ME504 Industrial Instrumentation Assignment Solution: Paper B

This document presents a solution to an Industrial Instrumentation assignment, covering topics such as differential equations, Laplace transforms, temperature sensors, radiation calculations, sensitivity analysis, and piezoelectric crystals. Specific problems addressed include determining time constants, temperature responses, incident radiation, and voltage sensitivity. The solutions involve applying relevant formulas and principles to calculate various parameters related to industrial instrumentation. Desklib offers a wide range of solved assignments and past papers to support students in their studies.

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Date of submission

Question 1
solution
The differential equation is given as
τd ∆ t
dt +∆ t=∆ tf
In Laplace transform, it is given as,
G ( s ) = ∆ T s
∆ T f ( s ) = 1
τs+1 =
1
τ
s + 1
τ
G ( t ) =1− e− ( t
τ )
Part i
G ( t ) =1− e− ( t
τ )
0.85=1− e− ( t
15 )
Introducing ln
ln ( e− ( t
15 ) ) =ln ( 0.15 )
− ( t
15 )=− 1.871
t=9.486 seconds
part ii
τd ∆ t
dt +∆ t=∆ tf
∆ tf =∆ T (1− e
− t
τ )
35+ ( 130 −35 ) ( 1 −e
−t
15 )
T ( t )=130 −35 e
−t
15
at t=25
T ( t )=130 −35 e
−25
15

T ( t ) =1 23.39 seconds
Amplitude attenuation=35
Phase lag=95.450
Question 2
The natural frequency is given in second order system
natural frequency=11000 Hz
damping ratio= C
Cc
=0.4 ∧ 0.3
s2 +2 ζ ωn s +ωn
2=0
ωn= √ k
m 11000= √ k
m
ζ = b
2 √ km 0.4= b
2 √km
ω peak=ωn √ 1− 2 ζ 2
ω peak =11000 √1 −2 ( 0.4 )2=7480 Hz
ω peak=11000 √ 1 −2 ( 0.3 ) 2=9020 Hz
Question 3
First order temperature sensor
T ( t )=105 −20 e
−t
0.2
t=25
25=105− 20 e
−t
0.2
Introducing ln,
1
20 ∗ ln ( 80 ) =ln ( e
−t
0.2 )
−t
0.2 =0.219
t=− 0.2925 seconds
T=0.3 seconds

Question 4
Calculate the incident radiation in watts at 3um
Consider Df to be 1Hz in the detectivity equation. Detectivity is given by
D= F ( Ad B )0.5
NEP
D= F ( Ad B )0.5
NEP =500= 6 ∗1011 ∗79 ∗10−9
NEP =94.8
The incident radiation is given as 94.8 Watts
Question 5
solution
The sensitivity (S) is defined as:
S= dE
dR = Ei Ri
( Ri+R )2
S= ( 12 ∗16 )
( 16+80 )2 = 192
9216 =0.02083
Question 6

What do times 1, 2, 3, 4 represent?
Time 1= rise time
Time 2= peak time & Overshoot
Time 3=
Time 4= Settling time
Question 7
The gapped core stores more energy for the same coil or core material and dimension. Air has
permeability in the order of 10e3 times lower than the ferrite as it is able to fill the gap. The air gap
accounts for a larger part of the reluctance as seen by the produced magnetic field which leads to
the gap contained in all field energies. Energy prefers therefore to reside in thin air rather than in the
core material. The flux produced is less sensitive to changes in the current and temperature in the
environment and it reduces the core losses as a result there is a higher Q factor.
Question 8

A quartz piezoelectric crystal generates a voltage output of 285 V having a thickness of 3 mm and is
subjected to a pressure of 210 psi. Calculate the voltage sensitivity. (4 marks)
Voltage sensitivity is given as,
Voltage output =285 Volts
Thickness= 3mm
Pressure = 210 psi
Voltage sensitivity is given by,
V out =P ∗ g ∗ t
To find g,
V out
Pt =g= 285
3
1000 ∗210
=431.8182 [ dimensionless ]
Question 9

Solution
The manometer fluid density of 82 percent compared to that of water at 280c hence,
ρm =0.82 ρ+ w=0.82 ( 996 kg
m3 )=816.7 kg/m3
The local atmosphere is
ρa =29.3inHg=9.922 ∗104 Pa
The fluid is air and has a density above pressure of 301K of
ρf =ρa= P
RT = 9.922 x 104
287 ∗301 =1.1485 kg /m3
The pressure is measured as,
ρ❑ − ρa = g
ge
h ( PM − Pf )
9.807
1.0 ( 0.2 ) ( 816.7 −1.1485 )=1599.63 Pa
P=1599.63+ ( 9.922∗ 104 ) =1.0082∗ 105 Pa