Download Observation and Analysis of Applied Force and Vibrations on Road

Observation and Analysis of Applied
Force and Vibrations on Road Signs
due to Pressurized Release of Varying
Gases
Annie Chapman
CE 511-01
May 6, 2011
Introduction
The release of hazardous chemicals in the gaseous
state – especially dense gas – is a real danger

Dense gases are stored in pressurized vessels for
transportation or general storage

A small rupture in a vessel would cause fast-moving gas
flow causing not only health issues but possible smallscale damage

Objective
Observe the effect a pressurized vapor release would
have when directly contacting a road sign

Create an equation in order to analyze the varying sign
displacement when exposed to a gases of different
densities

Assess whether a pressurized gas release is realistically
a threat to small-scale “structures”

Procedure
Equipment:
 Air Compressor (capable of pressures > 100 psi)
 “Road Sign”
 High-Seed Camera
Procedure
1. Pressurize air compressor to around 140 psi
2. Release compressed air on various locations on sign with different
trajectories – observe which combination creates most observable data
3. Record sign movement with high-speed camera
4. Measure displacement, velocity, and acceleration of the point of most
movement on sign
5. Create control equation(s) to vary with different gas densities
6. Observe effect gas density has on sign displacement
Calculations
Measured Data:
 Maximum displacement of sign = 0.46992 in @ t = 2.416 seconds
 Velocity = 3.17515 in/s @ t = 2.416 seconds
 Acceleration = 4,762.725 in/s^2 @ t = 2.416 seconds
 Pressure = 140 psi
 Cross-sectional area of sign = 1/8 sq in
Dynamic Force = (ρAV^2)/2
ρ(compressed air) = P/RT = 0.000014 pci
F(air) = 0.5*(0.000014 pci)*(1/8 sq in)*(3.17515 in/s)^2
F(air) = 0.000009 lb
ρ(chlorine gas) = 0.0001156 pci
F(chlorine gas) = 0.000073 lb
ρ(ammonia gas) = 0.000031 pci
F(ammonia gas) = 0.00002 lb
Calculations
mx(double dot) + kx = F
Or
(mass)(acceleration) + (stiffness coeff)(max displacement) = F(air)
(0.5 lb)*(4,762.725 in/s^2) + k*(0.46992 in) = 0.000009 lb
k = 5,067.59 lb/in
x(double dot) + 10,135.2x = 0.000009
ω(n) = √(5,067.59 lb/in / 0.5 lb) = 100.674 rad/s
xh(t) = A*sin(100.674*t) + B*cos(100.674*t)
xp = c = 0.000009
xp(t) = 0 + (10,135.3)*(0.000009) = 0.091215
x(t) = A*sin(100.674*t) + B*cos(100.674*t) + 0.091215
x(2.416) = 0.46992
v(2.416) = 3.17515
A = -0.429
B = 0.144
X air(t) = (-0.429)sin(100.674t) + (0.144)cos(100.674t) + 0.091215
Calculations
X air(0) = 0.013 in
V air(0) = -43.19 in/s
X chlorine(t) = (-0.429)sin(100.674t) – (0.7269)cos(100.674t) + 0.7399
X chlorine(2.416) = 1.34 in
X ammonia(t) = (-0.429)sin(100.674t) – (0.19)cos(100.674t) + 0.203
X ammonia(2.416) = 0.67 in
Results
Type of
Gas
Density
[pci]
Force on 1/8
sq in
x(t) @
t=2.416 [in]
Pressure
[lb/sq in]
Force on
Whole Sign [lb]
Air
0.000014
0.000009
0.47
0.000072
0.031
Ammonia
0.0001156
0.000073
0.67
0.000584
0.252
Chlorine
0.000031
0.000020
1.34
0.000160
0.069
Conclusions
Dense gases exert a larger force and create a larger sign
displacement

Force on the sign and displacement of the sign are not large
enough to worry about small-scale damage on surrounding
“structures”

The main risk associated with pressurized hazardous gas
release is health associated
