Your boss needs to recommend either the in-situ soil or the borrow material. Although you do not have much background on soil be
havior, what is your recommendation?
In providing your recommendation, you want to consider constructability issues (such as ease of compaction, moisture content). Without doing any additional tests, and based on your experience with soils, what compaction conditions would you suggest to achieve low permeability? What compaction conditions would you recommend to reduce soil expansion (swelling)?
Table 1 Compaction Characteristics of Selected Soils.
Borrow Soil In-Situ Soil
STANDARD MODIFIED STANDARD MODIFIED
PROCTOR PROCTOR PROCTOR PROCTOR
Dry Unit Water Dry Unit Water Dry Unit Water Dry Unit Water
Weight Content Weight Content | Weight Content Weight content
Yd (kN/m³) w(%) Yd (kN/m³) w(%) Yd (kN/m³) w(%) Yd (kN/m³) w(%)
16.68 8.4 17.58 5.7 15.63 12.7 15.72 10.4
18.79 9.7 20.02 7.4 17.72 14.4 18.76 12.2
S3? S3? 20.60 9.3 17.46 16.0 18.64 14.0
18.62 13.4 19.62 11.3 17.15 16.7 18.15 15.0
18.05 15.0 18.93 13.0 16.30 18.2 17.48 16.4
In providing your recommendation, you want to consider constructability issues (such as ease of compaction, moisture content). Without doing any additional tests, and based on your experience with soils, what compaction conditions would you suggest to achieve low permeability? What compaction conditions would you recommend to reduce soil expansion (swelling)?
Table 1 Compaction Characteristics of Selected Soils.
Borrow Soil In-Situ Soil
STANDARD MODIFIED STANDARD MODIFIED
PROCTOR PROCTOR PROCTOR PROCTOR
Dry Unit Water Dry Unit Water Dry Unit Water Dry Unit Water
Weight Content Weight Content | Weight Content Weight content
Yd (kN/m³) w(%) Yd (kN/m³) w(%) Yd (kN/m³) w(%) Yd (kN/m³) w(%)
16.68 8.4 17.58 5.7 15.63 12.7 15.72 10.4
18.79 9.7 20.02 7.4 17.72 14.4 18.76 12.2
S3? S3? 20.60 9.3 17.46 16.0 18.64 14.0
18.62 13.4 19.62 11.3 17.15 16.7 18.15 15.0
18.05 15.0 18.93 13.0 16.30 18.2 17.48 16.4
1 answer:
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Answer:
The curve length (<em>L</em>) will be = 1218 ft
The elevations and stations for PVC and PVI
a. station of PVC = 103 + 91.00
b. station of PVI = 116 + 09.00
c. elevation of PVC = 432.18ft
d. elevation of PVI = 426.09ft
Explanation:
First calculate for the length (<em>L</em>)
To calculate the length, use the formula of "elevation at any point".
where, elevation at any point = 424.5.
and ∴ PVC Elevation = (420 + 0.01L)
Then, calculate for Station of PVC and PVI and elevation of PVC and PVI
Answer:
Initial mass = 5.91 kg
Final mass = 16.8 kg
Heat transfer Q = - 625.9 KJ
Explanation:
Given data
Tank volume V = 1
Entering outside air temperature
Entering outside air pressure = 15 bar
Initial tank pressure = 5 bar
Initial tank temperature = 295 K
Final pressure = 15 bar
Final temperature = 310 K
We know that
P V = m R T
(a). Initial mass is given by
Put all the values in given equation
(b). Final mass is given by
This is the final volume of the tank.
Δ U = Δ q + Δ
Specific internal energy at initial temperature & pressure = 210.5
Specific internal energy at final temperature & pressure = 221.25
Specific enthalpy is 295.17
Q = ( 16.8 × 210.48 - 5.91 × 210.49 )- 295.17 ( 16.8 - 5.91 )
Q = 2292.23 - 3214.4
= - 741.4 KJ
The heat transfer for the tank is given by
= m C
= 20 × 0.385 × ( 310-295 )
= + 115.5 KJ
Total heat transfer Q = +
Q = - 741.4 + 115.5
Q = - 625.9 KJ
This is the heat transfer to the surrounding from the tank.