ( w1 + w2 )T3
dt
The simplified model now consists only of Eq. 5.
Degrees of freedom for the simplified model:
Parameters : ρ, V
Solution Manual for Process Dynamics and Control, 2nd edition,
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp.
2-1
(5)
Variables : w1, w2, T1, T2, T3
NE = 1
NV = 5
Thus, NF = 5 – 1 = 4
Because w1, w2, T1 and T2 are determined by upstream units, we assume
they are known functions of time:
w1 = w1(t)
w2 = w2 (t)
T1 = T1(t)
T2 = T2(t)
Thus, NF is reduced to 0.
2.2
Energy balance:
Cp
d 1V 2T − Tref 3 
dt
= wC p (Ti − Tref ) − wC p (T − Tref ) − UAs (T − Ta ) + Q
Simplifying
1VC p
dT
= wC p Ti − wC p T − UAs 2T − Ta 3 + Q
dt
dT
1VC p
= wC p 2Ti − T 3 − UAs 2T − Ta 3 + Q
dt
b)
T increases if Ti increases and vice versa.
T decreases if w increases and vice versa if (Ti – T) < 0. In other words, if
Q > UAs(T-Ta), the contents are heated, and T >Ti.
2.3
a)
Mass Balances:
2-2
ρA1
dh1
= w1 − w2 − w3
dt
(1)
dh2
= w2
dt
(2)
ρA2
Flow relations:
Let P1 be the pressure at the bottom of tank 1.
Let P2 be the pressure at the bottom of tank 2.
Let Pa be the ambient pressure.
w2 =
Then
P1 − P2
ρg
=
(h1 − h2 )
R2
g c R2
(3)
P1 − Pa
ρg
=
h1
R3
g c R3
(4)
w3 =
b)
Seven parameters: ρ, A1, A2, g, gc, R2, R3
Five variables : h1, h2, w1, w2, w3
Four equations
Thus NF = 5 – 4 = 1
1 input = w1 (specified function of time)
4 outputs = h1, h2, w2, w3
2.4
Assume constant liquid density, ρ . The mass balance for the tank is
d (ρAh + m g )
dt
= ρ( q i − q )
Because ρ, A, and mg are constant, this equation becomes
2-3
A
dh
= qi − q
dt
(1)
The square-root relationship for flow through the control valve is


ρgh
q = C v  Pg +
− Pa 
gc


1/ 2
(2)
From the ideal gas law,
Pg =
(m g / M ) RT
(3)
A( H − h)
where T is the absolute temperature of the gas.
Equation 1 gives the unsteady-state model upon substitution of q from Eq. 2 and
of Pg from Eq. 3:
1/ 2
 (mg / M ) RT ρ gh

dh
+
− Pa 
A = qi − Cv 
dt
gc
 A( H − h)

(4)
Because the model contains Pa, operation of the system is not independent of Pa.
For an open system Pg = Pa and Eq. 2 shows that the system is independent of Pa.
2.5
a)
For linear valve flow characteristics,
Pd − P1
P − P2
, wb = 1
,
Ra
Rb
Mass balances for the surge tanks
wa =
dm1
= wa − wb ,
dt
wc =
P2 − Pf
Rc
dm2
= wb − wc
dt
where m1 and m2 are the masses of gas in surge tanks 1 and 2,
respectively.
If the ideal gas law holds, then
2-4
(1)
(2)
P1V1 =
m1
RT1 ,
M
P2V2 =
m2
RT2
M
(3)
where M is the molecular weight of the gas
T1 and T2 are the temperatures in the surge tanks.
Substituting for m1 and m2 from Eq. 3 into Eq. 2, and noticing that V1, T1,
V2, and T2 are constant,
V1M dP1
V2 M dP2
= wa − wb and
= wb − wc
RT1 dt
RT2 dt
(4)
The dynamic model consists of Eqs. 1 and 4.
b)
For adiabatic operation, Eq. 3 is replaced by
γ
V 
V
P1  1  = P2  2
 m1 
 m2
or
 PV γ
m1 =  1 1
 C




γ

 = C , a constant

1/ γ
and
 PV γ
m2 =  2 2
 C




(5)
1/ γ
(6)
Substituting Eq. 6 into Eq. 2 gives,
1
γ
 V1 γ

 C





1/ γ
1
γ
 V2 γ

 C





1/ γ
P1
(1− γ ) / γ
dP1
= wa − wb
dt
(1− γ ) / γ
dP2
= wb − wc
dt
P2
as the new dynamic model. If the ideal gas law were not valid, one would
use an appropriate equation of state instead of Eq. 3.
2.6
a)
Assumptions:
1. Each compartment is perfectly mixed.
2. ρ and C are constant.
3. No heat losses to ambient.
Compartment 1:
2-5
Overall balance (No accumulation of mass):
0 = ρq − ρq1
thus
q1 = q
(1)
Energy balance (No change in volume):
V11C
dT1
= 1qC 2Ti − T1 3 − UA2T1 − T2 3
dt
(2)
Compartment 2:
Overall balance:
0 = ρq1 − ρq2
thus
q2 = q1= q
(3)
Energy balance:
V21C
b)
dT2
= 1qC 2T1 − T2 3 + UA2T1 − T2 3 − U c Ac 2T2 − Tc 3
dt
(4)
Eight parameters: ρ, V1, V2, C, U, A, Uc, Ac
Five variables: Ti, T1, T2, q, Tc
Two equations: (2) and (4)
Thus NF = 5 – 2 = 3
2 outputs = T1, T2
3 inputs = Ti, Tc, q (specify as functions of t)
c)
Three new variables: ci, c1, c2 (concentration of species A).
Two new equations: Component material balances on each compartment.
c1 and c2 are new outputs. ci must be a known function of time.
2.7
Let the volume of the top tank be γV, and assume that γ is constant.
Then, an overall mass balance for either of the two tanks indicates that the
flow rate of the stream from the top tank to the bottom tank is equal to
q +qR. Because the two tanks are perfectly stirred, cT2 = cT.
2-6
Component balance for chemical tracer over top tank:
4V
dcT 1
= qcTi + qR cT − 2 q + qR 3cT 1
dt
(1)
Component balance on bottom tank:
(1 − 43V
dcT 2
= 2q + qR 3cT 1 − qR cT − qcT
dt
or
(1 − 43V
dcT
= 2 q + qR 32cT 1 − cT 3
dt
(2)
Eqs. 1 and 2 constitute the model relating the outflow concentration, cT, to
inflow concentration, cTi. Describing the full-scale reactor in the form of
two separate tanks has introduced two new parameters into the analysis, qR
and γ. Hence, these parameters will have to be obtained from physical
experiments.
2.8
Additional assumptions:
(i) Density of the liquid, ρ, and density of the coolant, ρJ, are constant.
(ii) Specific heat of the liquid, C, and of the coolant, CJ, are constant.
Because V is constant, the mass balance for the tank is:
ρ
dV
= q F − q = 0 ; thus q = qF
dt
Energy balance for tank:
ρVC
dT
0.8
= q F ρC (TF − T ) − Kq J A(T − TJ )
dt
(1)
Energy balance for the jacket:
ρ J VJ C J
dTJ
dt
= q J ρ J C J (Ti − TJ ) + Kq J
2-7
0.8
A(T − TJ )
(2)
where A is the heat transfer area (in ft2) between the process liquid and the
coolant.
Eqs.1 and 2 comprise the dynamic model for the system.
2.9
Additional assumptions:
i. The density ρ and the specific heat C of the process liquid are
constant.
ii. The temperature of steam Ts is uniform over the entire heat
transfer area
iii. Ts is a function of Ps , Ts = f(Ps)
Mass balance for the tank:
dV
= qF − q
dt
Energy balance for the tank:
1C
d V (T − Tref ) 
dt
= qF 1C 2TF − Tref 3 − q1C 2T − Tref 3
(1)
(2)
+UA(Ts − T )
where: Tref is a constant reference temperature
A is the heat transfer area
Eq. 2 is simplified by substituting for (dV/dt) from Eq. 1, and replacing Ts
by f(Ps), to give
ρVC
dT
= q F ρC (TF − T ) + UA[ f ( Ps ) − T ]
dt
Then, Eqs. 1 and 3 constitute the dynamic model for the system.
2-8
(3)
2.10
Assume that the feed contains only A and B, and no C. Component
balances for A, B, C over the reactor give.
dc A
= qi c Ai − qc A − Vk1e− E1 / RT c A
dt
(1)
dcB
= qi cBi − qcB + V (k1e − E1 / RT c A − k2e− E2 / RT cB )
dt
(2)
dcC
= − qcC + Vk2e − E2 / RT cB
dt
(3)
V
V
V
An overall mass balance over the jacket indicates that qc = qci because the
volume of coolant in jacket and the density of coolant are constant.
Energy balance for the reactor:
d (Vc A M A S A + VcB M B S B + VcC M C SC ) T 
dt
= ( qi c Ai M A S A + qi cBi M B S B ) (Ti − T )
−UA(T − Tc ) + (−∆H1 )Vk1e − E1 / RT c A + (−∆H 2 )Vk2 e− E2 / RT cB
(4)
where MA, MB, MC are molecular weights of A, B, and C, respectively
SA, SB, SC are specific heats of A, B, and C.
U is the overall heat transfer coefficient
A is the surface area of heat transfer
Energy balance for the jacket:
1 j S jV j
dTc
= 1 j S j qci 2Tci − Tc 3 + UA2T − Tc 3
dt
where:
ρj, Sj are density and specific heat of the coolant.
Vj
is the volume of coolant in the jacket.
Eqs. 1 - 5 represent the dynamic model for the system.
2-9
(5)
2.11
Model (i) :
Overall mass balance (w=constant= w ):
d ( ρV )
dh
= Aρ
= w1 + w2 − w
dt
dt
(1)
A component balance:
d (ρVx)
= w1 − wx
dt
or
Aρ
d (hx)
= w1 − wx
dt
(2)
Note that for Stream 2, x = 0 (pure B).
Model (ii) :
Mass balance:
d (1V 3
dh
= Aρ
= w1 + w2 − w
dt
dt
(3)
Component balance on component A:
d (ρVx)
= w1 − wx
dt
or
Aρ
d (hx)
= w1 − wx
dt
(4)
2-10
2.12
a)
Note that the only conservation equation required to find h is an overall
mass balance:
dm d (ρAh)
dh
=
= ρA = w1 + w2 − w
dt
dt
dt
Valve equation: w = C v′
ρg
h = Cv h
gc
where C v = C v′
ρg
gc
(1)
(2)
(3)
Substituting the valve equation into the mass balance,
dh
1
=
( w1 + w2 − C v h )
dt ρA
(4)
Steady-state model :
0 = w1 + w2 − C v h
b)
c)
Cv =
w1 + w2
h
=
2.0 + 1.2 3.2
kg/s
=
= 2.13 1/2
1.5
2.25
m
Feedforward control
2-11
(5)
Rearrange Eq. 5 to get the feedforward (FF) controller relation,
w2 = C v hR − w1
where hR = 2.25 m
w2 = (2.13)(1.5) − w1 = 3.2 − w1
(6)
Note that Eq. 6, for a value of w1 = 2.0, gives
w2 = 3.2 –1.2 = 2.0 kg/s
which is the desired value.
If the actual FF controller follows the relation, w2 = 3.2 − 1.1w1 (flow
transmitter 10% higher), w2 will change as soon as the FF controller is
turned on,
w2 = 3.2 –1.1 (2.0) = 3.2 – 2.2 = 1.0 kg/s
(instead of the correct value, 1.2 kg/s)
Then C v h = 2.13 h = 2.0 + 1.0
or
h=
3
= 1.408 and h = 1.983 m (instead of 2.25 m)
2.13
Error in desired level =
2.25 − 1.983
×100% = 11.9%
2.25
The sensitivity does not look too bad in the sense that a 10% error in flow
measurement gives ~12% error in desired level. Before making this
2-12
conclusion, however, one should check how well the operating FF
controller works for a change in w1 (e.g., ∆w1 = 0.4 kg/s).
2.13
a)
Model of tank (normal operation):
dh
= w1 + w2 − w3
dt
π (2) 2
A=
= π = 3.14 m 2
4
ρA
(800)(3.14)
(Below the leak point)
dh
= 120 + 100 − 200 = 20
dt
20
dh
=
= 0.007962 m/min
dt (800)(3.14)
Time to reach leak point (h = 1 m) = 125.6 min.
b)
Model of tank with leak and w1 , w2 , w3 constant:
1A
dh
= 56 − δ q4 = 56 − 12676583 h − 9 = 20 − 20 h − 1 , h ≥ 1
dt
To check for overflow, one can simply find the level hm at which dh/dt =
0. That is the maximum value of level when no overflow occurs.
0 = 20 − 20
hm − 1 or
hm = 2 m
Thus, overflow does not occur for a leak occurring because hm < 2.25 m.
2.14
Model of process
Overall material balance:
2-13
ρAT
dh
= w1 + w2 − w3 = w1 + w2 − C v h
dt
(1)
Component:
ρAT
d (hx3 )
= w1 x1 + w2 x2 − w3 x3
dt
ρAT h
dx3
dh
+ ρAT x3
= w1 x1 + w2 x2 − w3 x3
dt
dt
Substituting for dh/dt (Eq. 1)
ρAT h
dx3
+ x3 ( w1 + w2 − w3 ) = w1 x1 + w2 x2 − w3 x3
dt
ρAT h
dx3
= w1 ( x1 − x3 ) + w2 ( x 2 − x3 )
dt
dx3
1
=
[w1 ( x1 − x3 ) + w2 ( x2 − x3 )]
dt ρAT h
or
a)
(2)
(3)
At initial steady state ,
w3 = w1 + w2 = 120 + 100 = 220 Kg/min
220
Cv =
= 166.3
1.75
b)
If x1 is suddenly changed from 0.5 to 0.6 without changing flowrates, then
level remains constant and Eq.3 can be solved analytically or numerically
to find the time to achieve 99% of the x3 response. From the material
balance, the final value of x3 = 0.555. Then,
dx3
1
=
[120(0.6 − x3 ) + 100(0.5 − x3 )]
dt (800)(1.75)π
=
1
[ (72 + 50) − 220 x3 )]
(800)(1.75)π
= 0.027738 − 0.050020x3
Integrating,
2-14
x3 f
∫
x3 o
t
dx3
= dt
0.027738 − 0.050020 x3 ∫0
where x3o=0.5 and
x3f =0.555 – (0.555)(0.01) = 0.549
Solving,
t = 47.42 min
c)
If w1 is changed to 100 kg/min without changing any other input variables,
then x3 will not change and Eq. 1 can be solved to find the time to achieve
99% of the h response. From the material balance, the final value of the
tank level is h =1.446 m.
800π
dh
= 100 + 100 − Cv h
dt
dh
1 
=
200 − 166.3 h
dt 800π 


= 0.079577 − 0.066169 h
where ho=1.75 and
hf =1.446 + (1.446)(0.01) = 1.460
By using the MATLAB command ode45 ,
t = 122.79 min
Numerical solution of the ode is shown in Fig. S2.14
1.8
1.7
h(m)
1.6
1.5
1.4
0
50
100
150
200
time (min)
250
300
Figure S2.14. Numerical solution of the ode for part c)
2-15
d)
In this case, both h and x3 will be changing functions of time. Therefore,
both Eqs. 1 and 3 will have to be solved simultaneously. Since
concentration does not appear in Eq. 1, we would anticipate no effect on
the h response.
a)
The dynamic model for the chemostat is given by:
2.15
dX
= Vrg − FX
dt
or
dX
F
= rg −   X
dt
V 
(1)
Product: V
dP
= Vrp − FP
dt
or
dP
F
= rp −   P
dt
V 
(2)
Substrate:
V
Cells:
V
dS
1
1
= F (S f − S ) −
Vrg −
VrP
dt
YX / S
YP / S
or
1
1
dS  F 
rg −
rP
=  ( S f − S ) −
YX / S
YP / S
dt  V 
b)
At steady state,
then,
dX
=0
dt
∴
rg = DX
µX = DX
∴
µ=D
A simple feedback strategy can be implemented where the growth rate
is controlled by manipulating the mass flow rate, F.
c)
Washout occurs if dX/dt = 0 is negative for an extended period of time;
that is,
rg − DX < 0
or
µ<D
Thus, if µ < D the cells will be washed out.
d)
(3)
At steady state, the dynamic model given by Eqs. 1, 2 and 3 becomes:
2-16
(4)
0 = rg − DX
(5)
0 = rp − DP
(6)
0 = D(S f − S ) −
1
YX / S
rg −
1
YP / S
rP
(7)
From Eq. 5,
DX = rg
(8)
From Eq. 7
rg = Y X / S ( S f − S ) D +
YX / S
rP
YP / S
(9)
Substituting Eq. 9 into Eq. 8,
YX / S
rp
YP / S
From Eq. 6 and the definition of YP/S in (2-92),
DX = Y X / S ( S f − S ) D +
rp = DP = DYP / S ( S f − S )
From Eq. 4
S=
DK S
µ max − D
Substituting these two equations into Eq. 10,

DK S 
 D
DX = 2Y X / S  S f −
µ
−
D
max


2-17
(10)
1
DX (g/L.h)
0.8
0.6
MAXIMUM
PRODUCTION
0.4
0.2
WASHOUT
0
0
0.05
0.1
0.15
0.2
0.25
D (1/h)
Figure S2.15. Steady-state cell production rate DX as a function of dilution rate D.
From Figure S2.15, washout occurs at D = 0.18 h-1 while the maximum
production occurs at D = 0.14 h-1. Notice that maximum and washout points
are dangerously close to each other, so special care must be taken when
increasing cell productivity by increasing the dilution rate.
2.16
a)
We can assume that ρ and h are approximately constant. The dynamic
model is given by:
rd = −
dM
= kAc s
dt
(1)
Notice that:
M = ρV
∴
dM
dV
=ρ
dt
dt
(2)
V = πr 2 h
∴
dV
dr
dr
= (2πrh)
=A
dt
dt
dt
(3)
2-18
Substituting (3) into (2) and then into (1),
− ρA
dr
= kAc s
dt
−ρ
∴
dr
= kc s
dt
Integrating,
r
∫ r dr = −
o
kcs t
dt
1 ∫0
∴
r (t ) = ro −
kc s
t
ρ
(4)
Finally,
M = ρV = ρπhr 2
then
kc

M (t ) = ρπh ro − s
ρ

b)

t 

2
The time required for the pill radius r to be reduced by 90% is given by
Eq. 4:
0.1ro = ro −
kc s
t
ρ
∴
t=
0.9ro ρ (0.9)(0.4)(1.2)
=
= 54 min
kc s
(0.016)(0.5)
Therefore, t = 54 min .
2.17
For V = constant and F = 0, the simplified dynamic model is:
dX
S
= rg = µ max
X
dt
Ks + S
dP
S
= rp = YP / X µ max
X
dt
Ks + S
dS
1
1
rg −
rP
=−
dt
YX / S
YP / X
Substituting numerical values:
dX
SX
= 0 .2
dt
1+ S
2-19
dP
SX
= (0.2)(0.2)
dt
1+ S
dS
SX
= 0 .2
dt
1+ S
 1 0 .2 
 − 0 .5 − 0 .1 


By using MATLAB, this system of differential equations can be solved.
The time to achieve a 90% conversion of S is t = 22.15 h.
Figure S2.17. Fed-batch bioreactor dynamic behavior.
2-20
1234567898
3.1
a)
[
1e
− bt
∞
]
sin ωt = ∫ e
− bt
0
∞
sin ωt e dt = ∫ sin ωt e − ( s + b )t dt
− st
0

[− (s + b) sin ωt − ω cos ωt ]
= e − ( s + b ) t

( s + b ) 2 + ω2

0
ω
=
( s + b) 2 + ω2
∞
b)
[
1 e
− bt
∞
]
cos ωt = ∫ e
0
− bt
∞
cos ωt e dt = ∫ cos ωt e − ( s + b )t dt
− st
0

[− (s + b) cos ωt + ω sin ωt ]
= e − ( s + b ) t

( s + b) 2 + ω2

0
s+b
=
( s + b) 2 + ω2
∞
3.2
a)
The Laplace transform provided is
Y ( s) =
4
s + 3s + 4 s 2 + 6 s + 4
4
3
We also know that only sin ωt is an input, where ω =
X ( s) =
ω
2
=
2
s +ω
s2 + 2
2
( )
2
=
2 . Then
2
s +2
2
Since Y(s) = D-1(s) X(s) where D(s) is the characteristic polynomial (when
all initial conditions are zero),
Solution Manual for Process Dynamics and Control, 2nd edition,
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp.
3-1
Y (s) =
2 2
2
2
( s + 3s + 2) ( s + 2)
2
and the original ode was
d2y
dy
+ 3 + 2 y = 2 2 sin 2t
2
dt
dt
with y ′(0) = y (0) = 0
b)
This is a unique result.
c)
The solution arguments can be found from
Y (s) =
2 2 2
( s + 1)( s + 2) + ( s 2 + 2)
which in partial fraction form is
Y (s) =
α1
α
a s + a2
+ 2 + 12
s +1 s + 2
s +2
Thus the solution will contain four functions of time
e-t
,
e-2t ,
sin 2 t , cos 2 t
3.3
a)
Pulse width is obtained when x(t) = 0
Since x(t) = h – at
tω : h − atω = 0
or
tω = h/a
b)
h
slope = -a
slope = a
x(t)
x(t)
slope = -a
x(t) = hS(t) – atS(t) + a(t -tω) S(t-tω)
3-2
c)
h a ae − stω h e − stω − 1
X ( s) = − 2 + 2 = +
s s
s
s
s2
d)
Area under pulse = h tω/2
a)
f(t) = 5 S(t) – 4 S(t-2) – S(t- 6)
1
F (s ) = = ( 5 - 4e -2s - e -6s )
s
3.4
b)
x(t)
x1
a
a
x4
tr
2tr
3tr
-a
-a
x2
x3
x(t) = x1(t) + x2(t) + x3(t) + x4(t)
= at – a(t − tr)S(t − tr) − a(t −2tr)S(t − 2tr) + a(t − 3tr)S(t − 3tr)
following Eq. 3-101. Thus
X(s) =
[
a
1 − e −tr s − e − 2tr s + e −3tr s
2
s
]
by utilizing the Real Translation Theorem Eq. 3-104.
3-3
3.5
55
55
t S(t) –
(t-30) S(t-30)
30
30
20 55 1
55 1 −30 s 20 55 1
T (s) =
+
−
e
=
+
1 − e −30 s
2
2
2
s 30 s
30 s
s 30 s
T(t) = 20 S(t) +
(
)
3.6
a)
X ( s) =
α1 =
α2 =
α3 =
b)
α
α
α
s ( s + 1)
= 1 + 2 + 3
( s + 2)( s + 3)( s + 4) s + 2 s + 3 s + 4
s ( s + 1)
( s + 3)( s + 4)
s ( s + 1)
( s + 2)( s + 4)
s ( s + 1)
( s + 2)( s + 3)
=1
s = −2
= −6
s = −3
=6
s = −4
X (s) =
1
6
6
−
+
s+2 s+3 s+4
X (s) =
s +1
s +1
=
2
( s + 2)( s + 3)( s + 4) ( s + 2)( s + 3)( s + 2 j )( s − 2 j )
X ( s) =
α + jβ 3 α 3 − j β 3
α1
α
+ 2 + 3
+
s+2 s+3
s+2j
s+2j
α1 =
α2 =
s +1
( s + 3)( s 2 + 4)
s +1
( s + 2)( s 2 + 4)
α 3 + jβ 3 =
=−
s = −2
=
s = −3
x(t ) = e−2t − 6e −3t + 6e −4t
and
1
8
2
13
s +1
( s + 2)( s + 3)( s − 2 j )
3-4
=
s = −2 j
1− 2 j
− 3 + 11 j
=
− 40 − 8 j
208
1
2
 −3 
 11 
x(t ) = − e − 2t + e −3t + 2
 cos 2t + 2
 sin 2t
8
13
 208 
 208 
1
2
3
11
= − e − 2t + e −3t −
cos 2t +
sin 2t
8
13
104
104
c)
X ( s) =
α
α2
s+4
= 1 +
2
s + 1 ( s + 1) 2
( s + 1)
(1)
α 2 = ( s + 4) s =−1 = 3
In Eq. 1, substitute any s≠-1 to determine α1. Arbitrarily using s=0, Eq. 1
gives
4 α1 3
=
+
1 12
12
or
α1 = 1
1
3
+
and
x( t ) = e− t + 3te − t
2
s + 1 ( s + 1)
1
1
1
X (s) = 2
=
=
2
2
2
s + s +1 
1  3 (s + b ) + ω
s +  +
2 4

1
3
where b =
and ω =
2
2
X (s) =
d)
t
1
2 −2
3
x(t ) = e −bt sin ω t =
e sin
t
ω
2
3
e)
X(s) =
s +1
e −0.5 s
s ( s + 2)( s + 3)
To invert, we first ignore the time delay term. Using the Heaviside
expansion with the partial fraction expansion,
Xˆ ( s ) =
s +1
A
B
C
= +
+
s ( s + 2)( s + 3) s s + 2 s + 3
Multiply by s and let s → 0
3-5
1
1
=
(2)(3) 6
Multiply by (s+2) and let s→ −2
A=
B=
− 2 +1
−1
1
=
=
(−2)(−2 + 3) (−2)(1) 2
Multiply by (s+3) and let s→-3
C=
− 3 +1
−2
2
=
=−
(−3)(−3 + 2) (−3)(−1)
3
Then
1 6 1 2 −2 3
Xˆ ( s ) =
+
+
s s+2 s+3
xˆ (t ) =
1 1 − 2 t 2 −3t
+ e − e
6 2
3
Imposing shift theorem
x(t ) = xˆ (t − 0.5) =
1 1 − 2 (t −0.5) 2 −3( t − 0.5)
+ e
− e
6 2
3
for t ≥ 0.5
3.7
a)
Y (s) =
α
6( s + 1)
6 α
= 2 = 1 + 22
2
s
s ( s + 1) s
s
6
s2
6
Y (s) = 2
s
α2 = s2
b)
Y (s) =
=6
α1 = 0
s =0
12( s + 2) α1 α 2 s + α 3
=
+ 2
s
s ( s 2 + 9)
s +9
3-6
Multiplying both sides by s(s2+9)
12( s + 2) = α 1 ( s 2 + 9) + (α 2 s + α 3 )( s )
or
12 s + 24 = (α 1 + α 2 ) s + α 3 s + 9α 1
2
Equating coefficients of like powers of s,
s2: α1 + α2 = 0
s1: α3
= 12
0
s : 9α1
= 24
Solving simultaneously,
−8
3
 8

 − s + 12 
8 1  3

Y (s) =
+
3s
s2 + 9
α1 =
c)
α2 =
α3 =
α2 =
,
,
( s + 2)( s + 3)
( s + 5)( s + 6)
( s + 2)( s + 3)
( s + 4)( s + 6)
( s + 2)( s + 3)
( s + 4)( s + 5)
=1
s = −4
= −6
s = −5
=6
s = −6
Y (s) =
1
6
6
−
+
s+4 s+5 s+6
Y (s) =
[(s + 1)
=
α 3 = 12
α
α
α
( s + 2)( s + 3)
= 1 + 2 + 3
( s + 4)( s + 5)( s + 6) s + 4 s + 5 s + 6
Y (s) =
α1 =
d)
8
3
1
2
]
2
+ 1 ( s + 2)
=
1
( s + 2 s + 2) 2 ( s + 2)
2
α s + α4
α
α1 s + α 2
+ 2 3
+ 5
2
2
s+2
s + 2s + 2 ( s + 2s + 2)
3-7
Multiplying both sides by ( s 2 + 2s + 2) 2 ( s + 2) gives
1 = α1s4 + 4α1s3 + 6α1s2 +4α1s + α2s3 +4α2s2 +6α2s +4α2 + α3s2 +2α3s +
α4s + 2α4 + α5s4 + 4α5s3 + 8α5s2 + 8α5s + 4α5
Equating coefficients of like power of s,
s4 : α1 + α5 = 0
s3 : 4α1 + α2 + 4α5 = 0
s2 : 6α1 + 4α2 + α3 + 8α5 = 0
s1 : 4α1 + 6α2 + 2α3 + α4 + 8α5 = 0
s0 : 4α2 + 2α4 + 4α5 = 1
Solving simultaneously:
α1 = -1/4
Y (s) =
α2 = 0
α3=-1/2
α4=0
− 1 / 4s
− 1 / 2s
1/ 4
+ 2
+
2
s+2
s + 2 s + 2 ( s + 2 s + 2)
2
3.8
a)
From Eq. 3-100
t
 1
1  ∫ f (t * )dt *  = F ( s )
0
 s
 t −τ  1
1
we know that 1  ∫ e dτ = 21 e − τ =
s ( s + 1)
0
 s
[ ]
∴
Laplace transforming yields
2
s ( s + 1)
2
or (s2 + 3s + 1) X(s) =
s ( s + 1)
s2X(s) + 3X(s) + 2X(s) =
3-8
α5 = ¼
X(s) =
x(t) = 1 − 2te-t − e-2t
and
b)
2
s ( s + 1) 2 ( s + 2)
Applying the final Value Theorem
lim x(t ) = lim sX ( s ) = lim
t →∞
s →0
s →0
2
=2
( s + 1) ( s + 2)
2
[ Note that Final Value Theorem is applicable here]
3.9
a)
X (s) =
6( s + 2)
6( s + 2)
=
( s + 9s + 20)( s + 4) ( s + 4)( s + 5)( s + 4)
2
 6s ( s + 2) 
=0
x(0) = lim 
s →∞ ( s + 5)( s + 4) 2 


 6s ( s + 2) 
x(∞) = lim
=0
s →0 ( s + 5)( s + 4) 2 


x(t) is converging (or bounded) because [sX(s)] does not have a limit at
s = −4, and s = −5 only, i.e., it has a limit for all real values of s ≥ 0.
x(t) is smooth because the denominator of [sX(s)] is a product of real
factors only. See Fig. S3.9a.
b)
10 s 2 − 3
10 s 2 − 3
X (s) = 2
=
( s − 6 s + 10)( s + 2) ( s − 3 + 2 j ) ( s − 3 − 2 j )( s + 2)


10s 3 − 3s
x(0) = lim  2
 = 10
s →∞ ( s − 6 s + 10)( s + 2)


Application of final value theorem is not valid because [sX(s)] does not
have a limit for some real s ≥ 0, i.e., at s = 3±2j. For the same reason, x(t)
is diverging (unbounded).
x(t) is oscillatory because the denominator of [sX(s)] includes complex
factors. See Fig. S3.9b.
3-9
16 s + 5
16s + 5
=
2
( s + 9) ( s + 3 j ) ( s − 3 j )
X (s) =
16s 2 + 5s 
x(0) = lim  2
 = 16
s →∞
 ( s + 9) 
Application of final value theorem is not valid because [sX(s)] does not
have a limit for real s = 0. This implies that x(t) is not diverging, since
divergence occurs only if [sX(s)] does not have a limit for some real value
of s>0.
x(t) is oscillatory because the denominator of [sX(s)] is a product of
complex factors. Since x(t) is oscillatory, it is not converging either. See
Fig. S3.9c
0.4
0.35
0.3
0.25
x(t)
0.2
0.15
0.1
0.05
0
-0.05
0
0.5
1
1.5
2
2.5
Time
3
3.5
4
4.5
Figure S3.9a. Simulation of X(s) for case a)
8000
6000
4000
2000
x(t)
c)
0
-2000
-4000
-6000
-8000
-10000
-12000
0
0.5
1
1.5
2
2.5
Time
Figure S3.9b. Simulation of X(s) for case b)
3-10
5
20
15
10
x(t)
5
0
-5
-10
-15
-20
0
0.5
1
1.5
2
2.5
Time
3
3.5
4
4.5
5
Figure S3.9c. Simulation of X(s) for case c)
The Simulink block diagram is shown below. An impulse input should be
used to obtain the function’s behavior. In this case note that the impulse
input is simulated by a rectangular pulse input of very short duration. (At
time t = 0 and t =0.001 with changes of magnitude 1000 and –1000
respectively). The MATLAB command impulse might also be used.
Figure S3.9d. Simulink block diagram for cases a), b) and c).
3-11
3.10
a)
i)
ii)
iii)
iv)
Y(s) =
2
2
A B
C
= 2
= 2+ +
s s+4
s ( s + 4 s ) s ( s + 4) s
∴
y(t) will contain terms of form: constant, t, e-4t
Y(s) =
2
2
A
B
C
=
= +
+
s ( s + 4 s + 3) s ( s + 1)( s + 3) s s + 1 s + 3
∴
y(t) will contain terms of form: constant, e-t, e-3t
Y (s) =
2
2
A
B
C
=
= +
+
2
2
s ( s + 2)
s+2
s ( s + 4 s + 4) s ( s + 2)
∴
y(t) will contain terms of form: constant, e-2t , te-2t
Y (s) =
2
s ( s + 4 s + 8)
2
2
2
2
s 2 + 4 s + 8 = ( s 2 + 4 s + 4) + (8 − 4) = ( s + 2) 2 + 2 2
2
Y (s) =
s[(s + 2) 2 + 2 2 ]
∴
b)
y(t) will contain terms of form: constant, e-2t sin2t, e-2tcos2t
A
Bs
C
2( s + 1)
2( s + 1)
=
= + 2
+ 2
2
2
2
2
s ( s + 4) s ( s + 2 ) s s + 2
s + 22
2( s + 1) 1
A = lim 2
=
s → 0 ( s + 4)
2
Y (s) =
2(s+1) = A(s2+4) + Bs(s) + Cs
2s+2 = As2 + 4A + Bs2 + Cs
Equating coefficients on like powers of s
s2 :
0=A+B
→
B = −A = −
s1 :
2= C
→
s0 :
2 = 4A
→
C=2
1
A=
2
3-12
1
2
∴
Y(s) =
1 2 − (1 2) s
2
+ 2
+ 2
2
s
s +2
s + 22
y(t) =
1 1
2
− cos 2t + sin 2t
2 2
2
y(t) =
1
(1 − cos 2t ) + sin 2t
2
3.11
a)
Since convergent and oscillatory behavior does not depend on initial
dx 2 (0) dx(0)
conditions, assume
=
= x ( 0) = 0
dt
dt 2
Laplace transform of the equation gives
s 3 X ( s ) + 2 s 2 X ( s ) + 2 sX ( s ) + X ( s ) =
3
1
3
1
3
s ( s + 1)( s + +
j )( s + −
j)
2 2
2 2
Denominator of [sX(s)] contains complex factors so that x(t) is oscillatory,
and denominator vanishes at real values of s= −1 and -½ which are all <0
so that x(t) is convergent. See Fig. S3.11a.
2
s 2 X ( s) − X (s) =
s −1
2
2
X (s) =
=
2
( s − 1)( s − 1) ( s − 1) 2 ( s + 1)
X (s) =
b)
3
s
3
=
s ( s + 2 s 2 + 2 s + 1)
3
The denominator contains no complex factors; x(t) is not oscillatory.
The denominator vanishes at s=1 ≥0; x(t) is divergent. See Fig. S3.11b.
c)
s 3 X ( s) + X (s) =
X (s) =
1
s +1
2
1
=
( s + 1)( s 3 + 1)
2
1
1
3
1
3
( s + j )( s − j )( s + 1)( s − +
j )( s − −
j)
2 2
2 2
The denominator contains complex factors; x(t) is oscillatory.
The denominator vanishes at real s = 0, ½; x(t) is not convergent. See Fig.
S3.11c.
3-13
s 2 X ( s ) + sX ( s ) =
4
s
4
4
= 2
s ( s + s ) s ( s + 1)
X (s) =
2
The denominator of [sX(s)] contains no complex factors; x(t) is not
oscillatory.
The denominator of [sX(s)] vanishes at s = 0; x(t) is not convergent. See
Fig. S3.11d.
3.5
3
2.5
x(t)
2
1.5
1
0.5
0
-0.5
0
1
2
3
4
5
time
6
7
8
9
10
Figure S3.11a. Simulation of X(s) for case a)
700
600
500
400
x(t)
d)
300
200
100
0
0
0.5
1
1.5
2
2.5
time
3
3.5
4
4.5
5
Figure S3.11b. Simulation of X(s) for case b)
3-14
80
60
x(t)
40
20
0
-20
-40
0
1
2
3
4
5
time
6
7
8
9
10
Figure S3.11c. Simulation of X(s) for case c)
18
16
14
12
x(t)
10
8
6
4
2
0
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
time
Figure S3.11d. Simulation of X(s) for case d)
3.12
Since the time function in the solution is not a function of initial
conditions, we Laplace Transform with
x ( 0) =
dx(0)
=0
dt
τ1τ2s2X(s) + (τ1+τ2)sX(s) + X(s) = KU(s)
3-15
X ( s) =
K
U (s)
τ1 τ 2 s + ( τ1 + τ 2 ) s + 1
2
Factoring denominator
X ( s) =
a)
K
U (s)
(τ1 s + 1)(τ 2 s + 1)
If u(t) = a S(t) then U(s)=
X a (s) =
a
s
Ka
s (τ1 s + 1)(τ 2 s + 1)
τ1 ≠ τ 2
xa(t) = fa( S(t), e-t/τ1, e– t/τ2)
b)
If u(t) = be-t/τ then U(s) =
X b ( s) =
bτ
τs +1
Kbτ
(τs + 1)(τ1 s + 1)(τ 2 s + 1)
τ ≠ τ1 ≠ τ 2
xb(t) = fb(e-t/τ , e-t/τ1, e– t/τ2)
c)
If u(t) =ce-t/τ where τ = τ1 , then U(s) =
X c ( s) =
τc
τ1 s + 1
Kcτ
(τ1 s + 1) 2 (τ 2 s + 1)
xc(t) = fc(e– t/τ1, t e– t/τ1, e– t/τ2)
d)
If u(t) = d sin ωt then U(s) =
X d (s) =
dω
s + ω2
2
Kd
( s + ω )(τ1 s + 1)(τ 2 s + 1)
2
2
xd(t) = fd(e– t/τ1, e– t/τ2, sin ωt, cos ωt)
3-16
3.13
dx3
+ 4 x = et
3
dt
a)
d 2 x(0) dx(0)
=
= x(0) = 0
dt 2
dt
with
Laplace transform of the equation,
s 3 X(s) + 4 X(s) =
X (s) =
=
1
s −1
1
1
=
3
( s − 1)( s + 4) ( s − 1)( s + 1.59)( s − 0.79 + 1.37 j )( s − 0.79 − 1.37 j )
α 3 + jβ 3
α 3 − jβ 3
α1
α2
+
+
+
s − 1 s + 1.59 s − 0.79 + 1.37 j s − 0.79 − 1.37 j
α1 =
α2 =
1
( s + 4)
=
3
s =1
1
5
1
( s − 1)( s − 0.79 + 1.37 j )( s − 0.79 − 1.37 j )
α 3 + jβ 3 =
1
( s − 1)( s + 1.59)( s − 0.79 − 1.37 j )
=−
s = −1.59
1
19.6
= −0.74 − 0.59 j
s = 0.79 −1.37 j
1
1
−
− 0.074 − 0.059 j − 0.074 + 0.059 j
X(s) = 5 + 19.6 +
+
s − 1 s + 1.59 s − 0.79 + 1.37 j
s − 0.79 − 1.37 j
1
1 −1.59t
x(t ) = e t −
e
− 2e 0.79t (0.074 cos 1.37t + 0.059 sin 1.37t )
5
19.6
dx
− 12 x = sin 3t
with
x(0) = 0
dt
b)
sX (s) − 12 X(s) =
X (s) =
=
3
s +9
2
3
3
=
( s + 9)( s − 12) ( s + 3 j )( s − 3 j )( s − 12)
2
α3
α 1 + j β1 α 1 − j β 1
+
+
s+3j
s−3j
s − 12
3-17
α 1 + j β1 =
α3 =
=
s = −3 j
3
1
4
=−
−
j
− 18 + 72 j
102 102
3
1
=
( s + 9 ) s =12 51
2
−
X (s) =
x(t ) = −
c)
3
( s − 3 j )( s − 12)
1
4
1
4
1
−
j −
+
j
102 102 + 102 102 + 51
s+3j
s −3j
s − 12
1
1
(cos 3t + 4 sin 3t ) + e12 t
51
51
d2x
dx
+ 6 + 25 x = e −t
2
dt
dt
dx(0)
= x(0) = 0
dt
with
s 2 X ( s ) + 6 sX ( s ) + 25X ( s ) + X ( s ) =
X ( s) =
α1 =
1
or
s +1
X( s ) =
1
( s + 1 )( s + 6 s + 25 )
2
α
α + β2 j α 2 − β2 j
1
= 1 + 2
+
( s + 1)( s + 3 + 4 j )( s + 3 − 4 j ) s + 1 s + 3 + 4 j s + 3 − 4 j
1
( s + 6 s + 25)
=
2
α 2 + jβ 2 =
s = −1
1
20
1
( s + 1)( s + 3 − 4 j )
=−
s = −3− 4 j
1
1
−
j
40 80
1
1
1
1
1
− −
j − −
j
X ( s ) = 20 + 40 80 + 40 80
s +1 s + 3 + 4 j
s + 3− 4 j
x(t ) =
d)
1 −t
1
1
e − e −3t ( cos 4t + sin 4t )
20
20
40
Laplace transforming (assuming initial conditions = 0, since they do not
affect results)
sY1(s) + Y2(s) = X1(s)
(1)
sY2(s) – 2Y1(s) + 3 Y2(s) = X2(s)
(2)
3-18
From (2),
(s+3) Y2(s) = X2(s) + 2Y1(s)
Y2 ( s ) =
1
2
X 2 (s) +
Y1 ( s )
s+3
s+3
Substitute in Eq.1
sY1(s) +
1
2
X 2 (s) +
Y1 ( s ) = X1(s)
s+3
s+3
We neglect X2(s) since it is equal to zero.
[s(s + 3) + 2]Y1 (s) = ( s + 3) X 1 (s)
( s 2 + 3s + 2)Y1 ( s ) = ( s + 3) X 1 ( s )
Y1 ( s ) =
s+3
s+3
X 1 (s) =
X 1 (s)
( s + 1)( s + 2)
s + 3s + 2
2
1
s +1
s+3
A
B
C
Y1 ( s ) =
=
+
+
2
2
( s + 1) ( s + 2) ( s + 1) ( s + 1)
s+2
Now if x1(t) = e-t then X1(s) =
∴
so that y1(t) will contain e-t/τ , te-t/τ, e–2t functions of time.
For Y2(s)
Y2 ( s ) =
2
A
B
C
=
+
+
2
s+2
( s + 1) ( s + 2) ( s + 1) ( s + 1)
2
so that y2(t) will contain the same functions of time as y1(t) (although
different coefficients).
3-19
3.14
d 2 y (t )
dy (t )
d ( x − 2)
+3
+ y (t ) = 4
− x(t − 2)
2
dt
dt
dt
Taking the Laplace transform and assuming zero initial conditions,
s2Y(s) + 3sY(s) + Y(s) = 4 e-2ssX(s) −e-2sX(s)
Rearranging,
Y (s)
− (1 − 4s )e −2 s
= G(s) = 2
X ( s)
s + 3s + 1
a)
(1)
The standard form of the denominator is : τ2s2 + 2ζτs + 1
From (1) , τ = 1 , ζ = 1.5
Thus the system will exhibit overdamped and non-oscillatory response.
b)
Steady-state gain
K = lim G ( s ) = −1
(from (1))
s →0
c)
For a step change in x
1.5
− (1 − 4 s )e −2 s 1.5
and Y(s) = 2
X(s) =
s
( s + 3s + 1) s
Therefore yˆ (t ) = −1.5 + 1.5e-1.5t cosh(1.11t) + 7.38e-1.5t sinh(1.11t)
Using MATLAB-Simulink, y(t)= yˆ (t − 2) is shown in Fig. S3.14
1.5
1
0.5
0
-0.5
-1
-1.5
0
5
10
15
20
25
30
Figure S3.14. Output variable for a step change in x of magnitude 1.5
3-20
3.15
f (t ) = hS (t ) − hS (t − 1 / h)
dx
+ 4 x = h[S (t ) − S (t − 1 / h)]
dt
x(0)=0
,
Take Laplace transform,
 1 e−s / h 

sX ( s ) + 4 X ( s ) = h −
s 
s
α 
1
α
X ( s ) = h(1 − e − s / h )
= h(1 − e − s / h )  1 + 2 
s ( s + 4)
 s s + 4
1
1
1
1
α1 =
=
, α2 =
=−
s + 4 s =0 4
s s =−4
4
h
1 
1
(1 − e − s / h )  −

4
s s + 4
X ( s) =
=
h 1 e −s / h
1
e −s / h 
−
−
+


4 s
s
s + 4 s + 4
0
h
(1 − e − 4t )
4
h − 4( t −1 / h )
e
− e −4t
4
x(t ) =
[
t <0
0 < t < 1/h
]
t > 1/h
1
h=1
h=10
h=100
0.9
0.8
0.7
x(t)
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.2
0.4
0.6
0.8
1
time
1.2
1.4
1.6
1.8
2
Figure S3.15. Solution for values h= 1, 10 and 100
3-21
3.16
a)
Laplace transforming
[s Y (s) − sy(0) − y′(0)]+ 6[sY (s) − y(0)] + 9Y (s) = s s+ 1
2
2
(s2 + 6s + 9)Y(s) − s(1) − 2 –(6)(1)=
(s2 + 6s + 9)Y(s) =
s
s +1
2
s
+s+8
s +1
2
s + s 3 + s + 8s 2 + 8
(s + 6s + 9)Y(s) =
s2 +1
2
Y(s) =
s 3 + 8s 2 + 2 s + 8
( s + 3) 2 ( s 2 + 1)
To find y(t) we have to expand Y(s) into its partial fractions
Y (s) =
A
B
Cs
D
+
+ 2
+ 2
2
( s + 3)
s + 3 s +1 s +1
y(t) = Ate-3t + Be-3t + C cost + D sint
b)
Y(s)=
s +1
s ( s + 4 s + 8)
Since
42
< 8 we know we will have complex factors.
4
2
∴ complete square in denominator
s2 + 4s + 8 = s2 + 4s + 4 + 8−4
= s2 + 4s + 4 + 4 = (s+2)2 + (2)2
∴ Partial fraction expansion gives
3-22
{ b = 2 , ω=2}
Y (s) =
A
B ( s + 2)
C
s +1
+ 2
+ 2
=
2
s s + 4 s + 8 s + 4 s + 8 s ( s + 4 s + 8)
Multiply by s and let s→0
A=1/8
Multiply by s(s2+4s+8)
A(s2+4s+8) + B(s+2)s + Cs = s + 1
As2 + 4As + 8A + Bs2 + 2Bs + Cs = s + 1
Y (s) =
→ B = −A = −
s2 :
A+B=0
s1 :
4A + 2B + C = 1
s0 :
8A = 1
→A=
1
1 3
C = 1 + 2  − 4  =
8
8 4
→
1
8
1
8
(This checks with above result)
1 / 8 (− 1 / 8)( s + 2)
3/ 4
+
+
2
2
s
( s + 2) + 2
( s + 2) 2 + 2 2
1 1
 3
y(t) =   −   e-2t cos 2t +   e-2t sin 2t
8 8
8
3.17
V
dC
+ qC = qCi
dt
Since V and q are constant, we can Laplace Transform
sVC(s) + qC(s) = q Ci(s)
Note that c(t = 0) = 0
Also, ci(t) = 0
ci(t) = ci
,
,
t ≤0
t>0
3-23
Laplace transforming the input function, a constant,
Ci ( s ) =
ci
s
so that
sVC(s) + qC(s) = q
ci
s
or
C(s) =
qci
( sV + q ) s
Dividing numerator and denominator by q
C(s) =
ci
V

 s +1 s
q

Use Transform pair #3 in Table 3.1 to invert (τ =V/q)
V
− t 


c(t) = ci 1 − e q 




Using MATLAB, the concentration response is shown in Fig. S3.17.
(Consider V = 2 m3, Ci=50 Kg/m3 and q = 0.4 m3/min)
50
45
40
35
c(t)
30
25
20
15
10
5
0
0
5
10
15
20
25
30
Time
Figure S3.17. Concentration response of the reactor effluent stream.
3-24
3.18
a)
If Y(s) =
KAω
s( s 2 + ω2 )
and input U(s) =
Aω
= 1 {A sin ωt}
(s + ω2 )
2
then the differential equation had to be
dy
= Ku (t )
dt
b)
Y(s) =
α1 =
with
y(0) = 0
α ω
α
α s
KAω
= 1+ 2 2 2 + 2 3 2
2
2
s( s + ω ) s s + ω
s +ω
KAω
s + ω2
2
=
s →0
KA
ω
Find α2 and α3 by equating coefficients
KAω= α1(s2+ω2) + α2s2+α3ωs
KAω = α1s2 + α1ω2 + α2s2 + α3ωs
s2 :
0 = α1 + α2
s:
0 = α3 ω
→ α2 = −α1 =
→ α3 = 0
KAω
KA / ω ( KA / ω) s
=
− 2
2
2
s
s( s + ω )
s + ω2
KA
(1 − cos ωt )
y(t) =
ω
∴ Y(s) =
3-25
− KA
ω
c)
A
u(t)
-A
2KA/ω
y(t)
0
Time
i)
We see that y(t) follows behind u(t) by 1/4 cycle = 2π/4= π/2 rad.
which is constant for all ω
ii)
The amplitudes of the two sinusoidal quantities are:
y : KA/ω
u: A
Thus their ratio is K/ω, which is a function of frequency.
3-26
1234567898
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
1234567898
5.1
a)
xDP(t) = hS(t) – 2hS(t-tw) + hS(t-2tw)
h
xDP (s) = (1 − 2e-tws + e-2tws)
s
b)
Response of a first-order process,
 K h
-t s
-2t s
Y (s) = 
 (1 − 2e w + e w )
 τs + 1  s
α 
α
or
Y(s) = (1 − 2e-tw s + e-2tw s)  1 + 2 
 s τs + 1
Kh
Kh
α1 =
= Kh
α2 =
τs + 1 s =0
s
s =−
1
τ
= − Khτ
 Kh Khτ 
Y(s) = (1 − 2e-tw s + e-2tw s) 
−
τs + 1 
 s
y(t) =
Kh(1−e-t/τ)
,
0 < t < tw
Kh(–1 – e-t/τ + 2e-(t-tw)/τ)
,
tw < t < 2tw
Kh(–e-t/τ + 2e-(t-tw)/τ − e-(t-2tw)/τ ) ,
2tw < t
Response of an integrating element,
Y (s) =
y(t) =
c)
K h
(1 − 2e-tw s + e-2tw s)
s s
Kht
,
0 < t < tw
Kh(-t + 2 tw)
,
tw < t < 2tw
0
,
2tw < t
This input gives a response, for an integrating element, which is zero after
a finite time.
Solution Manual for Process Dynamics and Control, 2nd edition,
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp.
5-1
5.2
a)
For a step change in input of magnitude M
y(t) = KM (1- e-t/τ) + y(0)
We note that KM = y(∞) – y(0) = 280 – 80 = 200°C
Then K =
200 1 C
= 400 °C/Kw
0.5Kw
At time t = 4,
230 − 80
= 1 − e − 4 / τ or τ = 2.89 min
280 − 80
Thus
T ′( s )
400
=
[°C/Kw]
P ′( s ) 2.89 s + 1
∴
For an input ramp change with slope a = 0.5 Kw/min
Ka = 400 × 0.5 = 200 °C/min
This maximum rate of change will occur as soon as the transient has died
out, i.e., after
5 × 2.89 min ≈ 15 min have elapsed.
1500
1000
T'
a)
y(4) = 230 °C
500
0
0
1
2
3
4
5
6
7
8
9
10
time(min)
Fig S5.2. Temperature response for a ramp input of magnitude 0.5 Kw/min.
5-2
5.3
The contaminant concentration c increases according to this expression:
c(t) = 5 + 0.2t
Using deviation variables and Laplace transforming,
c′(t ) = 0.2t
C ′( s ) =
or
0 .2
s2
Hence
C m′ ( s ) =
1
0 .2
⋅ 2
10 s + 1 s
and applying Eq. 5-21
cm′ (t ) = 2(e− t /10 − 1) + 0.2t
As soon as cm′ (t ) ≥ 2 ppm the alarm sounds. Therefore,
∆t = 18.4 s
(starting from the beginning of the ramp input)
The time at which the actual concentration exceeds the limit (t = 10 s) is
subtracted from the previous result to obtain the requested ∆t .
∆t = 18.4 − 10.0 = 8.4 s
2.5
2
c'm
1.5
1
0.5
0
0
2
4
6
8
10
12
14
16
18
20
time
Fig S5.3. Concentration response for a ramp input of magnitude 0.2 Kw/min.
5-3
5.4
a)
Using deviation variables, the rectangular pulse is
0
2
0
c ′F =
t<0
0≤t<2
2≤t≤∞
Laplace transforming this input yields
c ′F ( s ) =
(
2
1 − e −2s
s
)
The input is then given by
c ′( s ) =
8
8e −2 s
−
s (2s + 1) s (2s + 1)
and from Table 3.1 the time domain function is
c ′(t ) = 8(1 − e −t / 2 ) − 8(1 − e − (t −2 ) / 2 ) S (t − 2)
(1)
6
5
C'
4
3
2
1
0
0
2
4
6
8
10
12
14
16
18
20
time
Fig S5.4. Exit concentration response for a rectangular input.
b)
By inspection of Eq. 1, the time at which this function will reach its
maximum value is 2, so maximum value of the output is given by
5-4
c ′(2) = 8(1 − e −1 ) − 8(1 − e −0 / 2 ) S (0)
(2)
and since the second term is zero, c ′(2) = 5.057
c)
By inspection, the steady state value of c ′(t ) will be zero, since this is a
first-order system with no integrating poles and the input returns to zero.
To obtain c ′(∞) , simplify the function derived in a) for all time greater
than 2, yielding
c ′(t ) = 8(e − (t −2) / 2 − e −t / 2 )
(3)
which will obviously converge to zero.
Substituting c ′(t ) = 0.05 in the previous equation and solving for t gives
t = 9.233
5.5
a)
Energy balance for the thermocouple,
mC
dT
= hA(Ts − T )
dt
(1)
where m is mass of thermocouple
C is heat capacity of thermocouple
h is heat transfer coefficient
A is surface area of thermocouple
t is time in sec
Substituting numerical values in (1) and noting that
Ts = T
and
15
dT dT ′
=
,
dt
dt
dT ′
= Ts′ − T ′
dt
Taking Laplace transform,
T ′( s )
1
=
Ts′( s ) 15s + 1
5-5
Ts(t) = 23 + (80 − 23) S(t)
Ts = T = 23
From t = 0 to t = 20,
Ts′(t ) = 57 S(t)
T ′( s ) =
,
Ts′( s ) =
57
s
1
57
Ts′( s ) =
15s + 1
s (15s + 1)
Applying inverse Laplace Transform,
T ′(t ) = 57(1 − e −t / 15 )
Then
T (t ) = T ′(t ) + T = 23 + 57(1 − e −t / 15 )
Since T(t) increases monotonically with time, maximum T = T(20).
Maximum T(t) = T(20) = 23 + 57 (1-e-20/15) = 64.97 °C
50
45
41.97 º
40
35
30
25
T'
b)
20
15
10
5
0
0
5
10
15
20
25
30
35
40
time
Fig S5.5. Thermocouple output for part b)
5-6
45
50
5.6
a)
The overall gain of G is G s =0
=
b)
K1
K2
⋅
= K1 K 2
τ1 × 0 + 1 τ 2 × 0 + 1
If the equivalent time constant is equal to τ1 + τ2 = 5 + 3 = 8, then
y(t = 8)/KM = 0.632
for a first-order process.
5e −8 / 5 − 3e −8 / 3
y(t = 8)/KM = 1 −
= 0.599 ≠ 0.632
5−3
Therefore, the equivalent time constant is not equal to τ1 + τ2
c)
The roots of the denominator of G are
-1/τ1 and -1/τ2
which are negative real numbers. Therefore the process transfer function
G cannot exhibit oscillations when the input is a step function.
5.7
Assume that at steady state the temperature indicated by the sensor Tm is
equal to the actual temperature at the measurement point T. Then,
Tm′ ( s )
K
1
=
=
T ′( s ) τs + 1 1.5s + 1
Tm = T = 3501 C
Tm′ (t ) = 15sin ωt
where ω =2π × 0.1 rad/min = 0.628 rad/min
At large times when t/τ >>1, Eq. 5-26 shows that the amplitude of the
sensor signal is
5-7
Am =
A
ω2 τ2 + 1
where A is the amplitude of the actual temperature at the measurement
point.
Therefore
A = 15 (0.628)2 (1.5) 2 + 1 = 20.6°C
Maximum T = T + A =350 + 20.6 = 370.6
Maximum Tcenter = 3 (max T) – 2 Twall
= (3 × 370.6)−(2 × 200) = 711.8°C
Therefore, the catalyst will not sinter instantaneously, but will sinter if
operated for several hours.
5.8
a)
Assume that q is constant. Material balance over the tank,
A
dh
= q1 + q 2 − q
dt
Writing in deviation variables and taking Laplace transform
AsH ′( s ) = Q1′ ( s ) + Q2′ ( s )
H ′( s )
1
=
Q1′ ( s ) As
b)
q1′ (t ) = 5 S(t) – 5S(t-12)
5 5 −12 s
− e
s s
1
5/ A 5/ A
H ′( s ) =
Q1′ ( s ) = 2 − 2 e −12 s
As
s
s
Q1′ ( s ) =
5-8
5
5
t S(t) − (t − 12) S(t-12)
A
A
h ′(t ) =
4+
5
t = 4 + 0.177t
A
0 ≤ t ≤ 12
h(t) =
5

