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Figure 11-Tennessee Figure 12-Louisiana Figure 13-Texas Fertilizer Ratios Adopted w i t h o u t Use of Triangle h i n c i p l e at Recent S t a t e Conferences
In Figures 11, 12, and 13 are given the locations of the fertilizer ratios adopted in recent conferences called by the states of Tennessee, Louisiana, and Texas, respectively. definite They also indicate the desirability of having a scheme of selecting these ratios such as is permitted by use of the principle suggested. The triangle scheme has merit. There may be objections to it that may make it necessary to set the scheme aside. The chief reason for presenting the subject a t this time is to call forth all of the possible objections to it in order to determine to what extent these objections are well founded
or what modification could be made to the scheme to overcome them. Table IV-Fertilizer
TEXAS 2-4-2 2-6-2 3-10-3 4-10-7 0-5-2 34-5 2-10-2 2-6-o
2-4-3 2-5-1 0-6-2 6-10-7 1-6-1 3-10-8 3-8-3 2-8-2
..
Ratios Adopted by Individual Statesa LOUISIANA TENNESSEE 2-4-2 2-4-3 2-4-2 2-4-3 2-6-2 2-5-1 2-6-2 2-5-2 3-10-3 0-6-2 3-10-3 0-6-2 4-10-7 6-10-7 4-1&6 4-11-5 0-5-2 4-6-2 1-4-5 1-6-1 3-8-10 3-10-5 3-8-6 3-9-4 ... 5-10-3 3-3-3 2-9-5
5-5-0
. ..
...
...
...
4-6-2 . 0-7-5 . .. ... a Arranged in such manner as to show the similarity of the lists.
Computation of No-Filler Fertilizer Mixtures’ A. B. Beaumont and Harold R. Knudsen DEPARTMENT OF AGRONOMY, MASS.4CHUSETTS AGRICULTURAL COLLEGE, AMHERST, MASS.
ITH the development of high-analysis mixed fertilizers has come an increase in no-filler mixtures. The recent rapid increase in the number of concentrated fertilizer materials has made possible the compounding of grades of complete fertilizers double and triple the strength standard five years ago. These materials, together with a wider range of low- and medium-analysis organic and inorganic materials, make it easy to select the proper ingredients for no-filler mixtures of almost any desired grade. The purposes of this paper are to show how certain types of no-filler formulas for complete fertilizers may be computed and to indicate the possibilities and limitations of mixtures from different carriers. Either of the triangular systems proposed for calculation of ternary fertilizer mixtures is useful in ascertaining the ratios possible from any lot of materials. Merz and Ross2 have proposed the equilateral triangle, and Colbjornsen3has shown how the right isosceles triangle may be used for the same purpose with great facility. These authors, however, do not discuss the computation of no-filler mixtures. Figures 1 and 2 illustrate the use of the equilateral and right isosceles triangles in the determination of ratios. The vertices A , B, and C represent materials carrying nitrogen (N), phosphoric acid (P,Ob), and potash (&O), respectively. The designation 100 per cent simply meavs that all the plant food carried, or considered, in any particular material is of one 1 Presented by A. B. Beaumont as a part of the Symposium on “Concentrated Fertilizers and Fertilizer Materials” before the Division of Fertilizer Chemistry at the 76th Meeting of the American Chemical Society, Swampscott, Mass., September 10 to 14, 1928. 2 Merz and Ross, U. S. Dept. Agr., Bull. 1280 (1924). * Colbj6rnsen, IND.ENG.CHEM..18, 724 (1926).
kind only-e. g., nitrogen in sodium nitrate, phosphoric acid in superphosphate, or potash in muriate of potash. Materials containing two nutrients-e. g., ammonium phosphateare represented on the sides of triangles and materials or mixtures containing three nutrients by points within the triangle-e. g., a 4-10-6 or any other grade built on a 2:5:3 ratio is represented by the point G. Note-As used in this paper the figures of a fertilizer grade represent and potash (KzO),repercentage of nitrogen (N), phosphoric acid (PsOI), spectively. Also, following trade practice, dashes are used in grade designations, and should not be confused with minus signs.
