A 600 mm diameter RCC sewer is laid at a slope to developed a velocity...
Solution:
Given data,
Diameter of sewer (D) = 600 mm = 0.6 m
Velocity of flow at full capacity (V1) = 0.6 m/s
Manning's constant (n) = 0.013 (dimensionless)
Let's assume the depth of flow at half capacity (y) = 0.5D = 0.5 x 0.6 = 0.3 m
We know that the discharge (Q) in a sewer is given by the formula,
Q = (1/n) x A x R^2/3 x S^1/2
Where,
A = Cross-sectional area of flow
R = Hydraulic radius
S = Slope of sewer
At full capacity, the cross-sectional area can be calculated as follows,
A1 = π/4 x D^2 = π/4 x (0.6)^2 = 0.2827 m^2
The hydraulic radius can be calculated as follows,
R1 = A1/P
Where,
P = Wetted perimeter
P1 = π x D
R1 = 0.2827/1.88496 = 0.1499 m
Substituting the values in the formula of discharge, we get
0.6 = (1/0.013) x 0.2827 x (0.1499)^2/3 x S^1/2
S = 0.0012
Now, let's calculate the velocity of flow at half capacity,
At half capacity, the cross-sectional area can be calculated as follows,
A2 = π/4 x D^2 x y/ D = π/4 x (0.6)^2 x 0.5 = 0.1413 m^2
The hydraulic radius can be calculated as follows,
R2 = A2/P
Where,
P2 = π x D + 2 x (0.6/2) x cos^-1(1-2y/D)
P2 = π x 0.6 + 2 x (0.6/2) x cos^-1(1-2x0.3/0.6) = 1.88496 m
R2 = 0.1413/1.88496 = 0.0749 m
Substituting the values in the formula of discharge, we get
Q = (1/0.013) x 0.1413 x (0.0749)^2/3 x 0.0012^1/2
Q = 0.0568 m^3/s
The velocity of flow at half capacity can be calculated as follows,
V2 = Q/A2
V2 = 0.0568/0.1413 = 0.402 m/s
Therefore, the velocity of flow taking Manning's constant to be equal to 0.013 is nearly 0.4 m/s.
Hence, the correct option is (a) 0.3 m/s.
A 600 mm diameter RCC sewer is laid at a slope to developed a velocity...
A 600 mm diameter RCC sewer is laid at a slope to developed a velocity of flow of 0.6 m/s while just running full.When the sewer is running exactly half-full, the velocity of flow taking Manning's constant to be equal to 0.013 is nearly
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