Civil Engineering - Water Supply Engineering - Discussion

Discussion Forum : Water Supply Engineering - Section 2 (Q.No. 41)
41.
The water level in an open well was depressed by pumping 2.5 m and recuperated 2.87 m in 3 hours and 50 minutes. The yield of the well per minute is
0.0033
0.0044
0.0055
0.0066
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
10 comments Page 1 of 1.

Santosh Kumar said:   3 years ago
This question is incorrect.

Water can't recuperate above the original W/T level in any case.Hence most plausible scenario is that it's a misprint and the recuperated value should be 1.8 or 1.7 or 1.87 etc ( a value less than 2.5).

Solution
s1 =2.5
s2=2.5-1.8=0.7.
Now specific yield = 2.303/230 log(2.5/0.7),
=0.0055.

Note that since A and s values aren't given we can't find a yield of well, rather we can only find a specific yield of the well.
(5)

Deepali said:   1 decade ago
Please solve.
(2)

Mdpawar said:   10 years ago
How to solve this problems?
(2)

Gubendhiran said:   8 years ago
Yield=(2.3/T)*log(s1/s2),
T= time after the pumping is stopped=3.83hrs,
s1= depression head=2.5m,
s2= recuperation head=2.5-2.87=0.37m,
Y=(2.3/3.83)*log(2.5/0.37),
Y=0.49828 m2/hr,
Y=0.0083 m2/min.
(2)

Roy said:   8 years ago
K/A = yield = 2.3log(2.87/2.5)/230 =0.0055.
(1)

ANIL said:   6 years ago
What is 230?
(1)

Sukhesh S C said:   8 years ago
Can you explain it in detail?

Kajol said:   7 years ago
@Roy

could you please explain the solution clearly?

Hardik said:   6 years ago
@Anil.

3 hours = 180 minute +50 minute.
=230 minute.

Manu said:   4 years ago
0.0083 m^3/min.

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