Civil Engineering - Water Resources Engineering - Discussion

Discussion Forum : Water Resources Engineering - Section 1 (Q.No. 43)
43.
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
1.185 m2/minute
1.285 m2/minute
1.385 m2/minute
1.485 m2/minute
1.585 m2/minute.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
14 comments Page 1 of 2.

Hagos said:   7 years ago
Q=2π T(s1_s2)/2.3log(r2/r1),
Where Q=31.40,S1=1,S2=0.5,r2=80,r1=20.
Therefore substitute it then T=1.385.
(3)

Amrit Gajendra said:   7 years ago
The discharge should be taken in S.I. unit and the answer will come as 0.01385.
(3)

Saima said:   4 years ago
Can anyone explain in detail because I am not able to understand it.
(2)

Pihu said:   7 years ago
@Matiur.

After m2/s to m2/min answer not comes.
Please explain it in detail.
(1)

Anim said:   6 years ago
Thanks @Matiur Rahaman.
(1)

Kaushik das said:   1 decade ago
Please explain.

Deepak said:   10 years ago
Please explain, how it comes?

Swati said:   9 years ago
Explain it please.

M. A Sidi said:   9 years ago
By Thiem's formula.

Q = (2 * 3.142 * T(s1 - s2))/(2.3 * log(r2/r1)).

By making T the subject and substituting the given parameters we have T = -1.384.

Srinu said:   9 years ago
Apply formula.

q = 2piBK(S2 - S1)/ln(r2/r1).


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