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Given:

velocity of water on the bottom of the spillway (V1) = 12.8 m/s.

Depth of flow after the apron (d) = 3m

Width of Apron (b) = 50m

Discharge (Q) = 250 m3/s, So=0 and n = 0.025

It is given that hydraulic jump will occur.

Solution:

Discharge per unit width, q=Q/b= (V1xarea)/b

Area = d1xb,Thus cancelling b on both numerator and denominator we get,

q = V1x d1 where V1 and d1 are velocity and depth on the upstream (u/s) side

Therefore, q = 12.8x3 = 38.4 m2/s

Froude number on the u/s side, (Fe)1= V1/√(gx d1)=12.8/√(9.81x3)=2.35

Since Froude number is more than 1,it indicates that hydraulic jump occurs on the u/s side

In order to calculate the energy loss we will use the formula, hl =(d2- d1)3/4 d1 d2 ….Eqn (a)

Where hl = Loss of energy due to hydraulic jump, d1 and d2 are depth of hydraulic jump at u/s side and d/s side respectively

To calculate depth of hydraulic jump d2 we use the formula

d2= d1/2(√1+8(Fe)12-1)=3/2(√1+8(2.35)12-1)=3/2(√1+8(2.35)12-1)=3/2(√44.18)=1.5x√44.18=9.97m

Height of hydraulic jump = d2- d1=9.97-3=6.97 m

Therefore from Eqn (a) hl =(9.97-3)3/4x9.97x3=338.6/119.64=2.83m

The length of the hydraulic jump is Taken as 7 times the hydraulic jump=7x6.97=48.79

Hence to contain the hydraulic jump the length of the apron should be 48.79m
 

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