Civil Engineering - Soil Mechanics and Foundation Engineering - Discussion
Discussion Forum : Soil Mechanics and Foundation Engineering - Section 2 (Q.No. 23)
23.
For a homogeneous earth dam 50 m high having 2 m free broad, a flow net was constructed and the results were : Number of potential drops = 2.4 Number of flow channels = 0.4. If coefficiency of permeability of the dam mateiral is 3 x 10-3 cm3/sec, the discharge per metre length of dam, is
Discussion:
22 comments Page 1 of 3.
Abhesh kumar yadav said:
2 years ago
Number of potential drops = 2.4,
Number of flow channels = 0.4.
Coefficient of permeability of the dam material = 3 x 10^-3 cm^3/sec.
Height of the dam = 50 m.
Freeboard of the dam = 2 m.
Calculations:
Head loss, h = 50 m - 2 m = 48 m.
Discharge per meter length of the dam, Q = kh * (Nf/Nd).
= (3 x 10^-3 cm^3/sec) * 48 m * (0.4/2.4).
= 24 x 10^-5 m^3/sec.
Therefore, the discharge per meter length of the dam is 24 x 10^-5 m^3/sec.
Number of flow channels = 0.4.
Coefficient of permeability of the dam material = 3 x 10^-3 cm^3/sec.
Height of the dam = 50 m.
Freeboard of the dam = 2 m.
Calculations:
Head loss, h = 50 m - 2 m = 48 m.
Discharge per meter length of the dam, Q = kh * (Nf/Nd).
= (3 x 10^-3 cm^3/sec) * 48 m * (0.4/2.4).
= 24 x 10^-5 m^3/sec.
Therefore, the discharge per meter length of the dam is 24 x 10^-5 m^3/sec.
(4)
Hadiul Islam said:
4 years ago
Given data,
K = 3*10^-3cm3/sec.
(0.003/100) = 0.00003 m3/sec, OR.
3*10^-5m3/sec.
H = (50 -2 ) 48 m.
Nf = .4,
Nd = 2.4,
Then formula;
Q = kh* (Nf/Nd).
Q = .00003 * 48 * (.4/2.4)
Q = 0.00024 OR 24*10^-5 m3/sec.
K = 3*10^-3cm3/sec.
(0.003/100) = 0.00003 m3/sec, OR.
3*10^-5m3/sec.
H = (50 -2 ) 48 m.
Nf = .4,
Nd = 2.4,
Then formula;
Q = kh* (Nf/Nd).
Q = .00003 * 48 * (.4/2.4)
Q = 0.00024 OR 24*10^-5 m3/sec.
(2)
Er Amit Kumar said:
5 years ago
Given data,
K = 3*10^-3cm3/sec
(0.003/100) = 0.00003 m3/sec, OR
3*10^-5m3/sec.
H = (50 -2 ) 48 m.
Nf = .4
Nd = 2.4
Than formula,
Q = kh* (Nf / Nd)
Q = .00003 * 48 * (.4 / 2.4
Q = 0.00024 OR 24*10^-5 m3/sec.
K = 3*10^-3cm3/sec
(0.003/100) = 0.00003 m3/sec, OR
3*10^-5m3/sec.
H = (50 -2 ) 48 m.
Nf = .4
Nd = 2.4
Than formula,
Q = kh* (Nf / Nd)
Q = .00003 * 48 * (.4 / 2.4
Q = 0.00024 OR 24*10^-5 m3/sec.
(1)
Abhay said:
10 years ago
Discharge Q = kH(Nf/Nd).
Where k = Coefficient of permeability = 3x10^3 cm^2/sec.
H = Ht. of water or head = 50 - 2 = 48 m = 4800 cm.
= 3x10^3*4800*(0.4/2.4).
= 2.4 cm^3/s.
= 24x10 μ m^3/s.
Where k = Coefficient of permeability = 3x10^3 cm^2/sec.
H = Ht. of water or head = 50 - 2 = 48 m = 4800 cm.
= 3x10^3*4800*(0.4/2.4).
= 2.4 cm^3/s.
= 24x10 μ m^3/s.
Sushma Thota said:
7 years ago
The discharge PER METER is asked, so the calculated answer must be divided by 100
since 1m=100cm.
So, the given units and given answer are correct.
since 1m=100cm.
So, the given units and given answer are correct.
Aliabryne said:
9 years ago
K = 3 * 10^3 cm per sec (convert this to metre) = 3 * 10^5m per sec.
Q = Kh (Nf/Nd) , after putting the values, we get 24 * 10^5.
Q = Kh (Nf/Nd) , after putting the values, we get 24 * 10^5.
Raju raut said:
9 years ago
We know discharge (Q) = kH x number of flow channel/ number of Drops.
Where k = coeff of permeability and H = water head.
Where k = coeff of permeability and H = water head.
Siddhant Subedi said:
2 years ago
The value of the coefficient of permeability = 3x10^-3 cm/s not cm^3/s.
(1)
Vivek said:
7 years ago
K will be in cm/sec.
It is unit of the coefficient of permeability.
It is unit of the coefficient of permeability.
Roy said:
8 years ago
As I know the coefficient of permeability's unit is cm/s.
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