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.

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

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.