4 +  × 12  = 6.122
A

12 < t
2.5
2
h'(t)
1.5
1
0.5
0
0
5
10
15
20
25
30
35
40
45
50
time
Fig S5.8a. Liquid level response for part b)
c)
h = 6.122 ft at the new steady state t ≥ 12
d)
q1′ (t ) = 5 S(t) – 10S(t-12) + 5S(t-24) ; tw = 12
(
)
5
1 − 2e −12 s + e − 24 s
s
5 / A 10 / A
5/ A
H ′( s ) = 2 − 2 e −12 s + 2 e − 24 s
s
s
s
Q1′ ( s ) =
h(t) = 4 + 0.177tS(t) − 0.354(t-12)S(t-12) + 0.177(t-24)S(t-24)
For t ≥ 24
h = 4 + 0.177t − 0.354(t − 12) + 0.177(t − 24) = 4 ft at t ≥ 24
5-9
2.5
2
h'(t)
1.5
1
0.5
0
-0.5
0
5
10
15
20
25
30
35
40
time
Fig S5.8b. Liquid level response for part d)
5.9
a)
Material balance over tank 1.
A
dh
= C (qi − 8.33h)
dt
where A = π × (4)2/4 = 12.6 ft2
C = 0.1337
ft 3 /min
USGPM
AsH ′( s ) = CQi′ ( s ) − (C × 8.33) H ′( s )
H ′( s )
0.12
=
Qi′ ( s ) 11.28s + 1
For tank 2,
A
dh
= C (qi − q)
dt
5-10
45
50
AsH ′( s ) = CQi′( s )
b)
H ′( s ) 0.011
=
Qi′ ( s )
s
,
Qi′( s ) = 20 / s
For tank 1,
H ′( s ) =
2.4
2.4
27.1
=
−
s (11.28s + 1)
s 11.28s + 1
h(t) = 6 + 2.4(1 – e-t/11.28)
For tank 2,
H ′( s ) = 0.22 / s 2
h(t) = 6 + 0.22t
9
8
7
6
h'(t)
5
4
3
2
1
Tank 1
Tank2
0
0
5
10
15
20
25
30
35
40
time
Fig S5.9. Transient response in tanks 1 and 2 for a step input.
c)
d)
For tank 1,
h(∞) = 6 +2.4 – 0 = 8.4 ft
For tank 2,
h(∞) = 6 + (0.22 × ∞) = ∞ ft
For tank 1,
8 = 6 + 2.4(1 – e-t/11.28)
h = 8 ft at t = 20.1 min
8 = 6 + 0.22t
h = 8 ft at t = 9.4 min
For tank 2,
Tank 2 overflows first, at 9.4 min.
5-11
5.10
a)
The dynamic behavior of the liquid level is given by
d 2 h′
dh ′
+A
+ Bh ′ = C p ′(t )
dt
dt
where
A=
6µ
R 2ρ
B=
3g
2L
and C =
3
4ρL
Taking the Laplace Transform and assuming initial values = 0
s 2 H ′( s ) + AsH ′( s ) + BH ′( s ) = C P ′( s )
or H ′( s ) =
C/B
P ′( s )
1 2 A
s + s +1
B
B
We want the previous equation to have the form
H ′( s ) =
K
P ′( s )
τ s + 2ζτs + 1
2
Hence K = C/B =
1
2ρg
1
τ =
B
 2L 
then τ = 1 / B =  
 3g 
A
2ζτ =
B
3µ  2 L 
then ζ = 2  
R ρ  3g 
2
b)
2
1/ 2
1/ 2
The manometer response oscillates as long as 0 < ζ < 1 or
1/ 2
0 <
b)
3µ  2 L 
 
R 2ρ  3 g 
< 1
If ρ is larger , then ζ is smaller and the response would be more
oscillatory.
If µ is larger, then ζ is larger and the response would be less oscillatory.
5-12
5.11
Y(s) =
K
K2
KM
= 21 +
s (τs + 1)
s (τs + 1) s
2
K1τs + K1 + K2s = KM
K1 = KM
K2 = −K1τ = − KMτ
Hence
Y(s) =
or
KM
KMτ
−
2
s (τs + 1)
s
y(t) = KMt − KMτ (1-e-t/τ)
After a long enough time, we can simplify to
y(t) ≈ KMt - KMτ
(linear)
slope = KM
intercept = −KMτ
That way we can get K and τ
y(t)
Slope = KM
−ΚΜτ
Figure S5.11. Time domain response and parameter evaluation
5-13
5.12
a)
1y1 + Ky1 + 4 y = x
Assuming y(0) = y1 (0) = 0
Y (s)
1
0.25
= 2
=
2
X ( s ) s + Ks + 4 0.25s + 0.25 Ks + 1
b)
Characteristic equation is
s2 + Ks + 4 = 0
The roots are s =
− K ± K 2 − 16
2
-10 ≤ K < -4 Roots : positive real, distinct
Response : A + B e t / τ1 + C et / τ 2
K = -4
Roots : positive real, repeated
Response : A + Bet/τ + C et/τ
-4 < K < 0
Roots: complex with positive real part.
t
t
Response: A + eζt/τ (B cos 1 − ζ 2
+ C sin 1 − ζ 2 )
τ
τ
K=0
Roots: imaginary, zero real part.
Response: A + B cos t/τ + C sin t/τ
0<K<4
Roots: complex with negative real part.
t
t
Response: A + e-ζt/τ (B cos 1 − ζ 2
+ C sin 1 − ζ 2 )
τ
τ
K=4
Roots: negative real, repeated.
Response: A + Be-t/τ + C t e-t/τ
4 < K ≤ 10
Roots: negative real, distinct
Response: A + B e −t / τ1 + C e −t / τ 2
Response will converge in region 0 < K ≤ 10, and will not converge in
region –10 ≤ K ≤ 0
5-14
5.13
a)
The solution of a critically-damped second-order process to a step change
of magnitude M is given by Eq. 5-50 in text.
  t

y(t) = KM 1 − 1 + e −t / τ 
  τ

Rearranging
y
 t
= 1 − 1 + e − t / τ
KM
 τ

1 +

t  −t / τ
y
= 1−
e
τ
KM
When y/KM = 0.95, the response is 0.05 KM below the steady-state value.
KM
0.95KM
y
0
ts
 t s  −t / τ
= 1 − 0.95 = 0.05
1 + e
τ

 t  t
ln1 + s  − s = ln(0.05) = −3.00
τ τ

 t  t
Let E = ln1 + s  − s + 3
τ τ

5-15
time
and find value of
ts
that makes E ≈ 0 by trial-and-error.
τ
E
0.6094
-0.2082
0.2047
-0.0008
ts/τ
4
5
4.5
4.75
b)
∴
a value of t = 4.75τ is ts, the settling time.
Y(s) =
a
a a
a4
Ka
= 1 + 22 + 3 +
2
s s
τs + 1 (τs + 1) 2
s (τs + 1)
2
We know that the a3 and a4 terms are exponentials that go to zero for large
values of time, leaving a linear response.
a2 = lim
s →0
Define Q(s) =
Ka
= Ka
(τs + 1) 2
Ka
(τs + 1) 2
dQ − 2 Kaτ
=
ds (τs + 1) 3
Then a1 =
 − 2 Kaτ 
1
lim 

1! s →0  (τs + 1) 3 
(from Eq. 3-62)
a1 = − 2 Kaτ
∴ the long-time response (after transients have died out) is
y 2 (t ) = Kat − 2 Kaτ = Ka (t − 2τ)
= a (t − 2τ)
for K = 1
and we see that the output lags the input by a time equal to 2τ.
5-16
2τ
y
x=at
0
yl =a(t-2τ)
actual response
time
5.14
a)
11.2mm − 8mm
= 0.20mm / psi
31psi − 15psi
Gain =
12.7 mm − 11.2mm
= 0.47
11.2mm − 8mm
 − πζ 
 = 0.47
Overshoot = exp 
,
 1− ζ2 


 2πτ 
 = 2.3 sec
Period = 
 1− ζ2 


Overshoot =
τ = 2.3 sec ×
ζ = 0.234
1 − 0.234 2
= 0.356 sec
2π
R ′( s )
0.2
=
2
P ′( s ) 0.127 s + 0.167 s + 1
b)
(1)
From Eq. 1, taking the inverse Laplace transform,
11′ + 0.167 R1 ′ + R ′ = 0.2 P ′
0.127 R
11′ = R
11
R
R1 ′ = R1
R ′ = R-8
11 + 0.167 R1 + R = 0.2 P + 5
0.127 R
11 + 1.31 R1 + 7.88 R = 1.57 P + 39.5
R
5-17
P ′ = P-15
5.15
P ′( s )
3
=
2 2
T ′( s ) (3) s + 2(0.7)(3) s + 1
[ºC/kW]
Note that the input change p ′(t ) = 26 − 20 = 6 kw
Since K is 3 °C/kW, the output change in going to the new steady state
will be
T ′ = (31 C / kW )(6 kW ) = 18 1 C
t →∞
a)
Therefore the expression for T(t) is Eq. 5-51
0 .7 t 

 1 − ( 0 .7 ) 2
−

3 
T (t ) = 70 + 18 1 − e
cos
 
3

 
1
1

 1 − (0.7 ) 2
0. 7

t +
sin 
2


τ
1 − ( 0 .7 )


25
20
T'(t)
15
10
5
0
0
5
10
15
20
25
30
35
40
45
50
time
Fig S5.15. Process temperature response for a step input
b)
The overshoot can obtained from Eq. 5-52 or Fig. 5.11. From Figure 5.11
we see that OS ≈ 0.05 for ζ=0.7. This means that maximum temperature is
Tmax ≈ 70° + (18)(1.05) = 70 + 18.9 = 88.9°
From Fig S5.15 we obtain a more accurate value.
5-18
 
t  

 
The time at which this maximum occurs can be calculated by taking
derivative of Eq. 5-51 or by inspection of Fig. 5.8. From the figure we see
that t / τ = 3.8 at the point where an (interpolated) ζ=0.7 line would be.
∴
tmax ≈ 3.8 (3 min) = 11.4 minutes
5.16
For underdamped responses,

  1 − ζ2
− ζt / τ
cos
y (t ) = KM 1 − e
  τ

a)

 1 − ζ2
ζ

sin 
t +
2

 τ
1− ζ


 
t  

 
At the response peaks,
2

dy
 ζ −ζ t / τ   1 − ζ

= KM  e
cos 
dt
  τ
 τ
−e
−ζt / τ

 1− ζ2
ζ
t+
sin 

 τ
1 − ζ2


 1− ζ2
 1− ζ2
−
sin 
 τ
τ


 ζ
 1 − ζ2
t  + cos 
 τ
 τ



t 


  
t   = 0
 
 
Since KM ≠ 0 and e − ζt / τ ≠ 0
2
 ζ ζ   1− ζ
0 =  −  cos
 τ τ   τ
  ζ2
1 − ζ2
t + 
+
  τ 1 − ζ2
τ
 
 1− ζ2 
πτ
0 = sin
t  = sin nπ , t = n
 τ

1 − ζ2


where n is the number of the peak.
Time to the first peak,
b)
Overshoot, OS =
tp =
πτ
1 − ζ2
y (t p ) − KM
KM
5-19
  1 − ζ2
 sin 
  τ
 

t


(5-51)

ζ
 − ζt  
OS = − exp
sin(π)
 cos(π) +
 τ  
1 − ζ2

 − ζτπ 
 − πζ 
= exp
= exp 


2
2
 τ 1 − ζ 
 1 − ζ 
c)
Decay ratio, DR =
where y (t 3 p ) =
KM e
DR =
d)
y (t p ) − KM
3πτ
1− ζ2
− ζt 3 p / τ
KM e
y (t 3 p ) − KM
− ζt p / τ
is the time to the third peak.
 ζ  2πτ 
 ζ


= exp − (t 3 p − t p ) = exp − 
2

 τ  1 − ζ 
 τ



 −2πζ 
= exp 
 = (OS) 2
2
 1 − ζ 
Consider the trigonometric identity
sin (A+B) = sin A cos B + cos A sin B
 1− ζ2 
Let B = 
t  , sin A = 1 − ζ 2 ,
cos A = ζ
 τ





1
y (t ) = KM 1 − e −ζt / τ
1 − ζ 2 cos B + ζ sin B 

1− ζ2



e − ζt / τ
= KM 1 −
sin( A + B)

1 − ζ2

[
]
Hence for t ≥ t s , the settling time,
e − ζt / τ
1− ζ
2
≤ 0.05 , or
Therefore,
ts ≥
(
t ≥ − ln 0.05 1 − ζ 2
τ  20
ln
ζ  1 − ζ 2




5-20
)ζτ
5.17
a)
Assume underdamped second-order model (exhibits overshoot)
K=
1756456
10 − 6 ft
ft
=
= 0 .2
123456 140 − 120 gal/min
gal/min
Fraction overshoot =
11 − 10 1
= = 0.25
10 − 6 4
From Fig 5.11, this corresponds (approx) to ζ = 0.4
From Fig. 5.8 , ζ = 0.4 , we note that tp/τ ≈ 3.5
Since tp = 4 minutes (from problem statement), τ = 1.14 min
∴
G p(s) =
0.2
0 .2
=
2
(1.14) s + 2(0.4)(1.14) s + 1 1.31s + 0.91s + 1
2
2
b)
In Chapter 6 we see that a 2nd-order overdamped process model with a
numerator term can exhibit overshoot. But if the process is underdamped,
it is unique.
a)
Assuming constant volume and density,
5.18
Overall material balances yield: q2 = q1 = q
(1)
Component material balances:
dc1
= q (ci − c1 )
dt
dc
V2 2 = q (c1 − c 2 )
dt
(2)
V1
b)
(3)
Degrees of freedom analysis
3 Parameters : V1, V2, q
5-21
3 Variables : ci, c1, c2
2 Equations: (2) and (3)
NF = NV − NE = 3 − 2 = 1
Hence one input must be a specified function of time.
2 Outputs = c1, c2
1 Input = ci
c)
If a recycle stream is used
Overall material balances:
q1 = (1+r)q
(4)
q2 = q1 = (1+r)q
(5)
q3 = q2 – rq = (1+r)q − rq = q
(6)
Component material balances:
V1
dc1
= qci + rqc 2 − (1 + r )qc1
dt
(7)
V2
dc 2
= (1 + r )qc1 − (1 + r )qc 2
dt
(8)
Degrees of freedom analysis is the same except now we have
4 parameters : V1, V2, q, r
5-22
d)
If r → ∞ , there will a large amount of mixing between the two tanks as a
result of the very high internal circulation.
Thus the process acts like
ci
q
c2
q
Total Volume = V1 + V2
Model :
dc 2
= q (c i − c 2 )
dt
c1 = c2 (complete internal mixing)
(V1 +V2)
Degrees of freedom analysis is same as part b)
5.19
a)
For the original system,
dh1
h
= Cqi − 1
dt
R1
dh
h
h
A2 2 = 1 − 2
dt
R1 R2
A1
where A1 = A2 = π(3)2/4 = 7.07 ft2
ft 3 /min
gpm
h
2.5
ft
= 0.187 3
R1 = R2 = 1 =
Cqi 0.1337 × 100
ft /min
C = 0.1337
5-23
(9)
(10)
Using deviation variables and taking Laplace transforms,
H 1′ ( s )
=
Qi′ ( s )
C
=
CR1
0.025
=
A1 R1 s + 1 1.32 s + 1
1
R1
H 2′ ( s )
1 / R1
R2 / R1
1
=
=
=
1
H 1′ ( s )
A2 R2 s + 1 1.32 s + 1
A2 s +
R2
H 2′ ( s )
0.025
=
Qi′( s ) (1.32s + 1) 2
A1 s +
For step change in qi of magnitude M,
h1′max = 0.025M
h2′ max = 0.025M since the second-order transfer function
0.025
is critically damped (ζ=1), not underdamped
(1.32 s + 1) 2
2.5 ft
Hence Mmax =
= 100 gpm
0.025 ft/gpm
For the modified system,
A
dh
h
= Cq i −
dt
R
A = π(4) 2 / 4 = 12.6 ft 2
V = V1 + V2 = 2 × 7.07ft 2 × 5ft = 70.7ft3
hmax = V/A = 5.62 ft
R=
h
Cq i
H ′( s )
=
Qi′ ( s )
=
0.5 × 5.62
ft
= 0.21 3
0.1337 × 100
ft /min
C
As +
1
R
=
CR
0.0281
=
ARs + 1 2.64 s + 1
′ = 0.0281M
hmax
2.81 ft
Mmax =
= 100 gpm
0.0281 ft/gpm
5-24
Hence, both systems can handle the same maximum step disturbance in qi.
b)
For step change of magnitude M, Qi′( s ) =
M
s
For original system,
Q2′ ( s ) =
1
1
0.025 M
H 2′ ( s ) =
R2
0.187 (1.32 s + 1) 2 s
1

1.32
1.32
= 0.134M  −
−
2 
 s (1.32s + 1) (1.32s + 1) 
 
t  −t / 1.32 
q ′2 (t ) = 0.134 M 1 − 1 +
e

  1.32 

For modified system,
Q ′( s ) =
1
1
0.0281 M
2.64 
1
H ′( s ) =
= 0.134 M  −

R
0.21 (2.64 s + 1) s
 s 2.64 s + 1
[
q ′(t ) = 0.134 M 1 − e −t / 2.64
]
Original system provides better damping since q 2′ (t ) < q ′(t ) for t < 3.4.
5.20
a)
Caustic balance for the tank,
ρV
dC
= w1c1 + w2 c 2 − wc
dt
Since V is constant, w = w1 + w2 = 10 lb/min
For constant flows,
ρVsC ′( s ) = w1C1′ ( s ) + w2 C 2′ ( s ) − wC ′( s )
w1
C ′( s )
5
0.5
=
=
=
C1′ ( s ) ρVs + w (70)(7) s + 10 49s + 1
5-25
C m′ ( s )
K
,
=
C ′( s ) τs + 1
K = (3-0)/3 = 1
τ ≈ 6 sec = 0.1 min
,
(from the graph)
C m′ ( s )
1
0.5
0.5
=
=
C1′ ( s ) (0.1s + 1) (49s + 1) (0.1s + 1)(49s + 1)
b)
3
s
C1′ ( s ) =
1.5
s (0.1s + 1)(49 s + 1)


1
c ′m (t ) = 1.51 +
(0.1e −t / 0.1 − 49e −t / 49 )
 (49 − 0.1)

C m′ ( s ) =
c)
C m′ ( s ) =
0.5 3
1.5
=
(49 s + 1) s s (49 s + 1)
(
c ′m (t ) = 1.5 1 − e − t / 49
The responses in b) and c) are nearly the same. Hence the dynamics of the
conductivity cell are negligible.
1.5
1
Cm'(t)
d)
)
0.5
Part b)
Part c)
0
0
20
40
60
80
100
time
120
140
160
Fig S5.20. Step responses for parts b) and c)
5-26
180
200
5.21
Assumptions:
a)
1) Perfectly mixed reactor
2) Constant fluid properties and heat of reaction
Component balance for A,
V
dc A
= q (c A i − c A ) − Vk (T )c A
dt
(1)
Energy balance for the tank,
ρVC
dT
= ρqC (Ti − T ) + (− ∆H R )Vk (T )c A
dt
(2)
Since a transfer function with respect to cAi is desired, assume the other
inputs, namely q and Ti, are constant. Linearize (1) and (2) and note that
dc A dc ′A
dT dT ′
=
,
=
,
dt
dt
dt
dt
V
dc ′A
20000
= qc ′A i − (q + Vk (T ))c ′A − Vc A k (T )
T′
dt
T2
ρVC
(3)
dT ′
20000 

= − ρqC + ∆H RVc A k (T )
T ′ − ∆H RVk (T )c ′A (4)
dt
T2 

Taking 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 R )Vc A k (T ) T 2  T ′( s ) = (−∆H R )Vk (T )C A′ ( s ) (6)
Substituting C ′A (s ) from Eq. 5 into Eq. 6 and rearranging,
T ′( s )
=
C A′i ( s )
(−∆H R )Vk (T )q
20000 
20000

Vs + q + Vk (T )  ρVCs + ρqC − (−∆H R )Vc A k (T )
+ (−∆H R )V 2 c A k 2 (T )
2

T 
T2

(7)
c A is obtained from Eq. 1 at steady state,
5-27
cA =
qc Ai
= 0.001155 lb mol/cu.ft.
q + Vk (T )
Substituting the numerical values of T , ρ, C, –∆HR, q, V, c A into Eq. 7
and simplifying,
T ′( s )
11.38
=
C ′A i ( s ) (0.0722s + 1)(50s + 1)
For step response, C ′Ai ( s ) = 1 / s
T ′( s ) =
11.38
s (0.0722s + 1)(50s + 1)


1
T ′(t ) = 11.381 +
(0.0722e −t / 0.0722 − 50e −t / 50 ) 
 (50 − 0.0722)

A first-order approximation of the transfer function is
T ′( s )
11.38
=
C ′A i ( s ) 50s + 1
For step response, T ′( s ) =
[
11.38
or T ′(t ) = 11.38 1 − e −t / 50
s (50s + 1)
The two step responses are very close to each other hence the
approximation is valid.
12
10
T'(t)
8
6
4
2
Using transfer function
Using first-order approximation
0
0
20
40
60
80
100
time
120
140
160
180
200
Fig S5.21. Step responses for the 2nd order t.f and 1st order approx.
5-28
]
5.22
(τas+1)Y1(s) = K1U1(s) + Kb Y2(s)
(τbs+1)Y2(s) = K2U2(s) + Y1(s)
a)
(1)
(2)
Since the only transfer functions requested involve U1(s), we can let U2(s)
be zero. Then, substituting for Y1(s) from (2)
Y1(s) = (τbs+1)Y2(s)
(3)
(τas+1)(τbs+1)Y2(s) =K1U1(s) + KbY2(s)
(4)
Rearranging (4)
[(τas+1)(τbs+1) − Kb]Y2(s) =K1U1(s)
Y2 ( s )
K1
=
U 1 ( s ) (τ a s + 1)(τ b s + 1) − K b
Also, since
∴
(5)
Y1 ( s )
= τb s + 1
Y2 ( s )
(6)
From (5) and (6)
K 1 (τ b s + 1)
Y1 ( s ) Y2 ( s ) Y1 ( s )
=
×
=
U 1 ( s ) U 1 ( s ) Y2 ( s ) (τ a s + 1)(τ b s + 1) − K b
b)
(7)
The gain is the change in y1(or y2) for a unit step change in u1. Using the
FVT with U1(s) = 1/s.

K1
y 2 (t → ∞) = lim  s
s →0
 (τ a s + 1)(τ b s + 1) − K b
K1
1
=
s  1 − Kb
This is the gain of TF Y2(s)/U1(s).
Alternatively,


 Y (s) 
K1
K1
K = lim  2  = lim 
=

s →0 U ( s )
 1  s →0  (τ a s + 1)(τ b s + 1) − K b  1 − K b
For Y1(s)/U1(s)
5-29

K 1 (τ b s + 1)
y1 (t → ∞) = lim  s
s →0
 (τ a s + 1)(τ b s + 1) − K b
K1
1
=
s  1 − Kb
In other words, the gain of each transfer function is
c)
K1
1 − Kb
Y2 ( s )
K1
=
U 1 ( s ) (τ a s + 1)(τ b s + 1) − K b
(5)
Second-order process but the denominator is not in standard form, i.e.,
τ2s2+2ζτs+1
Put it in that form
Y2 ( s )
K1
=
2
U 1 ( s) τ a τ b s + (τ a + τ b ) s + 1 − K b
(8)
Dividing through by 1- Kb
K 1 /(1 − K b )
Y2 ( s )
=
τ a τ b 2 (τ a + τ b )
U 1 (s)
s +
s +1
1 − Kb
1 − Kb
(9)
Now we see that the gain K = K1/(1-Kb), as before
τ2 =
τa τ b
1 − Kb
2ζτ =
ζ=
τ=
τa τb
1 − Kb
(10)
τa + τb
, then
1 − Kb
1 τa + τb
2 1 − Kb
1 − K b  1 τa + τb 
=

τa τb
 2 τ a τ b 
1
1 − Kb
(11)
Investigating Eq. 11 we see that the quantity in brackets is the same as ζ
for an overdamped 2nd-order system (ζOD) [ from Eq. 5-43 in text].
ζ=
ζ OD
1 − Kb
where ζ OD =
1 τa + τ b
2 τa + τb
5-30
(12)
Since ζOD>1,
ζ>1, for all 0 < Kb < 1.
In other words, since the quantity in brackets is the value of ζ for an
overdamped system (i.e. for τa ≠ τb is >1) and 1 − K b <1 for any positive
Kb, we can say that this process will be more overdamped (larger ζ) if Kb
is positive and <1.
For negative Kb we can find the value of Kb that makes ζ = 1, i.e., yields a
critically-damped 2nd-order system.
ζ OD
ζ =1=
(13)
1 − K b1
ζ
or 1 = OD
1 − K b1
2
1 – Kb1 = ζOD2
Kb1 = 1 − ζOD2
(14)
where
Kb1 < 0 is the value of Kb that yields a critically-damped process.
Summarizing, the system is overdamped for 1 − ζOD2 < Kb < 1.
Regarding the integrator form, note that
Y2 ( s )
K1
=
2
U 1 ( s) τ a τ b s + (τ a + τ b ) s + 1 − K b
(8)
For Kb = 1
Y2 ( s )
K1
K1
=
=
2
U 1 ( s ) τ a τ b s + (τ a + τ b ) s s[τ a τ b s + (τ a + τ b )]
K 1 /(τ a + τ b )
 τ τ

s  a b s + 1
 τa + τ b

K 1′
which has the form =
( s indicates presence of integrator)
s (τ′s + 1)
=
5-31
d)
Return to Eq. 8
System A:
Y2 ( s )
K1
2K
1
=
= 2 1
= 2
2
U 1 ( s ) (2)(1) s + (2 + 1) s + 1 − 0.5 4s + 6s + 1 4s + 6s + 1
τ2 = 4
2ζτ = 6
→
→
τ=2
ζ = 1.5
System B:
For system
1
1
= 2
(2 s + 1)( s + 1) 2 s + 3s + 1
τ22 = 2
→
τ2 =
2ζ2τ2 = 3
→
ζ2 =
2
3
2 2
=
1.5
2
≈ 1.05
Since system A has larger τ (2 vs. 2 ) and larger ζ (1.5 vs 1.05), it will
respond slower. These results correspond to our earlier analysis.
5-32
1234567898
6.1
a)
By using MATLAB, the poles and zeros are:
Zeros: (-1 +1i) , (-1 -1i)
Poles: -4.3446
(-1.0834 +0.5853i)
(-1.0834 –0.5853i)
(+0.7557 +0.5830i)
(+0.7557 −0.5830i)
These results are shown in Fig E6.1
Figure S6.1. Poles and zeros of G(s) plotted in the complex s plane.
Solution Manual for Process Dynamics and Control, 2nd edition,
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp.
6-1
b)
c)
Process output will be unbounded because some poles lie in the right half
plane.
By using Simulink-MATLAB
8
2
x 10
0
-2
-4
Output
-6
-8
-10
-12
-14
-16
0
5
10
15
20
25
30
Time
Figure E6.1b. Response of the output of this process to a unit step input.
As shown in Fig. S6.1b, the right half plane pole pair makes the process
unstable.
6.2
a)
Standard form =
K (τ a s + 1)
(τ1 s + 1)(τ 2 s + 1)
b)
0.5(2s + 1)e −5 s
Hence G ( s ) =
(0.5s + 1)(2s + 1)
Applying zero-pole cancellation:
0.5e −5 s
G (s) =
(0.5s + 1)
c)
Gain = 0.5
Pole = −2
Zeros = No zeros due to the zero-pole cancellation.
6-2
d)
1/1 Pade approximation: e −5 s =
(1 − 5 / 2 s )
(1 + 5 / 2 s )
The transfer function is now
G (s) =
0.5
(1 − 5 / 2 s )
×
0.5s + 1 (1 + 5 / 2 s )
Gain = 0.5
Poles = −2, −2/5
Zeros = + 2/5
6.3
Y ( s ) K (τ a s + 1)
=
X ( s)
(τ1 s + 1)
X (s) =
,
M
s
From Eq. 6-13
a)
b)
  τ 

 τ − τ1 −t / τ1 
y(t) = KM 1 − 1 − a e −t / τ1  = KM 1 + a
e

τ1 
τ1


 

 τ − τ1  τ a
y (0 + ) = KM 1 + a
KM
=
τ1  τ1

Overshoot → y(t) > KM
 τ − τ1 −t / τ1 
KM 1 + a
e
 > KM
τ1


or τa − τ1 > 0 , that is, τa > τ1
y1 = − KM
c)
( τ a − τ1 )
τ1
2
e −t / τ1 < 0
for
KM > 0
Inverse response → y(t) < 0
 τ − τ1 −t / τ1 
KM 1 + a
e
<0
τ1


τ a − τ1
τa
< −e +t / τ1 or
< 1 − e +t / τ1 < 0
τ1
τ1
Therefore τa < 0.
6-3
at t = 0.
6.4
K (τ a s + 1)
Y (s)
=
, τ1>τ2,
X ( s ) (τ1 s + 1)(τ 2 s + 1)
X(s) = M/s
From Eq. 6-15
 τ − τ1 −t / τ1 τ a − τ 2 −t / τ2 
y (t ) = KM 1 + a
e
e
−

τ
−
τ
τ
−
τ
1
2
1
2


a)
Extremum → y1 (t ) = 0

1  τ − τ1  −t / τ1 1  τ a − τ 2
e
KM 0 −  a
+ 
τ
τ
−
τ
τ 2  τ1 − τ 2
1

1
2


1
1 
−t  − 
1 − τa / τ2
τ τ
= e  1 2 ≥1
1 − τ a / τ1
b)
 −t / τ 2 
e
=0


since
τ1>τ2
Overshoot → y (t ) > KM
 τ − τ1 −t / τ1 τ a − τ 2 −t / τ2 
KM 1 + a
e
−
e
 > KM
τ1 − τ 2
 τ1 − τ 2

 1
1
−t  − 
τ a − τ1
τ τ
> e  2 1  > 0 , therefore τa>τ1
τa − τ2
c)
Inverse response → y1 (t ) < 0 at t = 0+

1  τ − τ1  −t / τ1 1  τ a − τ 2  −t / τ2 
+
e
e
KM 0 −  a
+ 
 < 0 at t = 0
τ1  τ1 − τ 2 
τ 2  τ1 − τ 2 


1  τ − τ1  1  τ a − τ 2 
+ 
<0
−  a
τ1  τ1 − τ 2  τ 2  τ1 − τ 2 
1 1
τ a  − 
τ 2 τ1  < 0
τ1 −τ 2
6-4
Since τ1 > τ2, τa < 0.
d)
If an extremum in y exists, then from (a)
 1 1
− t  −
 τ1 τ 2



1 − τa τ2 

= 
 1 − τ a τ1 
1 − τa τ2 
ττ

t = 1 2 ln
τ1 − τ 2  1 − τ a τ1 
e
6.5
Substituting the numerical values into Eq. 6-15
Case (i) : y(t) = 1 (1 + 1.25e-t/10 − 2.25e-t/2)
Case (ii(a)) : y(t) = 1 (1 − 0.75e-t/10 − 0.25e-t/2)
Case (ii(b)) : y(t) = 1 (1 − 1.125e-t/10 + 0.125e-t/2)
Case (iii) : y(t) = 1 (1 − 1.5e-t/10 + 0.5e-t/2)
1.6
case(i)
case(ii)a
case(ii)b
case(iii)
1.4
1.2
1
y(t)
0.8
0.6
0.4
0.2
0
-0.2
0
5
10
15
20
25
30
35
40
45
50
Time
Figure S6.5. Step response of a second-order system with a single zero.
6-5
Conclusions:
τa > τ1 gives overshoot.
0 < τa < τ1 gives response similar to ordinary first-order process
response.
τa < 0 gives inverse response.
6.6
Y (s) =
K1
K2
K2 
K
U (s) +
U ( s) =  1 +
U ( s )
s
τs + 1
 s τs + 1
Y ( s ) K 1 τs + K 1 + K 2 s ( K 1τ + K 2 ) s + K 1
=
=
U ( s)
s (τs + 1)
s (τs + 1)
Put in standard K/τ form for analysis:


K 
K 1  τ + 2  s + 1
K1 
Y (s)


G( s) =
= 
U ( s)
s (τs + 1)
a)
Order of G(s) is 2 (maximum exponent on s in denominator is 2)
b)
Gain of G(s) is K1. Gain is negative if K1 < 0.
c)
Poles of G(s) are: s1 = 0 and s2 = –1/τ
s1 is on imaginary axis; s2 is in LHP.
d)
Zero of G(s) is:
sa =
If
− K1
−1
=

K  K1τ + K 2
 τ + 2 
K1 

K1
< 0 , the zero is in RHP.
K1τ + K 2
6-6
Two possibilities: 1. K1<0 and K1τ + K2 >0
2. K1 > 0 and K1τ + K2 < 0
e)
Gain is negative if K1 < 0
Then zero is RHP if K1τ + K2 > 0
This is the only possibility.
f)
Constant term and e-t/τ term.
g)
If input is M/s, the output will contain a t term, that is, it is not bounded.
6.7
a)
2
s
−3
−3 2
Q ′( s ) =
P ′( s ) =
20 s + 1
20 s + 1 s
p ′(t ) = (4 − 2) S (t )
, P ′( s ) =
Q ′(t ) = −6(1 − e −t / 20 )
b)
R ′( s ) + Q ′( s ) = Pm′ ( s )
r ′(t ) + q ′(t ) = p ′m (t ) = p m (t ) − p m (0)
r ′(t ) = p m (t ) − 12 + 6(1 − e − t / 20 )
K=
r ′(t = ∞)
18 − 12 + 6(1 − 0)
=
=6
p (t = ∞) − p (t = 0)
4−2
Overshoot,
OS =
r ′(t = 15) − r ′(t = ∞) 27 − 12 + 6(1 − e −15 / 20 ) − 12
=
= 0.514
r ′(t = ∞)
12
6-7
 − πζ
OS = exp
 1− ζ2


 = 0.514


,
ζ = 0 .2
Period, T, for r ′(t ) is equal to the period for pm(t) since e-t/20 decreases
monotonically.
c)
Thus,
T = 50 − 15 = 35
and
τ=
T
1 − ζ 2 = 5.46
2π
Pm′ ( s )
K
K′
= 2 2
+
P ′( s ) τ s + 2ζτs + 1 τ′s + 1
(K ′τ )s
=
2
d)
+ ( Kτ′ + 2 K ′ζτ) s + ( K + K ′)
(τ s 2 + 2ζτs + 1)(τ′s + 1)
2
2
Overall process gain =
Pm′ ( s )
P ′( s )
= K + K′ = 6 −3 = 3
s =0
%
psi
6.8
a)
Transfer Function for blending tank:
Gbt ( s ) =
K bt
τ bt s + 1
where K bt =
τ bt =
qin
≠1
∑ qi
2m 3
= 2 min
1m 3 / min
Transfer Function for transfer line
Gtl ( s ) =
K tl
(τ tl s + 1)5
where K tl = 1
τ tl =
6-8
0.1m 3
= 0.02 min
5 × 1m 3 / min
′ (s)
C out
K bt
=
C in′ ( s ) (2s + 1)(0.02s + 1) 5
∴
a 6th-order transfer function.
b)
Since τbt >> τtl [ 2 >> 0.02] we can approximate
1
by e-θ s
5
(0.02 s + 1)
5
where θ = ∑ (0.02) = 0.1
i =1
′ ( s ) K bt e −0.1s
C out
≈
C in′ ( s )
2s + 1
∴
c)
Since τbt ≈ 100 τtl , we can imagine that this approximate TF will yield
results very close to those from the original TF (part (a)). We also note
that this approximate TF is exactly the same as would have been obtained
using a plug flow assumption for the transfer line. Thus we conclude that
investing a lot of effort into obtaining an accurate dynamic model for the
transfer line is not worthwhile in this case.
[ Note that, if τbt ≈ τtl , this conclusion would not be valid]
By using Simulink-MATLAB,
1.2
1
0.8
Output/Kbt
d)
0.6
0.4
0.2
0
Exact model
Approximate model
-0.2
0
5
10
15
Time
20
25
30
Fig S6.8. Unit step responses for exact and approximate model.
6-9
6.9
a), b) Represent processes that are (approximately) critically damped. A step
response or frequency response in each case can be fit graphically or
numerically.
c)
θ = 2, τ = 10
d)
Exhibits strong overshoot. Can’t approximate it well.
e)
θ = 0.5, τ = 10
f)
θ = 1, τ = 10
g)
Underdamped (oscillatory). Can’t approximate it well.
h)
θ = 2, τ = 0
By using Simulink-MATLAB, models for parts c), e), f) and h) are
compared: (Suppose K = 1)
Part c)
1
0.9
0.8
0.7
Output
0.6
0.5
0.4
0.3
0.2
0.1
Exact model
Approximate model
0
0
5
10
15
20
25
time
30
35
40
45
50
Figure S6.9a. Unit step responses for exact and approximate model in part c)
6-10
Part e)
1
0.9
0.8
0.7
Output
0.6
0.5
0.4
0.3
0.2
0.1
Exact model
Approximate model
0
0
5
10
15
20
25
Time
30
35
40
45
50
Figure S6.9b. Unit step responses for exact and approximate model in part e)
Part f)
1
0.9
0.8
0.7
Output
0.6
0.5
0.4
0.3
0.2
0.1
Exact model
Approximate model
0
0
5
10
15
20
25
Time
30
35
40
45
50
Figure S6.9c. Unit step responses for exact and approximate model in part f)
6-11
Part h)
1.5
1
Output
0.5
0
-0.5
Exact model
Approximate model
-1
0
5
10
15
20
25
30
35
40
45
50
Time
Figure S6.9d. Unit step responses for exact and approximate model in part h)
6.10
a)
The transfer function for each tank is
C i′( s )
1
=
C i′−1 ( s )  V 
  s + 1
q
i = 1, 2, …, 5
,
where i represents the ith tank.
co is the inlet concentration to tank 1.
V is the volume of each tank.
q is the volumetric flow rate.
5
 C ′(s)   1 
C 5′ ( s )
= ∏ i
 ,
=
C 0′ ( s ) i =1  C i′−1 ( s )   6s + 1 
5
Then, by partial fraction expansion,
6-12

 t 1  t  2 1  t 3 1  t 4 
−t / 6 
c5 (t ) = 0.60 − 0.15 1 − e 1 + +   +   +   
 6 2!  6  3!  6  4!  6  

Using Simulink,
b)
0.6
c5
c4
0.58
c3
c2
Concentration
0.56
c1
0.54
0.52
0.5
0.48
0.46
0.44
0
5
10
15
20
25
time
30
35
40
45
50
Figure S6.10. Concentration step responses of the stirred tank.
The value of the expression for c5(t) verifies the simulation results above:


52 53 54 
c5 (30) = 0.60 − 0.15 1 − e −5 1 + 5 + + +  = 0.5161
2! 3! 4! 


6.11
a)
Y (s) =
− τa s + 1 E A B
C
= + 2 +
2
τ1 s + 1 s
s s
τ1 s + 1
We only need to calculate the coefficients A and B because Ce −t / τ1 → 0
for t >> τ1. However, there is a repeated pole at zero.
6-13
 E (− τ a s + 1) 
B = lim 
=E
s →0
 τ1 s + 1 
Now look at
E (− τ a s + 1) = As (τ1 s + 1) + B (τ1 s + 1) + Cs 2
− Eτ a s + E = Aτ1 s 2 + As + Bτ1 s + B + Cs 2
Equate coefficients on s:
− Eτ a = A + Bτ1
A = − E ( τ a + τ1 )
Then the long-time solution is
y (t ) ≈ Et − E (τ a + τ1 )
Plotting
(τa+τ1)
y
y(t)=Et
=Et-E(τa+τ1)
actual response
-E(τa+τ1)
time
b)
For a LHP zero, the apparent lag would be τ1 − τa
c)
For no zero, the apparent lag would be τ1
6-14
6.12
a)
Using Skogestad’s method
G ( s ) approx =
b)
5e − ( 0.5+0.2) s
5e −0.7 s
=
(10s + 1)((4 + 0.5) s + 1) (10s + 1)(4.5s + 1)
By using Simulink-MATLAB
5
4
Output
3
2
1
0
Exact model
Approximate model
-1
0
5
10
15
20
25
Time
30
35
40
45
50
Figure S6.12a. Unit step responses for exact and approximate model.
c)
Using MATLAB and saving output data on vectors, the maximum error is
Maximum error = 0.0521 at = 5.07 s
This maximum error is graphically shown in Fig. S6.12b
6-15
Exact model
Approximate model
0.9
0.8
0.7
Output
0.6
0.5
0.4
0.3
0.2
0.1
0
-0.1
0
1
2
3
4
Time
5
6
7
8
Figure S6.12b. Maximum error between responses for exact and approximate
model.
6.13
From the solution to Problem 2-5 (a) , the dynamic model for isothermal
operation is
V1 M dP1 Pd − P1 P1 − P2
=
−
RT1 dt
Ra
Rb
(1)
V2 M dP2 P1 − P2 P2 − Pf
=
−
RT2 dt
Rb
Rc
(2)
Taking Laplace transforms, and noting that Pf′ ( s ) = 0 since Pf is constant,
K b Pd′ ( s ) + K a P2′ ( s )
τ1 s + 1
K P ′( s )
P2′ ( s ) = c 1
τ2 s + 1
P1′( s ) =
where
6-16
(3)
(4)
K a = Ra /( Ra + Rb )
K b = Rb /( Ra + Rb )
K c = Rc /( Rb + Rc )
τ1 =
V1 M Ra Rb
RT1 ( Ra + Rb )
τ2 =
V2 M Rb Rc
RT2 ( Rb + Rc )
Substituting for P1′( s ) from Eq. 3 into 4,
 Kb Kc 


1
−
K
K
Kb Kc
P2′( s )
a
c 

=
=
Pd′ ( s ) (τ1 s + 1)(τ 2 s + 1) − K a K c  τ1 τ 2  2  τ1 + τ 2

 s + 
1
−
K
K
a
c 

1− Ka Kc
(5)

 s + 1

Substituting for P2′ ( s ) from Eq. 5 into 4,
 Kb 

 (τ2 s + 1)
1 − Ka Kc 
P1′( s )

=
Pd′ ( s )  τ1τ2  2  τ1 + τ2

s +
 1 − Ka Kc 
 1 − Ka Kc

 s +1

(6)
To determine whether the system is over- or underdamped, consider the
denominator of transfer functions in Eqs. 5 and 6.
 τ1 τ 2
τ 2 = 
1 − Ka Kc

τ + τ2
 , 2ζτ = 1
1− Ka Kc

Therefore,
ζ=
τ 
1 (τ1 + τ 2 ) (1 − K a K c ) 1  τ1
1
= 
+ 2 
2 (1 − K a K c )
2  τ2
τ1  (1 − K a K c )
τ1 τ 2
Since x + 1/x ≥ 2 for all positive x,
6-17
ζ≥
1
(1 − K a K c )
Since KaKc ≥ 0,
ζ ≥1
Hence the system is overdamped.
6.14
a)
For
Y (s) =
X (s) =
M
s
KM
A
B
C
= +
+
s (1 − s )(τs + 1) s 1 − s τs + 1
A = lim =
KM
= KM
(1 − s )(τs + 1)
B = lim =
KM
KM
=
s (τs + 1) τ + 1
s →0
s →1
 KM 
− KM τ2
KM
C = lim 
=
=
s →−1/ τ s (1 − s ) 
τ +1

  − 1 1 + 1 



 τ  τ 
Then,

et
τ −t / τ 
y1 (t ) = KM 1 −
−
e 
 τ +1 τ +1

For M =2 , K = 3, and τ = 3, the Simulink response is shown:
6-18
8
0
x 10
-2
Output
-4
-6
-8
-10
-12
0
2
4
6
8
10
12
14
16
18
Time
Figure S6.14a. Unit step response for part a).
b)
If
G2 ( s) =
Ke −2 s
(1 − s )(τs + 1)
then,
 e t −2
τ −( t − 2 ) / τ 
y 2 (t ) = KM 1 −
−
e
 S (t − 2)
 τ +1 τ +1

Note presence of positive exponential term.
c)
Approximating G2(s) using a Padé function
G2 ( s) =
K (1 − s )
K
=
( s + 1)(τs + 1)(1 − s ) ( s + 1)(τs + 1)
Note that the two remaining poles are in the LHP.
d)
For
X (s) =
Y (s) =
M
s
KM
s ( s + 1)(τs + 1)
Using Table 3.1
τ1 = 1
,
τ2 = τ
6-19
20
1


y3 (t ) = KM 1 +
(e −t − τe −t / τ )
 τ −1

Note that no positive exponential term is present.
e)
Instability may be hidden by a pole-zero cancellation.
f)
By using Simulink-MATLAB, unit step responses for parts b) and c) are
shown below: (M = 2 , K = 3 , τ = 3)
7
0
x 10
-1
-2
-3
Output
-4
-5
-6
-7
-8
-9
-10
0
2
4
6
8
10
12
14
16
18
20
Time
Figure S6.14b. Unit step response for part b).
6
5
Output
4
3
2
1
0
0
2
4
6
8
10
Time
12
14
16
18
Figure S6.14c. Unit step response for part c) .
6-20
20
6.15
From Eq. 6-71 and 6-72,
ζ=
Since x +
R2 A2 + R1 A1 + R2 A1
2 R1 R2 A1 A2
=
1  R1 A1
+
2  R2 A2
R2 A2
R1 A1
 1
+
 2

R2 A1
R1 A2
1
≥ 2 for all positive x and since R1, R2, A1, A2 are positive
x
ζ≥
1
(2) + 1 R2 A1 ≥ 1
2
2 R1 A2
6.16
a)
If w1 = 0 and ρ = constant
ρA2
dh2
= w0 − w2
dt
w2 =
1
h2
R2
[ Note: could also define R2 by q 2 =
1
ρ
h2 → w2 = ρq 2 =
h2 ]
R2
R2
Substituting,
ρA2
dh2
1
= w0 −
h2
dt
R2
or
ρA2 R2
dh2
= R2 w0 − h2
dt
Taking deviation variables and Laplace transforming
ρA2 R2 sH 2′ ( s ) + H 2′ ( s ) = R2W0′( s )
6-21
H 2′ ( s )
R2
=
W0′( s ) ρA2 R2 s + 1
Since W2′ ( s ) =
1
H 2′ ( s )
R2
W2′ ( s )
R2
1
1
=
=
W0′( s ) R2 ρA2 R2 s + 1 ρA2 R2 s + 1
Let τ2 = ρA2R2
W2′ ( s )
1
=
W0′( s ) τ 2 s + 1
b)
ρ = constant
dh
ρA1 1 = − w1
dt
w1 =
c)
1
(h1 − h2 )
R1
ρA2
dh2
= w0 + w1 − w2
dt
w2 =
1
h2
R2
Since this clearly is an interacting system, there will be a single zero. Also,
we know the gain must be equal to one.
∴
τ s +1
W1′( s )
= 2 2 a
W0′( s ) τ s + 2ζτs + 1
W2′ ( s )
1
= 2 2
W0′( s ) τ s + 2ζτs + 1
or
τa s + 1
W1′( s )
=
W0′( s ) (τ1′ s + 1)(τ′2 s + 1)
W2′ ( s )
1
=
W0′( s ) (τ1′ s + 1)(τ′2 s + 1)
where τ1′ and τ′2 are functions of the resistances and areas and can
only be obtained by factoring.
f)
Case b will be slower since the interacting system is 2nd-order, "including"
the 1st-order system of Case a as a component.
6-22
6.17
The input is Ti′(t ) = 12 sin ωt
2π radians
where
ω=
= 0.262 hr −1
24 hours
The Laplace transform of the input is from Table 3.1,
Ti′( s ) =
12ω
s + ω2
2
Multiplying the transfer function by the input transform yields
Ti′( s ) =
(−72 + 36 s )ω
(10 s + 1)(5s + 1)( s 2 + ω 2 )
To invert, either (1) make a partial fraction expansion manually, or (2) use
the Matlab residue function. The first method requires solution of a system
of algebraic equations to obtain the coefficients of the four partial
fractions. The second method requires that the numerator and denominator
be defined as coefficients of descending powers of s prior to calling the
Matlab residue function:
Matlab Commands
>> b = [ 36*0.262 −72*0.262]
b=
9.4320 −18.8640
>> a = conv([10 1], conv([5 1], [1 0 0.262^2]))
b=
50.0000
15.0000
4.4322
>> [r,p,k] = residue(b,a)
r=
6.0865 − 4.9668i
6.0865 + 4.9668i
38.1989
−50.3718
6-23
1.0297
0.0686
p=
−0.0000 − 0.2620i
−0.0000 + 0.2620i
−0.2000
−0.1000
k=
[]
Note: the residue function recomputes all the poles (listed under p). These
are, in reverse order: p1 = 0.1( τ1 = 10) , p2 = 0.2( τ 2 = 5) , and the two
purely imaginary poles corresponding to the sine and cosine functions.
The residues (listed under r) are exactly the coefficients of the
corresponding poles, in other words, the coefficients that would have been
obtained via a manual partial fraction expansion. In this case, we are not
interested in the real poles since both of them yield exponential functions
that go to 0 as t→ ∞.
The complex poles are interpreted as the sine/cosine terms using Eqs. 3-69
and 3-74. From (3-69) we have:
α1= 6.0865, β1 = 4.9668, b = 0, and ω=0.262.
Eq. 3-74 provides the coefficients of the periodic terms:
y (t ) = 2α1e − bt cos ωt + 2β1e −bt sin ωt + ...
Substituting coefficients (because b= 0, the exponential terms = 1)
y (t ) = 2(6.068) cos ωt + 2(4.9668) sin ωt + ...
or
y (t ) = 12.136 cos ωt + 9.9336 sin ωt + ...
The amplitude of the composite output sinusoidal signal, for large values,
of t is given by
A = (12.136) 2 + (9.9336) 2 = 15.7
Thus the amplitude of the output is 15.7° for the specified 12° amplitude
input.
6-24
6.18
a)
Taking the Laplace transform of the dynamic model in (2-7)
[( γVs + (q + q R )]CT′ 1 ( s) = qCTi′ ( s) + q R CT′ (s)
(1)
[(1 − γ)Vs + (q + q R )]CT′ ( s) = (q + q R )CT′ 1 ( s)
(2)
Substituting for CT′ (s ) from (2) into (1),
CT′ 1 ( s )
q[(1 − γ )Vs + (q + q R )]
=
CTi′ ( s ) [γVs + (q + q R )][(1 − γ )Vs + (q + q R )] − q R (q + q R )
 (1 − γ )V 

s + 1
(q + q R ) 

=
 γ (1 − γ )V 2  2 V 

s +  s + 1
q
 q(q + q R ) 
(3)
Substituting for CT′ 1 ( s ) from (3) into (2),
CT′ ( s )
1
==
CTi′ ( s )
 γ (1 − γ )V 2  2 V 

s +  s + 1
q
 q(q + q R ) 
b)
Case (i), γ → 0
 V 

s + 1
q + qR 
CT′ 1 ( s )

==
CTi′ ( s )
V 
 q s + 1
 
CT′ ( s )
1
==
CTi′ ( s )
V 
 q s + 1
 
Case (ii), γ → 1
CT′ ( s )
C ′ ( s)
1
=
= T1
CTi′ ( s ) V 
CTi′ ( s )
 q s + 1
 
6-25
(4)
Case (iii), q R → 0
CT′ ( s )
1
==
,
CTi′ ( s )
 γ (1 − γ )V 2  2 V 

s +  s + 1
q2
q


CT′ 1 ( s )
1
==
CTi′ ( s )
 γV 
 q s + 1
 
Case (iv), q R → ∞
CT′ ( s )
C ′ ( s)
1
=
= T1
CTi′ ( s ) V 
CTi′ ( s )
 q s + 1
 
c)
Case (i), γ → 0
This corresponds to the physical situation with no top tank. Thus the
dynamics for CT are the same as for a single tank, and CT′ 1 ≈ CTi′ for small
qR.
Case (ii), γ → 1
Physical situation with no bottom tank. Thus the dynamics for CT 1 are the
same as for a single tank, and CT = CT 1 at all times.
Case (iii), q R → 0
Physical situation with two separate non-interacting tanks. Thus, top tank
dynamics, CT 1 , are first order, and bottom tank, CT , is second order.
Case (iv), q R → ∞
Physical situation of a single perfectly mixed tank. Thus, CT = CT 1 , and
both exhibit dynamics that are the same as for a single tank.
d)
In Eq.(3),
 (1 − γ )V 

≥0
(
q
+
q
)

R 
Hence the system cannot exhibit an inverse response. From the
denominator of the transfer functions in Eq.(3) and (4),
6-26
1V
ζ=
2q
 γ (1 − γ )V 2 


 q(q + q R ) 
−
1
2
1
 (q + q R )  2
=

 4qγ (1 − γ ) 
Since γ (1 − γ ) ≤ (0.5)(1 − 0.5) for 0 ≤ γ ≤ 1 ,
1
 (q + q R )  2
ζ=
 ≥1
 q 
Hence, the system is overdamped and cannot exhibit overshoot.
e)
Since ζ ≥ 1 , the denominator of transfer function in Eq.(3) and (4) can be
written as (τ1 s + 1)(τ 2 s + 1) where, using Eq. 5-45 and 5-46,
1
 γ (1 − γ )V 2  2


 q( q + q R ) 
τ1 =
1
2
 (q + q R ) 
 (q + q R )

 4qγ (1 − γ )  −  4qγ (1 − γ ) − 1




1
2
1
 γ (1 − γ )V 2  2


 q(q + q R ) 
τ2 =
1
1
 (q + q R )  2  (q + q R )
2
+
−
1
 4qγ (1 − γ ) 
 4qγ (1 − γ ) 




It is given that
CTi′ ( s ) =
[
]
h
h h
1 − e −t w s = − e − t w s
s
s s
Then using Eq. 5-48 and (4)
 τ e− t / τ1 − τ2 e− t / τ2 
cT (t ) = S (t )h 1 − 1

τ1 − τ2


 τ1e − (t −t w ) / τ1 − τ 2 e − (t −tw ) / τ 2 
− S (t − t w )h 1 −

τ1 − τ 2


6-27
And using Eq. 6-15 and (3)
 τ − τ1 −t / τ1 τ a − τ 2 −t / τ2 
CT 1 (t ) = S (t )h 1 + a
e
+
e

