What is a chute?
A chute is a canal of very steep slope.
What is an aqueduct?
An aqueduct is a
canal to transport water for a specific use,
typically over terrain, or elevated over a valley, stream or road.
What is a culvert?
A culvert is a covered conduit (or conduits) of relatively short length, usually flowing partially full, to enable a stream to cross a highway or other embankment.
What is potamology?
Potamology is the scientific study of rivers.
What is sinuosity in connection with rivers?
Sinuosity is
the ratio of stream length to valley length.
What is a hydraulically wide channel?
A channel is hydraulically wide when the
wetted perimeter P can be approximated by the top width T.
What ratio of top width to hydraulic depth can be considered hydraulically wide?
A channel may be considered hydraulically wide if the top width T
is greater than or equal to 10 times the hydraulic depth D.
What is the typical range of the exponent β for trapezoidal channels?
Typical values of β
for trapezoidal channels lie approximately in the range 1.35 ≤ β ≤ 1.65.
What is the most common type of current meter in the United States?
The most common type of cup meter is the Price current meter, which has six cups mounted on a vertical axis.
Why is the true velocity head greater than the velocity head computed based on mean
velocity?
Due to the nonuniform velocity distribution over a cross section,
the true velocity head is greater than the velocity head computed based on the mean velocity.
How does energy differ from momentum?
Energy is the integral of a force over a distance; momentum is the integral of
the force over time. Energy is steady; momentum is unsteady.
How does the pressure distribution in the vertical vary under parallel flow?
Under parallel flow, the pressure distribution
is essentially hydrostatic, with the pressure varying as a linear function of partial flow depth.
How does the pressure distribution in the vertical vary under convex curvilinear flow?
Under convex curvilinear, the pressure distribution is nonhydrostatic,
with the pressure along the flow depth varying as a nonlinear function of partial flow depth.
A piezometer located at the partial depth below the water surface would rise to an elevation
which is lower than the water surface elevation.
How does the pressure distribution in the vertical vary under concave curvilinear flow?
Under convex curvilinear, the pressure distribution is nonhydrostatic,
with the pressure along the flow depth varying as a nonlinear function of partial flow depth.
A piezometer located at the partial depth below the water surface would rise to an elevation
which is higher than the water surface elevation.
What is a channel of large slope?
A channel with slope greater than 10% is referred to as a channel of large slope.
For a slope greater than 10%, the error in taking the vertical depth y in lieu of the pressure rise h is more than 1%.
What is the discharge per unit of width if the discharge is 24 m3/s
and the channel top width is T = 8 m?
q = Q / T = 24 / 8 = 3 m2/s
Given: α = 0.4; β = 1.55; A = 45.6 m2. What is the discharge Q?
Q = α Aβ = 0.4 (45.6)1.55 = 149.09 m3/s
Given culvert diameter do = 1 m; what is the flow area for θ = 300°? (See Table 2-1).
A = (θ - sin θ ) (do2 / 8) = { [(5/6) 2 π ] - sin 300°} (12/8) = 0.7627 m2
What is the force F developed by a discharge Q = 10 ft3/s at a velocity V = 1 ft/s, at a cross section
with a Boussinesq coefficient β = 1.05?
F = β ρ Q V = β (γ / g) Q V = (1.05) (62.4 lb/ft3 / 32.17 ft/s2 ) (10 ft3/s) (1 ft/s) = 20.367 lbs.
