1. Find: hL [m] 0.63 0.84 1.02 1.31 rectangular d channel b b=3 [m]
hydraulic
jump
Find the headloss, in meters, across the hydraulic jump. [pause] In this problem, --- d2 Q d1
d1=0.5 [m]
d2=2.0 [m]

2. Find: hL [m] 0.63 0.84 1.02 1.31 rectangular d channel b b=3 [m]
hydraulic
jump
a hydraulic jump occurs in a rectangular channel. d2 Q d1
d1=0.5 [m]
d2=2.0 [m]

3. Find: hL [m] 0.63 0.84 1.02 1.31 rectangular d channel b b=3 [m]
hydraulic
jump
The channel width is provided, as well as the initial and final water depths. [pause] d2 Q d1
d1=0.5 [m]
d2=2.0 [m]

4. Find: hL [m] + + ρ*g P y hT= v2 d 2*g b b=3 [m] hydraulic jump d2 Q d1
Bernoulli’s equation states the total head equals, --- d2 Q d1
d1=0.5 [m]
d2=2.0 [m]

5. Find: hL [m] + + ρ*g P y hT= v2 d 2*g b pressure b=3 [m] head
hydraulic
jump
the pressure head, --- d2 Q d1
d1=0.5 [m]
d2=2.0 [m]

6. Find: hL [m] + + ρ*g P y hT= v2 d 2*g b pressure b=3 [m] head
hydraulic
elevation head
jump
plus the elevation head, --- d2 Q d1
d1=0.5 [m]
d2=2.0 [m]

7. Find: hL [m] + + ρ*g P y hT= v2 d 2*g b pressure pressure velocity
b=3 [m]
head
head
head
hydraulic
elevation head
jump
plus the velocity head. d2 Q d1
d1=0.5 [m]
d2=2.0 [m]

8. Find: hL [m] + + ρ*g P y hT= v2 d 2*g b pressure pressure velocity
b=3 [m]
head
head
head
hydraulic
elevation head
jump
Since open channel flow is not pressurized flow, --- d2 Q d1
d1=0.5 [m]
d2=2.0 [m]

9. Find: hL [m] + + ρ*g P y hT= v2 d 2*g b pressure pressure velocityP v2 + y +
hT= d
ρ*g
2*g b
pressure
pressure
velocity
b=3 [m]
head
head
head
hydraulic
elevation head
jump
the pressure head for this problem, is zero, and the equation simplifies. d2 Q d1
d1=0.5 [m]
d2=2.0 [m]

10. Find: hL [m] + + + ρ*g P y hT= hT= y v2 d 2*g v2 b 2*g b=3 [m]P v2 + y +
hT= d
ρ*g
2*g v2 b
hT= y +
2*g
b=3 [m]
hydraulic
jump
The headloss across the hydraulic jump, is equal to --- d2 Q d1
d1=0.5 [m]
d2=2.0 [m]

11. Find: hL [m] + + + ρ*g P y hT= hT= y hL= h1 – h2 v2 d 2*g v2 b 2*gP v2 + y +
hT= d
ρ*g
2*g v2 b
hT= y +
2*g
b=3 [m]
hydraulic
hL= h1 – h2
jump
the initial total head, minus the final total head, which can be written out in terms of the --- d2 Q d1
d1=0.5 [m]
d2=2.0 [m]

12. Find: hL [m] + + + + - + ρ*g P y hT= hT= y hL= h1 – h2 hL= y1 y2 v2 dP v2 + y +
hT= d
ρ*g
2*g v2 b
hT= y +
2*g
b=3 [m]
hydraulic
hL= h1 – h2
jump
initial and final elevation and velocity heads. [pause] The elevation heads, v1 v2 d2 2 2
hL= y1 + - y2 + Q
2*g
2*g d1
d1=0.5 [m]
d2=2.0 [m]

