For fully developed laminar pipe flow in a circular pipe, the velocity profile is given by \(u(r)=2\left(1-r^{2} / R^{2}\right)\) in \(\mathrm{m} / \mathrm{s},\) where \(R\) is the inner radius of the pipe. Assuming that the pipe diameter is \(4 \mathrm{cm},\) find the maximum and average velocities in the pipe as well as the volume flow rate.

Short Answer

Expert verified
The maximum velocity in the pipe is 2 m/s, the average velocity is 1.33 m/s and the volume flow rate is 0.00000534 cubic metres per second.

Step by step solution

01

Determine Maximum Velocity

The maximum velocity \( u_{max} \) occurs at the center of the pipe where \( r = 0 \). So, substitute \( r = 0 \) into the velocity equation: \( u(r)=2\left(1-0^{2} / R^{2}\right) = 2 \, m/s \).
02

Calculate Average Velocity

The average velocity \( u_{avg} \) is given by: \( u_{avg} = {2 \over R^2 } * {2 \over 3} * R^2 = {4 \over 3} \, m/s\).
03

Find the Volume Flow Rate

The volume flow rate \( Q \) is got by multiplying the average velocity by the pipe's cross-sectional area (\( πR^{2} \)): \( Q = u_{avg} * πR^{2} = {4 \over 3} * π * (0.02)^2 = 0.00000534 \, m^3/s \).

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