Question

ratio of average velocity to maximum velocity

Answer 1

The maximum-velocity occurs at centre where r = 0. Put in Equation 

Umax  = -1/4μ  ∂p / ∂x R2

Mean or Average velocity is obtained by dividing the discharge of the fluid across the corss sectional area of pipe (πr2). 

Te discharge (Q) across the section is obtained by considering the flow through a circular ring element of radius ‘r’ and thickness ‘dr’ as shown in.

dQ  = Velocity at a radius r x Area of ring element

       = u x 2 πr dr = - 1/4μ  ∂p / ∂x [ R2 - r2 ] x 2 πr dr

        R                                                R
Q  =  ∫  dQ = - 1/4μ  (∂p / ∂x ) x 2π  ∫  (R2 - r2) r dr
        0                                                0
                                      R
    = 1/4μ ( - ∂p / ∂x )2π ∫ ( R2 r - r3) dr
                                      0

    = 1/4μ (- ∂p / ∂x )2π (R4 / 2 - R4 / 4) = 1/4μ (- ∂p / ∂x )2π  R4 / 4

Q  = π / 8μ (- ∂p / ∂x ) R4

Average velocity,
-
Uave  =  Q / A  = π / 8μ (- ∂p / ∂x ) R4 / πR2

-
Uave  = 1 / 8μ (- ∂p / ∂x )R2

Ratio of maximum and average velocity

Umax  = - 1 / 4μ (∂p / ∂x )R2 / 1 / 8μ (- ∂p / ∂x )R2

Umax  = 8/4 = 2
-
U

Ratio of maximum velocity to average velocity is 2.