A stream of flow area A = 100 m2 is divided into three sections: (1) left overbank, with 20% of the flow area, and velocity 0.2 m/s;
(2) inbank center, with 70% of the flow area, and velocity 1 m/s; and (3) right overbank, with 10% of the flow area, and velocity 0.1 m/s. Calculate the Coriolis coefficient α
and the Boussinesq coefficient β.
| Segment |
ΔA |
V |
V 2 |
V 3 |
V ΔA |
V 2 ΔA |
V 3 ΔA |
| Left overbank |
20 |
0.2 |
0.04 |
0.008 |
4.0 |
0.8 |
0.16 |
| Inbank center |
70 |
1.0 |
1.00 |
1.000 |
70.0 |
70.0 |
70.0 |
| Right overbank |
10 |
0.1 |
0.01 |
0.001 |
1.0 |
0.1 |
0.01 |
| Total |
100 |
0.75 |
- |
- |
75.0 |
70.90 |
70.17 |
V = ∑ V ΔA / ∑ ΔA = 75.0 / 100 = 0.75 m/s
α = ∑ V 3 ΔA / (V 3 A) = 70.17 / [(0.75)3 × 100] = 1.66
β = ∑ V 2 ΔA / (V 2 A) = 70.90 / [(0.75)2 × 100] = 1.26
Calculate the velocity distribution coefficients
for the following canal data, with flow area
A = 2768 ft2.
| Increment | Velocity v
(ft/s)
| Incremental
flow area ΔA (%)
| 1 | 3.5
| 0.5
| 2 | 4.0
| 2.9
| 3 | 4.5
| 10.3
| 4 | 5.0
| 23.5
| 5 | 5.5
| 52.7
| 6 | 6.0
| 10.1
| | | | | | |
Compare the results with those of approximate logarithmic
formulas.
| Increment
| Velocity v (ft/s)
| Incremental
flow area ΔA (%)
| Incremental
flow area ΔA (ft2)
| v ΔA
| v2 ΔA
| v3 ΔA
| 1 | 3.5 | 0.5 | 13.840
| 48.440 | 169.54 | 593.39
| | 2 | 4.0 | 2.9 | 80.272
| 321.088 | 1284.352 | 5137.408
| | 3 | 4.5 | 10.3 | 285.104
| 1282.968 | 5773.356 | 25980.102
| | 4 | 5.0 | 23.5 | 650.480
| 3254.4 | 16262 | 81310
| | 5 | 5.5 | 52.7 | 1458.736
| 8023.048 | 44126.764 | 242697.202
| | 6 | 6.0 | 10.1 | 279.568
| 1677.408 | 10064.448 | 60386.688
| | Sum | 5.277 | 100.0
| 2768.000 | 14607.352 | 77680.46 |
416104.79
| |
vm = ∑ vΔA / ∑ ΔA = 14607.352 / 2768 = 5.277
α = ∑ v3ΔA / (vm3 A) = 416104.79 / [(5.277)3 (2768)] =
1.023 ANSWER.
β = ∑ v2ΔA / (vm2 A) = 77680.46 / [(5.277)2 (2768)] =
1.0078 ANSWER.
ε = (vmax/vm) - 1 = (6.0/5.277) - 1 = 0.137
By formulas:
α = 1 + 3ε2 - 2ε3 = 1.051 ANSWER.
β = 1 + ε2 = 1.019 ANSWER.
Calculate the discharge
in a trapezoidal channel with flow depth y = 1 m;
bottom width b = 2 m;
side slope z1 = 1;
side slope z2 = 0.5; Manning's n = 0.015; and bottom slope S = 0.001.
Calculate average side slope z = 0.75.
Flow area A = (b + zy) y = 2.75 m2
Wetted perimeter P = b + y(1 + z1)1/2 +
y(1 + z2)1/2 = 4.532 m
Hydraulic radius R = A / P = 2.75 / 4.532 = 0.6068 m
Discharge Q = (1/n) A R 2/3 S 1/2 = 4.155 m3
A spillway is designed for a discharge q = 5 m2/s with a flow depth d = 0.5 m.
What is the minimum radius of curvature of the spillway cross section to ensure that the pressure does not fall below 50% of hydrostatic?
V = q / d = 5 / 0.5 = 10 m/s
The pressure drop c = (d V 2) / (gr )
c /d = V 2 / (gr ) ≤ 0.5
r ≥ 2 V 2 / (g ) = 2 × 102 / (9.806) = 20.39 m