Using the Pulse functions for calculating materials balance in continuous stirred tank .

Using the Pulse functions for calculating materials balance in continuous stirred tank .
Pulse function:
start from t = a

Start from t = 0



Exam Q1: A tank having a time constant of 1 min and a resistance of 1/9 ft.min / ft 3 is operating at steady state with a level of 10 ft. At time t =0, the set point of level is increased in Pulse function to 19 ft over a period from zero to 0.1 min. (See Fig.). Determine level of thank at t = ? and the level at t= 3 min.



Solution:
A: cross-sectional area of the tank (ft2). R: Valve resistance which is attached to the output flow (ft.min/ ft3).
q : The input volumetric flowrate (ft3 /min) qo : The output volumetric flowrate (ft3 /min) through the resistance, is related to the head h by the linear relationship (q?= h/R ).
h: level tank (ft)
Mass flow in - mass flow out = rate of accumulation of mass in the tank

Let: ? =AR ; h = Y ; q = X

Taken Laplace for both sides of equation
Frist order systm equation .

Pulse function :


A = 19 – 10 = 9 ft
? = 1 min
a = 0.1 min
R = 1/9 ft/ ft 3min

level of thank at t = ?.

?Y(t) = [(1-0.367)-(1-0.406)] = [0.633-0.594]=0.039 ft
?Y(t) = y(t) - y(0) = 0.039 + 10 = 10.039 ft
level at t = 3 min

?Y(t) = [(1-0.05)-(1-0.055)] = [0.95 - 0.945]=0.005 ft
?Y(t) = y(t) - y(0) = 0.005 + 10 = 10.005 ft
The Time Delay
The most commonly used to describe the dynamics of chemical process is first order plus time delay model. This model can be representing the dynamic of many industrial processes such as time delay or dead-times between input and outputs are very common industrial processes, engineering systems and biological systems. Any delay in measuring is called transportation delay or dead time, and it always reduces the stability of a system and limits the achievable time of the system.

qi(t) = Input to dead-time element.
qo(t) = Output from dead-time element.
The simplest dead-time (DT’s) approximation can be obtained graphically or by physical representation. In the first order system, the dynamics of time delay process is:

In general, the transfer function of a time delay combined with a first-order process is:

Example: Thermal system

If measured at T1 this can be modeled as:

Due to the delay time the temperature T2 represented by:

Sinsoidal Input
and
, first order system equation

This equation can be solved for y (t) by means of a partial fraction expansion as described in previous lectures.




As t=? then e-t/? = 0, the last term on the right side of main equation vanishes and leaves only the ultimate periodic solution, which is sometimes called the steady-state solution.
Example: A mercury thermometer having a time constant of 0.1 min is placed in a temperature bath at 100oF and allowed to come to equilibrium with the bath. At time t= 0, the temperature of the bath begins to vary sinusoid ally about its average temperature of l00oF with an amplitude of 2°F If the frequency of oscillation is 10/? cycles/min, Determine the ultimate response of the thermometer reading at time t= ? and t=0.05 min?
? = 0.1 min
f = 10/? cycles/min
To = l00oF
A = 2°F
Determine response of the thermometer reading at time t= ? and t=0.05 min?
Solution:

X(t) = A sin(wt) = 2 sin(20t)
W =2 ? x 10/? = 20 rad/min






a) At t = ?

y(t) = 100 + 1.419 = 101.419 oC
It is different from step change takes 36% of output steady state
?Y(t)= 0.63 x A = 0.63 x 2 = 1.26oC
?Y(t) = y(t) – y(0) = 1.26oC
y(t) = 100+ 1.26= 101.26oC
b) At t=0.05min

Y(t)= 0.515