Dairy Manure and Effluent
as Phosphorus Fertilizer
Phosphorus Fertilizer
Terminology—P to P2O5
In calculating manure or effluent application
rates based on phosphorus, it is extremely important to know if you are working with units of
phosphorus (P) or phosphorus pentoxide (P2O5).
TIAER
Example
Sample LX001 from Table 2 has a
percent P value of 0.0057%. This
equals 0.0131% P2O5 or
0.0057% P x 2.29 = 0.0131% P2O5
Historically, the phosphorus content of most
commercial fertilizers is expressed in terms of
phosphorus pentoxide (P2O5). Generally, crop
phosphorus nutrient recommendations are presented in terms of P2O5 to match the phosphorus
units of commercial fertilizer.
Example
Using sample MX001 from Table 2, the
manure is 0.56% P. This equals 1.28%
P2O5 or
0.56% P x 2.29 = 1.28% P2O5
Crop Phosphorus Requirements—
Percent to Pounds
As crop nutrient requirements are generally
expressed in terms of pounds rather than a percent, the next important step is to convert the percent P2O5 to pounds.
Dairy Manure and Effluent as Phosphorus Fertilizer
In contrast, when you send a manure or effluent
sample off for analysis, most testing laboratories
express the results in terms of percent phosphorus.
Effluent values are often expressed in terms
of P2O5 per 1,000 gallons for application. For
Equation
1 a percent P O of 0.0131%, this
our example
with
2 5
equals 1.1 pounds of P2O5 per 1,000 gallons of
[(0.0131%P2conversion
O5/100) x (62.4lb/ft
effluent assuming
factors3 effluent)
of 62.4 lb of
3
/
(7.48
gal/ft
)]
x
1000
=
(1.1lbP
O
1000
gal
2
5) /
effluentDairy
per cubic
foot
and
7.48 gallons
per
cubic
Manure
and
Effluent
as Phosphorus
Fertilizer
foot or:
Equation
1
0.0131%P
2O5 x 83.4 = 1.1lbP2O5 per 1000 gal
x (62.4lb/ft
effluent)
For solid[(0.0131%P
manure,2Othe
percent
P2O53 is
generally
5/100)
Equation
2gal/ft3)] x 1000 = (1.1lbP
/
(7.48
O
)
gala
converted to represent the number of2 pounds
5 / 1000in
( ( 1.28%P2ORemembering
⁄ ( ton )there
)
5 ) ⁄ 100 ) × ( ( 2000lb ) that
ton of manure.
are 2000
25.6 and
lb of Pusing
ton ofexample
dry manureof 1.28% P2O5,
lb in a =ton,
2O5 in aour
we find0.0131%P
that a ton
of
manure
contains 25.6 lb. To
2O5 x 83.4 = 1.1lbP2O5 per 1000 gal
get this result, multiply the fraction of P2O5 in
Note that the portion after the equals sign is not an equ
manure (1.28% P2O5/100) by 2000 lb or
Equation 2
( ( 1.28%P23O5 ) ⁄ 100 ) × ( ( 2000lb ) ⁄ ( ton ) )
Equation
= 25.6dry
lb ofmatter/
P2O5 in a2lb
tonmoist)
of dry manure
(1.2lb
x 100 = 60% dry matter
or
One further adjustment is needed for solid
To convert the percent P in effluent or manure
100
– 60%
dry
= 40%
moisture
Note
thatyou
thematter
portion
after
the
equals sign
is not an eq
manure
before
are
ready
to
calculate
manure
to percent P2O5 multiply the percent P by 2.29.
The conversion factor 2.29 is based on the atomic application rates based on crop nutrient require3 in a frame:
weights of phosphorus and oxygen in P and P2O5. ments. Equation
Equation
3
(1.2lb dry matter/ 2lb moist) x 100 = 60% dry matter
(1.2lb dry matter/ 2lb moist) x 100 = 60% dry matter
or
3 effluent)
[(0.0131%P
(62.4lb/ftas
Dairy Manure
andx Effluent
Phosphorus Fertilizer
2O5/100)
3
/ (7.48 gal/ft )] x 1000 = (1.1lbP2O5) / 1000 gal
SolidEquation
Manure
Application—
Calculating P Fertilizer
1
Application Rates for Manure
Adjusting
for Moisture Content
0.0131%P 2O5 x 83.4 = 1.1lbP2O5 per 1000
gal
[(0.0131%P2O5/100) x (62.4lb/ft3 effluent)
Most testing
laboratories present the nutrient & Effluent
/ (7.48 gal/ft3)] x 1000 = (1.1lbP2O5) / 1000 gal
analyses for solid manure in terms of 100 percent
Question 1
Equation 2
dry matter. Samples are oven dried prior to analyUsing our example, how much solid
( ( 1.28%P
O5 ) all
⁄ 100moisture.
