Evaluation method of cigarette mainstream smoke ingredient sensing style index
A technology of ingredients and smoke, applied in the field of information technology and automation, to achieve the effect of predicting the residual sum of squares is small, objectively evaluating the sensory characteristics of cigarettes, and eliminating the influence of subjective factors
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Embodiment 1
[0031] a. Select 30 cigarettes of different brands and different prices, and set up a 7-person smoking evaluation team to evaluate and smoke according to the requirements of YC / T 497 "Chinese Cigarette Style Sensory Evaluation Method", and obtain the fruity flavor score vector, Y t =(0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,0.9,1,1.1,1.2,1.3,1.3,1.4,1.5,1.6,1.7,1.8,1.9,1.9,2,2.1,2.2, 2.3,2.4,2.5,2.6,2.7,2.8);
[0032] b. After the above-mentioned cigarette samples and the Cambridge filter are placed in a constant temperature and humidity environment (20±2°C, 60±2%) to balance for 48 hours, according to GB / T 16450 "Definition and Standard Conditions of Smoking Machines for Routine Analysis" It is required to smoke each cigarette sample, then put the Cambridge filter with the total particle phase trapped in a 250mL Erlenmeyer flask with a stopper, and add 1.0mL internal standard solution (isopropanol solution containing 200μg / mL n-heptadecane) in sequence ) and 50 mL of dichloromethane,...
Embodiment 2
[0043] a. Select 30 cigarettes of different brands and different prices, and set up a 7-person smoking evaluation team to evaluate and smoke according to the requirements of YC / T 497 "Chinese Cigarette Style Sensory Evaluation Method", and obtain the fragrance score vector, Y t =(1.7,1.3,1.2,1.5,1.5,1.0,2.3,2.2,2.0,1.2,0.7,1.0,1.8,2.1,0.9,1.4,1.9,1.9,1.3,2.2,1.6,0.9,2.5,1.6, 2.4,1.4,0.6,2.1,1.1,0.8);
[0044] Steps b to d are the same as in Example 1.
[0045] e. Write the PLS algorithm with Matlab software, for X s and Y t Fitting is performed to determine the number of factors f 1 is 15, and obtain the regression coefficient vector β s = (0.018, 0.029, 0.039, -0.0016, ..., 0.097).
[0046] f. Take β s the absolute value of the vector vector β abss = (0.018, 0.029, 0.039, 0.0016, ..., 0.097), β abss The average value of the elements is 0.0311, and the elements less than 0.03111×15%=0.00467 in the vector are eliminated, and the contents of the remaining 91 compounds (c...
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