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Question

# The velocity profile in fully developed laminar flow in a pipe of diameter D is given by u=u0(1−4r2/D2), where r is the radial distance from the center. If the viscosity of the fluid is μ, the pressure drop across a length L of the pipe is

A
μu0LD2
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B
4μu0LD2
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C
8μu0LD2
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D
16μu0LD2
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Solution

## The correct option is D 16μu0LD2Method I: Given velocity profile, u=u0(1−4r2D2) At r = 0, u = umax ∴ umax=u0 We know that the pressure drop across a length L of the pipe, p1−p2=32μ¯uLD2 where ¯u = Average velocity ¯u=umax2=u02 ∴ p1−p2=32μu0L2×D2=16μu0LD2 Method II: For the laminar flow through circular pipe, Shear stress, τ=−r2(∂p∂x) μ∂u∂r=−r2ΔpL −r2ΔpL=μ∂∂r(u0(1−4r2D2)) ∵ r2ΔpL=μ(8u0rD2) r2ΔpL=(8μu0rD2) Pressure drop, Δp=16μu0LD2

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