A History of Research and a Review of Recent Developments
A History of Research and a Review of Recent Developments
A History of Research and a Review of Recent Developments
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116<br />
Structural loading from local explosions<br />
shock tube experiments at the US Army Ballistics <strong>Research</strong> Laboratory were<br />
carried out by Coulter [5.26] in which chamber volumes, entrance areas,<br />
pressures <strong>and</strong> geometries were varied. It was shown that the pressure wave<br />
exp<strong>and</strong>ing into the chamber decays in amplitude <strong>and</strong> reflects from internal<br />
surfaces. The attenuation <strong>of</strong> the instantaneous peak pressure was inversely<br />
proportioned to the opening diameter, <strong>and</strong> its positive duration was directly<br />
proportional to the diameter. For non-circular openings the diameter was<br />
replaced by a ‘mean opening dimension’. Tests at the US Army Waterways<br />
Experiment Station in the early 1980s, using C-4 <strong>and</strong> TNT charges, showed<br />
that side-on peak pressures in the test chambers (p max) were related to the<br />
peak pressure in the opening (entrance tunnel) (p 0) by the equation<br />
p max /p 0 =0.65(1–0.25α)(R/D) –1.35 ,<br />
(5.32)<br />
where R is the distance <strong>of</strong> the centre <strong>of</strong> the opening to the pressure gauge in<br />
the chamber <strong>and</strong> D is entrance tunnel diameter. a is the angle in radians between<br />
the normal to the centre <strong>of</strong> the opening <strong>and</strong> the gauge location. For small<br />
angles it was sufficient to ignore a <strong>and</strong> within the units <strong>of</strong> experimental scatter<br />
to simplify the above equation to p max/p 0=0.65(R/D) –1.35 . The values <strong>of</strong> p max<br />
are due to the first side-on peak at the gauge as the result <strong>of</strong> incident pressure<br />
<strong>and</strong> do not take account <strong>of</strong> internally reflected waves. If information about<br />
reflected waves is required, then a more complex analysis involving a shock<br />
diffraction model is needed. Path lengths <strong>of</strong> ‘rays’ <strong>of</strong> successively higher order<br />
reflections are generated, <strong>and</strong> arrival times calculated; then shock wave<br />
attenuation with distance is found. The procedure is fully described in ref.<br />
[5.24]. The combined pressure pulse in a chamber, found in this way, can be<br />
calculated using the code CHAMBER, originally for a mainframe computer<br />
but converted to run on a desktop computer. Predictions for the initial peak<br />
reflected shock were shown to agree well with test results.<br />
It is clear that the levels <strong>of</strong> pressure from an internal explosion, whether due to<br />
a high-explosive detonation or a vapour or dust cloud deflagration, can be reduced<br />
by the judicious use <strong>of</strong> vents. The design <strong>of</strong> venting systems is governed by the<br />
pressure reductions that are required, so there has been a considerable research<br />
effort in this area. Going back to the work <strong>of</strong> Christopherson [4.6], which we<br />
mentioned earlier, we noted that he analysed the effect on the pressure in a rigid<br />
rectangular chamber if one end were vented. The ‘partial enclosure’ was the subject<br />
<strong>of</strong> an empirical relationship in which the results <strong>of</strong> a small number <strong>of</strong> observations<br />
were recorded in terms <strong>of</strong> the charge weight <strong>of</strong> TNT, the volume (V) in cubic feet,<br />
<strong>and</strong> the face-on blast impulse on the venting side wall <strong>of</strong> a cubical enclosure (A T),<br />
in lb m sec/in 2 . His results are plotted in Figure 5.19 on the axes A T/W 1/3 <strong>and</strong> V/<br />
W(the volume <strong>of</strong> the chamber per lb <strong>of</strong> charge). For initial design purposes, the<br />
relationship took the approximate form (Imperial units):<br />
(A T /W 1/3 )(V/W)=15, (V in ft 3 ).<br />
(5.33)