INJP-3060 EZ-FLO Fertilizer Injector 3/4" Garden Hose Bib Connector Plastic

Description

INJP-3060: EZ-FLO Fertilizer Injector Replacement 3/4" Female x Male Garden Hose Bib Connector, Plastic
Replacement Hose Bib Connector and a component for EZ-FLO and similar/other Fertilizer Injection Systems.
EASY TO INSTALL - Equipped with male and
female garden hose threads, t
his injector adapter
may be installed on hose bibs, spigots, and
the
female
end of a garden hoses.
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Irrigation-Mart
800 SAY-RAIN
FERTIGATION: Applying fertilizer via Irrigation
An injector takes a fertilizer solution out of a concentrate tank and injects it into irrigation water. To determine the amount of fertilizer being applied, one must know (1) the concentration of
fertilizer
in the tank (
lb of fertilizer per gallon of concentrate
) and (2) the rate or ratio at which the concentrate is being injected into the irrigation water (
gallon of concentrate per gallon of irrigation water
).
For example, if 50 lbs of a 15-10-20 soluble fertilizer is dissolved in water making 30 gallons of concentrate, the concentration of fertilizer is 50 lb/30 gallons
in the concentrate
.  And, the concentration of N
in the concentrate
is 7.5 lb/30 gallons.  Likewise the concentration of phosphate, P
2
0
5
, is 5 lb/30 gal, and that of potassium oxide or potash, K
2
0, is 10 lb/30 gal.  The concentration of P is 2.2 lb/30 gal (43.7% of P
2
0
5
), and that of K is 8.3 lb/30 gal (83.0% of K
2
0).  [The values 43.7% and 83.0% are explained below under Fertilizer Calculations.]
Further, if the rate that this concentrate is being injected into the irrigation water is 1 gallon of concentrate for each 80 gallons of irrigation water, the injection ratio is 1 to 80, or
1/80
.  Thus the concentration of the
fertilizer
in the irrigation water
is (50 lb/30 gal) X (
1/80
) = 50 lb/2400 gal =
1 lb/48 gal
.  And, the concentration of N
in the irrigation water
is (7.5 lb/30 gal) X (
1/80
) = 7.5 lb/2400 gal =
1 lb/320 gal
.  Likewise the concentration of
phosphate
in the irrigation water
is 5 lb/30 gal) X (
1/80
) = 5 lb/2400 gal =
1 lb/480 gal
, and that of potash is 10 lb/2400 gal = 1 lb/240 gal.  The concentration of
P
in the irrigation water
is 2.2 lb/2400 gal =
1 lb/1100 gal
, and that of
K
is 8.3 lb/2400 gal =
1 lb/289 gal
.
Injection ratios vary even for similar injectors due to variations in manufacturing, wear, and operating conditions. Thus, injectors should be tested (at least once per year) to determine their true injection ratio. Additionally, non-proportional type injectors like canister injectors (e.g., EZ-Flo and Rainbow) and siphon injectors (e.g., Hozon, Syphonject, and Mazzei) should be tested using the same operating conditions that they will experience in use.
US Liquid Equivalents
1 lb = 16 oz
1 oz = 2 tbsp
1 cup = 8 oz = 16 tbsp = ½ lb
1 pint = 2 cups = 1 lb
1 quart = 2 pints = 2 lb
1 gal = 4 quarts = 8 pints = 128 oz = 4 lb
1 gal = 3.785411784 liter
1 liter = 0.2641721 gal = 1.056688209 qt = 2.113376418 pt = 33.8140227 oz = 2.1133764188 lb
Fertilizer Calculations
Plant fertilizer nutrients/elements are present in various compounds (e.g., urea, ammonium nitrate, phosphoric acid, calcium phosphate, potassium chloride). The composition as a
percentage by weight
of each of the ‘big 3’ elements present in a fertilizer must be stated on the bag/container.  This is referred to as the fertilizer
guarantee
, which expresses each of elemental N, phosphate, and potash as a
percent by weight
of the fertilizer.
