Mean velocity in a circular pipe

[kbaumen]

Homework Statement

Frictionless, inviscous flow in a circular pipe.

Velocity profile, $v(r) = 6 - 6r^{1.828}$
Volume flowrate, $Q = 9 \frac{\text{m}^3}{\text{s}}$
Pipe radius, $R = 1$ m
Given velocities, $v(0) = 6$ m/s, $v(R) = 0$ m/s.

Find mean velocity $v_{av}$

2. The attempt at a solution
If I just divide the flowrate by area, I get the correct answer - 9/$\pi$ (correct according to the tutorial solutions anyway). It also seems to make sense.

However, if I integrate the velocity along $r$ from 0 to $R$ and divide everything by $R$, I get a different value.

$$v_{av} = \frac{1}{R} \int_0^R (6 - 6r^{1.828}) \mathrm{d}r = 3.858$$

Can anyone explain the discrepancy? To me both approaches make sense but I can't work out why the results are different.

 

[tiny-tim]

hi kbaumen!

for an average, shouldn't the integral be (1/πR²) ∫ 2πrdr etc ?

 

[kbaumen]

hi kbaumen!

for an average, shouldn't the integral be (1/πR²) ∫ 2πrdr etc ?

Cheers, that actually makes sense. Integrate over area and divide by area.