I N D U S T R I A L A N D ENGINEERING CHEMISTRY
724
Vol. 18, No.
i
Geometrical Calculation of Ternary Fertilizer Mixtures Using Triangular Coordinates’ By B j a r n e Colbjornsen STOCKHOLMS SUPERFOSFAT FABRIKS AKTIEBOLAC, STOCKHOLM, SWEDEN
HERE is in America an active campaign to raise the analysis of mixed fertilizers. The manufacture of fixed nitrogen and of phosphoric acid by the volatilization method promises to provide the means for making many high-analysis fertilizer materials, which may have wide application in the manufacture of high-analysis mixtures. It is common for the raw materials in the fertilizer industry to contain two fertilizing elements, and in the near future they may contain all three fundamental fertilizing elements. Monoammonium phosphate ( N H ~ H z P O diammonium ~), phosphate [(NH4)2HP04], monopotassium phosphate (KHzPOI), and sodium ammonium phosphate (NaNH4HP04) are examples of raw materials containing two fertilizing elements; )Z] potassium ammonium phosphate H [ K N H ~ ( H ~ P O ~and cottonseed meal contain all three elements. The fertilizer industry will therefore soon face the problem of manufacturing its fertilizer mixtures from high-analysis fertilizer materials, containing one, two, or even three fertilizing elements and will have to work out new fertilizer formulas and ratios. Merz and Ross2 have recently handled the problem of characterizing ternary fertilizer mixtures by the use of triangular coordinates. They point out that in substituting for the many low-grade raw materials now employed, highanalysis synthetic chemicals containing two or more fertilizing constituents the number of possible combinations is limited, but by successively combining these materials in pairs with ammonia, phosphoric acid, and potash salts any fertilizer ratio may be obtained by varying the materials. These authors have also demonstrated the general algebraic formulas for the computation of ternary fertilizer mixtures from materials containing two fertilizer materials. The algebraic method of computation is, however, rather laborious and time-wasting and affords many chances for miscalculations, especially if the raw materials contain three fertilizing elements.
T
These formulas are considerably simplified if the materials contain only one or two fertilizing elements. In order to ascertain the true quantities of the various raw materials, it will be necessary to divide the calculated percentages by the sum of the percentages of the fertilizing elements and multiply by 100. Suppose we wish to manufacture a fertilizer mixture carrying the fertilizing elements “3, P205, and K20 in the ratio 20:30: 50 from monoammonium phosphate, monopotassium phosphate, and potassium nitrate. Table I-Percentage Composition of Fertilizer Raw Material -Total Material-Fertilizing ElementsRaw Material ”8 PzOp RzO “8 PzOs Kz0 NH4HzPOd 14.8 61.7 0.0 19.3 80.7 0.0
KHzPOd KNOs
0.0 16.9
52.2 0.0
34.6 46.6
0.0 26.6
60.1 0.0
39.9 73.4
In this case accordingly: A = 20, B = 30, C = a = 19.3, b = 80.7, c = a’ = 0, b‘ = 6 0 . 1 , ~ ’= a” = 26.6, b” = 0, c “ =
50% 0% 39.970 73.475
I t is required to show X, Y , and Z in percentages. If the above values are inserted in Formulas 1, 2, and 3, we will get the following equations:
+ +
x = 100
+
20(60.1-0) O(0-30) 26.6(30-60.1) 19.3(60.1-0) O(0-80.7) 26.6(80.7-60.1) lOO(20-26 6)-23.5(19.3-26.6) = 18,4 Y = 0-26.6 (100 X 20) - (18.4 X 0) - (23.5 X 19.3) = 58,1 Z = 26.6
+
- 23,j. -
And for checking only:
X fY
+ 2 = 23.5 4- 18.4 + 58.1 = 100.0
Algebraic Calculation
The general algebraic formulas for the computation of fertilizer mixtures from materials containing all three fertilizing elements are as follows:
x + Y+2 = where
X
100
(4)
= weight of raw material, which contains a per cent c per cent KzO NHa, b per cent PzOE,
Y = weight of raw material, which contains a’ per cent c’ per cent Kz0 NHa, b’ per cent P~OS, 2 = weight of raw material, which contains a’’ per cent NHa, b“ per cent P 2 0 6 , c” per cent K20
and t h e percentages of the fertilizing elements are calculated on a 100 per cent basis. A , B , and C are the desired percentages of NH,, P205, and K?O in the finished mixture, also calculated on a 100 per cent basis. 1 2
Received February 4, 1926. U. S. Depl. Agr., BuiE. 1280 (1924).