τ 2 − τ1
 τ1 − τ 2

 τ − τ1 −(t −tw ) / τ1 τ a − τ 2 −(t −t w ) / τ 2 
e
e
− S (t − t w )h 1 + a
+

τ 2 − τ1
 τ1 − τ 2

where
 (1 − γ )V 
τa = 

 (q + q R ) 
The pulse response can be approximated reasonably well by the impulse
response in the limit as t w → 0 , keeping htw constant.
6.19
Let
VR = volume of each tank
A1 = ρ1Cp1VR
A2 = ρ2Cp2VR
B1 = w1Cp1
B2 = w2Cp2
K = UA
Then energy balances over the six tanks give
dT8
= B2 (T6 − T8 ) + K (T3 − T8 )
dt
dT
A2 6 = B2 (T4 − T6 ) + K (T5 − T6 )
dt
dT4
A2
= B2 (T2 − T4 ) + K (T7 − T4 )
dt
A2
(1)
(2)
(3)
A1
dT7
= B1 (T5 − T7 ) + K (T4 − T7 )
dt
(4)
A1
dT5
= B1 (T3 − T5 ) + K (T6 − T5 )
dt
(5)
6-28
A1
dT3
= B1 (T1 − T3 ) + K (T8 − T3 )
dt
(6)
Define vectors
T ′( s ) = [T8′( s ), T7′ ( s ), T6′ ( s ), T5′( s ), T4′ ( s ), T3′( s )]
T
T ′ ( s )
*
T (s) =  2 
T1′( s ) 
Using deviation variables, and taking the Laplace transform of Eqs.1 to 6,
we obtain an equation set that can be represented in matrix notation as
s I T ′( s ) = A T ′( s ) + B T ( s )
*
(7)
where I is the 6 × 6 identity matrix
 − K − B2
 A
2


0



0

A=

0



0


K

 A1

0 0 0 0
B=
0 0 0 0

B2
A2
0
0
0
− K − B2
A2
K
A1
B1
A1
K
A2
− K − B1
A1
K
A1
B2
A2
K
A2
0
0
− K − B2
A2
0
0
0
0
− K − B1
A1
0
0
0

0

B1 
A1 
B2
A2
0
From Eq. 7,
T ′( s ) = ( s I − A) −1 B T ( s )
*
6-29
0




0



0

B1 

A1 

0


− K − B1 
A1 
K
A2
Then
T8′( s ) 
T ′ ( s ) =
 7 
1 0 0 0 0 0
*
−1
0 1 0 0 0 0 ( s I − A) B T ( s )


6.20
The dynamic model for the process is given by Eqs. 2-45 and 2-46,
which can be written as
dh
1
=
( wi − w)
dt ρA
(1)
w
dT
Q
= i (Ti − T ) +
dt ρAh
ρAhC
(2)
where h is the liquid-level
A is the constant cross-sectional area
System outputs: h , T
System inputs : w, Q
Hence assume that wi and Ti are constant. In Eq. 2, note that the nonlinear
 dT 
term  h
 can be linearized as
 dt 
or
h
dT ′ dT
+
h′
dt
dt
h
dT ′
since
dt
dT
=0
dt
Then the linearized deviation variable form of (1) and (2) is
dh′
1
=−
w′
dt
ρA
dT ′ − wi
1
=
T′+
Q′
dt
ρAh
ρAh C
6-30
Taking Laplace transforms and rearranging,
H ′( s )
=0 ,
Q ′( s )
H ′( s ) K 1
=
W ′( s )
s
,
where K 1 = −
1
;
ρA
and K 2 =
Unit-step change in Q: h(t ) = h
K2
T ′( s )
=
Q ′( s ) τ 2 s + 1
T ′( s )
=0 ,
W ′( s )
1
wi C
,
, τ2 =
ρAh
wi
T (t ) = T + K 2 (1 − e − t / τ 2 )
Unit step change in w: h(t ) = h + K 1t , T (t ) = T
6.21
Additional assumptions:
(i) The density, ρ, and the specific heat, C, of the process liquid are
constant.
(ii) The temperature of steam, Ts, is uniform over the entire heat transfer
area.
(iii) The feed temperature TF is constant (not needed in the solution).
Mass balance for the tank is
dV
= qF − q
dt
(1)
Energy balance for the tank is
ρC
d [V (T − Tref )]
dt
= q F ρC (TF − Tref ) − qρC (T − Tref ) + UA(Ts − T ) (2)
where Tref is a constant reference temperature
A is the heat transfer area
dV
from Eq. 1. Also, replace
dt
V by AT h (where AT is the tank area) and replace A by pT h
(where pT is the perimeter of the tank). Then,
Eq. 2 is simplified by substituting for
6-31
AT
dh
= qF − q
dt
ρCAT h
(3)
dT
= qF ρC (TF − T ) + UpT h(Ts − T )
dt
(4)
Then, Eqs. 3 and 4 constitute the dynamic model for the system.
a)
Making Taylor series expansion of nonlinear terms in (4) and introducing
deviation variables, Eqs. 3 and 4 become:
AT
dh ′
= q ′F − q ′
dt
ρCAT h
(5)
dT ′
= ρC (TF − T )qF′ − (ρCqF + UpT h )T ′
dt
+ UpT hTs′ + UpT (Ts − T )h′
(6)
Taking Laplace transforms,
H ′( s ) =
1
1
QF′ ( s ) −
Q′( s )
AT s
AT s
 ρCAT h

 ρCqF + UpT h
(7)


 ρC (TF − T ) 
 s + 1 T ′( s ) = 
 QF′ ( s )

 ρCqF + UpT h 



 UpT (Ts − T ) 
UpT h
+
 Ts′( s ) + 
 H ′( s )
ρ
Cq
+
Up
h
ρ
Cq
+
Up
h

F
T 

F
T 
(8)
Substituting for H ′( s ) from (7) into (8) and rearranging gives

ρCAT h
 ρCqF + UpT h
[ AT s ] 


 ρC (TF − T )

AT s  QF′ ( s )
 s + 1 T ′( s ) = 

 ρCqF + UpT h


 UpT hAT s 
 UpT (Ts − T ) 
+
 Ts′( s ) + 
 [QF′ ( s ) − Q′( s )]
ρ
Cq
+
Up
h
ρ
Cq
+
Up
h

F
T 

F
T 
Let τ =
ρCAT h
ρCqF + UpT h
Then from Eq. 7
6-32
(9)
H ′( s )
1
=
QF′ ( s ) AT s
H ′( s )
1
=−
Q′( s )
AT s
,
,
H ′( s )
=0
Ts′ ( s )
And from Eq. 9
 UpT (Ts − T )   ρC (TF − T ) AT

 
T ′( s )  ρCqF + UpT h   UpT (Ts − T )
=
QF′ ( s )
( AT s )( τs + 1)


 s + 1


 UpT (Ts − T ) 
−
ρCqF + UpT h 
T ′( s )

=
Q′( s )
( AT s )( τs + 1)


UpT h


T ′( s )  ρCqF + UpT h 
=
Ts′( s )
τs + 1
Note:
τ2 =
ρC (TF − T ) AT
UpT (Ts − T )
is the time constant in the numerator.
Because TF − T < 0 (heating) and Ts − T > 0 , τ2 is negative.
We can show this property by using Eq. 2 at steady state:
ρCqF (TF − T ) = −UpT h (Ts − T )
or ρC (TF − T ) =
−UpT h (Ts − T )
qF
Substituting
τ2 = −
hAT
qF
Let V = hAT so that τ2 = −
For
V
= − (initial residence time of tank)
qF
T ′( s )
T ′( s )
and
the “gain” in each transfer function is
QF′ ( s )
Q′( s )
6-33
 Up (T − T ) 
T
s

K =
 AT ( ρCqF + UpT h ) 
and must have the units temp/volume .
(The integrator s has units of t-1).
To simplify the transfer function gain we can substitute
UpT (Ts − T ) = −
ρCqF (TF − T )
h
from the steady-state relation. Then
K=
−ρCqFT (TF − T )
hAT ( ρCqF + UpT h )
or K =
T − TF
 Up h 
V 1 + T 
 ρCqF 
and we see that the gain is positive since T − TF > 0 .
Further, it has dimensions of temp/volume.
(The ratio
b)
UpT h
is dimensionless).
ρCqF
h − qF transfer function is an integrator with a positive gain. Liquid level
accumulates any changes in qF , increasing for positive changes and viceversa.
h − q transfer function is an integrator with a negative gain. h accumulates
changes in q, in opposite direction, decreasing as q increases and vice
versa.
h − Ts transfer function is zero. Liquid level is independent of Ts , and of
the steam pressure Ps .
T − q transfer function is second-order due to the interaction with liquid
level; it is the product of an integrator and a first-order process.
6-34
T − qF transfer function is second-order due to the interaction with liquid
level and has numerator dynamics since qF affects T directly as well if
TF ≠ T .
T − Ts transfer function is simple first-order because there is no interaction
with liquid level.
c)
h − qF : h increases continuously at a constant rate.
h − q : h decreases continuously at a constant rate.
h − Ts : h stays constant.
T − qF : for TF < T , T decreases initially (inverse response) and then
increases. After long times, T increases like a ramp function.
T − q : T decreases, eventually at a constant rate.
T − Ts : T increases with a first-order response and attains a new steady
state.
6.22
a)
The two-tank process is described by the following equations in deviation
variables:
dh1'
1  ' 1 '

=
w1 − (h1 − h2' 

dt ρA1 
R

dh2'
1
=
dt ρA2
1 '
' 
 R (h1 − h2 
(1)
(2)
Laplace transforming
ρA1 RsH1' ( s ) = RWi ' ( s ) − H1' ( s ) + H 2' ( s )
(3)
ρA2 RsH 2' ( s ) = H1' ( s ) − H 2' ( s )
(4)
6-35
From (4)
(ρA2 Rs + 1) H 2' ( s ) = H1' ( s )
(5)
H 2' ( s )
1
1
=
=
'
H1 ( s ) ρA2 Rs + 1 τ2 s + 1
(6)
or
where τ2 = ρA2 R
Returning to (3)
(ρA1 Rs + 1) H1' ( s ) − H 2' ( s ) = RWi ' ( s )
(7)
Substituting (6) with τ1 = ρA1 R

1  '
'
(τ1s + 1) −
 H1 ( s ) = RWi ( s )
τ
s
+
1

2

(8)
or
(τ1τ2 ) s 2 + (τ1τ2 ) s  H1' ( s ) = R (τ2 s + 1)Wi ' ( s )
H1' ( s )
R(τ2 s + 1)
=
'
W1 ( s ) s [ τ1τ2 s + (τ1 + τ2 )]
(9)
(10)
Dividing numerator and denominator by (τ1 + τ2 ) to put into standard form
H1' ( s ) [ R /(τ1 + τ2 )](τ2 s + 1)
=
W1' ( s )
 ττ

s  1 2 s + 1
 τ1 + τ2

(11)
Note that
K=
R
R
1
1
=
=
=
τ1 + τ2 ρA1 R + ρA2 R ρ( A1 + A2 ) ρA
since A = A1 + A2
Also, let
6-36
(12)
τs =
τ1τ2
ρ2 R 2 A1 A2
ρRA1 A2
=
=
τ1 + τ2 ρR( A1 + A2 )
A
(13)
so that
H1' ( s ) K (τ2 s + 1)
=
Wi ' ( s ) s (τ3 s + 1)
(14)
and
H 2' ( s ) H 2' ( s ) H1' ( s )
1
K (τ2 s + 1)
= '
=
'
'
Wi ( s ) H1 ( s ) Wi ( s ) (τ2 s + 1) s (τ3 s + 1)
K
=
s (τ3 s + 1)
(15)
Transfer functions (6), (14) and (15) define the operation of the two-tank
process.
The single-tank process is described by the following equation in
deviation variables:
dh'
1 '
=
wi
dt ρA
(16)
Note that ω , which is constant, subtracts out.
Laplace transforming and rearranging:
H ' ( s ) 1/ ρA
=
Wi ' ( s )
s
(17)
Again
K=
1
ρA
H ' ( s) K
=
Wi ' ( s ) s
(18)
which is the expected integral relationship with no zero.
6-37
b)
For A1 = A2 = A / 2
τ2 = ρAR / 2 

τ3 = ρAR / 4 
(19)
Thus τ2 = 2τ3
We have two sets of transfer functions:
One-Tank Process
Two-Tank Process
H ' ( s) K
=
Wi ' ( s ) s
H i' ( s ) K (2τ3 s + 1)
=
Wi ' ( s )
s (τ3 s + 1)
H 2' ( s )
K
=
'
Wi ( s ) s (τ3 s + 1)
Remarks:
-
The gain ( K = 1/ ρA) is the same for all TF’s.
Also, each TF contains an integrating element.
However, the two-tank TF’s contain a pole (τ3 s + 1) that will “filter
out” changes in level caused by changing wi(t).
On the other hand, for this special case we see that the zero in the first
tank transfer function ( H i' ( s ) / Wi ' ( s )) is larger than the pole
2 τ3 > τ 3
and we should make sure that amplification of changes in h1(t) caused
by the zero do not more than cancel the beneficial filtering of the pole
so as to cause the first compartment to overflow easily.
Now look at more general situations of the two-tank case:
H1' ( s )
K (ρA2 Rs + 1)
K (τ2 s + 1)
=
=
'
Wi ( s )
 ρRA1 A2
 s (τ3 s + 1)
s
s + 1
 A

'
H 2 (s)
K
=
'
Wi ( s ) s (τ3 s + 1)
For either A1 → 0 or A2 → 0 ,
τ3 =
ρRA1 A2
→0
A
6-38
(20)
(21)
Thus the beneficial effect of the pole is lost as the process tends to
“look” more like the first-order process.
c)
The optimum filtering can be found by maximizing τ3 with respect
to A1 (or A2)
ρRA1 A2 ρRA1 ( A − A1 )
=
A
A
∂τ
ρR
Find max τ3 : 3 =
[ ( A − A1 ) + A1 (−1)]
∂A1
A
τ3 =
Set to 0:
A − A1 − A1 = 0
2 A1 = A
A1 = A / 2
Thus the maximum filtering action is obtained when A1 = A2 = A / 2.
The ratio of τ2 / τ3 determines the “amplification effect” of the zero on
h1 (t ).
τ2
ρA2 R
A
=
=
ρ
RA
A
A1
τ3
1 2
A
τ
As A1 goes to 0, 2 → ∞
τ3
Therefore the influence of changes in wi (t ) on h1 (t ) will be very large,
leading to the possibility of overflow in the first tank.
Summing up:
The process designer would like to have A1 = A2 = A / 2 in order to obtain
a
the maximum filtering of h1 (t ) and h2 (t ). However, the process response
should be checked for typical changes in wi (t ) to make sure that h1 does
not overflow. If it does, the area A1 needs to be increased until that is not
problem.
Note that τ2 = τ3 when A1 = A , thus someone must make a careful study
(simulations) before designing the partitioned tank. Otherwise, leave well
enough alone and use the non-partitioned tank.
6-39
6.23
The process transfer function is
Y ( s)
K
= G( s) =
2
U ( s)
(0.1s + 1) (4 s 2 + 2 s + 1)
where K = K1K2
We note that the quadratic term describes an underdamped 2nd-order
system since
τ2 = 4
→
τ=2
a)
2ζτ = 2
→
ζ = 0 .5
For the second-order process element with τ2 = 2 and this degree of
underdamping (ζ = 0.5) , the small time constant, critically damped 2ndorder process element (τ1 = 0.1 ) will have little effect.
In fact, since 0.1 << τ2 (= 2) we can approximate the critically damped
element as e −2τ1 so that
G (s) ≈
b)
Ke −0.2 s
4s 2 + 2s + 1
From Fig. 5.11 for ζ = 0.5 , OS ≈ 0.15 or from Eq. 5-53
 − πζ
Overshoot = exp 
 1− ζ2


 = 0.163


Hence ymax = 0.163 KM + KM = 0.163 (1) (3) + 3 = 3.5
c)
From Fig. 5.4, ymax occurs at t/τ = 3K or tmax = 6.8 for underdamped 2ndorder process with ζ = 0.5 .
Adding in effect of time delay t ′ = 6.8 + 0.2 = 7.0
d)
By using Simulink-MATLAB
6-40
τ1 = 0.1
3.5
3
2.5
Output
2
1.5
1
0.5
0
Exact model
Approximate model
-0.5
0
5
10
15
Time
20
25
30
Fig S6.23a. Step response for exact and approximate model ; τ1 = 0.1
τ1 = 1
3.5
3
2.5
Output
2
1.5
1
0.5
0
Exact model
Approximate model
-0.5
0
5
10
15
20
25
30
Time
Fig S6.23b. Step response for exact and approximate model ; τ1 = 1
6-41
τ1 = 5
3.5
3
2.5
Output
2
1.5
1
0.5
0
Exact model
Approximate model
-0.5
0
5
10
15
20
25
30
Time
Fig S6.23c. Step response for exact and approximate model ; τ1 = 5
As noted in plots above, the smaller τ1 is, the better the quality of the
approximation. For large values of τ1 (on the order of the underdamped
element's time scale), the approximate model fails.
6.24
0
-0.2
-0.4
Output
-0.6
-0.8
-1
-1.2
-1.4
0
50
100
150
200
250
300
350
400
Time
Figure S6.24. Unit step response in blood pressure.
6-42
The Simulink-MATLAB block diagram is shown below
-1
40s+1
Step
Transfer Fcn
Transport
Delay1= 30 s
Scope
-0.4
40s+1
Step1
Transfer Fcn1
Transport
Delay = 75 s
It appears to respond approx. as a first-order or overdamped second-order
process with time delay.
6-43
1234567898
7.1
In the absence of more accurate data, use a first-order transfer function as
T '( s ) Ke −θs
=
Qi '( s ) τs + 1
o
T (∞) − T (0) (124.7 − 120)
F
=
= 0.118
∆qi
540 − 500
gal/min
θ = 3:09 am – 3:05 am = 4 min
K=
Assuming that the operator logs a 99% complete system response as “no
change after 3:34 am”, 5 time constants elapse between 3:09 and 3:34 am.
5τ = 3:34 min − 3:09 min = 25 min
τ = 25/5 min = 5 min
Therefore,
T '( s ) 0.188e−4 s
=
Qi '( s )
5s + 1
To obtain a better estimate of the transfer function, the operator should log
more data between the first change in T and the new steady state.
7.2
h(5.0) − h(0) (6.52 − 5.50)
min
=
= 0.336 2
∆qi
30.4 × 0.1
ft
Output at 63.2% of the total change
Process gain, K =
a)
= 5.50 + 0.632(6.52-5.50) = 6.145 ft
Interpolating between h = 6.07 ft
and
h = 6.18 ft
Solution Manual for Process Dynamics and Control, 2nd edition,
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp.
7-1
τ = 0.6 +
(0.8 − 0.6)
(6.145 − 6.07) min = 0.74 min
(6.18 − 6.07)
b)
dh
h(0.2) − h(0) 5.75 − 5.50 ft
ft
≈
=
= 1.25
dt t = 0
0.2 − 0
0.2
min
min
Using Eq. 7-15,
τ=
c)
KM
0.347 × (30.4 × 0.1)
=
= 0.84 min
1.25
 dh 


 dt t =0 
 h(t i ) − h(0) 
The slope of the linear fit between ti and z i ≡ ln 1 −
 gives an
 h ( ∞ ) − h ( 0) 
approximation of (-1/τ) according to Eq. 7-13.
Using h(∞) = h(5.0) = 6 .52, the values of zi are
ti
0.0
0.2
0.4
0.6
0.8
1.0
1.2
zi
0.00
-0.28
-0.55
-0.82
-1.10
-1.37
-1.63
ti
1.4
1.6
1.8
2.0
3.0
4.0
5.0
zi
-1.92
-2.14
-2.43
-2.68
-3.93
-4.62
-∞
Then the slope of the best-fit line, using Eq. 7-6 is
 1  13Stz − St S z
slope =  −  =
2
 τ  13Stt − ( St )
(1)
where the datum at ti = 5.0 has been ignored.
Using definitions,
St = 18.0
S z = −23.5
Stt = 40.4
Stz = −51.1
Substituting in (1),
 1
 −  = −1.213
 τ
τ = 0.82 min
7-2
d)
6.8
6.6
6.4
6.2
6
Experimental data
Model a)
Model b)
Model c)
5.8
5.6
5.4
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
Figure S7.2. Comparison between models a), b) and c) for step response.
7.3
a)
T1′( s )
K1
=
Q′( s ) τ1s + 1
T2′( s )
K2
=
T1′( s ) τ2 s + 1
T2′( s )
K1 K 2
K1 K 2 e −τ2 s
=
≈
Q′( s ) (τ1s + 1)(τ2 s + 1)
(τ1s + 1)
(1)
where the approximation follows from Eq. 6-58 and the fact that τ1>τ2 as
revealed by an inspection of the data.
T1 (50) − T1 (0) 18.0 − 10.0
=
= 2.667
∆q
85 − 82
T (50) − T2 (0) 26.0 − 20.0
K2 = 2
=
= 0.75
T1 (50) − T1 (0) 18.0 − 10.0
K1 =
Let z1, z2 be the natural log of the fraction incomplete response for T1,T2,
respectively. Then,
7-3
 T (50) − T1 (t ) 
18 − T1 (t ) 
= ln 
z1 (t ) = ln  1


8

 T1 (50) − T1 (0) 
 T (50) − T2 (t ) 
 26 − T2 (t ) 
z2 (t ) = ln  2
= ln 


6

 T2 (50) − T2 (0) 
A graph of z1 and z2 versus t is shown below. The slope of z1 versus t line
is –0.333 ; hence (1/-τ1)=-0.333 and τ1=3.0
From the best-fit line for z2 versus t, the projection intersects z2 = 0 at
t≈1.15. Hence τ2 =1.15.
T1 ' ( s ) 2.667
=
Q ' ( s ) 3s + 1
T2 ' ( s )
0.75
=
T1 ' ( s ) 1.15s + 1
(2)
(3)
0.0
-1.0
0
5
10
15
20
-2.0
z 1,z 2
-3.0
-4.0
-5.0
-6.0
-7.0
-8.0
time,t
Figure S7.3a. z1 and z2 versus t
By means of Simulink-MATLAB, the following simulations are obtained
28
26
24
22
20
T1 , T2
b)
18
16
T1
T2
T1 (experimental)
T2 (experimental)
14
12
10
0
2
4
6
8
10
12
14
16
18
20
22
time
Figure S7.3b. Comparison of experimental data and models for step change
7-4
7.4
Y (s) = G( s) X ( s) =
2
1.5
×
(5s + 1)(3s + 1)( s + 1) s
Taking the inverse Laplace transform
y (t ) = -75/8*exp(-1/5*t)+27/4*exp(-1/3*t)-3/8*exp(-t)+3
a)
Fraction incomplete response
 y (t ) 
z (t ) = ln 1 −
3 

0.0
-1.0 0
10
20
30
40
50
-2.0
z(t)
-3.0
-4.0
-5.0
-6.0
-7.0
z(t) = -0.1791 t + 0.5734
-8.0
-9.0
time,t
Figure S7.4a. Fraction incomplete response; linear regression
From the graph, slope = -0.179 and intercept ≈ 3.2
Hence,
-1/τ = -0.179 and τ = 5.6
θ = 3.2
G (s) =
b)
2e −3.2 s
5.6 s + 1
In order to use Smith’s method, find t20 and t60
y(t20)= 0.2 × 3 =0.6
y(t60)= 0.6 × 3 =1.8
Using either Eq. 1 or the plot of this equation, t20 = 4.2 , t60 = 9.0
Using Fig. 7.7 for t20/ t60 = 0.47
ζ= 0.65 , t60/τ= 1.75, and τ = 5.14
7-5
(1)
G (s) ≈
2
26.4 s + 6.68s + 1
2
The models are compared in the following graph:
2.5
2
1.5
y(t)
Third-order model
First order model
Second order model
1
0.5
0
0
5
10
15
20
25
30
35
40
time,t
Figure S7.4b. Comparison of three models for step input
7.5
The integrator plus time delay model is
K
G(s) e −θs
s
In the time domain,
y(t) = 0
t<0
y(t)= K (t-θ)
t≥0
Thus a straight line tangent to the point of inflection will approximate the
step response. Two parameters must be found: K and θ (See Fig. S7.5 a)
1.- The process gain K is found by calculating the slope of the straight
line.
1
K=
= 0.074
13.5
2.- The time delay is evaluated from the intersection of the straight line
and the time axis (where y = 0).
θ = 1.5
7-6
Therefore the model is G(s) =
0.074 −1.5 s
e
s
y(t)
Slope = KM
θ
Figure S7.5a. Integrator plus time delay model; parameter evaluation
From Fig. E7.5, we can read these values (approximate):
Time
0
2
4
5
7
8
9
11
14
16.5
30
Data
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Model
-0.111
0.037
0.185
0.259
0.407
0.481
0.555
0.703
0.925
1.184
2.109
Table.- Output values from Fig. E7.5 and predicted values by model
A graphical comparison is shown in Fig. S7.5 b
1
0.9
0.8
Output
0.7
0.6
0.5
0.4
Experimental data
Integrator plus time delay model
0.3
0.2
0.1
0
0
5
10
15
20
25
30
Time
Figure S7.5b. Comparison between experimental data and integrator plus time
delay model.
7-7
7.6
a)
b)
Drawing a tangent at the inflection point which is roughly at t ≈5, the
intersection with y(t)=0 line is at t ≈1 and with the y(t)=1 line at t ≈14.
Hence θ =1 , τ = 14−1=13
e−s
G1 ( s ) ≈
13s + 1
Smith’s method
From the graph, t20 = 3.9 , t60 = 9.6 ; using Fig 7.7 for t20/ t60 = 0.41
ζ = 1.0 ,
G (s) ≈
t60/τ= 2.0 ,
hence τ = 4.8 and τ1 = τ2 = τ = 4.8
1
(4.8s + 1) 2
Nonlinear regression
From Figure E7.5, we can read these values (approximated):
Time
0.0
2.0
4.0
5.0
7.0
8.0
9.0
11.0
14.0
17.5
30.0
Output
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Table.- Output values from Figure E7.5
In accounting for Eq. 5-48, the time constants were selected to minimize
the sum of the squares of the errors between data and model predictions.
Use Excel Solver for this Optimization problem:
τ1 =6.76 and
G (s) ≈
τ2 = 6.95
1
(6.95s + 1)(6.76s + 1))
The models are compared in the following graph:
7-8
1
0.9
0.8
0.7
Output
0.6
0.5
0.4
0.3
0.2
Non linear regression model
First-order plus time delay model
Second order model (Smith's method)
0.1
0
0
5
10
15
20
25
time
30
35
40
45
50
Figure S7.6. Comparison of three models for unit step input
7.7
a)
From the graph, time delay θ = 4.0 min
Using Smith’s method,
from the graph, t20 + θ ≈ 5.6 , t60 + θ ≈ 9.1
t20 = 1.6 , t60 = 5.1 , t20 / t60 = 1.6 / 5.1 = 0.314
From Fig.7.7 , ζ = 1.63 , t60 / τ = 3.10 , τ = 1.645
Using Eqs. 5-45, 5-46, τ1 = 4.81 , τ2 = 0.56
b)
Overall transfer function
10e −4 s
G (s) =
, τ1 > τ2
(τ1s + 1)(τ2 s + 1)
Assuming plug-flow in the pipe with constant-velocity,
7-9
G pipe ( s ) = e
−θ p s
, θp =
3
1
×
= 0.1min
0.5 60
Assuming that the thermocouple has unit gain and no time delay
GTC ( s ) =
1
(τ2 s + 1)
since τ2 << τ1
Then
10e −3 s
GHE ( s ) =
, so that
(τ1s + 1)
 10e −3 s  −0.1s  1 
G ( s ) = GHE ( s )G pipe ( s )GTC ( s ) = 
 (e ) 

 τ1s + 1 
 τ2 s + 1 
7.8
a)
To find the form of the process response, we can see that
Y (s) =
K
K
M
K M
U ( s) =
=
s (τs + 1)
s (τs + 1) s (τs + 1) s 2
Hence the response of this system is similar to a first-order system with a
ramp input: the ramp input yields a ramp output that will ultimately cause
some process component to saturate.
b)
By applying partial fraction expansion technique, the domain response for
this system is
A B
C
+ 2+
hence y(t) = -KMτ + KMt − KMτe-t/τ
s s
τs + 1
In order to evaluate the parameters K and τ, important properties of the
above expression are noted:
Y(s) =
1.- For large values of time (t>>τ) ,
2.- For t = 0, y′(0) = −KMτ
y(t) ≈ y′(t ) = KM (t-τ)
These equations imply that after an initial transient period, the ramp input
yields a ramp output with slope equal to KM. That way, the gain K is
7-10
obtained. Moreover, the time constant τ is obtained from the intercept in
Fig. S7.8
y(t)
Slope = KM
−ΚΜτ
Figure S7.8. Time domain response and parameter evaluation
7.9
For underdamped responses,

  1 − ζ2
y (t ) = KM 1 − e − ζt / τ cos
  τ

a)

 1 − ζ2
ζ
t +
sin 

 τ
1 − ζ2


 
t  

 
At the response peaks,

  1 − ζ2
dy
ζ
= KM  e −ζ t / τ cos 
dt
  τ
 τ

 1− ζ2
ζ
t+
sin 

 τ
1 − ζ2


 1− ζ2
 1− ζ2
−e −ζt / τ  −
sin 
 τ
τ


 ζ
 1 − ζ2
t  + cos 
 τ
 τ



t 


  
t   = 0
 
 
Since KM ≠ 0 and e − ζt / τ ≠ 0
2
 ζ ζ   1− ζ
0 =  −  cos
 τ τ   τ
  ζ2
1 − ζ2
t + 
+
  τ 1 − ζ2
τ
 
7-11
  1 − ζ2
 sin 
  τ
 

t


(5-51)
 1− ζ2 
0 = sin
t  = sin nπ ,
 τ



where n is the number of peak.
Time to the first peak,
b)
tp =
t= n
πτ
1 − ζ2
πτ
1 − ζ2
Graphical approach:
Process gain,
K=
wD (∞) − wD (0) 9890 − 9650
lb
=
= 80
hr
∆Ps
95 − 92
psig
Overshoot =
a 9970 − 9890
=
= 0.333
b 9890 − 9650
From Fig. 5.11,
ζ ≈ 0.33
tp can be calculated by interpolating Fig. 5.8
For ζ ≈ 0.33 , tp ≈ 3.25 τ
Since tp is known to be 1.75 hr , τ = 0.54
G (s) =
K
80
=
2
τ s + 2ζτs + 1 0.29s + 0.36 s + 1
2 2
Analytical approach
The gain K doesn’t change: K = 80
lb
hr
psig
To obtain the ζ and τ values, Eqs. 5-52 and 5-53 are used:
Overshoot =
a 9970 − 9890
=
= 0.333 = exp(-ζπ/(1-ζ2)1/2)
b 9890 − 9650
Resolving, ζ = 0.33
7-12
tp =
πτ
1− ζ2
G (s) =
c)
= 1.754
hence
τ = 0.527 hr
K
80
=
2
τ s + 2ζτs + 1 0.278s + 0.35s + 1
2 2
Graphical approach
From Fig. 5.8, ts/τ = 13 so ts = 2 hr (very crude estimation)
Analytical approach
From settling time definition,
y = ± 5% KM
so
9395.5 < y < 10384.5
(KM ± 5% KM) = KM[ 1-e(-0.633)[cos(1.793ts)+0.353sin(1.793ts)]]
1 ± 0.05 = 1 – e(0.633 ts) cos(1.793 ts) + 0.353e (-0.633 ts) sin(1.7973 ts)
Solve by trial and error……………………
ts ≈ 6.9 hrs
7.10
a)
T '( s )
K
= 2 2
W '( s ) τ s + 2ζτ + 1
K=
o
T (∞) − T (0) 156 − 140
C
=
= 0.2
∆w
80
Kg/min
From Eqs. 5-53 and 5-55,
a 161.5 − 156
=
= 0.344 = exp(-ζπ (1-ζ2)1/2
b
156 − 140
By either solving the previous equation or from Figure 5.11, ζ= 0.322
(dimensionless)
Overshoot =
7-13
There are two alternatives to find the time constant τ :
1.- From the time of the first peak, tp ≈ 33 min.
One could find an expression for tp by differentiating Eq. 5-51 and
solving for t at the first zero. However, a method that should work
(within required engineering accuracy) is to interpolate a value of
ζ=0.35 in Figure 5.8 and note that tp/τ ≈ 3
Hence τ ≈
33
≈ 9.5 − 10 min
3 .5
2.- From the plot of the output,
Period = P =
2πτ
1− ζ2
= 67 min and hence τ =10 min
Therefore the transfer function is
G (s) =
After an initial period of oscillation, the ramp input yields a ramp output
with slope equal to KB. The MATLAB simulation is shown below:
160
158
156
154
152
Output
b)
T ' (s)
0.2
=
2
W ' ( s ) 100 s + 6.44s + 1
150
148
146
144
142
140
0
10
20
30
40
50
time
60
70
80
Figure S7.10. Process output for a ramp input
7-14
90
100
We know the response will come from product of G(s) and Xramp = B/s2
KB
Then Y ( s ) = 2 2 2
s (τ s + 2ζτs + 1)
From the ramp response of a first-order system we know that the response
will asymptotically approach a straight line with slope = KB. Need to find
the intercept. By using partial fraction expansion:
Y (s) =
α s + α4
KB
α α
= 1 + 22 + 2 2 3
s (τ s + 2ζτs + 1) s s
τ s + 2ζτs + 1
2
2 2
Again by analogy to the first-order system, we need to find only α1 and
α2. Multiply both sides by s2 and let s→ 0, α2 = KB (as expected)
Can’t use Heaviside for α1, so equate coefficients
KB = α1s (τ2 s 2 + 2ζτs + 1) + α 2 (τ2 s 2 + 2ζτs + 1) + α 3 s 3 + α 4 s 2
We can get an expression for α1 in terms of α2 by looking at terms
containing s.
s: 0 = α1+α22ζτ
→ α1 = -KB2ζτ
and we see that the intercept with the time axis is at t = 2ζτ. Finally,
presuming that there must be some oscillatory behavior in the response,
we sketch the probable response (See Fig. S7.10)
7.11
a)
Replacing τ by 5, and K by 6 in Eq. 7-34
y (k ) = e−∆t / 5 y (k − 1) + [1 − e−∆t / 5 ]6u (k − 1)
b)
Replacing τ by 5, and K by 6 in Eq. 7-32
y (k ) = (1 −
∆t
∆t
) y (k − 1) + 6u (k − 1)
5
5
In the integrated results tabulated below, the values for ∆t = 0.1 are shown
only at integer values of t, for comparison.
7-15
t
0
1
2
3
4
5
6
7
8
9
10
y(k)
(exact)
3
2.456
5.274
6.493
6.404
5.243
4.293
3.514
2.877
2.356
1.929
y(k)
(1t=1)
3
2.400
5.520
6.816
6.653
5.322
4.258
3.408
2.725
2.180
1.744
y(k)
(1t=0.1)
3
2.451
5.296
6.522
6.427
5.251
4.290
3.505
2.864
2.340
1.912
Table S7.11. Integrated results for the first order differential equation
Thus ∆t = 0.1 does improve the finite difference model bringing it closer
to the exact model.
7.12
To find a1′ and b1 , use the given first order model to minimize
10
J = ∑ ( y (k ) −a1′ y (k − 1) − b1 x(k − 1)) 2
n =1
10
∂J
= ∑ 2( y (k ) −a1′ y (k − 1) − b1 x(k − 1))(− y (k − 1) = 0
∂a1′ n =1
10
∂J
= ∑ 2( y (k ) −a1′ y (k − 1) − b1 x(k − 1))(− x(k − 1)) = 0
∂b1 n =1
Solving simultaneously for a1′ and b1 gives
10
a1′ =
10
∑ y (k )y (k − 1) − b1 ∑ y (k − 1)x(k − 1)
n =1
n =1
10
∑ y (k − 1)
2
n =1
10
b1 =
10
10
10
∑ x(k − 1) y(k )∑ y(k − 1)2 − ∑ y (k − 1)x(k − 1)∑ y(k − 1) y(k )
n =1
n =1
n =1
n =1


x(k − 1) ∑ y (k − 1) −  ∑ y (k − 1)x(k − 1) 
∑
n =1
n =1
 n =1

10
10
10
2
2
7-16
2
Using the given data,
10
10
∑ x(k − 1) y(k ) = 35.212 ,
∑ y(k − 1) y(k ) = 188.749
n =1
10
n =1
10
∑ x(k − 1) 2 = 14
∑ y(k − 1)
,
n =1
2
= 198.112
n =1
10
∑ y (k − 1) x(k − 1) = 24.409
n =1
Substituting into expressions for a1′ and b1 gives
a1′ = 0.8187
,
b1 = 1.0876
Fitted model is
y (k + 1) = 0.8187 y (k ) + 1.0876 x(k )
or
y (k ) = 0.8187 y (k − 1) + 1.0876 x(k − 1)
(1)
Let the first-order continuous transfer function be
Y ( s)
K
=
X ( s ) τs + 1
From Eq. 7-34, the discrete model should be
y (k ) = e −∆t / τ y (k − 1) + [1 − e −∆t / τ ]Kx(k − 1)
Comparing Eqs. 1 and 2, for ∆t=1, gives
τ=5
and K = 6
Hence the continuous transfer function is 6/(5s+1)
7-17
(2)
8
actual data
fitted model
7
6
y(t)
5
4
3
2
0
1
2
3
4
5
6
7
8
9
10
time,t
Figure S7.12. Response of the fitted model and the actual data
7.13
To fit a first-order discrete model
y (k ) = a1′ y (k − 1) + b1 x(k − 1)
Using the expressions for a1′ and b1 from the solutions to Exercise 7.12,
with the data in Table E7.12 gives
a1′ = 0.918
, b1 = 0.133
Using the graphical (tangent) method of Fig.7.5 .
K = 1 , θ = 0.68 , and τ = 6.8
The response to unit step change for the first-order model given by
e −0.68 s
6.8s + 1
is
y (t ) = 1 − e −( t −0.68) / 6.8
7-18
y(t)
1
0,9
0,8
0,7
0,6
0,5
0,4
0,3
0,2
0,1
0
actual data
fitted model
graphical method
0
2
4 time,t 6
8
10
Figure S7.13- Response of the fitted model, actual data and graphical method
7-19
1234567898
8.1
a)
For step response,
M
s
 τD s +1 
 τ s +1 
Ya′ ( s ) = K c  D

U ′( s ) = K c M 
 s (ατ D s + 1) 
 ατ D s + 1
input is u ′(t ) = M
Ya′ (s ) =
U ′( s ) =
,
K c Mτ D
KcM
+
ατ D s + 1 s (ατ D s + 1)
Taking inverse Laplace transform
y ′a (t ) =
K c M −t /( ατ D )
e
+ K c M (1 − e −t /( ατ D ) )
α
As α →0
∞
e − t /( ατ D )
dt + K c M
α
t =0
y a′ (t ) = K c Mδ(t ) ∫
y a′ (t ) = K c Mδ(t )τ D
+ KcM
Ideal response,
KM
 τ s + 1
Yi′( s ) = Gi ( s )U ′( s ) = K c M  D
= KcMτD + c

s
 s 
yi′ (t ) = K c Mτ D δ(t ) + K c M
Hence y a′ (t ) → y i′ (t ) as α → 0
For ramp response,
input is u ′(t ) = Mt
,
U ′( s ) =
M
s2
Solution Manual for Process Dynamics and Control, 2nd edition,
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp.
8-1
 τDs +1 
 τ s +1 
Ya′ ( s ) = K c  D

U ′( s ) = K c M  2
 ατ D s + 1
 s (ατ D s + 1) 
Ya′ (s ) =
K c Mτ D
K M
+ 2 c
s (ατ D s + 1) s (ατ D s + 1)
 − ατ D 1
1
ατ D 
(ατ D ) 2 
K
M
= K c Mτ D  −
+
+
+


c

s 2 ατ D s + 1
 s ατ D s + 1
 s
Taking inverse Laplace transform
[
]
[
y a′ (t ) = K c Mτ D 1 − e − t /( ατ D ) + K c M t + ατ D (e − t /( ατ D ) − 1)
]
As α →0
y a′ (t ) = K c Mτ D + K c Mt
Ideal response,
K c Mτ D K c M
 τ s + 1
Yi′(s ) = K c M  D 2  =
+ 2
s
s
 s

yi′ (t ) = K c Mτ D
+ K c Mt
Hence y a′ (t ) → y i′ (t ) as α → 0
b)
It may be difficult to obtain an accurate estimate of the derivative for use
in the ideal transfer function.
c)
Yes. The ideal transfer function amplifies the noise in the measurement by
taking its derivative. The approximate transfer function reduces this
amplification by filtering the measurement.
8.2
a)
K1
K + K 2 τ1 s + K 2
P ′( s )
=
+ K2 = 1
E ( s ) τ1 s + 1
τ1 s + 1
8-2
 K 2 τ1

 K + K s + 1
2

= ( K1 + K 2 )  1
τ1 s + 1 





b)
Kc = K1 + K2
K2 = Kc − K1
→
τ1 = ατ D
τD =
K 2 τ1
K ατ
= 2 D
K1 + K 2 K1 + K 2
or
1=
K 2α
K1 + K 2
K1 + K 2 = K 2 α
K 1 = K 2 α − K 2 = K 2 (α − 1)
Substituting,
K 1 = ( K c − K 1 )(α − 1) = (α − 1) K c − (α − 1) K 1
Then,
 α −1
K1 = 
K c
 α 
c)
If Kc = 3
, τD = 2
K1 =
,
α = 0.1
− 0 .9
× 3 = −27
0 .1
K 2 = 3 − (−27) = 30
τ1 = 0.1 × 2 = 0.2
Hence
K1 + K2 = -27 + 30 = 3
K 2 τ1
30 × 0.2
=
=2
K1 + K 2
3
 2s + 1 
Gc ( s ) = 3

 0 .2 s + 1 
8-3
then,
8.3
a)
From Eq. 8-14, the parallel form of the PID controller is :


1
Gi ( s ) = K c′ 1 +
+ τ′D s 
 τ′I s

From Eq. 8-15, for α →0, the series form of the PID controller is:

1 
Ga ( s ) = K c 1 +
[τ D s + 1]
 τI s 
 τ

1
= K c 1 + D +
+ τ D s
 τI τI s



 τ D 
τDs
1


= K c 1 +
1+
+


τI 
 τ 
 τ

 1 + D τ I s 1 + D
τI
τI 
 





 
 
Comparing Ga(s) with Gi(s)
 τ 
K c′ = K c 1 + D 
τI 

 τ 
τ′I = τ I 1 + D 
τI 

τ′D =
b)
τD
τ
1+ D
τI
 τ
Since 1 + D
 τI
K c ≤ K c′ ,

 ≥ 1 for all τD, τI, therefore

τ I ≤ τ′I and τ D ≥ τ′D
c)
For Kc = 4, τI=10 min , τD =2 min
d)
K c′ = 4.8 , τ′I = 12 min , τ′D = 1.67 min
Considering only first-order effects, a non-zero α will dampen all
responses, making them slower.
8-4
8.4
Note that parts a), d), and e) require material from Chapter 9 to work.
a)
System I (air-to-open valve) : Kv is positive.
System II (air-to-close valve) : Kv is negative.
b)
System I : Flowrate too high → need to close valve → decrease controller
output → reverse acting
System II: Flow rate too high → need to close valve → increase
controller output → direct acting.
c)
System I : Kc is positive
System II : Kc is negative
d)
System I :
System II :
Kc
Kv
+
−
+
−
Kp Km
+
+
+
+
Kc and Kv must have same signs
e)
Any negative gain must have a counterpart that "cancels" its effect. Thus,
the rule:
# of negative gains to have negative feedback = 0 , 2 or 4.
# of negative gains to have positive feedback = 1 or 3.
8.5
a)
From Eqs. 8-1 and 8-2,
[
p(t ) = p + K c y sp (t ) − y m (t )
]
(1)
The liquid-level transmitter characteristic is
ym(t) = KT h(t)
(2)
where h is the liquid level
KT > 0 is the gain of the direct acting transmitter.
8-5
The control-valve characteristic is
q(t) = Kvp(t)
(3)
where q is the manipulated flow rate
Kv is the gain of the control valve.
From Eqs. 1, 2, and 3
[
]
[
q(t ) − q = K v p (t ) − p = K V K c y sp (t ) − K T h(t )
KV K c =
]
q (t ) − q
y sp − K T h(t )
For inflow manipulation configuration, q(t)> q when ysp(t)>KTh(t). Hence
KvKc > 0
then for "air-to-open" valve (Kv>0), Kc>0 :
and for "air-to-close" valve (Kv<0), Kc<0 :
reverse acting controller
direct acting controller
For outflow manipulation configuration, KvKc <0
then for "air-to-open" valve, Kc<0 :
and for "air-to-close" valve, Kc>0 :
b)
direct acting controller
reverse acting controller
See part(a) above
8.6
For PI control
t


1

p (t ) = p + K c  e(t ) + ∫ e(t*)dt * 
τI 0


t


1
p ′(t ) = K c  e(t ) + ∫ e(t*)dt * 
τI 0


Since
e(t) = ysp – ym
and
ym= 2
8-6
Then
e(t)= -2
t



1
2
p ′(t ) = K c  − 2 + ∫ (−2)dt *  = K c  − 2 −
τI 0
τI




t 

Initial response = − 2 Kc
Slope of early response = −
− 2 Kc = 6
−
2K c
= 1.2 min-1
τI
2K c
τI
→ Kc = -3
→ τI = 5 min
8.7
a)
To include a process noise filter within a PI controller, it would be placed
in the feedback path
b)

1 

K c 1 +
 τI s 
1
1 f s 23
c)
The TF between controller output P ′(s ) and feedback signal Ym(s) would
be
8-7
P ′( s ) − K c (τ I s + 1)
=
Ym ( s ) τ I s (τ f s + 1)
M
s
For Ym ( s ) =
P ′( s ) =
The
Negative sign comes from comparator
− KcM
τI
 τI s +1  − KcM
 2
=
τI
 s (τ f s + 1) 
A B
C 
 2 + +

s τ f s + 1
 s
C
term gives rise to an exponential.
τ f s +1
To see the details of the response, we need to obtain B (= τI - τf) and A(=1)
by partial fraction expansion.
The response, shown for a negative change in Ym, would be
Slope = KcM/τI
"Ideal" PI
y
Filtered PI
-KcM
-KcM(1-τf/τI)
time
d)
τf
→ 0 , the two responses become the same.
τI
If the measured level signal is quite noisy, then these changes might still
be large enough to cause the controller output to jump around even after
filtering.
Note that as
One way to make the digital filter more effective is to filter the process
output at a higher sampling rate (e.g., 0.1 sec) while implementing the
controller algorithm at the slower rate (e.g., 1 sec).
A well-designed digital computer system will do this, thus eliminating the
need for analog (continuous) filtering.
8-8
8.8
a)
From inspection of Eq. 8-26, the derivative kick = K c
b)
Proportional kick = K c ∆r
c)
e1 = e2 = e3 = …. = ek-2 = ek-1 = 0
τD
∆r
∆t
ek = ek+1 = ek+2 = …= ∆r
p k −1 = p


τ
∆t
p k = p + K c ∆r + ∆r + D ∆r 
τI
∆t 


∆t 
p k +i = p + K c ∆r + (1 + i ) ∆r  ,
τI


Kc
pk
τD
∆r
∆t
K c ∆r
Kc
p
k-1
c)
i = 1, 2, …
k
k+1
∆t
∆r
τI
k+2
k+3
To eliminate derivative kick, replace (ek – ek-1) in Eq. 8-26 by (yk-yk-1).
8-9
8.9
a)
The digital velocity P algorithm is obtained by setting 1/τI = τD = 0 in Eq.
8-28 as
∆pk = Kc(ek – ek-1)
[
]
= K c ( y sp − y k ) − ( y sp − y k −1 )
= K c [ y k −1 − y k ]
The digital velocity PD algorithm is obtained by setting 1/τI = 0 in Eq. 828 as
τ
∆pk = Kc [(ek – ek-1) + D (ek – 2ek-1 + ek-2)]
∆t
τ
= Kc [ (-yk + yk-1) + D (-yk – 2yk-1 + yk-2) ]
∆t
In both cases, ∆pk does not depend on y sp .
b)
c)
For both these algorithms ∆pk = 0 if yk-2 = yk-1 = yk. Hence steady state is
reached with a value of y that is independent of the value of y sp . Use of
these algorithms is inadvisable if offset is a concern.
∆t
( y sp − y k ) .
τI
Thus, at steady state when ∆pk = 0 and yk-2 = yk-1 = yk , yk = y sp and the
offset problem is eliminated.
If the integral mode is present, then ∆pk contains the term Kc
8.10
a)

τDs 
P ′( s )
1

= K c 1 +
+
E (s)
 τ I s ατ D s + 1 
= Kc
( τ I s(ατ D s + 1) + ατD s + 1 + τD sτ I s )
τ I s (ατ D s + 1)
1 + (τ I + ατ D ) s + (1 + α)τ I τ D s 2 
= Kc 

τ I s (ατ D s + 1)


8-10
Cross- multiplying
(
)
(ατ I τ D s 2 + τ I s ) P ′( s ) = K c 1 + (τ I + ατ D ) s + (1 + α )τ I τ D s 2 E ( s )
ατ I τ D
b)
d 2 e(t ) 
d 2 p ′(t )
dp ′(t )
de(t )


(
1
)
+
τ
=
K
e
(
t
)
+
(
τ
+
ατ
)
+
+
α
τ
τ

I
c
I
D
I D
dt
dt 2 
dt 2
dt

 τ s + 1  τ D s 
P ′( s )


= K c  I
E (s)
 τ I s  ατ D s + 1 
Cross-multiplying
τ I s 2 (ατ D s + 1) P ′( s ) = K c ((τ I s + 1)(τ D s + 1) ) E ( s )
ατ I τ D
c)
d 2 e(t ) 
d 2 p ′(t )
dp ′(t )
de(t )


+
τ
=
K
e
(
t
)
+
(
τ
+
τ
)
+
τ
τ

I
c
I
D
I D
dt
dt
dt 2
dt 2 

We need to choose parameters in order to simulate:
e.g.,
Kc = 2 ,
τI = 3
, τ D = 0.5
,
α = 0 .1
,
M=1
By using Simulink-MATLAB
Step Response
22
Parallel PID with a derivative filter
Series PID with a derivative filter
20
18
16
14
p'(t)
12
10
8
6
4
2
0
2
4
Time
6
8
10
Figure S8.10. Step responses for both parallel and series PID controllers
with derivative filter.
8-11
8.11
a)
 τ s +1
P ′( s )
(τ D s + 1)
= K c  I
E (s)
 τI s 
τ I s P ′( s ) = K c ((τ I s + 1)(τ D s + 1) ) E ( s )
d 2 e(t ) 
dp′(t ) K c 
de(t )

+ τI τD
=
 e(t ) + (τ I + τ D )
dt
dt
τI 
dt 2 
b)
With the derivative mode active, an impulse response will occur at t = 0.
Afterwards, for a unit step change in e(t), the response will be a ramp
with slope = K c (τ I + τ D ) / τ I and intercept = K c / τ I for t > 0 .
Impulse at t=0
p'
slope =
Kc
τI
t
8-12
K c (τ I + τ D )
τI
1234567898
9.1
a)
Flowrate pneumatic transmitter:
 15 psig - 3 psig 
qm(psig)= 
 (q gpm - 0 gpm) + 3 psig
 400 gpm-0 gpm 

psig 
=  0.03
 q (gpm) + 3 psig
gpm 

Pressure current transmitter:
 20 mA - 4 mA 
Pm(mA)= 
 ( p in.Hg − 10 in.Hg) + 4 mA
 30 in.Hg - 10 in.Hg 

mA 
=  0.8
 p (in.Hg) − 4 mA
in.Hg 

Level voltage transmitter:
 5 VDC - 1 VDC 
hm(VDC)= 
 (h(m) - 0.5m) + 1 VDC
 20 m - 0.5 m 
VDC 

=  0.205
 h(m) + 0.897 VDC
m 

Concentration transmitter:
 10 VDC - 1 VDC 
Cm(VDC)= 
 (C (g/L)-2 g/L)+1 VDC
 20 g/L - 2 g/L 

VDC 
=  0.5
 C (g/L)
g/L 

b)
The gains, zeros and spans are:
Solution Manual for Process Dynamics and Control, 2nd edition,
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp
9-1
GAIN
ZERO
SPAN
PNEUMATIC
0.03psig/gpm
0gal/min
400gal/min
VOLTAGE
CURRENT
VOLTAGE
0.8mA/in.Hg 0.205 VDC/m 0.5VDC/g/L
10 in.Hg
0.5m
2g/L
20 in.Hg
19.5m
18g/L
*The gain is a constant quantity
9.2
a)
The safest conditions are achieved by the lowest temperatures and
pressures in the flash vessel.
VALVE 1.- Fail close
VALVE 2.- Fail open
VALVE 3.- Fail open
VALVE 4.- Fail open
VALVE 5.- Fail close
Setting valve 1 as fail close prevents more heat from going to flash drum
and setting valve 3 as fail open to allow the steam chest to drain. Setting
valve 3 as fail open prevents pressure build up in the vessel. Valve 4
should be fail-open to evacuate the system and help keep pressure low.
Valve 5 should be fail-close to prevent any additional pressure build-up.
b)
Vapor flow to downstream equipment can cause a hazardous situation
VALVE 1.- Fail close
VALVE 2.- Fail open
VALVE 3.- Fail close
VALVE 4.- Fail open
VALVE 5.- Fail close
Setting valve 1 as fail close prevents more heat from entering flash drum
and minimizes future vapor production. Setting valve 2 as fail open will
allow the steam chest to be evacuated, setting valve 3 as fail close prevents
vapor from escaping the vessel. Setting valve 4 as fail open allows liquid
to leave, preventing vapor build up. Setting valve 4 as fail-close prevents
pressure buildup.
c)
Liquid flow to downstream equipment can cause a hazardous situation
VALVE 1.- Fail close
VALVE 2.- Fail open
VALVE 3.- Fail open
VALVE 4.- Fail close
VALVE 5.- Fail close
9-2
Set valve 1 as fail close to prevent all the liquid from being vaporized
(This would cause the flash drum to overheat). Setting valve 2 as fail open
will allow the steam chest to be evacuated. Setting valve 3 as fail open
prevents pressure buildup in drum. Setting valve 4 as fail close prevents
liquid from escaping. Setting valve 5 as fail close prevents liquid build-up
in drum
9.3
a)
Assume that the differential-pressure transmitter has the standard range of
3 psig to 15 psig for flow rates of 0 gpm to qm(gpm). Then, the pressure
signal of the transmitter is
 12 
PT = 3 +  2 q 2
 qm 
dP  24 
KT = T =  2  q
dq  qm 
2.4/qm
,
q = 10% of qm
12/qm
,
q = 50% of qm
18/qm
,
q = 75% of qm
21.6/qm
,
q = 90% of qm
KT =
b)
Eq. 9-2 gives
1/ 2
 ∆P 
q = Cv f (1)  v 
 gs 
= qm f ( 1 )
For a linear valve,
f (1) = 1 = αP , where α is a constant.
KV =
dq
= qm α
dP
Hence, linear valve gain is same for all flowrates
9-3
For a square-root valve,
f ( 1 ) = 1 = αP
q α 1
q αq
dq
1
KV =
= qm α
= m
= m m
dP
2 1
2 q
2 p
5qmα
,
q = 10% of qm
qmα
,
q = 50% of qm
0.67qmα
,
q = 75% of qm
0.56qmα
,
q = 90% of qm
KV =
For an equal-percentage valve,
f (1) = R 1 −1 = R αP −1
KV =
 q 
dq
= qm αR 1 −1 ln R = qm α ln R  
dP
 qm 
0.1qmαlnR
,
q = 10% of qm
0.5qmαlnR
,
q = 50% of qm
0.75qmαlnR
,
q = 75% of qm
0.9qmαlnR
,
q = 90% of qm
KV =
c)
The overall gain is
KTV = KTKV
Using results in parts a) and b)
For a linear valve
2.4α
,
q = 10% of qm
12α
,
q = 50% of qm
18α
,
q = 75% of qm
21.6α
,
q = 90% of qm
KTV =
9-4
For a square-root valve
KTV = 12α
for all values of q
For an equal-percentage valve
0.24αlnR
,
q = 10% of qm
6.0αlnR
,
q = 50% of qm
13.5αlnR
,
q = 75% of qm
19.4αlnR
,
q = 90% of qm
KTV =
The combination with a square-root valve gives linear characteristics over
the full range of flow rate. For R = 50 and α = 0.067 values, a graphical
comparison is shown in Fig. S9.3
7
Linear valve
Square valve
% valve
6
5
K
TV
4
3
2
1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
q/qm
Fig. S9.3.- Graphical comparison of the gains for the three valves
d)
In a real situation, the square-root valve combination will not give an
exactly linear form of the overall characteristics, but it will still be the
combination that gives the most linear characteristics.
9-5
9.4
Nominal pressure drop over the condenser is 30 psi
∆Pc = Kq2
30 = K (200)2
∆Pc =
,
K=
3
psi
4000 gpm 2
3
q2
4000
Let ∆Pv be the pressure drop across the valve and ∆P v , ∆P c be the
nominal values of ∆Pv , ∆Pc, respectively. Then,
(
)
(
)
∆Pv = ∆ P v + ∆ Pc −∆Pc = 30 + ∆ Pv −
3
q2
4000
(1)
Using Eq. 9-2
 ∆P
q = C v f (1) v
 gs



