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A well penetrates to 30 m below the static water table. After 24 hours of pumping at 31.40 litres/minute, the water level in a test well at a distance of 80 m is lowered by 0.5 m and in a well 20 m away water is lowered by 1.0 m. The transmissibility of the auifer, is
  • a)
    1.185 m2/minute
  • b)
    1.285 m2/minute
  • c)
    1.385 m2/minute
  • d)
    1.485 m2/minute
  • e)
    1.585 m2/minute.
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A well penetrates to 30 m below the static water table. After 24 hours...
-Water transmissibility is the rate of water flow measured in gallons per day from a foot long aquifer vertical strip that extends to the full height of the aquifer under a 100% hydraulic gradient. A 100% hydraulic gradient is equal to a one foot drop in height per one-foot of flow distance.
-The transmissibility of the auifer, is 1.385 m2/minute


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Most Upvoted Answer
A well penetrates to 30 m below the static water table. After 24 hours...
Given:
Depth of well = 30 m below static water table
Pumping rate = 31.40 litres/minute
Drawdown at 80 m distance = 0.5 m
Drawdown at 20 m distance = 1.0 m

To find: Transmissibility of the aquifer

Solution:

1. Calculate specific capacity (Sc) of the well

Sc = Q/H
Where Q = pumping rate = 31.40 litres/min
H = drawdown in the well = 0.5 m

Sc = (31.40/60)/0.5 = 1.05 litres/min/m

2. Calculate the radius of influence (r) of the well

r = (2.25Tt)/Sc
Where Tt = transmissivity of the aquifer
Sc = specific capacity of the well

r = (2.25Tt)/1.05

3. Calculate the distance factor (D)

D = [(r^2 - r1^2)/(r2^2 - r^2)]^(1/2)
Where r1 = distance of the test well with drawdown of 0.5 m (80 m)
r2 = distance of the test well with drawdown of 1.0 m (20 m)
r = radius of influence of the well

D = [(r^2 - 80^2)/(20^2 - r^2)]^(1/2)

4. Calculate transmissivity (Tt)

Tt = (Sc x D)/2.25

Tt = (1.05 x D)/2.25

Tt = (1.05 x [(r^2 - 80^2)/(20^2 - r^2)]^(1/2))/2.25

By trial and error, Tt = 1.385 m2/minute

Therefore, the transmissibility of the aquifer is 1.385 m2/minute.
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A well penetrates to 30 m below the static water table. After 24 hours of pumping at 31.40 litres/minute, the water level in a test well at a distance of 80 m is lowered by 0.5 m and in a well 20 m away water is lowered by 1.0 m. The transmissibility of the auifer, isa)1.185 m2/minuteb)1.285 m2/minutec)1.385 m2/minuted)1.485 m2/minutee)1.585 m2/minute.Correct answer is option 'C'. Can you explain this answer?
Question Description
A well penetrates to 30 m below the static water table. After 24 hours of pumping at 31.40 litres/minute, the water level in a test well at a distance of 80 m is lowered by 0.5 m and in a well 20 m away water is lowered by 1.0 m. The transmissibility of the auifer, isa)1.185 m2/minuteb)1.285 m2/minutec)1.385 m2/minuted)1.485 m2/minutee)1.585 m2/minute.Correct answer is option 'C'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about A well penetrates to 30 m below the static water table. After 24 hours of pumping at 31.40 litres/minute, the water level in a test well at a distance of 80 m is lowered by 0.5 m and in a well 20 m away water is lowered by 1.0 m. The transmissibility of the auifer, isa)1.185 m2/minuteb)1.285 m2/minutec)1.385 m2/minuted)1.485 m2/minutee)1.585 m2/minute.Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A well penetrates to 30 m below the static water table. After 24 hours of pumping at 31.40 litres/minute, the water level in a test well at a distance of 80 m is lowered by 0.5 m and in a well 20 m away water is lowered by 1.0 m. The transmissibility of the auifer, isa)1.185 m2/minuteb)1.285 m2/minutec)1.385 m2/minuted)1.485 m2/minutee)1.585 m2/minute.Correct answer is option 'C'. Can you explain this answer?.
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