13. Find: hL [m] + + + + - + ρ*g P y hT= hT= y hL= h1 – h2 hL= y1 y2 v2 dP v2 + y +
hT= d
ρ*g
2*g v2 b
hT= y +
2*g
b=3 [m]
hydraulic
hL= h1 – h2
jump
y1 and y2, are simply the water depths given in the problem statement, d1 and d2. v1 v2 d2 2 2
hL= y1 + - y2 + Q
2*g
2*g d1
d1=0.5 [m]
d2=2.0 [m]

14. Find: hL [m] + + + + - + ρ*g P y hT= hT= y hL= h1 – h2 hL= y1 y2 v2 dP v2 + y +
hT= d
ρ*g
2*g v2 b
hT= y +
2*g
b=3 [m]
hydraulic
hL= h1 – h2
jump
The value of g, is a constant, but we don’t know the velocities, --- v1 v2 d2 2
hL= y1 + - 2 y2 + Q
2*g
2*g d1
d1=0.5 [m] m
9.81
d2=2.0 [m] s2

15. Find: hL [m] + + + + - + ρ*g ? ? P y hT= hT= y hL= h1 – h2 hL= y1 y2P v2 + y +
hT= d
ρ*g
2*g v2 b
hT= y +
2*g
b=3 [m] ? ?
hydraulic
hL= h1 – h2
jump
v1 and v2. [pause] Velocities 1 and 2 equal, --- v1 v2 d2 2 2
hL= y1 + - y2 + Q
2*g
2*g d1
d1=0.5 [m] m
9.81
d2=2.0 [m] s2

16. Find: hL [m] + - + ? ? v1= v2= hL= y1 y2 Q A1 d Q A2 b b=3 [m]
hydraulic
jump
the flowrate, divided by areas 1 and 2, respectively, --- v1 v2 d2 2
hL= y1 + - 2 y2 + Q
2*g
2*g d1
d1=0.5 [m] m
9.81
d2=2.0 [m] s2

17. Find: hL [m] + - + ? ? v1= = v2= = hL= y1 y2 Q Q A1 b*d1 d Q Q A2 b*d2
b=3 [m] ? ?
hydraulic
jump
where the areas 1 and 2, equal the channel base times the depths 1 and 2, respectively. v1 v2 d2 2 2
hL= y1 + - y2 + Q
2*g
2*g d1
d1=0.5 [m] m
9.81
d2=2.0 [m] s2

18. Find: hL [m] + - + ? ? v1= = v2= = hL= y1 y2 Q Q A1 b*d1 d Q Q A2 b*d2
b=3 [m] ? ?
hydraulic
jump
Since we already know the base width and the depths, we need to solve for the flowrate, --- v1 v2 d2 2 2
hL= y1 + - y2 + Q
2*g
2*g d1
d1=0.5 [m] m
9.81
d2=2.0 [m] s2

19. Find: hL [m] + - + ? ? ? ? v1= = v2= = hL= y1 y2 Q Q A1 b*d1 d Q Q A2
b=3 [m] ? ?
hydraulic
jump
Q. [pause] An equation which relates the conjugate depths across --- v1 v2 d2 2
hL= y1 + - 2 y2 + Q
2*g
2*g d1
d1=0.5 [m] m
9.81
d2=2.0 [m] s2

20. Find: hL [m] ? ? v1= = v2= = d2 1 = 1 + 8 * Fr1 -1 d1 2 Q Q A1 b*d1 d
b=3 [m] d2 1 = *
1 + 8 * Fr1 -1 2
hydraulic d1 2
jump
a hydraulic jump, is, d2 divided by d1, equals, 1/2 times the quantity, the square root of 1 plus eight times the upstream froude number squared, minus 1. For a rectangular channel, the Froude number equals --- d2 Q d1
d1=0.5 [m]
d2=2.0 [m]

21. Find: hL [m] ? v1= = v v2= = Fr= d2 1 = 1 + 8 * Fr1 -1 d1 2 Q Q A1
b*d1 d Q Q v
v2= =
Fr= A2
b*d2
g*d b
b=3 [m] d2 1 = *
1 + 8 * Fr1 -1 2
hydraulic d1 2
jump
the velocity, divided by the square root of the gravitational acceleration times the depth, where the velocity equals, --- d2 Q d1
d1=0.5 [m]
d2=2.0 [m]