1/ 2
(2)
and
q  ∆ Pv 
Cv =


f (l )  g s 
−1/ 2
200  ∆ P v 
=


0.5  1.11 
−1/ 2
(3)
Substituting for ∆Pv from(1) and Cv from(3) into (2) ,
 ∆ Pv 
q = 400 

 1.11 
a)
−1/ 2
3

2 
 30 + ∆ P v − 4000 q 
f (1) 

1.11




∆ Pv = 5
Linear valve: f (1) = 1 , and Eq. 4 becomes
q  35 − 0.00075q 2

l=
188.5 
1.11



−1 / 2
9-6
−1/ 2
(4)
Equal % valve: f (1) = R 1 −1 = 20 1 −1 assuming R=20
 q  35 − 0.00075q 2  −1 / 2 

 
ln 
1.11
188.5 
 
l = 1+
ln 20
250
200
q(gpm)
150
Linear valve
Equal % valve
100
50
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
l (valve lift)
Figure S9.4a. Control valve characteristics for ∆ P v = 5
b)
∆ P v = 30
Linear valve: f (1) = 1 , and Eq. 4 becomes
q  60 − 0.00075q 2

l=
76.94 
1.11



−1 / 2
Equal % valve: f (1) = 20 1 −1 ; Eq. 4 gives
 q  60 − 0.00075q 2  −1 / 2 

 
ln 
1.11
 76.94 
 
l = 1+
ln 20
9-7
300
Linear valve
Equal % valve
250
q (gpm)
200
150
100
50
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
l (valve lift)
Figure S9.4b. Control valve characteristics for ∆ P v = 30
c)
∆ P v = 90
Linear valve: f (1) = 1 , and Eq. 4 becomes
q  120 − 0.00075q 2

l=
44.42 
1.11



−1 / 2
Equal % valve: f (1) = 20 1 −1 ; Eq. 4 gives
 q  120 − 0.00075q 2  −1 / 2 

 
ln 
1.11
 44.42 
 
l = 1+
ln 20
9-8
300
250
Equal % valve
200
150
100
Linear valve
Equal % valve
50
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
l (valve lift)
Figure S9.4c. Control valve characteristics for ∆ P v = 90
Conclusions from the above plots:
1) Linearity of the valve
For ∆ P v = 5, the linear valve is not linear and the equal % valve is
linear over a narrow range.
For ∆ P v = 30, the linear valve is linear for very low 1 and equal
% valve is linear over a wider range of 1 .
For ∆ P v = 90, the linear valve is linear for 1 <0.5 approx., equal %
valve is linear for 1 >0.5 approx.
2) Ability to handle flowrates greater than nominal increases as ∆ P v
increases, and is higher for the equal % valve compared to that for the
linear valve for each ∆ P v .
3)
The pumping costs are higher for larger ∆ P v . This offsets the
advantage of large ∆ P v in part 1) and 2)
9-9
9.5
Let ∆Pv/∆Ps = 0.33 at the nominal q = 320 gpm
∆Ps = ∆PB + ∆Po = 40 + 1.953 × 10-4 q2
∆Pv= PD - ∆Ps = (1 –2.44 × 10-6 q2)PDE – (40 + 1.953 × 10-4 q2)
(1 - 2.44 × 10 -6 × 320 2 )PDE - (40 + 1.953 × 10 -4 × 320 2 )
= 0.33
(40 + 1.953 × 10 -4 × 320 2 )
PDE = 106.4 psi
Let qdes = q = 320 gpm
For rated Cv, valve is completely open at 110% qdes i.e., at 352 gpm or the
upper limit of 350 gpm
 ∆p
C v = q v
 qs



−
1
2
 (1 − 2.44 × 10 × 350 )106.4 − (40 + 1.953 × 10 × 350 ) 
= 350 

0.9


−6
2
Then using Eq. 9-11
 q  66.4 − 4.55 × 10 − 4 q 2  −1 / 2 

 
ln 
0.9
101.6 
 
l = 1+
ln 50
9-10
−4
2
−
1
2
400
350
300
q (gpm)
250
200
150
Cv = 101.6
Cv = 133.5
100
50
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
l (valve lift)
Figure S9.5. Control valve characteristics
From the plot of valve characteristic for the rated Cv of 101.6, it is evident
that the characteristic is reasonably linear in the operating region 250 ≤ q
≤ 350.
The pumping cost could be further reduced by lowering the PDE to a value
that would make ∆Pv/∆Ps = 0.25 at q = 320 gpm. Then PDE = 100.0 and
for qdes = 320 gpm, the rated Cv = 133.5. However, as the plot shows, the
valve characteristic for this design is more nonlinear in the operating
region. Hence the selected valve is Cv = 101.6
9-11
9.6
a)
1
0.9
0.8
0.7
f
0.6
0.5
0.4
0.3
Linear
0.2
"Square root"
0.1
"Square"
0
0
0.2
0.4
0.6
0.8
1
l
The "square" valve appears similar to the equal percentage valve in Fig. 9.8
b)
Quick open
Linear
Slow open
/ d1 )
1 =0
1 =0.5
1 =1
1/ 2 1
∞
0.707
0.5
1
1
1
1
21
0
1
2
Gain ( df
Valve
The largest gain for quick opening is at 1 =0 (gain = ∞), while largest for
slow opening is at 1 =1 (gain = 2). A linear valve has constant gain.
c)
q = C v f (1)
For
∆Pv
gs
gs = 1
, ∆Pv = 64 , q = 1024
Cv is found when f (1) =1 (maximum flow):
9-12
Cv =
d)
∆Pv g s
=
1024 gal/min 1024
gal.in
=
=128
8
min.(lb)1/2
64 lb/in 2
1 in terms of applied pressure
1 =0
1 =1
when p = 3 psig
when p = 15 psig
Then 1 =
e)
q
(1 − 0)
1
( p − 3) =
p − 0.25
(15 − 3)
12
q = 128 1 2 ∆Pv
for slow opening ("square") valve
2
1

= 128 ∆Pv  p − 0.25 
 12

128
2
=
∆Pv ( p − 3) = 0.8889 ∆Pv ( p − 3) 2
144
p=3
,
q = 0 for all ∆Pv
p =15
,
q = 128 ∆Pv
= 0 for ∆Pv = 0
= 1024 for ∆Pv = 64
looks O.K
9.7
Because the system dynamic behavior would be described using deviation
variables, all that is important are the terms involving x, dx/dt and d2x/dt2.
Using the values for M, K and R and solving the homogeneous o.d.e:
0.3
d 2x
dx
+ 15,000 + 3600 x = 0
2
dt
dt
This yields a strongly overdamped solution, with ζ=228, which can be
approximated by a first order model by ignoring the d2x/dt2 ter
9-13
9.8
A control system can incorporate valve sequencing for wide range along
with compensation for the nonlinear curve (Shinskey, 1996). It features a
small equal-percentage valve driven by a proportional pH controller. The
output of the pH controller also operates a large linear valve through a
proportional-plus-reset controller with a dead zone. The system is shown
in Fig. E9.8
Reagent
Linear
pHC
Percent
Influent
Figure S9.8. Schematic diagram for pH control
Equal-percentage valves have an exponential characteristic, similar to the
pH curve. As pH deviates from neutrality, the gain of the curve decreases;
but increasing deviation will open the valve farther, increasing its gain in a
compensating manner. As the output of the proportional controller drives
the small valve to either of its limits, the dead zone of the two-mode
controller is exceeded. The large valve is moved at a rate determined by
the departure of the control signal from the dead zone and by the values of
proportional and reset. When the control signal reenters the dead zone, the
large valve is held in its last position. The large valve is of linear
characteristic, because the process gain does not vary with flow, as some
gains do.
9-14
Note: in the book’s second printing, the transient response in this problem will be
modified by adding 5 minutes to the time at which each temperature reading was taken.
We wish to find the model:
Tm′ ( s )
Km
=
T ′( s ) τm s + 1
where Tm is the measurement
T is liquid temperature
From Eq. 9-1,
Km =
range of instrument output 20 mA - 4 mA 16 mA
mA
=
=
=0.04 o
o
o
o
range of instrument input
400 C - 0 C
400 C
C
From Fig. 5.5, τ can be found by plotting the thermometer reading vs.
time and the transmitter reading vs. time and drawing a horizontal line
between the two ramps to find the time constant. This is shown in Fig.
S9.9.
Hence, ∆τ = 1.33 min = 80 sec
To get τ, add the time constant of the thermometer (20 sec) to ∆τ to get
τ = 100 sec.
122
120
118
T (deg F)
9.9
<Time constant>
116
114
112
110
Thermometer
Transmitter
108
106
2
2.5
3
3.5
time (min)
4
4.5
5
Figure S9.9. Data test from the Thermometer and the Transmitter
9-15
9.10
precision =
0.1 psig
= 0.5% of full scale
20 psig
accuracy is unknown since the "true" pressure in the tank is unknown
resolution =
0.1 psig
= 0.5% of full scale
20 psig
repeatability =
±0.1 psig
=±0.5% of full scale
20 psig
9.11
Assume that the gain of the sensor/transmitter is unity. Then,
Tm′ ( s )
1
=
T ′( s ) ( s + 1)(0.1s + 1)
where T is the quantity being measured
Tm is the measured value
T ′ (t) = 0.1 t °C/s , T ′ (s) =
Tm′ ( s ) =
0 .1
s2
1
0.1
× 2
( s + 1)(0.1s + 1) s
Tm′ (t ) = −0.0011e −10t + 0.111e − t + 0.1t − 0.11
Maximum error occurs as t→∞ and equals |0.1t − (0.1t − 0.11)| = 0.11 °C
If the smaller time constant is neglected, the time domain response is a bit
different for small values of time, although the maximum error (t→∞)
doesn't change.
9-16
2
1.8
1.6
1.4
T', Tm' (C)
1.2
1
0.8
0.6
Tm'(t)
T'(t)
0.4
0.2
0
0
2
4
6
8
10
12
14
16
18
20
t(s)
Figure S9.11. Response for process temperature sensor/transmitter
9-17
123456789 8
Rev: 12-6-03
10.1
According to Guideline 6, the manipulated variable should have a large
effect on the controlled variable. Clearly, it is easier to control a liquid
level by manipulating a large exit stream, rather than a small stream.
Because R/D>1, the reflux flow rate R is the preferred manipulated
variable.
10.2
Exit flow rate w4 has no effect on x3 or x4 because it does not change the
relative amounts of materials that are blended. The bypass fraction f has a
dynamic effect on x4 but no steady-state effect because it also does not
change the relative amounts of materials that are blended. Thus, w2 is the
best choice.
10.3
Both the steady-state and dynamic behavior needs to be considered. From
a steady-state perspective, the reflux stream temperature TR would be a
poor choice because it is insensitive to changes in xD, due to the small
nominal value of 5 ppm. For example, even a 100% change from 5 to 10
ppm would result in a negligible change in TR. Similarly, the temperature
of the top tray would be a poor choice. An intermediate tray temperature
would be more sensitive to changes in the tray composition but may not be
representative of xD. Ideally, the tray location should be selected to be the
highest tray in the column that still has the desired degree of sensitivity to
composition changes.
The choice of an intermediate tray temperature offers the advantage of
early detection of feed disturbances and disturbances that originate in the
stripping (bottom) section of the column. However, it would be slow to
respond to disturbances originating in the condenser or in the reflux drum.
But on balance, an intermediate tray temperature is the best choice.
Solution Manual for Process Dynamics and Control, 2nd edition,
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp
10-1
10.4
For the flooded condenser in Fig. E10.4, the area available for heat
transfer changes as the liquid level changes. Consequently, pressure
control is easier when the liquid level is low and more difficult when the
level is high. By contrast, for the conventional process design in Fig. 10.5,
the liquid level has a very small effect on the pressure control loop. Thus,
the flooded condenser is more difficult to control because the level and
pressure control loops are more interacting than they are for the
conventional process design in Fig. 10.5.
10.5
(a)
The larger the tank, the more effective it will be in “damping out”
disturbances in the reactor exit stream. A large tank capacity also provides
a large feed inventory for the distillation column, which is desirable for
periods where the reactor is shut down. Thus a large tank is preferred from
a process control perspective. However a large tank has a high capital cost,
so a small tank is appealing from a steady-state, design perspective. Thus,
the choice of the storage tank size involves a tradeoff of control and
design objectives.
(b)
After a set-point change in reactor exit composition occurs, it would be
desirable to have the exit compositions for both the reactor and the storage
tank change to the new value as soon as possible. But the concentration in
the storage tank will change gradually due to its liquid inventory. The time
constant for the storage tank is proportional to the mass of liquid in the
tank (cf. blending system models in Chapters 2 and 4). Thus, a large
storage tank will result in sluggish responses in its exit composition, which
is not desirable when frequent set-point changes are required. In this
situation, the storage tank size should be smaller than for case (a).
10.6
Variables : q1, q2,…. q6, h1, h2
NV = 8
Equations :
3 flow-head relations:
10-2
q3 = Cv1 h1
q5 = Cv 2 h2
q 4 = K (h1 − h2 )
2 mass balances:
1A1
dh1
= 12 q1 + q6 − q3 − q4 3
dt
dh
1A2 2 = 12 q2 + q4 − q5 3
dt
Thus NE = 5
Degrees of freedom: NF = NV – NE = 8 − 5 = 3
Disturbance variable : q6
ND = 1
NF = NFC + ND
NFC = 3 − 1 = 2
10.7
Consider the following energy balances assuming a reference temperature
of Tref = 0 :
Heat exchanger:
Cc (1 − f ) wc (TC 0 − TC1 ) = Ch wh (Th1 − Th 2 )
(1)
C c wc (TC 2 − TC1 ) = C h wh (Th1 − Th 2 )
(2)
wc = (1 − f ) wc + fwc
(3)
Overall:
Mixing point:
Thus,
10-3
NE =3 ,
NV = 8 ( f , wc , wh , Tc1 , Tc 2 , Tc 0 , Th1 , Th 2 )
NF =NV − NE = 8 − 3 = 5
NFC =2 (f, wh)
also
ND = NF − NFC = 3 (wc, Tc1, Tc2)
The degrees of freedom analysis is identical for both cocurrent and
countercurrent flow because the mass and energy balances are the same
for both cases.
10.8
The dynamic model consists of the following material balances:
Mass balance on the tank:
1A
dh
= 24 − f 3 w1 + w2 − w3
dt
(1)
Component balance on the tank:
1A
d (hx3 )
= 24 − f 3 x1w1 + x2 w2 − x3 w3
dt
(2)
Mixing point balances:
w4 = w3 + fw1
(3)
x4w4 = x3w3 + fx1w1
(4)
Thus,
NE = 4
(Eqs.1-4)
NV = 10
(h, f , w1 , w2 , w3 , w4 , x1 , x2 , x3 , x4 )
NF = NV − NE = 6
Because two variables ( w2 and f ) can be independently adjusted, it
would appear that there are two control degrees of freedom. However, the
10-4
fraction of bypass flow rate, f , has no steady-state effect on x4. To
confirm this assertion, consider the overall steady-state component
balance for the tank and the mixing point:
x1 w1 + x 2 w2 = x 4 w4
(5)
This balance does not depend on the fraction bypassed, f, either directly or
indirectly,
Conclusion : NFC = 1 (w2)
10.9
Ci
C
Vessel
q
q
Let Ci = concentration of N2 in the inlet stream = 100%
C = concentration in the vessel = exit concentration (perfect mixing)
Assumptions:
1. Perfect mixing
2. Initially, the vessel contains pure air, that is, C(0) = 79%.
N2 balance on the vessel:
V
dC
= q (C i − C )
dt
(1)
Take Laplace transforms and let τ=V/q:
56 sC 2 s 3 − C 2t = 738 =
Ci
− C 2s3
s
Rearrange,
10-5
C ( s) =
Ci
C (t = 0)
+
s (5s + 43
5s + 4
Take inverse Laplace transforms (cf. Chapter 3),
C (t ) = Ci (1 − e −t / 5 ) + C (t = 0)e−t 9 5
(2)
Also,
V  20, 000 L   1 m3 
5= =

=
q  0.8 m3 / min   1000 L 
Substitute for τ, Ci and C(0) into (2) and rearrange


21%
t = (25 min) ln 

100% − C (t ) 
(3)
Let C(t) = 98% N2 (i.e., 2% O2). From (3),
t = 58.7 min
10.10
Define k as the number of sensors that are working properly. We are
interested in calculating P (k ≥ 2) , when P(E) denotes the probability that
an event, E, occurs.
Because k = 2 and k = 3 are mutually exclusive events,
P (k ≥ 2) = P (k = 2) + P (k = 3)
(1)
These probabilities can be calculated from the binomial distribution 1
and the given probability of a sensor functioning properly (p = 0.99):
 3
1
P(k = 2) =   ( 0.01) (0.99)2 = 0.0294
2
 
 3
0
P(k = 3) =   ( 0.01) (0.99)3 = 0.9703
 3
10-6
 n
where the notation,   , refers to the number of combinations of n objects
r
taken r at a time, when the order of the r objects is not important. Thus
 3
 3
  = 3 and   = 1 . From Eq.(1),
 2
 3
P (k ≥ 2) = 0.0294 + 0.9703 = 0.9997
1
See any standard probability or statistics book, e.g., Montgomery D.C and G.C. Runger,
Applied Statistics and Probability for Engineers, 3rd ed., John Wiley, NY (2003).
10.11
Assumptions:
1. Incompressible flow.
2. Chlorine concentration does not affect the air sample density.
3. T and P are approximately constant.
The time tT that is required to detect a chlorine leak in the processing area
is given by:
tT = ttube + tA
where:
ttube is the time that the air sample takes to travel through the tubing
tA is the time that the analyzer takes to respond after chlorine first
reaches it.
The volumetric flow rate q is the product of the velocity v and the crosssectional area A:
q = vA
∴
then:
10-7
v=
q
A
π D 2 3.14 ( 6.35 − 0.762 )
A=
=
= 24.5 mm 2
4
4
3
10 cm / s
v=
= 40.8 cm / s
24.5 × 10−2 cm 2
2
Thus,
ttube =
4000 cm
= 98.1 s
40.8 cm / s
Finally,
tT = 98.1 + 5 = 103.1 s
Carbon monoxide (CO) is one of the most widely occurring toxic gases,
especially in confined spaces. High concentrations of carbon monoxide
can saturate a person’s blood in a matter of minutes and quickly lead to
respiratory problems or even death. Therefore, this amount of time is not
acceptable if the hazardous gas is CO.
10.12
The key safety concerns include:
1. Early detection of any leaks to the surroundings
2. Over pressurizing the flash drum
3. Maintain enough liquid level so that the pumps do not cavitate.
4. Avoid having liquid entrained in the gas.
These concerns can be addressed by the following instrumentation.
1. Leak detection: sensors for hazardous gases should be located in
the vicinity of the flash drum.
2. Over pressurization: Use a high pressure switch (PSH) to shut
off the feed when a high pressure occurs.
3. Liquid inventory: Use a low level switch (LSL) to shut down
the pump if a low level occurs.
10-8
4. Liquid entrainment: Use a high level alarm to shut off the feed
if the liquid level becomes too high.
This SIS system is shown below with conventional control loops for
pressure and liquid level.
Figure S10.12.
10-9
10.13
The proposed alarm/SIS system is shown in Figure S10.13:
The solenoid-operated valves are normally open. If the column pressure
exceeds a specified limit, the high pressure switch (PSH) shuts down both
the feed stream and the steam flow to the reboiler. Both actions tend to
reduce the pressure in the column.
10-10
12345678998
11.1
11.2

1 

Gc ( s ) = K c 1 +
 τI s 
The closed-loop transfer function for set-point changes is given by Eq. 11
1 
 ,
36 with Kc replaced by K c 1 +
 τI s 
H ′( s )
′ (s)
H sp
H ′( s )
′ (s)
H sp

1  1

K c K v K p K m 1 +
τ I s  (τs + 1)

=

1  1

1 + K c K v K p K m 1 +
 τ I s  (τs + 1)
(τ I s + 1)
=
2
τ 3 s + 2ζ 3 τ 3 s + 1
where ζ3,τ3 are defined in Eqs. 11-62, 11-63 , Kp = R = 1.0 min/ft2 ,
and τ = RA = 3.0 min
Solution Manual for Process Dynamics and Control, 2nd edition,
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp
11-1

ft 3 / min  min  psi 
1.0 2 1.7
K OL = K c K v K p K m = (4) 0.2
 = 1.36
psi 
ft 
ft 

τ3 =
2
τ τ I (3 min)(3 min)
=
= 6.62 min 2
K OL
1.36
 1 + K OL
2 ζ3 τ3 = 
 K OL

2.36
τ I =
× 3 = 5.21 min
1.36

H ′( s )
3s + 1
1
=
=
′ ( s ) (3.0s + 1) + (2.21s + 1) 2.21s + 1
H sp
For H sp′ ( s ) =
(3 − 2) 1
=
s
s
h′(t ) = 1 − e −t / 2..21
t = −2.21ln[1 − h′(t )]
h(t ) = 2.5 ft
h ′(t ) = 0.5 ft
t = 1.53 min
h(t ) = 3.0 ft
h ′(t ) = 1.0 ft
t →∞
Therefore,
h(t = 1.53 min) = 2.5ft
h(t → ∞) = 3.0 ft
11-2
11.3
Gc ( s ) = K c = 5 ma/ma
Assume τm = 0, τv = 0, and K1 = 1, in Fig 11.7.
a)
Offset = Tsp′ (∞) − T ′(∞) = 5 1 F − 4.14 1 F = 0.86 1 F
b)
 K 
K m K c K IP K v  2 
T ′( s )
 τs + 1 
=
Tsp′ ( s )
 K 
1 + K m K c K IP K v  2 
 τs + 1 
Using the standard current range of 4-20 ma,
Km =
20 ma − 4 ma
= 0.32 ma/ 1 F
50 1 F
K v = 1.2 , K IP = 0.75 psi/ma , τ =5 min , Tsp′ ( s ) =
T ′( s ) =
7.20 K 2
s (5s + 1 + 1.440 K 2 )
T ′(∞) = lim sT ′( s ) =
s →0
T ′(∞) = 4.14 1 F
c)
7.20 K 2
(1 + 1.440 K 2 )
K 2 = 3.34 1 F / psi
From Fig. 11-7, since Ti′ = 0
Pt′(∞) K v K 2 = T ′(∞) ,
and
Pt′(∞) = 1.03 psi
Pt′K v K 2 + Ti K 1 = T , Pt = 3.74 psi
Pt (∞) = Pt − Pt′(∞) = 4.77 psi
11-3
5
s
11.4
a)
b)
Gm ( s) = K m e − θm s
assuming τm = 0
(20 − 4)ma − 2 s 
ma  − 2 s
e
e =  2.67
lb sol
lb sol/ft 3 

(9 − 3) 3
ft

1 

Gc ( s ) = K c 1 +
 τI s 
Gm ( s) =
G IP ( s ) = K IP = 0.3 psi/ma
Gv ( s ) = K v =
(10 − 20) USGPM
USGPM
= −1.67
(12 − 6) psi
psi
Overall material balance for the tank,
USgallons  dh

= q1 + q 2 − C v h
 7.481
A
ft 3

 dt
(1)
Component balance for the solute,
7.481 A
d (hC 3 )
= q1c1 + q 2 c 2 − (C v h )c3
dt
Linearizing (1) and (2) gives
11-4
(2)
 C
dh ′
= q 2′ −  v
dt
2 h
7.481 A

h′


(3)
 C
dc′ 
 dh′
7.481 A c3
+ h 3  = c2 q′2 + q2c′2 − c3  v
dt 
 dt
2 h
(
Subtracting (3) times c3 from the above equation gives
7.481 Ah
(
)
dc3′
= (c 2 − c3 ) q ′2 + q 2 c ′2 − C v h c3′
dt
Taking Laplace transform and rearranging gives
K1
K2
Q2′ ( s ) +
C 2′ ( s )
τs + 1
τs + 1
C 3′ ( s ) =
where
K1 =
K2 =
τ=
c 2 − c3
lb sol/ft 3
= 0.08
USGPM
h
Cv
q2
= 0.6
Cv h
7.481A h
= 15 min
Cv
since A = πD 2 / 4 = 12.6 ft 2 , and
q
h =  3
 Cv
2

 q + q2
 =  1

 Cv
2

 = 4 ft

Therefore,
G p ( s) =
0.08
15s + 1
Gd ( s) =
0 .6
15s + 1
11-5
)

h′ − Cv h c3′

c)
The closed-loop responses for disturbance changes and for setpoint
changes can be obtained using block diagram algebra for the block
diagram in part (a). Therefore, these responses will change only if any of
the transfer functions in the blocks of the diagram change.
i.
c 2 changes. Then block transfer function G p (s ) changes
due to K1. Hence Gc(s) does need to be changed, and
retuning is required.
ii.
Km changes. Block transfer functions do change. Hence
Gc(s) needs to be adjusted to compensate for changes in
block transfer functions. The PI controller should be
retuned.
iii.
Km remains unchanged. No block transfer function changes.
The controller does not need to be retuned.
11.5
a)
One example of a negative gain process that we have seen is the liquid
level process with the outlet stream flow rate chosen as the manipulated
variable
c
LT
p
h
ω
With an "air-to-open" valve, w increases if p increases. However, h
decreases as w increases. Thus Kp <0 since ∆h/∆w is negative.
b)
KcKp must be positive. If Kp is negative, so is Kc. See (c) below.
c)
If h decreases, p must also decrease. This is a direct acting controller
whose gain is negative [ p ′(t ) = K c (r ′(t ) − h ′(t ) ]
11-6
11.6
For proportional controller, Gc ( s ) = K c
Assume that the level transmitter and the control valve have negligible
dynamics. Then,
Gm ( s) = K m
Gv ( s ) = K v
The block diagram for this control system is the same as in Fig.11.8.
Hence Eqs. 11-26 and 11-29 can be used for closed-loop responses to
setpoint and load changes, respectively.
The transfer functions G p (s ) and Gd (s ) are as given in Eqs. 11-66 and
11-67, respectively.
a)
Substituting for Gc, Gm, Gv, and Gp into Eq. 11-26 gives
 1 
KmKcKv −

Y
1
As 

=
=
Ysp
τs + 1
 1 
1+ Kc Kv −
K m
 As 
A
where τ = −
Kc Kv Km
For a step change in the setpoint, Ysp ( s ) = M / s
M / s
Y (t → ∞) = lim sY ( s ) = lim s 
=M
s →0
s →0
 τs + 1
Offset = Ysp (t → ∞) − Y (t → ∞) = M − M = 0
b)
Substituting for Gc, Gm, Gv, Gp , and Gd into (11-29) gives
 −1 
 1 


 
K c K v K m 
Y (s)
As 


=
=
D( s)
τs + 1
 1 
1+ Kc Kv −
K m
 As 
where τ is given by Eq. 1.
11-7
(1)
For a step change in the disturbance, D( s ) = M / s
 − M /( K c K v K m ) 
−M
Y (t → ∞) = lim sY ( s ) = lim s 
=

s →0
s →0
s (τs + 1)

 Kc Kv Km
 −M
Offset = Ysp (t → ∞) − Y (t → ∞) = 0 − 
 Kc Kv Km

 ≠ 0

Hence, offset is not eliminated for a step change in disturbance.
11.7
Using block diagram algebra
Y = G d D + G pU
[
(1)
(
~
U = Gc Ysp − Y − G pU
U=
From (2),
)]
(2)
Gc Ysp − Gc Y
~
1− Gc G p
Substituting for U in Eq. 1
[1 + G (G
c
p
]
~
~
− G p ) Y = Gd (1 − Gc G p ) D + G p Gc Ysp
Therefore,
G p Gc
Y
=
~
Ysp 1 + Gc (G p − G p )
and
Gd (1 − Gc G1 p )
Y
=
D 1 + Gc (G p − G1 p )
11-8
11.8
The available information can be translated as follows
1. The outlets of both the tanks have flow rate q0 at all times.
2. To ( s ) = 0
3. Since an energy balance would indicate a first-order transfer function
between T1 and Q0 ,
T ′(t )
= 1 − e −t / τ1
T ′(∞)
2
= 1 − e −12 / τ1 , τ1 = 10 min
3
or
Therefore
T1 ( s ) 31 F /(−0.75 gpm)
4
=
=−
Q0 ( s )
10 s + 1
10 s + 1
T3 ( s ) (5 − 3) 1 F /( −0.75 gpm)
2.67
=
=−
Q0 ( s )
τ 2s +1
τ 2s +1
for T2(s) = 0
T1 ( s ) (78 − 70) 1 F /(12 − 10)V
4
=
=
4.
V1 ( s )
10 s + 1
10 s + 1
T3 ( s ) (90 − 85) 1 F /(12 − 10)V
2.5
=
=
V2 ( s )
10 s + 1
10 s + 1
5. 5τ2 =50 min or τ2 = 10 min
Since inlet and outlet flow rates for tank 2 are q0
T3 ( s ) q0 / q 0
1
=
=
T2 ( s ) τ 2 s + 1 10.0s + 1
6.
V3 ( s )
= 0.15
T3 ( s )
 30 
7. T2 (t ) = T1  t −  = T1 (t − 0.5)
 60 
11-9
T2 ( s )
= e −0.5 s
T1 ( s )
Using these transfer functions, the block diagrams are as follows.
a)
Q0
1
V1
-2.67
+
+
T1
-4
10s+1
T3sp
V3sp
+
+
V2
+-
0.15
T2
e-0.5s
Gc
1
+
+
2.5
T3
10s+1
V3
0.15
b)
V2
-2.5
Q0
+
+
-2.67
1
T3sp
V3sp
0.15
V1
+-
Gc
+
+
T1
-4
10s+1
V3
0.15
11-10
e-0.5s
T2
+
+
1
10s+1
T3
c)
The control configuration in part a) will provide the better control. As is
evident from the block diagrams above, the feedback loop contains, in
addition to Gc, only a first-order process in part a), but a second-orderplus-time-delay process in part b). Hence the controlled variable responds
faster to changes in the manipulated variable for part a).
11.9
The given block diagram is equivalent to
For the inner loop, let
Gc
P
= Gc′ =
~
~
E
1 + Gc G * (1 − e − θs )
In the outer loop, we have
Gd G
Y
=
D 1 + Gc′ G
Substitute for G c′ ,
Y
=
D
Gd G
Gc G
1+
~
~*
1 + Gc G (1 − e − θs )
(
)
~
~
Gd G 1 + Gc G * (1 − e − θs )
Y
=
~
~*
D 1 + Gc G
(1 − e − θs ) + Gc G
11-11
11.10
a)
Derive CLTF:
Y = Y3 + Y2 = G3 Z + G2 P
Y = G3 ( D + Y1 ) + G2 K c E
Y = G3 D + G3G1 K c E + G2 K c E
Y = G3 D + (G3 G1 K c + G2 K c ) E
E = − K mY
Y = G3 D − K c (G3 G1 + G2 ) K mY
G3
Y
=
D 1 + K c (G3 G1 + G2 ) K m
b)
Characteristic Equation:
1 + K c (G3G1 + G2 ) K m = 0
4 
 5
1 + Kc 
+
=0
 s − 1 2 s + 1
 5(2 s + 1) + 4( s − 1) 
1 + Kc 
=0
 ( s − 1)(2 s + 1) 
( s − 1)(2 s + 1) + K c [5(2 s + 1) + 4( s − 1)] = 0
2 s 2 − s − 1 + K c (10 s + 5 + 4 s − 4) = 0
2 s 2 + (14 K c − 1) s + ( K c − 1) = 0
Necessary conditions: K c > 1 / 14 and K c > 1
For a 2nd order characteristic equation, these conditions are also sufficient.
Therefore, K c > 1 for closed-loop stability.
11-12
11.11
a)
c)
Transfer Line:
Volume of transfer line = π /4 (0.5 m)2(20m)= 3.93 m3
Nominal flow rate in the line = q A + q F = 7.5 m 3 / min
Time delay in the line =
3.93 m 3
= 0.52 min
7.5 m 3 /min
GTL ( s ) = e −0.52 s
Composition Transmitter:
Gm ( s) = K m =
(20 − 4) ma
ma
= 0.08
3
(200 − 0) kg/m
kg/m 3
Controller
From the ideal controller in Eq. 8.14
[
]

1 
~
 E ( s ) + K c τ D s C sp′ ( s ) − C m′ ( s )
P ′( s ) = K c 1 +
 τI s 
~
In the above equation, set C sp′ ( s ) = 0 in order to get the derivative on the
process output only. Then,
11-13

1 

G PI ( s ) = K c 1 +
 τI s 
G D (s) = − K c τ D s
with Kc >0 as the controller should be reverse-acting, since P(t) should
increase when Cm(t) decreases.
I/P transducer
K IP =
(15 − 3) psig
psig
= 0.75
(20 − 4) ma
ma
Control valve
Gv ( s ) =
Kv
τv s + 1
5τ v = 1
,
Kv =
dq A
dp v
τ v = 0.2 min
= 0.03(1 / 12)(ln 20)(20)
pv = pv
q A = 0.5 = 0.17 + 0.03(20)
0.03(20)
pv −3
12
pv − 3
12
= 0.5 − 0.17 = 0.33
K v = (1 / 12)(ln 20)(0.33) = 0.082
Gv ( s ) =
pv −3
12
m 3 /min
psig
0.082
0 .2 s + 1
Process
Assume cA is constant for pure A. Material balance for A:
V
dc
= q A c A + q F c F − ( q A + q F )c
dt
11-14
(1)
Linearizing and writing in deviation variable form
V
dc ′
= c A q ′A + q F c ′F − (q A + q F )c ′ − c q ′A
dt
Taking Laplace transform
[Vs + (q A + q F )]C ′(s) = (c A − c )Q ′A ( s) + q F C F′ ( s)
(2)
From Eq. 1 at steady state, dc / dt = 0 ,
c = (q A c A + q F c F ) /(q A + q F ) = 100 kg/m 3
Substituting numerical values in Eq. 2,
[5s + 7.5]C ′( s) = 700 Q ′A ( s) + 7 C F′ ( s)
[0.67s + 1]C ′( s) = 93.3 Q′A ( s) + 0.93 C F′ (s)
93.3
0.67 s + 1
0.93
Gd (s ) =
0.67 s + 1
G p ( s) =
11.12
The stability limits are obtained from the characteristic Eq. 11-83. Hence
if an instrumentation change affects this equation, then the stability limits
will change and vice-versa.
a)
The transmitter gain, Km, changes as the span changes. Thus Gm(s)
changes and the characteristic equation is affected. Stability limits would
be expected to change.
b)
The zero on the transmitter does not affect its gain Km. Hence Gm(s)
remains unchanged and stability limits do not change.
c)
Changing the control valve trim changes Gv(s). This affects the
characteristic equation and the stability limits would be expected to
change as a result.
11-15
11.13
a)
Ga ( s) =
Kc K
(τs + 1)( s + 1)
b)
Gb ( s ) =
K c K (τ I s + 1)
τ I s (τs + 1)( s + 1)
For a)
D( s ) + N ( s ) = (τs + 1)( s + 1) + K c K = τs 2 + (τ + 1) s + 1 + K c K p
Stability requirements:
1 + Kc K p > 0
∞ > K c K p > −1
or
For b)
D( s ) + N ( s ) = τ I (τs + 1)( s + 1) + K c K (τ I s + 1)
= τ I τs 3 + τ I (τ + 1) s 2 + τ I (1 + K c K p ) s + K c K p
Necessary condition: K c K p > 0
Sufficient conditions (Routh array):
τI τ
τ I (1 + K c K p )
τ I (τ + 1)
Kc K p
τ I (τ + 1)(1 + K c K p ) − τ I τK c K p
2
τ I (τ + 1)
Kc K p
Additional condition is:
τ I (τ + 1)(1 + K c K p ) − τ( K c K p ) > 0
(since τ I and τ are both positive)
11-16
τ I (τ + 1) + τ I (τ + 1) K c K p − τK c K p > 0
[τ I (τ + 1) − τ]K c K p > −τ I (τ + 1)
Note that RHS is negative for all positive τ I and τ
(∴ RHS is always negative)
Case 1:
If
τ I (τ + 1) − τ > 0
τ 

i.e., τ I > τ + 1


 − τ I (τ + 1) 
KcKp > 0 > 

 τ I (τ + 1) − τ 
In other words, this condition is less restrictive than KcKp >0 and doesn't
apply.
then
Case 2:
If
then
τ I (τ + 1) − τ < 0
τ 

i.e., τ I < τ + 1


 − τ I (τ + 1) 
KcKp < 

 τ I (τ + 1) − τ 
In other words, there would be an upper limit on KcKp so the
controller gain is bounded on both sides
0
c)
< KcKp <
− τ I (τ + 1)
τ I (τ + 1) − τ
Note that, in either case, the addition of the integral mode decreases the
range of stable values of Kc.
11-17
11.14
From the block diagram, the characteristic equation is obtained as

 4  
 (0.5) s + 3   2
 1 

 
1 + Kc 
=0

 4    s − 1  s + 10 

1 + (0.5) s + 3  


that is,
 2  2  1 
1 + Kc 


=0
 s + 5   s − 1  s + 10 
Simplifying,
s 3 + 14 s 2 + 35s + (4 K c − 50) = 0
The Routh Array is
1
35
14
4Kc-50
490 − (4 K c − 50)
14
4Kc – 50
For the system to be stable,
490 − (4 K c − 50)
>0
14
and
or
4 K c − 50 > 0 or Kc > 12.5
Therefore
12.5 < Kc < 135
11-18
Kc < 135
11.15
a)
Kc K
Kc K
K K /(1 + K c K )
Y ( s)
= 1 − τs =
= c
K K 1 − τs + K c K
τ
Ysp ( s )
−
s +1
1+ c
1 + Kc K
1 − τs
For stability
−
τ
>0
1+ KcK
Since τ is positive, the denominator must be negative, i.e.,
1+ Kc K < 0
K c K < −1
K c < −1 / K
Kc K
K CL =
1+ Kc K
Note that
b)
If K c K < −1 and 1 + K c K is negative,
then CL gain is positive. ∴ it has the proper sign.
c)
K = 10 and τ = 20
and we want
or
−
τ
= 10
1+ KcK
− 20 = 10 + (10)(10) K c
− 30 = 100 K c
K c = −0.3
Offset: K CL =
(−0.3)(10)
−3
=
= 1.5
1 + (−0.3)(10) − 2
∴ Offset = +1 − 1.5 = − 50% (Note this result implies overshoot)
11-19
d)
KcK
(1 − τs )(τ m s + 1)
Kc K
Y ( s)
=
=
Kc K
Ysp ( s )
(1 − τs )(τ m s + 1) + K c K
1+
(1 − τs )(τ m s + 1)
=
=
Kc K
− ττ m s + (τ m − τ) s + 1 + K c K
2
K c K /(1 + K c K )
ττ m
τ −τ
−
s2 + m
s +1
1 + Kc K
1 + Kc K
(standard form)
For stability,
(1)
−
ττ m
>0
1+ Kc K
(2)
1+ Kc K < 0
K c K < −1
1
Kc < −
K
From (1)
Since
From (2)
Since 1 + K c K < 0
For
K = 10 ,
Y ( s)
=
Ysp ( s )
τm − τ
>0
1+ Kc K
τm − τ < 0
− τ < −τ m
τ > τm
τ = 20 , Kc = –0.3 , τm = 5
1 .5
1 .5
=
2
(20)(5) 2 (5 − 20)
50 s + 2.5s + 1
−
s +
s +1
1− 3
(1 − 3)
Underdamped but stable.
11-20
11.16

1 

Gc ( s ) = K c 1 +
 τI s 
Gv ( s ) =
Kv
− 1 .3
=
(10 / 60) s + 1 0.167 s + 1
G p ( s) = −
1
1
=−
As
22.4 s
since A = 3 ft 2 = 22.4
gal
ft
Gm ( s) = K m = 4
Characteristic equation is

1   − 1.3  − 1 
 
1 + K c 1 +

 ( 4) = 0
 τ I s   0.167 s + 1  22.4 s 
(3.73τ I ) s 3 + (22.4τ I ) s 2 + (5.2 K c τ I ) s + (5.2 K c ) = 0
The Routh Array is
3.73τ I
5.2 K c τ I
22.4τ I
5.2 K c
5.2 K c τ I − 0.867 K c
5.2 K c
For stable system,
τI > 0 ,
5.2 K c τ I − 0.867 K c > 0
That is,
Kc > 0
τ I > 0.167 min
11-21
Kc > 0
11.17
 τ s + 1 
5

GOL ( s ) = K c  I
2
 τ I s  (10s + 1)
 N ( s)
 =
 D( s)
D( s ) + N ( s ) = τ I s (100 s 2 + 20 s + 1) + 5 K c (τ I s + 1) = 0
= 100τ I s 3 + 20τ I s 2 + (1 + 5 K c )τ I s + 5 K c = 0
a)
Analyze characteristic equation for necessary and sufficient conditions
Necessary conditions:
5K c > 0
→
(1 + 5 K c )τ I > 0 →
Kc > 0
τI > 0
and
Kc > −
Sufficient conditions obtained from Routh array
100τ I
(1 + 5 K c )τ I
20τ I
5K c
20τ I (1 + 5 K c ) − 500τ I K c
20τ I
2
5K c
Then,
20τ I (1 + 5K c ) − 500τ I K c > 0
2
τ I (1 + 5 K c ) > 25 K c
b)
or
τI >
25K c
1 + 5K c
Sufficient condition is appropriate. Plot is shown below.
11-22
1
5
7
6
Stability region
5
τI
4
3
2
1
0
0
1
2
3
4
5
6
7
Kc
c)
Find τ I as K c → ∞
 25K c 
 25 
lim 
 = Kclim

=5
→
∞
1 + 5K c 
1 / K c + 5 
Kc → ∞
∴ τ I > 5 guarantees stability for any value of Kc. Appelpolscher is
wrong yet again.
11.18
Gc ( s ) = K c
KV
GV ( s ) =
τV s + 1
Kv =
dws
dp
=
p =12
5τ v = 20 sec
G p ( s) =
0.6
2 12 − 4
= 0.106
lbm/sec
ma
τ v = 4 sec
−s
2.5e
10s + 1
Gm ( s) = K m =
(20 − 4) ma
ma
= 0 .4 1
1
(160 − 120) F
F
Characteristic equation is
11-23
−s
 0.106  2.5e 
(0.4) = 0
1 + ( K c )

 4 s + 1  10s + 1 
a)
(1)
Substituting s=jω in (1) and using Euler's identity
e-jω=cosω – j sin ω
gives
-40ω2 +14jω + 1 + 0.106 Kc (cosω – jsinω)=0
Thus
and
-40ω2 + 1 + 0.106 Kc cosω = 0
(2)
14ω - 0.106Kc sinω =0
(3)
From (2) and (3),
tan ω =
14ω
40ω 2 − 1
(4)
Solving (4), ω = 0.579 by trial and error.
Substituting for ω in (3) gives
Kc = 139.7 = Kcu
Frequency of oscillation is 0.579 rad/sec
b)
Substituting the Pade approximation
e −s ≈
1 − 0 .5 s
1 + 0 .5 s
into (1) gives
20 s 3 + 47 s 2 + (14.5 − 0.053K c ) s + (1 + 0.106 K c ) = 0
The Routh Array is
20
14.5 –0.053 Kc
47
1+ 0.106 Kc
14.07 – 0.098 Kc
1 + 0.106 Kc
11-24
For stability,
and
14.07 − 0.098 K c > 0
or Kc < 143.4
1 + 0.106 K c > 0
or Kc > -9.4
Therefore, the maximum gain, Kcu = 143.4, is a satisfactory approximation
of the true value of 139.7 in (a) above.
11.19
a)
G (s) =
4(1 − 5s )
(25s + 1)(4 s + 1)(2 s + 1)
Gc ( s ) = K c
D( s ) + N ( s ) = (25s + 1)(4s + 1)(2 s + 1) + 4 K c (1 − 5s ) = 0
100 s 2 + 29 s + 1
2s + 1
200 s 3 + 58s 2 + 2 s
100 s 2 + 29 s + 1
200 s 3 + 158s 2 + 31s + 1 + 4 K c − 20 K c s = 0
200 s 3 + 158s 2 + (31 − 20 K c ) s + 1 + 4 K c = 0
Routh array:
200
31-20 Kc
158
1+4 Kc
4898 –3160Kc –200 –800Kc
158(31-20Kc)-200(1+4Kc)
=
158
158
1+ 4 Kc
∴ 4698 –3960 Kc > 0
11-25
or
Kc < 1.2
b)
(25s + 1)(4 s + 1)(2 s + 1) + 4 K c = 0
Routh array:
200 s 3 + 158s 2 + 31s + (1 + 4 K c ) = 0
200
31
158
1 + 4 Kc
158 (31) − 200(1+4Kc) = 4898 –200 –800Kc
1+ 4 Kc
∴ 4698 –800 Kc > 0
c)
or
Kc < 5.87
Because Kc can be much higher without the RHP zero present, the process
can be made to respond faster.
11.20
The characteristic equation is
1+
a)
0.5 K c e −3s
=0
10 s + 1
(1)
Using the Pade approximation
e −3 s ≈
1 − (3 / 2) s
1 + (3 / 2) s
in (1) gives
15s 2 + (11.5 − 0.75 K c ) s + (1 + 0.5 K c ) = 0
For stability,
and
11.5 − 0.75 K c > 0
or
K c < 15.33
1 + 0.5 K c > 0
or
K c > −2
11-26
Therefore − 2 < K c < 15.33
b)
Substituting s = jω in (1) and using Euler's identity.
e −3 jω = cos(3ω) − j sin(3ω)
gives
10 jω + 1 + 0.5 K c [cos(3ω) − j sin(3ω)] = 0
Then,
and
1 + 0.5K c cos(3ω) = 0
(2)
10ω − 0.5 K c sin(3ω) = 0
(3)
From (3), one solution is ω = 0, which gives Kc = -2
Thus, for stable operation Kc > -2
From (2) and (3)
tan(3ω) = -10ω
Eq. 4 has infinite number of solutions. The solution for the range π/2 < 3ω
< 3π/2 is found by trial and error to be ω = 0.5805.
Then from Eq. 2, Kc = 11.78
The other solutions for the range 3ω > 3π/2 occur at values of ω for which
cos(3ω) is smaller than cos(3 × 5.805). Thus, for all other solutions of ω,
Eq. 2 gives values of Kc that are larger than 11.78. Hence, stability is
ensured when
-2 < Kc < 11.78
11-27
11.21
a)
To approximate GOL(s) by a FOPTD model, the Skogestad approximation
technique in Chapter 6 is used.
Initially,
3K c e − (1.5+0.3+ 0.2 ) s
3K c e −2 s
GOL ( s ) =
=
(60 s + 1)(5s + 1)(3s + 1)(2s + 1) (60s + 1)(5s + 1)(3s + 1)(2s + 1)
Skogestad approximation method to obtain a 1st -order model:
Time constant ≈ 60 + (5/2)
Time delay ≈ 2 +(5/2) + 3 + 2 =9.5
Then
GOL ( s ) ≈
b)
3K c e −9.5 s
62.5s + 1
The only way to apply the Routh method to a FOPTD transfer function is
to approximate the delay term.
e −9.5 s ≈
− 4.75s + 1
4.75s + 1
(1st order Pade-approximation)
Then
GOL ( s ) ≈
3K c (−4.75s + 1)
N ( s)
≈
D( s ) (62.5s + 1)(4.75s + 1)
The characteristic equation is:
D( s ) + N ( s ) = (62.5s + 1)(4.75s + 1) + 3K c (−4.75s + 1)
297 s 2 + 67.3s + 1 − 14.3K c s + 3K c = 0
297 s 2 + (67.3 − 14.3K c ) s + (1 + 3K c ) = 0
11-28
Necessary conditions:
67.3 − 14.3K c > 0
1 + 3K c > 0
− 14.3K c > −67.3
K c < 4.71
Range of stability:
c)
3 K c > −1
K c > −1 / 3
− 1 / 3 < K c < 4.71
Conditional stability occurs when K c = K cu = 4.71
With this value the characteristic equation is:
297 s 2 + (67.3 − 14.3 × 4.71) s + (1 + 3 × 4.71) = 0
297 s 2 + 15.13 = 0
s2 =
− 15.13
297
We can find ω by substituting jω → s
ω = 0.226
at the maximum gain.
11-29
Revised: 1-3-04
Chapter 12
12.1
For K = 1.0, τ1=10, τ2=5, the PID controller settings are obtained using
Eq.(12-14):
1 τ1 + τ 2 15
=
τc
K τc
ττ
τ D = 1 2 = 3.33
τ1 + τ 2
Kc =
,
τI = τ1+τ2=15
,
The characteristic equation for the closed-loop system is
 
 

1
1.0 + α
1 +  Kc 1 +
+ τ D s  
=0
s
s
s
τ
(10
1)(5
1)
+
+


I




Substituting for Kc, τI, and τD, and simplifying gives
τc s + (1 + α) = 0
Hence, for the closed-loop system to be stable,
τc > 0
and
(1+α) > 0
or α > −1.
(a)
Closed-loop system is stable for α > −1
(b)
Choose τc > 0
(c)
The choice of τc does not affect the robustness of the system to changes in
α. For τc ≤0, the system is unstable regardless of the value of α. For τc > 0,
the system is stable in the range α > −1 regardless of the value of τc.
Solution Manual for Process Dynamics and Control, 2nd edition,
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp
12-1
12.2
G = GvG p Gm =
−1.6(1 − 0.5s )
s (3s + 1)
The process transfer function contains a zero at s = +2. Because the
controller in the Direct Synthesis method contains the inverse of the
process model, the controller will contain an unstable pole. Thus, Eqs.
(12-4) and (12-5) give:
Gc =
( 3s + 1)
1 1
=−
G τc s
2τc (1 − 0.5s )
Modeling errors and the unstable controller pole at s = +2 would render
the closed-loop system unstable.
Modify the specification of Y/Ysp such that Gc will not contain the
offending (1-0.5s) factor in the denominator. The obvious choice is
 Y


 Ysp

1 − 0.5s
 =

d τc s + 1
Then using Eq.(12-3b),
Gc = −
3s + 1
2τc + 1
which is not physically realizable because it requires ideal derivative
action. Modify Y/Ysp,
 Y

 Ysp


1 − 0.5s
 =
2

d (τc + 1)
Then Eq.(12-3b) gives
Gc = −
3s + 1
2
2τc s + 4τc + 1
which is physically realizable.
12-2
12.3
K = 2 , τ = 1, θ = 0.2
(a)
Using Eq.(12-11) for τc = 0.2
Kc = 1.25 , τI = 1
(b)
Using Eq.(12-11) for τc = 1.0
Kc = 0.42 , τI = 1
(c)
From Table 12.3 for a disturbance change
KKc = 0.859(θ/τ)-0.977 or
τ/τI = 0.674(θ/τ)-0.680 or
(d)
Kc = 2.07
τI = 0.49
From Table 12.3 for a setpoint change
KKc = 0.586(θ/τ)-0.916 or
τ/τI = 1.03 −0.165(θ/τ) or
Kc = 1.28
τI = 1.00
(e)
Conservative settings correspond to low values of Kc and high values of τI.
Clearly, the Direct Synthesis method (τc = 1.0) of part (b) gives the most
conservative settings; ITAE of part (c) gives the least conservative
settings.
(f)
A comparison for a unit step disturbance is shown in Fig. S12.3.
1.2
1
Controller for (b)
Controller for (c)
0.8
0.6
y
0.4
0.2
0
-0.2
0
3
6
9
12
15
time
Fig S12.3. Comparison of part (e) PI controllers for unit step disturbance.
12-3
12.4
The process model is,
Ke −θs
G(s) =
s
(1)
Approximate the time delay by Eq. 12-24b,
e − θs = 1 − θ s
(2)
Substitute into (1):
K (1 − θs )
G ( s ) =
s
(3)
Factoring (3) gives G + ( s ) = 1 − θs
and
~
G− ( s) = K / s .
The DS and IMC design methods give identical controllers if,
 Y

Y
 sp

~
 =G
+ f

d
(12-23)
For integrating process, f is specified by Eq. 12-32:
C=
f =
dG +
ds
= −θ
(4)
s =0
(2τc − C ) s + 1
(τc s + 1)
2
=
(2τc + θ) s + 1
(τc s + 1) 2
(5)
~
Substitute G+ and f into (12-23):
 Y

 Ysp


 = (1 − θs )

d
 (2τc + θ) s + 1 

2 
 (τc s + 1) 
The Direct Synthesis design equation is:
12-4
(6)
  Y 

 

Y
1   sp  d
Gc = ~ 
G  Y 

1− 
  Ysp 
d
 







(12-3b)
Substitute (3) and (6) into (12-3b):
 (2τ + θ) s + 1 
(1 − θs )  c
2 


s
 (τc s + 1) 

Gc = 

 (2τ + θ) s + 1 
 K (1 − θs ) 
1 − (1 − θs )  c
2 
 (τc s + 1) 
(7)
or
Gc =
(2τc + θ) s + 1
s
2
K (τc s + 1) − (1 − θs ) [ (2τc + θ) s + 1]
(8)
1 (2τc + θ) s + 1
1 (2τc + θ) s + 1
=
Ks τc 2 + 2τcθs + θ 2 Ks (τc + θ)2
(9)
Rearranging,
Gc =
The standard PI controller can be written as
Gc = K c
τI s + 1
τI s
(10)
Comparing (9) and (10) gives:
τ I = 2τc + θ
(11)
Kc 1
1
=
K ( τ + θ )2
τI
c
(12)
Substitute (11) into (12) and rearrange gives:
Kc =
1 2τc + θ
K ( τ + θ )2
c
(13)
Controller M in Table 12.1 has the PI controller settings of Eqs. (11) and
(13).
12-5
12.5
Assume that the process can be modeled adequately by a first-order-plustime-delay model as in Eq. 12-10. Then using the given step response
data, the model fitted graphically is shown in Fig. S12.5,
18
17
16
Output
15
(mA)
14
13
12
0
2
4
6
8
10
12
Time (min)
Figure S12.5 Process data; first order model estimation.
This gives the following model parameters:


