Civil Engineering - Water Resources Engineering - Discussion

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

[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.

Answer: Option C

Explanation:

No answer description available for this question.

Kaushik Das said: (Jan 27, 2015)  
Please explain.

Deepak said: (Jan 29, 2016)  
Please explain, how it comes?

Swati said: (Aug 8, 2016)  
Explain it please.

Srinu said: (Oct 14, 2016)  
Apply formula.

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

Srinu said: (Oct 14, 2016)  
Transmissibility T = BK.

M. A Sidi said: (Oct 15, 2016)  
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.

Matiur Rahaman said: (Jan 14, 2017)  
Q = 31.40lit/min = 31.40/60 * 1000 = 0.000523mitre cube per second.

Where
K = 0.000523 * 2.3 * log(80/20)/3.14 * (29.5 * 29.5-29 * 29).
=0.00000788 miter per second.

T = kb = 0.00000788 * 30.
= 0.000236 meter square per second.
=1.385 meter square per minute.

Mahesh said: (Nov 4, 2017)  
29.5*29.5 - 29*29 how it will came?

Saranya said: (Dec 7, 2017)  
@Mahesh.
Here,
30-0.5= 29.5,
30-1=29.

Pihu said: (Apr 10, 2018)  
@Matiur.

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

Hagos said: (Aug 27, 2018)  
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.

Amrit Gajendra said: (Nov 17, 2018)  
The discharge should be taken in S.I. unit and the answer will come as 0.01385.

Anim said: (Aug 22, 2019)  
Thanks @Matiur Rahaman.

Post your comments here:

Name *:

Email   : (optional)

» Your comments will be displayed only after manual approval.