Civil Engineering - Water Resources Engineering - Discussion
Discussion Forum : Water Resources Engineering - Section 2 (Q.No. 42)
42.
A well is sunk in an unconfined aquifer having a saturated depth of 100 m. Assuming the equilibrium flow conditions and a homogeneous aquifer and radius of influence to be same, the ratio of discharges at 20 m and 40 m draw downs, is
Discussion:
7 comments Page 1 of 1.
Hassan Bilal said:
5 years ago
Thanks all for explaining.
Anirban Roy said:
5 years ago
But the drawdowns are 20m and 40m respectively. So s1=20m,
Hence h1=100-20=80m.
s2=40m,
Hence h2=100-40=60m.
So, q1/q2=(H^2-h1^2)÷(H^2-h2^2).
=(100^2-80^2)÷(100^2-60^2) = 45/16.
Hence h1=100-20=80m.
s2=40m,
Hence h2=100-40=60m.
So, q1/q2=(H^2-h1^2)÷(H^2-h2^2).
=(100^2-80^2)÷(100^2-60^2) = 45/16.
(3)
Thakur said:
5 years ago
Please explain the formula.
Raaj said:
7 years ago
Thank you all for the solution.
(1)
D.pal said:
8 years ago
(100^2-20^2)/(100^2-40^2),
= (100+20)(100-20)/(100+40)(100-40),
= 120*80/140*60,
= 8/7.
= (100+20)(100-20)/(100+40)(100-40),
= 120*80/140*60,
= 8/7.
(5)
Awais ahmed mirza said:
8 years ago
100^2-20^2 divide by 100^2 - 40^2 = 8/7.
(2)
Harsha said:
8 years ago
Can anyone give a clear explanation of the answer?
(1)
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