K = KIP Kv Kp Km =  0.75
psi  
psi   16.9 − 12.0 mA 

 = 1.65
  0.9
mA  
psi   20 − 18 psi 
θ = 1.7 min
θ + τ = 7.2 min or
(a)
τ = 5.5 min
Because θ/τ is greater than 0.25, a conservative choice of τc = τ / 2 is
used. Thus τc = 2.75 min.
Settling θc = θ and using the approximation e-θs ≈ 1 -θs, Eq. 12-11 gives
Kc =
(b)
1 τ
= 0.75 ,
K θ + τc
τI = τ = 5.5 min,
From Table 12.3 for PID settings for set-point change,
KKc = 0.965(θ/τ)-0.85
or
τ/τI = 0.796 − 0.1465 (θ/τ) or
or
τD/τ = 0.308 (θ/τ)0.929
12-6
Kc = 1.58
τI = 7.33 min
τD = 0.57 min
τD = 0
(c)
From Table 12.3 for PID settings for disturbance input,
KKc = 1.357(θ/τ)-0.947 or Kc = 2.50
τ/τI = 0.842 (θ/τ)-0.738 or τI = 2.75 min
τD/τ = 0.381 (θ/τ)0.995 or τD = 0.65 min
12.6
Let G be the open-loop unstable process. First, stabilize the process by
using proportional-only feedback control, as shown below.
D
Ysp
+-
E
Gc
P
+-
K c1
G
+
+
Y
Then,
Y
Ysp
K c 1G
1 + K c 1G
Gc G ′
=
=
K c 1G
1 + Gc G ′
1 + Gc
1 + K c 1G
Gc
K c 1G
1 + K c 1G
Then Gc is designed using the Direct Synthesis approach for the stabilized,
modified process G ′ .
where G ′ =
12.7
(a.i)
The model reduction approach of Skogestad gives the following
approximate model:
e −0.028 s
G ( s) =
( s + 1)(0.22 s + 1)
12-7
Applying the controller settings of Table 12.5 (notice that τ1 ≥ 8θ)
Kc = 35.40
τI = 0.444
τD = 0.111
(a.ii)
By using Simulink, the ultimate gain and ultimate period are found:
Kcu = 30.24
Pu = 0.565
From Table 12.6:
Kc = 0.45Kcu = 13.6
τI = 2.2Pu = 1.24
τD = Pu/6.3 = 0.089
(b)
0.08
0.07
Controller (i)
Controller (ii)
0.06
0.05
0.04
y
0.03
0.02
0.01
0
-0.01
0
0.5
1
1.5
time
2
2.5
3
3.5
4
Figure S12.7. Closed-loop responses to a unit step change in a disturbance.
12.8
From Eq.12-39:
1
p (t ) = p + K c bysp (t ) − ym (t )  + K c 
 τI
12-8
t
∫ 0 e(t*)dt * − τ D
dym 

dt 
This control law can be implemented with Simulink as follows:
CONTROLLER
WEIGHTING FACTOR
SET POINT
+
PROPORTIONAL
ACTION
+-
b
KC
+
CONTROLLER
OUTPUT
+
INTEGRAL
ACTION
-
CONTROLLER
INPUT
Closed-loop responses are compared for b = 1, b = 0.7, b = 0.5 and
b = 0.3:
4
b=1
b=0.7
b=0.5
b=0.3
3.5
3
2.5
y
2
1.5
1
0.5
0
0
50
100
150
200
250
300
Time
Figure S12.8. Closed-loop responses for different values of b.
As shown in Figure E12.8, as b increases, the set-point response becomes
faster but exhibits more overshoot. The value of b = 0.5 seems to be a
good choice. The disturbance response is independent of the value of b.
12-9
12.9
In order to implement the series form using the standard Simulink form of
PID control (the expanded form in Eq. 8-16), we first convert the series
controller settings to the equivalent parallel settings.
(a)
From Table 12.2, the controller settings for series form are:
 τ′ 
K c = K c′  1 + D  = 0.971
τ′I 

τ I = τ′I + τ′D = 26.52
τD =
τ′I τ′D
= 2.753
τ′I + τ′D
By using Simulink, closed-loop responses are shown in Fig. S12.9:
3
2.5
Parallel form
Series form
2
y
1.5
1
0.5
0
0
50
100
150
200
250
Time
Figure S12.9. Closed-loop responses for parallel and series form.
12-10
300
The closed-loop responses to the set-point change are significantly
different. On the other hand, the responses to the disturbance are slightly
closer.
(b)
By changing the derivative term in the controller block, Simulink shows
that the system becomes more oscillatory as τD increases. For the parallel
form, system becomes unstable for τD ≥5.4; for the series form, system
becomes unstable for τD ≥4.5.
12.10
(a)
X1'
X'sp
Km
X'sp
(mA)
E
+-
(mA)
P'
GC
(mA)
Gv
Gd
W'2
(Kg/min)
Gp
+
+
X'm
(mA)
(b)
Gm
Process and disturbance transfer functions:
Overall material balance: w1 + w2 − w = 0
Component material balance: w1x1 + w2 x2 − wx = ρV
(1)
dx
dt
Substituting (1) into (2) and introducing deviation variables:
12-11
(2)
X'
w1x1′ + w2′ x2 − w1x′ − w2 x − w2′ x = ρV
dx′
dt
Taking the Laplace transform,
w1 X1′ (s) + (x 2 − x)W2′ (s) = (w1 + w 2 + ρVs)X ′(s)
Finally:
x2 − x
x2 − x
w + w2
X ′( s )
G p ( s) =
=
= 1
1 + τs
W2′ ( s) w1 + w2 + ρVs
w1
w1
w + w2
X ′( s)
Gd ( s ) =
=
= 1
X1′ ( s) w1 + w2 + ρVs
1 + τs
where τ ρV
w1 + w2
Substituting numerical values:
G p ( s) =
2.6 × 10 −4
1 + 4.71s
Gd ( s) =
0.65
1 + 4.71s
Composition measurement transfer function:
Gm ( s) =
20 − 4 − s
e = 32e − s
0.5
Final control element transfer function:
Gv ( s ) =
15 − 3
300 / 1.2
187.5
×
=
20 − 4 0.0833s + 1 0.0833s + 1
Controller:
Let
G = Gv G p G m =
187.5
2.6 × 10 −4
32e − s
0.0833s + 1 1 + 4.71s
12-12
then
G=
1.56e − s
(4.71s + 1)(0.0833s + 1)
For a process with a dominant time constant, τc = τ dom / 3 is
recommended.
Hence τc = 1.57. From Table 12.1,
Kc = 1.92
τI = 4.71
(c)
By using Simulink,
0.04
0.035
0.03
0.025
y
0.02
0.015
0.01
0.005
0
0
5
10
15
20
25
30
Time
Figure S12.10c. Closed-loop response for step disturbance.
12-13
35
(d)
By using Simulink
0
-0.02
-0.04
y
-0.06
-0.08
-0.1
-0.12
0
5
10
15
20
Time
Figure S12.10d. Closed-loop response for a set-point change.
The recommended value of τc = 1.57 gives very good results.
(e)
Improved control can be obtained by adding derivative action: τ D = 0.4 .
0
-0.02
-0.04
y
-0.06
-0.08
-0.1
-0.12
0
5
10
Time
15
20
Figure S12.10e. Closed-loop response by adding derivative action.
12-14
(e)
For θ =3 min, the closed-loop response becomes unstable. It's well known
that the presence of a large process time delay limits the performance of a
conventional feedback control system. In fact, a time delay adds phase lag
to the feedback loop which adversely affects closed-loop stability (cf. Ch.
14). Consequently, the controller gain must be reduced below the value
that could be used if smaller time delay were present.
0.6
0.4
0.2
0
y
-0.2
-0.4
-0.6
-0.8
0
5
10
15
Time
20
25
30
35
Figure S12.10f. Closed-loop response for θ =3min.
12.11
The controller tuning is based on the characteristic equation for standard
feedback control.
1 + GcGI/PGvGpGm = 0
Thus, the PID controller will have to be retuned only if any of the transfer
functions, GI/P, Gv, Gp or Gm, change.
(a)
Km changes. The controller may have to be retuned.
(b)
The zero does not affect Gm. Thus, the controller does not require retuning.
(c)
Kv changes. Retuning may be necessary.
(d)
Gp changes. Controller may have to be retuned.
12-15
12.12
(a)
Using Table 12.4,
Kc =
0.14 0.28τ
+
K
Kθ
τI = 0.33θ +
6.8θ
10θ+τ
(b)
Comparing to the Z-N settings, the H-A settings give much smaller Kc and
slightly smaller τI, and are therefore more conservative.
(c)
The Simulink responses for the two controllers are compared in
Fig. S12.12. The controller settings are:
H-A:
Kc = 0.49 , τI =1.90
Cohen-Coon: Kc = 1.39 , τI =1.98
2
1.8
Hagglund-Astrom
Cohen and Coon
1.6
1.4
1.2
y
1
0.8
0.6
0.4
0.2
0
0
10
20
30
40
50
60
Time
Fig. S12.12. Comparison of Häggland-Åström and Cohen-Coon
controller settings.
12-16
From Fig. S12.12, it is clear that the H-A parameters provide a better setpoint response, although they produce a more sluggish disturbance
response.
12.13
From the solution to Exercise 12.5, the process reaction curve method
yields
K = 1.65
θ = 1.7 min
τ = 5.5 min
(a)
Direct Synthesis method:
From Table 12.1, Controller G:
Kc =
1 τ
1
5.5
=
= 0.94
K τc + θ 1.65 (5.5 / 3) + 1.7
τI = τ = 5.5 min
(b)
Ziegler-Nichols settings:
G (s ) =
1.65e −1.7 s
5 .5 s + 1
In order to find the stability limits, consider the characteristic equation
1 + GcG = 0
1 − 0.85s
Substituting the Padé approximation, e − s ≈
, gives:
1 + 0.85s
1.65K c (1 − 0.85s )
1 + GcG = 1 +
4.675s 2 + 6.35s + 1
or
4.675s2 + (6.35 –1.403Kc)s + 1 + 1.65Kc = 0
Substitute s = jωu and Kc = Kcu,
− 4.675 ωu2 + j(6.35 − 1.403Kcu)ωu + 1 +1.65Kcu = 0 + j0
Equating real and imaginary coefficients gives,
12-17
(6.35 − 1.403Kcu)ωu = 0 , 1+ 1.65Kcu − 4.675 ωu2 = 0
Ignoring ωu = 0, Kcu = 4.526 and ωu = 1.346 rad/min. Thus,
Pu =
2π
= 4.67 min
ωu
ThePI settings from Table 12.6 are:
ZieglerNichols
Kc
τI (min)
2.04
3.89
The ultimate gain and ultimate period can also be obtained using
Simulink. For this case, no Padé approximation is needed and the results
are:
Kcu = 3.76
Pu = 5.9 min
The PI settings from Table 12.6 are:
ZieglerNichols
Kc
τI (min)
1.69
4.92
Compared to the Z-N settings, the Direct Synthesis settings result in
smaller Kc and larger τI. Therefore, they are more conservative.
12.14
2e − s
Gv G p Gm =
5s + 1
To find stability limits, consider the characteristic equation:
or
1 + GcGvGpGm = 0
1+
2 K c (1 − 0.5s )
2.5s 2 + 5.5s + 1
=0
12-18
Substituting a Padé approximation, e − s ≈
1 − 0.5s
, gives:
1 + 0.5s
2.5s2 + (5.5 –Kc)s + 1 + 2Kc = 0
Substituting s = jωu and Kc = Kcu.
− 2.5 ωu2 + j(5.5 − Kcu)ωu + 1 +2Kcu = 0 + j0
Equating real and imaginary coefficients,
(5.5 − Kcu)ωu = 0 , 1+ 2Kcu − 2.5 ωu2 = 0
Ignoring ωu= 0, Kcu = 5.5 and ωu= 2.19. Thus,
Pu =
2π
= 2.87
ωu
Controller settings (for the Padé approximation):
Kc
τI
τD
Ziegler-Nichols
3.30
1.43
0.36
Tyreus-Luyben
2.48
6.31
0.46
The ultimate gain and ultimate period could also be found using Simulink.
For this approach, no Padé approximation is needed and:
Ku = 4.26
Pu = 3.7
Controller settings (exact method):
Kc
τI
τD
Ziegler-Nichols
2.56
1.85
0.46
Tyreus-Luyben
1.92
8.14
0.59
The set-point responses of the closed-loop systems for these controller
settings are shown in Fig. S12.14.
12-19
2
1.8
Hagglund-Astrom
Cohen and Coon
1.6
1.4
1.2
y
1
0.8
0.6
0.4
0.2
0
0
10
20
30
40
50
60
Time
Figure S12.14. Closed-loop responses for a unit step change in the set point.
12.15
Eliminate the effect of the feedback control loop by opening the loop. That
is, operate temporarily in open loop by switching the controller to the
manual mode. This action provides a constant controller output signal. If
oscillations persist, they must be due to external disturbances. If the
oscillations vanish, they were caused by the feedback loop.
12.16
The sight glass observation confirms that the liquid level is actually rising.
Since the controller output is saturated in response to the rising level, the
controller is working properly. Thus, either the actual feed flow is higher
than recorded, or the actual liquid flow is lower than recorded, or both.
Because the flow transmitters consist of orifice plates and differential
pressure transmitters, a plugged orifice plate could lead to a higher
recorded flow. Thus, the liquid-flow-transmitter orifice plate would be the
prime suspect.
12-20
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  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, Â 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
Revised: 1-3-04
Chapter 14
14.1
Let GOL(jωc) = R + jI
where ωc is the critical frequency. Then, according to the Bode stability
criterion
| GOL(jωc)| = 1 = R 2 + I 2
∠GOL(jωc) = -π = tan –1 (I/R)
Solving for R and I: R = -1 and I = 0
Substituting s = jωc into the characteristic equation gives,
1 + GOL(jωc) = 0
I + R + jI = 0
or
R = -1 , I = 0
Hence, the two approaches are equivalent.
14.2
Because sustained oscillations occur at the critical frequency
2π
= 0.628 min −1
10 min
Using Eq. 14-7,
ωc =
(a)
1 = (Kc)(0.5)(1)(1.0) or Kc = 2
(b)
Using Eq. 14-8,
– π= 0 + 0 +(-θωc) + 0
or θ =
π
ωc
= 5 min
Solution Manual for Process Dynamics and Control, 2nd edition,
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp
14-1
14.3
(a)
From inspection of the Bode diagrams in Tables 13.4 and 13.5, the
transfer function is selected to be of the following form
G(s) =
K (τ a s + 1)
s (τ1 s + 1)(τ 2 s + 1)
where τa, τ1, τ2 correspond to frequencies of ω = 0.1, 2, 20 rad/min,
respectively.
Therefore, τa = 1/0.1 = 10 min
τ1=1/2 = 0.5 min
τ2= 1/20 = 0.05 min
For low frequencies, AR ≈ |K/s| = K/ω
At ω = 0.01, AR = 3.2, so that K = (ω)(AR) = 0.032
Therefore,
G(s) =
(b)
0.032(10s + 1)
s(0.5s + 1)(0.05s + 1)
Because the phase angle does not cross -180°, the concept of GM is
meaningless.
14.4
The following process transfer can be derived in analogy with Eq. 6-71:
H1 ( s )
R1
=
Q1 ( s ) ( A1R1 A2 R2 ) s 2 + ( A1R1 + A2 R1 + A2 R2 ) s + 1
For R1=0.5, R2 = 2, A1 = 10, A2 = 0.8:
14-2
Gp(s) =
0.5
8s + 7 s + 1
For R2 = 0.5:
(a)
(1)
2
Gp(s) =
0.5
2s + 5.8s + 1
(2)
2
For R2 = 2
 − 7ωc 
∠Gp= tan-1 
2 
1 − 8ωc 
For Gv = Kv = 2.5,
For Gm =
1.5
,
0.5s + 1
ϕv=0,
,
|Gp| =
0.5
(1 − 8ωc ) + (7ωc ) 2
2 2
|Gv| = 2.5
ϕm= -tan-1(0.5ω) , |Gm| =
1 .5
(0.5ωc ) 2 + 1
Kcu and ωc are obtained using Eqs. 14-7 and 14-8:
 − 7ωc 
-180° = 0 + 0 + tan-1 
− tan-1(0.5ωc)
2 
1 − 8ωc 
Solving, ωc = 1.369 rad/min.

0.5
1 = ( K cu )(2.5)

2 2
2
 (1 − 8ωc ) + (7ωc )


1.5


  (0.5ω ) 2 + 1 
c


Substituting ωc = 1.369 rad/min, Kcu = 10.96, ωcKcu = 15.0
For R2=0.5
 − 5.8ωc 
∠Gp = tan-1 
2 
1 − 2ωc 
,

0.5
|Gp| = 

2 2
2
 (1 − 2ωc ) + (5.8ωc )
 − 5.8ωc 
− tan-1(0.5ωc)
-180° = 0 + 0 + tan-1 
2 
1 − 2ωc 
Solving, ωc = 2.51 rad/min.
Substituting ωc = 2.51 rad/min, Kcu = 15.93, ωcKcu = 40.0
14-3




(a)
From part (a), for R2=2,
ωc = 1.369 rad/min,
2π
= 4.59 min
Pu =
ωc
Kcu = 10.96
Using Table 12.6, the Ziegler-Nichols PI settings are
Kc = 0.45 Kcu = 4.932
, τI= Pu/1.2 = 3.825 min
Using Eqs. 13-63 and 13-62 ,
ϕc= -tan-1(-1/3.825ω)
2
 1 
|Gc| = 4.932 
 +1
 3.825ω 
Then, from Eq. 14-7
 − 7ωc 
 −1 
-1
− tan-1(0.5ωc)
-180° = tan-1 
 + 0 + tan 
2 
1 − 8ωc 
 3.825ωc 
Solving, ωc = 1.086 rad/min.
Using Eq. 14-8,
Ac = AROL|ω=ωc =


1

=  4.932 

 3.825ω c



1.5


 (0.5ω ) 2 + 1 
c


2



0 .5
 + 1 (2.5)

2 2
2


 (1 − 8ω c ) + (7ωc )

= 0.7362
Therefore, gain margin GM =1/Ac = 1.358.
Solving Eq.(14-16) for ωg
AROL|ω=ωc = 1
at
ωg = 0.925
14-4




Substituting into Eq. 14-7 gives ϕg=ϕ|ω=ωg = −172.7°.
Therefore, phase margin PM = 180+ ϕg = 7.3°.
14.5
(a)
K=2 , τ = 1 , θ = 0.2 , τc=0.3
Using Eq. 12-11, the PI settings are
1 τ
=1
K θ + τc
Kc =
τI = τ = 1 min,
,
Using Eq. 14-8 ,
 −1
-180° = tan-1   − 0.2ωc − tan-1(ωc) = -90° − 0.2ωc
 ωc 
or
ωc =
π/2
= 7.85 rad/min
0.2
Using Eq. 14-7,
Ac = AR OL
ω= ωc
=
1
ωc
2

 2
2
=
+1 
= 0.255

 ω
2
c
ω
+
1
c


From Eq. 14-11, GM = 1/Ac = 3.93.
(b)
Using Eq. 14-12,
ϕg = PM − 180° = − 140 ° = tan-1(-1/0.5ωg) − 0.2ωg − tan-1(ωg)
Solving by trial and error, ωg = 3.04 rad/min
AR OL
ω= ω g
= 1 = Kc
 1

 0.5ω
g

2

 +1




2




2
 ωg + 1 


Substituting for ωg gives Kc = 1.34. Then from Eq. 14-8
14-5
 −1
−180° = tan-1 
 0.5ωc

 − 0.2ωc − tan-1(ωc)

Solving by trial and error, ωc =7.19 rad/min.
From Eq. 14-7,
Ac = AR OL
ω = ωc
 1
= 1.34 
 0.5ωc
2

 + 1



2
 = 0.383



2
 ωc + 1 
From Eq. 14-11, GM = 1/Ac = 2.61
(c)
By using Simulink-MATLAB, these two control systems are compared for
a unit step change in the set point.
1.4
1.2
1
0.8
y
part (a)
part (b)
0.6
0.4
0.2
0
0
1
2
3
4
5
time
Fig S14.5. Closed-loop response for a unit step change in set point.
The controller designed in part a) (Direct Synthesis) provides better
performance giving a first-order response. Part b) controller yields a large
overshoot.
14-6
14.6
(a)
Using Eqs. 14-7 and 14-8,
AR OL =


Ym 
4ω 2 + 1  
2
0.4
= Kc


 (1.0)
2
  0.25ω 2 + 1   ω 25ω 2 + 1 
Ysp 
+
0.01
ω
1


ϕ= tan-1(2ω) − tan-1(0.1ω) − tan-1(0.5ω) – (π/2) − tan-1(5ω)
Bode Diagram
2
10
0
AR/Kc
10
-2
10
-4
10
-90
Phase (deg)
-135
-180
-225
-270
-2
-1
10
0
10
10
1
10
Frequency (rad/sec)
Figure S14.6a. Bode plot
(b)
Using Eq.14-12
ϕg = PM – 180° = 30°− 180° = −150°
ϕg = -150° at ωg = 1.72 rad/min
From the plot of ϕ vs. ω:
14-7
2
10
From the plot of
Because AR OL
ω= ω g
=1 ,
AR OL
Kc
Kc =
= 0.144
ω= ω g
1
= 6.94
0.144
From the phase angle plot:
ϕ = -180° at ωc = 4.05 rad/min
From the plot of
AR OL
vs ω,
Kc
Ac = AR OL
ω = ωc
AR OL
Kc
= 0.0326
ω = ωc
= 0.326
From Eq. 14-11, GM = 1/Ac = 3.07.
Bode Diagram
2
10
0
AR/Kc
10
-2
10
-4
10
-90
-135
Phase (deg)
(c)
AR OL
vs ω:
Kc
-180
-225
-270
-2
10
-1
0
10
10
1
10
Frequency (rad/sec)
Figure S14.6b. Solution for part (b) using Bode plot.
14-8
2
10
Bode Diagram
2
10
0
AR/Kc
10
-2
10
-4
10
-90
Phase (deg)
-135
-180
-225
-270
-2
-1
10
10
0
1
10
2
10
10
Frequency (rad/sec)
Figure S14.6c. Solution for part (c) using Bode plot.
14.7
(a)
For a PI controller, the |Gc| and ∠ Gc from Eqs. 13.62 and 13.63 need to
be included in the AR and ϕ given for GvGpGm to obtain AROL and ϕOL.
The results are tabulated below
ω
AR
|Gc|/Kc
AROL/Kc
ϕ
∠Gc
ϕOL
0.01
0.10
0.20
0.50
1.00
2.00
5.00
10.00
15.00
2.40
1.25
0.90
0.50
0.29
0.15
0.05
0.02
0.01
250
25.020
12.540
5.100
2.690
1.601
1.118
1.031
1.014
600
31.270
11.290
2.550
0.781
0.240
0.055
0.018
0.008
-3
-12
-22
-41
-60
-82
-122
-173
-230
-89.8
-87.7
-85.4
-78.7
-68.2
-51.3
-26.6
-14.0
-9.5
-92.8
-99.7
-107.4
-119.7
-128.2
-133.3
-148.6
-187.0
-239.5
From Eq. 14-12, ϕg = PM – 180° = 45°− 180° = -135°.
Interpolating the above table, ϕOL= -135° at ωg = 2.5 rad/min and
14-9
AR OL
Kc
= 0.165
ω= ω g
Because AR OL
(b)
=1 ,
ω= ω g
Kc =
1
= 6.06
0.165
From the table above,
ϕOL= -180° at ωc = 9.0 rad/min and
Ac = AR OL
ω= ωc
= 0.021
AR OL
Kc
= 0.021
ω = ωc
Kc = 0.127
From Eq. 14-11,
GM = 1/Ac = 1/0.127 = 7.86
(c)
From the table in part (a),
ϕOL= -180° at ωc = 10.5 rad/min and AR ω=ω = 0.016.
c
Therefore, Pu =
1
2π
= 0.598 min and Kcu =
= 62.5.
ωc
AR ω=ω
c
Using Table 12.6, the Ziegler-Nichols PI settings are
Kc = 0.45 Kcu = 28.1, τI = Pu/1.2 = 0.50 min
Tabulating AROL and ϕOL as in part (a) and the corresponding values of M
using Eq. 14-18 gives:
ω
0.01
0.10
0.20
0.50
1.00
2.00
5.00
10.00
15.00
|Gc|
5620
563.0
282.0
116.0
62.8
39.7
30.3
28.7
28.3
∠Gc
-89.7
-87.1
-84.3
-76.0
-63.4
-45.0
-21.8
-11.3
-7.6
AROL
13488
703
254
57.9
18.2
5.96
1.51
0.487
0.227
Therefore, the estimated value is Mp =1.64.
14-10
ϕOL
-92.7
-99.1
-106.3
-117.0
-123.4
-127.0
-143.8
-184.3
-237.6
M
1.00
1.00
1.00
1.01
1.03
1.10
1.64
0.94
0.25
14.8
Kcu and ωc are obtained using Eqs. 14-7 and 14-8. Including the filter GF
into these equations gives
-180° = 0 + [-0.2ωc − tan-1(ωc)]+[-tan-1(τFωc)]
Solving,
ωc = 8.443
ωc = 5.985
for
for
τF = 0
τF = 0.1
Then from Eq. 14-8,



2
1


1 = (K cu )



2
2
2
 ωc + 1  τ F ωc + 1 
Solving for Kcu gives,
Kcu = 4.251
Kcu = 3.536
for
for
τF = 0
τF = 0.1
for
for
τF = 0
τF = 0.1
Therefore,
ωcKcu = 35.9
ωc Kcu= 21.2
Because ωcKcu is lower for τF = 0.1, filtering the measurement results in
worse control performance.
14.9
(a)
Using Eqs. 14-7 and 14-8,




5
1
1
(1.0)




1
AR OL =  K c
+
2



2
2
25
ω
 100ω + 1  ω + 1 

ϕ = tan-1(-1/5ω) + 0 + (-2ω − tan-1(10ω)) + (- tan-1(ω))
14-11
Bode Diagram
2
10
1
AR/Kc
10
0
10
-1
10
-2
10
-100
Phase (deg)
-150
-200
-250
-300
-350
-2
10
-1
0
10
10
1
10
2
10
Frequency (rad/sec)
Figure S14.9a. Bode plot
(b)
Set ϕ = 180° and solve for ω to obtain ωc = 0.4695.
Then AR OL
= 1 = Kcu(1.025)
ω = ωc
Therefore, Kcu = 1/1.025 = 0.976 and the closed-loop system is stable for
Kc ≤ 0.976.
(c)
For Kc = 0.2, set AROL = 1 and solve for ω to obtain ωg = 0.1404.
Then ϕg = ϕ ω=ω = -133.6°
g
From Eq. 14-12, PM = 180° + ϕg = 46.4°
(d)
From Eq. 14-11
14-12
GM = 1.7 =
From part (b),
Therefore,
1
1
=
Ac
AR OL ω=ω
AR OL
ω= ωc
c
= 1.025 Kc
1.025 Kc = 1/1.7
or
Kc = 0.574
Bode Diagram
2
10
1
AR/Kc
10
0
10
-1
10
-2
10
-150
Phase (deg)
-180
-200
-250
-300
-350
-2
10
-1
0
10
10
1
10
2
10
Frequency (rad/sec)
Figure S14.9b. Solution for part b) using Bode plot.
Bode Diagram
2
10
1
AR/Kc
10
0
10
-1
10
-2
10
-150
Phase (deg)
-180
-200
-250
-300
-350
-2
10
-1
0
10
10
1
10
2
10
Frequency (rad/sec)
Figure S14.9c. Solution for part c) using Bode plot.
14-13
14.10
(a)
Gv ( s ) =
0.047
5.264
× 112 =
0.083s + 1
0.083s + 1
G p ( s) =
2
(0.432s + 1)(0.017 s + 1)
Gm ( s ) =
0.12
0.024 s + 1
Using Eq. 14-8
-180° = 0 − tan-1(0.083ωc) − tan-1(0.432ωc) − tan-1(0.017ωc)
− tan-1(0.024ωc)
Solving by trial and error, ωc = 18.19 rad/min.
Using Eq. 14-7,

 

5.624
2
⋅

1 = ( K cu )
 (0.083ω ) 2 + 1   (0.432ω ) 2 + 1 (0.017 ω ) 2 + 1 
c
c
c

 



0.12

×
 (0.024ω ) 2 + 1 
c


Substituting ωc=18.19 rad/min, Kcu = 12.97.
Pu = 2π/ωc = 0.345 min
Using Table 12.6, the Ziegler-Nichols PI settings are
Kc = 0.45 Kcu = 5.84 ,
(b)
τI=Pu/1.2 = 0.288 min
Using Eqs.13-62 and 13-63
ϕc = ∠ Gc = tan-1(-1/0.288ω)= -(π/2) + tan-1(0.288ω)
2
|Gc| = 5.84
 1 

 +1
 0.288ω 
Then, from Eq. 14-8,
14-14
-π = − (π/2) + tan-1(0.288ωc) − tan-1(0.083ωc) − tan-1(0.432ωc)
− tan-1(0.017ωc) − tan-1(0.024ωc)
Solving by trial and error, ωc = 15.11 rad/min.
Using Eq. 14-7,
2
 


5.264
⋅ 


Ac = AR OL ω=ωc
1
+

  (0.083ω ) 2 + 1 

c

 

 

2
0.12




×
⋅
 (0.432ω ) 2 + 1 (0.017ω ) 2 + 1   (0.024ω ) 2 + 1 
c
c
c

 


= 5.84



1

 0.288ωc
= 0.651
Using Eq. 14-11, GM = 1/Ac = 1.54.
Solving Eq. 14-7 for ωg gives
AR OL
ω= ω g
=1
at ωg = 11.78 rad/min
Substituting into Eq. 14-8 gives
ϕg = ϕ ω=ω = − (π/2) + tan-1(0.288ωg) − tan-1(0.083ωg) −
g
tan-1(0.432ωg) − tan-1(0.017ωg) − tan-1(0.024ωg) = -166.8°
Using Eq. 14-12,
PM = 180° + ϕg = 13.2 °
14.11
(a)
 10 

1.5
| G |= 

 (1)



2
2
+
+
ω
ω
1
100
1



ϕ = − tan-1(ω) − tan-1(10ω) − 0.5ω
14-15
Bode Diagram
2
10
1
AR
10
0
10
-1
10
0
Phase (deg)
-90
-180
-270
-2
10
-1
0
10
10
1
10
Frequency (rad/sec)
Figure S14.11a. Bode plot for the transfer function G=GvGpGm.
(b)
From the plots in part (a)
∠G = -180° at ωc = 1.4 and |G|ω=ωc = 0.62
AR OL
ω= ωc
= 1= (- Kcu) |G|ω=ωc
Therefore, Kcu = -1/0.62 = -1.61 and
Pu = 2π/ωc = 4.49
Using Table 12.6, the Ziegler-Nichols PI-controller settings are:
Kc = 0.45Kcu = -0.72
, τI = Pu/1.2 = 3.74
Including the |Gc| and ∠Gc from Eqs. 13-62 and 13-63 into the results of
part (a) gives
AR OL
2


15
 1 

= 0.72 
 + 1 

2
2
 3.74ω 
 ω + 1 100ω + 1 
14-16
=
2.89 14.0ω 2 + 1
ω 2 + 1 100ω 2 + 1 ω
ϕ = tan-1(-1/3.74ω) − tan-1(ω) − tan-1(10ω) − 0.5ω
Bode Diagram
4
10
2
AR
10
0
10
-2
10
90
Phase (deg)
0
-90
-180
-270
-360
-2
-1
10
0
10
10
1
10
Frequency (rad/sec)
Figure S14.11b. Bode plot for the open-loop transfer function GOL=GcG.
(c)
From the graphs in part (b),
ϕ = -180° at ωc=1.15
AR OL
ω= ωc
= 0.63 < 1
Hence, the closed-loop system is stable.
14-17
Bode Diagram
4
10
2
AR
10
0
10
-2
10
Phase (deg)
-90
-180
-270
-2
-1
10
0
10
10
1
10
Frequency (rad/sec)
`
Figure S14.11c. Solution for part (c) using Bode plot.
(d)
From the graph in part b),
AR OL
ω= 0.5
= 2.14 =
amplitude of ym (t )
amplitude of ysp (t )
Therefore, the amplitude of ym(t) = 2.14 × 1.5 = 3.21.
(e)
From the graphs in part (b), AR OL
ω= 0.5
= 2.14 and ϕ ω=0.5 =-147.7°.
Substituting into Eq. 14-18 gives M = 1.528. Therefore, the amplitude of
y(t) = 1.528 × 1.5 = 2.29 which is the same as the amplitude of ym(t)
because Gm is a time delay.
(f)
The closed-loop system produces a slightly smaller amplitude for ω = 0.5.
As ω approaches zero, the amplitude approaches one due to the integral
control action.
14-18
14.12
(a)
Schematic diagram:
TC
Hot fluid
TT
Cold fluid
Mixing Point
Sensor
Block diagram:
(b)
GvGpGm = Km = 6 mA/mA
GTL = e-8s
GOL = GvGpGmGTL = 6e-8s
If GOL = 6e-8s,
| GOL(jω) | = 6
∠ GOL (jω) = -8ω rad
Find ωc: Crossover frequency generates − 180° phase angle = − π radians
-8ωc = -π
or
ωc = π/8 rad/s
14-19
2π 2π
=
= 16s
ωc π / 8
1
1
Find Kcu: Kcu =
= = 0.167
| G p ( jω c ) | 6
Find Pu:
Pu =
Ziegler-Nichols ¼ decay ratio settings:
PI controller:
Kc = 0.45 Kcu = (0.45)(0.167) = 0.075
τI = Pu/1.2 = 16/1.2 = 13.33 sec
PID controller:
Kc = 0.6 Kcu = (0.6)(0.167) = 0.100
τI = Pu/2 = 16/2 = 8 s
τD = Pu/8 = 16/8 = 2 s
(c)
1.4
1.2
1
y
0.8
PID control
PI control
0.6
0.4
0.2
0
0
30
60
90
120
t
Fig. S14.12. Set-point responses for PI and PID control.
14-20
150
(d)
Derivative control action reduces the settling time but results in a more
oscillatory response.
(a)
From Exercise 14.10,
14.13
Gv ( s ) =
5.264
0.083s + 1
2
(0.432s + 1)(0.017 s + 1)
0.12
Gm ( s) =
(0.024s + 1)
1 

The PI controller is Gc ( s ) = 51 +

 0.3s 
Hence the open-loop transfer function is
G p ( s) =
GOL = Gc Gv G p Gm
Rearranging,
GOL =
6.317 s + 21.06
1.46 × 10 s + 0.00168s 4 + 0.05738s 3 + 0.556s 2 + s
−5
5
14-21
By using MATLAB, the Nyquist diagram for this open-loop system is
Nyquist Diagram
1
0.5
0
Imaginary Axis
-0.5
-1
-1.5
-2
-2.5
-3
-3.5
-4
-3
-2.5
-2
-1.5
-1
-0.5
0
Real Axis
Figure S14.13a. The Nyquist diagram for the open-loop system.
Gain margin = GM =
(b)
1
AR c
where ARc is the value of the open-loop amplitude ratio at the critical
frequency ωc. By using the Nyquist plot,
Nyquist Diagram
1
0.5
0
Imaginary Axis
-0.5
-1
-1.5
-2
-2.5
-3
-3.5
-4
-3
-2.5
-2
-1.5
-1
-0.5
Real Axis
Figure S14.13b. Graphical solution for part (b).
14-22
0
θ = -180
⇒
ARc = | G(jωc)| = 0.5
Therefore the gain margin is GM = 1/0.5 = 2.
14.14
1
, we must calculate Mp based on the CLTF
ω
Mp
with IMC controller design. In order to determine a reference Mp, we
~
assume a perfect process model (i.e. G − G = 0 ) for the IMC controller
design.
To determine max | em | <
∴
C
*
= Gc G
R
Factoring,
~ ~ ~
G = G+ G−
~
G+ = e − s
∴
*
Gc =
,
2s + 1
f
10
10
~
G− =
2s + 1
Filter Design: Because τ = 2 s, let τc = τ/3 = 2/3 s.
⇒
*
f =
1
2 3s +1
2s + 1
1
2s + 1
=
10 2 3 s + 1 20 3 s + 10
∴
Gc =
∴
 2 s + 1   10e − s 
10e − s
C
*

 
= Gc G = 
=
R
 20 3 s + 10   2 s + 1  20 3s + 10
∴
M p =1
The relative model error with K as the actual process gain is:
14-23
 Ke − s  10e − s 
~ 
−

G − G  2s + 1  2s + 1 K − 10
∴
em = ~ =
=
10
10e − s
G
2s + 1
K − 10
<1
Since Mp = 1, max | em | =
ω
10
⇒
∴
K − 10
<1
10
⇒
K < 20
K − 10
> −1
10
⇒
K > 0
0 < K < 20
for guaranteed closed-loop stability.
14.15
Denote the process model as,
~ 2e −0.2 s
G=
s +1
and the actual process as:
G=
2e −0.2 s
τs + 1
The relative model error is:
∴
~
G ( s ) − G ( s ) (1 − τ) s
∆( s) =
=
~
τs + 1
G ( s)
Let s = jω. Then,
∴
∆ =
(1 − τ) jω | (1 − τ)ω |
=
τjω + 1
| τjω + 1 |
14-24
(1)
or
∆ =
| (1 − τ ) | ω
τ 2ω 2 + 1
Because | ∆ | in (1) increases monotonically with ω,
max | ∆ | = lim | ∆ | =
ω
ω→∞
|1− τ |
τ
(2)
Substituting (2) and Mp = 1.25 into Eq. 14-34 gives:
|1− τ |
< 0.8
τ
This inequality implies that
1− τ
< 0.8
τ
⇒
1 < 1.8τ
⇒
τ > 0.556
τ −1
< 0.8
τ
⇒
0.2τ < 1
⇒
τ<5
and
Thus, closed-loop stability is guaranteed if
0.556 < τ < 5
14-25
Revised: 1-3-04
Chapter 15
15.1
For Ra=d/u
Kp =
∂Ra
d
=− 2
∂u
u
which can vary more than Kp in Eq. 15-2, because the new Kp depends on
both d and u.
15.2
By definition, the ratio station sets
(um – um0) = KR (dm – dm0)
Thus
u − u m0 K 2u 2 K 2  u 
=
=
KR = m
 
d m − d m0 K1d 2 K1  d 
2
(1)
For constant gain KR, the values of u and d in Eq. 1 are taken to be at the
desired steady state so that u/d=Rd, the desired ratio. Moreover, the
transmitter gains are
K1 =
(20 − 4)mA
Sd
2
K2 =
,
(20 − 4)mA
Su
2
Substituting for K1, K2 and u/d into (1) gives:
KR =
Su
2
Sd
2
Rd
2
 S 
=  Rd d 
Su 

2
Solution Manual for Process Dynamics and Control, 2nd edition,
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp
15-1
15.3
(a)
The block diagram is the same as in Fig. 15.11 where Y ≡ H2, Ym ≡ H2m,
Ysp ≡ H2sp, D ≡ Q1, Dm ≡ Q1m, and U ≡ Q3.
b)
(A steady-state mass balance on both tanks gives
0 = q1 – q3
or
Q1 = Q3 (in deviation variables)
(1)
From the block diagram, at steady state:
Q3 = Kv Kf Kt Q1
From (1) and (2), Kf =
1
K v Kt
(2)
c)
(No, because Eq. 1 above does not involve q2.
(b)
From the block diagram, exact feedforward compensation for Q1 would
result when
15.4
Q1 + Q2 = 0
Substituting
Gf = −
Q2 = KV Gf Kt Q1,
1
K v Kt
15-2
(c)
Same as part (b), because the feedforward loop does not have any dynamic
elements.
(d)
For exact feedforward compensation
Q4 + Q3 = 0
From the block diagram,
(1)
Q2 = KV Gf Kt Q4
(2)
Using steady-state analysis, a mass balance on tank 1 for no variation in q1
gives
Q2 − Q3 = 0
(3)
Substituting for Q3 from (3) and (2) into (1) gives
Q4 + KV Gf Kt Q4 = 0
Gf = −
or
1
K v Kt
For dynamic analysis, find Gp1 from a mass balance on tank 1,
A1
dh1
= q1 + q2 − C1 h1
dt
15-3
Linearizing (4), noting that q1′ = 0, and taking Laplace transforms:
A1
dh′
C
= q2′ − 1 h1′
dt
2 h
1
or
(2 h1 / C1)
H1′ ( s)
=
Q2′ ( s) (2 A h / C ) s + 1
1 1 1
q3 = C1 h1
Since
q3′ =
C1
2 h1
or
h1′
(5)
Q3′ ( s )
C
= 1
H1′ ( s) 2 h
1
(6)
From (5) and (6),
Q3′ ( s)
1
=
= GP1
Q2′ ( s) (2 A h / C ) s + 1
1 1 1
Substituting for Q3 from (7) and (2) into (1) gives
Q4 + (
or
1
) K v G f Kt Q4 = 0
(2 A1 h1 / C1 ) s + 1
Gf = −
1
[(2 A1 h1 / C1 ) s + 1]
K v Kt
15.5
(a)
For a steady-state analysis:
Gp=1,
Gd=2,
From Eq.15-21,
Gf =
(b)
− Gd
−2
=
= −2
Gv Gt G p (1)(1)(1)
Using Eq. 15-21,
15-4
Gv = Gm = Gt =1
(7)
−2
− Gd
−2
( s + 1)(5s + 1)
=
Gf =
=
5s + 1
G v Gt G p
 1 
(1)(1)

 s +1
(c)
Using Eq. 12-19,
where
1
~
~ ~
G = Gv G p G m =
= G+ G−
s +1
1
G + = 1, G − =
s +1
For τc=2, and r=1, Eq. 12-21 gives
f=
1
2s + 1
From Eq. 12-20
 1  s +1
Gc* = G − −1 f = ( s + 1) 
=
 2s + 1  2s + 1
From Eq. 12-16
s +1
s +1
= 2s + 1 =
Gc =
∗~
2s
1 − Gc G 1 − 1
2s + 1
Gc
(d)
∗
For feedforward control only, Gc=0. For a unit step change in disturbance,
D(s) = 1/s.
Substituting into Eq. 15-20 gives
Y(s) = (Gd+GtGfGvGp)
1
s
For the controller of part (a)

2
 1  1
+ (1)(−2)(1)
Y(s) = 

 s + 1  s
 ( s + 1)(5s + 1)
15-5
Y(s) =
or
5 / 2 − 25 / 2 2.5
− 2.5
− 10
=
−
=
+
( s + 1)(5s + 1) s + 1 5s + 1 s + 1 s + 1 / 5
y(t) = 2.5 (e-t – e- t/5)
For the controller of part (b)

2
 − 2   1  1
+ (1)
Y(s) = 
(1)
 = 0
 5s + 1   s + 1   s
 ( s + 1)(5s + 1)
or
y(t) = 0
The plots are shown in Fig. S15.5a below.
Figure S15.5a. Closed-loop response using feedforward control only.
(e)
Using Eq. 15-20:
For the controller of parts (a) and (c),
2

 1 
 ( s + 1)(5s + 1) + (1)(−2)(1)  s + 1   1


Y(s) = 
 s +1  1 

s
1+ 
 (1) 
 (1)


 2s   s + 1 
15-6
5
25 / 3 − 40 / 3
− 20s
=
+
+
( s + 1)(5s + 1)(2s + 1) s + 1 5s + 1 2 s + 1
5
20 / 3
5/3
=
−
+
s + 1 s + 1/ 2 s + 1/ 5
or
Y(s) =
or
y(t) = 5e-t –
20 -t/2 5 - t/5
e + e
3
3
and for controllers of parts (b) and (c)

2
 − 2   1 
 ( s + 1)(5s + 1) + (1) 5s + 1 (1) s + 1   1

 

=0
Y(s) = 

 s +1  1 
s
1+ 
(1)
(1)


 2s   s + 1 


or
y(t) = 0
The plots of the closed-loop responses are shown in Fig. S15.5b.
Figure S15.5b. Closed-loop response for feedforward-feedback control.
15.6
(a)
The steady-state energy balance for both tanks takes the form
0 = w1 C T1 + w2 C T2 − w C T4 + Q
15-7
where Q is the power input of the heater
C is the specific heat of the fluid.
Solving for Q and replacing unmeasured temperatures and flow rates by
their nominal values,
Q = C ( w1T1 + w 2 T 2 − wT 4 )
(1)
Neglecting heater and transmitter dynamics,
Q = Kh p
(2)
T1m = T1m0 + KT(T1-T10)
(3)
wm = wm0 + Kw(w-w0)
(4)
Substituting into (1) for Q, T1, and w from (2), (3), and (4), gives
p=
(b)
C
1
1
[ w1 (T10 +
(T1m − T1m0 )) + w2 T2 − T4 ( w0 +
( wm − wm0 ))]
Kh
KT
Kw
Dynamic compensation is desirable because the process transfer function
Gp= T4(s)/P(s) is different from each of the disturbance transfer functions,
Gd1= T4(s)/T1(s), and Gd2= T4(s)/w(s); this is more so for Gd1 which has a
higher order.
15.7
(a)
(b)
A steady-state material balance for both tanks gives,
15-8
0 = q1 + q2 + q4 − q5
Because q ′2 = q ′4 = 0, the above equation gives
0 = q1′ – q5′ or 0 = Q1 – Q5
(1)
From the block diagram,
Q5 = Kv Gf Kt Q1
Substituting for Q5 into (1) gives
0 = Q1 − Kv Gf Kt Q1
(c)
or
Gf =
1
Kv Kt
To find Gd and Gp, the mass balance on tank 1 is
A1
dh1
= q1 + q 2 − C1 h1
dt
where A1 is the cross-sectional area of tank 1.
Linearizing and setting q 2′ = 0 leads to
A1
dh1′
C
= q1′ − 1 h1′
dt
2 h1
Taking the Laplace transform,
H1′ ( s)
R1
=
Q1′ ( s) A1R1s + 1
Linearizing q3 = C1 h1
q3′ =
1
h1′
R1
or
where R1 ≡
C1
(2)
gives
Q3′ ( s) 1
=
H1′ ( s) R1
Mass balance on tank 2 is
A2
2 h1
dh2
= q3 + q 4 − q 5
dt
15-9
(3)
Using deviation variables, setting q ′4 = 0, and taking Laplace transform
A2 sH 2′ ( s ) = Q3′ ( s ) − Q5′ ( s )
H 2′ ( s)
1
=
Q3′ ( s ) A2 s
(4)
and
H 2′ ( s )
1
=−
= G p ( s)
Q5′ ( s)
A2 s
Gd ( s) =
H 2′ ( s) H 2′ ( s) Q3′ ( s) H1′ ( s )
1
=
=
Q1′ ( s ) Q3′ ( s ) H1′ ( s ) Q1′ ( s ) A2 s ( A1R 1 s + 1)
upon substitution from (2), (3), and (4).
Using Eq. 15-21,
1
− Gd
A2 s ( A1 R1 s + 1)
Gf =
=
Gt Gv G p
K t K v (−1 / A2 s )
−
=+
1
1
K v K t A1 R1 s + 1
15.8
For the process model in Eq. 15-22 and the feedforward controller in Eq.
15-29, the correct values of τ1 and τ2 are given by Eq. 15-42 and (15-43).
Therefore,
τ1 − τ2 = τp − τL
(1)
for a unit step change in d, and no feedback controller, set D(s)=1/s, and
Gc= 0 in Eq. 15-20 to obtain
[
Y(s) = Gd + Gt G f Gv G p
]1s
Setting Gt = Gv = 1, and using Eqs. 15-22 and 15-29,
15-10
 K
 K p  1
d + (1)  − K d / K P (τ1s + 1)  (1)
Y (s) = 


 
τ2s + 1
 τ d s + 1

  τ p s + 1   s
1
(τ1 − τ p )τ p 1 
τ
1 τ (τ − τ ) 1
= Kd  − d − − 2 1 2
−

τ p − τ 2 τ p s + 1 
 s τ d s + 1 s (τ 2 − τ p ) τ 2 s + 1

τ1 − τ p −t / τ p 
(τ − τ )
y(t) = K d  −e −t / τ − 1 2 e −t / τ 2 −
e

τ2 − τ p
τ p − τ2


or
∞
∫0
e(t )dt = ∫
=
∞
0

τ (τ − τ ) τ p (τ1 − τ p ) 
y (t )dt = − K d  τ d + 2 1 2 +

τ2 − τ p
τ p − τ 2 

−Kd 
τ d τ 2 − τ d τ p + τ 2 τ1 − τ 2 2 − τ p τ1 + τ p 2 + (τ p τ 2 − τ p τ 2 ) 


τ2 − τ p
= − K d  (τ1 − τ 2 ) − (τ p − τ d ) 
= 0 when (1) holds.
15.9
(a)
For steady-state conditions
Gp=1,
Gd=2,
Gv = Gm = Gt =1
Using Eq. 15-21,
Gf =
(b)
− Gd
−2
=
= −2
Gv Gt G p (1)(1)(1)
Using Eq. 15-21,
15-11
− 2e − s
− Gd
−2
( s + 1)(5s + 1)
=
Gf =
=
G v Gt G p
 1  − s 5s + 1
(1)(1)
e
 s + 1
(c)
Using Eq. 12-19,
e −s
~
~ ~
G = Gv G p G m =
= G+ G−
s +1
1
~
~
,
where
G+ = e − s
G− =
s +1
For τc=2, and r = 1, Eq. 12-21gives
f=
1
2s + 1
From Eq. 12-20
1
s +1
1
Gc * = ~ f = ( s + 1)
=
2s + 1 2s + 1
G−
From Eq. 12-16
s +1
2s + 1 = s + 1
Gc =
~=
2s
1 − Gc * G 1 − 1
2s + 1
Gc *
(d)
For feedforward control only, Gc=0. For a unit step disturbance,
D(s) = 1/s.
Substituting into Eq. 15-20 gives
Y(s) = (Gd+GtGfGvGp)
1
s
For the controller of part (a)

 e − s  1
2e − s

+ (1)(−2)(1)
Y(s) = 
 s + 1  s
 ( s + 1)(5s + 1)
15-12
=
or
− 10e − s
( s + 1)(5s + 1)
y(t) = 2.5 (e-(t-1) – e-( t-1)/5)S(t-1)
For the controller of part (b)
−s

2e − s
 − 2   e  1
 = 0
+ (1)
Y(s) = 
(1)
 5s + 1   s + 1   s
 ( s + 1)(5s + 1)
or
y(t) = 0
The plots are shown in Fig. S15.9a below.
Figure S15.9a. Closed-loop response using feedforward control only.
(e)
Using Eq. 15-20:
For the controllers of parts (a) and (c),

 e −s  
2e − s

+ (1)(−2)(1)

s + 1   1
( s + 1)(5s + 1)


Y(s) =
−s

s
 s + 1  e 
(1)
1+ 


(1)
 2s   s + 1 


15-13
and for the controllers of parts (b) and (c),

2
 − 2   1 
 ( s + 1)(5s + 1) + (1) 5s + 1 (1) s + 1   1


 
=0
Y(s) = 

 s + 1  1 
s
1+ 
(1)
(1)


 2s   s + 1 


or
y(t) = 0
The plots of the closed-loop responses are shown in Fig. S15.9b.
Figure S15.9b. Closed-loop response for the feedforward-feedback control.
15.10
(a)
For steady-state conditions
Gp=Kp,
Gd=Kd,
Gv = Gm = Gt =1
Using Eq. 15-21,
15-14
Gf =
− Gd
− 0.5
=
= −0.25
Gv Gt G p (1)(1)(2)
(b)
Using Eq. 15-21,
(c)
− 0.5e −30 s
− Gd
(95s + 1) −10 s
60 s + 1
=
= −0.25
e
Gf =
− 20 s
(60s + 1)
G v Gt G p
 2e


(1)(1)
 95s + 1 
Using Table 12.1, a PI controller is obtained from equation G,
Kc =
τ
1
1
95
=
= 0.95
K p τ c + θ 2 (30 + 20)
τ I = τ = 95
(d)
As shown in Fig.S15.10a, the dynamic controller provides significant
improvement.
(e)
0.08
Controller of part (a)
Controller of part (b)
0.06
0.04
y(t)
0.02
0
-0.02
-0.04
0
100
200
300
400
500
time
Figure S15.10a. Closed-loop response using feedforward control only.
15-15
0.06
Controllers of part (a) and (c)
Controllers of part (b) and (c)
0.04
0.02
y(t)
0
-0.02
-0.04
-0.06
0
100
200
300
400
500
time
Figure S15.10b. Closed-loop response for feedforward-feedback control.
f)
As shown in Fig. S15.10b, the feedforward configuration with the
dynamic controller provides the best control.
15.11
Energy Balance:
ρVC
dT
= wC (Ti − T ) − U (1 + q c ) A(T − Tc ) − U L AL (T − Ta ) (1)
dt
Expanding the right hand side,
dT
= wC (Ti − T ) − UA(T − Tc )
dt
− UAqc T + UAqc Tc − U L AL (T − Ta )
ρVC
(2)
Linearizing,
q cT ≈ q cT + q cT ′ + T qc′
Substituting (3) into (2), subtracting the steady-state equation, and
introducing deviation variables,
15-16
(3)
dT ′
= wC (Ti′ − T ′) − UAT ′ − UAT q c′ − UAq c T ′
dt
+ UATc qc′ − U L ALT ′
ρVC
(4)
Taking the Laplace transform and assuming steady-state at t = 0 gives,
ρVCsT ′( s) = wCTi′( s) + UA(Tc − T − )Qc′ ( s )
− ( wC + UA + UAq c + U L AL )T ′( s)
(5)
Rearranging,
T ′( s ) = GL ( s )Ti′( s ) + G p ( s )Qc′ ( s )
(6)
where:
Gd ( s) =
G p ( s) =
KL
τs + 1
Kp
τs + 1
wC
Kd =
K
UA(Tc − T )
Kp =
K
ρVC
τ=
K
K = wC + UA + UAq c + U L AL
(7)
The ideal FF controller design equation is given by,
GF =
−Gd
Gt GvG p
(17-27)
But, Gt = K t e − θs and Gv=Kv
(8)
Substituting (7) and (8) gives,
GF =
− wCe + θs
K t K vUA(Tc − T )
(9)
In order to have a physically realizable controller, ignore the e+θs term,
15-17
− wC
K t K vUA(Tc − T )
GF =
(10)
15.12
a)
A component balance in A gives:
V
dc A
= qc Ai − qc A − Vkc A
dt
(1)
At steady state,
0 = q c Ai − q c A − Vkc A
(2)
Solve for q ,
q=
kVC A
C Ai − C A
(3)
For an ideal FF controller, replace C Ai by C Ai , q by q1 and C A by C Asp :
q=
b)
kVC Asp
C Ai − C Asp
Linearize (1):
V
dc A
= q ciA+ qc′Ai + c Ai q′ − q c A − qc′A − c Aq′ − Vkc A
dt
Subtract (2),
V
dc′A
= qc′iA+ c Ai q′ − qc′A − c Aq′ − Vkc′A
dt
Take the Laplace transform,
sVc′A ( s ) = qc′iA( s ) + c Ai Q′( s ) − qc′A ( s) − c AQ′( s) − Vkc′A ( s)
Rearrange,
15-18
C ′A ( s) =
c −c
q
C ′Ai ( s ) + Ai A Q′( s)
sV + q + Vk
sV + q + Vk
(6)
or
C ′A ( s ) = Gd ( s )C ′Ai ( s ) + G p ( s )Q′( s )
(7)
The ideal FF controller design equation is,
GF ( s) = −
Gd ( s)
Gv ( s )G p ( s )Gt ( s )
(8)
Substitute from (6) and (7) with Gv(s)=Kv and Gt(s)=Kt :
GF ( s) = −
.
q
K v (c Ai − c A ) Kt
(9)
Note: G F ( s) = P ′( s) / C ′Ai m ( s ) where P is the controller output and cAim
is the measured value of cAi.
15.13
(a)
Steady-state balances:
0 = q5 + q1 − q3
(1)
0 = q3 + q 2 − q 4
(2)
0
0 = x5 q5 + x1 q1 − x3 q3
(3)
0 = x3 q 3 + x 2 q 2 − x 4 q 4
(4)
Solve (4) for x3 q3 and substitute into (3),
0 = x5 q5 + x 2 q 2 − x 4 q 4
Rearrange,
15-19
(5)
q2 =
x 4 q 4 − x5 q 5
x2
(6)
In order to derive the feedforward control law, let
x4 → x4 sp ,
x2 → x2 (t ),
x5 → x5 (t ),
and
q 2 → q 2 (t )
Thus,
q 2 (t ) =
x 4 sp q 4 − x5 (t )q5 (t )
x2
(7)
Substitute numerical values:
q 2 (t ) =
(3400) x 4 sp − x5 (t )q5 (t )
0.990
(8)
or
q 2 (t ) = 3434 x 4 sp − 1.01x5 (t )q 5 (t )
(9)
Note: If transmitter and control valve gains are available, then an
expression relating the feedforward controller output signal, p(t), to the
measurements , x5m(t) and q5m(t), can be developed.
(b)
Dynamic compensation: It will be required because of the extra
dynamic lag preceding the tank on the left hand side. The stream 5
disturbance affects x3 while q3 does not.
15-20
123456789 8
16.1
The difference between systems A and B lies in the dynamic lag in the
measurement elements Gm1 (primary loop) and Gm2(secondary loop). With
a faster measurement device in A, better control action is achieved. In
addition, for a cascade control system to function properly, the response of
the secondary control loop should be faster than the primary loop. Hence
System A should be faster and yield better closed-loop performance than
B.
Because Gm2 in system B has an appreciable lag, cascade control has the
potential to improve the overall closed-loop performance more than for
system A. Little improvement in system A can be achieved by cascade
control versus conventional feedback.
Comparisons are shown in Figs. S16.1a/b. PI controllers are used in the
outer loop. The PI controllers for both System A and System B are
designed based on Table 12.1 (τc = 3). P controllers are used in the inner
loops. Because of different dynamics the proportional controller gain of
System B is about one-fourth as large as the controller gain of System A
System A: Kc2 = 1
System B: Kc2 = 0.25
τI=15
τI=15
Kc1=0.5
Kc1=2.5
0.7
Cascade
Standard feedback
0.6
0.5
Output
0.4
0.3
0.2
0.1
0
0
10
20
30
40
50
60
70
80
90
100
time
Figure S16.1a. System A. Comparison of D2 responses (D2=1/s) for cascade
control and conventional PI control.
Solution Manual for Process Dynamics and Control, 2nd edition,
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp
16-1
In comparing the two figures, it appears that the standard feedback results
are essentially the same, but the cascade response for system A is much
faster and has much less absolute error than for the cascade control of B
0.7
Cascade
Standard feedback
0.6
0.5
Output
0.4
0.3
0.2
0.1
0
0
10
20
30
40
50
time
60
70
80
90
100
Figure S16.1b. System B .Comparison of D2 responses (D2=1/s) for cascade
control and conventional PI control.
Figure S16.1c. Block diagram for System A
16-2
Figure S16.1d. Block diagram for System B
16.2
a)
The transfer function between Y1 and D1 is
Y1
=
D1
Gd 1


