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.
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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Question 94 - The resolution of separation, Rs for chromatography is given by the formula Rs = (difference in retention time) / (average width at the base). In a chromatogram, 3 peaks a, b and c are found. Average widths W at the bases of the solutes are : Wa = 20 s, Wb = 40 s, Wc = 30 s. Resolutions of separation, Rs for solutes b and c in comparison to a are 2 and 4 respectively. The differences in retention times T for b and c in comparison to a are (Tb - Ta) and (Tc - Ta), Ta = Tc - Tb : (a) Form 2 equations involving Rs as a function of Wa, Wb, Wc, Ta, Tb and Tc. (b) Find the values of Ta, Tb and Tc.
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