Exam #3 SP15Solution
3
Name: By signing, I am signifying that I have neither received nor given help on any portion of this test. = ; = 2 = 2 ; β = β 2 ; =β 2 2 ; = 2 ; = , =[ 1 β 2 1 + 2 ] 2 ; = 2 1 [ 2 1 1 + 2 ] 2 p1 of 3 EXAM 3, Thursday, April 9, 2015 NW 407/507: CHARACTERIZATION METHODS IN NONWOVENS Instructions: This exam is closed book and closed notes. Read each question carefully. Units are important, so be sure to include them (where appropriate) in your answers. Show your calculations. Problem 1 (30 pts): (van Wyk Compression Model) The van Wyk Model assumes that changes occur only in the - direction (thickness) during the compression of a nonwoven assembly. Meanwhile, the length and width (- and - directions) of the structure remain constant. 1. (10 pts) With the strain defined as = (β 0 ) 0 and the final packing density defined as = , show that the strain can also be defined as = 0 β . . = β 0 0 = 0 β1= 0 β1= 0 β1= 0 β 2. (10 pts) What is the relationship, per van Wyk Model, between the compression stress or pressure and both the initial and final packing densities? = = (β 0 ) 3 3. (10 pts) Explain why the experimental and the model curves are on either side of =1 as shown on the graph above. The packing density, , cannot realistically be more than 1 when the total volume of fiber is that of the structure.. 0 Initial Assembly Stressed Assembly
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Transcript of Exam #3 SP15Solution
Name: By signing, I am signifying that I have neither received nor given help on any portion of this test.
π =
π» ; π =
2π
π= 2ππ ;
βπ₯
=
βπ
2π ; π¦π = β
2πΌ
πππ2 ; =2π
; π§ = ππ,
πΌπ
πΌπ= [
π§1βπ§2
π§1+π§2]
2
; πΌπ‘
πΌπ=
π§2
π§1[
2π§1
π§1+π§2]
2
p1 of 3
EXAM 3, Thursday, April 9, 2015
NW 407/507: CHARACTERIZATION METHODS IN NONWOVENS
Instructions: This exam is closed book and closed notes. Read each question carefully. Units are
important, so be sure to include them (where appropriate) in your answers. Show your calculations.
Problem 1 (30 pts): (van Wyk Compression Model)
The van Wyk Model assumes that changes occur only in the π§- direction
(thickness) during the compression of a nonwoven assembly. Meanwhile, the
length and width (π₯- and π¦- directions) of the structure remain constant.
1. (10 pts) With the strain defined as π =(π§βπ§0)
π§0 and the final packing density
defined as π =ππ
π₯π¦π§, show that the strain can also be defined as π =
π0βπ
π.
.
π =π§ β π§0
π§0=
π§
π§0β 1 =
π₯π¦π§ππ
π₯π¦π§0
ππ
β 1 =π0
πβ 1 =
π0 β π
π
2. (10 pts) What is the relationship, per van Wyk Model, between the compression stress π or
pressure π and both the initial and final packing densities?
π = π = πΆπ π‘(πβπ0)3
3. (10 pts) Explain why the experimental and the model curves are on
either side of π = 1 as shown on the graph above.
The packing density, π, cannot realistically be more than 1 when the
total volume of fiber is that of the structure..
π₯
π§ π§ 0
Initial Assembly
Stressed Assembly
Name: By signing, I am signifying that I have neither received nor given help on any portion of this test.
π =
π» ; π =
2π
π= 2ππ ;
βπ₯
=
βπ
2π ; π¦π = β
2πΌ
πππ2 ; =2π
; π§ = ππ,
πΌπ
πΌπ= [
π§1βπ§2
π§1+π§2]
2
; πΌπ‘
πΌπ=
π§2
π§1[
2π§1
π§1+π§2]
2
p2 of 3
Problem 2 (40 pts): (Sound Wave Analysis)
A sinusoidal sound wave with an intensity of 0.01W.m-2
and a frequency of 7.5 kHz travels in air at
a speed of 375 m.s-1
. Air density is 1.12 kg.m-3
.
1. (10 pts) Find the wave length of such sound.
= π»π =ππ
=375
7.5 Γ 103= 0.05π = 5ππ
2. (5 pts) Find the distance between two points with a phase difference of 45Β°.
βπ₯ = βπ
2π= 0.05π
45 Γ2π
3602π
= 0.0063π
3. (10 pts) What is the maximum particle displacement subjected to such sound?
π¦π = β2πΌ
πππ2= β
2 Γ 0.01
1.12 Γ 375 Γ (2π Γ 7.5 Γ 103)2= 0.146πΈ β 6π = 0.146ππ
4. (15 pts) A particle displacement when subjected to such sound is in the form of:
π¦ = πΆ1π ππ (πΆ2π₯ β πΆ3π‘)
Find the values and units of all 3 constants.
π¦ = π¦ππ ππ (ππ₯ β ππ‘)
πΆ1 =___0.146__(ππ ) πΆ2 =__375__(m.s-1
) πΆ3 =_2π Γ 7500 = 47,123(π ππ
π ππ)
5. (10 pts) Calculate the impedance of air described in this exercise.
π§ = ππ = 1.12 Γ 375 = 420 ππ. π /π
Name: By signing, I am signifying that I have neither received nor given help on any portion of this test.
π =
π» ; π =
2π
π= 2ππ ;
βπ₯
=
βπ
2π ; π¦π = β
2πΌ
πππ2 ; =2π
; π§ = ππ,
πΌπ
πΌπ= [
π§1βπ§2
π§1+π§2]
2
; πΌπ‘
πΌπ=
π§2
π§1[
2π§1
π§1+π§2]
2
p3 of 3
Problem 3 (30 pts): (Sound Interaction with Media)
The vibrating air described in Problem 2 is propagating into a plane of water under which the
density is 1,000 kg.m-3
and the speed is 1480 m.s-1
.
1. (5 pts) Describe what happened to incidental signal and its
frequency when it reaches the plane of water.
Frequency assumed to remain unchanged for the signal portions.
Part of the signal is reflected back into air and another component is
absorbed by Water.
2. (10 pts) Calculate the signal component that remains in the air.
π§π = 0.42 πππ.π
π and π§π = 1000 Γ 1480 = 1480 πππ. π /π
πΌπ
πΌπ= [
π§π΄ β π§π
π§π΄ + π§π]
2
= [0.420 β 1480
0.420 + 1480]
2
= 0.9989 ππ 99.89%
3. (10 pts) Calculate the signal component that is contained by the water.
πΌπ‘
πΌπ=
π§π
π§π΄[
2π§π΄
π§π΄ + π§π]
2
=1480
. 42[
2 Γ 0.42
0.42 + 1480]
2
= .0011 = 1 β 0.9989 = 0.11%
4. (5 pts) The incidental sound wave described in Problem 2 can be assessed in terms of its dB Level
according to:
πΏdB = 10 Γ ππππΌ
πΌ0
a. What is πΌ0 and why are sounds identified in terms of their dB level and not their pressure or
intensities?
πΌ0 = 10β12 π
π2: Intensity of faintest sound detected by human
Human response to sound varies logarithmically rather than linearly.
b. Calculate the dB level of such sound before it reaches the water plane. Based on its dB level,
what might have caused such sound?
πΏdB = 10 Γ πππ10β2
10β12= 100 ππ
Most loud machinery and regular sustained exposure may cause permanent damage
Incidental
Wave
Air
Water
