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A rectangular channel 3m wide carries uniform flow of water at a rate of 7 cumec/sec at a depth of 1.5m. If there is a local rise of 0.5m in the bed, calculate the change in upstream depth of flow?
- a)0.125m
- b)0.2m
- c)0.25m
- d)No change in upstream depth
Correct answer is option 'A'. Can you explain this answer?
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Given data:
Width of the rectangular channel, b = 3 m
Flow rate, Q = 7 cumec/sec
Depth of flow, y1 = 1.5 m
Local rise in the bed, Δz = 0.5 m
Change in upstream depth of flow, Δy = ?
We can use the equation of continuity to find the upstream depth of flow.
Continuity equation: Q = A1V1 = A2V2
where A1 and A2 are the cross-sectional areas of the flow at sections 1 and 2 respectively, and V1 and V2 are the velocities of the flow at sections 1 and 2 respectively.
Cross-sectional area of flow at section 1: A1 = b*y1 = 3*1.5 = 4.5 sq.m
Velocity of flow at section 1: V1 = Q/A1 = 7/4.5 = 1.56 m/sec
Cross-sectional area of flow at section 2: A2 = b*(y1 + Δy) = 3*(1.5 + Δy) = 4.5 + 3Δy sq.m
Velocity of flow at section 2: V2 = Q/A2 = 7/(4.5 + 3Δy) m/sec
Applying the Bernoulli's equation between sections 1 and 2, we get:
P1/γ + y1 + V1^2/(2g) = P2/γ + (y1 + Δy) + V2^2/(2g) + Δz
where P1/γ and P2/γ are the pressure heads at sections 1 and 2 respectively, and γ is the unit weight of water.
Since the channel is open to the atmosphere, P1/γ = P2/γ = 0
Substituting the values, we get:
1.5 + 1.56^2/(2*9.81) = 1.5 + Δy + (7/(4.5 + 3Δy))^2/(2*9.81) + 0.5
Simplifying the equation, we get:
3.06 = Δy + (0.328 + 0.077Δy + 0.054Δy^2)
Rearranging and solving the quadratic equation, we get:
Δy = 0.125 m
Therefore, the change in upstream depth of flow is 0.125 m. Hence, option (a) is correct.
Width of the rectangular channel, b = 3 m
Flow rate, Q = 7 cumec/sec
Depth of flow, y1 = 1.5 m
Local rise in the bed, Δz = 0.5 m
Change in upstream depth of flow, Δy = ?
We can use the equation of continuity to find the upstream depth of flow.
Continuity equation: Q = A1V1 = A2V2
where A1 and A2 are the cross-sectional areas of the flow at sections 1 and 2 respectively, and V1 and V2 are the velocities of the flow at sections 1 and 2 respectively.
Cross-sectional area of flow at section 1: A1 = b*y1 = 3*1.5 = 4.5 sq.m
Velocity of flow at section 1: V1 = Q/A1 = 7/4.5 = 1.56 m/sec
Cross-sectional area of flow at section 2: A2 = b*(y1 + Δy) = 3*(1.5 + Δy) = 4.5 + 3Δy sq.m
Velocity of flow at section 2: V2 = Q/A2 = 7/(4.5 + 3Δy) m/sec
Applying the Bernoulli's equation between sections 1 and 2, we get:
P1/γ + y1 + V1^2/(2g) = P2/γ + (y1 + Δy) + V2^2/(2g) + Δz
where P1/γ and P2/γ are the pressure heads at sections 1 and 2 respectively, and γ is the unit weight of water.
Since the channel is open to the atmosphere, P1/γ = P2/γ = 0
Substituting the values, we get:
1.5 + 1.56^2/(2*9.81) = 1.5 + Δy + (7/(4.5 + 3Δy))^2/(2*9.81) + 0.5
Simplifying the equation, we get:
3.06 = Δy + (0.328 + 0.077Δy + 0.054Δy^2)
Rearranging and solving the quadratic equation, we get:
Δy = 0.125 m
Therefore, the change in upstream depth of flow is 0.125 m. Hence, option (a) is correct.
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A rectangular channel 3m wide carries uniform flow of water at a rate of 7 cumec/sec at a depth of 1.5m. If there is a local rise of 0.5m in the bed, calculate the change in upstream depth of flow?a)0.125mb)0.2mc)0.25md)No change in upstream depthCorrect answer is option 'A'. Can you explain this answer?
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