Water flows in a 8-m-wide rectangular channel that has a longitudinal slope of 0
ID: 1844999 • Letter: W
Question
Water flows in a 8-m-wide rectangular channel that has a longitudinal slope of 0.0001. The channel has an equivalent sand roughness of 2 mm. Calculate the uniform flow depth in the channel when the flow rate is 15 m3/s. Use the Darcy-Weisbach equation. Water flows in a 8-m-wide rectangular channel that has a longitudinal slope of 0.0001. The channel has an equivalent sand roughness of 2 mm. Calculate the uniform flow depth in the channel when the flow rate is 15 m3/s. Use the Darcy-Weisbach equation.Explanation / Answer
Given data:
Width of the rectangular channel = 8 m.
Slope (s) = 0.0001.
Equivalent sand roughness = 2 mm (or) 0.002 m.
Flow rate = 15 m3/s
Solution:
Darcy - Weisbach law is generally applicable for Head loss in closed channel flow, but still there is a relation between manning's coefficient and Darcy - Weisbach friction coefficient.
Darcy - Weisbach Equation for head loss = (1/f) *(L/D)*(U2/2g)
Here,
f - Friction loss coefficient
L - Length of the channel
D - Diameter
U - velocity of fluid
g - gravitational force
Mannings formulae relates velocity of fluid in open channel,
U = (1/n)*(R(2/3))*(S(1/2))
Here, n - Mannings coefficient,
R - Hydraulic mean depth,
S - slope of the channel,
In darcy-weisbach equation, Friction coeficient (f) = 1/(1.2+(2.03*log(R/Equivalent sand size))2).
If we compare friction coefficient with mannings coefficient (n),
(1/n) = (1/(4*f)(1/2))*(2*g),
From above equation,
n = (0.038*(d(1/6)))
For 2 mm Equivalent sand roughness, n = 0.0135.
R = (A/P)
A = 8*D, Here, D - Depth of flow,
P - Wetted Perimeter - (8 +(2*D))
R = ( 8*D)/(8+(2*D)
U = Q/A
Q/A = (1/0.0135)*(8d/(8+(2*D)))(2/3)* (0.0001)(1/2)
Q/A = (15/ 8*D)
(15/ 8*D) = (1/0.0135)*(8d/(8+(2*D)))(2/3)* (0.0001)(1/2)
By solving the above equation by trial and error method, by using calculator with a configuration of fx-991 ES (or) fx 991 MS (or) equivalent,
We will get Depth of 2.06 m
Conclution: Uniform flow depth in the given open channel flow = 2.06 m
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