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Two pipes, each of diameter d, converge to form a pipe of diameter D. What should be the relation between d and D such that the flow velocity in the third pipe becomes double of that in each of the two pipes?
- a)D = d
- b)D = 2d
- c)D = 3d
- d)D = 4d
Correct answer is option 'A'. 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 = D2v
D = d.
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 = D2v
D = d.
Most Upvoted Answer
Two pipes, each of diameter d, converge to form a pipe of diameter D. ...
To find the relation between the diameters of the pipes, we can use the principle of continuity equation, which states that the mass flow rate of a fluid is constant in a closed system. In this case, we can assume that the fluid is incompressible and the flow is steady.
Let's consider the flow in each of the two smaller pipes separately. The mass flow rate in each pipe can be given by the equation:
m1 = ρ1 * A1 * V1
where m1 is the mass flow rate, ρ1 is the density of the fluid, A1 is the cross-sectional area of the smaller pipe, and V1 is the flow velocity in the smaller pipe.
Similarly, for the third pipe, the mass flow rate can be given by:
m3 = ρ3 * A3 * V3
where m3 is the mass flow rate, ρ3 is the density of the fluid, A3 is the cross-sectional area of the larger pipe, and V3 is the flow velocity in the larger pipe.
According to the given condition, the flow velocity in the third pipe is double that in each of the two smaller pipes. Therefore, we can write:
V3 = 2 * V1
Since the mass flow rate is constant, we can equate the mass flow rates in the two smaller pipes and the larger pipe:
m1 = m3
ρ1 * A1 * V1 = ρ3 * A3 * V3
Substituting V3 = 2 * V1, we get:
ρ1 * A1 * V1 = ρ3 * A3 * 2 * V1
ρ1 * A1 = ρ3 * A3 * 2
Dividing both sides by A3, we get:
(ρ1 * A1) / A3 = ρ3 * 2
Since the cross-sectional area of a pipe is directly proportional to the square of its diameter, we can write:
(ρ1 * d1^2) / d3^2 = ρ3 * 2
Rearranging the equation, we get:
d3^2 = (ρ1 * d1^2) / (ρ3 * 2)
Taking the square root of both sides, we get:
d3 = √((ρ1 * d1^2) / (ρ3 * 2))
Since the density of the fluid is constant, we can simplify the equation to:
d3 = √(d1^2 / 2)
This can be further simplified to:
d3 = d1 / √2
Therefore, the relation between the diameters is:
D = d = d1 / √2
Hence, the correct option is D = d.
Let's consider the flow in each of the two smaller pipes separately. The mass flow rate in each pipe can be given by the equation:
m1 = ρ1 * A1 * V1
where m1 is the mass flow rate, ρ1 is the density of the fluid, A1 is the cross-sectional area of the smaller pipe, and V1 is the flow velocity in the smaller pipe.
Similarly, for the third pipe, the mass flow rate can be given by:
m3 = ρ3 * A3 * V3
where m3 is the mass flow rate, ρ3 is the density of the fluid, A3 is the cross-sectional area of the larger pipe, and V3 is the flow velocity in the larger pipe.
According to the given condition, the flow velocity in the third pipe is double that in each of the two smaller pipes. Therefore, we can write:
V3 = 2 * V1
Since the mass flow rate is constant, we can equate the mass flow rates in the two smaller pipes and the larger pipe:
m1 = m3
ρ1 * A1 * V1 = ρ3 * A3 * V3
Substituting V3 = 2 * V1, we get:
ρ1 * A1 * V1 = ρ3 * A3 * 2 * V1
ρ1 * A1 = ρ3 * A3 * 2
Dividing both sides by A3, we get:
(ρ1 * A1) / A3 = ρ3 * 2
Since the cross-sectional area of a pipe is directly proportional to the square of its diameter, we can write:
(ρ1 * d1^2) / d3^2 = ρ3 * 2
Rearranging the equation, we get:
d3^2 = (ρ1 * d1^2) / (ρ3 * 2)
Taking the square root of both sides, we get:
d3 = √((ρ1 * d1^2) / (ρ3 * 2))
Since the density of the fluid is constant, we can simplify the equation to:
d3 = √(d1^2 / 2)
This can be further simplified to:
d3 = d1 / √2
Therefore, the relation between the diameters is:
D = d = d1 / √2
Hence, the correct option is D = d.
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Two pipes, each of diameter d, converge to form a pipe of diameter D. What should be the relation between d and D such that the flow velocity in the third pipe becomes double of that in each of the two pipes?a)D = db)D = 2dc)D = 3dd)D = 4dCorrect answer is option 'A'. Can you explain this answer?
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