) × ( ( 2000lb ) ⁄As
( tonmost
))
sis to
drive2off
solid manure
manure would be necessary to meet the
0.0131%P
83.4
1.1lbP
per 1000 gal
= 25.6
lb of P2Oas
in5100
axton
of=dry
manure
is not
applied
percent
dry
a further
25O
2O5 matter,
crop P needs for a yield goal of two - three
step involves adjusting the phosphorus content
cuttings for dry land Coastal?
of the Equation
manure 2for the moisture content of the
Note that the portion after the equals sign is not an equation.
manure when it is applied. A rough estimate of
Answer 1
( ( 1.28%P2O5 ) ⁄ 100 ) × ( ( 2000lb ) ⁄ ( ton ) )
percent dry matter
versus percent moisture can
= 25.6 lbby
of dry manure
2O5 in a ton
100 lb P2O5 per acre required by Coastal
be obtained
a sample
and weighing it,
Equation
3 of Ptaking
(from Table 1) / 15.4 lb P2O5 available
spreading
sample
out and
letting
dry
in the
(1.2lb drythe
matter/
2lb moist)
x 100
= 60% it
dry
matter
sun forNote
a day
or
two,
then
weighing
the
sample
that the portion after the equals sign is not an equation. per ton of manure as is = 6.5 tons of
or
manure as is should be applied per acre
again. Percent dry matter is calculated as the ratio
100 – 60% dry matter = 40% moisture
of the
moist weight to the dry weight. For examEquation
ple if the
moist3 weight is 2 pounds and the dry
Question 2
Equation
3dry
inpounds
amatter/
frame: the
weight
is 1.2
percent
dry
matter
is 60
(1.2lb
2lb moist)
x 100
= 60%
dry matter
Assuming a yield goal of four to six cutpercentoror
tings of Coastal for a field receiving irri100 dry
– 60%
dry matter
= 40%
moisture
(1.2lb
matter/
2lb moist)
x 100
= 60% dry matter
gated effluent, how much effluent using
or
the example above would be needed to
100% – 60% dry matter = 40% moisture
Equation 3 in a frame:
meet the crop P needs?
The pounds of P2O5 in the manure on a dry
Equation
matter
basis
then
adjusted
for xthe
(1.2lb4is
dry
matter/
2lb moist)
100moisture
= 60% dryloss
matter
by the
oven
drying
at
the
laboratory
by
multiplyor
25.6 lb of P2O5 in a ton of dry manure
– 60% dryof
matter
40% moisture
ing it by100%
the fraction
dry =matter
in the manure.
x (60% dry matter / 100)
In our example, the adjusted pounds of P2O5 in a
ton of
moist
manure
lb of Pon
2O5aper
ton=of15.4
manure
moist
basis
is 15.4
or
Equation 4
Equation 4 in a frame
25.6 lb of P2O5 in a ton of dry manure
Answer 2
130 lb P2O5 per acre required by Coastal
(from Table 1) / 1.1 pounds of P2O5
per 1,000 gallons = 118,000 gallons of
effluent applied per acre
x (60% dry matter / 100)
= 15.4 lb of P2O5 per ton of moist manure
Equation 4 in a frame
Note: These examples assume 100 % availability of the phosphorus in the manure for fields that have received manure or effluent in the past.
For a field that is receiving manure for the first time a lower availability value may be applicable. These application rates are also specific to
Draftthe manure and effluent analyses presented as examples. Your application rates will vary depending on your
9 manure and effluent analyses.
Please check with the TSSWCB, TAEX, or NRCS for guidance prior to applying manure or effluent if you are uncertain about the appropriate
nutrient application rates for your operation.
Produced by Anne McFarland, Research Scientist, TIAER at Tarleton State University
in cooperaDraft
9
tion with the Texas State Soil and Water Conservation Board and the United States Environmental
Protection Agency as part of a Section 319 Nonpoint Source Pollution Control Program project.
Fact Sheet 0002