CIVIL ENGINEERING FORMULAS

(Frankie) #1

322 CHAPTER TWELVE


The pressure force Fdeveloped in hydraulic jump is

(12.86)


whered 1 depth before jump, ft (m)
d 2 depth after jump, ft (m)
wunit weight of water, lb/ft^3 (kg/m^3 )


The rate of change of momentum at the jump per foot width of channel equals


(12.87)


whereMmass of water, lbs^2 /ft (kgs^2 /m)
V 1 velocity at depth d 1 , ft/s (m/s)
V 2 velocity at depth d 2 , ft/s (m/s)
q discharge per foot width of rectangular channel, ft^3 /s (m^3 /s)
tunit of time, s
gacceleration due to gravity, 32.2 ft/s^2 (9.81 kg/s^2 )


Then (12.88)


(12.89)


(12.90)


The head loss in a jump equals the difference in specific-energy head before
and after the jump. This difference (Fig. 12.17) is given by


(12.91)


whereHe 1 specific-energy head of stream before jump, ft (m); and He 2 
specific-energy head of stream after jump, ft (m).
The depths before and after a hydraulic jump may be related to the critical
depth by


(12.92)

whereqdischarge, ft^3 /s (m^3 /s) per ft (m) of channel width; and dccritical
depth for the channel, ft (m).
It may be seen from this equation that if d 1 dc,d 2 must also equal dc.
Figure 12.18 shows how the length of hydraulic jump may be computed using
the Froude number and the L/d 2 ratio.


d 1 d 2

d 1 d 2
2




q^2
g

d^3 c

HeHe 1 He 2 

(d 2 d 1 )^3
4 d 1 d 2

d 1 

d 2
2




B


2 V^22 d 2
g




d^22
4

d 2 

d 1
2




B


2 V^21 d 1
g




d^21
4

V^21 


gd 2
2 d 1

(d 2 d 1 )

F


MV 1 MV 2


t




qw
g

(V 1 V 2 )


F


d^22 w
2




d 12 w
2
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