### Lecture two

```‫بسم هللا الرحمن الرحيم‬
Lecture two
1- A modeling procedure (Marlin, Chapter 3)
2- Empirical modeling (Smith & Corripio, Chapter 7)
3- Control valve: Action, characteristics and capacity (Smith & Corripio, Chapter 5)
Lecturer: M. A. Fanaei
1
Modeling (relation between inputs and outputs of process)
We can tune the controller only after the process steady-state and
dynamic characteristics are known.
Types of model
• White box (first principles) n black box (empirical)
• Linear n non-linear
• Static n dynamic
• Distributed n lumped
• Time domain n frequency domain
• Continuous n discrete
: Roffel & Beltlem, “Process dynamics and control”, Wiley, 2006
2
A modeling procedure
1. Define goals
5. Analyze results
Specific design decisions
Check results for correctness
Numerical values
Functional relationships
Accuracy of numerical method
Required accuracy
Interpret results
2. Prepare information
Plot solution
Sketch process and identify system
Characteristic behavior
Identify variables of interest
Relate results to data and assumptions
State assumptions and data
Evaluate sensitivity
3. Formulate model
6. Validate model
Conservation balances
Select key values for validation
Constitutive equations
Compare with experimental results
Rationalize
Compare with results from more
Check degrees of freedom
complex model
Dimensionless form
4. Determine Solution
Analytical
Numerical
3
Example 1. Isothermal CSTR
F
Define Goals
F
V
CAo
1.
Dynamic response of a CSTR to
a step in the inlet concentration.
2.
The reactant concentration
should never go above 0.85
mole/m3
3.
When the concentration reaches
0.83 mole/m3, would a person
have enough time to respond?
What would a correct response
be?
CA
1.
The system is the liquid in
the tank (as shown in Fig.).
2.
The important variable is
the reactant concentration in
the reactor.
Prepare Information
4
Example 1. Isothermal CSTR
Prepare Information …
F
F
CAo
V
CA
3.
Assumptions
•
Well-mixed vessel
•
Constant density
•
Constant flow in
•
Constant temperature
4.
Data
•
F = 0.085 m3/min , V = 2.1 m3
•
(CAo)initial = 0.925 mole/m3 , DCAo = 0.925 mole/m3
•
The reaction rate is rA = -kCA , with k = 0.04 min-1
5
Example 1. Isothermal CSTR
F
Formulate Model
1.
F
V
dC A
dt
2.
 FC Ao  FC A  kVC
CA
A
Rationalize :
dC A
dt
3.
V
CAo
Material balance:

1

CA 
F
V
C Ao
where
 
V
F  VK
Degrees-of-freedom: One equation, one variable(CA), two external
variables (F and CAo) and two parameters (V and k).
Therefore the DOF is zero, and the model is exactly specified.
6
Example 1. Isothermal CSTR
F
F
CAo
V
CA
Analytical Solution
C A  ( C A ) init  K p [ C Ao  ( C A ) init ](1  e
where
K
p

F
F  Vk
 0 . 503 ,
t /
)
  12 . 4 min
7
Example 1. Isothermal CSTR
8
Empirical Modeling (Step Testing)
Step
Change
Record
m(t), %
Final
Control
Element
Process
Sensor/
Transmitter
c(t) , %
first order plus dead time :
C (s)

M (s)
K e
 t0 s
 s 1
Process Gain:
K 
Dcs
Dm
9
FOPDT Model
Fit 1 :
10
FOPDT Model
Fit 2 :
11
FOPDT Model
Fit 3 :
 
3
2
( t 2  t1 )
,
t0  t2  
12
Control Valve
m(t)
vp(t)
Control
Valve Action
Cv(t)
Control Valve
Characteristics
f(t)
Control valve
Capacity
13
Control Valve
1. Control Valve Action is selected based on safety consideration
•
•
Fail-Closed (FC) or Air-to-Open (AO) :  v
Fail-Open (FO) or Air-to-Close (AC) :
v
dv p ( t )
dt
dv p ( t )
dt
 v p (t ) 
m (t )
100
 v p (t )  1 
m (t )
100
τv : Time constant of valve actuator
(3-6 sec for pneumatic actuator)
The gain of FC (AO) valve is positive
The gain of FO (AC) is negative
14
Control Valve
2. Control Valve Characteristics
•
Linear
•
Quick-opening
•
Equal percentage
C v ( t )  C v , max v p ( t )
C v ( t )  C v , max 
v p ( t ) 1
Rangeability parameter
(50 or 100)
15
Control Valve
2. Control Valve Characteristics : How we must select the
correct valve characteristics (Linear or Equal percentage)
The correct selection requires a detailed analysis of the installed
characteristics
As a rule of thump:
 Choose a linear valve if at design conditions the valve is taking more
than half of the total pressure drop (Δpv > 0.5 Δpo ).
 Choose an equal percentage valve if at design conditions the valve is
taking less than half of the total pressure drop (Δpv < 0.5 Δpo ).
 Equal percentage valves are probably the most common ones.
16
Control Valve
3. Control Valve Capacity
The control valve capacity is : The flow in U.S. gallons per minute (gpm) of
water that flows through a valve at a pressure drop of 1 psi across the valve
Liquid Flow:
f (t )  C v (t )
Dpv
Gf
Where: f(t) = volume flow rate (gpm)
Δpv = presuure drop across the valve (psi)
Gf = specicific gravity
17
Control Valve
3. Control Valve Capacity
Gas Flow
• Subcritical flow: f s ( t )  836 C v ( t ) C f
• Critical flow:
f s ( t )  836 C v ( t ) C f
Where:
y
1 . 63
Dpv
Cf
p1
p1
( y  0 . 148 y )
3
for y  1 . 5
GT
p1
for y  1 . 5
GT
fs (t) = Gas volume flow at standard conditions,14.7 psia & 60 oF (scfh)
Cf = Critical flow factor (0.6 – 0.95 , typically 0.9)
p1 = Pressure at valve inlet (psia), T = Tempreture at valve inlet (oR)
G = Gas specific gravity
18
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