Varnish Blending Graph A graphical method is described for determining weight-per cent and weight-volume relationships of the volatile and nonvolatile components of oleoresinous varnishes.
R. C. SHUEY Bakelite Corporation, Bloomfield, N. J.
T
HE blending of two or more varnishes often provides the most satisfactory means of solving specific coating problems with a minimum of experimental effort. This is a procedure as old as the varnish industry. With the advent of synthetics and the increasing variety of products and uses, the varnish formulator is resorting more and more to the blending of tried and proved materials to secure intermediate effects. The operation itself is exceedingly simple but the calculations necessary to produce a desired blend of two varnishes are somewhat complicated by the fact that the oil length of a varnish is expressed in gallons of oil per 100 pounds of resin and that in plant practice pounds and gallons are both frequently used in the same formula. As a result approximations are frequently resorted to, in the interest of simplification, and still the continuous repetition of similar calculations remains laborious. Since the mathematics involved is simple, it has been found easier to handle the problems graphically, and the graph included here is an attempt to assemble all of the necessary dataso interrelated that any type of problem involving these variations can be solved. With a little care the accuracy of the work can be made to exceed that ordinarily obtained with the slide rule, and in many cases the operation is much more rapid because several variables can be handled a t once. Operations involved fall into two classes: (1) weightpercentage relationships involving the nonvolatile portion and (2) weight-volume relationships including also the volatile components or thinners. Each of these classes of operation requires its own ruling; therefore actually two complete sets of curves are required. For the sake of Convenience these two sets of curves have been placed on a single system of two-component rulings. Since the vertical scales in right and left margins are identical and used with both the upper and lower marginal rulings, they have received no designation letter. The weight-percentage relationships are handled horizontally on the top margin by the scales marked A and B. These two scales read in opposite directions and for any given point total 100 per cent. Each diagonal line is marked as to which scale applies t o it. The weight-volume dilution relationships are handled on the bottom scale marked C, which reads in gallons of total volume per 10 pounds of nonvolatile. The choice of 10 pounds as unity was made because resin is the heaviest component of varnish arid some resins (e. g., Bakelite BR-254
and BR-820) weigh 10 pounds per gallon. By using the heaviest component as unity, there is no crossing of the unity line. The advantage of choosing a decimal unity is obvious.
Construction of the Graph The varnish length spacing shown in the side margins mas obtained by scaling off vertically the resin content by weight of the various lengths on the basis of total nonvolatile components. The horizontal percentage spacing in the top margin therefore makes this portion of the graph a simple blending nomograph. By derivation, the percentage resin by weight in the varnishes of the various lengths is therefore expressed as a straight-line function and this line is drawn to be read on scale B. Theother component of the varnishnonvolatile content is oil. Therefore the percentage weight of the oil would equal 100 per cent minus the percentage resin and can be directly read from scale A a t the point where the percentage resin is read. However, for cost calculations it is often convenient to express oil in gallons rather than by weight. So the percentage oil divided by 7.84 (the weight of a gallon of oil) would give the volume of oil equivalent to that percentage weight. The resultant values are scaled off as the oil volume line (to be read on scale B ) on the basis of 100 per cent equals 1 gallon of oil in 10 pounds of nonvolatile. However, varnishes of over 46 gallons length (78.4 per cent oil) contain more than 1 gallon of oil, so that the line is continued downward from that point by turning back toward the left, and the reading is shifted to scale A (for convenience in not crossing the chart). This requires the addition of 1 gallon to the A reading over this portion of the line. This oil volume line is a composite, connecting weight to gallons, and not a weight percentage line as the others have been. Scale C a t the bottom of the chart is labeled weight-volume factor and expresses in gallons the total volume of 10 pounds of nonvolatile diluted. Dilution in percentages of nonvolatile by weight is expressed in the side margins by the values 100 to 44 formerly used to express varnish length. The weight-volume correction curves (or density coriversion curves) are scaled off on the basis that 10 pounds of resin bulk 1 gallon, and 7.84 pounds of oil or 6.5 pounds of thinner bulk 1 gallon. The reciprocals of these values or the gallons per 10 pounds of nonvolatile are spotted along the 100 per cent nonvolatile line for common varnish lengths from straight resin t o straight oil. 391
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INDUSTRIAL AND ENGINEERING CHEMISTRY