The true quantities of the respective raw materials are obtained by dividing the above values by the sum of the percentages of the fertilizing constituents and multiplying by 100. In the foregoing example, which is calculated on the basis of chemically pure materials, the values 23.5, 18.4, and 58.1 should be divided by 76.5, 86.8, and 63.5, respectively, and multiplied by 100. The true values are therefore 30.7, 21.2, and 91.6, respectively, or together 143.5, and calculated on a 100 per cent basis 21.4, 14.7, and 63.9 per cent, respectively. The most concentrated mixture it is possible to manufacture from the raw materials in question may be calculated by dividing the ratio 20:30:50 by 143.5 and multiplying by 100. The ratio will be 14:21:35. If still lower percentages are desired, filler may be added as usual. The algebraic method of calculating fertilizer mixtures is rather laborious, as the foregoing example shows. Still more troublesome are the calculating operations if the raw materials contain all three fundamental fertilizing elements, as when such raw materials as potassium ammonium phosphate, tankage, and cottonseed meal are employed, or when remanufacturing older complete mixtures.
I N D U S T R I A L dh’D ESGI,VEERING CHEMISTRY
July, 1926
Graphical Treatment with Triangular Coordinates
If one or more of the raw materials contains two or three fertilizing elements, a geometrical method of calculation may be employed to a d ~ a n t a g e . ~ Schreiner and Skinner4 have used the triangular diagram A
725
versals: If the angular transversals of a triangle intersect one another in one single point, the sum of the ratios of the foot pieces to their transversals will equal 1. Thus
GD > x
+ GE + ZGF = 1 YE F A
100% P?Os
100% Kv0
100% NHa
100% s“
(Figure 3)
(5)
100% PnOs
F
100% KzO.
Figure 1
Figure 2
in their experimental work on the effects upon plants of different ratios of ammonia, phosphoric acid, and potash. Merz and Ross2 have used triangular coordinates for characterizing ternary fertilizer mixtures. In both cases the equiangular form was used (Figures 1 and 2 ) . The fundamental theorem in this case is that the sum of the perpendiculars from a point inside an equiangular tric = h, angle t o the sides is equal to the median line (a 6 Figure I ) . The corners of the triangle represent 100 per cent of animonia. phosphoric acid, and potash. Each point within the triangle mill represent a definite mixture of all three constituents in such proportion as t o total 100 per cent. Each side of the triangle represents binary mixtures of the two constituents represented by the adjacent corners.2 It is easily shown, however, that the equiangular form may be replaced by the common rect’angular coordinate diagram. The equiangular form offers difficulty to many people in reading off the percentages arid in marking the points according to the coordinates. Furthermore, the triangular coordinate diagrams are not so easily obtainable as the common rectangular coordinate sheets. The scale of the third component, the median line from B to AC, however, is in this -: 1 = 0.7071, but this is of case abbreviated at the ratio .\/a
Application of Transversal Theorem to Mixing Problems
+ +
2
no calculatory or constructive importance. 4 s was the case in the equiangular form, all binary mixtures are situated on the side lines (Figure 5 ) , all ternary mixtures inside the triangle, and any line drawn between two points in the triangle represents all possible mixtures between the mixtures represented by the two points. The fundamental t>heorem used in solving problems of mixing binary or ternary mixtures is the theorem of trans:In the spring of 1922, in connection with some work at Massachusetts Institute of Technology. the author considered the possibility of using triangular coordinates for characterizing ternary fertilizer mixtures. At the same time a geometrical method of calculation was touched upon. but in the autumn of 1924 the method was finally worked out, after having studied the very deserving work by Philip: “Algebraical, Graphical, and Geometrical Calculation of Ternary Mixtures,” Ingenidrsvetenskaps Akademiens Handlingar, No. 20 (1923), Stockholm. ‘J. A m . Soc. Agron., 10, 225 (1918).
