Two reservoirs open to the atmosphere are connected by a pipe 800 metres long. The pipe goes over a hill whose height is 6 m above the level of water in the upper reservoir. The pipe diameter is 300 mm and friction factor f = 0.032. The difference in water levels in the two reservoirs is 12.5 m. If the absolute pressure of water anywhere in the pipe is not allowed to fall below 1.2 m of water in order to prevent vapour formation, calculate the length of pipe in the portion between the upper reservoir and the hill summit. Neglect bend losses.Correct answer is 'Range: 180 to 200'. Can you explain this answer?

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Two reservoirs open to the atmosphere are connected by a pipe 800 metres long. The pipe goes over a hill whose height is 6 m above the level of water in the upper reservoir. The pipe diameter is 300 mm and friction factor f = 0.032. The difference in water levels in the two reservoirs is 12.5 m. If the absolute pressure of water anywhere in the pipe is not allowed to fall below 1.2 m of water in order to prevent vapour formation, calculate the length of pipe in the portion between the upper reservoir and the hill summit. Neglect bend losses.

Correct answer is 'Range: 180 to 200'. Can you explain this answer?

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Two reservoirs open to the atmosphere are connected by a pipe 800 met...

Let the length of pipe upstream of C be L₁ and that of the downstream be L₂ It is given L₁ + L₂= 800 m Considering the entry, friction and exit losses, The total loss from A to C

Applying Bernoulli’s equation between ! and B, we have

Δ H = h_f

which gives h_f1 = 2.99 m Using the value of h_f1= 2.99 m, and V = 1.68 m/s in equation ① we get

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Two reservoirs open to the atmosphere are connected by a pipe 800 met...

Given information:
- Two reservoirs are connected by a pipe 800 meters long.
- The pipe goes over a hill with a height of 6 meters above the level of water in the upper reservoir.
- Pipe diameter = 300 mm
- Friction factor (f) = 0.032
- Difference in water levels = 12.5 meters
- Minimum pressure to prevent vapor formation = 1.2 meters of water

To find:
- Length of the pipe between the upper reservoir and the hill summit.

Assumptions:
- Steady flow conditions
- Incompressible and frictional flow
- Neglecting bend losses

Calculations:

1. Head Loss due to elevation:
The head loss due to elevation can be calculated using the equation:

h_elevation = h_hill + h_pipe

where,
h_elevation = head loss due to elevation
h_hill = height of the hill = 6 meters
h_pipe = head loss due to pipe length

2. Head Loss due to friction:
The head loss due to friction can be calculated using the Darcy-Weisbach equation:

h_friction = f * (L / D) * (V^2 / 2g)

where,
h_friction = head loss due to friction
f = friction factor = 0.032
L = length of the pipe = 800 meters
D = diameter of the pipe = 300 mm = 0.3 meters
V = velocity of the flow
g = acceleration due to gravity = 9.81 m/s²

3. Total head loss:
The total head loss is the sum of the head loss due to elevation and the head loss due to friction:

h_total = h_elevation + h_friction

4. Length of pipe between upper reservoir and hill summit:
The length of pipe between the upper reservoir and the hill summit can be calculated by rearranging the above equation:

L_upper = (h_total - h_hill) / (f * (V^2 / 2g))

5. Solving for velocity:
The velocity can be found using the equation:

V = Q / A

where,
Q = volumetric flow rate
A = cross-sectional area of the pipe

6. Solving for cross-sectional area:
The cross-sectional area can be calculated using the equation:

A = π * (D/2)^2

7. Solving for volumetric flow rate:
The volumetric flow rate can be calculated using the equation:

Q = A * V

8. Substituting values and calculating:
- Convert the diameter from mm to meters: D = 0.3 m
- Calculate the cross-sectional area: A = π * (0.3/2)^2 = 0.0707 m²
- Rearrange the equation for volumetric flow rate: Q = A * V
- Rearrange the equation for velocity: V = Q / A
- Substitute the values of D and A to

Answered by Engineers Club · View profile

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Two reservoirs open to the atmosphere are connected by a pipe 800 metres long. The pipe goes over a hill whose height is 6 m above the level of water in the upper reservoir. The pipe diameter is 300 mm and friction factor f = 0.032. The difference in water levels in the two reservoirs is 12.5 m. If the absolute pressure of water anywhere in the pipe is not allowed to fall below 1.2 m of water in order to prevent vapour formation, calculate the length of pipe in the portion between the upper reservoir and the hill summit. Neglect bend losses.Correct answer is 'Range: 180 to 200'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about Two reservoirs open to the atmosphere are connected by a pipe 800 metres long. The pipe goes over a hill whose height is 6 m above the level of water in the upper reservoir. The pipe diameter is 300 mm and friction factor f = 0.032. The difference in water levels in the two reservoirs is 12.5 m. If the absolute pressure of water anywhere in the pipe is not allowed to fall below 1.2 m of water in order to prevent vapour formation, calculate the length of pipe in the portion between the upper reservoir and the hill summit. Neglect bend losses.Correct answer is 'Range: 180 to 200'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two reservoirs open to the atmosphere are connected by a pipe 800 metres long. The pipe goes over a hill whose height is 6 m above the level of water in the upper reservoir. The pipe diameter is 300 mm and friction factor f = 0.032. The difference in water levels in the two reservoirs is 12.5 m. If the absolute pressure of water anywhere in the pipe is not allowed to fall below 1.2 m of water in order to prevent vapour formation, calculate the length of pipe in the portion between the upper reservoir and the hill summit. Neglect bend losses.Correct answer is 'Range: 180 to 200'. Can you explain this answer?.

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