To each 10 pounds of nonvolatile must be added a certain weight of volatile thinner to give the desired dilution. This value is divided by 6.5 to give the equivalent number of gallons of thinner which, added t o the volume of 10 pounds of nonvolatile, gives the total bulk in gallons. The spacing for these curves was derived by scaling off from unity on scale C (the volume of 10 pounds of resin) the volume of thinner corresponding t o a given dilution and spotting on that dilution line. The curve connecting these spots therefore gives the diluted volume of 10 pounds of resin. In like manner, curves were drawn for the diluted volumes of 5-, lo-, 2 6 , 50-, and 100-gallon varnishes and straight oil. The error produced by interpolation for other lengths is generally negligible because the nonvolatile content of the component varnishes is seldom known beyond the accuracy obtained. Specific gravities have not been used or shown but they can be obtained if desired. If 10 pounds of resin equals 1 gallon (specific gravity, 1.200), it is possible t o find the specific gravity of any varnish from the weight-volume factor (see example 6). Concerning errors due to choice of constants, the fact that ester gum, the lightest common resin, weighs only 9 pounds per gallon introduces a measurable error in very short oil varnishes. A 10-gallon, 50 per cent nonvolatile, ester gum varnish should have a factor of 2.74 while the graph shows 2.66, making the error in percentage nonvolatile about 1 per cent. At 50 gallons it is less than one-third per cent. Thus in most cases it can be completely ignored, since the actual nonvolatile content is seldom known within an accuracy of one per cent. Although raw linseed oil bulks 0.1288 gallon per pound and raw China wood oil 0.1277, the China wood oil remains practically constant during heat treatment, and linseed approaches China wood oil value ;On slight heat treatment and passes it if heavy-bodied. Since linseed oil is practically always given some body for varnish use, the error due to partial eubstitution of linseed oil can therefore be ignored. The error due to a substitution of such thinners as turpentine (1 pound = 0.1385 gallon) for mineral spirits (1 pound = 0.1539 gallon) is also measurable but easily compensated. If in thinning a varnish 9 gallons of turpentine are substituted for 10 gallons of mineral spirits, no error in weight percentage is introduced, but the total volume is 1 gallon less. If 20 per cent turpentine is substituted in a 50-gallon, 50 per cent nonvolatile varnish, the 10 per cent correction on that part of the thinner is 0.03 gallon in the weight-volume factor, and this changes the nonvolatile percentage reading by less than 0.5 per cent. Therefore, the practice of substituting 9 gallons of turpentine for 10 gallons of mineral spirits produces the necessary compensation. The above examples illustrate practically the maximum magnitude of errors. In most varnishes the errors are under 0.5 per cent. The operative procedure is shown by the following examples. They illustrate the solution of some of the common problems of the varnish maker. Many others, as they occur, can be solved with equal facility. Example 1 Given a 70-gallon varnish containing 70 per cent nonvolatile and a 10-gallon varnish containing 50 per cent nonvolatile, what volume of each is required to produce a blend 30 gallons in length? Also, what is the nonvolatile content (by weight) of the blend? SOLUTION.Locate the 70-gallon length value in the lefthand margin and the 10-gallon length in the right-hand margin. Place a straight edge across the graph connecting these two points. Follow the 30-gallon length line from the margin horizontally until it strikes the straight edge, then vertically to the percentage scale a t the top. The reading is 64.5 per
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cent A and 35.5 per cent B. These are the weights of nonvolatile content of the 70- and 10-gallon varnishes, respectively, required to make the blend. However, A contains 70 per cent by weight of nonvolatile. To obtain the volume required a t 70 per cent nonvolatile and 70 gallons length, follow the 70 line (now used for 70 per cent nonvolatile) horizontally until it meets the weight-volume or density correction group of curves. Stop a t 70 gallons length (approximately halfway between the 50- and 100gallon curves). Drop perpendicularly from this point to the bottom margin, scale C, giving a reading of 1.90 which represents the volume required t o deliver one weight per cent of nonvolatile. Multiply this by 64.5, the weight percentage of A , t o give the total volume of this component required (122.9 gallons of the 70-gallon varnish). Likewise B contains 50 per cent by weight of nonvolatile; therefore, follow the 50 per cent nonvolatile line horizontally to the 10-gallon weight-volume correction curve, then down to the value 2.66 gallons required per unit weight. Multiply this by 35.5 (the percentage B obtained in the first operation) which makes 94.4 gallons of the 10-gallon varnish. Therefore, these two varnishes mixed in the proportion of 122.9 gallons of the 70-gallon varnish and 94.4 gallons of the 10-gallonvarnish will make 217.3 gallons of a 30-gallon varnish. To obtain the nonvolatile content of this varnish, the operation is now reversed. The value of 217.3 gallons represents 100 units. Therefore divide by 100 and follow the 2.17 weight-volume factor line vertically until it meets the 30gallon correction point (approximately the 25-gallon correction curve), thence horizontally to either margin where this varnish is found to be 61 per cent nonvolatile. Example 2 Given a blend of 56.6 per cent by volume of a 70-gallon varnish of 70 per cent nonvolatile content and 43.40 per cent by volume of a 10-gallon varnish of 50 per cent nonvolatile content, what is the oil length and percentage nonvolatile content of the blend? SOLUTION.First find the weight-volume factor of each component. For the 70-gallon component follow the 70 per cent nonvolatile line horizontally to the 70-gallon correction point, thence down to 1.90 gallons per 10 pounds of nonvolatile. In like manner follow the 50 per cent nonvolatile line to the 10-gallon correction curve and down to 2.66 gallons per 10 pounds of nonvolatile in the short component. Divide the volume percentages used by these factors to obtain the relative weights. The sum of these relative weights is 100 per cent of the nonvolatile (N. V,), Divide 100 by this sum to obtain the weight-volume (W-V) factor of the blend (2.17). We now require the percentage weights. These are obtained by dividing the relative weights by the sum weight. The calculations are: 70% N. V.
