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Question

# The velocity profile of a fully developed laminar flow in a straight circular pipe, as shown in the figure, given by the expression u(r)=R24μ(∂p∂x)(1−r2R2) where ∂p∂x is a constant The average velocity of fluid in the pipe is

A
R28μ(px)
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B
R24μ(px)
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C
R22μ(px)
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D
R2μ(px)
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Solution

## The correct option is A −R28μ(∂p∂x) The velocity profile of a fully developed laminar flow in a straight circular pipe, as shown in figure, is given by the expression u(r)=−R24μ(dpdx)[1−r2R2] =Umax[1−r2R2] where Umax=−R24μ(dpdx) where dpdx is a constant The average velocity of fluid in the pipe, Vavg=Umax2=−R28μ(dpdx) Alternatively: Given velocity profile expression, u(r)=−R24μ(dpdx)[1−r2R2] Consider a small circular ring element of radius r and thickness dr as shown in figure, The fluid flowing per second through small elementary ring, dQ = Velocity at radius r × Area of a small ring element) =u×2πrdr =−14μR2(dpdx)[1−r2R2]×2πrdr =−π2μR2(dpdx)[r−r3R2]dr Net discharge flow through pipe, Q=R∫0−π2μR2(dpdx)[r−r3R2]dr =−π2μR2(dpdx)R∫0[r−r3R2]dr =−π2μR2(dpdx)[r22−r44R2]R0 =−π2μR2(dpdx)[R22−R44R2] =−π2μR2(dpdx)[R22−R24] =−π2μR2(dpdx)×R24=−π8μR4(dpdx) ∴ Average velocity, ¯u=QπR2=−π8μR4(dpdx)πR2 =−18μ(dpdx)R2

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