22. Find: hL [m] ? v1= = v= v v2= = Fr= d2 1 = 1 + 8 * Fr1 -1 d1 2 Q Q QA1
b*d1 A d Q Q v
v2= =
Fr= A2
b*d2
g*d b
b=3 [m] d2 1 = *
1 + 8 * Fr1 -1 2
hydraulic d1 2
jump
the flowrate, divided by the area, and the area equals, --- d2 Q d1
d1=0.5 [m]
d2=2.0 [m]

23. Find: hL [m] ? v1= = v= v v2= = Fr= d2 1 = 1 + 8 * Fr1 -1 d1 2 A=b*d Q
g*d b
b=3 [m] d2 1 = *
1 + 8 * Fr1 -1 2
hydraulic d1 2
jump
the base width times the depth. [pause] After these substitutions are made, the hydraulic jump equation --- d2 Q d1
d1=0.5 [m]
d2=2.0 [m]

24. Find: hL [m] ? v1= = v= v v2= = Fr= d2 1 = 1 + 8 * Fr1 -1 d1 2 d2 1 =
A=b*d Q Q ? Q
v1= = v= A1
b*d1 A d Q Q v
v2= =
Fr= A2
b*d2
g*d b
b=3 [m] d2 1 = *
1 + 8 * Fr1 -1 2
hydraulic d1 2
jump
is effectively 1 equation and 1 unknown, and that unknown value is our flowrate, Q. [pause] After a few lines of algebra, --- d2 d2 1 Q 2 Q =
1 + 8 * -1 d1 * d1 2
g*b2*d1 3
d1=0.5 [m]
d2=2.0 [m]

25. Find: hL [m] ? +1 -1 v1= = v= v v2= = Fr= d2 1 = 1 + 8 * Fr1 -1 d1 2
A=b*d Q Q ? Q
v1= = v= A1
b*d1 A d Q Q v
v2= =
Fr= A2
b*d2
g*d b
b=3 [m] d2 1 = *
1 + 8 * Fr1 -1 2
hydraulic d1 2
jump
the flowrate is simply a function of the gravitational acceleration, and the given data. d2 d2 1 Q 2 Q =
1 + 8 * -1 d1 * d1 2
g*b2*d1 3
d1=0.5 [m] 2 d2
g*b2*d1 * 2 3 Q= +1 -1 *
d2=2.0 [m] d1 8

26. Find: hL [m] +1 -1 v1= = v2= = 2 d2 d1 Q Q A1 b*d1 d Q Q A2 b*d2 b
b=3 [m]
hydraulic
jump m
Plugging in the values for g, b, d1 and d2, the flowrate computes to, ---
9.81 d2 s2 Q d1
d1=0.5 [m] 2 d2
g*b2*d1 * 2 3 Q= +1 -1 *
d2=2.0 [m] d1 8

27. Find: hL [m] +1 -1 v1= = v2= = Q=10.50 2 d2 d1 Q Q A1 b*d1 d Q Q A2
b=3 [m] m3
Q=10.50 s
hydraulic
jump m
10.50 meters cubed per second. [pause] To solve for the upstream and downstream velocities, ---
9.81 d2 s2 Q d1
d1=0.5 [m] 2 d2
g*b2*d1 * 2 3 Q= +1 -1 *
d2=2.0 [m] d1 8

28. Find: hL [m] +1 -1 v1= = v2= = Q=10.50 2 d2 d1 Q Q A1 b*d1 d Q Q A2
b=3 [m] m3
Q=10.50 s
hydraulic
jump m
the flowrate, base width, and depths are substituted in, and we get ---
9.81 d2 s2 Q d1
d1=0.5 [m] 2 d2
g*b2*d1 * 2 3 Q= +1 -1 *
d2=2.0 [m] d1 8

