Answer:
Time constant = 15.34 seconds
The thermometer shows an error of 0.838°
Explanation:
Given
t = 1 minute = 60 seconds
c(t) = 98% = 0.98
According to the question, the thermometer is a first order system.
The first order system transfer function is given as;
C(s)/R(s) = 1/(sT + 1).
To calculate the time constant, we need to calculate the step response.
This is given as
r(t) = u(t) --- Take Laplace Transformation
R(s) = 1/s
Substitute 1/s for R(s) in C(s)/R(s) = 1/(sT + 1).
We have
C(s)/1/s = 1/(sT + 1)
C(s) = 1/(sT + 1) * 1/s
C(s) = 1/s - 1/(s + 1/T) --- Take Inverse Laplace Transformation
L^-1(C(s)) = L^-1(1/s - 1/(s + 1/T))
Since, e^-t <–> 1/(s + 1) --- {L}
1 <–> 1/s {L}
So, the unit response c(t) = 1 - e^-(t/T)
Substitute 0.98 for c(t) and 60 for t
0.98 = 1 - e^-(60/T)
0.98 - 1 = - e^-(60/T)
-0.02 = - e^-(60/T)
e^-(60/T) = 0.02
ln(e^-(60/T)) = ln(0.02)
-60/T = -3.912
T = -60/-3.912
T = 15.34 seconds
Time constant = 15.34 seconds
The error signal is given as
E(s) = R(s) - C(s)
Where the temperature changes at the rate of 10°/min; 10°/60 s = 1/6
So.
E(s) = R(s) - 1/6 C(s)
Calculating C(s)
C(s) = 1/s - 1/(s + 1/T)
C(s) = 1/s - 1/(s + 1/15.34)
Remember that R(s) = 1/s
So, E(s) becomes
E(s) = 1/s - 1/6(1/s - 1/(s + 1/15.34))
E(s) = 1/s - 1/6(1/s - 1/(s + 0.0652)
E(s) = 1/s - 1/6s + 1/(6(s+0.0652))
E(s) = 5/6s + 1/(6(s+0.0652))
E(s) = 0.833/s + 1/(6(s+0.0652)) ---- Take Inverse Laplace Transformation
e(t) = 1/6e^-0.652t + 0.833
For a first order system, the system attains a steady state condition when time is 4 times of Time constant.
So,
Time = 4 * 15.34
Time = 61.36 seconds
So, e(t) becomes
e(t) = 1/6e^-0.652t + 0.833
e(t) = 1/(6e^-0.652(61.36)) + 0.833
e(t) = 0.83821342824942664566211
e(t) = 0.838 --- Approximated
Hence, the thermometer shows an error of 0.838°
Answer:
error shown by the thermometer = 0.836⁰
Explanation:
time required by thermometer = 1 minute to indicate 98% response = 0.98
changes linearly at = 10⁰/min
since the thermometer is a first order system we will apply the first order transfer function
= [tex]\frac{C(s)}{R(s)} = \frac{1}{ST +1}[/tex] equation 1
applying Laplace transform to step response R(t)
R(s) = 1/s putting this back into equation 1 and simplifying it
C(s ) = [tex]\frac{1}{S} -\frac{1}{s+\frac{1}{T} }[/tex] equation 2
taking Laplace transform of this equation
[tex]e^{-t} has a laplace = \frac{1}{s+1}[/tex]
1 Laplace = [tex]\frac{1}{s}[/tex]
therefore response time e(t) = 1 - [tex]e^{\frac{-t}{T} }[/tex]
= 0.98 = 1 - [tex]e^{\frac{-60}{T} }[/tex]
therefore T = [tex]\frac{-60}{ln 0.02}[/tex] = 15.33 sec
back to equation 2
C(s) = [tex]\frac{1}{s} - \frac{1}{s + \frac{1}{15.33} }[/tex]
the temperature changes at 10⁰/min = 10/60 = 1/6
To calculate error E(s) = R(s) - C(s)
E(s) = R(s) - 1/6 ( C(s) )
E(s) = [tex]\frac{1}{s} -\frac{1}{6s}+ \frac{1}{6(s+0.0652)}[/tex] equation 3
taking the inverse Laplace transform equation 4 becomes
e(t) = [tex]\frac{1}{6} e^{-0.0652t} + 0.833[/tex] equation 4
note to attain steady state condition T will have to be multiplied by 4
T = 61.32 therefore equation 4 becomes
e(t) = [tex]\frac{1}{6} e^{-0.065*61.32} +0.833[/tex]
hence e(t) = 0.836
