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Find the discharge through a circular channel section having diameter of 5m, the value of Chezy’s constant is 90 and the bed slope is 1 in 4000.
- a)13.61 m3/s
- b)14.61 m3/s
- c)15.61 m3/s
- d)16.61 m3/s
Correct answer is option 'C'. Can you explain this answer?
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Find the discharge through a circular channel section having diameter ...
The discharge through a circular channel can be calculated using the Chezy-Manning equation:
Q = (1/n) * A * R^(2/3) * S^(1/2)
Where:
Q = Discharge (m^3/s)
n = Manning's roughness coefficient (dimensionless)
A = Cross-sectional area of the channel (m^2)
R = Hydraulic radius (m)
S = Slope of the channel (m/m)
In this case, the diameter of the circular channel is given as 5m. The cross-sectional area can be calculated as:
A = π * (D/2)^2
A = π * (5/2)^2
A = 19.63 m^2
The hydraulic radius can be calculated as:
R = A / P
R = A / (π * D)
R = 19.63 / (π * 5)
R = 1.25 m
Now, let's assume a value for the Manning's roughness coefficient (n). This value depends on the type of channel material and flow conditions. Let's assume a value of 0.03, which is commonly used for concrete channels.
Finally, let's assume a slope (S) of 0.001, which means a 0.1% slope.
Plugging in these values into the equation, we get:
Q = (1/0.03) * 19.63 * (1.25)^(2/3) * (0.001)^(1/2)
Q ≈ 218.52 m^3/s
Therefore, the discharge through the circular channel section is approximately 218.52 m^3/s.
Q = (1/n) * A * R^(2/3) * S^(1/2)
Where:
Q = Discharge (m^3/s)
n = Manning's roughness coefficient (dimensionless)
A = Cross-sectional area of the channel (m^2)
R = Hydraulic radius (m)
S = Slope of the channel (m/m)
In this case, the diameter of the circular channel is given as 5m. The cross-sectional area can be calculated as:
A = π * (D/2)^2
A = π * (5/2)^2
A = 19.63 m^2
The hydraulic radius can be calculated as:
R = A / P
R = A / (π * D)
R = 19.63 / (π * 5)
R = 1.25 m
Now, let's assume a value for the Manning's roughness coefficient (n). This value depends on the type of channel material and flow conditions. Let's assume a value of 0.03, which is commonly used for concrete channels.
Finally, let's assume a slope (S) of 0.001, which means a 0.1% slope.
Plugging in these values into the equation, we get:
Q = (1/0.03) * 19.63 * (1.25)^(2/3) * (0.001)^(1/2)
Q ≈ 218.52 m^3/s
Therefore, the discharge through the circular channel section is approximately 218.52 m^3/s.
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Find the discharge through a circular channel section having diameter of 5m, the value of Chezy’s constant is 90 and the bed slope is 1 in 4000.a)13.61 m3/sb)14.61 m3/sc)15.61 m3/sd)16.61 m3/sCorrect answer is option 'C'. Can you explain this answer?
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