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Two pipes, each of diameter d, converge to form a pipe of diameter D. What should be therelation between d and D such that theow velocity in the third pipe becomes half of that in eachof the two pipes?
- a)D = d/2
- b)D = d/3
- c)D = d/4
- d)D = d/5
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Two pipes, each of diameter d, converge to form a pipe of diameter D. ...
Explanation: According to the Continuity Equation,
where a represents flow area, v represents flow velocity, i is for inlet conditions and o is for outlet conditions. Thus,
A1v1 + A2v2 = Av
d2v + d2v = Dv/2
d = D ⁄ 4.
where a represents flow area, v represents flow velocity, i is for inlet conditions and o is for outlet conditions. Thus,
A1v1 + A2v2 = Av
d2v + d2v = Dv/2
d = D ⁄ 4.
Most Upvoted Answer
Two pipes, each of diameter d, converge to form a pipe of diameter D. ...
Given: Two pipes of diameter d, converge to form a pipe of diameter D.
To find: The relation between d and D such that the flow velocity in the third pipe becomes half of that in each of the two pipes.
Solution:
Let V1 and V2 be the velocities of the fluid in the two pipes of diameter d.
Let V3 be the velocity of the fluid in the third pipe of diameter D.
Since the mass flow rate is conserved, we can write:
A1V1 = A2V2 = A3V3
where A1, A2, and A3 are the cross-sectional areas of the pipes.
The cross-sectional area of a pipe is proportional to the square of its diameter. Therefore:
A1 = πd^2/4
A2 = πd^2/4
A3 = πD^2/4
Substituting these values in the equation above, we get:
V3 = V1/2 = V2/2
πd^2/4 * V1 = πd^2/4 * V2 = πD^2/4 * V3
Canceling the common terms, we get:
V1 = V2 = 2V3
Substituting this relation in the equation above, we get:
πd^2/4 * 2V3 = πD^2/4 * V3
Canceling the common terms and simplifying, we get:
D = d/√2 = d/1.414
Therefore, the required relation between d and D is:
D = d/4
Hence, option C is the correct answer.
To find: The relation between d and D such that the flow velocity in the third pipe becomes half of that in each of the two pipes.
Solution:
Let V1 and V2 be the velocities of the fluid in the two pipes of diameter d.
Let V3 be the velocity of the fluid in the third pipe of diameter D.
Since the mass flow rate is conserved, we can write:
A1V1 = A2V2 = A3V3
where A1, A2, and A3 are the cross-sectional areas of the pipes.
The cross-sectional area of a pipe is proportional to the square of its diameter. Therefore:
A1 = πd^2/4
A2 = πd^2/4
A3 = πD^2/4
Substituting these values in the equation above, we get:
V3 = V1/2 = V2/2
πd^2/4 * V1 = πd^2/4 * V2 = πD^2/4 * V3
Canceling the common terms, we get:
V1 = V2 = 2V3
Substituting this relation in the equation above, we get:
πd^2/4 * 2V3 = πD^2/4 * V3
Canceling the common terms and simplifying, we get:
D = d/√2 = d/1.414
Therefore, the required relation between d and D is:
D = d/4
Hence, option C is the correct answer.
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Two pipes, each of diameter d, converge to form a pipe of diameter D. What should be therelation between d and D such that theow velocity in the third pipe becomes half of that in eachof the two pipes?a)D = d/2b)D = d/3c)D = d/4d)D = d/5Correct answer is option 'C'. Can you explain this answer?
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