1. ## application of integration

in a cylindrical tube of radius R and length l, the velocity v of the fluid flowing through the tube depends upon the distance r from the central axis according to:

v(r) = (P / (4nl)) * (R^2 - r^2) , for 0<=r<=R

where P is the pressure difference between the ends of the tube
and n is the fluid viscosity

compare the average velocity with the maximum velocity over 0<=r<=R

the answer i got for average velocity was (PR^2) / (6nl)
and for maximum velocity was (PR^2) / 4nl

can someone verify if these are the correct answers and if not provide details on the correct solution

2. Originally Posted by razorfever
in a cylindrical tube of radius R and length l, the velocity v of the fluid flowing through the tube depends upon the distance r from the central axis according to:

v(r) = (P / (4nl)) * (R^2 - r^2) , for 0<=r<=R

where P is the pressure difference between the ends of the tube
and n is the fluid viscosity

compare the average velocity with the maximum velocity over 0<=r<=R

the answer i got for average velocity was (PR^2) / (6nl)
and for maximum velocity was (PR^2) / 4nl

can someone verify if these are the correct answers and if not provide details on the correct solution

$\displaystyle v_{avg} = \frac{P}{4Rnl} \int_0^R R^2 - r^2 \, dr = \frac{PR^2}{6nl}$
$\displaystyle v_{max}$ occurs when $\displaystyle r = 0$ ... $\displaystyle v(0) = \frac{PR^2}{4nl}$