Civil Engineering - Water Resources Engineering - Discussion
Discussion Forum : Water Resources Engineering - Section 1 (Q.No. 40)
40.
If the viscosity of ground water is 1.00, the Slitcher's constant is 400, the effective size of soil particles in acquifer is 0.5 mm and hydraulic gradient is 1 in 80, the velocity of flow is
Discussion:
10 comments Page 1 of 1.
Abhishek said:
1 decade ago
Can you provide explanation for this question?
Am getting different answer by substituting in the formula.
Am getting different answer by substituting in the formula.
Purvi said:
10 years ago
Please someone explain?
Eng.kate said:
9 years ago
Anyone have explain please.
Swati said:
9 years ago
Anybody can explain it please.
Snehal Wankhede said:
9 years ago
Slitcher's formula v = 400 * i * D^10/4.
Santhikrishna said:
9 years ago
Slichter formula v = K * s * D^2/m.
400 * 1/80 * 0.5^2/1 = 1.25.
400 * 1/80 * 0.5^2/1 = 1.25.
Bidyut Barman said:
8 years ago
Slichter's formula(V) =K'I(D10 power 2/u).
Where, V=Velocity of flow, K=Slichter Constant.
I=Slope of hydraulic gradient, D10= Patrice size, u= Viscosity of water.
Then, V=K'I(D10 Power 2/u) =Put the value=1.25 m/day.
Where, V=Velocity of flow, K=Slichter Constant.
I=Slope of hydraulic gradient, D10= Patrice size, u= Viscosity of water.
Then, V=K'I(D10 Power 2/u) =Put the value=1.25 m/day.
Ghost said:
8 years ago
V = K*I (D^2)/u.
K = 400.
I = 1/80.
D = 0.5.
U = 1.
D10 is only D don't multiply 10 with it.
Put the values u will get the answer.
K = 400.
I = 1/80.
D = 0.5.
U = 1.
D10 is only D don't multiply 10 with it.
Put the values u will get the answer.
(1)
Dastaan said:
5 years ago
By using Slitcher's formula:
Velocity, v = {400 * i * (D10)^2}/u (unit= m/day);
Where,
i = hydraulic gradient.
D10 = effective size of particle in mm.
u = dynamic viscosity.
Therefore,
v = {400 * (1/80) * 0.5^2}/1.
= 1.25 m/day.
Velocity, v = {400 * i * (D10)^2}/u (unit= m/day);
Where,
i = hydraulic gradient.
D10 = effective size of particle in mm.
u = dynamic viscosity.
Therefore,
v = {400 * (1/80) * 0.5^2}/1.
= 1.25 m/day.
(1)
Vikash said:
3 years ago
V = (K'id10^2)/u
V = (400*1/80*0.5^2)/1 =1.25 m/day.
V = (400*1/80*0.5^2)/1 =1.25 m/day.
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