# Civil Engineering - Water Resources Engineering - Discussion

### Discussion :: 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

 [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.

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.