A 300 acre watershed just south of Chicago is currently undeveloped (woods). The soils are primarily clay, the hydraulic length of the watershed is6560 ft, and the average lank slope is 1%. Determine the SCS synthetic UH before development and finish the table below.Solution:from Table 11.7;1. For clay soil with woods cover, CN=2. The soil moisture deficit S =in;3. For a watershed with hydraulic length, time of concentration Te=hrmin;4. Time to peak Tp=min:5. Peak discharge Op=cfsjin;6. Storm duration is AD=hr=min;7. The-min UH is displayed below using Table 11.11;FlowTimeTime RatiosFlow Ratios(min)(cfs)(t/tp)(9/ap)21130.10.240420.310.40.660.60,930.81251210.931.21021461.40.78167730.561.6511880.391.8

The SCS curve number method is generally an important and most helpful method for calculating the…

A 300 acre watershed just south of Chicago is currently undeveloped (woods). The soils are primarily clay, the hydraulic length of the watershed is 6560 ft, and the average lank slope is 1%. Determine the SCS synthetic UH before development and finish the table below. Solution: 1. For clay sol with woods cover, CN = from Table 11.7; 2. The soil moisture deficit S= in: hr= min 3. For a watershed with hydraulic length, time of concentration T 4. Time to peak T,= hr min 5. Peak discharge Q,= cfs/in 6. Storm duration is AD = hr3 min 7. The -min UH is displayed below using Table 11.11; Time Flow Time Ratios Flow Ratios (min) (cfs) (t/tp) (q/ap)

A 300 acre watershed just south of Chicago is currently undeveloped (woods). The soils are primarily clay, the hydraulic length of the watershed is 6560 ft, and the average lank slope is 1%. Determine the SCS synthetic UH before development and finish the table below. Solution: 1. For clay sol with woods cover, CN = from Table 11.7; 2. The soil moisture deficit S= in: hr= min 3. For a watershed with hydraulic length, time of concentration T 4. Time to peak T,= hr min 5. Peak discharge Q,= cfs/in 6. Storm duration is AD = hr3 min 7. The -min UH is displayed below using Table 11.11; Time Flow Time Ratios Flow Ratios (min) (cfs) (t/tp) (q/ap)

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Concept explainers

Sediment transport

Sediment transport or Soil transport is the movement of solid particles, usually due to the combination of gravitational forces acting on the sediment, and the movement of the liquid at which the sediment enters. Soil erosion occurs in natural systems where the particles are sandstones (sand, rocks), mud, or clay; the liquid is air, water, or ice; and the gravitational force that works to move particles to the landing surface. Sediment transportation or soil migration due to the movement of liquids occurs in rivers, lakes, lakes, oceans, and other bodies of water as a result of currents and waves. Transportation is also caused by glaciers as they flow, and to places on earth under the influence of wind. Soil erosion due to gravity alone is possible in most sloping areas, including hills, scarves, cliffs, and the continental shelf — the continental boundary of the slope.

Question

A 300 acre watershed just south of Chicago is currently undeveloped (woods). The soils are primarily clay, the hydraulic length of the watershed is
6560 ft, and the average lank slope is 1%. Determine the SCS synthetic UH before development and finish the table below.
Solution:
from Table 11.7;
1. For clay soil with woods cover, CN=
2. The soil moisture deficit S =
in;
3. For a watershed with hydraulic length, time of concentration Te=
hr
min;
4. Time to peak Tp=
min:
5. Peak discharge Op=
cfsjin;
6. Storm duration is AD=
hr=
min;
7. The
-min UH is displayed below using Table 11.11;
Flow
Time
Time Ratios
Flow Ratios
(min)
(cfs)
(t/tp)
(9/ap)
21
13
0.1
0.2
40
42
0.31
0.4
0.66
0.6
0,93
0.8
125
121
0.93
1.2
102
146
1.4
0.78
167
73
0.56
1.6
51
188
0.39
1.8

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