I was analyzing ISO 1988-Hard Coal Sampling and I got stuck somewhere while calculating the amount of coal that would be taken by a single operation of an automatic sampler.
According to the tables I need to take 32 increments for ash and 16 increments for moisture. Thus the calculated figure (from 3.3.2) for the minimum mass of increment is 3 kg for my case (for 40 mm pieces)
Since my consignment is 14400 t/day I multiply the number of increments with the factor sqrt(14400/1000), thus I find that I need to get 5 ash and 3 moisture increments per hour. Totally I need 8 increments.
I calculated before that the minimum amount of increments should be 3 kg, thus I multiply 3 by 8 and get 24 which is the amount of coal that my sampler need to take in a single operation.
However, at page 60 of the standard, there's an example and it goes like this:
800 t/hr stream wet coal so needs 32 increments for moisture and 16 for general analysis. It multiplies 16 by 6 and says it needs totally 96 increments for general analysis. I can get how it gets 6 ({(800 t/hr*24hr)/1000}*32=6) but I want to know why it did not multiply 32 by 6 as well. Or is it wrong why I think 16 is multiplied by 6.
It is totally a specific subject but I wish there's someone that could make me understand the case better which is used to calculate the amount of coal that an automatic sampler will take.
ISO 1988 Sampling Question
Hi all,
I was analyzing ISO 1988-Hard Coal Sampling and I got stuck somewhere while calculating the amount of coal that would be taken by a single operation of an automatic sampler.
According to the tables I need to take 32 increments for ash and 16 increments for moisture. Thus the calculated figure (from 3.3.2) for the minimum mass of increment is 3 kg for my case (for 40 mm pieces)
Since my consignment is 14400 t/day I multiply the number of increments with the factor sqrt(14400/1000), thus I find that I need to get 5 ash and 3 moisture increments per hour. Totally I need 8 increments.
I calculated before that the minimum amount of increments should be 3 kg, thus I multiply 3 by 8 and get 24 which is the amount of coal that my sampler need to take in a single operation.
However, at page 60 of the standard, there's an example and it goes like this:
800 t/hr stream wet coal so needs 32 increments for moisture and 16 for general analysis. It multiplies 16 by 6 and says it needs totally 96 increments for general analysis. I can get how it gets 6 ({(800 t/hr*24hr)/1000}*32=6) but I want to know why it did not multiply 32 by 6 as well. Or is it wrong why I think 16 is multiplied by 6.
It is totally a specific subject but I wish there's someone that could make me understand the case better which is used to calculate the amount of coal that an automatic sampler will take.
Thanks in advance
Engin ■