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News
How to Compute the Mean Particle Diameter from a LISST Volume Distribution
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March 26, 2010
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How to Compute the Mean Particle Diameter from a LISST Volume Distribution
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| [Sequoia, March 26, 2010] |
In the table below, the computation of the mean particle size from a LISST volume distribution is outlined.
Column A is the size class number, from 1-32. Column B is the MID POINT of the size bin in µm. In this example we are using the size classes associated with a LISST type B randomly shaped inversion (LISST Bin Sizes). Column C is the Volume Distribution in µl/l.
Each row in column D (dSum) is computed as the size class # for that row multiplied with the volume concentration in the size class. For example, for size class 8, dSum is equal to 8 * 3.035 = 24.28.
| A | B | C | D | Size class
| Size class mid point (µm)
| Volume Distribution (µl l-1)
| dSum (Size class # * VC in size class)
| | 1 | 1.09
| 0.190
| 0.190 | | 2 | 1.28 | 0.256 | 0.512 | | 3 | 1.51
| 0.360 | 1.080 | | 4 | 1.79 | 0.616 | 2.464 | | 5 | 2.11 | 0.853 | 4.265 | | 6 | 2.49 | 1.041 | 6.246 | | 7 | 2.93 | 1.429 | 10.003 | | 8 | 3.46 | 3.035 | 24.280
| | 9 | 4.09 | 4.071 | 36.639 | | 10 | 4.82 | 4.419 | 44.190 | | 11 | 5.69 | 4.818 | 52.998 | | 12 | 6.71 | 5.714 | 68.568 | | 13 | 7.92 | 6.218 | 80.834 | | 14 | 9.35 | 6.252 | 87.528 | | 15 | 11.03
| 6.151 | 92.265 | | 16 | 13.02 | 6.360 | 101.76 | | 17 | 15.36 | 6.288 | 106.896 | | 18 | 18.13 | 6.311 | 113.598 | | 19 | 21.39 | 6.732 | 127.908 | | 20 | 25.25 | 7.726 | 154.52 | | 21 | 29.79 | 8.252 | 173.292 | | 22 | 35.16
| 9.573
| 210.606 | | 23 | 41.49 | 10.666 | 245.318 | | 24 | 48.96 | 11.581 | 277.944 | | 25 | 57.77 | 11.843 | 296.075 | | 26 | 68.18 | 12.214 | 317.564 | | 27 | 80.45 | 11.177 | 301.779 | | 28 | 94.94 | 9.434 | 264.152 | | 29 | 112.04 | 6.993 | 202.797 | | 30 | 132.21 | 4.750 | 142.500 | | 31 | 156.02 | 3.128 | 96.968 | | 32 | 184.11 | 2.300 | 73.600 | | | | ?VD = VC
| ?dSum | | | | 180.751 | 3719.339 |
By definition, the mean particle size in terms of size class number is now ?dSum/?VD = ?dSum/VC, in this example 3719.339 / 180.751 = 20.57714.
We can now see that the mean particle size, in terms of size class numbers is somewhere between size class 20 and 21, i.e. somewhere between 25.25 and 29.79 µm.
Now round down ?dSum/VC to the nearest integer, in this case it would be 20.
Compute the remainder; in this case it would be 20.57714 – 20 = 0.57714.
The mean particle size is now computed as the midpoint of size class 20 (from rounding down ?dSum/VC) multiplied by the ratio of the bin midpoints (200^(1/32) = 1.1801) raised to the power of the remainder: 25.25 * 1.1801^0.57714 = 27.78µm. |
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