ENGINEERING MATHEMATICS - EXAMPLE 8.1 : A local utility burns coal having the fo

ENGINEERING MATHEMATICS - EXAMPLE 8.1 : A local utility burns coal having the following composition on a dry basis : Carbon (C) 83.05 %, hydrogen (H) 4.45 %, oxygen (O) 3.36 %, nitrogen (N) 1.08 %, sulfur (S) 0.7 % and ash 7.36 %. Calculate the ash free composition of the coal with reference to C, H, O, N and S.

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ENGINEERING MATHEMATICS - EXAMPLE 8.1 : A local utility burns coal having the following composition ..

Answer / kangchuentat

ENGINEERING MATHEMATICS - ANSWER 8.1 : Total percentage without ash is 83.05 + 4.45 + 3.36 + 1.08 + 0.7 = 92.64 %. For ash free composition percentage calculations, C = 83.05 / 92.64 x 100 % = 89.65 %. H = 4.45 / 92.64 x 100 % = 4.80 %. O = 3.36 / 92.64 x 100 % = 3.63 %. N = 1.08 / 92.64 x 100 % = 1.17 %. S = 0.7 / 92.64 x 100 % = 0.76 %. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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