1. Causes of delay in process systems
Dynamic processes
Causes of delay in process systems
Whenever material or energy flows into or out of certain system, it takes
time. Thus, the level of a liquid, the temperature of the vessel or the position of the solid mass cannot change suddenly. They are subjects to delays which are dependent on the magnitude of the capacity and the resistance to the inflow of material or energy.
Heater process:
Electric hot water boiler has electric power Pin (w).
We assume that the metal parts of the systems (resistor and casing) are small compared to mass of water.
Thermal power Q = c * m * ΔT (1)
c = heat capacity of water
m = mass
ΔT = Twater - Tambient = temperature difference
Picture 1
Twater
Tambient
Pin ΔT

2. First let's assume that insulation of the boiler is very good
First let's assume that insulation of the boiler is very good. There are no heat transfer from boiler.
Heater resistor is switched on with constant power.
Next block diagram describes the case
STEP
Pin
ΔTout
WATER
Temperature difference
Power/w
Picture 2
Equation (1) gives ΔT = Q/(c * m) (2)
Energy of vessel = Q = W = ∫Pdt (3)
Equations (2) and (3) gives:
ΔT = ∫Pdt/(c * m) (4)
We have got temperature difference to the process, if input power is step function.
With constant power equation (4) gives:
ΔT = P * t/(c * m)
Time-domain plot is growing strait line.

3. 1/S ∫Pdt = W integrator Pin Laplace transform:
Laplace transforms are used to solve dynamic systems.
Time-domain signals are transferred to s-domain signals.
s = jω, ω = angular velocity
Computer based transient analyses are using L-transform.
Integral change to multiplication and gives easy handling for equations.
Matlab simulink can utilize L-transforms.
Time-domain
Time-domain
Time-domain
Pin
∫Pdt = W
integrator
1/S
s-domain
STEP
1/c*m
GAIN ΔT
Picture 3
Time P ΔT
Transient Analyse

4. - + 1/S Pout Pout Tambient Pin ∫Pdt = W integrator Twater Pin
Let´s assume that isolation of the boiler is not perfect.
Heat is transferred through the walls of vessel.
Q = U * A * ΔT (5)
Q = Transferred heat w.
U = Conductance w/m2K
ΔT = Tout - Tin = Temperature difference
A = Area of casing m2
Pin
∫Pdt = W
integrator
1/S
s-domain
STEP
1/c*m
GAIN ΔT + -
U * A = 0.2
U * A * ΔT
Picture 4

5. Temperature difference of the boiler is not any more strait line
Temperature difference of the boiler is not any more strait line. It is exponential and it reaches certain maximum level after a long time.
In this example power loses are huge and temperature inside of the vessel doesn't get over 100 degrees. ΔT
Pin
Time
Transient analyse
We can edit the block diagram and substitute the feedback:

6. Pin Block diagram in picture 4 can be more simple.
Feed-back block diagram with integrator is substituted to one transfer function.
Step response will remain the same figure and magnitude.
Pin
STEP
5/(5s+1) ΔT
Transfer function
Picture 5
Transfer function formula can be written: A/(τs + 1)
A = Gain of the transfer function = ΔT/Pin , K/w
τ = Time constant , second

7. Next graph describes common step response
Next graph describes common step response. Any process with one "energy store" is indentical. The process has 1st order delay.
Output = A * (1 - e-t/τ)
τ = Time constant
tangent A
Time/s
Step function
63,2
Exponent function of
1 st order delay

8. in out - mass R + C out in out in RC-circuit, τ = R * C Heat power
Next processes have similar responses:
Heat power in
mass
ΔT =Tmass - Tambient
out + -
RC-circuit, τ = R * C R C in
out
Velocity
+Force
- Force, example wind force
out in

9. water Cattle Electric heater plate Heat resistor Resistor Heater plate
SECOND ORDER DELAYS
Always we cannot describe the process with one delay.
Example of heat process:
water
Cattle
Electric heater plate
Heat resistor
Resistor
Heater plate
Cattle
Water
Power/w τ1 τ2 τ3 τ4
Tresist
Tplate
Tcattle
Twater

10. Step response 1...4 delays, τ1 = τ2 = τ3 = τ4 = 1 s
input step = 100
time in
out

11. velocity = v Could water TRANSPORT DELAY
Delays are also caused by the time taken for material to travel along pipes or conveyors from one place to another. This type delay, shown in next figure.
Temperature mixed water
Hot water
velocity = v
Distance to measuring point = L
Time delay = L/v
Could water
Step response of the prosess
Transport delay = 2 s, Capacitive delay = 3s

12. RELATIVE DIFFICULTY OF THE PROCESS
Step response of the process is run to the open-loop system:
steepest tangent
Step
Step response
in output signal
Relative change Tt Tr A B
Controllability
S = Tr/Tt
S < 4 difficult to control
S > 4 easy to control
Sketch deepest tangent to the response and mark the points A and B.
A = minimum level of response.
B = maximum level of response (after a long period)

13. Dynamic Process Simulation Information
What is the difference between mathematical models and simulations?
On this site we will use the term "mathematical models" to represent sets of equations
that mathematically describe the process. The term "simulator" refers to a computer program or
a digital system running a computer program that implements the mathematical model.
The simulator may be connected to the control system or may be embedded within the control system.
What types of simulations used in process engineering?
Static
Static simulations, typically used in process design, simulate the process at steady state conditions,
usually at the design operating conditions. Time is not a variable.
Dynamic
Dynamic models consider time as a variable and simulate the process over a period of time.
A dynamic simulation can be used to estimate or illustrate the response, over time, to a change in the process. This primary concern of this site is dynamic models.