Question

Measurements

#6. The relation between input Ti and output X of a liquid bulb thermometer is given by: dx 23.0dt + 2.0-2.0 Ti measured in mm and Ti is the input measured in is 0 mm. The meter is under a step input from 0 oC (at 0) to 90 oC (t> 0). Find the reading of the where X is output oC. Find time constant and static sensitivity of the instrument (3 +2 pts). At 0 oC the X thermo thermometer at t 24 seconds. (show all the steps of your calculation and draw the approximate response curve). (10 + 5 =15 pts)

Answer 1

#6.

The given equation is

$23.0\frac{dx}{dt}+2.0x=2.0T_i$

Or

$\frac{dx}{dt}+0.08696x=0.08696T_i$

The above is in the form

$\frac{dx}{dt}+\frac{1}{\tau}x=KT_i$

Hence the time constant is

$\tau=\frac{1}{0.08696}=11.5\: s$

The static sensitivity is

$K=0.08696$

The differential equation is with step input of 90⁰ C is

$\frac{dx}{dt}+0.08696x=0.08695(90), t>0$

Solving

$x=ce^{-0.08696t}+90$

At t=0, x=0

Hence

$c=-90$

Therefore

$x=90(1-e^{-0.08696t})$

At t= 24 s

$x=90(1-e^{-0.08696\times24})=78.84\: mm$

Hence reading of thermometer is 78.84 mm

The plot is given below:

90 80 70 60 50 40 30 20 10 0 5 10 15 2025 30 35 40 45 50 Time-S