No-filler mixing problems may be solved by using simultaneous algebraic equations wholly or partially. Following is a presentation of types of problems and suggested methods of solution. Type 1, the simplest, may be solved by simple arithmetical computation as readily as by algebra. With other types arithmetical solutions must be by trial and error and therefore very tedious. Type 1-Mixtures
Composed of Three Single-Element Components
By single-element components are meant such materials as sodium nitrate, superphosphate (acid phosphate), and muriate of potash. From such carriers a mixture of any ratio of nitrogen (N), phosphoric acid (PZOS), and potash (K,O) is possible. Any grade, up to a certain maximum, can be built on any ratio, with filler, but there is only one no-filler grade possible for any ratio, and this may or may not be integral. The maximum grade, in this case, is the no-filler grade. Question. What no-filler grade may be mixed on the 2: 5: 3 ratio, using ammonium sulfate (20.5 per cent N),acid phos-
INDUSTRIAL A N D ENGINEERING CHEMISTRY
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Vol. 21, No. 4
+
0.155X 0.46 Y = 160 (pounds of nitrogen required) phate (16 per cent PzO~), and muriate of potash (50 per cent KaO)? Whence X = 427.4 pounds sodium nitrate Y = 203.8 pounds urea Arithmetic calculation in the customary way shows that the weights of the components, ammonium sulfate, superIn general, for single-component mixtures of four or more phosphate, and muriate of potash, needed to make a ton of a 2-5-3 mixture amounts to 195.12, 625.0, and 120.0 components, but no more than two for any one nutrient, where pounds, respectively. Multiplication of the ratio 2:5:3 by hi-j = grade 2000 A I and At = components carrying nitrogen shows the grade to be 4.25-10.64-6.38. X = pounds component A1 required for 2000 pounds mixed 195.12 + 625 + 120 fertilizer I n general, when X, Y,and 2 = pounds of three singleY = pounds component A2 required for 2000 pounds mixed element components needed to make 2000 pounds of an fertilizer 2000 a1 = per cent plant food in A1 A-B-C grade, and AI-BI-CI = no-filler grade, a2 = per cent plant food in A2 X+Y+Z B = pounds of other components required (A-B-C) =AI-B1-C1. Type 2-Mixtures
of Four or More Single-Element Components
Then, X and
+u Y~ X=+ 2000 -B azY = 2000 h
Whence X =
2000 h
- 2000 a2 + UZB
a1 - a2 CASE1-From the preceding solutions it has been proved Y = 2000 - B - X that a 4-10-6 no-filler grade is impossible with the materials used. Question. W h a t additional material is needed to T h e recent addition of a number of concentrated m a k e a no-filler 4-10-6 fertilizer materials t o a constantly growing list of lowgrade? a n d medium-analysis organic a n d inorganic materials Since the materials used h a s made i t easy t o select t h e proper ingredients for are slightIy too concentrated no-filler mixtures of almost a n y grade. T h e triangular as proved by the above solusystems proposed by Merz a n d Ross a n d more recently tions, it appears that an adby Colbjornsen are useful in ascertaining t h e ratio ditional bulky nitrogenous possibilities of m a n y materials, b u t neither is suffim a t e r i a l like nitrogenous cient for computing no-filler mixtures. In this paper tankage would make a nothere is proposed a method for the computation of nofiller mixture possible. filler mixtures based o n t h e use of simultaneous alge-
