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ICHCA International <strong>Safe</strong>ty Panel Briefing Pamphlet No 30<br />
where α is the coefficient <strong>of</strong> cubical expansion, tr is the reference<br />
temperature to be taken as 50ºC and tf is the filling temperature,<br />
in this case 15 o C.<br />
Step 3 Calculate the coefficient <strong>of</strong> cubical expansion (see A2.2.5):<br />
α =<br />
α =<br />
d15 – d50<br />
35d50<br />
0.812 – 0.772<br />
35 x 0.772<br />
α = 0.0014803<br />
Step 4 Enter the coefficient <strong>of</strong> cubical expansion into the formula<br />
obtained at Step 2:<br />
95<br />
Degree <strong>of</strong> filling =<br />
1 + 0.0014893(50-15)<br />
Degree <strong>of</strong> filling = 90.29%<br />
Solution: The maximum degree <strong>of</strong> filling permitted by the IMDG Code <strong>of</strong> a<br />
tank container laden with this substance is 90.3%.<br />
This means for a 26,000 litre tank container, filled with<br />
ACRYLONITRILE, STABLIZED, .only 90.29% <strong>of</strong> its capacity can<br />
be used which equates to 23,478 litres (26,000 x 0.0929<br />
(90.29%)).<br />
The shipper originally wanted to ship 22,500 kg <strong>of</strong> this<br />
substance, but this weight <strong>of</strong> the cargo was:<br />
1. equated to a volume larger than the tank available<br />
(27,709 litres – Step 1)<br />
2. is restricted to 23,478 litres by the IMDG Code and the<br />
available tank container.<br />
The shipper an convert the volume back to net mass by<br />
multiplying the available volume by the liquid density:<br />
23,478 litres x 0.812 = 19064 kg (80.72% <strong>of</strong> the original cargo<br />
mass)<br />
Page 51 ©ICHCA International Limited
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