Development of T.F for first order system .

Development of T.F for first order system .
Example: Response of thermocouple sensor in coffee cup
Thermometer: It is a measuring device used to measure the temperature of a stream.
Temperature is often measured with a thermocouple sensor based on the fact that electric properties are affected by temperature. We have a thermocouple and a coffee cup and perform the following experiments:
1. Initially, we hold the thermocouple sensor in the air (such that it measures the air temperature).
2. We put the thermocouple into the coffee (step change) and keep it there for some time so that the thermocouple’s temperature is almost the same as the coffee’s temperature.
3. We remove it from the coffee (the temperature will decrease and eventually approach the temperature of air – actually, it may temporarily be lower than the air temperature because of the heat required for evaporation of remaining coffee drops).




The energy balance around the thermocouple is the supplied heat to the thermocouple from the surroundings is:
Q = UA(To ? T ) (1)
We have
Q = mcp (dT/dt) (2)
The energy balance then becomes
mcp (dT/dt) = UA(To ? T ) (3)
where
• T – temperature of thermocouple [K]
• To – temperature of surroundings (coffee or air) [K]
• m – mass of thermocouple [kg]
• cp – specific heat capacity of thermocouple [J/kg K]
• A – area of thermocouple [m2]
• U – heat transfer coefficient from surroundings to thermocouple [W/m2K]
Eq. (3) can be rewritten as
(4)
Let :
(5)

This is in standard form 1st ODE
(6)
Taken Laplace for the equation

(7)
(8)
Step Change

At s=0 ; ?o = A
At s= -1/? ; A= ?o(-?/ ?+1)- ?1/ ? ; then ?1= -A ?


(9)
Several features of this response, worth remembering, are:
? The value of y(t) reaches 63.2 % of its ultimate value when the time elapsed is equal to one time constant ?.
? When the time elapsed is 2?, 3?, and 4?, the percent response is 86.5%, 95%, and 98%, respectively.
Example: A thermometer having a time constant (?) of 0.3 s is at a steady state temperature of 20 oC., the thermometer is placed in a cup of coffee maintained at 50 oC (step change). Determine the time needed for the thermometer to read 45 oC.


Solution:
At s.s. the xo = yo = 20 oC
A= 50 - 20 = 30 oC
By apply eq. 9
Y(t) = A(1- e-t/? )
Y(t) = 30(1- e-3.33t )
Substitute Y(t)= y(t) – y(o)= 45-20=25 oC
25 = 30(1- e-3.33t )
0.1666= e-3.33t
ln (0.1666)= ln (e-3.33t )
-3.33t = ln (0.166)
t= -0.3 ln (0.166) = 0.537 s
Example: Temperature dynamics in continuous stirred tank.
Consider the continuous process in Figure below where a liquid stream of 1 kg/s (constant) flows through a mixing tank with constant volume 1.2 m3. The density of the liquid is 1000 kg/m3 (constant) and the heat capacity is 4 kJ/kg K. perfect mixing in the tank is assumed. The process is initially operated at steady state such that the inlet temperature Tin is 50 oC and the outlet temperature Tout = T is 50oC (so we assume no heat loss). Suddenly, the temperature of the inflow is changed to 60oC (step change). The outlet temperature will also“eventually” reach 60oC. Determine:
a- The time constant, and what outlet stream’s temperature at t=? ?
b- The outlet stream’s temperature after time period of 30 min?

Solution: Since the mass in the tank is constant, the mass balance gives wout = win =
w = 1 kg/s. The energy balance for the tank is (liquid). With the assumption of constant heat capacity cp, this gives:


With Y = T and X = Tin we see that this is in standard form (1st ODE)
a)

In other words, it will take ? = 1200 s = 20 min. the outlet stream’s temperature reaches 56.3oC (because the temperature has increased by:
?Y(t)= 0.63 x A = 0.63 x 10 = 6.3oC
?Y(t)= y(t) –y(0)=6.3oC
Y(t)= 6.3+50 = 56.3oC
b) (1)

Taken Laplace for the equation



(2)

(3)
At s.s. the xo = yo = 50 oC
A= 60 - 50 = 10 oC
Y(t) = A(1- e-t/?)
Y(t) = 10(1- e-30/20) = 10*0.7768
?Y(t)=y(t)-y(0)=7.768
Y(t)=50+7.768= 57.768 oC