Question

A thermometer requires 1 minute to indicate 98% of the response to a unit step input. Assuming the thermometer to be a first order system, find the time constant. If the thermometer is placed in a bath, the temperature of which i changing linearly at a rate of 10◦/min, how much error does the thermometer show?

Answer 1

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

= $\frac{C(s)}{R(s)} = \frac{1}{ST +1}$     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 ) = $\frac{1}{S} -\frac{1}{s+\frac{1}{T} }$  equation 2

taking Laplace transform of this equation

$e^{-t} has a laplace = \frac{1}{s+1}$

1 Laplace = $\frac{1}{s}$

therefore response time   e(t) = 1 - $e^{\frac{-t}{T} }$

= 0.98 = 1 - $e^{\frac{-60}{T} }$

therefore T = $\frac{-60}{ln 0.02}$  = 15.33 sec

back to equation 2

C(s) = $\frac{1}{s} - \frac{1}{s + \frac{1}{15.33} }$

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) = $\frac{1}{s} -\frac{1}{6s}+ \frac{1}{6(s+0.0652)}$  equation 3

taking the inverse Laplace transform equation 4 becomes

e(t) = $\frac{1}{6} e^{-0.0652t} + 0.833$  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) = $\frac{1}{6} e^{-0.065*61.32} +0.833$

hence e(t) = 0.836

Answer 2

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°