Method for estimating composition of product obtained by liquefaction of coal

A method of estimating yields of products of liquefaction at 465° C. of coal or its analogues, comprising:

• (a) obtaining spectral data of coal or its analogues by using a solid state 13C nuclear magnetic resonance spectrometer;
• (b) deconvoluting said spectral data into 24 peaks according to Table 1 below and classifying said 24 peaks into 9 carbon groups according to Table 1 below,
• TABLE 1 Half-value Assignment Peak No. Peak Position (ppm) width (ppm) of Carbon 1 11.3 <12 CH3 2 18.8 9 3 24-26 7 CH2 4 30-35 <10 5 38.9 15 6 49 7 7 54-56 <10 OCH3 8 62.5 <10 9 72-75 <10 R—OH 10 80 9 C—O 11 84 7 12 89 4 13 100 <10 Ar—H 14 105 5 15 114-116 <13 16 124 <12 17 131 12 Ar—C 18 140-141 <12 19 145-146 <12 Ar—O 20 150-153 <14 21 160 <14 22 175 15 COOH 23 180 15 24 200 15 C═O
• said 9 groups consisting of carbonyl group (C═O), carboxyl group (COOH), aromatic carbon bonded to an oxygen atom (Ar—O), aromatic carbon bonded to a carbon atom (Ar—C), aromatic carbon bonded to a hydrogen atom (Ar—H), aliphatic carbon bonded to a hydroxyl group or an etheric oxygen (R—OH or C—O), methoxyl group (OCH3), methylene group (CH2) and methyl group (CH3);
• (c) determining distribution of carbons of the CH3, CH2, C═O, COOH, Ar—O and OCH3 in terms of wt. %, and aromaticity fa from the peak intensity of the spectral data obtained in step (b);
• and
• (d) calculating an estimated yield of at least one liquefaction product of the coal or its selected from the group consisting of hydrocarbon gas, COx gas, water, oil and residues according to the following equations:
• Hydrocarbon Gas (wt. %)=0.33×(CCH2+CCH3)+6.15
• CO×Gas (wt. %)=1.43×(CC═O+CCOOH)−1.45
• Water (wt. %)=2.03×(CAr—O−COCH3)−10.84
• Oil (wt. %)=0.86×CCH2+27.5
• Residues (wt. %)=122.0×fa−49.2
• wherein CCH2, CCH3, CC═O, CCOOH, CAr—O and COCH3 represent distribution, in terms of wt. %, of carbons of the CH2, CH3, C═O, COOH, Ar—O and OCH3, respectively.