29. Find: hL [m] +1 -1 v1= = =7.00 v2= = Q=10.50 2 d2 d1 Q Q A1 b*d1 d Q Qs A1
b*d1 d Q Q
v2= = A2
b*d2 b
b=3 [m] m3
Q=10.50 s
hydraulic
jump m
velocity 1 equals 7 meters per second, and ---
9.81 d2 s2 Q d1
d1=0.5 [m] 2 d2
g*b2*d1 * 2 3 Q= +1 -1 *
d2=2.0 [m] d1 8

30. Find: hL [m] +1 -1 v1= = =7.00 v2= = =1.75 Q=10.50 2 d2 d1 Q Q A1 b*d1s A1
b*d1 d Q Q m
v2= =
=1.75 s A2
b*d2 b
b=3 [m] m3
Q=10.50 s
hydraulic
jump m
velocity 2 equals 1.75 meters per second. [pause] By knowing our velocities, ---
9.81 d2 s2 Q d1
d1=0.5 [m] 2 d2
g*b2*d1 * 2 3 Q= +1 -1 *
d2=2.0 [m] d1 8

31. Find: hL [m] + - + ? ? v1=7.00 v2=1.75 hL= y1 y2 d b b=3 [m] hydraulics d m
v2=1.75 s b
b=3 [m] ? ?
hydraulic
jump
we can return to our equation, for headloss across the hydraulic jump, --- v1 v2 d2 2
hL= y1 + - 2 y2 + Q
2*g
2*g d1
d1=0.5 [m] m
9.81
d2=2.0 [m] s2

32. Find: hL [m] - + + v1=7.00 v2=1.75 hL= y1 y2 d b b=3 [m] hydraulics d m
v2=1.75 s b
b=3 [m]
hydraulic
jump
plug in the values for velocity, --- v1 v2 d2 2 - 2
hL= y1 + y2 + Q
2*g
2*g d1
d1=0.5 [m] m
9.81
d2=2.0 [m] s2

33. Find: hL [m] - + + v1=7.00 v2=1.75 hL= y1 y2 d b b=3 [m] hydraulics d m
v2=1.75 s b
b=3 [m]
hydraulic
jump
plug in the values for acceleration and the depths, and the headloss computes to --- v1 v2 d2 2 - 2
hL= y1 + y2 + Q
2*g
2*g d1
d1=0.5 [m] m
9.81
d2=2.0 [m] s2

34. Find: hL [m] + - + hL= 0.841 [m] v1=7.00 v2=1.75 hL= y1 y2 d b b=3 [m]s d m
v2=1.75 s b
hL= [m]
b=3 [m]
hydraulic
jump
0.841 meters. v1 v2 d2 2
hL= y1 + - 2 y2 + Q
2*g
2*g d1
d1=0.5 [m] m
9.81
d2=2.0 [m] s2

35. Find: hL [m] + - + hL= 0.841 [m] v1=7.00 v2=1.75 0.63 0.84 1.02 1.31s d m
v2=1.75
0.63
0.84
1.02
1.31 s b
hL= [m]
b=3 [m]
hydraulic
jump
Looking over the possible solutions, --- v1 v2 d2 2
hL= y1 + - 2 y2 + Q
2*g
2*g d1
d1=0.5 [m] m
9.81
d2=2.0 [m] s2

36. Find: hL [m] + - + hL= 0.841 [m] AnswerB v1=7.00 v2=1.75 0.63 0.84
1.02
1.31 s b
hL= [m]
b=3 [m]
AnswerB
hydraulic
jump
the answer is B. v1 v2 d2 2 2
hL= y1 + - y2 + Q
2*g
2*g d1
d1=0.5 [m] m
9.81
d2=2.0 [m] s2

37. ? Index σ’v = Σ γ d γT=100 [lb/ft3] +γclay dclay 1
Find: σ’v at d = 30 feet
(1+wc)*γw
wc+(1/SG)
σ’v = Σ γ d d
Sand
10 ft
γT=100 [lb/ft3]
100 [lb/ft3]
10 [ft]
20 ft
Clay
= γsand dsand
+γclay dclay A W S
V [ft3]
W [lb]
40 ft
text
wc = 37% ? Δh
20 [ft]
(5 [cm])2 * π/4