Type 3-Mixtures of One Double-Element Compon e n t a n d One Single-Elem e n t Component
This combination allows a very restricted number of ratios. For example, take commercial a m m o n i u m phosphate carrying 11 per cent N and 48 per cent PzO5 or a plant-food distribution of 18.6 per cent and 81.40 per cent. I n Figure 3 this material is represented by the braic equations containing two or more unknowns. point D and all mixtures of Then, let X = pounds amVarious types of mixtures a r e discussed a n d methods it with muriate of potash lie monium sulfate (20.5 per cent for solving problems related t o t h e m suggested. along the line DC and have N) Specific problems a r e presented a n d solved. T h e Y = pounds nitrogenous the same ratio of N:PzOs as tankage (7.4 per cent N) limitations a n d adaptations of certain classes of main the unmixed ammonium From Type 1, solution 1, it terials a r e pointed out. In general, single-element phosphate. will be obvious that components widen t h e ratio possibilities, and double1250 pounds superphosQuestion1. Can a 4-10-6 element components increase t h e no-filler possibilities. phate and no-filler grade be made from 240 pounds muriate of ammonium phosphate and potash are needed muriate of Dotash? 1490 pounds total This grade has a 2:5:3 or 20: 50:30 ratio. Since all grades of fertilizer made from materials in question must have an Then 2000 - 1490 = 510 = X Y (1) and 0.205X 4- 0.074Y = 80 (Dounds N in 1 ton 4-10-6) (2) N:PZOs ratio of 18.6:81.4, the 4-10-6 is impossible, with or Solving (1)'and (2) as simdianeous equations without filler; for in Figure 3 the point G, representing the X = 322.6 4-10-6, does not fall on the line CD. Y = 187.4 Question 2. What no-filler grade with 4 per cent nitrogen CASE 2-What single-element components may be used t o can be made from ammonium phosphate and muriate of potash? mix a no-filler 8-20-12? By an algebraic process similar to that of solution 2 of This calls for 160 pounds nitrogen, 400 pounds PZOs,and 240 pounds KzO. From the preceding solutions i t is evident type 1, the type equations for this problem are b u n d to be: that low- and medium-analysis materials such as have been z = c - - CX ( 2 ) y = -bX used cannot be used in this case. If superphosphate is used a (1) as the carrier of phosphoric acid, 2500 pounds of that alone in which X = per cent N, Y = per cent Pl0s,Z = per cent KzO would be needed. Therefore, some concentrated materials in mixed fertilizer. a = per cent N, b = per cent PtOa, c = per cent must be used. Try sodium nitrate (15.5 per cent N), urea K 2 0 in components. InthiscaseX = 4, a = 11, b = 48, c = 50 (46.0 per cent N), a triple superphosphate (45per cent PzOS), Therefore, substituting in (1) and (2): muriate of potash (50 per cent K?O). Y = 17.44 Z = 31.8
+
Let X = pounds of sodium nitrate required Y = pounds urea required (g)2000
=
888.8 pounds triple superphosphate required
2000 = 480.0 pounds muriate of potash required 1368.8 pounds total 2000 - 1368.8 = 631.2 X Y = 631.2
()'
+
The no-filler grade is therefore 4-17.44-31.8. Type 4-Mixtures of One Double-Element a n d Two o r More $ingle-Element Components
These materials allow a degree of latitude for no-filler mixtures not hitherto shown. For example, consider AmmoPhos (11 per cent N, 48 per cent Pzos),muriate of potash, and nitrate of soda. The ratio possibilities are included within the triangle ADC, Figure 3.
