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

a)
`Re = (4\cdot \rho \cdot Q)/(\pi\cdot \mu\cdot D)`, b)
`Re = (4\cdot \dot m)/(\pi\cdot \mu\cdot D)`, c) 1600

Step-by-step explanation:

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

`Re = (\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 = (\rho \cdot (4Q)/(\pi\cdot D^(2))\cdot D )/(\mu)`

`Re = (4\cdot \rho \cdot Q)/(\pi\cdot \mu\cdot D)`

b) The Reynolds Number has this alternative form:

`Re = (4\cdot \dot m)/(\pi\cdot \mu\cdot D)`

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