Answer:
Time constant = 15.34 seconds
The thermometer indicates an error margin of 0.838°
Explanation:
Given
t = 1 minute = 60 seconds
c(t) = 98% = 0.98
According to the details provided, the thermometer functions as a first-order system.
The transfer function of such a system is expressed as;
C(s)/R(s) = 1/(sT + 1).
To determine the time constant, the step response must be evaluated.
This is defined as
r(t) = u(t) --- Applying Laplace Transformation
R(s) = 1/s
Replacing 1/s back into C(s)/R(s) = 1/(sT + 1).
What we have
C(s)/1/s = 1/(sT + 1)
C(s) = 1/(sT + 1) * 1/s
C(s) = 1/s - 1/(s + 1/T) --- Taking Inverse Laplace Transformation
L^-1(C(s)) = L^-1(1/s - 1/(s + 1/T))
Given that e^-t <–> 1/(s + 1) --- {L}
1 <–> 1/s {L}
Thus, the unit response c(t) = 1 - e^-(t/T)
Substituting 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
It follows that the time constant = 15.34 seconds
The error signal is characterized by
E(s) = R(s) - C(s)
Where the temperature shifts at 10°/min; which equals 10°/60 s = 1/6
Thus,
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)
Note that R(s) = 1/s
Thus, E(s) turns into
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)) ---- Taking Inverse Laplace Transformation
e(t) = 1/6e^-0.652t + 0.833
In a first-order system, a steady state condition is reached when the time is four times the time constant.
Thus,
Time = 4 * 15.34
Time = 61.36 seconds
Consequently, 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 --- Rounded off
Thus, the thermometer reveals an error of 0.838°