For example, suppose a fertilizer (guarantee) has the numbers 10-5-8. This fertilizer contains 10% (1
st
number) elemental nitrogen, 5% (2
nd
number) available phosphate (P
2
O
5
) and 8% (3
rd
number) water soluble potash (K
2
O). The remainder of the fertilizer material is comprised of other elements and filler. The filler helps to assure accurate/uniform application/spreading of the small amounts of the nutrients to relatively large crop areas. The filler often includes ground limestone, to offset the acid potential of the fertilizer.
In order to make sure that these values are understood, let’s calculate the amount of elemental N, P and K in a 100 pound bag of 10-5-8 fertilizer.
Begin with N, the easier calculation. The 10-5-8 fertilizer is 10% N by weight. Convert 10% to a decimal (0.1) and compute the weight of N in the 100 lb bag of 10-5-8 fertilizer:
100 lb X 0.1 =
10 lb
N
Likewise there is 5 lb of N in a 50 lb bag of 10-5-8, and 20 lb of N in a 50 lb bag of 40-5-8.
Elemental P is more difficult, requiring another step. The guarantee (5%) is expressed as percent by weight of phosphate (P
2
O
5
). We need to find out how much P is in P
2
O
5
.
Atomic weights are: P = 31 and O = 16
P
2
O
5
has two atoms of P, so 2 X 31 = 62
P
2
O
5
has five atoms of O, so 5 X 16 = 80
Atomic weight for P
2
O
5
is [62 + 80] = 142
Therefore, the proportion of P in P
2
O
5
is [62 / 142] = 0.437 =
43.7%
Thus the 100 pound bag of 10-5-8 fertilizer contains 5 lb of P
2
O
5
of which P is 5 lb X 0.437 =
2.18 lb
.
Elemental K requires a step similar to P. The guarantee (8%) is expressed as percent by weight of potash (K
2
O). We need to find out how much K is in K
2
O.
Atomic weights are: K = 39 and O = 16
K
2
O
has two atoms of K, so 2 X 39 = 78
K
2
O
has one atom of O, so 1 X 16 = 16
Atomic weight for K
2
O is [78 + 16] = 94
Therefore, the proportion of K in K
2
O is [78 / 94] = 0.830 =
83.0%
Thus the 100 pound bag of 10-5-8 fertilizer contains 8 lb of K
2
O of which K is 8 lb X 0.830 =
6.64 lb
.
Fertilizer Application Rates for Individual Plants/Trees
How many ounces of a 33-15-15 fertilizer should be applied to a tree/area to supply 2 oz of elemental N?  Algebraically this can be written as:
0.33 X F = 2 oz, where F is the amount of fertilizer needed.  Thus,
F = 2 oz / 0.33 = 6.06 oz
About 6 oz or 12 tablespoons of 33-15-15 fertilizer provides 2 oz N.
Now, if 6.06 oz of 33-15-15 fertilizer is added to a tree, how much elemental P and K are also added?  First, 6.06 oz of 33-15-15 fertilizer contains 15%P
2
O
5
or 0.15 X 6.06 oz = 0.909 oz, and P
2
O
5
contains 43.7% P.  Thus 0.437 X 0.909 oz = 0.397 oz of elemental P is applied when 2 oz of elemental N is applied.  Likewise, the 6.06 oz of 33-15-15 fertilizer contains 15% K
2
O or 0.15 X 6.06 oz = 0.909 oz, and K
2
O
contains 83.0% K.  Thus 0.830 X 0.909 oz = 0.754 oz of elemental K is applied when 2 oz of elemental N is applied.
And, how much of 46-0-0 fertilizer would need to be applied to a plant/area to provide 5 oz of N?  (Note that the amount of P and K applied would be none.)
Answer: F = 5 oz / 0.46 = 10.87 oz or 22 tablespoons = 0.7 lb = 0.7 pint = 1.4 cups = 1/3 L
Availability of Plant Nutrients vs. Soil pH
Most plants/crops like a soil pH near 6.5.  This is because most plant nutrients are readily available to plants at 6.5 pH.