Question

What is the maximum flow rate of water in a smooth pipe of 8cm diameter if the flow is to be laminar

Answer 1

Let's first find the average velocity of water in the pipe:

$$R_e=\dfrac{\rho v_{avg}d}{\mu},$$

here, $R_e=2000$ is the Reynolds number, $\mu=10^{-3}$ is the viscosity of water.

Then, we get:

$$v_{avg}=\dfrac{R_e\mu}{\rho d}=\dfrac{2000\cdot10^{-3}\ Pa\cdot s}{1000\ \dfrac{kg}{m^3}\cdot0.08\ m}=0.025\ \dfrac{m}{s}.$$

The maximum velocity of water in the pipe will be twice the average velocity:

$$v_{max}=2v_{avg}=2\cdot0.025\ \dfrac{m}{s}=0.05\ \dfrac{m}{s}=5.0\ \dfrac{cm}{s}.$$