Agriculture Reference
In-Depth Information
The final/target EC of the mixture, EC
m
=
8.0
We know,
C
m
=
C
1
×
r
1
+
C
2
×
r
2
(A)
and,
r
1
+
r
2
=1
Or,
r
2
=1
r
1
Putting the values in Eq. (A), 8
−
=
2
×
r
1
+
15(1
−
r
1
)
Or, 13
r
1
=
7
Thus,
r
1
=
0.53
Then,
r
2
=
0.47
Thus the ratio would be 53:47 (Ans.).
(1-r
1
)
=
Example 8.4
The salt concentrations of two solutions are 10 and 50 mg/l. Find the salt con-
centration of mixed water, if the proportions of the solutions are 70 and 30%,
respectively.
Solution
We know,
C
x
=
C
1
×
r
1
+
C
2
×
r
2
Given,
Concentration of 1st solution,
C
1
=
10 mg/l
Proportion of 1st solution,
r
1
=
70%
=
0.70
Concentration of 2nd solution,
C
2
=
50 mg/l
Proportion of 2nd solution,
r
2
=
0.3
Putting the above values, salt concentration of the mixed water,
C
x
30%
=
=
(10
×
0.7)
+
(50
×
0.3)
=
22 mg/l (Ans.)
Example 8.5
A cereal crop having 120 days growing period has the following data from field
condition:
Total ET: 80 cm
Irrigation cycle: 10 days
Irrigation time: 6 h
Average infiltration rate: 1.0 cm/h
EC of drainage water: 1.5 dS/m
Assuming no drainage limitation, calculate leaching fraction and permissible
electrical conductivity of irrigation water.
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