Question

Consider incompressible flow in a circular channel. Derive general expressions for Reynolds number in terms of (a) volume flow rate Q and tube diameter D and (b) mass flow rate mp and tube diameter. The Reynolds number is 1800 in a section where the tube diameter is 6 mm. (c) Find the Reynolds number for the same flow rate in a section where the tube diameter is 6 mm.

Answer 1

Answer:

a) $Re = \frac{4\cdot \rho \cdot Q}{\pi\cdot \mu\cdot D}$, b) $Re = \frac{4\cdot \dot m}{\pi\cdot \mu\cdot D}$, c) 1600

Explanation:

a) The Reynolds Number is modelled after the following formula:

$Re = \frac{\rho \cdot v \cdot D}{\mu}$

Where:

$\rho$ - Fluid density.

$\mu$ - Dynamics viscosity.

$D$ - Diameter of the tube.

$v$ - Fluid speed.

The formula can be expanded as follows:

$Re = \frac{\rho \cdot \frac{4Q}{\pi\cdot D^{2}}\cdot D }{\mu}$

$Re = \frac{4\cdot \rho \cdot Q}{\pi\cdot \mu\cdot D}$

b) The Reynolds Number has this alternative form:

$Re = \frac{4\cdot \dot m}{\pi\cdot \mu\cdot D}$

c) Since the diameter is the same than original tube, the Reynolds number is 1600.