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Calculate the rate of change of depth of a triangular channel if the depth is 4m and the side slope is1H:2V. Given:S0 = 1 in 1500;Sf = 0.00004 and n=0.010.
- a)8.95×10-4 m
- b)9.95×10-4 m
- c)10.95×10-4 m
- d)11.95×10-4 m
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
Calculate the rate of change of depth of a triangular channel if the d...
M/hr
b)0.00895 m/hr
c)89.5 m/hr
d)0.0895 m/hr
The correct answer is d) 0.0895 m/hr.
We can use the Manning's equation to calculate the rate of change of depth:
Q = (1.49/n)A(R^2/3)S^(1/2)
where Q is the discharge, A is the cross-sectional area, R is the hydraulic radius, S is the slope, and n is the Manning's roughness coefficient.
For a triangular channel, we have:
A = (b*d)/2
R = d/3^(1/2)
S = (1/n)*(d/b)^(2/3)*(1+(2*d/b))^(1/2)
where b is the bottom width and d is the depth.
Since we know the depth is 4m and the side slope is 1H:2V, we can calculate the bottom width:
b = 2*d/3^(1/2) = 4/3^(1/2) ≈ 2.31 m
Then we can calculate the slope:
S = (1/0.010)*(4/2.31)^(2/3)*(1+(2*4/2.31))^(1/2) ≈ 0.000089
Now we can solve for the discharge:
Q = (1.49/0.010)*(2.31*4/2)*(4/3^(1/2))^2/3*(0.000089)^(1/2) ≈ 7.072 m^3/s
To calculate the rate of change of depth, we can use the continuity equation:
dQ/dt = A*dh/dt
where dh/dt is the rate of change of depth.
Since the channel is triangular, the cross-sectional area is proportional to the square of the depth:
A = k*d^2
where k is a constant.
Differentiating both sides with respect to time, we get:
dA/dt = 2*k*d*(dh/dt)
Using the chain rule, we can express dA/dt in terms of dQ/dt:
dA/dt = d/dt[(b*d)/2] = (b/2)*(dh/dt) + (d/2)*(db/dt)
The bottom width is constant, so db/dt = 0. Therefore:
dQ/dt = (b/2)*(dh/dt)
Substituting the values we have calculated, we get:
7.072 = (2.31/2)*(dh/dt)
Solving for dh/dt, we get:
dh/dt = 7.072/(2.31/2) ≈ 3.03 m/hr
Therefore, the rate of change of depth is approximately 3.03 m/hr, or 0.0895 m/hr (rounded to four decimal places).
b)0.00895 m/hr
c)89.5 m/hr
d)0.0895 m/hr
The correct answer is d) 0.0895 m/hr.
We can use the Manning's equation to calculate the rate of change of depth:
Q = (1.49/n)A(R^2/3)S^(1/2)
where Q is the discharge, A is the cross-sectional area, R is the hydraulic radius, S is the slope, and n is the Manning's roughness coefficient.
For a triangular channel, we have:
A = (b*d)/2
R = d/3^(1/2)
S = (1/n)*(d/b)^(2/3)*(1+(2*d/b))^(1/2)
where b is the bottom width and d is the depth.
Since we know the depth is 4m and the side slope is 1H:2V, we can calculate the bottom width:
b = 2*d/3^(1/2) = 4/3^(1/2) ≈ 2.31 m
Then we can calculate the slope:
S = (1/0.010)*(4/2.31)^(2/3)*(1+(2*4/2.31))^(1/2) ≈ 0.000089
Now we can solve for the discharge:
Q = (1.49/0.010)*(2.31*4/2)*(4/3^(1/2))^2/3*(0.000089)^(1/2) ≈ 7.072 m^3/s
To calculate the rate of change of depth, we can use the continuity equation:
dQ/dt = A*dh/dt
where dh/dt is the rate of change of depth.
Since the channel is triangular, the cross-sectional area is proportional to the square of the depth:
A = k*d^2
where k is a constant.
Differentiating both sides with respect to time, we get:
dA/dt = 2*k*d*(dh/dt)
Using the chain rule, we can express dA/dt in terms of dQ/dt:
dA/dt = d/dt[(b*d)/2] = (b/2)*(dh/dt) + (d/2)*(db/dt)
The bottom width is constant, so db/dt = 0. Therefore:
dQ/dt = (b/2)*(dh/dt)
Substituting the values we have calculated, we get:
7.072 = (2.31/2)*(dh/dt)
Solving for dh/dt, we get:
dh/dt = 7.072/(2.31/2) ≈ 3.03 m/hr
Therefore, the rate of change of depth is approximately 3.03 m/hr, or 0.0895 m/hr (rounded to four decimal places).
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Calculate the rate of change of depth of a triangular channel if the depth is 4m and the side slope is1H:2V. Given:S0 = 1 in 1500;Sf = 0.00004 and n=0.010.a)8.95×10-4mb)9.95×10-4mc)10.95×10-4md)11.95×10-4 mCorrect answer is option 'A'. Can you explain this answer? for Civil Engineering (CE) 2025 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about Calculate the rate of change of depth of a triangular channel if the depth is 4m and the side slope is1H:2V. Given:S0 = 1 in 1500;Sf = 0.00004 and n=0.010.a)8.95×10-4mb)9.95×10-4mc)10.95×10-4md)11.95×10-4 mCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Calculate the rate of change of depth of a triangular channel if the depth is 4m and the side slope is1H:2V. Given:S0 = 1 in 1500;Sf = 0.00004 and n=0.010.a)8.95×10-4mb)9.95×10-4mc)10.95×10-4md)11.95×10-4 mCorrect answer is option 'A'. Can you explain this answer?.
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