April, 1929
INDUSTRIAL AND ENGINEERING CHEMISTRY
If to these materials there be added a third single-element material as superphosphate (acid phosphate), in effect the triangle CDB is added, and the entire triangle ABC includes the possible ratios, which are all ratios. Type 5-Mixtures of Two or More Double-Element Components and One or More Single-Element Components
The possibilities with this combination depend on the two elements present in the double-element components. CASE1-If there are two nitrogen-phosphorus compounds, say 11-48 and 16.5-20 Ammo-Phos and muriate of potash, then the triangle DEC, Figure 3, will include all possible
s loo%&s
A
ratios among which is the 4-10-6 (point G). I n Figure 3, the point D represents the 11-48 grade (18.6 per cent N and 81.4 per cent P205),and E , the 16.5-20 grade (45.2 per cent Tu’, 54.8 per cent P205); DC represents all possible combinations of muriate of potash (or any other single-element potash carrier) with the 11-48 grade, EC with the 16.5-20 grade, and the area DEC all possible combinations of the two grades of ammonium phosphate and any other single-element potash carrier. CASE 2-If potassium nitrate, 1 1 4 8 Ammo-Phos, and sodium nitrate be used, the area in which possible ratios may be found is ADE, Figure 4. The commercial potassium nitrate considered carries 14 per cent N and 44 per cent KzO, thus having a nutrient distribution of 24.1 per cent N and 75.9 per cent K20. I t is therefore represented as point E of Figure 4. The 11-48 grade of Ammo-Phos falls a t point D and muriate of potash a t C. The triangle DEC therefore includes all the ratios possible by mixing Ammo-Phos (11 per cent N, 48 per cent P206), potassium nitrate, and muriate of potash. By inspection it may be seen that the 4-10-6 (2:5:3 ratio) grade, represented by point G, falls within the triangle DEC and is therefore possible with the materials t hand. Further, it may be seen that 4-6-10 (2:3:5 ratio), point GI, also falls within DEC, and is possible, b u t a 6-10-4 (3:5:2), point G2, falls without DEC and is therefore not possible I I I I I I I )oo$i>Cwith the mate1m%N Figure2 rials considered.
387
Type 6-Mixtures of Three or More Double-Element Components
Monoammonium phosphate, potassium nitrate, and monopotassium phosphate are represented by the points D, E , and F of Figure 5, and possible ratios fall within the triangle DEF. Cases and examples of possibilities might be multiplied almost indefinitely. For example, if to the three components just considered, a fourth, the 16.5-20 grade of Ammo-Phos, be added, the quadrilateral DHEF, Figure 5 , includes the possible ratios. Three-element components such as cottonseed meal, potassium-ammonium phosphate, nitrophoska, etc., might also be considered here. Determination of Possible Ratios
In making use of the triangles for determining ratios possible from certain components, it should be distinctly remembered that the triangles give no information as to what no-filler grades are possible. That can be obtained only by a mathematical process. In general, single-element components widen the ratio possibilities and double-element compounds widen the nofiller possibilities. The two kinds of compounds work together to make all no-filler ratios possible.
The grade possibilities involve still different considerations. While all ratios may be possible with certain materials as indicated, it does not follow that all grades are likewise possible. I n general, high-analysis grades call for a predominance of high-analysis materials in the mixture and medium- and low-analysis grades for corresponding materials. No-filler problems coming under Types 4, 5, and 6 may be solved by the use of simultaneous linear equations involving two or more unknowns. Some examples are given to illustrate the principle and method involved in working problems under Types 4, 5, and 6. Let the problem be to mix a no-filler 8-20-12 (2:5:3 ratio). This being a rather high-analysis grade, it is apparent that some of the components to be used must be rather concentrated. By inspection of Figure 3 it becomes evident that the 2:5:3 ratio is possible from sodium nitrate (15.5 per cent N), Ammo-Phos (11 per cent N, 48 per cent P206),and muriate of potash (50 per cent K,O), the point G, which represents the 2 :5:3 ratio, falling within the triangle DCA. Example 1. Let
X Y
pounds sodium nitrate (15.5 per cent N) pounds Ammo-Phos (11 per cent N, 48 per cent P105) In 2000 pounds 8-20-12 mixture there must be: = =
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160 pounds N, 400 pounds PzOS,240 pounds K 2 0 240 t 0.50 = 480 pounds muriate of potash 2000 - 480 = 1520 pounds of other components X Y = 1520 0.155X 0.1 1Y = 160eounds nitrogen Solving, 0.045X = -7.20
+ +
Therefore, fmding a positive quantity equaling a negative quantity, one knows that it is impossible to make a no-filler 8-20-12 from the components used.
