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At what distance r from center of a pipe of radius R the average velocity is indicated as the local velocity in laminar flow ?
- a)r = 0.36 R
- b)r = 0.45 R
- c)r = 0.59 R
- d)r = 0.707 R
Correct answer is option 'D'. Can you explain this answer?
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To determine the distance from the center of a pipe where the average velocity is indicated as the local velocity in laminar flow, we can refer to the concept of the velocity profile in laminar pipe flow.
Laminar flow refers to a smooth, orderly flow of fluid with layers moving parallel to each other. In the case of laminar flow in a pipe, the velocity profile is parabolic, with the maximum velocity occurring at the center of the pipe and decreasing towards the pipe walls.
The velocity profile can be mathematically described by the Hagen-Poiseuille equation, which relates the average velocity (V_avg) to the local velocity (V_loc) at a specific radial distance (r) from the pipe center:
V_avg = 2/3 * V_loc
Based on this equation, we can determine the value of r at which the average velocity is equal to the local velocity.
Solution:
Let's solve the equation for r:
V_avg = 2/3 * V_loc
Substituting the definition of average velocity in terms of volumetric flow rate (Q) and pipe cross-sectional area (A), we have:
Q/A = 2/3 * V_loc
Rearranging the equation, we get:
V_loc = 3/2 * Q/A
Now, let's consider the velocity profile in a pipe. The maximum velocity occurs at the center of the pipe (r = 0), so we can substitute r = 0 in the equation:
V_loc_max = 3/2 * Q/A
Since we're looking for the distance at which the average velocity is equal to the local velocity, we set V_loc = V_avg:
V_avg = 3/2 * Q/A
Now, let's solve for r using the relationship between the cross-sectional area of the pipe (A) and the radius of the pipe (R):
A = π * R^2
Substituting this into the equation, we have:
V_avg = 3/2 * Q / (π * R^2)
Simplifying the equation further:
V_avg = (3/2π) * (Q / R^2)
Now, we can solve for r by setting V_avg = V_loc and substituting the values:
V_loc = (3/2π) * (Q / R^2)
V_loc = (3/2π) * (Q / (0.7071R)^2)
Simplifying the equation, we find:
V_loc = (3/2π) * (Q / 0.5R^2)
V_loc = (3/π) * (Q / R^2)
Comparing this equation with the Hagen-Poiseuille equation, we can conclude that the distance r from the center of the pipe where the average velocity is indicated as the local velocity is r = 0.7071R.
Hence, the correct answer is option D: r = 0.707 R.
Laminar flow refers to a smooth, orderly flow of fluid with layers moving parallel to each other. In the case of laminar flow in a pipe, the velocity profile is parabolic, with the maximum velocity occurring at the center of the pipe and decreasing towards the pipe walls.
The velocity profile can be mathematically described by the Hagen-Poiseuille equation, which relates the average velocity (V_avg) to the local velocity (V_loc) at a specific radial distance (r) from the pipe center:
V_avg = 2/3 * V_loc
Based on this equation, we can determine the value of r at which the average velocity is equal to the local velocity.
Solution:
Let's solve the equation for r:
V_avg = 2/3 * V_loc
Substituting the definition of average velocity in terms of volumetric flow rate (Q) and pipe cross-sectional area (A), we have:
Q/A = 2/3 * V_loc
Rearranging the equation, we get:
V_loc = 3/2 * Q/A
Now, let's consider the velocity profile in a pipe. The maximum velocity occurs at the center of the pipe (r = 0), so we can substitute r = 0 in the equation:
V_loc_max = 3/2 * Q/A
Since we're looking for the distance at which the average velocity is equal to the local velocity, we set V_loc = V_avg:
V_avg = 3/2 * Q/A
Now, let's solve for r using the relationship between the cross-sectional area of the pipe (A) and the radius of the pipe (R):
A = π * R^2
Substituting this into the equation, we have:
V_avg = 3/2 * Q / (π * R^2)
Simplifying the equation further:
V_avg = (3/2π) * (Q / R^2)
Now, we can solve for r by setting V_avg = V_loc and substituting the values:
V_loc = (3/2π) * (Q / R^2)
V_loc = (3/2π) * (Q / (0.7071R)^2)
Simplifying the equation, we find:
V_loc = (3/2π) * (Q / 0.5R^2)
V_loc = (3/π) * (Q / R^2)
Comparing this equation with the Hagen-Poiseuille equation, we can conclude that the distance r from the center of the pipe where the average velocity is indicated as the local velocity is r = 0.7071R.
Hence, the correct answer is option D: r = 0.707 R.
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At what distance r from center of a pipe of radius R the average velocity is indicated as the local velocityin laminar flow ?a)r = 0.36 Rb)r = 0.45 Rc)r = 0.59 Rd)r = 0.707 RCorrect answer is option 'D'. Can you explain this answer?
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