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