To solve this, we first need to break down the given information:

• The measurement system exhibits a first-order response, which implies the dynamics can be modeled using a simple first-order differential equation.
• The 90% risetime is given as 8 seconds, meaning that the system takes 8 seconds to reach 90% of its final value in response to a step change.
• The static sensitivity is given as 5 mV/°C, meaning that for every 1°C change in temperature, the output changes by 5 mV.

Let’s address each part of the question:

Part A: Find the time constant of the measurement system.

For a first-order system, the relationship between the risetime and the time constant $\tau$ is known. The risetime $t_{r90}$ to reach 90% of the final value is approximately:

$t_{r90} \approx 2.3 \times \tau$

Given that $t_{r90} = 8$ seconds, we can solve for $\tau$:

$\tau = \frac{8}{2.3} \approx 3.48 \text{ seconds}$

Thus, the time constant $\tau$ of the system is approximately 3.48 seconds.

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Part B: Find the output voltage at $t = 2$ seconds after the transient occurs.

When the temperature changes from 20°C to 100°C, the system will experience a step input. The output of a first-order system subjected to a step input changes exponentially, following this equation:

$V(t) = V_{\infty} \left( 1 - e^{-t/\tau} \right)$

Where:

• $V(t)$ is the output voltage at time $t$.
• $V_{\infty}$ is the final output voltage after the system reaches steady state.
• $t$ is the time elapsed since the step input was applied.
• $\tau$ is the time constant of the system.

Step 1: Calculate $V_{\infty}$ (the final output voltage):

The final temperature change is from 20°C to 100°C, so:

$\Delta T = 100°C - 20°C = 80°C$

Given the static sensitivity of 5 mV/°C, the final output voltage will be:

$V_{\infty} = 80°C \times 5 \, \text{mV/°C} = 400 \, \text{mV}$

Step 2: Calculate $V(2)$ (the output voltage at $t = 2$ seconds):

Now, we substitute $t = 2 \, \text{s}$, $\tau = 3.48 \, \text{s}$, and $V_{\infty} = 400 \, \text{mV}$ into the first-order response equation:

$V(2) = 400 \left( 1 - e^{-2/3.48} \right)$

Let’s calculate this value.

import math

Given values

V_infinity = 400 # mV tau = 3.48 # seconds t = 2 # seconds

Calculate the output voltage at t = 2 seconds

V_2 = V_infinity * (1 - math.exp(-t / tau)) V_2The output voltage at $t = 2$ seconds after the temperature transient occurs is approximately 174.85 mV.

Summary:

• A. The time constant $\tau$ of the measurement system is approximately 3.48 seconds.
• B. After the temperature changes from 20°C to 100°C, the output voltage at $t = 2$ seconds is approximately 174.85 mV.

Do you need more details or have any questions?

Related Questions:

1. How does the time constant affect the speed of response in a first-order system?
2. What is the significance of static sensitivity in a temperature measurement system?
3. How would a larger time constant affect the output voltage at $t = 2$ seconds?
4. What would the output voltage be at $t = 5$ seconds?
5. How would the system behave if it had a second-order response instead of first-order?

Tip:

In a first-order system, a shorter time constant results in a faster response to changes in input, whereas a larger time constant leads to a slower response.