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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

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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

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)

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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