FORD FLTM BI 102-05-2001 VOLUME SOLIDS DETERMINATION OF COATINGS - THEORETICAL (Replaces FLTM BI 002-05)《涂层中的固体物体积的理论性测定 替代FLTM BI 002-05》.pdf

1、 FORD LABORATORY TEST METHOD BI 102-05 Date Action Revisions 2001 03 13 Revised Editorial no technical change A. Cockman 1996 10 02 Printed copies are uncontrolled Page 1 of 2 Copyright 2001, Ford Global Technologies, Inc. VOLUME SOLIDS DETERMINATION OF COATINGS THEORETICAL Application This test met

2、hod describes the procedure to calculate the theoretical volume of solids in a coating or paint sample. The procedure uses values of weight percent solids of the coating and weight per liter as determined by Ford Laboratory Test Method BI 102-01 and ASTM D 1475, respectively. Apparatus Required Equi

3、pment and procedures for required determinations are specified in FLTM BI 102-01 and ASTM D 1475. Conditioning and Test Conditions All test values indicated herein are based on material conditioned in a controlled atmosphere of 23 +/- 2 C and 50 +/- 5 % relative humidity for not less than 24 h prior

4、 to testing and tested under the same conditions unless otherwise specified. Procedure To calculate the theoretical volume solids of a paint sample, the following information is required: 1. Weight per liter of the coating (ASTM D 1475). 2. Percent weight solids of the coating (FLTM BI 102-01). 3. W

5、eight per liter of each solvent in the subject coating. 4. Percent of each solvent by weight in the subject coating (based on 100 %). Calculations A. Weight of the solvent = (weight/liter of coating) x (100 % - weight % of solids) B. Calculate the average weight per liter of the solvents in the coat

6、ing. FORD LABORATORY TEST METHOD BI 102-05 Page 2 of 2 Copyright 2001, Ford Global Technologies, Inc. C. Volume of the solvents = Weight of the Solvent Average wt/L of the Solvents D. Volume of solids = (1.00 - volume of the solvents). Sample Calculation Information Needed 1. Coating wt/L = 1.2 kg/L

7、 2. Solids = 50 % (0.50) 3. Solvent A = 1.08 kg/L Solvent B = 0.96 kg/L Solvent C = 0.84 kg/L 4. Solvent A = 50 % (0.50) Solvent B = 25 % (0.25) Solvent C = 25 % (0.25) A. Weight of solvent = 1.2 x (1.00 - 0.50) = 0.6 kg/L B. Solvent A = 0.50 x 1.08 = 0.54 Solvent B = 0.25 x 0.96 = 0.24 Solvent C =

8、0.25 x 0.84 = 0.21 0.99 wt/L of solvents C. 5.00 = 0.606 = 60.6 % volume of the solvents 8.25 0.60 = 0.606 = 60.6 % volume of the solvents 0.99 D. Percent volume of solids = (1.00 - 0.606) x 100 = 39.4 % Chemicals, materials, parts, and equipment referenced in this document must be used and handled properly. Each party is responsible for determining proper use and handling in its facilities.