Respuesta :
Solution :
It is given that the soil consists of :
Sand -- 50 %
Silt -- 25 %
Clay -- 25 %
Clay is present in the soil and so it has a very slow infiltration rates when the soil is thoroughly wetted.
a). Therefore the type of soil bis HSG Group D.
b). It is give that :
Industrial building is 25 acres, i.e.
CN = 93
adjusted CN = 93 x 1 = 93
I acre lots is 125 acres, i.e.
CN = 84
Adjusted CN = 84 x 1.07 = 89.88
= 90 (approx.)
60 acres of open or park space, i.e.
CN = 80
Adjusted CN = 80 x 1.14 = 91.2
= 91 (approx.)
225 acres of the meadows
CN = 78
Adjusted CN = 78 x 1.14 = 88.92
= 89 (approx.)
Commercial development of 40 acres, i.e.
CN = 95
Adjusted CN = 95 x 1 = 95
Therefore the curve number is given by :
[tex]$=\frac{(25 \times 93)+(125 \times 90)+(60 \times 91)+(40 \times 95)+(225 \times 78)}{25+125+60+40+225}$[/tex]
[tex]$= 85.02$[/tex]
[tex]$\sim 85$[/tex]
Thus, the curve number for the site for AMC III = 85
Answer:
It is given that the soil consists of :
Sand -- 50 %
Silt -- 25 %
Clay -- 25 %
Clay is present in the soil and so it has a very slow infiltration rates when the soil is thoroughly wetted.
a). Therefore the type of soil bis HSG Group D.
b). It is give that :
Industrial building is 25 acres, i.e.
CN = 93
adjusted CN = 93 x 1 = 93
I acre lots is 125 acres, i.e.
CN = 84
Adjusted CN = 84 x 1.07 = 89.88
= 90 (approx.)
60 acres of open or park space, i.e.
CN = 80
Adjusted CN = 80 x 1.14 = 91.2
= 91 (approx.)
225 acres of the meadows
CN = 78
Adjusted CN = 78 x 1.14 = 88.92
= 89 (approx.)
Commercial development of 40 acres, i.e.
CN = 95
Adjusted CN = 95 x 1 = 95
Therefore the curve number is given by :
Thus, the curve number for the site for AMC III = 85
Explanation:
