Civil Engineering - GATE Exam Questions - Discussion
Discussion Forum : GATE Exam Questions - Section 4 (Q.No. 4)
4.
Water emerges from an ogee spillway with velocity 13.72 m/s and depth = 3.0 m at its toe. The tail water depth required to form a hydraulic jump at the toe is
Discussion:
7 comments Page 1 of 1.
Sangay Tenzin said:
7 years ago
I also got 9.334.
y2/y1 = 1/2*(√(1+8.F1^2) - 1)
y1 = depth of water before hydraulic jump
y2 = depth of water after hydraulic jump
F1 = froude number of flow before hydraulic jump i.e F=V/√(g*depth) = 13.72/(√(9.81*3)) = 2.52.
y2 = 3* [1/2*( sqrt(1+8*2.52^2) - 1)].
y2 = 9.334m.
y2/y1 = 1/2*(√(1+8.F1^2) - 1)
y1 = depth of water before hydraulic jump
y2 = depth of water after hydraulic jump
F1 = froude number of flow before hydraulic jump i.e F=V/√(g*depth) = 13.72/(√(9.81*3)) = 2.52.
y2 = 3* [1/2*( sqrt(1+8*2.52^2) - 1)].
y2 = 9.334m.
(1)
RAKESH KUMAR said:
9 years ago
Can anyone explain this?
Hitesh said:
8 years ago
Answer should be 9.8m as per calculation.
Savy said:
7 years ago
9.34 m is the answer.
Noni said:
7 years ago
How do you calculate? Explain.
DPR said:
5 years ago
Yes, you are right @SANGAY TENZIN.
9.334 m is the correct answer.
The direct formula without getting frouds no is;
y2/y1=0.5*(-1+{1+((8v1^2)/g*y)}^0.5).
9.334 m is the correct answer.
The direct formula without getting frouds no is;
y2/y1=0.5*(-1+{1+((8v1^2)/g*y)}^0.5).
Jay Garachh said:
4 years ago
Put 0.3m instead of 3m you will get an answer.
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