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A vertical water pipe, 1.5 m long, diverges from 75 mm diameter at the bottom to 150 mm diameter at the top and the rate of flow is 50 L/s upwards. If the pressure at the bottom end is 150 kN/m2 , the pressure at the top will be nearly
(!) 195.2 kN⁄m2
- a)191.4 kN⁄m2
- b)187.6 kN⁄m2
- c)183.8 kN⁄m2
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A vertical water pipe, 1.5 m long, diverges from 75 mm diameter at th...
p1 = 150 × 103
Q = 50 L/s
Applying continuity equation at 1 and 2
V1 x A1 = V2 x A2
V1/V2 = (150/75)2
V1=4 V2
Applying Bernoulli equation and 1 and 2
V2 = 2.83m/sec
Most Upvoted Answer
A vertical water pipe, 1.5 m long, diverges from 75 mm diameter at th...
Solution:
Given data:
Length of the pipe, L = 1.5 m
Diameter of the pipe at the bottom, D1 = 75 mm
Diameter of the pipe at the top, D2 = 150 mm
Rate of flow, Q = 50 L/s
Pressure at the bottom end, P1 = 150 kN/m2
To find: Pressure at the top end, P2
Assumptions:
The flow is steady, incompressible, and inviscid.
The density of the fluid is constant throughout the flow.
The pipe is straight and uniform.
The frictional losses are negligible.
Calculation:
1. Cross-sectional areas at the bottom and top of the pipe:
Area at the bottom, A1 = π/4 (D1)2 = π/4 (0.075)2 = 4.418 × 10-3 m2
Area at the top, A2 = π/4 (D2)2 = π/4 (0.150)2 = 1.767 × 10-2 m2
2. Velocity at the bottom of the pipe:
Velocity at the bottom, V1 = Q/A1 = (50 × 10-3)/4.418 × 10-3 = 11.32 m/s
3. Velocity at the top of the pipe:
Using the principle of continuity, the velocity of the fluid at the top can be calculated as:
A1V1 = A2V2
V2 = A1V1/A2 = (4.418 × 10-3 × 11.32)/(1.767 × 10-2) = 2.828 m/s
4. Pressure at the top of the pipe:
Using Bernoulli's equation between the bottom and top of the pipe, neglecting the frictional losses:
P1/ρg + ½ V1²/g = P2/ρg + ½ V2²/g
where ρ is the density of water and g is the acceleration due to gravity.
Substituting the given values, we get:
150 × 103/ (1000 × 9.81) + ½ (11.32)²/9.81 = P2/ (1000 × 9.81) + ½ (2.828)²/9.81
P2 = 195.2 kN/m2
Therefore, the pressure at the top end of the pipe is 195.2 kN/m2.
Hence, option A is the correct answer.
Given data:
Length of the pipe, L = 1.5 m
Diameter of the pipe at the bottom, D1 = 75 mm
Diameter of the pipe at the top, D2 = 150 mm
Rate of flow, Q = 50 L/s
Pressure at the bottom end, P1 = 150 kN/m2
To find: Pressure at the top end, P2
Assumptions:
The flow is steady, incompressible, and inviscid.
The density of the fluid is constant throughout the flow.
The pipe is straight and uniform.
The frictional losses are negligible.
Calculation:
1. Cross-sectional areas at the bottom and top of the pipe:
Area at the bottom, A1 = π/4 (D1)2 = π/4 (0.075)2 = 4.418 × 10-3 m2
Area at the top, A2 = π/4 (D2)2 = π/4 (0.150)2 = 1.767 × 10-2 m2
2. Velocity at the bottom of the pipe:
Velocity at the bottom, V1 = Q/A1 = (50 × 10-3)/4.418 × 10-3 = 11.32 m/s
3. Velocity at the top of the pipe:
Using the principle of continuity, the velocity of the fluid at the top can be calculated as:
A1V1 = A2V2
V2 = A1V1/A2 = (4.418 × 10-3 × 11.32)/(1.767 × 10-2) = 2.828 m/s
4. Pressure at the top of the pipe:
Using Bernoulli's equation between the bottom and top of the pipe, neglecting the frictional losses:
P1/ρg + ½ V1²/g = P2/ρg + ½ V2²/g
where ρ is the density of water and g is the acceleration due to gravity.
Substituting the given values, we get:
150 × 103/ (1000 × 9.81) + ½ (11.32)²/9.81 = P2/ (1000 × 9.81) + ½ (2.828)²/9.81
P2 = 195.2 kN/m2
Therefore, the pressure at the top end of the pipe is 195.2 kN/m2.
Hence, option A is the correct answer.
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A vertical water pipe, 1.5 m long, diverges from 75 mm diameter at the bottom to 150 mm diameter at the top and the rate of flow is 50 L/s upwards. If the pressure at the bottom end is 150 kN/m2 , the pressure at the top will be nearly(!) 195.2 kN⁄m2a) 191.4 kN⁄m2b) 187.6 kN⁄m2c) 183.8 kN⁄m2Correct answer is option 'A'. Can you explain this answer?
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