Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 2 (Q.No. 12)
12.
The discharge per metre width at the foot of a spillway is 10 m3/s at a velocity of 20 m/s. A perfect free hydraulic jump will occur at the foot of the spillway when the tail water depth is approximately equal to
Discussion:
4 comments Page 1 of 1.
Apcivilian said:
4 years ago
The approximate answer is 6.14.
(1)
Mohsin said:
8 years ago
@All.
Discharge = velocity x area.
area= width x height.
In question discharge per meter width is given as 10 cms, therefore,
Unit width discharge = velocity x height.
Discharge = velocity x area.
area= width x height.
In question discharge per meter width is given as 10 cms, therefore,
Unit width discharge = velocity x height.
(1)
Abhay said:
10 years ago
Discharge Q = Velocity * Head (or water depth).
Q = 10 m^3/s.
v = 20 m/s.
h1 = 10/20 = 0.5 m.
The jump will occur if and only if the level of inflowing (supercritical) water level (h_0) satisfies the condition:
h0/h1 = [-1 + √(1 + 8v^2/(9.81 * h1))]/2.
h0/0.5 = [-1 + √(1 + 8*20^2/(9.81*0.5))]/2 = 3.32.
h0 = 6.64 m.
Q = 10 m^3/s.
v = 20 m/s.
h1 = 10/20 = 0.5 m.
The jump will occur if and only if the level of inflowing (supercritical) water level (h_0) satisfies the condition:
h0/h1 = [-1 + √(1 + 8v^2/(9.81 * h1))]/2.
h0/0.5 = [-1 + √(1 + 8*20^2/(9.81*0.5))]/2 = 3.32.
h0 = 6.64 m.
(3)
DANESH said:
1 decade ago
V*Y = q.
From here y1 = 0.5.
Now, y2 = (y1/2)*((sq rt of 1+8F^2)-1).
From here y1 = 0.5.
Now, y2 = (y1/2)*((sq rt of 1+8F^2)-1).
(2)
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