Prove that the DarcyWeisbach
friction factor is related to Manning's n by the following relation:
f_{D} = 8 g n^{ 2} / (k^{ 2} R^{ 1/3})
in which f_{D}
= DarcyWeisbach friction factor, g = gravitational acceleration,
R = hydraulic radius,
and k = constant specific for the system of units,
equal to 1 in SI units and 1.486 in U.S. Customary Units.
Express the relation in SI and U.S. Customary units.
Chezy equation:
V = C (RS)^{1/2}
V^{ 2} = C^{ 2} RS = ( C^{ 2} /g ) gRS = ( 8 / f_{D} ) gRS [1]
Manning equation:
V = (k/n) R^{ 2/3} S^{ 1/2}
V^{ 2} = (k^{ 2}/n^{ 2}) R^{ 4/3} S [2]
Equating [1] and [2]: (8 / f_{D} ) g =
(k^{ 2}/n^{ 2}) R^{ 1/3}
f_{D} = 8 g n^{ 2} / (k^{ 2} R^{ 1/3}) ANSWER.
In SI Units: f_{D} = 8 (9.806) n^{ 2} / (1^{2}
R^{ 1/3}) = 78.45 n^{ 2} / R^{ 1/3} ANSWER.
In U.S. Customary Units: f_{D} = 8 (32.17) n^{ 2} /
(1.486^{ 2} R^{ 1/3}) =
116.55 n^{ 2} / R^{ 1/3} ANSWER.
Calculate the discharge Q using the Manning
equation, given: flow area A = 23.5 ft^{2}; hydraulic radius R = 5.6 ft;
channel slope S = 0.0025; Manning's n = 0.035.
Q = 157.29 ft^{3}/s ANSWER.
Calculate the discharge Q using the Manning
equation, given: flow area A = 45 m^{2}; hydraulic radius R = 6 m;
channel slope S = 0.003; Manning's n = 0.04.
Q = 203.46 m^{3}/s ANSWER.
Given f = 0.0025,
calculate the discharge Q for a flow area A = 12.4 m^{2}, hydraulic radius R = 2.1 m;
and channel slope S = 0.0015.
Q = V A
V = C R^{1/2} S^{1/2} = (C/g^{1/2}) g^{1/2}R^{1/2} S^{1/2}
f = g / C^{ 2}
C/g^{1/2} = (1/f )^{1/2} = (1/0.0025)^{1/2} = 20
Q = V A = 20 [ (9.806) (2.1) (0.0015) ]^{1/2} (12.4)
Q = 43.586 m^{3}/s ANSWER.
Given f = 0.0035,
calculate the discharge Q for a flow area A = 18 ft^{2}, hydraulic radius R = 4.5 ft;
and channel slope S = 0.0018.
Q = V A
V = C R^{1/2} S^{1/2} = (C/g^{1/2}) g^{1/2}R^{1/2} S^{1/2}
f = g / C^{ 2}
C/g^{1/2} = (1/f )^{1/2} = (1/0.0035)^{1/2} = 16.9
Q = V A = 16.9 [ (32.17) (4.5) (0.0018) ]^{1/2} (18)
Q = 155.28 ft^{3}/s ANSWER.
Use
ONLINE CHANNEL 01 to calculate normal depth, velocity, and Froude number
for the following case:
Q = 150 m^{3}/s, b = 10 m, z = 2, S_{o} = 0.0005, n = 0.025.
Normal depth y_{n} = 4.452 m,
normal velocity v_{n} = 1.783 m,
normal Froude number F_{n} = 0.327 ANSWER.
Use
ONLINE CHANNEL 01 to calculate normal depth, velocity, and Froude number
for the following case: Q = 250 cfs, b = 20 ft, z = 1, S_{o} = 0.001,
n = 0.030.
Normal depth y_{n} = 3.476 ft,
normal velocity v_{n} = 3.064 ft/s,
normal Froude number F_{n} = 0.31 ANSWER.
Using ONLINE CHANNEL 15,
calculate the discharge for a prismatic channel with b = 20 ft, y = 3 ft, z = 2, n = 0.025, S = 0.0016.
Fig. 518 Definition sketch for a trapezoidal channel.


Q = 326.3 ft^{3}/s ANSWER.
Using ONLINE CHANNEL 15,
calculate the discharge for a prismatic channel with b = 6 m, y = 1 m, z = 1.5, n = 0.015, S = 0.0002.
Q = 5.995 m^{3}/s ANSWER.
A recent
flood on Clearwater Creek has left observable water marks on a certain river reach.
To estimate the flood magnitude, hydraulic data has been measured
at two cross sections A and B, a distance of 1,850 ft apart.
The reach fall between the cross sections is 9.1 ft and
the average Manning's n is 0.035.
The upstream flow area, wetted perimeter, and Coriolis α coefficient are
550 ft^{2}, 55 ft, and 1.17;
the downstream flow area, wetted perimeter, and α coefficient are
620 ft^{2}, 52 ft, and 1.10.
Use SLOPE AREA to calculate the flood discharge.
Run SLOPE AREA with the given data to calculate the flood discharge Q = 8,841 cfs. ANSWER.
Calculate the unitwidth discharge in an overland flow plane, under laminar flow, with mean depth of 1.5 cm and slope of 0.001.
Assume water temperature T = 20^{o}C. Report discharge in L/s/m.
q = [ g S / (3 ν) ] y_{m}^{3}
For T = 20^{o}C: ν = 1. ×
10^{6} m^{2}/s
q = [ ( 9.81 × 0.001 ) / (3 × 1. × 10 ^{6}) ] (0.015)^{3}
q = 0.01103 m^{2}/s
q = 0.01103 m^{3}/s/m
q = 11.03 L/s/m ANSWER.