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Water flows at the rate of 12 m3/s in a 6 m wide rectangular channel. A hydraulic jump is formed in the channel at a point where the upstream depth is 30 cm (just before the jump). Considering acceleration due to gravity as 9.81 m/s2 and density of water as 1000 kg/m3, the energy loss in the jump is
- a)114.2 kW
- b)141.2 J/s
- c)141.2 h.p.
- d)114.2 MW
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Water flows at the rate of 12 m3/s in a 6 m wide rectangular channel. ...
Assuming channle bed to be horizontal and frictionless.
q = 12/6 = 2m3/s/m
Initial Froude No.
From Belenger’s Momentum equation for a rectangular channel
= 5.018
∴ Y2 = 5.018 × 0.3 = 1.505 m
Head loss in the jump (hL) =
= 0.968 m
Power lost in the jump = γwQhL
= (9.81 × 12 × 0.968) kW
= 114.04 kW
q = 12/6 = 2m3/s/m
Initial Froude No.
From Belenger’s Momentum equation for a rectangular channel
= 5.018
∴ Y2 = 5.018 × 0.3 = 1.505 m
Head loss in the jump (hL) =
= 0.968 m
Power lost in the jump = γwQhL
= (9.81 × 12 × 0.968) kW
= 114.04 kW
Most Upvoted Answer
Water flows at the rate of 12 m3/s in a 6 m wide rectangular channel. ...
Given information:
- Flow rate of water, Q = 12 m^3/s
- Width of the rectangular channel, B = 6 m
- Upstream depth before the hydraulic jump, h1 = 30 cm = 0.3 m
- Acceleration due to gravity, g = 9.81 m/s^2
- Density of water, ρ = 1000 kg/m^3
To find: Energy loss in the hydraulic jump
1. Determine the upstream velocity:
- The cross-sectional area of flow, A1 = B * h1
- The upstream velocity, V1 = Q / A1
2. Determine the downstream depth:
- The downstream depth after the hydraulic jump, h2 = 4 * h1
3. Determine the downstream velocity:
- The cross-sectional area of flow, A2 = B * h2
- The downstream velocity, V2 = Q / A2
4. Calculate the energy head before the hydraulic jump:
- The energy head before the jump, H1 = (V1^2) / (2g) + (h1)
5. Calculate the energy head after the hydraulic jump:
- The energy head after the jump, H2 = (V2^2) / (2g) + (h2)
6. Calculate the energy loss during the hydraulic jump:
- The energy loss, H_loss = H1 - H2
7. Convert the energy loss to kilowatts:
- Power = H_loss * Q * g * ρ / 1000
Detailed Calculation:
1. Upstream velocity:
- A1 = B * h1 = 6 m * 0.3 m = 1.8 m^2
- V1 = Q / A1 = 12 m^3/s / 1.8 m^2 = 6.67 m/s
2. Downstream depth:
- h2 = 4 * h1 = 4 * 0.3 m = 1.2 m
3. Downstream velocity:
- A2 = B * h2 = 6 m * 1.2 m = 7.2 m^2
- V2 = Q / A2 = 12 m^3/s / 7.2 m^2 = 1.67 m/s
4. Energy head before the hydraulic jump:
- H1 = (V1^2) / (2g) + (h1) = (6.67 m/s)^2 / (2 * 9.81 m/s^2) + 0.3 m = 2.84 m
5. Energy head after the hydraulic jump:
- H2 = (V2^2) / (2g) + (h2) = (1.67 m/s)^2 / (2 * 9.81 m/s^2) + 1.2 m = 1.33 m
6. Energy loss during the hydraulic jump:
- H_loss = H1 - H2 = 2.84 m - 1.33 m = 1.51 m
7. Energy loss in kilowatts:
- Power = H_loss * Q * g * ρ / 1000 = 1.51 m * 12 m^3/s * 9.81 m/s^2 * 100
- Flow rate of water, Q = 12 m^3/s
- Width of the rectangular channel, B = 6 m
- Upstream depth before the hydraulic jump, h1 = 30 cm = 0.3 m
- Acceleration due to gravity, g = 9.81 m/s^2
- Density of water, ρ = 1000 kg/m^3
To find: Energy loss in the hydraulic jump
1. Determine the upstream velocity:
- The cross-sectional area of flow, A1 = B * h1
- The upstream velocity, V1 = Q / A1
2. Determine the downstream depth:
- The downstream depth after the hydraulic jump, h2 = 4 * h1
3. Determine the downstream velocity:
- The cross-sectional area of flow, A2 = B * h2
- The downstream velocity, V2 = Q / A2
4. Calculate the energy head before the hydraulic jump:
- The energy head before the jump, H1 = (V1^2) / (2g) + (h1)
5. Calculate the energy head after the hydraulic jump:
- The energy head after the jump, H2 = (V2^2) / (2g) + (h2)
6. Calculate the energy loss during the hydraulic jump:
- The energy loss, H_loss = H1 - H2
7. Convert the energy loss to kilowatts:
- Power = H_loss * Q * g * ρ / 1000
Detailed Calculation:
1. Upstream velocity:
- A1 = B * h1 = 6 m * 0.3 m = 1.8 m^2
- V1 = Q / A1 = 12 m^3/s / 1.8 m^2 = 6.67 m/s
2. Downstream depth:
- h2 = 4 * h1 = 4 * 0.3 m = 1.2 m
3. Downstream velocity:
- A2 = B * h2 = 6 m * 1.2 m = 7.2 m^2
- V2 = Q / A2 = 12 m^3/s / 7.2 m^2 = 1.67 m/s
4. Energy head before the hydraulic jump:
- H1 = (V1^2) / (2g) + (h1) = (6.67 m/s)^2 / (2 * 9.81 m/s^2) + 0.3 m = 2.84 m
5. Energy head after the hydraulic jump:
- H2 = (V2^2) / (2g) + (h2) = (1.67 m/s)^2 / (2 * 9.81 m/s^2) + 1.2 m = 1.33 m
6. Energy loss during the hydraulic jump:
- H_loss = H1 - H2 = 2.84 m - 1.33 m = 1.51 m
7. Energy loss in kilowatts:
- Power = H_loss * Q * g * ρ / 1000 = 1.51 m * 12 m^3/s * 9.81 m/s^2 * 100
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Water flows at the rate of 12 m3/s in a 6 m wide rectangular channel. A hydraulic jump is formed in the channel at a point where the upstream depth is 30 cm (just before the jump). Considering acceleration due to gravity as 9.81 m/s2 and density of water as 1000 kg/m3, the energy loss in the jump isa)114.2 kWb)141.2 J/sc)141.2 h.p.d)114.2 MWCorrect answer is option 'A'. Can you explain this answer?
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Water flows at the rate of 12 m3/s in a 6 m wide rectangular channel. A hydraulic jump is formed in the channel at a point where the upstream depth is 30 cm (just before the jump). Considering acceleration due to gravity as 9.81 m/s2 and density of water as 1000 kg/m3, the energy loss in the jump isa)114.2 kWb)141.2 J/sc)141.2 h.p.d)114.2 MWCorrect answer is option 'A'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Water flows at the rate of 12 m3/s in a 6 m wide rectangular channel. A hydraulic jump is formed in the channel at a point where the upstream depth is 30 cm (just before the jump). Considering acceleration due to gravity as 9.81 m/s2 and density of water as 1000 kg/m3, the energy loss in the jump isa)114.2 kWb)141.2 J/sc)141.2 h.p.d)114.2 MWCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Water flows at the rate of 12 m3/s in a 6 m wide rectangular channel. A hydraulic jump is formed in the channel at a point where the upstream depth is 30 cm (just before the jump). Considering acceleration due to gravity as 9.81 m/s2 and density of water as 1000 kg/m3, the energy loss in the jump isa)114.2 kWb)141.2 J/sc)141.2 h.p.d)114.2 MWCorrect answer is option 'A'. Can you explain this answer?.
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