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nitrogen, one can either arbitrarily decide on a given amount of nitrogen to be so derived, say 10 pounds per ton of mixture, and treat the urea in the same way that muriate of potash was treated in the calculation; or one can decide that the ureal nitrogen shall bear a certain ratio to the nitric nitrogen, to the ammonic nitrogen, to that from tankage, or to the whole amount of nitrogen. The following general equation may be used for working no-filler problems in which no more than three unknowns are employed:
X
A required to mix E pounds of h-i-j fertilizer Y = pounds component B required to mix E pounds of h i - j fertilizer Z = pounds component C required to mix E pounds of h-i-j fertilizer U = pounds other component (or components) containing only one nutrient required E = pounds total mixture And a and a1 = per cent N and PZOSin component A b and bl = per cent N and PZOSin component B c and c1 = per cent N and PSOSin component C d = per cent KzO in component U Then: Let
= pounds component
Material Lbs.
X
Y Z U
Example 2:- With the addition of superphosphate (16 per to the three components just considered, the 2:5:3 cent PzOS) ratio is still possible, and the no-filler possibilities are greater. Trying this combination as above, we fmd that 2000 pounds of no-filler 8-20-12 can be made with 505.1 pounds sodium nitrate, 742.4 pounds Ammo-Phos, 272.5 pounds superphosphate, and 480 pounds muriate of potash. Example 3. The 8-20-12 mixture of sodium nitrat; Ammo-Phos, acid phosphate, and muriate, being all inorganic, may be unsatisfactory from a physical standpoint. Therefore, let us try another combination containing a low-analysis organic component such as tankage. The 2:5:3 ratio can be obtained if to the three components first tried, tankage (9 per cent N, 5 per cent available P203 be added, and the addition of this comparatively low-analysis component will increase the no-filler possibilities.
N Lbs. aX bY
P206
KxO
Lbs. alX bi Y
Lbs.
..
ClZ
cz
..
... ... ...
dU
+ + + + + + abl + alc + bcl - alb - acl - blc
( E - U ) (bcl - cbl) E(ci blh - bi - hct) ab1 alc bcl - alb - acl - blc y = -( E - U ) (alc - ac,) E(c& ai - alh - ci)
X =
Z = E - X - Y - U
B
Let X = pounds sodium nitrate (15.5 per cent N) Y = pounds tankage (9 per cent N, 5 per cent PzOS) Z = pounds Ammo-Phos (11 per cent N, 48 per cent P2OS) In 2000 pounds of 8-20-12 mixture there must be: 160 pounds N, 400 pounds P~OS, 240 pounds K20 240 + 0.50 = 480 pounds muriate of potash 2000 - 480 = 1520 pounds of other components Then, X Y 2 = 1520 0.155X 0.09Y 0.112 = 160 pounds N 0.05Y 0.482 = 400 pounds PZOS Solving these simultaneous equations, we find X = 120.8 Y = 631.7 2 = 767.5 For proof:
+ + + ++
COMPONENTSCOMPOSITION N Per cent Lbs. Lbs. 120.8 15.5 18.7 X = lbs. sodium nitrate 631.7 9.0-5.0 56.9 Y = lbs. tankage 767 5 11.0-48.0 84.4 Z = lbs. Ammo-Phos 480.0 50.0 ... Muriate of potash
Total
2000.0
160.0
P2Os
R20
Lhs.
Lbs.
31'6 368.4
::: ...
...
240
400.0
240
Example 4. There are many possible variations of this problem. For example, urea may be substituted, entirely or in part, for the sodium nitrate. If it is desired to displace all the nitrate nitrogen by that from urea, simply substitute 56 per cent for 15.5 per cent in the above calculations. If i t is desired to substitute urea for only part of the nitrate
I n case of Example 3 : a1 = 0, b = 0.09, bl = 0.05, c = 0.11, c1 = 0.48, d = 0.50, h = 0.08, i = 0.20, j = 0.12, E = 2000 Substituting these values in the general equations, we have: (2000) = 480 E - U = 1520 U = (''l2) 0.50 X = 120.7 Y = 631.7 2 = 2000 - 120.7 - 631.7 - 480 = 767.6
a = 0.155,
In using this method, if a positive number is found equal to a negative quantity, it is proof that the no-filler mixture desired cannot be made from the components used.
