Agriculture Reference
In-Depth Information
of a solution is its resistance to pH change when acid or alkali is added. An ex-
ample of pH buffering capacity is given in section 4.6.3.3.
Examples of How to Calculate
N
min
.
for Different Organic N Sources
Appendix 4
The instantaneous rate of net mineralization of soil organic N is given by
dN
kN
(A4.1)
dt
The amount of mineral N released over a period of time is calculated using
the integral form of equation A4.1, that is
N
t
N
o
exp(
kt
) (A4.2)
where
N
o
and
N
t
are the amounts of organic N in a defined volume of soil at time
zero and time
t
, respectively. The amount of
N
min
. produced over the time in-
terval is therefore given by
N
min.
(
N
o
N
t
) (A4.3)
N
o
(1
exp(
kt
)) (A4.4)
This equation should be used if
N
min
. is to be calculated for a relatively short
period, up to 3-4 months. An example follows. Suppose that for a green manure
crop, the decay coefficient
k
is 1.0 yr
1
and 1.5 t dry matter/ha is incorporated
(containing 40% C and a C:N ratio of 20). From this, we calculate that the value
of
N
o
for the green manure is
(0.4
1500/20)
30 kg N/ha (A4.5)
From equation A4.4, we calculate that the value of
N
min
. produced in 3
months is
N
min.
0.22
N
o
(A4.6)
6.6 kg N/ha (A4.7)
In other words, in 3 months, about 22% of the green manure N should min-
eralize.
For longer time intervals, when a large amount of organic N is involved,
N
min
.
can be calculated using an approximate solution of equation A4.1, which does not
involve an exponential function, that is,
N
min.
N
kN
o
t
(A4.8)
Consider the rate of mineralization of soil organic matter (
SOM
) in a vine-
yard soil over 1 year. The value of
N
o
is large, but the decay coefficient is much
smaller than that of a green manure. Suppose that for
SOM
,
k
0.005 yr
1
and
the amount of soil organic N is 2000 t/ha. From equation A4.8, over 1 year, we
have
N
min.
N
min.
N
min
.
0.005
2000
(A4.9)
N
min
.
10 kg N/ha
(A4.10)
