A temperature sensor has a 5.0s time constant and is subjected to rapid change of temperature from 70°F to 200°F. What will be the temperature, in °F, reported by the sensor after 1.3 s?

Answer 1

Hi Alexander,

The concept of a "time constant" uses an exponential equation to describe how quickly something changes, or how quickly something responds to a change in input. The time constant (always measured in an amount of time) refers to the how long it takes for a certain amount of that change to take place. A short time constant means that the response is quick, while a long time constant means it takes a long time for that change to take place.

The way that we are going to set up the equation is going to depend on the type of exponential change we're talking about - either exponential buildup or exponential decay. For example, let's say we're talking about exponential decay - starting at a particular value, V₀, and exponentially decreasing toward 0. The remaining amount (V) at any particular time would be:

V = V₀ * e^{-(# of time constants that have passed)} (for exponential decay)

After 1 time constant has passed, the amount would be:

V = V₀ * e^-1 = ~0.368V₀

After 2 time constants have passed, the amount would be:

V = V₀ * e^-2 = ~0.135V₀

If we know the specific time constant value, τ, we can change the exponent into the fraction (t/τ), so that we can plug in specific values for time, t.

V(t) = V₀ * e^-(t/τ) (for exponential decay)

At the beginning, when t = 0, e^-0 = 1, so the initial value V(0) = V₀. As time goes by, t gets larger, so e^-t/τ becomes a smaller and smaller number, eventually approaching 0.

For an exponential build-up from 0 to a final value, V_f, the equation will change to:

V(t) = V_f * (1 - e^{-(# of time constants that have passed)}) (for exponential build-up)

And again, if we know the specific time constant value, τ, we can change the exponent into the fraction (t/τ), so that we can plug in specific values for time, t :

V(t) = V_f * (1 - e^-(t/τ))

At the beginning, when t = 0, e^-0 = 1, so V(0) = V_f * (1-1) = 0. As time goes by, t becomes larger, causing e^-(t/τ) to become smaller, causing the (1-e^-(t/τ)) term to become larger. The larger t becomes, the more V(t) gets closer and closer to V_f.

In the case of this problem, we're starting at one value, and exponentially increasing to another value, so we have a to be a little creative with how to set up the equation. We know that the time constant is 5 seconds, so τ = 5. We need to set up an equation where the initial value is 70, and the final value is 200. There are a couple different ways we could do this, but I think that the easiest would probably be:

Temperature (as a function of time) = (Final value) - (Difference between final and initial value) * (a factor that would start at 1, and get smaller and smaller as time goes by, eventually approaching 0)

T(t) = 200 - (200 - 70)*e^-(t/τ)

T(t) = 200 - 130*e^-(t/5)

Before we use this to find our answer, let's double check to make sure the results will match what we need. At the beginning, t=0, so:

T(0) = 200 - 130*e^-0 = 200-130(1) = 70

This makes sense - at the beginning, the sensor should be reading 70. Now let's try an infinitely large number - after a long time, the sensor should eventually read the final value of 200:

T(∞) = 200 - 130*e^-∞ = 200 - 130*0 = 200

Perfect. So our equation for the temperature indication is:

T(t) = 200 - 130*e^-(t/5)

We can plug in any amount of elapsed time, t, to calculate what the temperature reading would be.

After 1.3 seconds, T(1.3) = 200 - 130e^-(1.3/5) ≈ 99.8°F

I hope this helps!

Andy