The relations between the fractions of the theorem and the weights of the mixtures in the problem may be deduced as follo\vs:3 X , Y , and Z are the weights of the ternary mixtures to be made over into the ternary mixture G. Equation 5 of the theorem multiplied by G will take the form: GD GE GF GX - +D G - + G - = G YE ZF
X = G XD kg. GE and in the same way Y = G - kg. = G YE GF and Z = G - kg. ZF Hence X + Y + Z = G GD 4 - G -G+E G XD YE
GF = G ZF
(7)
If t h e t h r e e r a w X materials X,Y , and %, each c o i i t a i n i n g a l l three fundamental fertilizing e l e m e n t s i n known quantities, should be combined to make a mixture G of a definite percentage, the method of operation is : D e n o t e the points X , Y , and Z (Figure 4) according to percentages; construct interD z Figure 3 section points D, E , and F of transversals with sides; measure transversals and foot pieces.
x
= 100
g;
GE Y = 100 -; YE
z
= 100
GF ZF
-
IhTD USTRIAL A N D ENGINEERING CHEMISTRY
726
%, being the difference from 100, is calculated only from the point of control. Returning to the ultimate mixing problem, the geometric solution is as follows (Figure 5): The points X , Y , and Z represent monoammonium phosphate, monopotassium phosphate, and potassium nitrate, respectively, calculated on a 100 per cent basis. The point G represents a mixture containing the three elements "3, P~OF,,and KzO in the ratio 20:30:50. The point Gis located inside the triangle X Y Z , and the mixture represented by this point may therefore be manufactured from the raw materials in question. A
100% PnOs
Vol. 18, S o . 7
be manufactured from either monoammonium phosphate, potassium phosphate, and cottonseed meal, or from monoammonium phosphate, potassium nitrate, and cottonseed meal. In this special case, however, all four raw materials should be used, and monoammonium phosphate and cottonseed meal in a mixture, represented by the point X u , located on the straight line between the points X and U. The point G1 is located inside the triangle X,YZ, and the mixture represented by this point may therefore be manufactured from the raw materials represented by the three points X u , Y , and %. A 100% PnOs
A 100% PnOa
100%Kz0 100% NHs
100% NHs Figure 4
Z Figure 5
The intersection points D, E, and F of transversals with sides are constructed, the transversals and foot pieces are measured, and the values found are inserted in Equation 6; G is considered as 100. GD - = 100 21.9 - = 23.5 XD 93.2 7.0 Y = 100 GE = 100 - = 18.4 YE: 38.0 z = 100 F-zGF = 100 53.0 - = 58.1
Draw the transversals through the point GI, and measure them and the foot pieces.
X = 100
X
+ Y 4-Z
91.1
= 23.5
+ 18.4 + 58.1 = 100.0
The operations consist in denoting four points in a triangular diagram, drawing the transversals, and measuring the transversals and the foot pieces. The calculations are extremely simple, and are reduced to one fraction for each value. Combination of More than Three Raw Materials
If one wishes to combine more than three raw materials, the geometric method of calculation may further simplify the work. Assume that a mixture should be manufactured which contains the three fundamental fertilizing elements in the ratio 40:40:20. This mixture is represented in Figures 5 and 6 by the point GI. This point is situated outside the triangle X Y Z , the corners of which represent monoammonium phosphate, monopotassium phosphate, and potassium nitrate. The mixture in question cannot, therefore, be manufactured from these three materials alone. The point U (Figure 6) represents cottonseed meal, containing 7.5 per cent "3, 2.8 per cent PzOS, and 1.7 per cent K20,or calculated on the basis of 100 per cent plant food, 62.5, 23.3, and 14.2 per cent, respectively. The point G1 is located inside the triangles X Y U and X Z U , and the mixture in question may therefore
100% Kz0
Figure 6
Y= Z
GIE 5.5 = 16.3 YE 33.8 GIF 5.85 100 - = 100 - = 8.1 ZF 72.25 100 - = 100
X and U are calculated according to the lever rule in a similar way:
X
=
xu XU
17 1 75.6 = - 75.6 = 21.8 59.3
and consequently:
U
= 75.6-21.8
= 53.8
In order to obtain the true amounts of the different raw materials, the above values should be divided by 0.765, 0.868, 0.635, and 0.12, respectively. The true amounts are accordingly 28.5, 18.8, 12.8, and 449.2, respectively, and calculated on a 100 per cent basis, 5.6, 3.7, 2.5, and 88.2 per cent, respectively. The most concentrated mixture ratio to be obtained from the raw materials in question is: 100
28.5
+ 18.8 + 12.8 + 449.2 40 : 40 :20 = 7.9
: 7.9 : 3.95
Conclusions
As may be readily seen, the geometrical method of calculation is very simple and rapid. The algebraical solution requires a considerably longer time and offers opportunities for miscalculations and other difficulties. It is hoped that the geometrical method may be of some value to the fertilizer industry, and that its use may be extended to other industries as well.