Gc 2Gv
1 + Gc1 
 G p Gm1
1
+
G
G
G
c2 v m2 

and that between Y1 and D2 is
G p Gd 2
Y1
=
D2 1 + Gc 2Gv Gm 2 + Gc 2Gv Gm1Gc1G p
using Gv =
Gp =
5
s +1
,
Gd 2 = 1 ,
4
, Gm1 = 0.05 ,
(2 s + 1)(4s + 1)
16-3
Gd 1 =
1
,
3s + 1
Gm 2 = 0.2
For Gc1 = Kc1 and Gc2 = Kc2, we obtain
Y1
8s 3 + (14 + 8 K c 2 ) s 2 + (7 + 6 K c 2 ) s + K c 2 + 1
=
D1 24 s 4 + (50 + 24 K c 2 ) s 3 + [10 + K c 2 (9 + 3K c1 )]s + (35 + 26 K c 2 ) s 2 + K c 2 (1 + K c1 ) + 1
Y1
4( s + 1)
= 3
2
D2 8s + (14 + 8K c 2 ) s + (7 + 6 K c 2 ) s + K c 2 (1 + K c1 ) + 1
The figures below show the step load responses for Kc1=43.3 and for
Kc2=25. Note that both responses are stable. You should recall that the
critical gain for Kc2=5 is Kc1=43.3. Increasing Kc2 stabilizes the controller,
as is predicted.
-3
1
12
x 10
0.8
10
0.6
8
0.4
6
Output
Output
0.2
0
4
-0.2
2
-0.4
0
-0.6
-0.8
0
5
10
15
time
20
25
30
-2
0
5
10
15
time
20
25
30
Figure S16.2a. Responses for unit load change in D1 (left) and D2 (right)
b)
The characteristic equation for this system is
1+Gc2GvGm2+Gc2GvGm1Gc1Gp = 0
(1)
Let Gc1=Kc2 and Gc2=Kc2. Then, substituting all the transfer functions into
(1), we obtain
8s 3 + (14 + 8K c 2 ) s 2 + (7 + 6 K c 2 ) s + K c 2 (1 + K c1 ) + 1
(2)
Now we can use the Routh stability criterion. The Routh array is
Row 1
Row 2
8
14 + 8 K c 2
24 K c 2 + 66 K c 2 + 45 − 4 K c1 K c 2
7 + 4K c2
1 + K c 2 (1 + K c1 )
7 + 6K c2
1 + K c 2 (1 + K c1 )
2
Row 3
Row 4
16-4
0
For 1 ≤ Kc2≤ 20, there is no impact on stability by the term 14+8Kc2 in the
second row. The critical Kc1 is found by varying Kc2 from 1 to 20, and
using
24 K c 2 + 66 K c 2 + 45 − 4 K c1 K c 2 ≥ 0
1 + K c 2 (1 + K c1 ) ≥ 0
2
(3)
(4)
Rearranging (3) and (4), we obtain
24 K c 2 + 66 K c 2 + 45
K c1 ≤
4K c2
(5)
 K + 1

K c1 ≥ − c 2
 K c2 
(6)
2
Hence, for normal (positive) values of Kc1 and Kc2,
24 K c 2 2 + 66 K c 2 + 45
K c1,u =
4Kc 2
The results are shown in the table and figure below. Note the nearly linear
variation of Kc1 ultimate with Kc2. This is because the right hand side is
very nearly 6 Kc2+16.5. For larger values of Kc2, the stability margin on
Kc1 is higher. There don’t appear to be any nonlinear effects of Kc2 on Kc1,
especially at high Kc2.
There is no theoretical upper limit for Kc2, except that large values may
cause the valve to saturate for small set-point or load changes.
Kc1,u
33.75
34.13
38.25
43.31
48.75
54.38
60.11
65.91
71.75
77.63
83.52
89.44
95.37
101.30
107.25
113.20
119.16
125.13
131.09
137.06
160.00
140.00
120.00
Kc1, ultimate
Kc2
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
100.00
80.00
60.00
40.00
20.00
0.00
0
5
10
Kc2
Figure S16.2b. Effect of Kc2 on the critical gain of Kc1
16-5
15
20
c)
With integral action in the inner loop,
Gc1 = K c1
1

Gc 2 = 51 + 
 5s 
Substitution of all the transfer functions into the characteristic equation
yields
1 5
1 5


1 + 51 + 
(0.2) + 51 + 
(0.05) K c1
 5s  s + 1
 5s  s + 1
4
=0
(4s + 1)(2s + 1)
Rearrangement gives
8s 4 + 54 s 3 + 45s 2 + (12 + 5 K c1 ) s + K c1 + 1 = 0
The Routh array is:
Row 1
8
45
1 + K c1
Row 2
54
12 + 5 K c1
0
Row 3
Row 4
1167 − 20 K c1
27
2
− 100 K c1 + 4137 K c1 + 12546
1167 − 20 K c1
1 + K c1
0
1 + K c1
Row 5
Using the Routh array analysis
Row 3:
1167 − 20 K c1 > 0
∴
K c1 < 58.35
1 + K c1 > 0
∴
K c1 > −1
Row 4: Since 1167 − 20 K c1 is already positive,
− 100 K c1 + 4137 K c1 + 12546 > 0
Solving for the positive root, we get K c1 < 43.3
2
16-6
The ultimate K c1 is 43.3, which is the same result as for proportional only
control of the secondary loop.
With integral action in the outer loop only,
1

Gc1 = K c1 1 + 
 5s 
Gc 2 = 5
Substituting the transfer functions into the characteristic equation.
1+ 5
5
5
1
4

( 0 .2 ) + 5
(0.05) K c1 1 + 
=0
s +1
s +1
 5s  (4 s + 1)(2 s + 1)
∴ 8s 4 + 54 s 3 + 37 s 2 + (6 + 5 K c1 ) s + K c1 = 0
The Routh array is
Row 1
8
Row 2
54
Row 3
Row 4
37
6 + 5K c1
975 − 20 K c1
27
2
− 100 K c1 + 3297 K c1 + 5850
975 − 20 K c1
Row 5
K c1
0
K c1
0
K c1
Using the Routh array analysis,
Row 3:
975 − 20 > 0
∴
K c1 < 48.75
K c1 > 0
Row 4: Since 975 − 20 K c1 is already positive,
− 100 K c1 + 3297 K c1 + 5850 > 0
Solving for the positive root, we get K c1 < 34.66
2
Hence, Kc1<34.66 is the limiting constraint. Note that due to integral
action in the primary loop, the ultimate controller gain is reduced.
16-7
Calculation of offset:
For
Y1
=
D1

1 
Gc1 = K c1 1 +

 τ I 1s 
Gc 2 = K c 2
,
,
(τI 2 = ∞)
Gd 1 (1 + K c 2Gv Gm 2 )

1 
1 + K c 2Gv Gm 2 + K c 2Gv Gm1 K c1 1 +
 Gp
 τ I 1s 
Y1
( s = 0) = 0
D1
Since Gc1 contains integral action, a step-change in D1 does not produce an
offset in Y1.
Y1
=
D2
G p Gd 2

1 
1 + K c 2GvGm 2 + K c 2Gv Gm1 K c1  1 +
 Gp
 τ I 1s 
Y1
( s = 0) = 0
D2
Thus, for the same reason as before, a step-change in D2 does not produce
an offset in Y1.
For
Gc1 = K c1
(ie. τ I 1 = ∞)
,

1 
Gc 2 = K c 2 1 +

 τI 2 s 

1 
Gd 1 (1 + K c 2  1 +
 Gv Gm 2 )
τI 2 s 
Y1

=
D1


1 
1 
1 + K c 2 1 +
 Gv Gm 2 + K c 2Gv Gm1 K c1 1 +
 Gp
 τI 2 s 
 τI 2 s 
Y1
( s = 0) ≠ 0
D1
Therefore, when there is no integral action in the outer loop, a primary
disturbance produces an offset.
Thus, there is no offset for a step-change in the secondary disturbance.
.
16-8
Y1
=
D2
G p Gd 2


1 
1 
1 + K c 2 1 +
 Gv Gm 2 + K c 2Gv Gm1 K c1 1 +
 Gp
 τI 2 s 
 τI 2 s 
Y1
( s = 0) = 0
D2
Thus, there is no offset for a step-change in the secondary disturbance.
16.3
Tuning the slave loop:
The open-loop transfer function is
K c2
G (s) =
(2 s + 1)(5s + 1)( s + 1)
Since a proportional controller is used, a high Kc2 reduces the steady-state
offset. The highest Kc2 which satisfies the bounds on the gain and phase
margins is 5.3. For this Kc2, the gain margin is 2.38, and the phase margin
is 30.7°.
By using MATLAB, the Bode plot of G(s) with Kc2 = 5.3 is shown below.
Bode Diagram
6
Gain margin graphical solution
Magnitude (abs)
5
Phase margin graphical solution
4
3
2
1
0
0
-45
Phase (deg)
a)
-90
-135
-180
-225
-270
-2
10
-1
0
10
10
1
10
Frequency (rad/sec)
Figure S16.3a. Bode plot for the inner open-loop; gain and phase margins.
16-9
Tuning the master loop:
The input-output transfer function of the inner loop is
Gi n ( s ) =
5.3( s + 1)
10 s + 17 s 2 + 8s + 6.3
(with Kc2 = 5.3)
3
The ultimate gain Kc1,u can be found by simulation. In doing so,
Kc1,u = 3.2491
The corresponding period of oscillation is
Pu = 2π/ω = 8.98 time units.
The Ziegler-Nichols tuning criteria for a PI-controller yield
Kc1 = Kc1,u / 2.2 = 1.48
τ I 1 = Pu / 1.2 = 7.48
The closed-loop response with these tuning constant values (Kc1=1.48,
τI 1 = 7.48 , Kc2 = 5.3) is shown below.
1.4
1.2
1
0.8
Output
b)
0.6
0.4
0.2
0
0
10
20
30
40
50
Time
60
70
80
90
100
Fig S16.3b. Closed-loop response for a unit step set-point change.
16-10
16.4
For the inner controller (Slave controller), IMC tuning rules are used
Gc 2 * =
1
(2 s + 1)(5s + 1)( s + 1)
=
−
G2
(τc 2 s + 1)3
Closed-loop responses for different values of τc2 are shown below. A τc2 value of
3 yields a good response.
For the Master controller,
Gc 1 * =
1
G1
G1− =
where
−
(2 s + 1)(5s + 1)( s + 1)
1
3
(τc1s + 1)
(10 s + 1)
This higher-order transfer function is approximated by first order plus time delay
using a step test:
1
0.9
0.8
0.7
Output
0.6
0.5
0.4
0.3
0.2
0.1
0
0
10
20
30
time
40
50
60
Figure S16.4. Reaction curve for the higher order transfer function
−
Hence G1 ≈
e −0.38 s
(15.32s + 1)
15.32
and τi = 15.32
τc1 + 0.38
Closed-loop responses are shown for different values of τc1. A τc1 value of 7
yields a good response.
From Table 12.1: (PI controller, Case G): K c =
16-11
1.4
0.9
Tauc1=3
Tauc1=5
Tauc1=7
Tauc2=0.5
Tauc2=3
Tauc2=7
Tauc2=5
0.8
1.2
0.7
1
0.6
0.8
Output
Output
0.5
0.4
0.6
0.3
0.4
0.2
0.2
0.1
0
0
0
10
20
30
time
40
50
60
0
10
20
30
40
50
time
60
70
80
90
100
Figure S16.4b. Closed-loop response for τc2 Figure S16.4c. Closed-loop response for τc1
Hence for the master controller, Kc = 2.07
and
τI = 15.32
16.5
a) The T2 controller (TC-2) adjusts the set-point, T1sp, of the T1 controller (TC1). Its output signal is added to the output of the feedforward controller.
Figure S16.5a. Schematic diagram for the control system
b)
This is a cascade control system with a feedforward controller being used
to help control T1. Note that T1 is an intermediate variable rather than a
disturbance variable since it is affected by V1.
16-12
c)
Block diagram:
Figure S16.5b. Block diagram for the control system in Exercise 16.5.
16.6
a)
For the inner loop, the characteristic equation reduces to:
1 + K inner
s +1
=0
s−3
1
s(1+ Kinner) –3 + Kinner = 0
Hence, s =
1
s − 3 + Kinner s + Kinner = 0
3 − K inner
1 + K inner
The inner loop will be stable if this root is negative. Thus, we conclude
that this loop will be stable if either Kinner>3 or Kinner<−1.
b)
The servo transfer function for the outer loop is:
Gc ( s ) K inner G p ( s )
Y (s)
=
Ysp ( s ) 1 + K inner G p ( s ) + Gc ( s ) K inner G p ( s )
16-13
The complex closed-loop poles arise when the characteristic polynomial
is factored. This polynomial is
(s2 + s + 0.313) = (s + 0.5 + 0.25 i) (s + 0.5 −0.25i)
1+ 6
 τ s +1  s +1
s +1
+ Kc  I
=0
6
s −3
 τI s  s − 3
1 ( τ I + 6τ I + K c 6 τ I ) s 2
+ (−3τ I + 6τ I + 6τ I K c + 6 K c ) s
+ Kc 6 = 0
The poles are also the roots of the characteristic equation:
Hence, the PI controller parameters can be found easily:
K c = 0.052
τ I = 0.137
16.7
Using MATLAB-Simulink, the block diagram for the closed-loop system
is shown below.
Figure S16.7a. Block diagram for Smith predictor
16-14
represents the time-delay term e-θs.
where the block
The closed-loop response for unit set-point and disturbance changes are
shown below. Consider a PI controller designed by using Table 12.1(Case
A) with τc = 3 and set Gd = Gp. Note that no offset occurs,
Servo response
Regulatory response
1.2
1
Output
0.8
0.6
0.4
0.2
0
0
5
10
15
time
20
25
30
Figure S16.7b. Closed-loop response for setpoint and disturbance changes.
16.8
The block diagram for the closed-loop system is
Figure S6.8. Block diagram for the closed-loop system
 1 + τI s 
where Gc = K c 
−θs 
 1 + τI s − e 
16-15
and
Gp =
K p e−θs
1 + τs
a)
 1 + τ I s  e −θs
Kc K p 

Gc G p
1 + τ I s − e −θs  1 + τs
Y

=
=
Ysp 1 + Gc G p
 1 + τ I s  e−θs
1 + Kc K p 
−θs 
 1 + τ I s − e  1 + τs
1
Since K c =
and
τI = τ
Kp


e −θs

−θs 
1 + τI s − e 
Y
e−θs
= 
=
Ysp

 1 + τ I s − e −θs + e −θs
e −θs
1+ 
−θs 
 1 + τI s − e 
Hence dead-time is eliminated from characteristic equation:
Y
e −θs
=
Ysp 1 + τ I s
b)
The closed-loop response will not exhibit overshoot, because it is a first
order plus dead-time transfer function.
16.9
For a first-order process with time delay, use of a Smith predictor and
proportional control should make the process behave like a first-order
system, i.e., no oscillation. In fact, if the model parameters are accurately
known, the controller gain can be as large as we want, and no oscillations
will occur.
Appelpolscher has verified that the process is linear, however it may not
be truly first-order. If it were second-order (plus time delay), proportional
control would yield oscillations for a well-tuned system. Similarly, if there
are errors in the model parameters used to design the controller even when
the actual process is first-order, oscillations can occur.
16-16
16.10
a)
Analyzing the block diagram of the Smith predictor
Gc G′p e −θs
Y
=
Ysp 1 + Gc G2 ′p (1 − e −θs ) + Gc G′p e −θs
1
=
Gc G′p e −θs
1 + Gc G2 ′p + Gc G′p e−θs − Gc G2 ′p e −θs
1
~
Note that the last two terms of the denominator can when G ′p = G ′p and
θ = θ2
The characteristic equation is
= 1 + Gc G2 ′p + Gc G′p e−θs − Gc G2 ′p e −θs = 0
1
b)
The closed-loop responses to step set-point changes are shown below for
the various cases.
Figure S16.10a. Simulink diagram block; base case
16-17
1.4
1
0.9
1.2
0.8
1
0.7
0.8
Output
Output
0.6
0.5
0.6
0.4
0.3
0.4
0.2
0.2
0.1
0
0
0
5
10
15
20
25
time
30
35
40
45
50
0
5
10
Figure S16.10b. Base case
15
20
25
time
30
35
Figure S16.10c.
1
40
45
50
Kp = 2.4
1.4
0.9
1.2
0.8
0.7
1
0.6
Output
Output
0.8
0.5
0.6
0.4
0.3
0.4
0.2
0.2
0.1
0
0
5
10
15
20
25
time
30
35
40
45
0
50
0
5
10
15
20
25
time
30
35
40
45
50
Figure S16.10e. τ = 6
Figure S16.10d. Kp = 1.6
1.4
6
1.2
4
1
2
Output
Output
0.8
0.6
0
-2
0.4
-4
0.2
0
0
5
10
15
20
25
time
30
35
40
45
50
Figure S16.10f. τ = 4
-6
0
5
10
15
20
25
time
Figure S16.10g.
16-18
30
35
40
θ=2.4
45
50
25
20
15
10
Output
5
0
-5
-10
-15
-20
-25
0
5
10
15
20
25
time
30
35
40
45
50
θ = 1.6
Figure S16.10h.
It is immediately evident that errors in time-delay estimation are the most
serious. This is because the terms in the characteristic equation which
contain dead-time do not cancel, and cause instability at high controller
gains.
When the actual process time constant is smaller than the model time
constant, the closed–loop system may become unstable. In our case, the
error is not large enough to cause instability, but the response is more
oscillatory than for the base (perfect model) case. The same is true if the
actual process gain is larger than that of the model. If the actual process
has a larger time constant, or smaller gain than the model, there is no
significant degradation in closed loop performance (for the magnitude of
the error, ± 20% considered here). Note that in all the above simulations,
the model is considered to be
2e −2 s and the actual process parameters
5s + 1
have been assumed to vary by ± 20% of the model parameter values.
c)
The proportional controller was tuned so as to obtain a gain margin of 2.0.
This resulted in Kc = 2.3. The responses for the various cases are shown
below
1.4
0.9
0.8
1.2
0.7
1
0.6
0.8
Output
Output
0.5
0.4
0.6
0.3
0.4
0.2
0.2
0.1
0
0
5
10
15
20
25
time
30
35
40
45
50
Base case
0
0
5
10
15
20
25
time
Kp = 3
16-19
30
35
40
45
50
0.7
0.9
0.8
0.6
0.7
0.5
0.6
0.4
Output
Output
0.5
0.4
0.3
0.3
0.2
0.2
0.1
0.1
0
0
5
10
15
20
25
time
30
35
40
45
0
50
0
5
10
15
20
25
time
30
35
40
45
50
τ=1
Kp = 1
1.4
1.4
1.2
1.2
1
1
0.8
Output
Output
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
0
5
10
15
20
25
time
30
35
40
45
0
50
τ = 2.5
0
5
10
15
20
25
time
30
35
40
45
50
θ=3
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
5
10
15
20
25
30
35
40
45
50
θ=1
Nyquist plots were prepared for different values of Kp, τ and θ, and
checked to see if the stability criterion was satisfied. The stability regions
when the three parameters are varied one to time are.
Kp ≤ 4.1
τ ≥ 2.4
(τ = 5 ,
θ = 2)
(Kp=2 , θ = 2)
θ ≤ 0.1 and 1.8 ≤ θ ≤ 2.2
16-20
(Kp = 2 ,
τ = 5)
16.11
From Eq. 16-24,
(
∗
−θs
Y Gd 1 + GcG (1 − e )
=
D
1 + Gc G ∗
)
that is,

2 −3 s  K c + K c τ I s 2
e 1 +
1 − e −3 s ) 
(
s
τI s
Y s


=
Kc + Kc τI s 2
D
1+
s
τI s
Using the final value theorem for a step change in D:
lim y (t ) = lim sY ( s )
t →∞
s →0
then

2 −3 s  K c + K c τ I s 2
1 − e −3 s ) 
e 1 +
(
τI s
s
s

1
lim sY ( s ) = lim s
s →0
s →0
K + Kc τI s 2
s
1+ c
τI s
s
2 −3 s 
2

e  τ I s + ( K c + K c τ I s ) (1 − e−3 s ) 
s
s


= lim
s →0
2
τ I s + ( K c + K c τ I s)
s
Multiplying both numerator and denominator by s2,
= lim
(
2e −3 s τ I s 2 + ( K c + K c τ I s )2 (1 − e −3 s )
)
τ I s + ( K c + K c τ I s )2s
3
s →0
Applying L'Hopital's rule:
= lim
s →0
(
−6e−3 s τ I s 2 + ( K c + K c τ I s )2 (1 − e−3 s )
)
3τ I s 2 + 2( K c + 2 K c τ I s )
−3 s
−3 s
−3 s
−3 s
+ 2e (2τ I s + 6 K c e 2+ 2 K c τ I − 2 K c τ I e + 6 K c τ I se ) = 6
3τ I s + 2( K c + 2 K c τ I s )
16-21
Therefore
lim y (t ) = lim sY ( s ) = 6
t →∞
s →0
and the PI control will not eliminate offset.
16.12
For a Smith predictor, we have the following system
Figure S16.12. Smith Predictor diagram block
where the process model is Gp(s) = Q(s) e-θs
For this system,
Gc′ G p
Y
=
Ysp 1 + Gc′ G p
where Gc’ is the transfer function for the system in the dotted box.
Gc′ =
Gc
1 + Gc Q(1 − e −θs )
Gc G p
∴
1 + Gc Q(1 − e −θs )
Y
=
Gc G p
Ysp
1+
1 + Gc Q(1 − e −θs )
Simplification gives
16-22
Y Gc Qe −θs
=
= P( s )e −θs
Ysp 1 + Gc Q
where P( s ) =
Gc Q
1 + Gc Q
If P(s) is the desired system performance (after the time delay has elapsed)
under feedback control, then we can solve for Gc in terms of P(s).
Gc =
P( s)
Q( s )(1 − P( s ))
The IMC controller requires that we define
G2 + = e −θs
~
G− = Q (s)
(the invertible part of Gp)
Let the filter for the controller be f(s) =
1
τF s + 1
Therefore, the controller is
f ( s)
~ −1
Gc = G − f ( s ) =
Q( s)
The closed-loop transfer function is
Y
e−θs
= Gc G p =
= G2 + f
Ysp
1 + τF s
Note that this is the same closed-loop form as analyzed in part (a), which
led to a Smith Predictor type of controller. Hence, the IMC design also
provides time-delay compensation.
16-23
16.13
Referring to Example 4.8, if q, the flowrate, and Ti, the inlet temperature,
are know and are constant, then the Laplace transform models in (4-79)
and (4-80) are
( s − a11 )C ′A ( s ) = a12T ′( s )
(4-79)
( s − a 22 )T ′( s ) = a12 C ′A ( s ) + b2Ts′( s )
(4-80)
where Ts′( s ) is the coolant temperature. Using Eq. 4-86, we can directly
compute concentration from the temperature signal, i.e.,
C ′A ( s ) =
a12
T ′( s )
s − a11
which is a first-order filter operating on T ′( s )
So inferential control of concentration using temperature would be
feasible in this case. If q and Ti varied, a more general expression for the
linearized model would be necessary, but there would still be a direct way
to infer CA from T.
16.14
One possible solution would be to use a split range valve to handle the
100≤ p≤ 200 and higher pressure ranges. Moreover, a high-gain controller
with set-point = 200 psi can be used for the vent valve. This valve would
not open while the pressure is less than 200 psi, which is similar to how a
selector operates.
Stephanopoulos (Chemical Process Control, Prentice-Hall, 1989) has
described many applications for this so-called split-range control. A
typical configuration consists of 1 controller and 2 final control elements
or valves.
16-24
VENT
SPLIT RANGE
CONTROLLER
PT
INLET
OUTLET
REACTOR
Figure S16.14. Process instrumentation diagram
16.15
The amounts of air and fuel are changed in response to the steam pressure.
If the steam pressure is too low, a signal is sent to increase both air and
fuel flowrates, which in turn increases the heat transfer to the steam.
Selectors are used to prevent the possibility of explosions (low air-fuel
ratio). If the air flowrate is too low, the low selector uses that
measurement as the set-point for the fuel flow rate controller. If the fuel
flowrate is too high, its measurement is selected by the high selector as the
set-point for the air flow controller. This also protects against dynamic
lags in the set-point response.
16.16
TT
TC
TC
L
TT
FC
COOLING
WATER
CONDENSATE
Figure S16.16. Control condensate temperature in a reflux drum
16-25
16.17
Supposing a first-order plus dead time process, the closed-loop transfer
function is


1
+ τ D s  e −θs
1 +
τI s

Kc K p 
GcG p
(τ p s + 1)
GCL ( s ) =
∴ GCL ( s ) =
1 + Gc G p


1
+ τ D s  e−θs
1 +
τI s

1 + Kc K p 
(τ p s + 1)
Notice that Kc and Kp always appear together as a product. Hence, if we
want the process to maintain a specified performance (stability, decay
ratio specification, etc.), we should adjust Kc such that it changes inversely
with Kp; as a result, the product KcKp is kept constant. Also note, that since
there is a time delay, we should adjust Kc based upon the future estimate
of Kp:
K c (t ) =
Kc K p
Kˆ p (t + θ)
=
Kc K p
b
a+
Mˆ (t + θ)
where Kˆ p (t + θ) is an estimate of Kp θ time units into the future.
16.18
This is an application where self-tuning control would be beneficial. In
order to regulate the exit composition, the manipulated variable (flowrate)
must be adjusted. Therefore, a transfer function model relating flowrate to
exit composition is needed. The model parameters will change as the
catalyst deactivates, so some method of updating the model (e.g., periodic
step tests) will have to be derived. The average temperature can be
monitored to determine a significant change in activation has occurred,
thus indicating the need to update the model.
16-26
16.19
a)
GcG p
1 + G c Gp
=
1
τc s + 1
∴
Gc =
1
τc s + 1

1 
G p 1 −

 τc s + 1 
=
1 1
G p τc s
Substituting for Gp
1 τ1τ2 s 2 + (τ1 + τ2 ) s + 1
1
Gc ( s ) =
=
τc s
Kp
K p τc
1

(τ1 + τ2 ) + τ1τ2 s + s 
Thus, the PID controller tuning constants are
Kc =
(τ1 + τ2 )
K p τc
τ I = τ1 + τ2
τD =
τ1τ2
τ1 + τ2
(See Eq. 12-14 for verification)
b)
For τ1 = 3
and τ2 = 5 and τc = 1.5, we have
Kc = 5.333
τI = 8.0 and τD = 1.875
Using this PID controller, the closed-loop response will be first order
when the process model is known accurately. The closed-loop response to
a unit step-change in the set-point when the model is known exactly is
shown above. It is assumed that τc was chosen such that the closed loop
response is reasonable, and the manipulated variable does not violate any
bounds that are imposed. An approximate derivative action is used by
τ s
Simulink-MATLAB, namely D when β=0.01
1 + βs
16-27
Figure S16.19a. Simulink block diagram.
1.4
1200
1.2
1000
1
Manipulated variable
800
Output
0.8
0.6
600
400
0.4
200
0.2
0
0
0
5
10
15
20
time
25
30
35
-200
40
0
5
10
15
20
time
25
30
35
40
Figure S16.19b. Output (no model error) Figure S16.19c. Manipulated variable (no
model error)
1.4
1200
1.2
1000
1
Manipulated variable
800
Output
0.8
0.6
0.4
600
400
200
0.2
0
0
-200
0
5
10
15
20
time
25
30
35
40
Figure S16.19d. Output (Kp = 2)
0
5
10
15
20
time
25
30
35
40
Figure S16.19e. Manipulated variable
(Kp = 2)
16-28
1.4
1200
1.2
1000
1
Manipulated variable
800
Output
0.8
0.6
600
400
0.4
200
0.2
0
0
5
10
15
20
time
25
30
35
0
40
Figure S16.19f. Output (Kp = 0.5)
0
5
10
15
20
time
25
30
35
40
Figure S16.19g. Manipulated variable
(Kp =0.5)
1.4
1200
1.2
1000
1
Manipulated variable
800
Output
0.8
0.6
600
400
0.4
200
0.2
0
0
5
10
15
20
time
25
30
35
0
40
Figure S16.19h. Output (τ2 = 10)
0
5
10
15
20
time
25
30
35
40
Figure S16.19i. Manipulated variable
(τ2 = 10)
1200
1
0.9
1000
0.8
800
Msanipulated variable
0.7
Output
0.6
0.5
0.4
0.3
600
400
200
0.2
0
0.1
0
0
5
10
15
20
time
25
30
35
-200
40
Figure E16.9 j.- Output (τ2 = 1)
0
5
10
15
20
time
25
30
35
40
Figure E16.9 k.- Manipulated variable
(τ2 = 1)
(1)
The closed-loop response when the actual Kp is 2.0 is shown above. The
controlled variable reaches its set-point much faster than for the base case
(exact model), but the manipulated variable assumes values that are more
negative (for some period of time) than the base case. This may violate
some bounds.
16-29
(2)
When Kp = 0.5, the response is much slower. In fact, the closed-loop time
constant seems to be about 3.0 instead of 1.5. There do not seem to be any
problems with the manipulated variable.
(3)
If (τ2 = 10), the closed-loop response is no longer first-order. The settling
time is much longer than for the base case. The manipulated variable does
not seem to violate any bounds.
(4)
Both the drawbacks seen above are observed when τ2 = 1. The settling
time is much longer than for the base case. Also the rapid initial increase
in the controlled variable means that the manipulated variable drops off
sharply, and is in danger of violating a lower bound.
16.20
Based on discussions in Chapter 12, increasing the gain of a controller
makes it more oscillatory, increasing the overshoot (peak error) as well as
the decay ratio. Therefore, if the quarter-decay ratio is a goal for the
closed-loop response (e.g., Ziegler-Nichols tuning), then the rule proposed
by Appelpolscher should be satisfactory from a qualitative point of view.
However, if the controller gain is increased, the settling time is also
decreased, as is the period of oscillation. Integral action influences the
response characteristics as well. In general, a decrease in τI gives
comparable results to an increase in Kc. So, Kc can be used to influence the
peak error or decay ratio, while τI can be used to speed up the settling time
(a decrease in τI decreases the settling time). See Chapter 8 for typical
response for varying Kc and τI.
16.21
SELECTIVE CONTROL
Selectors are quite often used in forced draft combustion control system to
prevent an imbalance between air flow and fuel flow, which could result
in unsafe operating conditions.
For this case, a flow controller adjusts the air flowrate in the heater. Its setpoint is determined by the High Selector, which chooses the higher of the
two input signals:
.- Signal from the fuel gas flowrate transmitter (when this is too high)
16-30
.- Signal from the outlet temperature control system.
Similarly, if the air flow rate is too low, its measurement is selected by the
low selector as the set-point for the fuel-flow rate.
CASCADE CONTROLLER
The outlet temperature control system can be considered the master
controller that adjusts the set-point of the fuel/air control system (slave
controller). If a disturbance in fuel or air flow rate exists, the slave control
system will act very quickly to hold them at their set-points.
FEED-FORWARD CONTROL
The feedforward control scheme in the heater provides better control of
the heater outlet temperature. The feed flowrate and temperature are
measured and sent to the feedback control system in the outflow. Hence
corrective action is taken before they upset the process. The outputs of the
feedforward and feedback controller are added together and the combined
signal is sent to the fuel/air control system.
16.22
ALTERNATIVE A.Since the control valves are "air to close", each Kv is positive (cf. Chapter
9). Consequently, each controller must be reverse acting (Kc>0) for the
flow control loop to function properly.
Two alternative control strategies are considered:
Method 1: use a default feed flowrate when Pcc > 80%
Let :
Pcc = output signal from the composition controller (%)
~
Fsp = (internal) set point for the feed flow controller (%)
Control strategy:
~
~
If Pcc > 80% , Fsp = Fsp , low
~
where Fsp , low is a specified default flow rate that is lower than the normal
~
value, Fsp nom .
16-31
Method 2: Reduce the feed flow when Pcc > 80%
Control strategy:
~
~
If Pcc < 80%, Fsp = Fsp nom − K(Pcc – 80%)
where K is a tuning parameter (K > 0)
Implementation:
~
Fnom
80 %
Pcc
HS
+
-
K
-
+
~
Fsp
80 %
~
Note: A check should be made to ensure that 0 ≤ Fsp ≤ 100%
ALTERNATIVE B.A selective control system is proposed:
V--1
FT
>
V--2
FC
CC
CT
Figure S16.22. Proposed selective control system
Both control valves are A-O and transmitters are “direct acting”, so the
controller have to be “reverse acting”.
When the output concentration decreases, the controller output increases.
Hence this signal cannot be sent directly to the feed valve (it would open
16-32
the valve). Using a high selector that chooses the higher of these signals
can solve the problem
.- Flow transmitter
.- Output concentration controller
Therefore when the signal from the output controller exceeds 80%, the
selector holds it and sends it to the flow controller, so that feed flow rate is
reduced.
16.23
ALTERNATIVE A.Time delay.- Use time delay compensation, e.g., Smith Predictor
Variable waste concentration.- Tank pH changes occurs due to this
unpredictable changes. Process gain changes also (c,f. literature curve for
strong acid-strong base)
Variable waste flow rate.- Use FF control or ratio qbase to qwaste.
Measure qbase .- This suggests you may want to use cascade control to
compensate for upstream pressure changes, etc
ALTERNATIVE B.Several advanced control strategies could provide improved process
control. A selective control system is commonly used to control pH in
wastewater treatment .The proposed system is shown below (pH T = pH
sensor; pH C = pH controller)
V--4
V--3
FT
FT
S
FC
FC
pH T
pH C
`
T-1
Figure S16.23. Proposed selective control system.
16-33
S
where S represents a selector ( < or >, to be determined)
In this scheme, several manipulated variables are used to control a single
process variable. When the pH is too high or too low, a signal is sent to the
selectors in either the waste stream or the base stream flowrate controllers.
The exactly configuration of the system depends on the transmitter,
controller and valve gains.
In addition, a Smith Predictor for the pH controller is proposed due to the
large time delay. There would be other possibilities for this process such
as an adaptive control system or a cascade control system. However the
scheme above may be good enough
Necessary information:
.- Descriptions of measurement devices, valves and controllers; direct
action or reverse action.
.- Model of the process in order to implement the Smith Predictor
16.24
For setpoint change, the closed-loop transfer function with an integral
controller and steady state process (Gp = Kp) is:
1 K
G
G
τI s P
KP
1
C
P
Y
=
=
=
=
Ysp 1 + G G
1
τ
τI s + K P
1+
K
I
C P
s +1
τI s P
KP
Hence a first order response is obtained and satisfactory control can be
achieved.
For disturbance change (Gd = Gp):
Y
Gd
KP
K (τ s )
τI s
=
=
= P I =
D 1+ G G
1 + 1 K P τ I s + K P τI
C P
s +1
τI s
KP
Therefore a first order response is also obtained for disturbance change.
16-34
123456789 8
17.1
Using Eq. 17-9, the filtered values of xD are shown in Table S17.1
time(min)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
α=1
0
0.495
0.815
1.374
0.681
1.889
2.078
2.668
2.533
2.908
3.351
3.336
3.564
3.419
3.917
3.884
3.871
3.924
4.300
4.252
4.409
α = 0.8
0
0.396
0.731
1.245
0.794
1.670
1.996
2.534
2.533
2.833
3.247
3.318
3.515
3.438
3.821
3.871
3.871
3.913
4.223
4.246
4.376
α = 0.5
0
0.248
0.531
0.953
0.817
1.353
1.715
2.192
2.362
2.635
2.993
3.165
3.364
3.392
3.654
3.769
3.820
3.872
4.086
4.169
4.289
Table S17.1. Unfiltered and filtered data.
To obtain the analytical solution for xD, set F ( s ) =
1
in the given transfer
s
function, so that
1
5
5
1 
F ( s) =
= 5 −

10s + 1
s (10s + 1)
 s s + 1 10 
Taking inverse Laplace transform
X D ( s) =
xD(t) = 5 (1 − e-t/10)
A graphical comparison is shown in Fig. S17.1
Solution Manual for Process Dynamics and Control, 2nd edition,
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp
17-1
4.5
4
3.5
3
XD
2.5
2
1.5
1
noisy data
alpha = 0.5
alpha = 0.8
analytical solution
0.5
0
0
2
4
6
8
10
12
time (min)
14
16
18
20
Fig S17.1. Graphical comparison for noisy data, filtered data and analytical
solution.
As α decreases, the filtered data give a smoother curve compared to the
no-filter (α=1) case, but this noise reduction is traded off with an increase
in the deviation of the curve from the analytical solution.
17.2
The exponential filter output in Eq. 17-9 is
yF (k ) = αym (k ) + (1 − α) yF (k − 1)
(1)
Replacing k by k-1 in Eq. 1 gives
yF (k − 1) = αym (k − 1) + (1 − α) yF (k − 2)
(2)
Substituting for yF (k − 1) from (2) into (1) gives
yF (k ) = αym (k ) + (1 − α)αym (k − 1) + (1 − α ) 2 yF (k − 2)
Successive substitution of yF (k − 2) , yF (k − 3) ,… gives the final form
k −1
yF (k ) = ∑ (1 − α)i αym (k − i ) + (1 − α) k yF (0)
i=0
17-2
17.3
Table S17.3 lists the unfiltered output and, from Eq. 17-9, the filtered data
for sampling periods of 1.0 and 0.1. Notice that for sampling period of 0.1,
the unfiltered and filtered outputs were obtained at 0.1 time increments,
but they are reported only at intervals of 1.0 to preserve conciseness and
facilitate comparison.
The results show that for each value of ∆t, the data become smoother as α
decreases, but at the expense of lagging behind the mean output y(t)=t.
Moreover, lower sampling period improves filtering by giving smoother
data and less lagg for the same value of α.
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
α=1
0
1.421
1.622
3.206
3.856
4.934
5.504
6.523
8.460
8.685
9.747
11.499
11.754
12.699
14.470
14.535
15.500
16.987
17.798
19.140
19.575
α=0.8
0
1.137
1.525
2.870
3.659
4.679
5.339
6.286
8.025
8.553
9.508
11.101
11.624
12.484
14.073
14.442
15.289
16.647
17.568
18.825
19.425
∆t=1
α=0.5
0
0.710
1.166
2.186
3.021
3.977
4.741
5.632
7.046
7.866
8.806
10.153
10.954
11.826
13.148
13.841
14.671
15.829
16.813
17.977
18.776
α=0.2
0
0.284
0.552
1.083
1.637
2.297
2.938
3.655
4.616
5.430
6.293
7.334
8.218
9.115
10.186
11.055
11.944
12.953
13.922
14.965
15.887
α=0.8
0
1.381
1.636
3.227
3.916
4.836
5.574
6.571
8.297
8.688
9.741
11.328
11.770
12.747
14.284
14.662
15.642
16.980
17.816
19.036
19.655
∆t=0.1
α=0.5
0
1.261
1.678
3.200
3.973
4.716
5.688
6.664
8.044
8.717
9.749
11.078
11.778
12.773
14.051
14.742
15.773
16.910
17.808
18.912
19.726
α=0.2
0
0.877
1.647
2.779
3.684
4.503
5.544
6.523
7.637
8.533
9.544
10.658
11.556
12.555
13.649
14.547
15.544
16.605
17.567
18.600
19.540
Table S17.3. Unfiltered and filtered output for sampling periods of 1.0 and 0.1
17-3
Graphical comparison:
20
18
16
14
y(t)
12
10
8
6
α
α
α
α
4
2
0
0
2
4
6
8
10
time, t
12
14
16
=1
= 0.8
= 0.5
= 0.2
18
20
Figure S17.3a. Graphical comparison for ∆t = 1.0
20
18
16
14
y(t)
12
10
8
6
α=1
α=0.8
α=0.5
α=0.2
4
2
0
0
2
4
6
8
10
time, t
12
14
16
Figure S17.3b. Graphical comparison for ∆t = 0.1
17-4
18
20
17.4
Using Eq. 17-9 for α = 0.2 and α = 0.5, Eq. 17-18 for N* = 4, and Eq. 1719 for ∆y=0.5, the results are tabulated and plotted below.
α=1
0
1.50
0.30
1.60
0.40
1.70
1.50
2.00
1.50
t
0
1
2
3
4
5
6
7
8
(a)
(a)
α=0.2
0
0.30
0.30
0.56
0.53
0.76
0.91
1.13
1.20
α=0.5
0
0.75
0.53
1.06
0.73
1.22
1.36
1.68
1.59
(b)
N*=4
0
0.38
0.45
0.85
0.95
1.00
1.30
1.40
1.68
(c)
∆y=0.5
0
0.50
0.30
0.80
0.40
0.90
1.40
1.90
1.50
Table S17.4. Unfiltered and filtered data.
2.5
2
y(t)
1.5
1
α=1
α=0.2
α=0.5
0.5
N*=4
∆ y=0.5
0
0
1
2
3
4
time, t
5
6
7
8
Figure S17.4. Graphical comparison for filtered data and the raw data.
17-5
17.5
Let C denote the controlled output. Then
Gd
C ( s)
=
d ( s ) 1 + K c GvG p GmGF
d ( s) =
,
1
s +1
2
For τF = 0, GF = 1 and
1 /(5s + 1)
1
1
=
2
1 + 1 /(5s + 1) s + 1 (5s + 2)( s 2 + 1)
C ( s) =
For τF = 3, GF = 1/(3s+1) and
C ( s) =
1/(5s + 1)
1
3s + 1
=
2
2
1 + [1/(5s + 1)][1/(3s + 1) ] s + 1 (15s + 8s + 2)( s 2 + 1)
By using Simulink-MATLAB,
0.3
No filtering
Exponetial filtering
0.2
0.1
y(t)
0
-0.1
-0.2
-0.3
-0.4
0
2
4
6
8
10
time. t
12
14
16
18
20
Figure S17.5. Closed-loop response comparison for no filtering and for
an exponential filter (τF = 3 min)
17-6
17.6
Y (s) =
1
1 1
X (s) =
s +1
s +1 s
,
y(t) = 1 − e-t
then
For noise level of ± 0.05 units, several different values of α are tried in
Eq. 17-9 as shown in Fig. S17.6a. While the filtered output for α = 0.7 is
still quite noisy, that for α = 0.3 is too sluggish. Thus α = 0.4 seems to
offer a good compromise between noise reduction and lag addition.
Therefore, the designed first-order filter for noise level ± 0.05 units is α =
0.4, which corresponds to τF = 1.5 according to Eq. 17-8a.
Noise level = ± 0.05
1.4
α=1.0
α=0.7
α=0.4
α=0.3
1.2
1
y(t)
0.8
0.6
0.4
0.2
0
0
2
4
6
8
10
t
12
14
16
18
20
Figure S17.6a. Digital filters for noise level = ± 0.05
Noise level = ± 0.1
1.4
1.2
1
y(t)
0.8
0.6
α=1
α=0.3
α=0.2
α=0.15
0.4
0.2
0
0
5
10
15
20
t
Figure S17.6b. Digital filters for noise level = ± 0.1
17-7
Noise level = ± 0.01
1.4
1.2
1
y(t)
0.8
α=1
0.6
0.4
0.2
0
0
5
10
15
20
t
Figure S17.6c. Response for noise level = ± 0.01; no filter needed.
Similarly, for noise level of ± 0.1 units, a good compromise is α =0.2 or τF
= 4.0 as shown in Fig. S17.6b. However, for noise level of ±0.01 units, no
filter is necessary as shown in Fig. S17.6c. thus α=1.0, τF = 0
17.7
y(k) = y(k-1) − 0.21 y(k-2) + u(k-2)
k
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
u(k)
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
u(k-1)
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
17-8
u(k-2)
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
y(k)
0
0
1.00
1.00
0.79
0.58
0.41
0.29
0.21
0.14
0.10
0.07
0.05
0.03
0.02
0.02
0.01
0.01
0.01
0.00
Plotting this results
1.2
1
0.8
y
0.6
0.4
0.2
0
0
2
4
6
8
10
k
12
14
16
18
20
Fig S17.7. Graphical simulation of the difference equation
The steady state value of y is zero.
17.8
By using Simulink and STEM routine to convert the continuous signal to a
series of pulses,
12
10
8
Tm'(t)
a)
6
4
2
0
0
5
10
15
20
25
time
30
35
40
45
50
Figure S17.8. Discrete time response for the temperature change.
Hence the maximum value of the logged temperature is 80.7° C.
This maximum point is reached at t = 12 min.
17-9
17.9
a)
2.7 z −1 ( z + 3)
2.7 + 8.1z −1
Y ( z)
=
=
U ( z ) z 2 − 0.5 z + 0.06 z 2 − 0.5 z + 0.06
Dividing both numerator and denominator by z2
Y ( z)
2.7 z −2 + 8.1z −3
=
U ( z ) 1 − 0.5 z −1 + 0.06 z −2
Then
Y ( z )(1 − 0.5 z −1 + 0.06 z −2 ) = U ( z )(2.7 z −2 + 8.1z −3 )
or
y(k) = 0.5y(k-1) − 0.06y(k-2) + 2.7u(k-2) + 8.1u(k-3)
The simulation of the difference equation yields
k
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
u(k)
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
u(k-2)
0
0
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
17-10
u(k-3)
0
0
0
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
y(k)
0
0
2.70
12.15
16.71
18.43
19.01
19.20
19.26
19.28
19.28
19.28
19.29
19.29
19.29
19.29
19.29
19.29
19.29
19.29
19.29
b)
20
18
16
14
12
y
10
8
6
4
2
0
Difference equation
Simulink
0
2
4
6
8
10
k ∆t
12
14
16
18
20
Fig S17.9. Simulink response to a unit step change in u
c)
The steady state value of y can be found be setting z =1. In doing so,
y =19.29
This result is in agreement with data above.
17.10
1

Gc ( s ) = 2 1 + 
 8s 
Substituting s ≅ (1-z-1)/∆t and accounting for ∆t=1

 2.25 − 2 z −1
1
Gc ( z ) = 2 1 +
=
−1 
(1 − z −1 )
 8(1 − z ) 
By using Simulink-MATLAB, the simulation for a unit step change in the
controller error signal e(t) is shown in Fig. S17.10
17-11
70
60
50
b(k)
40
30
20
10
0
0
5
10
15
k
20
25
Fig S17.10. Open-loop response for a unit step change
17.11
a)
Y ( z)
5( z + 0.6)
= 2
U ( z ) z − z + 0.41
Dividing both numerator and denominator by z2
Y ( z)
5 z −1 + 3z −2
=
U ( z ) 1 − z −1 + 0.41z −2
Then Y ( z )(1 − z −1 + 0.41z −2 ) = U ( z )(5 z −1 + 3z −2 )
or
y(k) = y(k-1) − 0.41y(k-2) + 5u(k-1) + 3u(k-2)
b)
The simulation of the difference equation yields
17-12
30
u(k)
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
k
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
c)
u(k-1)
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
u(k-2)
0
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
y(k)
5
13.00
18.95
21.62
21.85
20.99
20.03
19.42
19.21
19.25
19.37
19.48
19.54
19.55
19.54
19.52
19.51
19.51
19.51
By using Simulink-MATLAB, the simulation for a unit step change in u
yields
25
Difference equation
Simulink
20
15
y
10
5
0
0
2
4
6
8
10
k∆t
12
14
16
18
20
Fig S17.11. Simulink response to a unit step change in u
d)
The steady state value of y can be found be setting z =1. In doing so,
y =19.51
This result is in agreement with data above.
17-13
17.12
a)
1
1 − z −1
7
6
5
Output
4
3
2
1
0
0
1
2
3
4
5
3
4
5
3
4
5
Time
b)
1
1 + 0.7 z −1
1
Output
0.8
0.6
0.4
0.2
0
0
1
2
Time
1
1 − 0.7 z −1
3
2.5
2
Output
c)
1.5
1
0.5
0
0
1
2
Time
17-14
d)
1
(1 + 0.7 z )(1 − 0.3 z −1 )
−1
1
Output
0.8
0.6
0.4
0.2
0
0
1
2
3
4
5
3
4
5
3
4
5
Time
e)
1 − 0.5 z −1
(1 + 0.7 z −1 )(1 − 0.3z −1 )
1
Output
0.8
0.6
0.4
0.2
0
0
1
2
Time
f)
1 − 0.2 z −1
(1 + 0.6 z −1 )(1 − 0.3 z −1 )
1
0.8
0.6
0.4
0.2
0
0
1
2
17-15
Conclusions:
.- A pole at z = 1 causes instability.
.- Poles only on positive real axis give oscillation free response.
.- Poles on the negative real axis give oscillatory response.
.- Poles on the positive real axis dampen oscillatory responses.
..- Zeroes on the positive real axis increase oscillations.
.- Zeroes closer to z = 0 contribute less to the increase in oscillations.
17.13
By using Simulink, the response to a unit set-point change is shown in Fig.
S17.13a
1.8
1.6
1.4
Output
1.2
1
0.8
0.6
0.4
0.2
0
0
5
10
15
20
Time
25
30
35
40
Fig S17.13a. Closed-loop response to a unit set-point change (Kc = 1)
Therefore the controlled system is stable.
The ultimate controller gain for this process is found by trial and error
17-16
8
7
6
Output
5
4
3
2
1
0
0
5
10
15
20
Time
25
30
35
40
Fig S17.13b. Closed-loop response to a unit set-point change (Kc =21.3)
Then Kcu = 21.3
17.14
By using Simulink-MATLAB, these ultimate gains are found:
∆t = 0.01
2
1.8
1.6
1.4
Output
1.2
1
0.8
0.6
0.4
0.2
0
0
1
2
3
Time
4
5
6
Fig S17.14a. Closed-loop response to a unit set-point change (Kc =1202)
17-17
∆t = 0.1
2
1.8
1.6
1.4
Output
1.2
1
0.8
0.6
0.4
0.2
0
0
5
10
15
Time
Fig S17.14b. Closed-loop response to a unit set-point change (Kc =122.5)
∆t = 0.5
2
1.8
1.6
1.4
Output
1.2
1
0.8
0.6
0.4
0.2
0
0
5
10
15
Time
Fig S17.14c. Closed-loop response to a unit set-point change (Kc =26.7)
Hence
∆t = 0.01
∆t = 0.1
∆t = 0.5
Kcu = 1202
Kcu = 122.5
Kcu = 26.7
As noted above, decreasing the sampling time makes the allowable
controller gain increases. For small values of ∆t, the ultimate gain is large
enough to guarantee wide stability range.
17-18
17.15
By using Simulink-MATLAB
Kc = 1
1.4
1.2
1
Output
0.8
0.6
0.4
0.2
0
0
5
10
15
20
25
time
30
35
40
45
50
Fig S17.15a. Closed-loop response to a unit set-point change (Kc =1)
Kc = 10
1.8
1.6
1.4
Output
1.2
1
0.8
0.6
0.4
0.2
0
0
5
10
15
20
25
time
30
35
40
45
50
Fig S17.15b. Closed-loop response to a unit set-point change (Kc =10)
17-19
Kc = 17
2.5
2
Output
1.5
1
0.5
0
-0.5
0
5
10
15
20
25
time
30
35
40
45
50
Fig S17.15c. Closed-loop response to a unit set-point change (Kc =17)
Thus the maximum controller gain is
Kcm = 17
17.16
Gv(s) = Kv = 0.1 ft3 / (min)(ma)
Gm(s) =
4
0 .5 s + 1
In order to obtain Gp(s), write the mass balance for the tank as
A
dh
= q1 + q 2 − q3
dt
Using deviation variables and taking Laplace transform
AsH ′( s ) = Q1′ ( s ) + Q2′ ( s ) − Q3′ ( s )
Therefore,
17-20
G p ( s) =
−1
H ′( s ) −1
=
=
Q3′ ( s ) As 12.6 s
By using Simulink-MATLAB,
Kc = -10
1.4
1.2
1
y(t)
0.8
0.6
0.4
0.2
0
0
5
10
15
20
25
time
30
35
40
45
50
Fig S17.16a. Closed-loop response to a unit set-point change (Kc = -10)
Kc = -50
1.8
1.6
1.4
1.2
y(t)
1
0.8
0.6
0.4
0.2
0
0
5
10
15
20
25
time
30
35
40
45
50
Fig S17.16b. Closed-loop response to a unit set-point change (Kc = -50)
17-21
Kc = -92
3.5
3
2.5
2
y(t)
1.5
1
0.5
0
-0.5
-1
-1.5
0
5
10
15
20
25
time
30
35
40
45
50
Fig S17.16c. Closed-loop response to a unit set-point change (Kc = -92)
Hence the closed loop system is stable for
-92 < Kc < 0
As noted above, offset occurs after a change in the setpoint.
17.17
a)
The closed-loop response for set-point changes is
Gc G ( s )
Y (s)
=
Ysp ( s ) 1 + Gc G ( s )
then
Gc ( z ) =
1 (Y / Ysp )
G 1 − (Y / Ysp )
We want the closed-loop system exhibits a first order plus dead time
response,
(Y / Ysp ) =
e − hs
λs + 1
or
(Y / Ysp ) =
Moreover,
17-22
(1 − A) z − N −1
1 − Az −1
where A = e-∆t/λ
G (s) =
e −2 s
3s + 1
or
G( z) =
0.284 z −3
1 − 0.716 z −1
Thus, the resulting digital controller is the Dahlin's controller Eq. 17-66.
Gc ( z ) =
(1 − A)
1 − 0.716 z −1
1 − Az −1 − (1 − A) z − N −1
0.284
(1)
If a value of λ=1 is considered, then A = 0.368 and Eq. 1 is
0.632
1 − 0.716 z −1
Gc ( z ) =
1 − 0.368 z −1 − 0.632 z −3
0.284
(2)
b)
(1-z-1) is a factor of the denominator in Eq. 2, indicating the presence of
integral action. Then no offset occurs.
c)
From Eq. 2, the denominator of Gc(z) contains a non-zero z-0 term. Hence
the controller is physically realizable.
d)
First adjust the process time delay for the zero-order hold by adding ∆t/2
to obtain a time delay of 2 + 0.5 = 2.5 min. Then obtain the continuos PID
controller tuning based on the ITAE (setpoint) tuning relation in Table
12.3 with K = 1, τ=3, θ = 2.5. Thus
KKc = 0.965(2.5/3) − 0.85
, Kc = 1.13
τ/τI = 0.796 + (-0.1465)(2.5/3) ,
, τI = 4.45
τD/τ = 0.308(2.5/3)0.929
, τD = 0.78
Using the position form of the PID control law (Eq. 8-26 or 17-55)


