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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