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The pre-jump and post-jump depth in a rectangular channel during a hydraulic jump are 0.25 m and 1.25 m, then the head loss in the jump is
- a)0.50 m
- b)0.60 m
- c)0.80 m
- d)1.25 m
Correct answer is option 'C'. Can you explain this answer?
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The pre-jump and post-jump depth in a rectangular channel during a hyd...
Head loss during a hydraulic jump
where y1 = pre-jump depth and y2 = post-jump depth
where y1 = pre-jump depth and y2 = post-jump depth
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The pre-jump and post-jump depth in a rectangular channel during a hyd...
The head loss in a hydraulic jump can be calculated using the energy equation. The energy equation relates the pre-jump and post-jump depths, velocities, and specific energies. In this case, the pre-jump and post-jump depths are given as 0.25 m and 1.25 m, respectively. Let's calculate the head loss using the energy equation:
Energy equation:
h1 + V1^2/2g = h2 + V2^2/2g + hL
where:
h1 = pre-jump depth (0.25 m)
V1 = pre-jump velocity
h2 = post-jump depth (1.25 m)
V2 = post-jump velocity
g = acceleration due to gravity (9.81 m/s^2)
hL = head loss
To calculate the head loss, we need to determine the velocities at the pre-jump and post-jump locations. The velocities can be calculated using the specific energy equation:
Specific energy equation:
E = h + V^2/2g
where:
E = specific energy
h = depth
V = velocity
Let's calculate the velocities at the pre-jump and post-jump locations:
Velocity at pre-jump location:
E1 = h1 + V1^2/2g
V1^2/2g = E1 - h1
V1^2 = 2g(E1 - h1)
V1 = sqrt(2g(E1 - h1))
Velocity at post-jump location:
E2 = h2 + V2^2/2g
V2^2/2g = E2 - h2
V2^2 = 2g(E2 - h2)
V2 = sqrt(2g(E2 - h2))
Substituting the values of h1, h2, V1, and V2 into the energy equation, we can solve for the head loss (hL):
h1 + V1^2/2g = h2 + V2^2/2g + hL
0.25 + (sqrt(2g(E1 - h1)))^2/2g = 1.25 + (sqrt(2g(E2 - h2)))^2/2g + hL
Since the specific energies (E1 and E2) are not given, we cannot directly calculate the head loss. Therefore, without additional information about the specific energies, we cannot determine the exact value of the head loss. Hence, the answer cannot be determined with the given information.
Energy equation:
h1 + V1^2/2g = h2 + V2^2/2g + hL
where:
h1 = pre-jump depth (0.25 m)
V1 = pre-jump velocity
h2 = post-jump depth (1.25 m)
V2 = post-jump velocity
g = acceleration due to gravity (9.81 m/s^2)
hL = head loss
To calculate the head loss, we need to determine the velocities at the pre-jump and post-jump locations. The velocities can be calculated using the specific energy equation:
Specific energy equation:
E = h + V^2/2g
where:
E = specific energy
h = depth
V = velocity
Let's calculate the velocities at the pre-jump and post-jump locations:
Velocity at pre-jump location:
E1 = h1 + V1^2/2g
V1^2/2g = E1 - h1
V1^2 = 2g(E1 - h1)
V1 = sqrt(2g(E1 - h1))
Velocity at post-jump location:
E2 = h2 + V2^2/2g
V2^2/2g = E2 - h2
V2^2 = 2g(E2 - h2)
V2 = sqrt(2g(E2 - h2))
Substituting the values of h1, h2, V1, and V2 into the energy equation, we can solve for the head loss (hL):
h1 + V1^2/2g = h2 + V2^2/2g + hL
0.25 + (sqrt(2g(E1 - h1)))^2/2g = 1.25 + (sqrt(2g(E2 - h2)))^2/2g + hL
Since the specific energies (E1 and E2) are not given, we cannot directly calculate the head loss. Therefore, without additional information about the specific energies, we cannot determine the exact value of the head loss. Hence, the answer cannot be determined with the given information.
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The pre-jump and post-jump depth in a rectangular channel during a hydraulic jump are 0.25 m and 1.25 m, then the head loss in the jump isa)0.50 mb)0.60 mc)0.80 md)1.25 mCorrect answer is option 'C'. Can you explain this answer?
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