You are observing the rise of a river during flood.
The width of the river at the observation point, and for
some distance upstream is 65 m, and according to the gage reading,
the discharge is Q = 70 m^{3}/s.
Estimate the discharge at a point located 15.875 km upstream,
if the water surface is rising at the rate of 9 mm/hr at your location
and at 12 mm/hr at the upstream cross section.
(∂Q / ∂x) + (∂A / ∂t) = 0
(∂Q / ∂x) + B (∂y / ∂t) = 0
r = ∂y/∂t
(∂Q / ∂x) + B r = 0
Q_{d}  Q_{u} =  B (Δx) r
Q_{u} = Q_{d} + B (Δx) r
Average r = 10.5 mm/hr
Q_{u} = 70 m^{3}/s + 65 m (15,875 m) (10.5 mm/hr / 3600 s/hr) (1 m / 1000 mm)
Q_{u} = 73 m^{3}/s ANSWER.
A hydraulically wide channel is operating at Froude number F = 0.22.
The unitwidth discharge is q = 2.8 m^{2}/s.
What are the two absolute Lagrange celerities?
F = v / (gy)^{ 1/2} = 0.22
v = 0.22 (gy)^{ 1/2}
F^{ 2} = v^{ 2} / (gy)
F^{ 2} = q^{ 2} / (gy^{3})
y = [ q^{ 2} / (gF^{ 2}) ]^{ 1/3}
y = [ 2.8^{2} / (9.806 × 0.22^{ 2}) ]^{ 1/3} = 2.547 m.
c_{r} = (gy)^{1/2} = (9.806 × 2.547) ^{1/2} = 4.998 m/s
v = 0.22 (gy)^{1/2}
v = 0.22 × 4.998 = 1.099 m/s
c_{d} = v + c_{r} = 1.099 + 4.998 = 6.097 m/s ANSWER.
c_{u} = v  c_{r} = 1.099  4.998 =  3.899 m/s ANSWER.
A flood wave is traveling inbank through a straight
river reach of top width T = 320 m wide and length L = 5625 m.
For every 1 cm of flood rise,
the discharge goes up 10 m^{3}/s. How long will it take
for a given discharge to travel the length of the reach?
Use the Seddon formula for flood wave speed (celerity):
c = (1/T) dQ / dy
c = (1/T) ΔQ / Δy
Assume T_{t} = travel time through the reach.
c = L / T_{t}
L / T_{t} = (1/T ) ΔQ / Δy
T_{t} = L T (Δy / ΔQ )
T_{t} = (5625 m) (320 m) (0.01 m) / (10 m^{3}/s)
T_{t} = 1800 seconds ANSWER.
T_{t} = 30 minutes ANSWER.
Calculate the exponent β of the dischargearea rating for
a triangular channel with Chezy friction.
Q = C A R^{1/2} S^{1/2}
Q = C A (A^{1/2} / P^{1/2}) S^{1/2}
Q = C (A^{3/2} / P^{1/2}) S^{1/2}
Q = K_{1} A^{3/2} / P^{1/2}
T = Kd
A = Kd^{ 2} / 2
P = 2d [ 1 + (K^{ 2}/4) ]^{1/2}
d = (2A / K )^{1/2}
P = 2 (2A / K )^{1/2} [ 1 + (K^{2}/4) ]^{1/2}
P = 2 (2 / K )^{1/2} [ 1 + (K^{ 2}/4) ]^{1/2} A^{1/2}
P^{1/2} = K_{2} A^{1/4}
Q = K_{1} A^{3/2} / (K_{2} A^{1/4})
Q = K_{3} A^{5/4}
β = 5/4 ANSWER.
Fig. 1023 Definition sketch for a triangular channel.


Calculate the exponent β of the dischargearea rating for
a triangular channel with Manning friction.
Q = (1/n) A R^{2/3} S^{1/2}
Q = (1/n) A (A^{2/3} / P^{ 2/3}) S^{1/2}
Q = C (A^{5/3} / P^{ 2/3}) S^{1/2}
Q = K_{1} A^{5/3} / P^{ 2/3}
T = Kd
A = Kd^{ 2} / 2
P = 2d [ 1 + (K^{ 2}/4) ]^{1/2}
d = (2A / K )^{1/2}
P = 2 (2A / K )^{1/2} [ 1 + (K^{2}/4) ]^{1/2}
P = 2 (2 / K )^{1/2} [ 1 + (K^{ 2}/4) ]^{1/2} A^{1/2}
P^{ 2/3} = K_{2} A^{1/3}
Q = K_{1} A^{5/3} / (K_{2} A^{1/3})
Q = K_{3} A^{4/3}
β = 4/3 ANSWER.
A flood hydrograph has the following data: timeofrise t_{r} = 2 hr, reference velocity V_{o} = 2 ft/s,
reference flow depth d_{o} = 6 ft, bottom slope S_{o} = 0.004. Determine is this wave is a kinematic wave.
t_{r} S_{o} V_{o} / d_{o} = (2 × 3600) (0.004) (2) / (6) = 9.6 < 85
This wave is NOT a kinematic wave. ANSWER.
A flood hydrograph has the following data: timeofrise t_{r} = 2 hr, reference velocity V_{o} = 2 ft/s,
reference flow depth d_{o} = 6 ft, bottom slope S_{o} = 0.004. Determine is this wave is a diffusion wave.
t_{r} S_{o} (g / d_{o})^{1/2} =
(2 × 3600) (0.004) (32.17 / 6)^{1/2} = 66.7 > 15
This wave is a diffusion wave. ANSWER.
A flood hydrograph has the following data: timeofrise t_{r} = 1 hr, reference velocity V_{o} = 2 m/s,
reference flow depth d_{o} = 2 m, bottom slope S_{o} = 0.0004. Determine is this wave is a diffusion wave.
t_{r} S_{o} (g / d_{o})^{1/2} =
(1 × 3600) (0.0004) (9.806 / 2)^{1/2} = 3.18 < 15
This wave is NOT a diffusion wave. ANSWER.
Using ONLINE ROUTING 04,
route a flood wave using the Muskingum method. The inflow hydrograph ordinates are [25 ordinates, starting at time = 0, to time = 24 hr]:
100, 130, 150, 180, 220, 250, 300, 360, 450, 550, 700, 550, 490, 370, 330, 310, 280, 230, 170, 150, 130,
120, 110, 105, 100.
Assume Δt = 1 hr, K = 1 hr, X = 0.3. Report the peak outflow and time of occurrence.
From the output table, the peak outflow is 651.46 m^{3}/s, and it occurs at time t = 11 hr. ANSWER.
Using ONLINE ROUTING 05,
route the same flood wave as the previous problem using the MuskingumCunge method. Use the following input data:
Q_{p} = 700 m^{3}/s, A_{p} = 400 m^{2},
T_{p} = 88 m,
Δt = 1 hr,
Δx = 9.6 km,
β = 1.65,
S_{o} = 0.0007. Report the peak outflow and time of occurrence.
From the output table, the peak outflow is 652.994 m^{3}/s, and it occurs at time t = 11 hr. ANSWER.