Civil Engineering - GATE Exam Questions - Discussion
Discussion Forum : GATE Exam Questions - Section 3 (Q.No. 16)
16.
Water flows at a depth of 0.1 m with a velocity of 6 m/s in a rectangular channel. The alternate dpeth is
Discussion:
11 comments Page 1 of 2.
Shaheen said:
1 decade ago
y2/y1 = 1/2[-1+sqrt(1+8Fr2)].
Hamza said:
7 years ago
The answer is "A" because on equating the equations E=Y+ (V)2/2g as a first equation and E=Y+(q/y)2/2g we get nearly .32.
Neeraj nautiyal said:
7 years ago
Can anyone please explain this?
Basith said:
7 years ago
0.1/2*(√(1+8*Fr^2)-1) = 0.808.
fr = 6/√(9.81*0.1).
fr = 6/√(9.81*0.1).
Keshav Kaushik said:
7 years ago
Fr = V/√(gL) = 6.
y2 = y1* 0.5* [ -1+ √( 1+8*Fr^2)].
y2 = 0.8 is the answer.
y2 = y1* 0.5* [ -1+ √( 1+8*Fr^2)].
y2 = 0.8 is the answer.
Anomiee said:
7 years ago
Here it is asking about alternate depth not sequent depth. For alternate depths, we have to equate specific energies at the two sections.
Replyfast said:
6 years ago
Can anyone explain how to calculate alternate depth for this numerical problem?
Mayuri jamdade said:
5 years ago
Y2 = 1/2 * Y1 * (((1+(8 * Fr^2))^0.5)-1).
Fr = (V/((gY1)^0.5)) = (6/((9.81*0.1)^0.5)) = 6.06
Y2 = 0.5 * 0.1*(((1+(8 * 6.06 * 6.06))^0.5)-1) = 0.808m
Y2 = 0.808m is approximately 0.81m.
Fr = (V/((gY1)^0.5)) = (6/((9.81*0.1)^0.5)) = 6.06
Y2 = 0.5 * 0.1*(((1+(8 * 6.06 * 6.06))^0.5)-1) = 0.808m
Y2 = 0.808m is approximately 0.81m.
(2)
Krishan Hatria said:
4 years ago
Here we need to calculate alternate depth (not conjugate depth).
So we can not use hydraulic jump formula y2/y1 = 1/2[-1+sqrt(1+8Fr2)] because this formula has driven from a specific force equation and it is used for finding conjugate depth only. But in question ask alternate depth which can be found by equating specific energy.
So we can not use hydraulic jump formula y2/y1 = 1/2[-1+sqrt(1+8Fr2)] because this formula has driven from a specific force equation and it is used for finding conjugate depth only. But in question ask alternate depth which can be found by equating specific energy.
Krishan Hatria said:
4 years ago
@All.
According to me the solution is
q=v*y1 = 0.1*6 = .6 cubic meter per meter width.
Now use formula drive from equating specific energy at 1-1 section and 2-2 section.
q^2÷ 2g = y1^2 y2^2 ÷ (y one + y two) after solving y two= 1.93m.
Correct me if I am wrong.
According to me the solution is
q=v*y1 = 0.1*6 = .6 cubic meter per meter width.
Now use formula drive from equating specific energy at 1-1 section and 2-2 section.
q^2÷ 2g = y1^2 y2^2 ÷ (y one + y two) after solving y two= 1.93m.
Correct me if I am wrong.
(2)
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