Question

A process temperature is known to fluctuate with a frequency of 0.1 Hz. The engineer needs to monitor this process using a first-order thermocouple and wishes to have no more than 2% amplitude reduction.

a. What time constant of the thermocouple is needed?

Answer 1

t ≤ 0.31s

Step-by-step explanation:

Given

Frequency = 0.1Hz

Dynamic Error:

σ = 1 - M(σ)

From the question

σ= 1 - M(σ) ≤ 2%

1 - M(σ) ≤ 0.02

The above corresponds to

M(σ) = 1/(1 + (wt))² ≥ 100% - 0.02

M(σ) = 1/(1 + (wt))² ≥ 1 - 0.02

M(σ) = 1/(1 + (wt))² ≥ 0.98

Since frequency = 0.1Hz

w = 2πf =

w = 0.628571428571428

w = 0.628rads

So, M(0.628) = 1/(1 + (0.628t)²) ≥ 0.98

Solving 1/(1 + (0.628t)²) ≥ 0.98 we get that

t ≤ 0.31s