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A triangular open channel has a vertex angle of 90° and carries flow at a critical depth of 0.30 m. The discharge in the channel is
- a)0.08 m3/s
- b)0.11 m3/s
- c)0.15 m3/s
- d)0.2 m3/s
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A triangular open channel has a vertex angle of 90° and carries flow ...
To determine the discharge in a triangular open channel, we can use the Manning's equation. Manning's equation relates the flow rate in an open channel to the cross-sectional properties of the channel and the roughness of the channel bed.
The Manning's equation for flow in an open channel is given by:
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 flow (m^2)
R = Hydraulic radius (m)
S = Slope of the channel bed (m/m)
Given that the vertex angle of the triangular channel is 90°, we can determine the cross-sectional properties of the channel.
1. Cross-sectional area (A):
For a triangular channel with a vertex angle of 90°, the cross-sectional area is given by:
A = (b * h) / 2
Where:
b = Base width of the triangular channel (m)
h = Depth of flow (m)
2. Hydraulic radius (R):
The hydraulic radius is defined as the ratio of the cross-sectional area to the wetted perimeter. For a triangular channel with a vertex angle of 90°, the wetted perimeter is given by:
P = b + 2 * h
The hydraulic radius can be calculated as:
R = A / P
3. Slope of the channel bed (S):
The slope of the channel bed can be determined based on the critical depth of flow. The critical depth occurs when the specific energy is minimized. For a triangular channel with a vertex angle of 90°, the critical depth is given by:
y_c = (2/5) * h
The slope of the channel bed can be calculated as:
S = (h - y_c) / L
Where:
y_c = Critical depth of flow (m)
L = Length of the channel (m)
Now, we can substitute the given values into the Manning's equation and solve for the discharge (Q).
Given:
Critical depth of flow (y_c) = 0.30 m
1. Cross-sectional area (A):
A = (b * h) / 2
Since the vertex angle is 90°, the base width (b) is equal to twice the critical depth (2 * y_c).
A = (2 * y_c * h) / 2
A = y_c * h
2. Hydraulic radius (R):
P = b + 2 * h
Since the vertex angle is 90°, the base width (b) is equal to twice the critical depth (2 * y_c).
P = 2 * y_c + 2 * h
R = A / P
R = (y_c * h) / (2 * y_c + 2 * h)
3. Slope of the channel bed (S):
S = (h - y_c) / L
Now, substitute the values of A, R, and S into the Manning's equation:
Q = (1/n) * A * R^(2/3) * S^(1/2)
Since we don't have the value of the Manning's roughness coefficient (n), we cannot directly calculate the discharge (Q). However, we can compare the given options and find the closest value to the calculated discharge using
The Manning's equation for flow in an open channel is given by:
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 flow (m^2)
R = Hydraulic radius (m)
S = Slope of the channel bed (m/m)
Given that the vertex angle of the triangular channel is 90°, we can determine the cross-sectional properties of the channel.
1. Cross-sectional area (A):
For a triangular channel with a vertex angle of 90°, the cross-sectional area is given by:
A = (b * h) / 2
Where:
b = Base width of the triangular channel (m)
h = Depth of flow (m)
2. Hydraulic radius (R):
The hydraulic radius is defined as the ratio of the cross-sectional area to the wetted perimeter. For a triangular channel with a vertex angle of 90°, the wetted perimeter is given by:
P = b + 2 * h
The hydraulic radius can be calculated as:
R = A / P
3. Slope of the channel bed (S):
The slope of the channel bed can be determined based on the critical depth of flow. The critical depth occurs when the specific energy is minimized. For a triangular channel with a vertex angle of 90°, the critical depth is given by:
y_c = (2/5) * h
The slope of the channel bed can be calculated as:
S = (h - y_c) / L
Where:
y_c = Critical depth of flow (m)
L = Length of the channel (m)
Now, we can substitute the given values into the Manning's equation and solve for the discharge (Q).
Given:
Critical depth of flow (y_c) = 0.30 m
1. Cross-sectional area (A):
A = (b * h) / 2
Since the vertex angle is 90°, the base width (b) is equal to twice the critical depth (2 * y_c).
A = (2 * y_c * h) / 2
A = y_c * h
2. Hydraulic radius (R):
P = b + 2 * h
Since the vertex angle is 90°, the base width (b) is equal to twice the critical depth (2 * y_c).
P = 2 * y_c + 2 * h
R = A / P
R = (y_c * h) / (2 * y_c + 2 * h)
3. Slope of the channel bed (S):
S = (h - y_c) / L
Now, substitute the values of A, R, and S into the Manning's equation:
Q = (1/n) * A * R^(2/3) * S^(1/2)
Since we don't have the value of the Manning's roughness coefficient (n), we cannot directly calculate the discharge (Q). However, we can compare the given options and find the closest value to the calculated discharge using
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Community Answer
A triangular open channel has a vertex angle of 90° and carries flow ...
Side slope: z horizontal to 1 vertical
When θ = 90o, z = 1
Critical depth, yc = 0.3 m
Qc = 0.11 m3/sec
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A triangular open channel has a vertex angle of 90° and carries flow at a critical depth of 0.30 m. The discharge in the channel isa)0.08 m3/sb)0.11 m3/sc)0.15 m3/sd)0.2 m3/sCorrect answer is option 'B'. Can you explain this answer?
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