Answer:
t=14ns
Explanation:
We make the relation between the specific access time and the memory percentage in each level, so
Average Access Time is 14 nsec.
2. A well of 0.1 m radius is installed in the aquifer of the preceding exercise and is pumped at a rate averaging 80 liter/min.
a. Estimate drawdown in this well if the radius of influence is 100 ft. b. Estimate the hydraulic gradient in the aquifer at distances of 20 and 75 ft, both directly upstream and downstream of the well.
1 answer:
Question:
The question is not complete. See the complete question and the answer below.
A well that pumps at a constant rate of 0.5m3/s fully penetrates a confined aquifer of 34 m thickness. After a long period of pumping, near steady state conditions, the measured drawdowns at two observation wells 50m and 100m from the pumping well are 0.9 and 0.4 m respectively. (a) Calculate the hydraulic conductivity and transmissivity of the aquifer (b) estimate the radius of influence of the pumping well, and (c) calculate the expected drawdown in the pumping well if the radius of the well is 0.4m.
Answer:
T = 0.11029m²/sec
Radius of influence = 93.304m
expected drawdown = 3.9336m
Explanation:
See the attached file for the explanation.
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Answer:
Throat diameter =28.60 mm
Explanation:
Bore diameter ⇒
Manometric deflection x=235 mm
Flow rate Q=240 Lt/min⇒ Q=.004
Coefficient of discharge =0.8
We know that discharge through venturi meter
=13.6 for Hg and
=1 for water.
h=2.961 m
Now by putting the all value in
⇒=28.60 mm
So throat diameter =28.60 mm
Answer:
The difference of head in the level of reservoir is 0.23 m.
Explanation:
For pipe 1
For pipe 2
Q=2.8 l/s
We know that Q=AV
head loss (h)
Now putting the all values
So h=0.23 m
So the difference of head in the level of reservoir is 0.23 m.