Answer:
Explanation:
Given that:
For a first order instrument with a sensitivity of .4 mV/K
constant c = 25 ms = 25 × 10⁻³ s
The initial temperature [tex]T_1[/tex] = 273 K
The final temperature [tex]T_2[/tex] = 473 K
The initial volume = 0.4 mV/K × 273 K = 109.2 V
The final volume = 0.4 mV/K × 473 K = 189.2 V
the instrument’s response as a function of time for a sudden temperature increase can be computed as follows:
Let consider y to be the function of time i.e y(t).
So;
y(t) = 109.2 + (189.2 - 109.2)( 1 - [tex]\mathbf{e^{-t/c}}[/tex])mV
y(t) = (109.2 + 80 ( 1 - [tex]\mathbf{e^{t/25\times 10^{-3}}}[/tex])) mV
Plot the response y(t) as a function of time.
The plot of y(t) as a function of time can be seen in the diagram attached below.
What are the units for y(t)?
The unit for y(t) is mV.
Find the 90% rise time for y(t90) and the error fraction,
The 90% rise time for y(t90) is as follows:
Initially 90% of 189.2 mV = 0.9 × 189.2 mV
= 170.28 mV
170.28 mV = (109.2 + 80 ( 1 - [tex]\mathbf{e^{t/25\times 10^{-3}}}[/tex])) mV
170.28 mV - 109.2 mV = 80 ( 1 - [tex]\mathbf{e^{t/25\times 10^{-3}}}[/tex])) mV
61.08 mV = 80 ( 1 - [tex]\mathbf{e^{t/25\times 10^{-3}}}[/tex])) mV
0.7635 mV = ( 1 - [tex]\mathbf{e^{t/25\times 10^{-3}}}[/tex])) mV
t = 1.44 × 25 × 10⁻³ s
t = 0.036 s
t = 36 ms
The error fraction = [tex]\dfrac{189.2-170.28 }{189.2}[/tex]
The error fraction = 0.1
The error fraction = 10%
