A thermometer requires 1 minute to indicate 98% of the response to a unit step input. Assuming the thermometer to be a first ord

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A thermometer requires 1 minute to indicate 98% of the response to a unit step input. Assuming the thermometer to be a first ord

er 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?

Engineering

2 answers:

Mrrafil [253] · Answer 1 · 2026-03-18T02:44:50+03:00

Answer:

The thermometer shows an error of 0.836°

Explanation:

The time taken by the thermometer to indicate is 1 minute to achieve 98% of the response, which equals 0.98

and changes at a linear rate of 10°/min

Since it’s regarded as a first-order system, we will use the first-order transfer function =

equation 1 $\frac{C(s)}{R(s)} = \frac{1}{ST +1}$ Applying the Laplace transform to the step response R(t)

R(s) = 1/s, and inserting this into equation 1 before simplifying

C(s ) =

equation 2

$\frac{1}{S} -\frac{1}{s+\frac{1}{T} }$ Taking the Laplace transform of the resulting equation

the inverse Laplace yields e(t), equating to 0.98 indicates

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

that T =

$\frac{1}{s}$

= 15.33 sec

$e^{\frac{-t}{T} }$

Returning to equation 2

$e^{\frac{-60}{T} }$ C(s) =

Where the temperature varies at 10°/min, or 10/60 = 1/6 $\frac{-60}{ln 0.02}$ To assess the error E(s) = R(s) - C(s)E(s) = R(s) - 1/6 ( C(s) )

Substituting and rearranging results in

equation 3

$\frac{1}{s} - \frac{1}{s + \frac{1}{15.33} }$ Taking the inverse Laplace transform reveals

equation 4

noting that a steady state condition will arise when T is multiplied by 4

Thus, T = 61.32 and equation 4 finalizes as

Then we find e(t) = 0.836 $\frac{1}{s} -\frac{1}{6s}+ \frac{1}{6(s+0.0652)}$

mote1985 [204] · Answer 2 · 2026-03-18T02:44:50+03:00

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°