 1 
Gc ( z ) = 1.13 1 + 0.225 
+ 0.78(1 − z −1 ) 
−1 
 1− z 


=
2.27 − 2.89 z −1 + 0.88 z −2
1 − z −1
By using Simulink-MATLAB, the controller performance is examined:
17-23
1.4
1.2
1
y(t)
0.8
0.6
0.4
0.2
0
0
5
10
15
20
25
Time
30
35
40
45
50
Fig S17.17. Closed-loop response for a unit step change in set point.
Hence performance shows 21% overshoot and also oscillates.
17.18
a)
The transfer functions in the various blocks are as follows.
Km = 2.5 ma / (mol solute/ft3)
Gm(s) = 2.5e-s
17-24
H(s)=
1 − e−s
s
Gv(s) = Kv = 0.1 ft3/min.ma
To obtain Gp(s) and Gd(s), write the solute balance for the tank as
V
dc3
= q1c1 + q2 (t )c2 (t ) − q3c3 (t )
dt
Linearizing and using deviation variables
V
dc3′
= q 2 c 2′ + c 2 q 2′ − q3 c3′
dt
Taking Laplace transform and substituting numerical values
30 sC 3′ ( s ) = 1.5Q2′ ( s ) + 0.1C 2′ ( s ) − 3C 3′ ( s )
Therefore,
b)
G p ( s) =
C 3′ ( s )
1.5
0.5
=
=
Q2′ ( s ) 30 s + 3 10s + 1
Gd ( s ) =
C3′ ( s )
0.1
0.033
=
=
C2′ ( s ) 30 s + 3 10 s + 1
G p ( z) =
C3 ( z)
0.05
=
Q2 ( z ) 1 − 0.9 z −1
A proportional-integral controller gives a first order exponential response
to a unit step change in the disturbance C2. This controller will also give a
first order response to setpoint changes. Therefore, the desired response
could be specified as
(Y / Ysp ) =
1
λs + 1
17-25
17.19
HG p ( z ) K m Gc ( z )
Y
=
Ysp 1 + HG p Gm ( z )Gc ( z )
Solving for Gc(z)
Gc ( z ) =
Y
Ysp
HG p ( z ) K m − HG p Gm ( z )
Y
Ysp
(1)
Since the process has no time delay, N = 0. Hence
Y

 Ysp

(1 − A) z −1
 =
1 − Az −1
d
Moreover
z −1
HG p ( z ) =
1 − z −1
HG p Gm ( z ) =
z −2
1 − z −1
Km = 1
Substituting into (1) gives
(1 − A) z −1
1 − Az −1
Gc ( z ) = −1
z
z −2 (1 − A) z −1
−
1 − z −1 1 − z −1 1 − Az −1
Rearranging,
(1 − A) − (1 − A) z −1
Gc ( z ) =
1 − Az −1 − (1 − A) z − 2
By using Simulink-MATLAB, the closed-loop response is shown for
different values of A (actually different values of λ) :
17-26
λ=3
λ=1
λ = 0.5
A = 0.716
A = 0.368
A = 0.135
5
4.5
4
3.5
y(t)
3
2.5
2
1.5
1
λ=3
λ=1
λ=0.5
0.5
0
0
5
10
15
20
25
Time
30
35
40
45
50
Fig S17.19. Closed-loop response for a unit step change in disturbance.
17.20
The closed-loop response for a setpoint change is
HG ( z ) K v Gc ( z )
Y
=
Ysp 1 + HG ( z ) K v K m ( z )Gc ( z )
Hence
Gc ( z ) =
Y
Ysp
1
HG ( z ) K − K K Y
v
v m
Ysp
The process transfer function is
17-27
G( s) =
2 .5
10 s + 1
or
HG ( z ) =
0.453 z −1
1 − 0.819 z −1
(θ = 0 so N = 0)
Minimal prototype controller implies λ = 0 (i.e., A → 0) . Then,
Y
= z −1
Ysp
Therefore the controller is
1 − 0.819 z −1
z −1
Gc ( z ) =
0.453z −1 0.2 − (0.2)(0.25) z −1
Simplifying,
Gc ( z ) =
z −1 − 0.819 z −2
1 − 0.819 z −1
=
0.091z −1 − 0.023z −2 0.091 − 0.023 z −1
17.21
a)
From Eq. 17-71, the Vogel-Edgar controller is
GVE =
(1 + a1 z −1 + a 2 z −2 )(1 − A)
(b1 + b2 )(1 − Az −1 ) − (1 − A)(b1 + b2 z −1 ) z − N −1
where A = e-∆t/λ = e –1/5 = 0.819
Using z-transforms, the discrete-time version of the second-order transfer
function yields
a1 = -1.625
a2 = 0.659
b1 = 0.0182
b2 = 0.0158
Therefore
GVE =
=
(1 − 1.625 z −1 + 0.659 z −2 )0.181
(0.0182 + 0.0158)(1 − 0.819 z −1 ) − 0.181(0.0182 + 0.0158 z −1 ) z −1
0.181 − 0.294 z −1 + 0.119 z −2
0.034 − 0.031z −1 − 0.003 z −2
17-28
By using Simulink-MATLAB, the controlled variable y(k) and the
controller output p(k) are shown for a unit step change in ysp.
Controlled variable y(k):
1
0.9
0.8
0.7
y(k)
0.6
0.5
0.4
0.3
0.2
0.1
0
0
5
10
15
20
25
k
Figure S17.21a. Controlled variable y(k) for a unit step change in ysp.
Controller output p(k):
5.5
5
4.5
4
p(k)
3.5
3
2.5
2
1.5
1
0.5
0
5
10
15
20
25
k
Figure S17.21b. Controlled output p(k) for a unit step change in ysp.
17-29
17.22
Dahlin's controller
From Eq. 17-66 with a1 = e-1/10=0.9, N=1, and A=e-1/1 = 0.37, the Dahlin
controller is
G DC ( z ) =
=
(1 − 0.37)
1 − 0.9 z −1
1 − 0.37 z −1 − (1 − 0.37) z −2 2(1 − 0.9)
3.15 − 2.84 z −1
3.15 − 2.84 z −1
=
1 − 0.37 z −1 − 0.63z −2 (1 − z −1 )(1 + 0.63z −1 )
By using Simulink, controller output and controlled variable are shown
below:
3.5
3
2.5
p(t)
2
1.5
1
0.5
0
0
5
10
15
20
25
time
30
35
40
45
50
Fig S17.22a. Controller output for Dahlin controller.
1.4
1.2
1
Output
0.8
0.6
0.4
0.2
0
0
5
10
15
20
25
time
30
35
40
45
50
Fig S17.22b. Closed-loop response for Dahlin controller.
17-30
Thus, there is no ringing (this is expected for a first-order system) and no
adjustment for ringing is required.
PID (ITAE setpoint)
For this controller, adjust the process time delay for the zero-order hold by
adding ∆t/2, and K=2, τ=10, θ=1.5 obtain the continuous PID controller
tunings from Table 12.3 as
KKc = 0.965(1.5/10) − 0.85
,
Kc = 2.42
τ/τI = 0.796 + (-0.1465)(1.5/10) ,
τD/τ = 0.308(1.5/10)0.929
,
,
τI = 12.92
τD = 0.529
Using the position form of the PID control law (Eq. 8-26 or 17-55)
1  1 


Gc ( z ) = 2.42 1 +
+ 0.529(1 − z −1 ) 

−1 
 12.92  1 − z 

=
3.89 − 4.98 z −1 + 1.28 z −2
1 − z −1
By using Simulink,
4
3.5
3
2.5
p(t)
2
1.5
1
0.5
0
-0.5
-1
0
5
10
15
20
25
time
30
35
40
45
50
Fig S17.22c. Controller output for PID (ITAE) controller
17-31
1.4
1.2
1
Output
0.8
0.6
0.4
0.2
0
0
5
10
15
20
25
Time
30
35
40
45
50
Fig S17.22d. Closed-loop response for PID (ITAE) controller.
Dahlin's controller gives better closed-loop performance than PID because
it includes time-delay compensation.
17.23
From Eq. 17-66 with a1 = e-1/5=0.819, N=5, and A=e-1/1 = 0.37, the Dahlin
controller is
G DC ( z ) =
=
(1 − 0.37)
1 − 0.819 z −1
1 − 0.37 z −1 − (1 − 0.37) z −6 1.25(1 − 0.819)
2.78 − 2.28 z −1
(1 − 0.37 z −1 − 0.63 z −6 )
By using Simulink-MATLAB, the controller output is shown in Fig.
S17.23
17-32
3
2.5
p(k)
2
1.5
1
0.5
0
5
10
15
20
25
k
Figure S17.23. Controller output for Dahlin controller.
As noted in Fig.S17.23, ringing does not occur. This is expected for a
first-order system.
17.24
Dahlin controller
Using Table 17.1 with K=0.5 , r =1.0, p =0.5,
G( z) =
0.1548 z −1 + 0.0939 z −2
1 − 0.9744 z −1 + 0.2231z −2
From Eq. 17-64, with λ = ∆t = 1, Dahlin's controller is
G DC ( z ) =
=
(1 − 0.9744 z −1 + 0.2231z −2 ) 0.632 z −1
0.1548 z −1 + 0.0939 z −2
1 − z −1
0.632 − 0.616 z −1 + 0.141z −2
(1 − z −1 )(0.1548 + 0.0939 z −1 )
From Eq. 17-63,
17-33
Y ( z)
0.632 z −1
=
Ysp ( z ) 1 − 0.368 z −1
y(k) = 0.368 y(k-1) + 0.632 ysp(k-1)
Since this is first order, no overshoot occurs.
By using Simulink-MATLAB, the controller output is shown:
5
4
p(k)
3
2
1
0
-1
0
5
10
15
20
25
k
Figure S17.24a. Controller output for Dahlin controller.
As noted in Fig. S17.24 a, ringing occurs for Dahlin's controller.
Vogel-Edgar controller
From Eq. 17-71, the Vogel-Edgar controller is
GVE ( z ) =
2.541 − 2.476 z −1 + 0.567 z −2
1 − 0.761z −1 − 0.239 z −2
Using Eq. 17-70 and simplifying,
Y ( z ) (0.393 z −1 + 0.239 z −2 )
=
Ysp ( z )
1 − 0.368 z −1
y(k) = 0.368 y(k-1) + 0.393 ysp(k-1) + 0.239 ysp (k-2)
Again no overshoot occurs since y(z)/ysp(z) is first order.
By using Simulink-MATLAB, the controller output is shown below:
17-34
2.6
2.4
2.2
2
p(k)
1.8
1.6
1.4
1.2
1
0.8
0
5
10
15
20
25
k
Figure S17.24b. Controller output for Vogel-Edgar controller.
As noted in Fig. S17.24 b, the V-E controller does not ring.
17.25
a)
Material Balance for the tanks,
dh1
1
= q1 − q2 − (h1 − h2 )
dt
R
dh
1
A2 2 = (h1 − h2 )
dt
R
A1
where A1 = A2 = π/4(2.5)2=4.91 ft2
Using deviation variables and taking Laplace transform
A1sH1′( s ) = Q1′( s ) − Q2′ ( s ) −
A2 sH 2′ ( s ) =
1
1
H1′( s ) + H 2′ ( s )
R
R
1
1
H1′( s ) − H 2′ ( s )
R
R
17-35
(1)
(2)
From (2)
H 2′ ( s ) =
1
H1′( s )
A2 Rs + 1
Substituting into (1) and simplifying
( A1 A2 R ) s 2 + ( A1 + A2 ) s  H1′( s ) = [ A2 Rs + 1][Q1′( s ) − Q2′ ( s )]
G p ( s) =
−( A2 Rs + 1)
−0.204( s + 0.102)
H1′( s )
=
=
2
Q2′ ( s ) ( A1 A2 R ) s + ( A1 + A2 ) s
s ( s + 0.204)
Gd ( s ) =
0.204( s + 0.102)
H1′( s )
A2 Rs + 1
=
=
2
Q1′( s ) ( A1 A2 R) s + ( A1 + A2 ) s
s ( s + 0.204)
Using Eq. 17-64, with N =0, A=e-∆t/λ and HG(z) = KtKvHGp(z), Dahlin's
controller is
GDC ( z ) =
1 (1 − A) z −1
HG (1 − z −1 )
Using z-transforms,
HG(z)=KtKvHGp(z)=
−0.202 z −1 + 0.192 z −2
(1 − z −1 )(1 − 0.9 z −1 )
Then,
GDC ( z ) =
(1 − z −1 )(1 − 0.9 z −1 )
(1 − A) z −1
⋅
(−0.202 z −1 + 0.192 z −2 ) (1 − z −1 )
=
b)
GDC =
(1 − A)(1 − 0.9 z −1 )
−0.202 + 0.192 z −1
(1 − A)(1 − 0.9 z −1 )
−0.202 + 0.192 z −1
By using Simulink-MATLAB,
17-36
0
-0.2
-0.4
-0.6
p(k)
-0.8
-1
-1.2
-1.4
-1.6
-1.8
-2
0
5
10
15
20
25
time
30
35
40
45
50
Figure S17.25. Controller output for Dahlin's controller.
As noted in Fig. S17.25, the controller output doesn't oscillate.
c)
This controller is physically realizable since the z-0 coefficient in the
denominator is non-zero. Thus, controller is physically realizable for all
value of λ.
d)
λ is the time constant of the desired closed-loop transfer function. From
the expression for Gp(s) the open-loop dominant time constant is 1/0.204 =
4.9 min.
A conservative initial guess for λ would be equal to the open-loop time
constant, i.e., λ = 4.9 min. If the model accuracy is reliable, a more bold
guess would involve a smaller λ, say 1/3 rd of the open-loop time constant.
In that case, the initial guess would be λ = (1/3) × 4.9 =1.5 min.
17.26
G f ( s) =
K (τ1s + 1) P ( s )
=
τ2 s + 1
E ( s)
Substituting s ≅ (1 − z −1 ) / ∆t
G f ( z) = K
into equation above:
(τ1 + ∆t ) − τ1 z −1
τ1 (1 − z −1 ) / ∆t + 1
τ1 (1 − z −1 ) + ∆t
K
K
=
=
(τ2 + ∆t ) − τ2 z −1
τ2 (1 − z −1 ) / ∆t + 1
τ2 (1 − z −1 ) + ∆t
17-37
Then,
G f ( z) =
b1 =
where
b1 + b2 z −1 P ( z )
=
1 + a1 z −1 E ( z )
K (τ1 + ∆t )
τ2 + ∆t
b2 =
,
− K τ1
τ2 + ∆t
a1 =
and
−τ2
τ2 + ∆t
Therefore,
(1 + a1 z −1 ) P ( z ) = (b1 + b2 z −1 ) E ( z )
Converting the controller transfer function into a difference equation form:
p(k ) = − a1 p (k − 1) + b1e(k ) + b2 e(k − 1)
Using Simulink-MATLAB, discrete and continuous responses are
compared : ( Note that b1=0.5 , b2 = −0.333 and a1= −0.833)
1
0.9
Output
0.8
0.7
0.6
0.5
Continuos response
Discrete response
0.4
0
5
10
15
20
25
Time
30
35
40
45
50
Figure S17.26. Comparison between discrete and continuous controllers.
17-38
17.27
Using Table 17.1 with K= -1/3, r = 1/3, p = 1/5,
G( z) =
(−0.131 − 0.124 z −1 ) z −5
≡ G( z)
(1 − 0.716 z −1 )(1 − 0.819 z −1 )
~
Since the zero is at z =-0.95, it should be included in G+ ( z ) . Therefore
(−0.131 − 0.124 z −1 ) z −5
~
G+ ( z ) =
= 0.514 z −5 + 0.486 z −6
(−0.131 − 0.124)
~
G− ( z ) =
(−0.131 − 0.124)
(1 − 0.716 z −1 )(1 − 0.819 z −1 )
For deadbeat filter, F(z) = 1
Using Eq. 17-77, the IMC controller is
(1 − 0.716 z −1 )(1 − 0.819 z −1 )
Gc *( z ) = G1 − −1 ( z ) F ( z ) =
(−0.131 − 0.124)
By using Simulink-MATLAB, the IMC response for a unit step in load at
t=10 is shown in Fig. S17.27
0.7
0.6
0.5
Output
0.4
0.3
0.2
0.1
0
0
5
10
15
20
25
30
35
40
45
Time
Fig. S17.27. IMC close-loop response for a unit step change in load at t=10.
17-39
123456789 8
19.1
From definition of xc, 0 ≤ xc ≤ 1
f(x) = 5.3 x e (-3.6x +2.7)
Let three initial points in [0,1] be 0.25, 0.5 and 0.75. Calculate x4 using
Eq. 19-8,.
x1
0.25
f1
8.02
x2
0.5
f2
6.52
x3
0.75
f3
3.98
x4
0.0167
For next iteration, select x4, and x1 and x2 since f1 and f2 are the largest
among f1, f2, f3. Thus successive iterations are
x1
0.25
0.25
0.25
0.25
f1
8.02
8.02
8.02
8.02
x2
0.5
0.5
0.334
0.271
xopt = 0.2799
f2
6.52
6.52
7.92
8.06
x3
0.017
0.334
0.271
0.280
f3
1.24
7.92
8.06
8.06
x4
0.334
0.271
0.280
not needed
7 function evaluations
19.2
As shown in the drawing, there is both a minimum and maximum value of
the air/fuel ratio such that the thermal efficiency is non- zero. If the ratio is
too low, there will not be sufficient air to sustain combustion. On the other
hand, problems in combustion will appear when too much air is used.
The maximum thermal efficiency is obtained when the air/fuel ratio is
stoichiometric. If the amount of air is in excess, relatively more heat will
be “absorbed” by the air (mostly nitrogen). However if the air is not
sufficient to sustain the total combustion, the thermal efficiency will
decrease as well.
Solution Manual for Process Dynamics and Control, 2nd edition,
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp
19-1
19.3
By using Excel-Solver, this optimization problem is quickly solved. The
selected starting point is (1,1):
Initial values
Final values
X1
X2
1
1
0.776344 0.669679
max Y= 0.55419
Constraints
0 ≤ X1 ≤ 2
0 ≤ X2 ≤2
Table S19.3. Excel solution
Hence the optimum point is ( X1*, X2* ) =(0.776, 0.700)
and the maximum value of Y is Ymax = 0.554
19.4
Let N be the number of batches/year. Then NP ≥ 300,000
Since the objective is to minimize the cost of annual production, only the
required amount should be produced annually and no more. That is,
NP = 300,000
a)
(1)
Minimize the total annual cost,
 $ 
 $   batch 
0.4  hr 
min TC = 400,000 

 +2P 
 50   N 
 batch 
 batch 
 hr   yr 
 $ 
+ 800 P0.7  
 yr 
Substituting for N from (1) gives
min TC = 400,000 + 3x107 P–0.6 + 800 P0.7
19-2
b)
There are three constraints on P
i)
ii)
P≥0
N is integer. That is,
(300,000/P) = 0, 1, 2,…
iii)
Total production time is 320 x 24 hr/yr
 batch 
 hr 
(2 P0.4 + 14) 
 ≤ 7680
 × N
 batch 
 yr 
Substituting for N from (1) and simplifying
6 × 105P-0.6 + 4.2 × 106P-1 ≤ 7680
c)
d (TC )
= 0 = 3 ×107 (−0.6) P −1.6 + 800(0.7) P −0.3
dP
1/1.3
 3 × 107 ( −0.6) 
lb
opt
P =
= 2931

batch
 −800(0.7) 
d 2 (TC )
= 3 ×107 (−0.6)(−1.6) P −2.6 + 800(0.7)(−0.3) P −1.3
2
dP
2
d (TC )
= 2.26 ×10−2 ⟩ 0 hence minimum
dP 2 P = Popt
Nopt = 300,000/Popt = 102.35 not an integer.
Hence check for Nopt = 102 and Nopt = 103
For Nopt = 102, Popt = 2941.2, and TC = 863207
For Nopt = 103, Popt = 2912.6, and TC = 863209
Hence optimum is 102 batches of 2941.2 lb/batch.
Time constraint is
6 ×105 P −0.6 + 4.2 ×106 P −1 = 6405.8 ≤ 7680 , satisfied
19-3
19.5
Let x1 be the daily feed rate of Crude No.1 in bbl/day
x2 be the daily feed rate of Crude No.2 in bbl/day
Objective is to maximize profit
max P = 2.00 x1 + 1.40 x2
Subject to constraints
gasoline :
kerosene:
fuel oil:
0.70 x1 + 0.31 x2 ≤ 6000
0.06 x1 + 0.09 x2 ≤ 2400
0.24 x1 + 0.60 x2 ≤ 12,000
By using Excel-Solver,
Initial values
Final values
x1
1
0
x2
1
19354.84
max P = 27096.77
Constraints
0.70 x1 + 0.31 x2
0.06 x1 + 0.09 x2
0.24 x1 + 0.60 x2
6000
1741.935
11612.9
Table S19.5. Excel solution
Hence the optimum point is (0, 19354.8)
Crude No.1 = 0 bbl/day
Crude No.2 = 19354.8 bbl/day
19-4
19.6
Objective function is to maximize the revenue,
max R = -40x1 +50x3 +70x4 +40x5 –2x1-2x2
(1)
*Balance on column 2
x2 = x4 + x5
(2)
* From column 1,
1 .0
x1 =
x 2 = 1.667( x 4 + x5 )
0.60
0 .4
x3 =
x 2 = 0.667( x 4 + x5 )
0.60
(3)
(4)
Inequality constraints are
x4 ≥ 200
x4 ≤ 400
x1 ≤ 2000
x4 ≥ 0 x5 ≥ 0
(5)
(6)
(7)
(8)
The restricted operating range for column 2 imposes additional inequality
constraints. Medium solvent is 50 to 70% of the bottoms; that is
0.5 ≤
x4
≤ 0.7 or
x2
0.5 ≤
x4
≤ 0.7
x4 + x5
Simplifying,
x4 –x5 ≥ 0
0.3 x4 –0.7x5 ≤ 0
(9)
(10)
No additional constraint is needed for the heavy solvent. That the heavy
solvent will be 30 to 50% of the bottoms is ensured by the restriction on
the medium solvent and the overall balance on column 2.
By using Excel-Solver,
19-5
Initial values
Final values
max R =
Constraints
x2 - x4 - x5
x1 - 1.667x2
x3 - 0.667x2
x4
x4
x1 - 1.667x2
x4 - x5
0.3x4 - 0.7x5
x1
1
1333.6
x2
1
800
x3
1
533.6
x4
1
400
x5
1
400
13068.8
0
7.467E-10
-1.402E-10
400
400
1333.6
0
-160
Table S19.6. Excel solution
Thus the optimum point is x1 =1333.6, x2 =800; x3=533.6, x4 = 400 and
x5 = 400.
Substituting into (5), the maximum revenue is 13,068 $/day, and the
percentage of output streams in column 2 is 50 % for each stream.
19.7
The objective is to minimize the sum of the squares of the errors for the
material balance, that is,
min E = (wA + 11.1 – 92.4)2 + (wA +10.8 –94.3)2 + (wA + 11.4 –93.8)2
Subject to wA ≥ 0
Solve analytically,
dE
= 0 = 2 (wA + 11.1 – 92.4) + 2(wA +10.8 –94.3)
dw A
+2(wA + 11.4 –93.8)
Solving for wA…
wA opt = 82.4 Kg/hr
Check for minimum,
d 2E
= 2 + 2 + 2 = 6 > 0 , hence minimum
2
dw A
19-6
19.8
a)
Income = 50 (0.1 +0.3xA + 0.0001S – 0.0001 xAS)
Costs = 2.0 + 10xA + 20 xA2 + 1.0 + 0.003 S + 2.0x10-6S2
f = 2.0 +5xA + 0.002S – 20xA2 – 2.0x10-6S2 – 0.005xAS
b)
Using analytical method
∂f
= 0 = 5 − 40 x A − 0.005S
∂x A
∂f
= 0 = 0.002 − 4.0 × 10 −6 − 0.005 x A
∂S
Solving simultaneously, xA = 0.074 ,
constraints.
S = 407 which satisfy the given
19.9
By using Excel-Solver
Initial values
Final values
TIME
0
1
2
3
4
5
6
7
8
9
10
EQUATION
0.000
0.066
0.202
0.351
0.490
0.608
0.703
0.778
0.835
0.879
0.911
τ1
τ2
1
0.5
2.991562 1.9195904
DATA
0.000
0.058
0.217
0.360
0.488
0.600
0.692
0.772
0.833
0.888
0.925
SQUARE ERROR
SUM=
19-7
0.00000000
0.00005711
0.00022699
0.00007268
0.00000403
0.00006008
0.00012252
0.00003428
0.00000521
0.00008640
0.00019150
0.00086080
Hence the optimum values are τ1=3 and τ2=1.92. The obtained model is
compared with that obtained using MATLAB.
1
Y/K
0.8
0.6
0.4
0.2
MATLAB
data
equation
0
0
5
10
15
time
20
25
30
Figure S19.9. Comparison between the obtained model with that obtained
using MATLAB
19.10
Let
x1 be gallons of suds blended
x2 be gallons of premium blended
x3 be gallons of water blended
Objective is to minimize cost
min C = 0.25x1 + 0.40x2
(1)
Subject to
x1 + x2 + x3 = 10,000
(2)
0.035 x1 + 0.050 x2 = 0.040 × 10,000
(3)
x1 ≥ 2000
(4)
x1 ≤ 9000
(5)
19-8
x2 ≥ 0
(6)
x3 ≥ 0
(7)
The problem given by Eqs. 1, 2, 3, 4, 5, 6, and 7 is optimized using ExcelSolver,
Initial values
Final values
min C =
Constraints
x1+x2+x3-10000
0.035x1+0.050x2- 400
x1- 2000
x1- 9000
x2
x3
x1
1
6666.667
x2
1
3333.333
x3
1
0
3000
0
0.0E+00
4666.667
-2333.333
3333.333
0
Table S19.10. Excel solution
Thus the optimum point is x1 = 6667 , x2 = 3333 and x3 = 0.
The minimum cost is $3000
19.11
Let
xA be bbl/day of A produced
xB be bbl/day of B produced
Objective is to maximize profit
max P = 10xA + 14xB
(1)
Subject to
Raw material constraint:
120xA+ 100xB ≤ 9,000
(2)
Warehouse space constraint: 0.5 xA + 0.5 xB ≤ 40
(3)
Production time constraint: (1/20)xA + (1/10)xB ≤ 7
(4)
19-9
Initial values
Final values
max P =
xA
1
20
xB
1
60
1040
Constraints
120xA+ 100xB
0.5 xA + 0.5 xB
(1/20)xA + (1/10)xB
8400
40
7
Table S19.11. Excel solution
Thus the optimum point is xA = 20 and xB = 60
The maximum profit = $1040/day
19.12
PID controller parameters are usually obtained by using either process
model, process data or computer simulation. These parameters are kept
constant in many cases, but when operating conditions vary, supervisory
control could involve the optimization of these tuning parameters. For
instance, using process data, Kc ,τI and τD can be automatically calculated
so that they maximize profits. Overall analysis of the process is needed in
order to achieve this type of optimum control.
Supervisory and regulatory control are complementary. Of course,
supervisory control may be used to adjust the parameters of either an
analog or digital controller, but feedback control is needed to keep the
controlled variable at or near the set-point.
19.13
Assuming steady state behavior, the optimization problem is,
max f = D e
Subject to
0.063 c –D e = 0
0.9 s e – 0.9 s c – 0.7 c – D c = 0
19-10
(1)
(2)
-0.9 s e + 0.9 s c + 10D – D s = 0
D, e, s, c ≥ 0
(3)
where f = f(D, e, c, s)
Excel-Solver is used to solve this problem,
c
D
e
s
1
1
1
1
0.479031 0.045063 0.669707 2.079784
Initial values
Final values
max f = 0.030179
Constraints
0.063 c –D e
0.9 s e – 0.9 s c – 0.7 c – Dc
-0.9 s e + 0.9 s c + 10D – Ds
2.08E-09
-3.1E-07
2.88E-07
Table S19.13. Excel solution
Thus the optimum value of D is equal to 0.045 h-1
19.14
Material balance:
Overall :
FA + FB = F
Component B:
FB CBF + VK1CA – VK2CB = F CB
Component A:
FA CAF + VK2CB – VK1CA = FCA
Thus the optimization problem is:
max (150 + FB) CB
Subject to:
0.3 FB + 400CA − 300CB = (150 + FB)CB
45 + 300 CB – 400 CA = (150 + FB) CA
FB ≤ 200
19-11
CA, CB, FB ≥ 0
By using Excel- Solver, the optimum values are
FB = 200 l/hr
CA = 0.129 mol A/l
CB = 0.171 mol B/l
19.15
Material balance:
Overall :
FA + FB = F
Component B:
FB CBF + VK1CA – VK2CB = F CB
Component A:
FA CAF + VK2CB – VK1CA = FCA
Thus the optimization problem is:
max (150 + FB) CB
Subject to:
0.3 FB + 3 × 106e(-5000/T)CA V − 6 × 106e(-5500/T)CB V = (150 + FB)CB
45 + 6 × 106e(-5500/T)CB V – 3 × 106e(-5000/T) CA V = (150 + FB) CA
FB ≤ 200
300 ≤ T ≤ 500
CA, CB, FB ≥ 0
By using Excel- Solver, the optimum values are
FB = 200 l/hr
CA = 0.104 molA/l
CB = 0.177 mol B/l
T = 311.3 K
19-12
123456789 8
20.1
a)
The unit step response is

 1 
2e − s
5
20 
1
  = 2e − s  +
Y ( s ) = G p ( s )U ( s ) = 
−
 s 5s + 1 10s + 1
 (10 s + 1)(5s + 1)  s 
Therefore,
[
y (t ) = 2 S (t − 1) 1 + e − (t −1) / 5 − 2e − (t −1) / 10
]
For ∆t = 1.0,
S i = y (i∆t ) = y (i ) = {0, 0.01811, 0.06572, 0.1344, 0.2174, 0.3096...}
b)
From the expression for y(t) in part (a) above
y(t) = 0.95 (2) at
t =37.8, by trial and error.
Hence N = 38, for 95% complete response.
20.2
Note that G ( s ) = Gv ( s )G p ( s )Gm ( s ) . From Figure 12.2,
a)
Ym ( s )
2(1 − 9 s )
= G( s) =
P(s )
(15s + 1)(3s + 1)
For a unit step change, P ( s ) = 1 / s , and (1) becomes:
Ym ( s ) =
1
2(1 − 9 s )
s (15s + 1)(3s + 1)
Partial Fraction Expansion:
Solution Manual for Process Dynamics and Control, 2nd edition,
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp
20-1
(1)
Ym ( s ) =
A
B
C
1
2(1 − 9 s )
+
+
=
s (15s + 1) (3s + 1) s (15s + 1)(3s + 1)
(2)
where
A=
B=
C=
2(1 − 9 s )
(15s + 1)(3s + 1)
2(1 − 9 s )
s (3s + 1)
=2
s =0
= −60
1
s =−
15
2(1 − 9 s )
s (15s + 1)
=6
s =−
1
3
Substitute into (2) and take inverse Laplace transform:
y m (t ) = 2 − 4e − t / 15 + 2e − t / 3
b)
(3)
The new steady-state value is obtained from (3) to be ym(∞)=2
For t = t99, ym(t)=0.99ym(∞) = 1.98. Substitute into (3)
1.98 = 2 − 4e − t99 / 15 + 2e − t99 / 3
(4)
Solving (4) for t99 by trial and error gives t99 ≈ 79.5 min
Thus, we specify that ∆t =79.5 min/40 ≈ 2 min
Sample No
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Si
-0.4739
-0.5365
-0.4106
-0.2076
0.0177
0.2393
0.4458
0.6330
0.8022
0.9482
1.0785
1.1931
1.2936
1.3816
1.4587
Sample No
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Table S20.2. Step response coefficients
20-2
Si
1.5263
1.5854
1.6371
1.6824
1.7221
1.7568
1.7871
1.8137
1.8370
1.8573
1.8751
1.8907
1.9043
1.9163
1.9267
Sample No
31
32
33
34
35
36
37
38
39
40
Si
1.9359
1.9439
1.9509
1.9570
1.9624
1.9671
1.9712
1.9748
1.9779
1.9807
20.3
From the definition of matrix S, given in Eq. 20-20, for P=5, M=1, with Si
obtained from Exercise 20.1,
 S1   0 
 S   0.01811
 2 

S =  S 3  = 0.06572
  

 S 4   0.1344 
 S 5   0.2174 
From Eq. 20-58
Kc = (STS)-1ST
Kc = [0 0.2589 0.9395 1.9206 3.1076] = Kc1T
Because Kc1T is defined as the first row of Kc .
Using the given analytical result,
Kc1T =
[S1
1
5
∑ (S
i =1
2
i
S2
S3
S4
S5 ]
)
=
1
[0 0.01811 0.06572 0.1344 0.2174]
0.06995
=
[0
0.2589 0.9395 1.9206 3.1076]
which is the same as the answer obtained above using (20-58)
20.4
The step response is obtained from the analytical unit step response as in
Example 20.1. The feedback matrix Kc is obtained using Eq. 20-57 as in
Example 20.5. These results are not reported here for sake of brevity. The
closed-loop response for set-point and disturbance changes are shown
below for each case. MATLAB MPC Toolbox was used for the
simulations.
20-3
i)
For this model horizon, the step response is over 99% complete as in
Example 20.5; hence the model is good. The set-point and disturbance
responses shown below are non-oscillatory and have long settling times
Outputs
1.5
1
y
0.5
0
0
10
20
30
40
50
60
70
80
90
100
70
80
90
100
Time
Manipulated Variables
2
1.5
u
1
0.5
0
0
10
20
30
40
50
60
Time
Figure S20.4a. Controller i); set-point change.
Outputs
0.8
0.6
y
0.4
0.2
0
0
10
20
30
40
50
60
70
80
90
100
70
80
90
100
Time
Manipulated Variables
0
-0.5
u
-1
-1.5
0
10
20
30
40
50
60
Time
Figure S20.4b. Controller i); disturbance change.
20-4
ii)
The set-point response shown below exhibits same overshoot, smaller
settling time and undesirable "ringing" in u compared to part i). The
disturbance response shows a smaller peak value, a lack of oscillations,
and faster settling of the manipulated input.
Outputs
1.5
1
y
0.5
0
0
10
20
30
40
50
60
70
80
90
100
70
80
90
100
Time
Manipulated Variables
15
10
5
u
0
-5
-10
0
10
20
30
40
50
60
Time
Figure S20.4c. Controller ii); set-point change.
Outputs
0.4
0.3
y
0.2
0.1
0
0
10
20
30
40
50
60
70
80
90
100
70
80
90
100
Time
Manipulated Variables
0
-0.5
u
-1
-1.5
0
10
20
30
40
50
60
Time
Figure S20.4d. Controller ii); disturbance change.
20-5
iii)
The set-point and disturbance responses shown below show the same
trends as in part i).
Outputs
1.5
1
y
0.5
0
0
10
20
30
40
50
60
70
80
90
100
70
80
90
100
Time
Manipulated Variables
10
5
u
0
-5
-10
0
10
20
30
40
50
60
Time
Figure S20.4e. Controller iii); set-point change.
Outputs
0.4
0.3
y
0.2
0.1
0
0
10
20
30
40
50
60
70
80
90
100
70
80
90
100
Time
Manipulated Variables
0
-0.5
u
-1
-1.5
0
10
20
30
40
50
60
Time
Figure S20.4f. Controller iii); disturbance change.
20-6
iv)
The set-point and load responses shown below exhibit the same trends as
in parts (i) and (ii). In comparison to part (iii), this controller has a larger
penalty on the manipulated input and, as a result, leads to smaller and less
oscillatory input effort at the expense of larger overshoot and settling time
for the controlled variable.
Outputs
1.5
1
y
0.5
0
0
10
20
30
40
50
60
70
80
90
100
70
80
90
100
Time
Manipulated Variables
4
3
2
u
1
0
-1
0
10
20
30
40
50
60
Time
Figure S20.4g. Controller iv); set-point change.
Outputs
0.5
0.4
0.3
y
0.2
0.1
0
0
10
20
30
40
50
60
70
80
90
100
70
80
90
100
Time
Manipulated Variables
0
-0.5
u
-1
-1.5
0
10
20
30
40
50
60
Time
Figure S20.4h. Controller iv); disturbance change.
20-7
20.5
There are many sets of values of M, P and R that satisfy the given
constraint for a unit load change. One such set is M=3, P=10, R=0.01 as
shown in Exercise 20.4(iii). Another set is M=3, P=10, R=0.1 as shown in
Exercise 20.4(iv). A third set of values is M=1, P=5, R=0 as shown in
Exercise 20.4(i).
20.6
(Use MATLAB Model Predictive Control Toolbox)
As shown below, controller a) gives a better disturbance response with a
smaller peak deviation in the output and less control effort. However,
controller (a) is poorer for a set-point change because it leads to
undesirable "ringing" in the manipulated input.
Outputs
1.5
1
y
0.5
0
0
10
20
30
40
50
60
70
80
90
100
70
80
90
100
Time
Manipulated Variables
15
10
5
u
0
-5
-10
0
10
20
30
40
50
60
Time
Figure S20.6a. Controller a); set-point change
20-8
Outputs
1.5
1
y
0.5
0
0
10
20
30
40
50
60
70
80
90
100
70
80
90
100
Time
Manipulated Variables
15
10
5
u
0
-5
-10
0
10
20
30
40
50
60
Time
Figure S20.6b. Controller a); disturbance change.
Outputs
0.4
0.3
y
0.2
0.1
0
0
10
20
30
40
50
60
70
80
90
100
70
80
90
100
Time
Manipulated Variables
0
-0.5
u
-1
-1.5
0
10
20
30
40
50
60
Time
Figure S20.6c. Controller b); set-point change.
20-9
Outputs
0.4
0.3
y
0.2
0.1
0
0
10
20
30
40
50
60
70
80
90
100
70
80
90
100
Time
Manipulated Variables
0
-0.5
u
-1
-1.5
0
10
20
30
40
50
60
Time
Figure S20.6d. Controller b); disturbance change.
20.7
The unconstrained MPC control law has the controller gain matrix:
Kc = (STQS+R)-1STQ
For this exercise, the parameter values are:
m = r = 1 (SISO), Q=I, R=1 and M=1
Thus (20-57) becomes
Kc = (STQS+R)-1STQ
Which reduces to a row vector: Kc =
[S1 S 2 S 3 ... S P ]
P
∑S
i =1
2
i
+1
20.8
Inequality constraints on the manipulated variables are usually satisfied if
the instrumentation and control hardware are working properly. However
the constraints on the controlled variables are applied to the predicted
outputs. If the predictions are inaccurate, the actual outputs could exceed
the constraints even though the predicted values do not.
20-10
20.9
(Use MATLAB Model Predictive Control Toolbox)
a) M=5 vs. M=2
2
XD
XB
1.5
1
y(t)
0.5
0
-0.5
0
20
40
60
80
100
120
140
Time (min)
0.2
R
S
0.1
u(t)
0
-0.1
-0.2
0
20
40
60
80
100
120
140
Time (min)
Figure S20.9a1. Simulations for P=10 , M=5 and R=0.1I.
2
XD
XB
1.5
1
y(t)
0.5
0
-0.5
0
20
40
60
80
100
120
140
Time (min)
0.2
R
S
0.1
u(t)
0
-0.1
-0.2
0
20
40
60
80
100
120
140
Time (min)
Figure S20.9a2. Simulations for P=10 , M=2 and R=0.1I.
20-11
b) R=0.1I .vs R=I
2
XD
XB
1.5
1
y(t)
0.5
0
-0.5
0
20
40
60
80
100
120
140
Time (min)
0.2
R
S
0.1
u(t)
0
-0.1
-0.2
0
20
40
60
80
100
120
140
Time (min)
Figure S20.9b1. Simulations for P=10 , M=5 and R=0.1I.
2
XD
XB
1.5
1
y(t)
0.5
0
-0.5
0
20
40
60
80
100
120
140
Time (min)
0.2
R
S
0.1
u(t)
0
-0.1
-0.2
0
20
40
60
80
100
120
140
Time (min)
Figure S20.9b2. Simulations for P=10 , M=5 and R=I.
Notice that the larger control horizon M and the smaller input weighting
R, the more control effort is needed.
20-12
20.10
The open-loop unit step response of Gp(s) is
 e −6 s 1 

10  
1
− ( t − 6 ) / 10
 = L-1  e −6 s  −
y (t ) = L 
  = S (t − 6) 1 − e
 s 10 s + 1  

 10s + 1 s 
[
-1
]
By trial and error, y(34) < 0.95, y(36) > 0.95.
Therefore N∆t =36 or N = 18
The coefficients {S i } are obtained from the expression for y(t) and the
predictive controller is obtained following the procedure of Example 20.5.
The closed-loop responses for a unit set-point change are shown below for
the three controller tunings
20.11
(Use MATLAB Model Predictive Control Toolbox)
a) M=5 vs. M=2
2
XD
XB
1.5
1
y(t)
0.5
0
-0.5
0
20
40
60
80
100
120
140
Time (min)
0.2
R
S
0.1
0
u(t)
-0.1
-0.2
-0.3
0
20
40
60
80
100
Time (min)
Figure S20.11a1. Simulations for P=10 , M=5 and R=0.1I.
20-13
120
140
2
XD
XB
1.5
1
y(t)
0.5
0
-0.5
0
20
40
60
80
100
120
140
Time (min)
0.2
R
S
0.1
0
u(t)
-0.1
-0.2
-0.3
0
20
40
60
80
100
120
140
Time (min)
Figure S20.11a2. Simulations for P=10 , M=2 and R=0.1I.
b) R=0.1I .vs R=I
2
XD
XB
1.5
1
y(t)
0.5
0
-0.5
0
20
40
60
80
100
120
140
Time (min)
0.2
R
S
0.1
0
u(t)
-0.1
-0.2
-0.3
0
20
40
60
80
100
120
140
Time (min)
Figure S20.11b1. Simulations for P=10 , M=5 and R=0.1I.
20-14
2
XD
XB
1.5
1
y(t)
0.5
0
-0.5
0
20
40
60
80
100
120
140
Time (min)
0.2
R
S
0.1
0
u(t)
-0.1
-0.2
-0.3
0
20
40
60
80
100
120
Time (min)
Figure S20.11b2. Simulations for P=10 , M=5 and R=I..
20-15
140
12345678998
22.1
Microwave Operating States
Condition
Fan
Open the door
Place the food inside
Close the door
Set the time
Heat up food
Cooking complete
Light Timer
Rotating Microwave
Base
Generator
Door
Switch
OFF
ON
OFF
OFF
OFF
ON
OFF
OFF
ON
OFF
OFF
OFF
ON
OFF
OFF
OFF
ON
OFF
OFF
OFF
ON
OFF
OFF
OFF
ON
OFF
OFF
OFF
OFF
OFF
Safety Issues:
o Door switch is always OFF before the microwave generator is turned ON.
o Fan always ON when microwave generator is ON.
22.2
Input Variables:
ON
STOP
EMERGENCY
Output Variables:
START
STOP
(1)
(0)
Truth Table
ON
1
0
1
0
1
0
1
0
STOP
1
1
0
0
1
1
0
0
EMERGENCY
1
1
1
1
0
0
0
0
START/STOP
0
0
0
0
0
0
1
0
Solution Manual for Process Dynamics and Control, 2nd edition,
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp
22-1
The truth state table is used to find the logic law that relates inputs with
outputs:
ON • STOP • EMERGENCY
Applying Boolean Algebra we can obtain an equivalent expression:
ON • ( STOP • EMERGENCY ) = ON • ( STOP + EMERGENCY )
Finally the binary logic and ladder logic diagrams are given in Figure
S22.2:
Binary Logic Diagram:
ON
STOP
EMERGENCY
Ladder Logic Diagram
Start
CR1
CR1
Stop
CR2
CR2
CR3
TH
CR3
M
Figure S22.2.
22-2
22.3
A
0
1
0
1
B
0
0
1
1
Y
1
1
0
1
From the truth table it is possible to find the logic operation that gives the
desired result,
A• B
Since a NAND gate is equivalent to an OR gate with two negated inputs,
our expression reduces to: A • B = A + B
Finally the binary logic diagram is given in Figure S22.3.
A
Y
B
Figure S22.3.
22-3
22.4
Information Flow Diagram
START
Inlet valve
open
Outlet valve
close
Stop
No
L>LH
Stirrer ON
Yes
Yes
Stop
No
T=Tsetpoint
Inlet valve
closed
Outlet valve
open
Stirrer OFF
Inlet valve
closed
Outlet valve
open
Stirrer OFF
Yes
Stop
L<LL
22-4
No
Ladder Logic Diagram
Start
CR1
LH
Tsetpoint
CR1
CR2
CR2
Stop
LH
LL
Tsetpoint
Sequential Function Chart
22-5
M1
M2
1
Fill the tank
Liquid Level = LH
1a
Open inlet valve
2
1b
Close exit valve
Heat liquid
Temperature = Tsetpoint
3
Empty tank
Liquid Level = LL
3a
Close inlet valve
3b
Figure S22.4.
22-6
Open exit valve
22.5
Information Flow Diagram
START
Open V1
P1 ON
L=L1
No
No
L=L0
Heat ON
Close V1
P1 OFF
No
Temperature>TH
Heat OFF
Open V2
P2 ON
No
L=L0
Close V2
P2 OFF
22-7
Ladder Logic Diagram:
R1= Pump 1 R2= Valve 2 R3= Heater
Start
CR2
R4= Pump 2
CR1
L1
CR1
CR3
TH
L0
CR3
CR4
TH
CR2
L0
CR4
Sequential Function Chart:
1
Initial Step
B
Heat
Fill
1
V1
Q
Fill
2
1
Temp
V1
L1
Full
TH
1
4
V1
L0
Figure S22.5.
22-8
22.6
Information Flow Diagram:
START
L<L1
L<L3
No
Open V1
Open V2
No
No
L<L2
L<L4
Close V1
Open W 1
Close V2
Open W 2
No
No
L<L1
L<L3
Close W 1
Close W 2
22-9
No
Sequential Function Chart:
Init
Init
Tank 1
Tank 2
B
Fill
2
Fill
2
Q
L2
V1
L4
Empty
1
Close V1
Empty
Close V2
Open W1
2
Open W2
L1
L3
Ladder Logic Diagram:
Valve 1
Start
CR1
L1
CR3
L3
CR1
CR1
Valve 2
CR2
CR4
CR2
CR2
W1
CR3
CR5
CR3
L2
CR3
W4
CR4
CR6
L4
CR4
L1
CR5
L3
CR6
Figure S22.6.
22-10
CR4
22.7
Information Diagram:
No
START
LS2
Start M
No
LS1
P=Ps
Stop M
Open V2
Yes
Close V2
Open V3
No
No
LS3
LS1
Close V2
Stop M
Close V3
Open V4
No
LS2
22-11
Ladder Logic Diagram:
R1= V1
R2= M
R3= V4
Start
CR3
R4= V2
CR1
LS2
CR1
CR2
LS2
LS1
CR2
CR3
LS3
LS2
CR3
CR3
LS2
Ps
LS2
Ps
CR5
CR4
CR3
CR5
LS1
CR5
Sequential Function Chart:
Init
Open V1
LS2
Mix
Start M
P = PS
pH
1
Open V2
L = LS3
P = PS
Close V2
Full
Drain
Close V2
Open V3
LS1
Reduce
Level
Open V4
Stop
Stop M
Close V3
LS2
Figure S22.7.
22-12
R5= V3
22.8
In batch processing, a sequence of one or more steps is performed in a
defined order, yielding a finished product of a specific quantity.
Equipment must be properly configured in unit operations in order to be
operated and maintained in a reasonable manner.
The discrete steps necessary to carry out this operation could be:
.- Open exit valve in tank car.
.- Turn on pump 1
.- Empty the tank car by using the pump and transfer the chemical
to the storage tank (assume the storage tank has larger capacity
than the tank car)
.- Turn off pump 1
.- Close tank car valve (to prevent backup from storage tank)
.- Open exit valve in storage tank.
.- Transfer the chemical to the reactor by using the second pump
.- Close the storage tank exit valve and turn off pump 2.
.- Wait for the reaction to reach completion.
.- Open the exit valve in the reactor.
.- Discharge the resulting product
Safety concerns:
Because a hazardous chemical is to be handled, several safety issues must
be considered:
.- Careful and appropriate transportation of the chemical, based
on safety regulation for that type of product.
.- Appropriate instrumentation must also be used. Liquid level
indicators could be installed so that pumps are turned off based on
level.
22-13
.- Chemical leak testing, detection, and emergency shut-down
.- Emergency escape plan.
Therefore, care should be exercised when transporting and operating
hazardous chemicals. First of all, tanks and units should be vented prior to
charging. Generally, materials should be stored in a cool dry, wellventilated location with low fire risk. In addition, outside storage tanks
must be located at minimum distances from property lines.
Pressure, level, flow and temperature control could be utilized in all units.
Hence, they must be equipped with instrumentation to monitor these
variables. For instance, tank levels can be measured accurately with a
float-type device, and storage temperatures could be maintained with
external heating pads operated by steam or electricity. It is possible for a
leak to develop between the tank car and storage tank, which could cause
high flow rates, so a flow rate upper limit may be desirable.
Valves and piping should have standard connections. Enough valves are
required to control flow under normal and emergency conditions.
Centrifugal pumps are often preferred for most hazardous chemicals. In
any case, the material of construction must take into account product
chemical properties.
Don't forget that batch process control often requires a considerable
amount of logic and sequencing for their operation. Besides, interlocks
and overrides are usually considered to analyze and treat possible failure
modes.
22-14
22.9
1.- Because there is no steady state for a batch reactor, a new linearization
point is selected at t = 0. Then,
Linearization point for batch reactor: t = 0 ≡ t *
2.- Available information:
k = 2.4 × 1015 e −20000 / T (min −1 )
C = 0.843
ρ = 52
where T is in o R
BTU
lb o F
V = 1336 ft 3
ft 3
min
mol
C Ai = 0.8 3
ft
lb
ft 3
q = 26
(−∆H ) = 500
kJ
mol
Ti = 150o F
UA = 142.03
Ts = 25o C
kJ
min o F
For continuous reactor, T = 150o F
Physical properties are assumed constant.
Problem solution:
A stirred batch reactor has the following material and energy balance
equations:
− kC A =
dC A
dt
(1)
(−∆H )kVC A + UA(Ts − T ) = V ρC
dT
dt
where k = k 0 e − E / RT
From Eqs. 1 and 2, linearization gives:
22-15
(2)
*
E

