Question

Prove that the maximum velocity in a circular pipe for viscous flow is equal to two times the average velocity of the flow

Answer 1

The equation of velocity distribution of laminar fluid flow through a pipe having radius R is given by,

u = – (1 / (4 μ)) (∂p/∂x) [R2 – r2]

where,

u = velocity of fluid

R = radius of pipe

r = distance from the centre of the pipe

• Maximum velocity occurs at the centre where r = 0. Putting this in above equation,

Umax = – (1 / (4 μ)) (∂p/∂x) [R2]

• Average velocity obtained by dividing discharge of the fluid across the cross sectional area of pipe (πR2)

dQ = Velocity at radus r x Area of ring element  

= u x 2πr dr

= – (1 / (4 μ)) (∂p/∂x) [R2] x 2πr dr

• Therefore,

Q = 0∫R – (1 / (4 μ)) (∂p/∂x) [R2] x 2πr dr

= (π/8μ) (- ∂p/∂x) R4

• Therefore, Average velocity,

u = Q/A

u = (1/8μ) (- ∂p/∂x) R2

• Ratio of maximum and average velocity  

Umax / u = 2

Answer 2

Answer:

Concept:

The maximum velocity in a circular pipe for viscous flow is equal to two times the average velocity of the flow

Explanation:

The equation of velocity distribution of  laminar fluid flow through a pipe having radius "r"

$u=-(1 /(4 \mu))\left(\frac{\partial P}{\partial x}\right)\left[R_{2}-r_{2}\right]$

u = velocity of the fluid

R = radius of the pipe

r = distance from the center of the pipe

The maximum velocity in a circular pipe occurs in the center which is

r = 0,

$i.e. U_{\max }=2 \times U_{a v g} \\ U_{\max }=\frac{1}{4 \mu}\left(-\frac{\partial P}{\partial x}\right) \cdot R^{2} \\ U_{a v g}=\bar{U}=\frac{1}{8 \mu}\left(-\frac{\partial P}{\partial x}\right) \cdot R^{2} \\Hence \frac{U_{\max }}{\bar{U}}=2$

Hence proved.

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