- 70-gallon (from graph) W-V factor
68.8 = 29.8 pacta 1.90 60% N. V.
43.4 = 2.66
i:.
-
N. V.
- 10-gallon (from graph) W-V factor
=
-
1.90
2 66
16.3 parts N. V. 2.17 = W-V factor of blend
= 64.6% by weight of 70-gallon N. V
i::: = 36.4%
by weight of 10-gallon N. V.
Place the rule from 70-gallon to 10-gallon varnish lengths on the left-hand and right-hand margins, respectively. Then drop from scale A - B (64.6 - 35.4 per cent) to the rule, thence left to the margin where the varnish length of the blend is found to be 30 gallons.
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VARNISH BLENDING GRAPH 70
60
50
40
30
20
30
40
50
60
70
80
10 90
5 0
WEIGHT VOLUME FACTOR (GALLONS
P € R lO*OFNON
3 VOlATIL€)
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Knowing the length to be 30 gallons, follow the volumeweight factor of the blend (2.17) vertically to the 30-gallon correction point, thence horizontally to the per cent nonvolatile which is found to be slightly less than 61 per cent. This problem is the reverse of example 1 and is a simple method of solving an unknown of two components for percentage composition of the nonvolatile and for oil length.
Example 3 Given a blend of 56.6 per cent by volume of a 70-gallon varnish of 70 per cent nonvolatile content, and 43.4 per cent by volume of a 10-gallon varnish of 50 per cent nonvolatile content, find the composition of each stated in pounds of resin per gallon, gallons of oil per gallon, and gallons of thinner per gallon. Then by adding these components, obtain the composition of the blend and finally determine the percentage composition and oil length of the blend. SOLUTION.Follow the 70 per cent nonvolatile line horizontally to the 70-gallon correction point, thence down to the weight-volume factor (scale C), reading 1.90 gallons per 10 pounds of nonvolatile diluted. Follow the 70-gallon varnish line horizontally to the percentage resin line, then up to scale B reading 15.4 per cent by weight. Then 15.4 per cent of 10 pounds = 1.54 pounds of resin in 1.90 gallons of varnish; 1.54/1.90 = 0.810 pound of resin per gallon. Follow the 70-gallon varnish length line horizontally to the volume of oil line, thence up to scale A (which applies to varnishes longer than 46 gallons because the volume is greater than 1 gallon). The reading is 8.0 per cent or 0.080 gallon plus 1 gallon = 1.080 gallons of oil in 1.90 gallons of varnish. Then 1.080/1.90 = 0.568 gallon of oil per gallon of varnish. The composition is as follows: 0 081 gallon of resin or 0 810 pound of resin 0 568 gallon of oil, 0 351 gallon of thinner (by difference)
1,000gallon total volume
In like manner follow the 50 per cent nonvolatile line to the 10-gallon correction curve and down to 2.66 gallons per 10 pounds of nonvolatile in the short component. Follow the 10-gallon varnish length line to the percentage resin line, thence up to scale B, reading 56 per cent by weight. Then 56 per cent of 10 pounds = 5.6 pounds of resin in 2.66 gallons of varnish; 5.6/2.66 = 2.105 pounds of resin per gallon. Follow the 10-gallon varnish length line horizontally to the volume oil line, thence up to scale B, reading 0.56 gallon of oil in 2.66 gallons of varnish. Then 0.56/2.66 = 0.2105 gallon of oil per gallon of varnish. The composition is as follows : 0.2105 gallon of resin or 2.105 pounds 0 . 2 1 0 5 gallon of oil, 0,5790 gallon of thinner (by difference)
-
1.0000 gallon total volume
The blend was 56.6 per cent of the 70-gallon and 43.4 per cent of the 10-gallon varnish; therefore: 0.566 X 0.061 gallon of resin = 0.0458 0.0914 0.434 X 0.2105 gallon of resin
--
-
0.1372 gallon resin
0.3215 0.434 X 0.2105 = 0 0 9 1 4 0.566 X 0.568
0.4129 gallon oil