 dC ′A
′
T
−  k *C A* + k *C A′ + C A*k0 e − E / RT
=
RT *2 
dt

(3)
*
E


(−∆H )V  k *C A* + k *C A′ + C A*k0 e− E / RT
T ′
*2
RT


+UA(Ts′ − T ′) = V ρC
dT ′
dt
(4)
Rearranging, the following equations are obtained:
b11C A′ + b12T ′ =
dC A′
dt
(5)
b21C A + b22T ′ + b23Ts′ =
dT ′
dt
(6)
where
b11 = −k 0 e − E / RT = −13.615
*
*
E 
*
b12 = − k 0 e − E / RT C A 
= −0.586
*2 
 RT 
(−∆H )k0 e− E / RT
b21 =
= 155.30
ρC
*
1
 E  UA
b22 =
(−∆H )k0 e − E / RT C A* 
−
= 6.66
*2 
ρC
 RT  ρVC
*
b23 =
UA
= 2.43 × 10 −3
ρVC
From Example 4.8, substituting values for continuous reactor
a11 = −13.636
a12 = −8.35 × 10 −4
a 21 = 155.27
22-16
a 22 = −0.0159
b2 = 2.43 × 10 −3
(Note that , from material balance,
C A = 0.00114 )
Hence the transfer functions relating the steam jacket temperature Ts′(s )
and the tank outlet concentration C ′A (s ) are:
Continuous reactor:
C A′ ( s )
−2.03 × 10−6
−5.86 × 10−6
=
=
Ts′( s ) s 2 + 13.651s + 0.3464 2.887 s 2 + 39.4s + 1
then τdom ≈ 35 min
Batch reactor:
C A′ ( s )
−1.424 × 10−3
−5.47 × 10−3
= 2
=
Ts′( s ) s + 6.931s + 0.26 3.84s 2 + 26.65s + 1
then τdom ≈ 25 min
As noted in transfer functions above, the time constant for the batch is
smaller than the time constant for the continuous reactor, but the gain is
much larger.
1
1
1
1
1
1
1
1
1
22-17
22.10
The reactor equations are:
dx1
= − k1 x1
dt
dx 2
= k1 x1 − k 2 x 2
dt
(1)
(2)
where k1= 1.335 × 1010e-75,000/(8.31 × T) and
k2= 1.149 × 1017e-125,000/(8.31 × T)
By using MATLAB, this differential equation system can be solved using
the command "ode45". Furthermore we need to apply the command
"fminsearch" in order to optimize the temperature. In doing so, the results
are:
a)
Isothermal operation to maximize conversion (x2(8)):
Top = 357.8 K
b)
and
x2max = 0.3627
Cubic temperature profile: the values of the parameters in T=a0 +
a1t + a2t2 + a3t3 that maximize x2(8) are:
a0 = 372.78
a1 = -10.44
a2 = 2.0217
a3 = -0.1316
and
x2max = 0.3699
The optimum temperature profile and the optimum isothermal operation
are shown in Fig. S22.10.
375
Temperature profile
370
T(K)
365
360
Isothermal operation
355
350
0
1
2
3
4
5
6
7
8
time
Figure S22.10. Optimum temperature for the batch reactor.
22-18
MATLAB simulation:
a) Constant temperature (First declare Temp as global variable)
1.- Define the differential equation system in a file called batchreactor.
function dx_dt=batchreactor(time_row,x)
global Temp
dx_dt(1,1)=-1.335e10*x(1)*exp(-75000/8.31/Temp);
dx_dt(2,1)=1.335e10*x(1)*exp(-75000/8.31/Temp) 1.149e17*x(2)*exp(-125000/8.31/Temp);
2.- Define a function called conversion that gives the final value of x2 (given a
value of the temperature)
function x2=conversion(T)
global Temp
Temp=T;
x_0=[0.7,0];
[time_row, x] = ode45('batchreactor', [0 8], x_0 );
x2=-(x(length(x),2));
3.- Find the optimum temperature by using the command fminsearch
[T,negative_x2max]=fminsearch('conversion', To)
where To is our initial value to find the optimum temperature.
b) Temperature profile (First declare a0 a1 a2 a3 as global variables)
1.- Define the differential equation system in a file called batchreactor2.
function dx_dt=batchreactor2(time_row,x)
global a0 a1 a2 a3
Temp=a0+a1*time_row+a2*time_row^2+a3*time_row^3;
dx_dt(1,1)=-1.335e10*x(1)*exp(-75000/8.31/Temp);
dx_dt(2,1)=1.335e10*x(1)*exp(-75000/8.31/Temp) 1.149e17*x(2)*exp(-125000/8.31/Temp);
2.- Define a function called conversion2 that gives the final value of x2
the values of the temperature coefficients)
(given
function x2b=conversion(a)
global a0 a1 a2 a3
a0=a(1);a1=a(2);a2=a(3);a3=a(4);x_0=[0.7,0];
[time_row, x] = ode45('batchreactor2', [0 8], x_0 );
x2b=-x(length(x),2);
3.- Find the optimum temperature profile by using the command fminserach
[T,negative_x2max]=fminsearch('conversion2', ao)
where ao is the vector of initial values to find the optimum temperature profile.
22-19
22.11
The intention is to run the reactor at the maximum feed rate of the gas to
minimize the time cycle, but the reactor is also cooling-limited. Therefore,
if the pressure controller calls for a gas flow that exceeds the cooling
capability of the reactor, the temperature will start to rise. The reaction
temperature is not critical, but it must not exceed some maximum
temperature. The temperature controller will then take over control of the
feed valve and reduce the feed rate. The output of the selector sets the
setpoint of a flow controller. The flow controller minimizes the effects of
supply pressure changes on the gas flow rate. So this is a cascade type
control system, with the primary controller being an override control
system.
In an override control system, one of the controllers is always in a standby
condition, which will cause that controller to saturate. Reset windup can
be prevented by feeding back the selector relay output to the setpoint of
each controller. Because the reset actions of both controllers have the
same feedback signal, control will transfer when both controllers have no
error. Then the outputs of both controllers will be equal to the signal in the
reset sections. Because neither controller has any error, the outputs of both
controllers will be the same. Particular attention must be paid to make sure
that at least one controller in an override control system will always be in
control. If not, then one of the controllers can wind up, and reset windup
protection is necessary.
FC
FT
EXTERNAL FEEDBACK
<
EXTERNAL FEEDBACK
PC
GAS
22-20
PT
TC
TT
22.12
Material balance:
( − rA ) = −
dC A
2
= kC A0 (1 − X )(Θ B − 2 X )
dt
Since
C A = C A0 (1 − X )
then
dX
1 dC A
=−
dt
C A0 dt
Therefore
dX
= kC A0 (1 − X )(Θ B − 2 X )
dt
(1)
Energy balance:
dT Q g − Qr
=
dt
NC p
where
(2)
Qg = kC AO 2 (1 − X )(Θ B − 2 X )V (∆H RX )
Qr = UA(T − 298)
Eqs. 1 and 2 constitute a differential equation system. By using MATLAB,
this system can be solved as long as the initial conditions are specified.
Command "ode45" is suggested.
A.- ISOTHERMAL OPERATION UP TO 45 MINUTES
We will first carry out the reaction isothermally at 175 °C up to the time
the cooling was turned off at 45 min.
Initial conditions : X(0) = 0 and T(0)= 448 K
Figure S22.12a shows the isothermal behavior for these first 45 minutes.
22-21
Conversion
449
0.03
448.5
0.02
448
0.01
447.5
---
Conversion
__
0
Temperature
0.04
0
5
10
15
20
25
30
Temperature
35
40
447
45
Time
Figure S22.12a. Isothermal behavior for the first 45 minutes
B.- ADIABATIC OPERATION FOR 10 MINUTES
The cooling is turned off for 45 to 55 min. We will now use the conditions
at the end of the period of isothermal operation as our initial conditions for
adiabatic operation period between 45 and 55 minutes.
t = 45 min
X = 0.033
T = 448
470
0.04
460
0.035
450
Conversion
Temperature
0.045
__
Temperature
Conversion
0.03
45
46
47
48
49
50
51
52
53
54
440
55
Time
Figure S22.12b. Adiabatic operation when the cooling is turned off.
22-22
C.- BATCH OPERATION WITH HEAT EXCHANGE
Return of the cooling occurs at 55 min. The values at the end of the period
of adiabatic operation are:
t = 55
T = 468 K X = 0.0423
1.5
1000
Temperature
Conversion
900
1
700
Temperature
Conversion
800
0.5
600
500
0
60
70
80
90
100
110
120
130
Time
Figure S22.12c. Batch operation with Heat Exchange; temperature
runaway.
As shown in Fig. S22.12c, the temperature runaway is finally unavoidable
under new conditions:
. Feed composition = 9.044 kmol of ONCB, 33.0 kmol of NH3, and 103.7
kmol of H20
. Shut off cooling to the reactor at 45 minutes and resume cooling reactor
at 55 minutes.
MATLAB simulation:
1.- Let's define the differential equation system in a file called reactor.
function dx_dt=reactor(t,x)
dx_dt(1,1)=((17e-5*exp(11273/1.987*(1/4611/x(2))))*1.767*(1-x(1))*(3.64-2*x(1)));
dx_dt(2,1)=((-(17e-5*exp(11273/1.987*(1/461-1/x(2))))*
122*(1-x(1))*(3.64-2*x(1))*5.119*(-5.9e5) 35.85*(x(2)-298))/2504 );
22-23
where dx_dt(2,1)the must be equal to 0 for the isothermal operation
2.- By using the command "ode45", system above can be solved
[times_row,x]=ode45('reactor',[to, tf],[X0,T0]);
plot(times_row,x(:,1),times_row,x(:,2));
where to, tf, X0 and T0 must be specified for each interval.
22.13
Tr = Reactor temperature profile
Tjsp = Jacket set-point temperature profile (manipulated variable)
PID controller:
Kc = 26.5381
τI = 2.8658
τD = 0.4284
140
Tr
Tjsp
120
100
80
T (C)
a)
60
40
20
0
0
20
40
60
80
100
120
time (min)
Figure S22.13a. Numerical simulation for PID controller.
22-24
b)
Batch unit
Kc = 10.7574
τI = 53.4882
120
Tr
Tjsp
100
T(C)
80
60
40
20
0
0
20
40
60
80
100
120
time(min)
Figure S22.13b. Numerical simulation for batch unit.
Batch unit with preload
Kc = 10.7574
τI = 53.4882
120
Tr
Tjsp
100
80
T(C)
c)
60
40
20
0
0
20
40
60
time(min)
80
100
120
Figure S22.13c. Numerical simulation for batch unit with preload.
22-25
Dual mode controller
1.- Full heating is applied until the reactor temperature is within 5% of its
set point temperature.
2.- Full cooling is then applied for 2.8 min
3.- The jacket temperature set point Tjsp of controller is then set to the
preload temperature (46 °C) for 2.4 min.
140
Tr
Tjsp
120
100
80
T (C)
d)
60
40
20
0
0
20
40
60
time (min)
80
100
120
Figure S22.13d. Numerical simulation for dual-mode controller.
MATLAB simulation:
1.- Define a file called brxn:
function dy=brxn(t,y)
%
% Batch reactor example
% Cott & Machietto (1989); "Temperature control
% of exothermic batch reactors using generic model
% control", I&EC Research, 28, 1177
%
% Parameters
cpa=18.0; cpb=40.0; cpc=52.0; cpd=80.0;
cp=0.45; cpj=0.45;
dh1=-10000.0; dh2=6000.0;
uxa=9.76*6.24;
rhoj=1000.0;
k11=20.9057; k12=10000;
k21=38.9057; k22=17000;
vj=0.6921;
tauj=3.0;
wr=1560.0;
22-26
dy=zeros(7,1);
ma=y(1); mb=y(2); mc=y(3); md=y(4); tr=y(5); tj=y(6);
tjsp=y(7);
k1=exp(k11-k12/(tr+273.15));
k2=exp(k21-k22/(tr+273.15));
r1=k1*ma*mb;
r2=k2*ma*mc;
qr=-dh1*r1-dh2*r2;
mr=ma+mb+mc+md;
cpr=(cpa*ma+cpb*mb+cpc*mc+cpd*md)/mr;
qj=uxa*(tj-tr);
dy(1)=-r1-r2;
dy(2)=-r1;
dy(3)=r1-r2;
dy(4)=r2;
dy(5)=(qr+qj)/(mr*cpr);
dy(6)=(tjsp-tj)/tauj-qj/(vj*rhoj*cpj);
dy(7)=0;
Note: The error between the reactor temperature and its set-point (e=cvsp-cv) is
computed at each sampling time. That is, control actions are computed in the
discrete-time. For the integral action, error is simply summated (se = se+e).
Controller output is estimated by mv=Kc*e+Kc/taui*se*st, where Kc =
proportional gain, taui=integral time, e=error, se=summation of error and
st=sampling time
2.- PID controller simulation
clear
clf
%
% batch reactor control system
% PID controller (velocity form)
%
% process initial values
ma=12.0; mb=12.0; mc=0; md=0; tr=20.0; tj=20.0;
tjsp=20.0;
y0=[ma,mb,mc,md,tr,tj,tjsp];
% controller initial values
kc=26.5381; taui=2.8658; taud=0.4284;
en=0; enn=0;
cvsp=92.83; mv=20;
% simulation
st=0.2;
t0=0; tfinal=120;
ntf=round(tfinal/st)+1;
cvt=zeros(1,ntf); mvt=zeros(1,ntf);
for it=1:ntf
[tt,y]=ode45('brxn',[(it-1)*st it*st],y0);
y0=y(length(y(:,1)),:);
cv=y0(5);
22-27
% PID control calculation
e=cvsp-cv;
mv=mv+kc*(e*st/taui+(e-en)+taud*(e-2*en+enn)/st);
if mv>120, mv=120; elseif mv<20, mv=20; end
enn=en; en=e;
y0(7)=mv;
cvt(it)=cv; mvt(it)=mv;
end
t=(1:it)*st;
plot(t,cvt,'-r',t,mvt,'--g')
3.- Batch unit simulation
% controller
kc=10.7574; taui=53.4882;
mh=120; ml=20; mq=46;
mv=20;
cvsp=92.83;
% simulation
st=0.2;
z=ml; al=exp(-st/taui);
t0=0; tfinal=120;
ntf=round(tfinal/st)+1;
for it=1:ntf
[tt,y]=ode45('brxn',[(it-1)*st,it*st],y0);
y0=y(length(y(:,1)),:);
cv=y0(5);
e=cvsp-cv;
m=kc*e+z;
if m>mh, m=mh;
end
f=m
z=al*z+(1-al)*f; [f z m]
y0(7)=m;
cvt(it)=cv;
mvt(it)=m;
end
t=(1:it)*st;
plot(t,cvt,'-r',t,mvt,'-g');
4.- Batch unit with preload simulation
% controller
kc=10.7574; taui=53.4882;
22-28
mh=120; ml=20; mq=46;
mv=20;
cvsp=92.83;
% simulation
st=0.2;
z=ml; al=exp(-st/taui);
t0=0; tfinal=120;
ntf=round(tfinal/st)+1;
for it=1:ntf
[tt,y]=ode45('brxn',[(it-1)*st,it*st],y0);
y0=y(length(y(:,1)),:);
cv=y0(5);
e=cvsp-cv;
m=kc*e+z;
if m>mh, m=mh; else if m<ml, m=ml
end
end
f=m
z=al*z+(1-al)*f; [f z m]
y0(7)=m;
cvt(it)=cv;
mvt(it)=m;
end
t=(1:it)*st;
plot(t,cvt,'-r',t,mvt,'-g');
5.- Dual-mode simulation
clear
clf
%
% batch reactor control system
% dual-mode controller
%
% initial values
ma=12.0; mb=12.0; mc=0; md=0; tr=20.0; tj=20.0;
tjsp=20.0;
y0=[ma,mb,mc,md,tr,tj,tjsp];
% controller initial values
kc=26.5381; taui=2.8658; taud=0.4284;
en=0; enn=0;
cvsp=92.83;
td1=2.8; td2=2.4; pl=46; Em=0.95;
mv=20;
is=0;
% simulation
st=0.2;
t0=0; tfinal=120;
ntf=round(tfinal/st)+1;
cvt=zeros(1,ntf); mvt=zeros(1,ntf);
for it=1:ntf
22-29
[tt,y]=ode45('brxn',[(it-1)*st it*st],y0);
y0=y(length(y(:,1)),:);
cv=y0(5);
if is==0
%
if cv<Em*cvsp
mv=120;
else
is=1;
tcool=it*st;
end
end
heat up stage
if is==1
%
cooling stage
if it*st<tcool+td1
mv=20;
else
is=2;
tpre=it*st;
end
end
if is==2
%
preload stage
if it*st<tpre+td2
e=cvsp-cv;
mv=pl;
else
is=3;
end
enn=en; en=e;
end
if is==3
%
control stage
e=cvsp-cv;
mv=mv+kc*(e*st/taui+(e-en)+taud*(e2*en+enn)/st);
if mv>120, mv=120; elseif mv<20, mv=20; end
enn=en; en=e;
end
y0(7)=mv;
cvt(it)=cv;
mvt(it)=mv;
end
t=(1:it)*st;
plot(t,cvt,'-r',t,mvt,'-g')
22-30
123456789 8
23.1
Option (a):
• Production rate set via setpoint of wA flow controller
• Level of R1 controlled by manipulating wC
• Ratio of wB to wA controlled by manipulating wB
• Level of R2 controlled by manipulating wE
• Ratio of wD to wC controlled by adjusting wD
Options (b)-(e) are developed similarly. See table below.
Option
a
b
c
d
e
•
Production
Rate Set With
wA
wA
wA
wA
wA
wB
wB
wB
wB
wB
wC
wC
wC
wC
wC
wD
wD
wD
wD
wD
wE
wE
wE
wE
wE
Control
Loop #
1
2
3
4
5
1
2
3
4
5
1
2
3
4
5
1
2
3
4
5
1
2
3
4
5
Type of
Controller
Flow
Ratio
Level
Ratio
Level
Flow
Ratio
Level
Ratio
Level
Flow
Ratio
Level
Ratio
Level
Flow
Ratio
Level
Ratio
Level
Flow
Ratio
Level
Ratio
Level
Controlled
Variable
wA,m
wB,m
HR1
wD,m
HR2
wB,m
wA,m
HR1
wD,m
HR2
wC,m
wB,m
HR1
wD,m
HR2
wD,m
wC,m
HR1
wB,m
HR2
wE,m
wD,m
HR2
wB,m
HR1
Manipulated
Variable
wA (V1)
wB (V2)
wC (V3)
wD (V4)
wE (V5)
wB (V2)
wA (V1)
wC (V3)
wD (V4)
wE (V5)
wC (V3)
wB (V2)
wA (V1)
wD (V4)
wE (V5)
wD (V4)
wC (V3)
wA (V1)
wB (V2)
wE (V5)
wE (V5)
wD (V4)
wC (V3)
wB (V2)
wA (V1)
Subscript m denotes “measurement”.
Solution Manual for Process Dynamics and Control, 2nd edition,
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp
23-1
In options c, d, and e, valves 1 and 2 can be used interchangeably. Thus, a
total of 8 options can be developed.
Advantages and Disadvantages:
Each option is equivalent in the sense that 5 control loops are required: 1
flow, 2 level, and 2 ratio. Since there is no cost or complexity advantage with
any option, the production rate should be set via the actual product rate, wE ,
i.e. option e.
WB
RC
2
A
FT
1
B
WD
V2
RC
4
Ratio
A
FT
2
B
Ratio
FC
1
V4
FT
4
FT
3
WA
WC
LC
3
V1
V3
LT
2
LT
1
LC
5
R1
FT
5
R2
WE
P1
P2
Figure S23.1. Solution for option a)
23-2
V5
23.2
a)
The level in the distillate (HD) will be controlled by manipulating the recycle
flow rate (D), and the level in the reboiler (HB) via the bottoms flow rate (B).
Thus, HD and HB are pure integrator elements.
Closed-loop TF development assuming PI controller:
τI s + 1
GCL ( s ) =
(
τI
)s2 + τI s + 1
Kc K p
Relations:
τI
= τ2
Kc K p
τ I = 2ζτ
K p = −1 for both HD and HB loops.
12345 = 1 for a critically damped response
Initial settings:
K c = −0.4
τ I = 10
Final tuning: changed to proportional control only to obtain a faster response
K c = −1
b)
The distillate composition (xD) will be controlled by manipulating the reflux
flow rate (R), and the bottoms composition (xB) via the vapor boilup (V).
Use a step response to determine an approximate first-order model
(calculations are shown on last page).
xD
0.0012
=
R 2.33s + 1
xB −0.000372
=
V
2.08s + 1
23-3
Using the Direct Synthesis method:
K
τs + 1
1
τ
Gc ( s ) =
(1 + )
τc K
τs
G (s) =
Choosing τc =
c)
1
τ , the settings are:
4
xD - R loop
{
xB - V loop
{
K c = 3333.3
τ I = 2.33
K c = −10649.6
τ I = 2.08
The reactor level (HR) will be controlled by manipulating the flow from the
reactor (F).
HR is a pure integrator element.
Using the same relations as in part a, initial controller settings are:
K c = −0.4
τ I = 10
After tuning:
K c = −1
τI = 5
23-4
Figure S23.2a. Simulink-MATLAB block diagram for case a)
1
D
2
0.3408
xD
KR
DxD/HR
KRz/HR
1
s
3
F0
5
F
Integrator z
1
s
flow sum
z
Level integrator
Sum-z
4
z0
2
F0z/HR
F0z0/HR
Dz/HR
1
HR
Figure S23.2b. Simulink-MATLAB block diagram for the CSTR block
23-5
2
R
R
V
23.5
xD
Hs
Hs
2
xD
1
D
D
y (i+1)
x(i-1)
1
HD
HD
Top 13-20
x(i-1)
L
yi
3
F
F
4
z
z
V
LF
H
0
xi
xi
y (i+1)
FeedTray 12
x(i-1)
yi
Hs
R
F
5
V
xB
V
6
B
B
3
xB
HB
Bottom 1-11
4
HB
Figure S23.2c. Simulink-MATLAB block diagram for the Tower block
23-6
600
300
280
260
560
HD (lb-mol)
D (lb-mol/hr)
580
540
240
220
520
500
200
0
5
10
15
20
25
180
30
460
5
10
15
20
25
30
0
5
10
15
time (h)
20
25
30
280
260
440
240
420
HB (lb-mol)
B (lb-mol/hr)
0
400
380
360
220
200
180
0
5
10
15
time (h)
20
25
30
160
Figure S23.2d. Step change in V (1600-1700) at t=5
23-7
0.955
xD (mole fraction A)
1250
1200
R (lbmol/hr)
0.95
1150
0.945
1100
1050
0.94
0
10
20
30
20
30
0
10
20
30
20
30
20
30
0.02
1750
0.018
0.016
V (lbmol/hr)
1700
1650
0.014
1600
1550
10
xB (mole fraction A)
1800
0
0.012
0
10
20
30
0.01
time (h)
560
260
D (lb-mol/hr)
240
HD (lb-mol)
540
220
520
500
200
0
10
20
30
180
340
500
320
480
0
10
300
460
440
10
HB (lb-mol)
B (lb-mol/hr)
520
0
280
0
10
20
30
time (h)
260
time (h)
Figure S23.2e. Step change in F (960-1060) at t=5
23-8
xD (mole fraction A)
1160
0.952
1140
R (lbmol/hr)
1120
0.948
1100
1080
0.95
0
10
20
30
10
20
30
0
10
20
30
0
10
20
30
0
10
20
30
0
10
20
30
0.014
V (lbmol/hr)
0.013
1650
0.012
1600
1550
0.011
0
10
20
30
0.01
540
240
HD (lb-mol)
D (lb-mol/hr)
530
220
520
200
510
500
0
10
20
180
30
290
470
285
HB (lb-mol)
B (lb-mol/hr)
475
465
280
460
455
275
0
10
20
270
30
1020
2440
2430
HR (lb-mol)
F (lb-mol/hr)
0
xB (mole fraction A)
1700
0.946
1000
2420
980
960
2410
0
10
20
30
time (h)
2400
time (h)
Figure S23.2f. Step change in F0 (460-506) at t=5
23-9
* Calculation of First-Order Model Parameters for xD and xB Loops
xD -R:
A step change in the reflux rate (R) of +10 lbmol/hr is made and the resulting
response is used to fit a first-order model:
K
τs + 1
∆x
0.9624 − 0.950
K= D =
= 0.0012
∆R
10
G (s) =
672489
4 432427 7243 44
0.632(∆xD ) = (0.632)(0.012) = 0.007584
τ = time( xD = 0.957584) = 12.33 − 10 = 2.33
xB -V:
Similarly, a step change in the vapor boilup (V) of +10 lbmol/hr is made:
K=
∆xB 0.00678 − 0.0105
=
= −0.00372
∆V
10
Use 63.2% of the response to find 0.632(∆xB ) = (0.632)(−0.00372) = −0.00235
τ = time( xB = 0.00815) = 12.08 − 10 = 2.08
0.966
0.964
0.962
xD (mol fraction A)
0.96
0.958
0.956
0.954
0.952
0.95
0.948
0
5
10
15
20
Time (hr)
Figure S23.2g. Responses for step change in the reflux rate R
23-10
25
-3
11
x 10
10.5
10
xB (mol fraction A)
9.5
9
8.5
8
7.5
7
6.5
0
5
10
15
20
25
Time (hr)
Figure S23.2h. Responses for step change in the vapor boilup V.
23-11
23.3
The same controller parameters are used from Exercise 23.2
Figure S23.3a. Simulink-MATLAB block diagram
23-12
0.96
R (lbmol/hr)
xD (mole fraction A)
1300
1200
0.95
1100
1000
0
10
20
30
40
20
30
40
50
0
10
20
30
40
50
0
10
20
30
40
50
0
10
20
30
40
50
0
10
20
30
40
50
V (lbmol/hr)
0.014
1600
0.012
0
10
20
30
40
50
0.01
700
400
HD (lb-mol)
D (lb-mol/hr)
650
300
600
200
550
0
10
20
30
40
100
50
340
500
320
HB (lb-mol)
B (lb-mol/hr)
520
480
300
460
440
10
0.016
1800
500
0
xB (mole fraction A)
2000
1400
0.94
50
280
0
10
20
30
40
260
50
2480
HR (lb-mol)
1200
F (lb-mol/hr)
2460
1100
2440
1000
900
2420
0
10
20
30
40
50
2400
time (h)
time (h)
Figure S23.3b. Step change in F0 (+10%) at t=5
23-13
xD (mole fraction A)
0.955
R (lbmol/hr)
1150
1100
0.95
1050
1000
0
10
20
30
40
50
0.945
0
10
20
30
40
50
0
10
20
30
40
50
0
10
20
30
40
50
0
10
20
30
40
50
0
10
20
30
40
50
-3
11
V (lbmol/hr)
xB (mole fraction A)
1700
1600
10
1500
1400
0
10
20
30
40
HD (lb-mol)
D (lb-mol/hr)
200
500
150
450
100
0
10
20
30
40
50
50
290
470
285
HB (lb-mol)
B (lb-mol/hr)
475
465
280
460
455
275
0
10
20
30
40
270
50
1000
2410
2400
HR (lb-mol)
F (lb-mol/hr)
9
8
50
550
400
x 10
950
2390
900
850
2380
0
10
20
30
40
50
2370
time (h)
time (h)
Figure S23.3c. Step change in z0 (-10%) at t=5
23-14
23.4
The flow controller on F, the column feed stream, should be simulated in
MATLAB as a constant flow. The controller parameters used are taken from
those derived in Exercise 23.2.
0.952
R (lbmol/hr)
xD (mole fraction A)
1140
1120
1100
1080
0
10
20
30
40
50
0.95
0.948
0
10
20
30
40
50
0
10
20
30
40
50
0
10
20
30
40
50
0
10
20
30
40
50
0
10
20
30
40
50
0.011
V (lbmol/hr)
xB (mole fraction A)
1610
1600
0.0105
1590
1580
0
10
20
30
40
50
0.01
200
500
180
HD (lb-mol)
D (lb-mol/hr)
520
480
160
460
440
140
0
10
20
30
40
120
50
340
500
320
HB (lb-mol)
B (lb-mol/hr)
520
480
300
460
440
280
0
10
20
30
40
520
3000
HR (lb-mol)
500
2800
480
460
260
50
F0 (lb-mol/hr)
a)
2600
0
10
20
30
40
50
time (h)
2400
time (h)
Figure S23.4a. Step change in F0 (+10%) at t=5
23-15
Using the approximate relation (23-17), a +10% step change in F0 will result
in a reactor holdup of:
z0
0.90
=
= 2800 lbmol
1 1
1
1
k R ( − ) 0.34(
−
)
506 960
F0 F
HR ≈
Using the exact relation (23-6, rearranged):
F 0 z 0 − Bx B (506)(0.9 − 0.0105)
=
= 2910 lbmol
(0.34)(0.455)
kR z
HR =
The value taken from the graph (2910) matches up with the expected value
from the equation without the approximation.
3000
2900
2800
HR (lb-mol)
b)
2700
2600
2500
2400
0
5
10
15
20
25
30
35
40
45
time (h)
Figure S23.4b. Step change in F0 (+10%) at t=5
23-16
50
23.5
a),b) Feedforward control is implemented using the HR-setpoint equation:
H R (t ) ≈
z0 (t )
1
1
− )
kR (
F0 (t ) F
Empirical adjustment of the feedforward equation is required because it is
not exact:
H R (t ) ≈
z0 (t )
+ 70
1
1
kR (
− )
F0 (t ) F
This adjustment matches the initial values of HR (i.e., with and without
feedforward control).
Parts a and b are represented graphically.
23-17
R (lbmol/hr)
xD (mole fraction A)
1120
1110
1100
1090
0
10
20
0.95
30
xB (mole fraction A)
V (lbmol/hr)
1600
1590
30
0
10
20
30
0
10
20
30
0
10
20
30
0
10
20
30
0.0095
0
10
20
30
190
HD (lb-mol)
D (lb-mol/hr)
500
180
490
170
480
0
10
20
160
30
490
300
HB (lb-mol)
B (lb-mol/hr)
480
290
470
280
460
0
10
20
270
30
500
2500
HR (lb-mol)
F0 (lb-mol/hr)
2400
400
2300
300
200
20
0.01
510
450
10
0.0105
1595
470
0
0.011
1605
1585
no control
FF control
2200
0
10
20
30
2100
time (h)
time (h)
Figure S23.5a. Step change in z0 (-10%) at t=5
23-18
c)
The controlled plant response is much faster with the feedforward controller
(~10 hours settling time versus ~20 hours without it).
d)
Advantage: Faster response.
Disadvantage: Have to measure or estimate two flow rates and one
concentration, therefore significantly more expensive.
a)
Use a flow controller to keep F constant (make F a constant in the
simulation).
b)
Use ratio control to set F. The ratio should be based on the initial steady state
values (960/460). Therefore, as F0 changes, F will be controlled to the
corresponding value set by the ratio.
23.6
Parts a) and b) show very different results for the two alternatives. With
alternative # 3, feedforward control is necessary to keep the level in the
distillate receiver from integrating. However, in alternative # 4, the control
structure without the feedforward loop is superior to that with feedforward
control.
Responses are displayed with controlled variables adjacent to their
corresponding manipulated variable.
23-19
1500
HD (lb-mol)
R (lb-mol/hr)
2500
2000
1500
1000
0
20
40
0
40
60
60
0
20
40
60
0
20
40
60
0
20
40
60
0
20
40
60
0
20
40
60
0.016
V (lbmol/hr)
0.014
0.012
0
20
40
60
0.01
0.52
D (lbmol/hr)
z (mole fraction A)
600
500
0.5
0
20
40
60
0.48
2800
HR (lb-mol)
1100
1050
F (lbmol/hr)
250
x B (mole fraction A)
20
2000
2600
1000
2400
0
20
40
60
2200
1
x D (lbmol/hr)
F0 (lbmol/hr)
520
500
0.95
480
460
40
HB (lb-mol)
B (lb-mol/hr)
0
2500
950
20
300
3000
400
0
350
500
1500
500
60
550
450
No control (a)
FF control (b)
1000
0.9
0
20
40
60
time (h)
time (h)
Figure S23.6a. Alternative #3 (with and without FF controller). Step change in F0
(+10%) at t=10
23-20
400
1200
300
HD (lb-mol)
R (lb-mol/hr)
1300
1100
1000
0
20
40
100
60
0
20
40
250
60
60
0
20
40
60
0
20
40
60
0
20
40
60
0
20
40
60
0
20
40
60
x B (mole fraction A)
V (lbmol/hr)
0.02
1800
0.015
1600
0.01
0
20
40
60
0.005
0.98
D (lbmol/hr)
550
x D (mole fraction A)
600
0.96
500
0
20
40
60
0.94
3000
HR (lb-mol)
1100
1050
2500
1000
950
0
20
40
60
0.6
F0 (lb-mol/hr)
520
500
480
460
2000
z (mol fraction A)
F (lb-mol/hr)
40
300
2000
450
20
HB (lb-mol)
500
1400
0
350
B (lb-mol/hr)
550
450
No control (a)
FF control (b)
200
0.5
0
20
40
60
time (h)
time (h)
Figure S23.6b. Alternative #4 (with and without FF controller). Step change in F0
(+10%) at t=10
23-21
23.7
Parts a) and b) can be satisfied by combining two or more of the previous
simulations into one to compare the results together. To compare how the
alternatives match up, in terms of the snowball effect, a set of arrays has been
constructed.
All arrays are of the form:
D
F
 0
D
z
 0
HR 
F0 

HR 
z0 
where the response of D or HR is analyzed as a result of a step change in F0
or z0.
In the notation below:
S represents the occurrence of the snowball effect (>20% change in steadystate output for a 10% change in input).
A represents an acceptable response (~10% change in steady-state output).
B represents the best possible response (no change in steady-state output).
Alternative #1
S B


S B
Alternative #2
B S 


 B A
Alternative #3
 A A


 B A
Alternative #4
 A A


 B A
These results indicate that Alternative #2 still exhibits a snowballing
characteristic, but in HR instead of D. Alternatives #3 and #4, on the other
hand, eliminate the effect altogether.
23-22
400
R (lb-mol/hr)
1400
HD (lb-mol)
300
1200
200
1000
0
20
40
100
60
60
Alternative #1
Alternative #2
Alternative #3
Alternative #4
300
0
20
40
250
60
0
20
40
60
0
20
40
60
0
20
40
60
0
20
40
60
40
60
1800
xD (mole fractionxB
A) (mole fraction A)
0.02
V (lbmol/hr)
2000
0.015
1600
1400
40
HB (lb-mol)
500
450
20
350
B (lb-mol/hr)
550
0
0.01
0
20
40
60
0.96
D (lbmol/hr)
700
0.005
600
0.94
z (mole fraction A)
500
400
0
20
40
60
3000
HR (lb-mol)
0.6
0.5
0.4
0.92
2500
0
20
40
60
2000
1200
500
1100
F (lb-mol/hr)
F0 (lb-mol/hr)
520
480
460
time (h)
1000
0
20
40
60
900
0
20
Figure S23.7a. Step change in F0 (+10%) at t=10
23-23
300
HD (lb-mol)
R (lb-mol/hr)
1150
1100
200
1050
1000
100
0
20
40
0
60
40
60
Alternative
Alternative
Alternative
Alternative
290
280
0
20
40
60
V (lbmol/hr)
1700
1600
1500
0
20
40
60
11
#1
#2
#3
#4
0
-3
x 10
20
40
60
0
20
40
60
0
20
40
60
0
20
40
60
40
60
10
9
8
0.96
D (lbmol/hr)
550
270
xB (mole fraction A)
xD (mole fraction A)
460
1400
20
HB (lb-mol)
480
440
0
300
B (lb-mol/hr)
500
500
0.95
450
400
0
20
40
60
2600
2400
HR (lb-mol)
z (mole fraction A)
0.6
2200
0
20
40
60
0.9
2000
time (h)
1000
F (lb-mol/hr)
z0 (mol fraction A)
0.5
0.85
0.8
0.94
0
20
40
60
950
900
850
0
20
Figure S23.7b. Step change in z0 (-10%) at t=10
23-24
23.8
Begin with a dynamic energy balance on the reactor:
CP
d ( H R (TR − TRef ))
dt
1
Q = UA(TR − TC )
= CP F0 (T0 − TRef ) + CP D (TD − TRef ) − C P F (TR − TRef ) − H R λ kz − Q1
This model can be simplified using the mass balance:
dH R
= F0 + D − F
dt
And, rearranging to get an equation for modeling the reactor temperature:
1
dTR
=
[ F0CP (T0 − TR ) + DCP (TD − TR ) − UA(TR − TC ) − H R λ kz ]
dt
CP H R
It is clear from the following figures that the temperature loop is much faster
than the interconnected level-flow loops. This characteristic allows the
reaction rate multiplier to settle before it can affect the other variables.
23-25
616.45
Thermal model
Constant temperature
616.44
z (mol fraction A)
0.55
TR (R)
616.43
616.42
0.5
0.45
616.41
616.4
30
40
50
1300
596
1250
595
1200
40
50
60
30
40
50
60
30
40
50
60
TC (R)
1150
593
1100
592
1050
591
1000
590
30
F (lb-mol/hr)
597
594
30
40
50
60
0.341
950
700
650
D (lb-mol/hr)
0.3405
KR (lb-mol/hr)
0.4
60
600
0.34
550
0.3395
500
0.339
30
40
50
60
450
Time (hr)
Figure S23.8a. Step change in F0 (+10%) at t=30 (Constant temperature
simulation does not include a thermal model)
23-26
616.45
Thermal model
Constant temperature
616.44
z (mol fraction A)
0.55
TR (R)
616.43
616.42
0.5
0.45
616.41
616.4
30
40
50
0.4
60
600
30
40
50
60
30
40
50
60
30
40
50
60
1000
F (lb-mol/hr)
598
TC (R)
596
594
950
900
592
590
30
40
50
60
0.341
850
550
500
D (lb-mol/hr)
KR (lb-mol/hr)
0.3405
0.34
450
0.3395
0.339
30
40
50
60
400
Time (hr)
Figure S23.8b. Step change in z0 (-10%) at t=30 (Constant temperature simulation
does not include a thermal model)
23-27
123456789 8
24.1
a)
i.
First model (full compositions model):
Number of variables: NV = 22
w1
xR,A
x2B
w2
xR,B
x2D
w3
xR,C
x4C
w4
xR,D
x5D
w5
xT,D
x6D
w6
VT
x7D
w7
HT
x8D
w8
Number of Equations: NE = 17
Eqs. 2-8, 9, 10, 12, 13, 15, 16, 18, 20(3X), 21, 22, 27, 28, 29, 31
Number of Parameters: NP = 4
VR, k, α, ρ
Degrees of freedom: NF = 22 – 17 = 5
Number of manipulated variables: NMV = 4
w1, w2, w6, w8
Number of disturbance variables: NDV = 1
x2D
Number of controlled variables: NCV = 4
x4A, w4, HT, x8D
nd
Solution Manual for Process Dynamics and Control, 2 edition,
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp
24-1
ii. Second model (simplified compositions model):
Number of variables: NV = 14
w1
xR,A
w4
w2
xR,B
x4A
w6
xR,D
x8D
w8
xT,D
HT
x2D
VT
Number of Equations: NE = 9
Eq. 2-33 through Eq. 2-41
Number of Parameters: NP = 4
VR, k, α, ρ
Degrees of freedom: NF = 14 – 9 = 5
Number of manipulated variables: NMV = 4
w1, w2, w6, w8
Number of disturbance variables: NDV = 1
x2D
Number of controlled variables: NCV = 4
x4A, w4, HT, x8D
iii.Third model (simplified holdups model):
Number of variables: NV = 14
w1
HR,A
w4
w2
HR,B
x4A
w6
HR,D
x8D
w8
HT,B
HT
x2D
HT,D
24-2
Number of Equations: NE = 9
Eq. 2-48 through Eq. 2-56
Number of Parameters: NP = 3
VR, k, α
Degrees of freedom: NF = 14 – 9 = 5
Number of manipulated variables: NMV = 4
w1, w2, w6, w8
Number of disturbance variables: NDV = 1
x2D
Number of controlled variables: NCV = 4
x4A, w4, VT, x8D
b)
Model 1:
The first model is left in an intermediate form, i.e., not fully reduced, so
the key equations for the units are more clearly identifiable. Also, such a
model is easier to develop using traditional balance methods because not
as much algebraic effort is expended in simplification.
Models 2 and 3:
Both of the reduced models are easier to simulate (fewer equations), yet
contain all of the dynamic relations needed to simulate the plant.
Model 3:
The “holdups model” has the further advantage of being easier to
analyze using a symbolic equation manipulator because of its more
symmetric organization. Also, it requires one less parameter for its
specification.
c)
Each model can be simulated using the equations given in Appendix E
of the text. Models 2 and 3 are simulated using the differential equation
editor (dee) in Matlab. An example can be found by typing dee at the
command prompt. Step changes are made in the manipulated variables
w1, w2, w6 and w8 and in disturbance variable x2D to illustrate the
dynamics of the entire plant.
24-3
Figure S24.1a. Simulink-MATLAB block diagram for first model
1
w3
Sum
input flows1
2
x8
1
s
1
w8
A
3
1
s
x1
Demux
4
B
Mux
1
s
w1
2
3000
C
5
-out
+ in
+generated
6
HR
1
s
x2
w2
x3
D
Sum
input flows
[-0.5 -0.5 1 0]
330
stoich. factors
k
3000
rho*VR
Demux
generation &
Accumulation
T1
T2
Figure S24.1b. Simulink-MATLAB block diagram for the reactor block
24-4
Product3
4
x4
3
2
Product2
Demux
1
x3
Product
w4
w3
2
Product5
x5
1
Product1
w5
Product4
Figure S24.1c. Simulink-MATLAB block diagram for the flash block
1
Purge Flow
1
2
w7
w5
3
2
x5
x7
Figure S24.1d. Simulink-MATLAB block diagram for purge block
24-5
1010
w1
w1
w1-save
xRA
xRA-save
w1-change
xRB
w2
xRB-save
w2-save
1100
xRD
w2
xRD-save
x2D
w2-change
Equations 2-(32-38)
VT-save
0.01
x2D-set
xTD
110
Purge Flow
HT
xTD-save
w6
w4
w6-save
w4-save
890
x4A
Recycle Flow
w8
E.2-32-38
x4A-save
w8-save
Figure S24.1e. Simulink-MATLAB block diagram for second model
24-6
HRA
HRA-save
HRB
1010
w1
w1
w1-save
HRB-save
HRD
1100
HRD-save
w2
w2
HTB
w2-save
HTB-save
Equations 2-(48-52)
HTD
HTD-save
0.01
w4
x2D
w4-save
110
Purge Flow
x4A
w6
x4A-save
w6-save
xTD
890
xTD-save
Recycle Flow
HT
w8
w8-save
E.2-48-52
HT-save
Figure S24.1f. Simulink-MATLAB block diagram for third model
24-7
2010
1014
2008
1013
2006
Full model
Reduced concentrations
Reduced holdups
w4
w1
1015
1012
2004
1011
2002
1010
0
10
20
30
40
1101
2000
0
10
20
30
40
0
10
20
30
40
0
10
20
30
40
0
10
20
30
40
0.104
1100.5
w2
0.102
xTD
1100
1099.5
1099
0.1
0
10
20
30
40
110.5
500
HT
550
w6
111
110
450
109.5
109
400
0
10
20
30
350
40
890.5
0.0101
x 4A
0.0102
w8
891
890
0.0101
889.5
889
0.01
0.01
0
10
20
30
40
Time (hr)
Time (hr)
Figure S24.1g. Step change in w1 (+5) at t=5
24-8
1010.5
2006
1010
2004
1009.5
1009
Full model
Reduced concentrations
Reduced holdups
w4
2008
w1
1011
2002
0
10
20
30
2000
40
0.102
1108
0.1
1106
0.098
10
20
30
40
0
10
20
30
40
0
10
20
30
40
20
30
40
w2
1110
0
0.096
x TD
1104
1102
1100
0.094
0
10
20
30
0.092
40
111
800
110.5
HT
w6
700
110
600
109.5
109
0
10
20
30
500
40
-3
890.5
10
890
9.95
889.5
889
x 10
x 4A
10.05
w8
891
9.9
0
10
20
30
9.85
40
Time (hr)
0
10
Time (hr)
Figure S24.1h. Step change in w2 (+10) at t=5
24-9
1011
2000
Full model
Reduced concentrations
Reduced holdups
2000
1010.5
w4
2000
w1
1010
2000
1009.5
1009
2000
0
10
20
30
40
2000
20
30
40
0
10
20
30
40
0
10
20
30
40
0
10
20
30
40
0.1
0.1
w2
1100.5
1100
0.1
1099.5
0.1
1099
10
x TD
1101
0
0
10
20
30
0.1
40
120
500
118
400
HT
116
w6
300
114
200
112
110
0
10
20
30
40
891
100
0.01
890.5
x 4A
w8
0.01
890
0.01
889.5
889
0
10
20
30
40
0.01
Time (hr)
Time (hr)
Figure S24.1i. Step change in w6 (+10) at t=5
24-10
1011
2008
w4
2006
w1
1010.5
1010
2004
1009.5
1009
2002
0
10
20
30
40
2000
0.1001
1100.5
0.1001
w2
x TD
1101
1100
0
10
20
30
40
0
10
20
30
40
10
20
30
40
10
20
30
40
0.1
1099.5
1099
Full model
Reduced concentrations
Reduced holdups
0.1
0
10
20
30
0.1
40
111
500
498
110.5
HT
w6
496
110
494
109.5
109
492
0
10
20
30
490
40
0
-3
10.01
898
10
9.99
w8
896
894
9.98
892
9.97
890
x 10
x 4A
900
0
10
20
30
40
9.96
0
Time (hr)
Time (hr)
Figure S24.1j. Step change in w8 (+10) at t=5
24-11
1011
2000.5
Full model
Reduced concentrations
Reduced holdups
1010.5
2000
w4
w1
1010
1999.5
1009.5
1009
0
10
20
30
40
1999
1101
0
10
20
30
40
0
10
20
30
40
0
10
20
30
40
0
10
20
30
40
0.16
1100.5
w2
x TD
0.14
1100
0.12
1099.5
1099
0
10
20
30
0.1
40
110.5
515
HT
520
w6
111
110
510
109.5
109
505
0
10
20
30
500
40
891
0.0105
w8
x4A
890.5
890
0.01
889.5
889
0
10
20
30
40
Time (hr)
Time (hr)
Figure S24.1k. Step change in x2D (+0.005) at t=5
24-12
24.2
To obtain a steady state (SS) gain matrix through the use of simulation,
step changes in the manipulated variables are made. The resulting matrix
should compare closely with that found in Eq. 24.1 of the text (or the
table below). The values calculated are:
Gain Matrix w1
w2
w6
w8
1.93
2.26 E-2
0
6.25 E-3
w4
8.8 E-4
-7.62 E-4
0
5.68 E-6
x8D
2.57
E-5
-1.14
E-5
0
-3.15 E-6
x4A
-0.918*
0.973*
-1*
-6.25 E-3*
HT
* For integrating variables: “Gain” = the slope of the variable vs. time
divided by the magnitude of the step change.
RGA
w4
x8D
x4A
HT
w1
0.9743
0
0.0257
0
w2
0.0135
0.9737
0.0128
0
w6
0
0
0
1
w8
0.0122
0.0263
0.9615
0
24.3
Controller parameters are given in Tables E.2.7 and E.2.8 in Appendix E
of the text. A transfer function block is placed inside each control loop
to slow down the fast algebraic equations, which otherwise yield large
“output spikes”. These blocks are of the form of a first-order filter.:
G f ( s) =
1
0.001s + 1
In principle, ratio control can provide tighter control of all variables.
However, it is clear from the x2D results that it offers no advantage for
this disturbance variable. For a step change in production rate, w4, one
would anticipate a different situation because a change in manipulated
variable w1 is induced. Using ratio control, w2 does change along with w1
to maintain a satisfactory ratio of the two feed streams. Thus, ratio
control does provide enhanced control for the recycle tank level, HT, and
composition, xTD, but not for the key performance variables, w4 and x4A.
This characteristic is likely a result of the particular features of the
recycle plant, namely the use of a splitter (instead of a flash unit) and the
lack of holdup in that vessel.
24-13
2004
No ratio control
Ratio control
w4
2002
2000
1998
1996
0
5
10
15
0
5
10
15
0
5
10
15
20
25
30
35
20
25
30
35
20
25
30
35
0.0101
x 4A
0.01
0.0099
Time (hr)
0.12
x TD
0.11
0.1
0.09
Figure S24.3a. Step change in x2D (+0.02) at t=5 (Corresponds to Fig.24.7)
2150
No ratio control
Ratio control
w4
2100
2050
2000
1950
0
5
10
15
0
5
10
15
0
5
10
15
20
25
30
35
20
25
30
35
20
25
30
35
0.01015
x 4A
0.0101
0.01005
0.01
0.00995
Time (hr)
0.104
x TD
0.102
0.1
0.098
Figure S24.3b. Step change in w4 (+100) at t=5 (Corresponds to Fig.24.8)
24-14
24.4
a)
A simple modification of the controller pairing is needed.
The settings for the modified controller setup are:
Loop
xTD-w6
HT-R
w4-w1
x4A-w8
Gain
Integral Time (hr)
-5000
1
0.002
1
2
1
-500000
1
(See Figure S24.4a)
b)
The RGA shows that flowrate w6 will not directly affect the composition
of D in the recycle tank, xTD, but the xTD-w6 loop will cause unwanted
interaction with the other control loops. The system can be controlled,
however, if the other three loops are tuned more conservatively and
“assist” the xTD-w6 loop.
c)
The manipulated variable, w6, is the rate of purge flow. Purging a stream
does not affect the compositions of its constituent species, only the total
flowrate. Therefore, purging the stream before the recycle tank will only
affect the level in the tank and not its compositions. The resulting RGA
yields a zero gain between xTD and w6.
d)
The RGA structure handles a positive 5% step change in the production
rate well, as it maintains the plant within the specified limits. The setup
with one open feedback loop defined by this exercise, however, goes out
of control. The xTD-w6 loop requires the interaction of the other loops to
maintain stability. When the x4A-w8 loop is broken, the system will no
longer remain stable.
(See Figure S24.4b)
e)
With a set point change of 10%, the controllers must be detuned to keep
variables within operating constraints. The HT-w6 loop in the RGA
structure must be more conservative (gain reduced to -1) to keep the
purge flow, w6, from hitting its lower constraint, zero. A 20% change
will create a problem within the system that these control structures
cannot handle. The new set point for w4 does not allow a steady-state
value of 0.01 for x4A. This will make the x4A-w8 control loop become
unstable. This outcome results for production rate step changes larger
than roughly 12% (for this system).
(See Figure S24.4c)
24-15
1060
2100
1040
2050
w4
2150
w1
1080
1020
1000
Exercise 24.4
RGA structure
2000
0
10
20
30
40
1950
1180
0
10
20
30
40
0
10
20
30
40
0
10
20
30
40
0
10
20
30
40
520
1160
510
HT
1140
w2
500
1120
490
1100
1080
0
10
20
30
480
40
140
0.102
x TD
120
w6
100
80
0.1
0.098
60
40
0
10
20
30
40
0.096
0.0102
1200
1100
w8
x 4A
0.0101
1000
0.01
900
800
0.0099
0
10
20
30
40
Time (hr)
Time (hr)
Figure S24.4a. Step change in w4 set point(+5%) at t=5
24-16
1100
2100
1050
2050
w4
2150
w1
1150
1000
950
2000
0
5
10
15
20
1950
600
1300
550
0
5
10
15
20
0
5
10
15
20
0
5
10
15
20
0
5
10
15
20
w2
HT
1400
1200
500
1100
1000
Exercise 24.4
RGA structure
450
0
5
10
15
400
20
160
0.115
0.11
x TD
140
0.105
w6
120
0.1
100
80
0.095
0
5
10
15
0.09
20
891
0.012
890.5
x 4A
w8
0.011
890
0.01
889.5
889
0
5
10
15
20
Time (hr)
0.009
Time (hr)
Figure S24.4b. Step change in w4 set point(+5%) at t=5 with one loop open.
24-17
1150
2300
2200
1100
w4
w1
2100
Exercise 24.4
RGA structure
1050
2000
1000
0
20
40
1900
60
1250
0
20
40
60
0
20
40
60
0
20
40
60
0
20
40
60
540
520
1200
w2
500
1150
HT
480
1100
1050
460
0
20
40
440
60
150
0.104
0.102
w6
x TD
100
0.1
0.098
50
0.096
0
0
20
40
60
0.094
0.0104
1400
0.0103
w8
x 4A
1600
1200
0.0102
1000
800
0.0101
0
20
40
60
Time (hr)
0.01
Time (hr)
Figure S24.4c. Step change in w4 set point(+10%) at t=5 with HT controller
detuned
24-18
24.5
a)
Gain
Matrix
w4
x8D
x4A
HT
HR
Using the same methods as described in solution 24.3, the resulting gain
matrix is:
w1
w2
5.762E-3
4.760E-3
5.554E-6
4.558E-6
-2.905E-6 -2.398E-6
-2.137
-1.927
1
1
All variables are integrating
w6
w8
w3
0
0
0
-1
0
5.831E-3
5.445E-6
-2.944E-6
-10.12
1
-1.285E-2
-9.130E-6
6.542E-6
8.829
-1
The resulting RGA does not provide useful insight for the preferred
controller pairing due to the nature of these integrating variables.
b)
Results similar to those obtained in Exercise 24.3 can be obtained with
an added loop for reactor level using the w3 flow rate as the manipulated
variable. Both P and PI controllers yield relatively constant reactor level.
The quality variable, x4A, cannot be controlled as tightly however. The
responses with P-only control are only slightly different as compared to
PI control, which means that zero-offset control on the reactor volume is
not necessary for reliable plant operation.
Controller parameters used for variable reactor holdup simulation:
Loop
w4-w1
xTD-w2
x4A-w8
HT-w6
HR-w3
Gain (Kc) 123456789
49 I)
1
1
-6300
1
-200000
1
-3.5
1
-10
1*
* For PI control
24-19
2120
Variable reactor level (P)
Variable reactor level (PI)
Constant reactor level
2100
2080
w4
2060
2040
2020
2000
1980
0
5
10
15
20
25
30
35
0
5
10
15
20
25
30
35
20
25
30
35
0.104
0.103
x TD
0.102
0.101
0.1
0.099
Time (hr)
0.0108
0.0106
x 4A
0.0104
0.0102
0.01
0.0098
0
5
10
15
Figure S24.5a. Step change in w4 set point (+100) at t=5
24-20
1200
2150
Variable reactor level (P)
Variable reactor level (PI)
Constant reactor level
2100
w4
w1
1100
2050
1000
900
2000
0
10
20
30
1950
40
0.104
1150
0.102
10
20
30
40
0
10
20
30
40
0
10
20
30
40
x TD
w2
1200
0
1100
1050
0
10
20
30
40
0.1
0.098
510
100
500
w6
HT
150
50
0
490
0
10
20
30
0.011
1600
1400
x 4A
w8
0.0105
1200
0.01
1000
800
0
10
20
30
40
3400
0.0095
0
10
20
30
3030
3020
w3
HR
3200
3010
3000
2800
480
40
3000
0
10
20
30
40
2990
0
10
20
Time (hr)
Time (hr)
Figure S24.5b. Step change in w4 setpoint (+100) at t=5
24-21
30
40
24.6
To simulate the flash/splitter with a non-negligible holdup, derive a mass
balance around the unit. Assume that components A and C are well
mixed and are held up in the flash for an average HF/w4 amount of time.
Also assume that the vapor components B and D are passed through the
splitter instantaneously.
dH F
= w3 − w4 − w5 = 0
dt
d ( H F xFA )
= w3 x3 A − w4 x4 A
dt
d ( H F xC )
= w3 x3C − w4 x4C
dt
Since the holdup is constant, the flows out of the splitter can be modeled
as:
w5 = w3 ( x3 B + x3 D )
w4 = w3 − w5
Use the component balances and output flow equations to simulate the
flash/splitter unit. This will add a dynamic lag to the unit which slows
down the control loops that have the splitter in between the manipulated
variable and the controlled variable. However, a 1000kg holdup only
creates a residence time of 0.5 hr. Considering the time scale of the
entire plant, this is very small and confirms the assumption of modeling
the flash/splitter as having a negligible holdup.
24-22
1014
2001
1010
2000
w4
2000.5
w1
1012
1008
1006
Negligible holdup flash
1000 kg holdup flash
1999.5
0
10
20
30
40
1200
1999
0
10
20
30
40
0
10
20
30
40
0
10
20
30
40
0
10
20
30
40
0.108
0.106
1150
x TD
w2
0.104
0.102
1100
0.1
1050
0
10
20
30
40
0.098
510
200
505
w6
HT
250
150
100
500
0
10
20
30
40
495
0.01
900
0.01
w8
x 4A
850
0.01
800
0.01
750
0.01
0
10
20
30
40
Time (hr)
Figure S24.6. Step change in x2D (+0.005) at t=5
24-23
24.7
The MPC controller achieves satisfactory results for step changes made
within the plant. The production rate can easily be maintained within
desirable limits and large set-point changes (20%) do not cause a
breakdown in the quality of this stream. A change in the kinetic
coefficient (k), occurring simultaneously with a 50% disturbance change
will, however, initially draw the product quality (composition of the
production stream) out of the required limits.
0.6
0.5
0.4
0.2
w4
w1
0
-0.5
0
0
10
20
30
40
-1
50
8
1.5
6
1
x4A
w2
-0.2
4
20
30
40
50
0
10
20
30
40
50
0
10
20
30
40
50
0
10
20
30
40
50
0
0
10
20
30
40
50
-0.5
2
0
0
-0.2
-2
-0.4
HT
w6
10
0.5
2
0
0
-4
-0.6
-6
-8
0
10
20
30
40
50
-0.8
60
0
40
xTD
w8
-2
20
-4
0
-20
0
10
20
30
40
50
-6
Figure S24.7a. Step change in x2D (+50%). All variables are recorded in
percent deviation.
24-24
6
-3
4
w4
w1
-2
-4
-5
2
0
10
20
30
40
0
50
2
10
20
30
40
50
0
10
20
30
40
50
0
10
20
30
40
50
0
10
20
30
40
50
1.5
0
1
-2
x4A
w2
0
-4
0
-6
-8
0.5
0
10
20
30
40
50
-0.5
0.1
HT
-5
-10
0
w6
-0.1
-15
-0.2
-20
0
10
20
30
40
50
-0.4
15
0.5
10
0
xTD
w8
-25
-0.3
5
-0.5
0
-5
-1
0
10
20
30
40
50
-1.5
Figure S24.7b. Step change in production rate w4 (+5%). All variables are
recorded in percent deviation.
24-25
1
6
0
4
-1
w4
w1
8
2
0
-2
0
10
20
30
40
-3
50
15
0
10
20
30
40
50
0
10
20
30
40
50
0
10
20
30
40
50
0
10
20
30
40
50
20
15
10
x4A
w2
10
5
0
-5
0
0
10
20
30
40
-5
50
20
1
0
0.5
-20
HT
w6
5
-40
-0.5
-60
-80
0
0
10
20
30
40
-1
50
150
0
100
xTD
w8
-5
50
-10
0
-50
0
10
20
30
40
50
-15
Figure S24.7c. Simultaneous step changes in x2D (+50%) and k(+20%). All
variables are recorded in percent deviation.
24-26
-5
25
20
-10
w1
w4
15
10
-15
5
-20
0
10
20
30
40
0
50
10
0
10
20
30
40
50
0
10
20
30
40
50
0
10
20
30
40
50
0
10
20
30
40
50
6
0
w2
4
x4A
-10
-20
0
-30
0
10
20
30
40
-2
50
-20
0.5
-40
0
HT
w6
-40
2
-60
-0.5
-80
0
10
20
30
40
50
-1.5
60
2
40
0
xTD
w8
-100
-1
20
0
-20
-2
-4
0
10
20
30
40
50
-6
Figure S24.7d. Step change in production rate w4 (+20%). All variables are
recorded in percent deviation.
24-27