Time for Spontaneous Ignition

Example 5.1 Animal feedstock, a distillery byproduct, is to be stored in a silo, a rectangular facility 12 m high and 3 m x 3 m square. For an ambient temperature of 20 °C do we expect a problem What would be the critical ambient temperature sufficient to cause spontaneous ignition in the silo The following oven data was obtained for cubes of side s(= 2r0) from feedstock samples. 

Solution Using the procedure just outlined with 8C = 2.52 for a cube, we can show that P = 39.88 in units of rG in m and E/R = 8404 K. Substituting these values into Equation (5.12) yields 

For the rectangular silo, we compute the corresponding critical Damkohler number, 

Note that this Sc is defined in terms of r0 = 1.5 m. Calculating 8 for rG = 1.5 m and Too = 20 °C from our feedstock equation gives 6 = 1.9. Since the 8 for this condition is greater than the critical value of 1.8, we will have spontaneous ignition. 

The prospect of successfully estimating the likelihood of spontaneous ignition depends on our ability to know 8C for our configuration and heating condition, and the bulk kinetic 

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The time for spontaneous ignition to occur will decrease as the Damkohler number... 

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