How do you determine PSI at point B from an explosion (PSI) at point A.?

Question

... Given the expantion of gasses happens on a flat surface and under ideal conditions.. excluding air resistance and energy absorbed by the ground. Lets use a Radius of 10 meters.

Answer 1

Assume adiabatic expansion for an ideal gas around ground zero. Using PV = nRT; we can assume an initial pressure P0 = nRT0/V0.

That initial pressure will decrement as the volume expands adiabatically from V0 to V; where V = 1/2 4/3 pi r^3 (The 1/2 accounts for a hemisphere.) So, the pressure upon expanding becomes P = nRT0/V; where T0 remains the same as this is an adiabatic expansion. (In a non-ideal case, aka reality, this would not be the case, but the P derived would serve as an upper bound with the expectation the real P would be somewhat less.)

Therefore P/P0 = nRT0/V//nRT0/V0 = V0/V and P = P0 (V0/V) = P0 (V0/[2/3 pi r^3]); where r = 10 meters. If you measure P0 at ground zero, say, r0 = 1 meter, you can write P = P0 (r0/r)^3 = P0 (1/r^3) when r0 = 1 meter where P0 was measured.

Bottom line, ideally, pressure decrements inversely with the cube of the distance from ground zero.

Answer 2

Imagine that the explosive has a mass confined within
within a strict spherical body. When the explosive
detonates a spherical wave is emitted from the
point of detonation. Assume that the crest of the
wave contains the entirety of the mass of the
bomb at any given point in time. As the wave moves
outward the surface area of the wave crest increases
and the density decreases. Using this model we
can imagine a lesser psi based on the ray drawn
from the initial center of the sphere out to the
edge of the wave crest. Therefore, the spheres
shape is ideally maintained and the ray's magnitude
increases with time. Surface area for a sphere is
4(pi)r^2 and r changes with respect to time in
our model. Using precalculus substitute r=1 and
r=10 and evaluate both.

p(r) where r=magnitude of the ray from center to edge of sphere
p(1)=4(pi)(1)^2=4(pi)
p(10)=4(pi)(10)^2=400(pi)

Therefore, the psi at 10 meters is 100 times less than
at the initial detonation point.


On a flat surface, perhaps you mean half a sphere.
Given our model no modification need be made,
but if we were examining energy then more of the
energy would be deflected to the emitted bisphere.

Other models could have been chosen, but the wave
crest model is easy and ideal.

Answer 3

Here is my Formula. It is complex, water is much simpler. At 10m you will probably want HENRYCH3.

VELOCITY=((((6*PRESSB)+(7*PRESSO))/(7*PRESSO))^.5)*SOUNDVEL
NEWDEN=(((6*PRESSB)+(7*PRESSO))/(PRESSB+(7*PRESSO)))*ORIGDEN
ZED=RADIUSM/(WGHTKG^.33333)
IF AND (0.02<=ZED, ZED<=0.3) THEN PRESSB=HENRYCH
IF AND (0.3 IF ZED>1 THEN PRESSB=HENRYCH3
HENRYCH=(14.072/ZED)+(5.540/(ZED^2))-(.357/(ZED^3))+(.00625/(ZED^4))
HENRYCH2=(6.194/ZED)-(.326/(ZED^2))+(2.132/(ZED^3))
HENRYCH3=(.662/ZED)+(4.05/(ZED^2))+(3.288/(ZED^3))
PRESSE=PRESSB*14.5
DYNPRESM=(5*(PRESSB^2))/(2*(PRESSB+(7*PRESSO)))
VELENG=VELOCITY*3.28
COMPRESS=-(1-(NEWDEN/ORIGDEN))*100
DYNPRES=DYNPRESM*14.5
IF DYNPRES>PRESSE THEN PRESSURE=DYNPRES ELSE PRESSE=PRESSURE
STRESS=(PRESSURE*OUTRAD)/WALL
RADIUS=RADIUSM*3.28
WEIGHT=WGHTKG*2.205
RADIUS2=RADIUS*12
PRESSURE=OBSERVE/6895000000
DBAIR=20*(LOG(OBSERVE/20.4))
Kpa=PRESSURE*6895

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