The composition of the blend is therefore: 0.1372 gallon resin or 1 , 3 7 2 pounds of resin 0.4129 gallon oil. 0.4499 gallon thinner (by difference)
1,0000gallon total volume
To find the oil length of the blend, first find the weight of the oil by finding the value 41.3 on scale B and dropping to the
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oil volume curve near the top of the chart; proceed horizontally to the resin weight curve, reading (scale -4)32.7 per cent oil and showing 3.27 pounds of oil in a gallon of varnish. From the composition given we know that there are 1.372 pounds of resin in the gallon of varnish, therefore:
+
3.27 1.37 = 4.64 pounds,, total N. 3.27/4.64 = 70.5 per cent oil 1.37/4.64 = 29.5 per cent resin
v.
From scale B and the percentage resin curve we find that 29.5 per cent resin corresponds to a 30.5-gallon varnish. This method carries composition in the terms used by the stock and accounting departments and is the same problem as example 2. Where commercial production is involved and cost figures are necessary or where the number of components is greater than two, this method is generally preferred.
Example 4 Given a 50-gallon varnish containing 70 per cent nonvolatile for pigmentation, find the volume of thinned varnish equivalent t o 1 gallon of nonvolatile. SOLUTIOX.Follow the 70 per cent nonvolatile line horizontally until it intersects the 50-gallon correction curve, drop perpendicularly and the volume is indicated to be 1.89 gallons. S o w return along the 30-gallon correction curve to 100 per cent nonvolatile (to subtract the thinner), or if preferred follow the 100 per cent nonvolatile line to the 50-gallon correction curve and drop t o the factor 1.22, the volume of nonvolatile contained in 1.89 gallons of 70 per cent varnish. Then 1.89/1.22 = 1.55 gallons, the bulk of thinned varnish required to give 1 gallon of nonvolatile. This is a convenient method where bulking factors and volumes are used in pigmentation calculations,
Example 5 Given a 50-gallon varnish containing 70 per cent nonvolatile, find the weight of thinned varnish equivalent to 1 gallon of nonvolatile. SOLUTION.Follow the 100 per cent nonvolatile line horizontally t o the 50-gallon correction curve and drop perpendicularly to the factor 1.22 gallons. Multiply this factor by 70 per cent, giving 0.854 gallon. Divide 100 by 8.54 t o give 11.7 pounds of diluted varnish, the weight required to give 1 gallon of nonvolatile.
Example 6 Find the specific gravity of a 50-gallon 60 per cent nonvolatile varnish. SOLUTION.Follow the 60 per cent nonvolatile line horizontally to the 50-gallon weight-volume curve, thence down to the weight-volume factor (reading 2.245). This factor, however, is 60 per cent nonvolatile so that 10 pounds of the diluted varnish (not the diluted 10 pounds of nonvolatile as before) bulk (0.60 X 2.245) = 1.347 gallons. Divide 1.200 (the specific gravity of the resin bulking 1gallon per 10 pounds) by 1.347, giving 0.891, the specific gravity of the varnish. If the specific gravity of a varnish containing no volatile is required, simply find the weight-volume factor corresponding t o the varnish length (on the 100 per cent nonvolatile line) and divide 1.200 by this value. If a blend of oil and any varnish is desired, it is a simple matter t o extend the percentage resin line and the (0 per cent B ) left-hand marginal line until they meet (a gallons long = 0 per cent resin) and use this point as the oil component in placing the straight edge across the chart for reading the required percentage blend. RECEIVEDJuly 19, 1935. Presented before the Division of Paint and Varnish Chemistry at the 90th Meeting of the American Chemioal Society, San Francisco, Calif., August 19 to 23, 1935.
