A well penetrates an unconfined aquifer. Prior to pumping, the water level (head) is 25 meters. After a long period of pumping a
t a constate rate of 0.05 cubic meters per second, the drawdowns at distances of 50 and 150 meters from the well were observed to be three and 1.2 meters respectively. Computer the hydraulic conductivity of the aquifer and the radius of influence of the pumping well.
1 answer:
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Answer:
Option D
Explanation:
Given information
Bulk unit weight of 107.0 lb/cf
Water content of 7.3%,=0.073
Specific gravity of the soil solids is 2.62
Specifications
Dry unit weight is 113 lb/cf
Water content is 6%.
Volume of embankment is 440,000-cy
Borrow material
Embankment
Considering that the volume of embankment is inversely proportional to the dry unit weight
Therefore,
Therefore, volume of borrow material is 498594-cy
(b)
The weight of water in embankment is found by multiplying the moisture content and dry unit weight.
Assuming that all the specifications are achieved, weight of water in embankment=0.06*113=6.78 lb/cf
Since
The embankment requires water of 6.78*27*440000= 80546400 lb
Borrow materials’ water will also be 0.073*99.72041=7.27959 lb/cf
Borrow material requires water of 7.27959*27*498594=97998120 lb
Extra water between borrow material and embankment=97998120 lb-80546400 lb=17451720 lb
1 gallon is approximately hence
That's approximately 4.2 gallons
Answer:
hello some parts of your question is missing attached below is the missing part ( the required fig and table )
answer : The solar collector surface area = 7133 m^2
Explanation:
Given data :
Rate of energy input to the collectors from solar radiation = 0.3 kW/m^2
percentage of solar power absorbed by refrigerant = 60%
Determine the solar collector surface area
The solar collector surface area = 7133 m^2
attached below is a detailed solution of the problem
